Mechanics: Newtonian, Classical, Relativistic Theory, Problems, Applications

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MECHANICS: NEWTONIAN, CLASSICAL, RELATIVISTIC: THEORY, PROBLEMS, APPLICATIONS Prasun Kumar Nayak

Asian Books Private Limited

MECHANICS: NEWTONIAN, CLASSICAL, RELATIVISTIC THEORY, PROBLEMS, APPLICATIONS

By

Prasun Kumar Nayak Lecturer and Head. Department of Mathematics, Bankura Christian CoI\ege, Bankura - W.B.

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PREFACE Mechanics plays an important role in the spheres of Applied Mathematics, Physics, and Engineering due to its inherent viabilities. The aim of this book is to give a rigorous and thorough analysis and applications of various aspects of Mechanics. Also, the present book has been designed in a lucid and coherent manner so that the Honours and Postgraduate students of various Universities may reap consIderable benefit out of it. I have chosen the topics with great care and have tried to present them systematically. The author expresses his sincere gratitude to his great teachers Prof. P.K. Chowdhury and Prof. P.R Chowdhury, Department of Applied Mathematics, Calcutta University, who taught him this course at PG level in 1992-1994. Author is thankful to his friends and colleagues, specially, Dr. G.B. Sural, Dr. A.P. Sinha of Bankura ChrIstian College, for their great help and valuable suggestions in the preparation of the book. Author also extends his thanks to Dr. Madhumangal Pal, Dept. of Applied Mathematics, Vidyasagar University, Dr. RRN. Bajpai, Principal, Bankura christian College, for their effective encouragement. This book could not have been completed without the loving support and encouragement of the parents, wife (Mousumi) and son (Bubai). I extend my thanks to my students, relatives, family members for their constant encouragement and support. Critical evaluation and suggestions for the improvement of the book will be appreciated and greatly acknowledged.

Prasun Kumar N ayak Bankura Christian College, Bankura, India, 722 101.

~

I

( About the Author) Prasun Kumar Nayak, a graduate (1992) of Calcutta University

(with Honours in Mathematics from R.K. Mission Residential College, Narendrapur), received his M.Sc. in Applied Mathematics (1994), from the same University. He is currently a voluntary research scholar at Department of Applied Mathematics, Vidyasagar University. His research interests are in Fuzzy Game Theory. At present, he has been serving as a Lecturer and Head of the Department of Mathematics, in Bankura Christian College, Bankura (West Bengal), a highly reputed constituent college (NAAC accredited B++) to the University ofBurdwan, India.

Contents Preface Bibliography 1 Mechanics of Particles 1.1 The Particle. . . . . 1.2 Newton's Laws of Motion 1.3 Conservative force field. . 1.4 Mechanics of a Particle. . 1.4.1 Conservation of linear momentum 1.4.2 Conservation of angular momentum 1.4.3 Conservation of energy. . 1.4.4 Equation of motion . . . . 1.5 Mechanics of System of Particles 1.5.1 Centre of mass . . . 1.5.2 Linear momentum .. 1.5.3 Angular Momentum . 1.5.4 Energy of the system 1.5.5 Kinetic energy . 1.5.6 Rigid body . . . . . 1.6 Motion of variable mass . . 1.7 Nonrealistic rocket motion . 1. 7.1 Single-stage rocket . 1.7.2 Multistage rockets . 1.8 Limitations of Newton's Law 1.9 Virial Theorem . 1.10 Exercise . . . . . . . . . 2

Generalised Co-ordinates 2.1 Constraints . . . . . . . . . . . . . 2.1.1 Holonomic Constraints . . . 2.1.2 Non-Holonomic Constraints 2.1.3 Scleronomic Constraints 2.1.4 Rheonomic Constraints 2.1.5 Bilateral Constraints . . 2.1.6 Unilateral Constraints . 2.1.7 Conservative Constraints (\'ii)

1

3 4 4 6 8 8 9 10 12 15 15 16 18 21 23 24 24 26 26 30 32 33 35

37 37 38 38 39 39 39 39 40,

(viii) 2.1.8 Dissipative Constraints 2.2 Generalised .Co-ordinates . . . . 2.2.1 Notation for GC . . . . 2.2.2 Transformation Equations 2.2.3 Configuration Space 2.2.4 Degrees of Freedom 2.3 Generalised Displacement 2.4 Generalised Velocity . . . . 2.5 Generalized Acceleration .. 2.6 Generalized Kinetic Energy 2.7 Generalised Momentum 2.8 Generalised Force . . . . . 2.9 Generalized Potential .. . 2.10 Principle of Virtual Work 2.10.1 Virtual Displacement. 2.10.2 Virtual Work . . . . . 2.10.3 Principle of Virtual Work 2.10.4 D'Alembert's Principle. 2.11 Exerci se . . . . . . .

CONTENTS

40 40 42 42 42 43 43

44 45 45

49 50 51 52 52 52 52 53 55

3 .Variational Principle 3.1 Properties of Functionals . . . . . . . . . . . . 3.2 'Euler's Equation . . . . . . . . . . . . . . . . . 3.3 Functionals depend on higher order derivatives 3.4 Functionals on several independent variables. 3.5 Variational problems with moving boundaries 3.6 Parametric representation 3.7 Exercise . . . . . .

57 58 60 66

4

77

Lagrangian Mechanics 4.1 Lagrange's equation of motion. 4.2 Dissipative system 4.3 Non-Holonomic System . . . . 4.4 Cyclic Co-ordinates. . . . . . . 4.4.1 Systems with ignorable co-ordinates 4.4.2 Separation of variables . . . . . 4.4.3 Liouville's class of Lagrangians . 4.5 Conservation Theorems . . . . . . . . . 4.5.1 Conservation theorem of energy . 4.5.2 Conservation theorem of linear momentum 4.5.3 Conservation theorem of angular momentum 4.5.4 Condition of invariance 4.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . .

68 71

74 75

77 89

91 95 96 99 100 104 104

105 106

109 116

CONTENTS

(ix)

5 Hamiltonian Mechanics 5.1 Phase Space . . . . . . . . . . . 5.2 Hamiltonian Function . . . . . 5.3 Hamilton's Equation of Motion 5.4 Hamilton's Principle . . . . . . 5.4.1 Modified Hamilton's principle. 5.5 Action Principles . . . . . . . . . . . . 5.5.1 Action as function of co-ordinates 5.5.2 Principle of Least Action .. 5.5.3 Second Form of Least Action Maupertui's Principle . . . . 5.5.4 5.6 Symmetry, Conservation Laws . . . . 5.6.1 Conservation of linear momentum 5.6.2 Conservation of angular momentum 5.6.3 Conservation of energy 5.6.4 Noether's Theorem 5.7 Conclusion 5.8 Exercise.......

119 119 120 123 133 135 139 139 140 143 146 149 150 151 151 153 154 154

6 Transformation Theory 6.1 Poisson Brackets .. 6.1.1 Properties of PB 6.1.2 Basic PBs . . . . 6.2 Lagrange Bracket . . . . 6.2.1 Angular Momentum and PB relations 6.3 Point Transformation . . . 6.4 Legendre Transformations 6.5 Canonical Transformation 6.6 Generating Functions. 6.6.1 First Form 6.6.2 Second Form 6.6.3 Third Form 6.6.4 Fourth Form 6.6.5 Cyclic relations 6.6.6 Examples of CT 6.7 Condition for Canonicality . 6.7.1 Exact Differential Form 6.7.2 Poisson Bracket Form 6.7.3 Lagrange Bracket Form 6.7.4 Bilinear Covariant Form 6.8 Properties of CT . . . . . . . . 6.9 Equation of Motion . . . . . . . 6.10 Infinitesimal Contact Transformation. 6.11 Finite Transformations . . . . . . . . .

157 157 158 161 162 163 165 165 166 168 170 174 175 176 178 178 182 182 184 186 186 189 196 201 204

(x)

CONTENTS 6.12 Liouville's Theorem 6.13 Exercise . . . . . . .

205 208

7 Hamilton-Jacobi Theory 7.1 Hamilton-Jacobi Equation . . . . . . . . . . . . . 7.2 Hamilton principal function . . . . . . . . . . . . 7.2.1 Connection with canonical transformation 7.2.2 Significance of S . . . . . . 7.3 Hamilton's characteristic function. 7.3.1 Significance of W . 7.4 Separation of Variables. 7.5 Action-Angle Variables.

211 211 212 213 214 218 220 222 228

8

239 239 241 242 243 246 247 248 264

Reference Frames 8.1 Translation motions 8.2 Rotating frames of reference . . . . . . . . 8.2.1 Rate of change of arbitrary vector 8.2.2 Velocity and Acceleration 8.2.3 Equation of motion .. 8.3 Lagrangian and Hamiltonian 8.4 Motion of rotating earth 8.5 Exercises

9 Rigid Body Motion 9.1 Moments, Products of Inertia . . . . . . . . 9.1.1 Axes theorem for moment of inertia 9.1.2 Radius of gyration. . . . . . . 9.1.3 MI of some symmetrical bodies 9.2 Independent Co-ordinates . . . . 9.2.1 Generalised Co-ordinates 9.2.2 Eulerian angles . . . . . . 9.3 Angular Velocity . . . . . . . . . 9.3.1 Precession and Nutation. 9.4 Angular Momentum 9.5 Kinetic Energy . . . . . . . 9.6 Euler's Theorem . . . . . . 9.7 Euler's Equation of Motion 9.8 Force/Torque-free motion . 9.8.1 Poinsot's geometrical representation 9.8.2 Polhode and Herpolhode 9.8.3 Body and Space cone 9.8.4 Binet Ellipsoid . . . 9.9 Motion of Symmetrical Top 9.9.1 Integrals of motion.

265 265 266 267 267 268 268 268 271 272 272 273 276 276 278 282 285 287 288 290 292

CONTENTS

(xi)

9.9.2 Precession without Nutation 9.9.3 Nutational Motion . . . . 9.9.4 Steady motion . . . . . . 9.9.5 Stability of steady motion 9.10 Exercise .

294 296 298 299 304

10 Central Force Field 10.1 Central force . . . . . . . . . 10.1.1 Angular momentum. 10.2 Centre of Mass . . . . . . . . 10.3 Two-Body Problem . . . . . . 10.4 Lagrange's equation of motion. 10.4.1 Expression for total energy 10.4.2 Potential energy . . . 10.4.3 Integrals of motion .. 10.5 General equation of the orbit 10.5.1 Equation of orbit . . . 10.5.2 Stability of orbits. . . 10.6 Motion under arbitrary potential field 10.7 Kepler's Law . . . . 10.8 Unbounded Motion . . . . 10.9 Scattering . . . . . . . . . 10.9.1 Elastic Scattering. 10.9.2 In-elastic scattering 10.9.3 Reaction . . . . 1O.10Direct/linear collisions .. . 10.10.1 Loss of KE . . . . . 10.10.2 Characteristic of direct collision . 10. 11 Oblique collisions. 10.11.1 Loss of KE 10. 12Exercises . . . . .

305 305 306 307 308 313 314 315 316 321 323 323 326 328 329 330 330 333 333 334 334 335 336 336 337

11 Theory of Oscillations 11.1 Equilibrium . . . . . . 11.1.1 Stable equilibrium .. 11.1.2 Unstable equilibrium. 11.2 Oscillatory problems . . . . . 11.2.1 Simple pendulum .. 11.2.2 Damped harmonic oscillation 11.2.3 Forced harmonic oscillation 11.3 Small oscillation . . . . . . . . . . . 11.3.1 Generalised co-ordinates . . . 11.3.2 Kinetic and Potential energy 11.4 Lagrange's Equation of Motion . . .

339 339 339 340 340 340 342 343 345 345 345 348

CONTENTS

11.4.1 The Eigenfunctions and Eigenfrequencies . . . 11.4.2 Normal co-ordinates, Frequencies of Vibration 11.4.3 Oscillation under constraint . . . . . . . . 11.4.4 Stationary property of the normal modes 11.5 Exercises 12 Relativistic Mechanics 12.1 Inertial frame of reference 12.2 Newtonian relativity .. 12.3 Galilean transformation . 12.4 Galilean invariance . . . 12.4.1 Invariance of space 12.4.2 Invariance of Time interval 12.4.3 Invariance of Velocity . . . 12.4.4 Invariance of acceleration . 12.4.5 Illvariance of Newton's Law 12.4.6 Conservation law of linear momentum 12.4.7 Conservation law of Energy 12.5 Special Theory of Relativity .. . 12.6 Lorentz Transformation . . . . . 12.6.1 Relativity of simultaneity 12.6.2 Length contraction. 12.6.3 Time dilation . . . . . . 12.7 Addition of velocities. . . . . . 12.8 Transformation of acceleration 12.9 Variation of mass with velocity. 12.lOMass-energy relation . . . . . . . 12.11Momentum and energy . . . . . . 12.12Transformation of energy and momentum 12.13Relativistic force . . . . . 12.14Equation of motion . . . . 12.14.1 Relativistic rocket 12.15Inelastic collision .. 12.16 Minkowski space .. 12.16.1 Proper time. 12.16.2 Four velocity 12.16.3Four momentum 12.16.4Four force . . . . 12.16.5 Kinetic energy . 12.17The relativistic Lagrangian 12.18Exercise . . . . . . . . . . .

(XIl)

348 351 361 362 370 373 373 373 374 375 376 376 376 377 377

378 378 380 381 384 385 '388 391 393 395 397 400 401 404 405 405 406 407 409 409 410 411 412 412 413

Bibliography [1] A. Sommerfield, Classical Mechanics, Academic Press, New York, 1964. [2] A. S. Gupta, Calculus of Variations with Applications, Prentice-Hall of India Private Limited, New Delhi, 2005. [3] A. S. Ramsey, Dynamics (part-2), CBS Publishers and Distributors. [4] C. R. Mondal, Classical Mechanics, Prentice-Hall of India Private Limited. [5] Gupta, Kumar and Sharma, Classical Mechanics, Pragati Prakashan. [6] E. T. Whittaker, A treatise on the Analytical Dynamics of Particles and Rigid bodies, 4/e, Dovar Publication.

[7] H. Goldstein, Classical Mechanics, Addision-Weseley Publishing Company. [8] J. Synge and B. Griffith, Principle of Mechanics, 2/e, McGraw Hill, New York, 1949. Prentice-Hall of India Private Limited. [9] L. Greiner, Classical Mechanics, System of Particles and Hamiltonian Dynamics, Springer International Edition. [10] N. C. Rana and P. S. Joag, Classical Mechanics, TMH Publishing Company Limited. [11] P. K. Nayak, Numerical Analysis (Theory and Applications), Asian Books Pvt. Limited. [12] R. G. Takwale and P. S. Puranik, Introduction to Classical Mechanics, Tata McGraw-Hill Publishing Company Limited.

1


Chapter 1

Mechanics of Particles Mechanics can be defined as that science which describes and predicts the conditions of rest or motion of material bodies around us under the action of forces where a body is said to be at rest, when it does not change its position with reference to the surrounding objects, but when it changes its position, we say that it is in motion. The topic of mechanics has three major sections 1. mechanics of rigid bodies 2. mechanics of deformable bodies

:t mechanics of fluids The mechanics of rigid bodies may be subclassified into three subsections (i) Kinematics: It deals with the all possible motion of material systems with reference to the agency which causes motion and gives the geometrical description of the body. Here we are to describe the motion of the material bodies in terms of quantities such as displacement, velocity, acceleration etc. (ii) Dynamics: It deals the causes of motion which determine, among all possible motions, which motion will actually take place in any particular system. Here we introduce the concept of force. (iii) Statics: It deals with the system of forces which actually gives no motion (rest) to the system relative to some given frame of reference. For equilibrium of the system the resultant effect of the forces should be zero. There are many fields of study in mechanics, such as classical, relativistic, quantum, continuum, space mechanics, and so on. Classical mechanics which is primarily supposed as the study of motions of physical objects such as motion of celestial bodies, is now considered as the part of mechanics dealing with the objects neither too big so that there exists a close agreement between theory and experiment, nor too small interacting object so that systems are considered on an atomic scale. 3

CHAPTER 1. MECHANICS OF PARTICLES

4

1.1

The Particle

By a term 'particle', we mean a point body having some mass or a body of some mass but having negligible physical dimensions in comparison with the dimensions of the other bodies involved in the systems under consideration. For example, in the planetary motion, the planet which rounds the sun can be treated as a particle. In atomic structure, au atom which can go too close to another charged body, such a molecule, may not be treated as a particle. Thus by a particle, we mean, an object whose size, internal structure, rotational and vibrational motions are negligible for the problem with which we are concerned.

1.2

Newton's Laws of Motion

Newton's laws of motion are the basis of the development of classical mechanics. These laws were formulated by Newton by considering mass, length and time as three fundamental quantities, which are invariant and independent of each other. Newton's laws of motion are stated in the following form: (i) Every body continues to be in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by external forces acting on it (Law of inertia). (ii) The time rate of change of momentmn of the particle is proportional to the external force and is in the direction of the force (Law of causality). (iii) To every action there is always an equal and opposite reaction. This law prescribes the general nature of forces of reaction in relation to the forces of action (Law of reciprocity). Note that, these laws of motion has meaning only with respect to a particular coordinate system. They are valid in a class of reference frames, called inertial frames. Thus these laws successfully explain the motion of celestial bodies, such as planets, stars, etc. From the laws we have the following discussions ; Result 1.2.1. The first law of motion introduces two important concepts Concept of inertia (ii) Concept of Force.

(i)

Deduction 1.2.1. Inertia: The first law gives the concept of state of rest or of uniform motion that are actually kinematical concepts. The term inertia means the tendency of a body to continue, i.e., to remain in a state of rest or of uniform motion for ever in absence of any external force. The property of a body to remain in the state of rest or of uniform motion when no external force acts on it is known as inertia and a frame of reference in which such a state is observed is known as inertial frame of reference. In fact, we can turn around and define the inertial frames to be those frames in which Newton's laws of motion are valid.

1.2. NEWTON'S LAWS OF MOTION

5

Deduction 1.2.2. Force: The first law of motion gives us the qualitative definition of 'force'. A force is something which changes or tends to change the state of rest or of uniform motion of a body in a straight line. Result 1.2.2. Measurement of force: The second law of motion gives us a quantitative definition of a force, i.e., it provides us with a measure of the applied ---+ force. Let a particle of mass m be moving with velocity 11. Let F be the applied force acting on the particle in the direction of motion. Thus the velocity changes, and produces acceleration 0:. Thus from the Newton's second law we get, ---+

F

---+

<X

dp ---+ dt; p = dp

----> m v = change momentum d

---+

---->

d7f

=> F = k dt = k dt (m v ) = km dt = km a

where k is the constant of proportionality. Let us define, the unit of force as that force which acting on a body of unit mass produces a unit acceleration. Thus, ---+ m = 1,1 F 1= 1, so, k = 1. Thus the expression of the second law becomes ----> d7f ---+ F =km- =1na. (1.1) dt Since the second law demands the priori knowledge of the total force that a body experiences, this total force must also include all the forces of reactions that it experiences. F is known as the imprelised force and m 0: is called the effective force. Now we have the following discussions: ---+

---+

(i) From the second law, it follows that when external force F = 0 ,i.e., if there is no impressed force, momentum m 7f is a constant, i.e., the body will continue to be in the state of uniform motion. We can consider the state of rest as a special case of the second law.

(ii) It follows from the equation that, 0: = ~ F, i.c., accelera.tion may be defined as the force per unit mass. ----> ----> (iii) The equation F = m a can be used to measure the mass of a body. Thus, ---+ let the same force F act on two bodies having ml and m2 so as to produce acceleration 0: 1 and 0: 2 respectively. Therefore, ----> 1

ml 1a 1

1 ml = m2 1----> a 2 => - = m2

10:2 1 1----> I' al

If, now, m2 is known or taken as unit mass, ml can be calculated by measuring :~~:. The mass of a body measured in the manner is called the inertial mass.

(iv) The second law states that the change of momentum produced by a force in a body is in the direction of the straight line in which the force acts. Thus, if two or more forces act on a body at the same time, then each produces an acceleration proportional to it in its own line of action independent of the motion of the body. This is the principle of the physical independent of forces.

6

CHAPTER 1. MECHANICS OF PARTICLES (v) Another way of measuring the mass of a body is the weighting it, i.e., by comparing the gravitational force acting on it with that on a standard mass. In this procedure, we make use of the fact that the gravitational field the weight of a body is exactly equal to the gravitational force acting on it. In -t --t this case, the equation F = m 71 becomes the weight W = m g, where 9 is the acceleration due to gravity. The mass of a body measured in this manner is called the gravitatio~al mass.

Deduction 1.2.3. The third law states that the mutual actions of any two bodies are equal and oppositely directed along the same straight line. This law can be applied only to the two isolated particles exerting forces on each other when forces due to all other particles are completely absent. The forces of action and reaction or of mutual interaction between the two particles (not in the same body) do so along the liue joining the two particles.

Ex 1.2.1. If (X, Y) be the component of forces acting in 2 dimension on a particles of unit mass, then the Newton's equation can be written as fx[(Y -X~)/~1-2X = o. SOLUTION: The given particle have unit mass, i.e., m second law of motion, we have,

1.3

= 1, hence by Newton's

Conservative force field

Conservative forces have a natural occurrence in classical mechanics. Typically, conservative forces are such that the work done by them, as the system moves from one configuration to another, depends only upon the initial and final co-ordinates of the particles. Gravitational and electrostatic fields are the common examples of conservative fields. This forces may be distinguished by anyone of the following equivalent features : (i) The work done by the force is independent of path.

1.3. CONSERVATIVE FORCE FIELD

7

(ii) Around any closed path in the force field, the work done is 0, i.e.,

f

---t

F .dr:> = O.

---t

(iii) If (Fx, FYI F z ) are the cartesian components of the force F, then Fxdx+Fydy+ Fzdz is an exact differential. ---t

---t

---t

---t

---t

(iv) F is only a function of position and \l x F = 0, i.e., one can write F as the gradient of some scalar point function. Since the gradient points towards the direction of increasing potential and forces cause the system to move towards ---t ---t the lower potential, so F = - \lV. Scalar point function V introduced here is called potential energy of the particle at that point. Thus a conservative force is also central in nature. (v) A potential energy function V exists that has a definite value at every point. (vi) At every point T

+V

=constant for a particle, where T is the kinetic energy. ---t

---t

Note that, in the equation F = - \lV, the potential energy V introduced makes it ---t --; --; clear that, it is not unique, as F = - \l V = - \l (V + C), where C is a coristant, so the zero level of V is arbitrary. Hence the absolute value of the PE has no meanin.g. But we can calculate the difference of PE by: B

WAB =

B

J

J

A

A -.

F.drt = -

VV.drt = VA - VB

where WAR is the work done by the force F in displacing the particle from position A to position B.

F

3

Ex 1.3.1. Prove that = (2xy + z3)i + x 2 + 3xz2k is a conservative force field. Find the scalar potential function V. Find also the work done in moving an object in this field from (1, -2, 1) to (3,1,4). SOLUTION: For the given force field, we have ---t

---t

k

J

\l x F -

..JL ..JL..JL ax a~ az . 2xy + z3 X 3xz 2 = (0 - O)i - (3z 2 - 3z 2)3 + (2x - 2x)k

= O.

Thus the given force field is conservative. Thus :3 a scalar potential function, V such ---t ---t that F = - \lV, Le.,

(2xy

+z =}

3

2

2

av

av

av

)i

+ x j + 3:z;z k = - [ax - i + !:lj + !:lk] uy uz

av ax

= -2xy - z3; V = _x 2y - xz 3 + 'lh(y, z)

A

av

!:l =

uy

av az

A

A

A

_x 2; V = _x 2y + 'ljJ2(X, z)

= -3xz2; V = -xz 3 + 1/J3(:t, y)

A

A


8

Thus the scalar potential function is Vex, y, z) = -x 2 y-xz 3 +c, where c is independent of x, y, z. Now the work done in moving an object in this field from A(I, -2, 1) to B(3, 1,4) is WAB = VA - VB = 202units.

1.4

Mechanics of a Particle

Let P be the position of the particle whose position vector is ~ with respect to some convenient origin O. The path of the particle may be described by some parameter l. If l depend on time t, then

~ = ~(t) = x(t)i + yet)] The velocity

+ z(t)k.

v is the time rate of change of the displacement of a particle, given by d--:t ~ v =-= r dt

----+

which is the vector tangent to the path l at PI and is directed in the sense in which the variable t increases along the path. The acceleration vector is similarly defined as the time rate of change of velocity vector

a

dv d2~ a =-=-2-'

----+

dt

dt

Newton's second law of motion is basic law of mechanics, which states that --+() F e

where

a = d;X

dv ----+ = -d (----+) Tn v = ' f n - = 'm a

dt

dt

(1.2)

is acceleration produced in the motion of the particle on applying --+

the external force F(e). The reference frame in which equation (1.2) is valid is called an inertial or Galilean system. Many of the important conclusions of mechanics can be expressed in the form of conservation theorems, which indicates under what conditions various mechanical quantities are constant in time. Now we derive the following conservation theorems for a particle in motion using Newtonian mechanics.

1.4.1

Conservation of linear momentum

Let Ii be the linear momentum of a body and p(e) be the external force applied on the particle. If the particle is of mass m, which is constant in time and is moving with velocity then

v,

-+

p

----+

= mv

d~ = m dt '

(1.3) --+

with the dimension [.MLT-Ij. According to Newton's second law of motion,F(e) = ->.

¥,: = Ii,

where

--+ F(e)

is the applied external force in the motion of the particle.

1.4. MECHANICS OF A PARTICLE

9

Suppose that there is no external force acting on the particle, then ----7

F(e)

dp

---+

= dt = 0 ::::}

---+

P

---+

= m v = a constant.

Thus if no external force acts on the free particle, then its linear momentum of the particle remains constant. It is the law of conservation of linear momentum. Also, if the mass of the particle is assumed to remain constant, it will continue to move along a straight line. This corresponds to a statement of Newton's first law of motion.

1.4.2

Conservation of angular momentum

A moment of momentum is angular momentum. Consider a particle P, whose position vector at any instant is r?, with respect to the base point O. If P is the ---+ linear momentum of the particle at a given instant, then its angular momentum L and the moment of force or torque -;;J about the point 0 is defined as ---+

---+

---+

L=rxPi

---+

---+

---+

(1.4)

T=TXF

---+

where P is the total force, experienced by the particle. Angular momentum has the dimension [AI L 2 T- 1]. The time rate of change of angular momentum L is given by ---+

dL dt

=

---+

d ---+ ---+ ) dt (r x p ---+

---+

=

---+

dr ---+ dt x p

---+

+

dP r x dt

---+

dp

=vxp+rxdt

=

---+

r

dp

x -

[as11 and

dt

p

are parallel, so11 x

p

---+

= 0]. ---+

Now, we are to show that, the time rate of change of angular momentum L is ----7

defined as torque. If an external force F(e) acts on the particle, then the external torque acting on the particle is defined as ----+

T(e)

= r?

----7

x p(e)

= r?

x

d---+

dL

dt

dt

--.!!... = -

=

. L.

Let us suppose that no external torque acts on the particle, then we have () T e

---+

=

dL dt

---+

---+

= 0 ::::} L = a constant vector.

Thus, if no external torque acts on a particle, then its angular momentum is conserved. It is the law of conservation of angular momentum. Planets moving around the sun is an example of this conservation law. Ex 1.4.1. The position vector of a particle of mass m 1rw'lring u:ndeT the influence of a force is given by r' = A sin wti + B cos wt] . Find the expressions for linear momentum and force. Also find the angular momentum particle about the origin.


10

= Asinwti + Bcoswt3.

SOLUTION: Here given that -:t at time t is given by --7

v

The velocity of the particle

d-:t = dt d [A' = dt smwti~ + B coswt]~'l = w[Acoswti - Bsinwt31.

Therefore, the linear momentum of the particle is given by

p = ml1 = mw[A cos wti -

Bsinwt31.

-+

The force F acting on the particle is given by

dp

-+

~

d

= dt = dt[mw(Acoswti -

F

~

Bsinwt.i)l

= -mw 2[Asinwti + Bcoswt31

= -mw 2-:t.

Now, the angular momentum of the particle about origin is,

L = 1" x p = [Asinwti + Bcoswt31 x mw[Acoswti ]

~

BSinwt31

k

Asinwt Bcoswt 0 mwAcoswt -mwAsinwt 0

= -mwAB[sin2 wt + cos 2 wt]k = -mwABk. 1.4.3

Conservation of energy

Energy conservation has most wide applicably, encompassing all branches of sciences equally. Consider that a particle of mass m is moved from a point A to B under --->

the action of an external force F(e). Then, work done by the force in moving the particle from A to B is given by

. =. j B

WAB

j' d .d-:t =. dt (m l1).d-:t B

--->

F(e)

A

A B

=m

J A

J B

dl1 -+ dt.d r

=m

A

J-+-+ B

d-:t d v.Tt -+

=m

v.d v

d-:t -+ as-= v dt

A

where v A, v n are the velocities at point A and n respectively, AND the scalar quantities TA, Tu are the kinetic energies of the particles in the positions A and B respectively. This relation is true for non-relativistic motions only.

1.4. MECHANICS OF A PARTICLE

11

(i) If TA > TB, WAB < 0, i.e., work is done by the particle against the force and the KE has decreased. The work done against dissipative forces like the frictional force always negative. (ii) If TB > TA, WAB > 0, i.e., work done by the force on the particle and the KE of the particle has increased.

Suppose that the external force acting on the particle is conservative in nature. Therefore, work done in moving the particle from the point A to B can also be written as B

WAB = ]

~.d7 =

A

B

]

-

~~ dr =

VA - VB

A

where VA and VB are the values of the potential energy of the particles at points A and B respectively. Thus WAB

= TB - TA = VA - VB =;. TA + VA = TB + TB

=;. T

+V =

constant .

Thus, if a particle moves under the action of conservative force, then the total energy of the particle is conserved, and has the dimension [ML 2T-2]. It is law of conservation of energy for a particle. A particle moving under the gravitational force is an example of this conservation law. However, if an given system frictional or other dissipative forces are present, then F .d7 is always positive and the closed path integral is non-zero. Such a system is called a non-conservative system. Ex 1.4.2. Find the period of oscillation for the potential V

= Vo tan 2 ax.

SOLUTION: Consider the conservation of energy in one dimension as

~mx2 + V(x) or, t =

= E

Vfin] 2:

dx

JE _ V(x)

+

constant.

For a moving particle, KE is always a positive definite quantity and as such there is the constraint V(x) < E. The period T of the oscillation is given by the time that the particle takes to travel from A to B and back. Thus, b

T

=

v'2m] JE-V(x)' dx a


12

where a, b are the roots of the equation V(x) = E, assuming E to be given. For the given potential V(x) = Vo tan 2 ax, the period of oscillation is

I a

T= v'2m

dx ; a= ~tan-l Vo tan 2 ax a

. VE -a

[E

YVa

a

= 2 ~J

YE+""Vo o Vk

cos ax dx; sin2 ax

k;2

=

2 -

E

E

+ Va

2 ~. -l[VE+Vo. ] 7r ~ = ;y E+""Vo sm E smaa =;y E+""Vo'

1.4.4

Equation of motion

The study of motion of a particle involves the relationship of space and time. When we consider the motion of a particle in an inertial frame, then we make the following basic assumptions: (i) The inertial frame is homogeneous: This assumption ensures that the equation of motion of the particle is independent of position of the origin of the coordinate system and orientation of the co-ordinate axes. (ii) The inertial space is isotropic: Under this assumption it follows that the physical properties are the same in all directions in the frame. (iii) The inertial frame time is homogeneous: This assumption ensures that a free particle which moves with a certain velocity in the inertial frame during a certain interval, does not move with a diffe'rent constant velocity during a later or an earlier time interval, Generally, the force acting on a particle may depend on its position, velocity and time and hence the equation of motion of the particle can be written in the form (1.5) This second order ordinary differential equation in space co-ordinates (7) and alter integrating it twice, we will get the equation of trajectory of the particle. There will be two constants of integration introduced and these will be determined by knowing the initial conditions of the particle. Taking the initial condition 7(to) = 11 0 , integrating equation (1.5) with respect to time, we get t

m,

t

J ~d' r

,f =

to

J--+P(--+ l' , ~ T ,t )d' ,t to

t

'* m[~ -

'Vo] =

JF to

(-;;7, ~,t)dt'.

(1.6)

13

1.4. MECHANICS OF A PARTICLE t~

.

J F (7, 7, t )dt'

The right hand integral

represents the impulse imparted to the

to

particle during time interval (t - to) and is equal to the change in the momentum of the particle. Integrating equation (1.6) again with respect to time, we get

J 'J ~F (~~ t

~ ( - to ) + m 1 r - ~ r a= ~ vat

t

dt

r , r ,t )d" t.

to

(1.7)

to

If the force Fe?, 7, t) is given, performing the integration in (1.7), we get the explicit form of the equation of the trajectory of the particle. For some given simple forms of the force F(7, 7, t) we shall discuss the nature of the trajectory.

(i) Let the given force F(7, ~

·ma.

7, t)

ticular, if F = mg. when considered constant.

g,

be of the form F(7, 7, t) = In parthe acceleration due to earth's gravity can be

~~~

~~~

~

(ii) Let the given force F ( .,. , 'r , t) be of the form F ( .,. , .,. , t) = F (t). For exam. pIe F (7, 7, t) be of the form F = F a sin wt which is experienced by charged particles in alternating fields. ~

~

~

~~~

~~~

~~

(iii) Let the given force F( r , r, t) be of the form F( r, r, t) = F( r). For exam. pIe F (7, 7, t) be of the form F = ~ 7 which is experienced by gravitational or electrostatic fields, under inverse square law. ~

~

~~~

~~~

~~

(iv) Let the given force F ( r , r ,t) be of the form F ( r , r ,t) = F ( r). This type of frictional force is experienced by a particle moving in a viscous medium. Ex 1.4.3. Forced Motion: For simplicity, let us consider an one dimensional motion along the x-axis. Let a time dependent force of the type F = Fa sin wt be acting on the particle. The equation of motion is

.. x with the initial condition t

=?

Fa. = -smwt m

= 0, x = Xa, v = Va.

:i: = Va

+

x = Xa

F + [va + -]t 'mw

Fa [1 - coswt];

mw

Fa

--2

'fnw

Thus we get integrating, sinwt.;

integrating

This problem is of interest in connection with scattering of electromagnetic radiation by free electrons such as those appearing in the ionosphere. Ex 1.4.4. Linear Harmonic Oscillator:

14


A block slides on a frictionless horizontal table. At equilibrium position, the spring is relaxed and exerts force on the body. When the body is displaced to the right, the direction of the force exerted by the spring on the body points to the left, and when the body is displaced to the left the direction of the force points to the right (restoring force). In each case, the magnitude of the force is proportional to the magnitude of the displacement of the body. When the mass is displaced through

Figure 1.1: Linear harmonic oscillator the distance x, the restoring force set up in the stretched spnng is F(x) The equation of motion can be written as

-kx.

(I2x dv F(x) = m dt 2 = m dt x =?

x

I F(x)dx = m I Xu

t

~~ dx = m I ~~ vdt to

Xo

t

X

~ I(-kx)d:r = m I :t(~v2)dt to

Xu

~

1

2

--kx + 2

1 .) 1 2 1 2 -k.1:6 = -mv - -mvo 2

2

2'

Let the mass be released initially from rest at a distance Xo from the origin. Therefore, the initial conditions are x = :1:0, v = Vo = 0 at t = O. Thus,

dx /'kV 2 2 v=-=V .!::.... x-x dt mO. ~

I

dx

CE

yf x 5-

x2

-

~A -

Xll

~ X

=

Xo

.' sm(wt + 2'7T )

m

It

dt .

0

= :1:0 cos wt;

w=~.

Deduction 1.4.1. In the above section, we have derived three immutable conservation laws of mechanics from Newton's laws of motion. These laws are the results of the first integral of equation of motion, and are thus called the first integrals of equations of motion. The range of validity of the conservation laws is quite broad, extending even to the domain of relativistic motions, where Newton's laws are replaced by laws of relativistic mechanics. These laws prove to be very powerful tools in solving mechanical problems in the following respects: (i) Conservation laws a.re independent of the details of the trajectory and often, of

1.5. MECHANICS OF SYSTEM OF PARTICLES

15

details of the particular force. These laws assure us many times that some aspects of motion are impossible and must left out. (ii) Conservation laws have an intimate relation with invariance. Their failure in certain cases may result in the discovery of new and not yet understood phenomena.

1.5

Mechanics of System of Particles

Here we extend the considerations of the previous section to a system of more than one particles. In mechanics, many time it is simpler to consider the motion of a body by dividing the body into large number of discrete points and by analyzing the motion of the latter. Hence, 'a system of particles' is a very useful concept. A system of particles is called a dynamical system.

1.5.1

Centre of mass

Consider a system of n particles, the ith particle of the system having mass mi and let its position vector with respect to some arbitrary base point is r\. The point G whose position vector is given by ----->

OG

~

n ~

~ - n - - L..J m 1 r i = m 1 i=l

1

=R=

2::

n 1~

~

M L..J mi r' 1 i=l

i=l

is called the centre of mass of the system of particles, where M is the total mass of all the particles. Physically, the centre of mass of the system can be thought of as a weighted average position of the system of particles. If the mass is continuously distributed, then the CM of bodies of simple geometric shape can be calculated by replacing summation by integration.

Theorem 1.5.1. The position of the CM is independent on the origin chosen. PROOF: To prove, the position of the centre of mass does not depend on the origin -----> ~ ~ ~ chosen, let us take some other point 0' as origin. Let 00' = a. If 0 P = r' ~, then

With 0' as origin, the centre of mass G', by definition, will be given by,

L

n

L

n

'i=l

i=l

~ 1 ~, 1 (~ o G = -M rn' '/' . = -M m' r' + ~) a il il 1 = !vI

L

~

'II

mi

1

ri

+M

1=1

----->

= OG

+~ Q =

~

n

L

mi

Q

i=l

----->

~

~

OG + 0 0 = 0 G.

This means that, G' and G are the same point. So, G, the centre of mass, is a fixed point whatever the co-ordinate system is used. Now

16

CHAPTER 1. MECHANICS OF PAR'UCLES

(i) If we take a co-ordinate system with the centre of mass (G) as origin (0'), then n

2::::

m~ T7:'

=

~

O. Here,

T7:'

is the position vector of the ith particle with centre

i=l

of mass as the origin. Thus, we may define the centre of mass of a system of particles as 'that point about which the vector sum of the first moments of all the masses vanishes.' (ii) A 'body' is a system of particles in which the mass is continuously distributed. If a body or any non-continuous system of particles occupies a region at all points of which the acceleration 9 due to gravity is the same, the centre of mass and centre of gravity coincide. In the absence of gravity, it is idle to speak of centre of gravity. It is better to speak of centre of mass.

1.5.2

Linear momentum

Consider a system of n particles, the ith particle of the system having mass mi and moving with velocities v\, 11 2 .... ,11 n respectively. T7~ is the position vector of the ith particle with respect to an arbitrary origin 0 and T7~ is the position vector of the same particle with respect to 0', the centre of mass, and R is the position vector of CM with respect to O. Then ~

.

where

Vi = T7 ~

.,

and v~.

= T7 i are the velocities of the ith particle with respect to 0 ~ ~ and R = V is the velocity of the CM with respect to O. The ---+

and 0' respectively total linear momentum P of the system of particles is given by

~ P = ""' ~ ---+ Pi = ""' ~ m~ ~ r i = ""' ~ mil ~ R + ~, r il d ""'~, ~ =M ~ V + dt ~ m'i r i = M V; ~

""'

---+,

as ~ mi r ~

=

~ 0

~

where we have assumed that mass of the individual particle does not change with time. The particles of the system exert force on each other due to mutual interaction (action and reaction). These forces are called inertial forces. In addition, the particles of the system may be acted upon by external forces due to sources outside the system. Thus, any particle will be under the action of the following two forces : Internal Force: If represents the interacting force on the ith particle due to the lh particle, then total inertial force on the ith particle will be ,

F;.

~

--+

F(t)",=LF J1.; j=1,2, ... ,n. J


17

Since a particle does not exert any force on itself, it follows that in the above summation j i= i. Therefore, the total inertial force on the ith particle due to all other particles should be expressed as

where the symbol L.j will imply that i = 1,2, ... , n, but j i= i. For example, if the particles are charged, the Coulomb forces are inertial forces. ~

External force: Let

F(e)

be the external force acting on the system, so that the

~

external force F(e)i on the ith particle. If Pl is the linear momentum of the ith particle, then according to Newton's second law of motion, the equation of motion for the ith particle is ~ l

~

or,

p(e)

d-t

---+

p(e)

+ p(i) =-.!!....:.. dt l

-

d-t

+ LF.1'i = %t

(1.8)

l

J

or,

L# + LFJi = L d~l; i,J .

For the system.

The equation (1.8) gives the equation of motion for the ith particle of the system. Now if F(t) be the total external force on the system, then

The factor ~ appears in the above equation, because in summing over both i and j, the pair of the ith and the lh particles appears twice, once for the ith particle due to lh and secondly for the lh particle due to the ith. Since both are same and should appear only once, the factor ~ has been included. Therefore, -t(t)

F

1~ -. -t ~ dPi +2~[PJi+PljJ=~&.

(1.9)

According to Newton's third law of motion, mutual action and reaction forces are -t -t equal and opposite ( such as gravitational field, electrostatic field), so P ji + p lJ = 0, i.e., the force of self-interaction is zero. This assumption is sometimes referred to as weak law of action and reaction. Thus, the simplified form of the equation (UJ) is (1.10) This equation shows that in the case of a system of n particles, under the action of external forces, the centre of mass of the system behaves like a particle whose

18


mass is equal to the total mass of the system and is acted upon the total external force. The motion of the eM of the system is independent of the inertial forces that exist among the particles of the system. Therefore, in an inertial reference frame the time rate of change of the total linear momentum of a system is equal to the total external force acting on the system. If the total external force acting on the ---t ---t system is zero, i.e., P (t) = 0, then the above equation becomes

dp t dt

~

L..,

---t

= 0

=>

~ ---t

a constant vector.

L.., P i =

i

i

Thus, if a system of particles is acted upon by no net external force, then the total linear momentum of the system is conserved. It is the law of conservation of linear momentum of a system of particles.

1.5.3

Angular Momentum

If -:;7 i be the position vector of the 'i th particle, then the torque acting on the particle is given by ~ ---t 'f' . t

x P (e)

L p.

-

+ ---t 'f' . X t

ith

d---t

Pi - ---t 'r t. x --' dt'

---t

(1.11)

Jt -

j

--->

The first factor -:ti x p(e) represents the torque on the ith particle due to external force acting on it, while the second factor -:ti x ",£Fji represents the torque on the j

particle due to internal forces acting on it due to all other particles of the system. If the internal forces between two particles in addition to being equal and opposite also lie along the line joining the particles - a condition known as the strong law of ith

--->

action and reaction, then all of the cross product vanish. Denoting -:tt x p(e) = the torque on the

ith

-n(e) ,

particle due to external force, the equation (1.11) becomes, ---t (e)

T.i

+ ---tr i

X

~ ---tp L..,

ji

=

---t

r i

X

dp i

dt

j

or,

L

-n(e)

+ I:[-:t i

x

I:F

ji ]

=

I:[-:t

i

x d!t]

j

or,

I:

-n(e)

+ L[-:t i

x

F ji ]

= 2)-:ti x d!i].

tJ

1

Now, the first factor of the (1.12)

(1.12)

i

"'£ T:;(e) 'i

system and the second factor ",£[-:;7i x

= -:f(e) , total external torque on the

Fji ] = -;P(i) , the sum of the torques due to

i,j

internal forces acting on the system. Thus, the equation (1.12) may be written as

-;pee) + -;P(i) = ~[-:ti t

x d!i].

(1.13)


19

-

The vector 7i - 7 j is a vector directed from the particle j to i. Since the force F ji is directed from particle i to j, the vectors 7 i - 7 j and Fji are parallel to each other and so -(') T ~ = O. Therefore, the equation (1.13) reduces to

-

-

- --

Let L ~ = 7i X Pi be the angular momentum of the ith particle, then the angular momentum of the system is ~ L = 2::: L ~ = 2:::[ l' i x p iJ. Thus, i

If the total external torque

-

7(e)

dL = -0 dt

i

acting on the system of particles is zero, then =}

-L

= a constant vector .

Thus, if total external torque acting on a system of particles is zero, then the angular momentum is conserved. It is the law of conservation of angular momentum. Theorem 1.5.2. Prove that if no external force acts on a system of particles, then the centre of mass of the system moves with constant velocity. PROOF: Let us consider a system of n particles of masses ml, m2, ... , mn with position vectors 7 1 , 7 2 , ... , 7 n respectively. If 11,1 = ml + m2 + ... + m n , then the

-

position vector R of the centre of mass of the system is given by


20 n

=?

= Lmt~t =

MV

t=1 -+

=?

where

dJ}

V is the

M

dV dt

n

LP\ i=1 n-+

n

= ~[LPi] = L dt

i=1

i=1

dp't dt

linear velocity of the centre of mass and

dr'

=

~i is the

linear velocity of the 'ith particle of the system. If F(t) is the total external force on the system, then according to Newton's law of motion, we get,

-+

--+

dX

where P = is the acceleration of the centre of mass of the system. If the total -+ -+ external force acting on the system is zero, i.e.,F(tot) = 0, then we have -+

dV -+ M dt = 0

-+

=?

V = a constant vector.

Thus, if no external force acts on a system of particles, then the centre of mass of the system moves with constant velocity. Theorem 1.5.3. Prove that angular momentum of a system of particles about the origin is equal to the sum of the angular momentum of the centre of mass of the system about the origin and the angular momentum of the constituting particles about the centre of mass. PROOF: Let us consider a system of n particles of masses ml, m2, ... , mn with position vectors -;;:'1, -;;:'2, ... , -;;:' n respectively. Let -;;:'~ be the position vector of the ith particle with respect to the centre of mass of the system, then -;;:' t = Ii + -;;:'~, -+ -+ -+ -+ where R. be the position vector of the centre of mass. Also, 11 i = V + v~, where

1/ t

-+

--+

= d

It'

and V

--+

= dd~

are the respective velocities of the

;t

ith

particle and the centre

d--+'

of mass, while 1/~ = is the velocity of the ith particle with respect to the centre of mass. The angular momentum of the system of particles is given by, 1

-+

L

-+ =~ L.J r i

-+~"""' -+

x mi v i

=

L.J (R

i

= ~-+ L.J R ~-+

-+

+ ~-+ L.J R

-+

=

x mt V

Now,L.J R x mi V

-+

~ -+

-+-+/

x m t (V

+

v i)

t

x mt ~

-+/ Vi

+ ~-+/ L.J r i x

-+

R x (L.J mt)V

i

+ -+/ r i)

-+

=

-+

mi V

+ ~-+/ L.J r i

-+/

x mi v i'1.14)

-+

R x MV

i

-+/

-+

~

/

d ~

-+

-+

L.J R x mi Vi = R x (L.... mi Vt ) = R x dt L.... mi( r i

t

-+

-+ i -

R)

i

-+

-+

= R x [M R - M R]

=

-+

~

0 as L.... m't -;;:' t

-+

= MR.

1.5. MECHANICS OF SYSTEM OF PARTICLES ~ -+1

~

---+

-+1

~ r i x mi V = ~ (mi r

i)

X

21

~

-+

-+

V = ~ mi ( r

i

-+

R)

i -

X

-+

V

i

-+

-+

= [M R - M R]

-+

X

V

-+

= O.

Substituting the values of the above factors in the equation (1.14), we have, -+

-+

L = R

-+

X

MV

-+1 +~ ~ r i

-+1

X

Pi'

i

-+

-+

The first factor R x M V represents the angular momentum of the centre of mass of the system about the origin, while the second factor L: T7~ x p~ represents the 1.

angular momentum of the constituting particles about the centre of mass. In case, -+ -+ the centre of mass of the system of particles is at rest, i.e., V = 0, or moves parallel -+ -+ -+ -+ -+ -+-+ to its position vector R, then R x MV = 0, so that L = L: r ~ x p~. Hence the t

angular momentum of the particles of the system about the centre of mass is same as the angular momentum of the particles of the system about the origin. Hence, the angular momentum of a system of particles is independent of the choice of reference point, in case the centre of mass is at rest or moves parallel to its position vector.

1.5.4

Energy of the system

The total force acting on a particle of a system is equal to the sum of the external force acting on it and the internal forces due to other particles of the system on it. If denotes the total force on the ith particle of the system, then

Pi

p. - p(ext) + ~p .. i ~

2 -

)2'

j

In case, the internal structure of the system does not change, the effect of total -+ -+ r force F 2 on the ill. particle will be to impart acceleration, i.e., F i = m 2 It i. Let us suppose that, the system of particles is moved from the configuration 1 to 2. Then, amount of work done is given by 2

W 12

=L t

JPi. dT7 1

1.

(1.15)


22

where Tl and T2 are the total KE of the system in configuration 1 and 2 respectively. A system of particles possesses following two types of potential energies. (i) Due to the interactions of the particles with each other, the system possesses internal potential energy. The internal potential energy of the ith particle due to its interaction with /h particle is denoted by ~~int), so that its internal potential energy due to interaction with all the particles of the system will be

v;,(int)

= L~~int). Since J

internal forces are conservative in nature, the internal forces on the to all other particles can be expressed as " ' F .. = _ ~ y(int) = _ "'~ y(int) ~

J1

1

~

t

1

J1

ith

particle due

•

j

J

The script i on the del operator indicates that derivatives implied by V -operator is to be carried out with respect to the co-ordinates of the ith particle. (ii) Due to the actions of external forces on the particles of the system, the system possesses external potential energy. The external potential energy of the ith particle due to its external force acting on it is denoted by v;,(ext). In case, the external forces acting on the particles of the system are also conservative, then ~F (ext) t

= _ ri .V(ext) v t . 1

Now, amount of work done in moving the system of particles from configuration 1 to 2 is given by

1 I: 1F~ext) 2

W12

= 2;= I

2

F 1.dT't

= I: 1[F~ext) + I:Fjt].dT'1

1

=

t

The first term

J

1

I

2

1 2

.dT'i

+ I:

1

F Jt .dT'1'

t,J 1

ext ~ L J2 --> F; ) .d T .,. represents change in potential energy of the system 1

t

_2~

due to external forces acting on it, while the second term

J F ji.dT't

L t,j 1

represents

the change in potential energy due to inertial forces between the particles of the system. Now,

I: .l F~ext) .dT'i = - 2;= .l ~ v;,(ext) .dT' 2

2

i

t

1

1

2

Further, "'1~ ~ F ji.d ~ r i = I,J 1

i

= V1(ext)

1

1"'1 "2 [~ 2

~

I,J 1

r i F ji. d ~

+~ F ij.d r J]. --J

-

V2(ext).


23

The factor ~ appears in the equation, because in summing over both i and j, the pair of the ith and lh particles appears twice, once for the ith particle due to the yth and secondly for the lh particle due to ith. Since both are same and should appear only once, the factor ~ has been included. Now, ~~int), so the above j{ = equation becomes,

F

Vi

2

= -~L:

/ Vi~~int).(d7i

- d7 j )

t,J 1

--

_~ ~ / 2~

2

Vy(mt) d-7 .. _ ji . r t

tJ -

V(int) _ V;(int) 1

2·

i,j 1

+ [Vj(int) - V;(mt)] =? T+ V = Constant.

=? WI2 = [V?xt) - V;(ext)] =? TI

+ VI = T2 + V2

= VI - V2

Thus, if the external and internal forces acting on a system of particles are conservative, then the total energy of the system of particles is conserved. It is the law of conservation of energy for a system of particle system.

1.5.5

Kinetic energy

Here we are to show that, the KE of a system of particles is equal to the sum of the KE of the motion of the centre of mass and the KE of motion of the particles with -7 -7 respect the centre of mass. Let V be the velocity of the centre of mass and v~, the -7 velocity of the ith particle with respect to the centre of mass, then 7J i = V + 7J~. The KE of a system of particle with reference to the origin of the co-ordinate frame is given by


24

1L: m-v-. n

1 2 = -MV +-2 2

12

t

t

i=l

Like angular momentum, we find the kinetic energy is also of two terms. Thus, the KE of a system of particles is equal to the sum of the KE of the motion of the mass particle Ai, i.e., centre of mass and the KE of motiol). of the particles with respect the centre of mass.

1.5.6

Rigid body

A system of particles is called rigid body, if the following conditions are satisfied (i) The distance between any two neighbouring particles is infinitesimally small. (ii) The distance between any two particles does not change due to any external force. (iii) As each particle exerts action on its neighbouring particle in contact with it, so equal reaction is introduced in the opposite direction. Then the resultant of all such actions and reactions must be zero.

1.6

Motion of variable mass

So far we have discussed the motion of bodies whose mass remain constant throughout the motion. In this section, we shall discuss the motion of the bodies whose mass changes during motion. Let a body of variable mass m( t) moving with a velocity v at time t receive an increment 8m of mass in an interval of time Ot. Let u be the component of velocity of this increment 8m in the direction of v just before union with m. Let F be the external impressed force acting on m, which may be due to gravitational and atmospheric resistance in the same direction and let v + 8v be the velocity of the body at time t + 6t. Hence the change in momentum in time ot is

= (m + om)(v + 8v) - (mv + uom) = m8v + vOm + 8m.8v - u8m. During the time interval 8t, the change in linear momentum of the main body Can be written as

.

mOv + 'v8m + 8m.ov - u8m

6t->O

Ot

= hm - - - - - - - - - - = m

dv dt

dm

+ vyt -

dm d dm uYt = dt(mv) - uYt.

Following the Newton's second law of motion, applied force will balance the rate of change of linear momentum of the system. Hence,

d dm dt(mv) -1J.Yt

=F

(1.16)

1.6. MOTION OF VARIABLE MASS

25

which is the equation of motion for variable mass. Below are some important examples of varying mass. Ex 1.6.1. A spherical raindrop of radius acms falls through a vertical height h, receiving throughout the motion an accumulation of condensed vapor at the rate of I< grammes per square cm. per second, no vertical force but gravity acting; show that when it reaches the ground its radius will be I< ~[1

+

VI +

2\a;'2]'

SOLUTION: Let the radius and the mass of the raindrop be respectively rand M, when it has fallen through a distance x in time t. If p be the density /c.c, then, 4 37fr3p.

=

M

Initially, t = 0, x = 0, r = a. Since, the vapor accumulates at the rate of kgm./sp.cm./sec., so, by the given condition, dM 2 dr dt = p.47fr .k =;. dt = k =;.

Using the relation, ftCMv) d dx dt(M dt)

T

+ a;

= kt

as t = 0, r = a.

= P, we have

=

d4 3 dx Mg =;. dt[3 7fr p dt]

=;.

~(r3dx) dt

=;. r3

= g1'2 =

dt

~; =

gp

ft r 3dr k

:k 1'4 -

43

= 37fr

.~:4;

dt

as r = a,

~; =

0

9 4 4 9 a4 d1' 4kT 3 (r - a ) = 4k2 (r - 1'3) dt 9 1'2 a4 ga 2 =;. x = 4k 2 [2' + 2r 2 ]- 4k 2 ; as l' = a,x = 0 g a4 g r2 - a 2 2 =;> x = 8k 2 [r2 + r2 - 2a ] = 8k 2 ( )2. r

dx

=;. dt

=

When x = h, we have,

{2h

2

2

r - 2ky T.r - a = 0 =;.

T

=

~[2k fih ± 2

YT

,-----

2 8k h

g

+ 4a 2 = k fih[1 +

YT

1+

ga 2 2hk 2 ]

where we are taking the positive sign. Ex 1.6.2. A mass M is fastened to a chain of mass m per unit length coiled upon (J, 1'0'//,.'111, hO'f"izontal plane, who8e coefficient of friction is p,. The mass is projected from the coil with velocity V. Show that it will be brought to rest in a distance M[(1 m

+ 3mV2)1/3 -1]. 2111/1,9


26

SOLUTION: Let x be the chain uncoiled at time t, then a mass of (M + mx) is moving with velocity v(say) and a mass of [M + m(x + 8x)] is moving with velocity v + 8v at time t + 8t. Hence, [1\1 + m(x

+ 8x)](v + 8v)

- (M + mx)v = change in momentum

= impulse of the acting forces = -/L[M + mx]g8t. In the limit, 8t

~

0, we have,

(M

dv

d

or, dt[(M +mx)v] or,2(M

dx

+ mx) dt + m dt v =

= -J.L(M +mx)g

+ mx)v d~ [(M + mx)v] = or,[(M + mx)v]

2

-211.(M + mx)2g

2J.Lg = --(M + mx) 3 + Cj 3m

where C is a constant of integration. Initially, x 21.J'l~J3 = C. Hence,

Now, let the velocity v

1. 7

=

°

when x

-J.L(M + mx)g

= Xo,

=

0, v

= V,

integrating,

so that C

= M2V2 +

so we have,

Nonrealistic rocket motion

A rocket fixed from the earth will always be affected by the gravitational pull of the earth. Other celestial objects are at great distances from the rocket and the effect of such objects on the motion of the rocket can be ignored. To simplify the problem tltill further, we are considering the free flight of the rocket based on the motion of a body of variable mass, the thrust being created by ejecting a part of the mass of the body, neglecting the effect of rotation and gravitational pull of the earth.

1.7.1

Single-stage rocket

Suppose that, a rocket (propelled by burning fuel) having a mass M(t) at the instant t and moving at a velocity v throws away a fuel mass dM' at a velocity u. It should be borne in mind that 1\1 and d1\1' are relativistic masses and the velocities v and u are

1.7. NONREALISTIC ROCKET MOTION

27

measured relative to the inertial reference frame in which the motion is considered (and not relative to the rocket). Then the law of conservation mass gives dM+dM'=O.

Obviously, dAf < 0 as the mass of the rocket decreases. To derive the equation of motion we are to use the conservation law of momentum. At the instant t, the total momentum of the system is Mv, while at the instant t + dt, it is given by (!vI + dJll) x (v + dv) + udM'. The momentum conservation law for this isolated system can be written in the form Mv

= (M + dM) x (v + dv) + udM' => Mdv

+ vdM + udM' = O.

The term dvdlvI has been neglected, as it is an infinitesimal of the second order of smallness. Taking dlvI + dM' = 0, into consideration, the equation of motion is dM !vIdv+vdM-udM , =o=> -d ( Mv ) = u dt dt

(1.17)

which is valid for both relativistic and non-relativistic cases. If the velocities are small, they can be added by using the formula from classical mechanics, and hence u can be written as u = u' + v, where u' is the velocity of the ejected mass relative to the rocket. Hence, d ( ) dt M v

=

(' u

) dM

+ v dt => M

dv dt

, dM

= u dt'

(1.18)

where 111 = Mp + Mf(t) + Ms; Mp denotes the mass of the payload, Mf denotes the mass of the fuel and Ms the mass of the structure or casing of the rocket. This is the equation describing the motion of rockets at non-relativistic velocities in the absence of external forces. If the rocket is subjected to the action of a force F, then the equation of motion takes the form M dv,.._ F dt -

+u

' dM dt .

(1.19)

Let M f + !vIs = Mo. Initially, let Mf(O) = EMo, so that Ms = (1 - E)Jl10. Here (1 - E) is called structural factor and the parameter E is the ratio of the initial fuel mass to the initial rocket mass. Obviously, 0 < E < 1 and it is preferable to have E, nearly equal to 1, in the sense that the structure should be as light as possible, such that we shall be able of carry more fuel. In practice, E lies in (0.7,0.9). Once the fuel is ignited, at any time, we have M = Mp + Mf(t) + (1 - E)Mo.

Deduction 1.7.1. Tsiolkovsky formula: Let us consider the acceleration of a rocket moving in a straight line, assuming the velocity of the ejected gases to be constant relative to the rocket. In absence of all external forces (including gravity and air resistance), the equation of motion of the rocket is M dv dt

=

_u,dM. dt


28

The negative sign on the RHS is due to the fact that in the case of acceleration, the exhaust velocity u' is directed against the velocity v. In other words, v, increases with the decrease of lvI. Let v and M t be the velocity and mass of the system at instant t respectively and vo, Mo those at t = 0 (before acceleration). In this case, v

/

Mt

dv

Vo

=

-u' / -dM => v M

t M = Vo - u ,log -.

Mo

Mo

Hence the change in speed of the rocket of the rocket in any interval of time is independent of ~'1, i.e., how the mass is released and depends only on the exhaust velocity 'u' and on the friction of mass exhausted during that time interval. When ~f~ = e ~ 2.72, then v - Vo = u' and ~~ = e 2 ~ 7.4, then v - Vo = -2u'. However, it is impossible to construct a rocket with mass ratio 7.4 : 1. Also

which is the Tsiolkovsky formula. The first form expresses the change in the velocity of the rocket, when it changes mass from Mo to M t , while the second formula expresses the mass of the rocket whose velocity changes from Vo to v. Deduction 1. 7.2. Characteristic velocity: Suppose that it is required to change the velocity of a rocket (acceleration, deceleration, change in direction of flight). In the reference frame, in which the rocket is at rest at the given instant of time, the problem can be reduced to imparting a certain velocity v to the rocket in a direction which would ensure the desired maneuver. The fuel consumption in this maneuver for a rocket flying outside the gravitational field can be taken into account with the help of the formula M = Moe-(v-vo)/u', with Vo = 0, where Mo is the mass of the rocket before maneuver. The velocity v required to be imparted to the rocket is called characteristic velocity of the maneuver. Deduction 1.7.3. Let us suppose that the fuel is burnt and ejected at constant rate d~f = -/-L and it lasts for time T. The quantity /-LU' is called the relative force. If the direction of u' is opposite to that of v, the rocket is accelerated. If the two directions coincide, the rocket is retarded. For any other relation between u' and v, the rocket's velocity changes not only in magnitude but also in direction. Let Ms be the mass of the structure and that of the fuel as t = 0 be Mf' then Mo = Ms + Mf. Hence at an instant t, the mass of the vehicle-fuel system can be written as

Suppose at t

= T(burn

out), the entire fuel is burnt, in which Mf(T)

T = ~~o. Hence the rocket mass at any time t in (0, T) as M(t) = Mp

=

0 so that

+ Mo -

/-Lt.


29

Using this value of M t , from equation (1.18), we have,

dv, J.Lt ) dt = -u J.L

+ Mo -

(Mp

'J t

v

J

dv = -u J.L

:!:::}

Mp

0

va

,

=> ~} = Vo + u log[

dt

+ Mf(O) + Ms -

Mp

J.Lt

+ Mf(O) + Ms - J.Lt + Ms 1

A1p + A1f(0)

= vo + u' log[1 - M J.L M tl p+ 0

(1.20)

which is the velocity acquired by the rocket in time t. From this equation, it is clear that, the performance of the rocket depends on (i) u', the exhaust velocity, (ii) (1 - e), the structural factor, (ii'i) ~ = Mf(~rtMs = mass nitio. Also, ,

v = Vo

t

x

=>

+ u + log[1 -

J

elx = Vo

o

t

J+ J elt

_

=> x - vot

log[1 -

u'

0

t

eMo

+ Mf(O) + Ms Tl

Mp

M:~~o ;ldt

0

_ T(Mp

+ Mo) u'[(I _

1I

elViO

M

eMo p

t ) 1 ~(1 _ c:Mo t) og e M 1I T p + lViO

+ lViO T 1I

+ 11

which is the distance covered by the rocket in time t.

Deduction 1.7.4. The most important problem is to attain the maximum velocity with the minimum consumption of fuel, i.e., for the minimum difference between Mo and M I,. However, the velocities of ejected gases are restricted. Therefore, the rocket attains maximum velocity at t = T, when all its entire fuel is burnt out and the velocity V max caused by the expenditure of the entire propellent is given by, V max

'[

eMo

= Vo + u log 1 - Mp + Mf(O) + Ms = Vo

[ + u'log 1-

1

fMO M M .1

p+

0

This shows that, the larger the value of the ratio ~, the greater will be the maximum velocity attained by the rocket. It is this logarithmic dependance on mass that requires enormous rockets to deliver small payloads.

Deduction 1.7.5. The rocket is moving vertically upward if the constant gravitational pull of the earth (neglecting air resistance) on it is assumed to be constant,

30


then the equation of motion of the single stage rocket can be written as

A1

dv = dt

_u

,dM _ Mg dt t

v

=* JdV = -UIJ.LJ 0

110

M: p+

=* v = Vo + u I log [1 Now , at t = T =

EMo 11

v

'

I

0 -

t

gJdt

J.Lt

0

J.Lt] 111, - gt. Mp+ 0

we have ,

= 'U -

= 'U I log[l -

'00

eNlo geMO ] - -Mp+Mo 1-"

which gives the velocity increment caused by the expenditure of the entire propellant. Also,

v = Vo

+ u I log [1 t

h

NIp

J.Lt] + Mo - gt

t

t

=* J dx = Vo J dt + u ' J log[l - Mp ~ Mo ]dt - 9 J tdt o

0

0

0

J.Lt J.Lt =* h = vot - -gt - -(Mp + Mo)[(l - 11'[ A1,) log(l - M 111, )]. 2 J.L p+ 0 p+ 0 1

u'

2

which is the expression for the height attained by the rocket at time t. Hence the change speed of a rocket in the gravitational field does depend (in construct to that in free space) on how the mass is released. The shorter the burn time, greater the change in velocity. Hence for greater gain in speed, the rocket fuel should burn as quickly &"l possible in the region of gravitational field (in absence of atmosphere). Also at burn out, T = E~O, hence at burn out, the height attainable by a single stage rocket is 2

voeMo 1g eM6 H -- - - - - -u ' [111,0 (1 J.L 2 2J.L2 J.L

1.7.2

-

eMo] . e ) + M]log [1 p Mp + Mo

Multistage rockets

Normally, the rockets are expected to carry some load, called the payload. Payload may be the load of a satellite which is to be placed in an orbit around the earth, or of bombs in the case of missiles. The payload and the body of the r~cket have a fixed mass. All the load carried by a rocket is not useful right up to the end of the flight. Ratio ~:, has practical limit and it cannot be increased indefinitely. Hence it is expedient to get rid of such a load as soon as possible. A single rocket (one stage rocket) will not attain high velocities that are required, and for this purpose, the multistage rockets are used.


31

Ex 1.7.1. Find the optimal parameters for a two-stage rocket in order to impart a given velocity to a given payload of mass 500kg, the final velocity 8km/ s, and the velocity of ejected gas 2km/s. Assume that according to the constructional requirements, the mass of the fuel in this stage. SOLUTION: Let MI and M2 be the masses of the fuel in the first and second stages of the rocket respectively and M be the mass of the payload. If EMI and EM2 be the masses of the structure of the empty rockets in the first and second stages respectively, then the total masses of the two stages are (1 + E)MI and (1 + e)M2' The total initial mass mo of the rocket is

rno

=

M + (1 + E)(MI + M 2);

e = 0.1.

Now, if VI and V2 be the speeds of the rocket at the end of consumption of fuel of the first and second stages respectively, then I

VI -

Vo

mo

= u log - ;

V2 -

ml

VI = 1l

I

log

lvI + (1 ~I

11'

+ E)M2 + EM 2

where the negative signs are dropped to represent the numerical value only. The initial velocity Vo = 0 and let ~7 = a, then we get,

M

mo -

=}

+ (1 + E)M2 M +cA12

mI'

[AI + EMI + (1 + E)AI2][M + cA12J M + (1 +E)M2

Also, for minimum value of mo, we have ~~2 C)(MI + M2)' Thus,

= 0,

-e

a

-

-a

= moe

i.e., ~~;

=

-1 as mo

= M + (1 +

[E~ + (1 + E)](AI + EM2) + [M + EMI + (1 + E)AI2Jc M + (1 + E)M2 _ [M + eMI + (1 + E)M2J(M + EM - 2)(1 + E) = 0 [M + (1 + e)M2J2 M2 =} MI = M2 + (1 + e) ;) . Using the relation mo = M + (1 + E)(MI + M2), we get e

a

= M+(I+E)(MI +M2) M+(I+E)M2+EMI

M+(I+E)M2 M+EM2

X ---'---'----

[M + 2(1 + E)M2 + (1 + E)2%!-][M + (1 + E)M2J 1If2

[M + (1 + 2E)M2 + E(I + e):xf][M + EM2J Now, a

= ~ = 4,e = O.I,M = 500kg,

tcM M i][500 + l.IM2J

[500 + 2.2M2 + e [500 + I.2M2 + 4 _

=}

lOe4

so

~'cici J[500 + 0.IM2J

11 M 5 0+0.1 2

= [500 + 2.2M2 + 0.00242MiJ [ 0

e4

----J. 500 + l.IM2

32


The mass ma = 500+2.2M2+O.00242Mi has its minimum value when the expression within the parentheses has its maximum value as a function of M 2 . The extremum condition for the expression within the parentheses gives the equality, e2

=

2 500 + 1.1Ah =? M2 = 500(e - 1) = 8846.795kg 500 + 0.IM2 1.1 - 0.le 2

+ 0.1) (8~~~)2 = 181039.6998kg. mo = 500 + 1.01 x (8847 + 781040) = 209375.7kg.

Ni} = 8847 + (1

The mass of the two stages and that at take-off are (1

+ e)Ml =

199144kg.; (1

+ e)M2 = 9732kg.; ma =

209376kg.

For a single stage rocket, v - Va = u/log[M.t~~~;f], where M f is the mass of the fuel. Using Va = 0 and the above data, we see that for V = 8km/ sec, 500 + 1.1Mf 8 log 500 + O.IMf = "2 =? Mf = -6147kg. But the amount of fuel cannot be negative. Thus, a single stage rocket cannot achieve above final speed. Also, M < < M f so, the mass ratio Mr is given by Mr

::::::J

M

+ (1 + e)Mf + eMf

M

::::::J

1+e

= 1+~

E

e

i.e., mass ratio in a rocket cannot be increased at will. Here e and with this mass, a rocket can attain a speed V

= 0.1 and so Mr = 11

1

= u 10g(1 + -) = 2.083km/ sec. c

which is much less than 8km/ sec. This can be increased further by constructing a lighter empty (e-smaller) rocket, but technically it is impossible to construct a rocket with e < 0.1.

1.8

Limitations of Newton's Law

The limitations for applicability of Newton's law are, (i) These laws are only valid in inertial frame. For non-inertial frame, the transformed v~rsion of.Newton's Laws are made because of the presence of pseudo force in that frame.

cit particle on which Newton's law is required to apply, particle velocity be close to that of light in vacuum, the by Einstein's relativity, mass will change with velocity.

(ii) The velocity

1.9.

VIRIAL THEOREM

33

But Newton's law only is applicable for stationary mass. Also in setting up the differential equations of motion we usually choose particular co-ordinate system from symmetry and boundary conditions of the particular problem which helps to simplify it. However, in Newtonian formulation, Newton's equations of motion no longer maintain their simple form form in all co-ordinate systems and in fact writing down of fiuch equations in different co-ordinate systems is not straight forward. Vector form (independent Cartesian Spherical Cylindrical of co-ordinate form) (x,y,z) (r,e, =1= m¢; Fz =1= mz Such a dependence on the co-ordmate system IS, however, undesIrable and we should be able to write the equations of motion without any specific reference to the coordinate system used. After Newton, these developments in mechanics were contributed by D'Alembet, Lagrange, Euler, Poinsot, Einstein and others.

me

1.9

Virial Theorem

Virial theorem was introduced by Rudolf Clausius (1870). Consider a system of n interacting particles that has achieved dynamical equilibrium through some kind of central force interactions. (i) Let be the applied forces on the ith particle of mass m~ with position vector -:t i. The virial of the system is defined as

Pi

which is connected with the total potential energy of the system. (ii) Let T be the total translation kinetic energy of the system, then d--+ - 2 ~ m~ dt . dt . n

--+

T-~"

.drt~

i=l

The virial theorem states that, for an n body system, which is bounded in size and velocity, the average value of the KE over a large time interval equals minus half times the average value of the PE of the system and is equal to minus times the total energy of the n body system. For this, taking the time derivative of the quantity n

G

= I: p~.ri,

which is a bounded function, we get,

1,=1

dG

It

n

n

n

i=l n

i=l

i=l

i=l

dt = LP~ri + LPiTi = LPirt + L = LFiri + 2T = W + 2T; i=l

miTiTi ~-:ti

as

mi--a:i2 =

~

Pi

_

= Fi


34

=?

d n -["'p-r'] dt L.J t 1. = W

+ 2T.

i=l

Thus the time average of a quantity over the time interval r is obtained by integrating both sides with respect to t from 0 to r, and diving by r, we get

J

JdI

T n T

1 :;:

d [L.JPirt] '" dt

o

= :;:1

dG dt

=W +2T

0

t=l

=?

W

-

+ 2T =

1 -[G(r) - G(O)], r

where Wand T stand for the time averaged values of Wand T respectively. If the motion is periodic with r is chosen as period, then G(r) - G(O) = O. For a stable system, -Y;\, the distance of the body and the velocity of that body are bounded and then also G(r) - G(O) ~ 0 as r ~ 00. When the motion is not periodic, the co-ordinates and velocities for all particles remain finite and hence there is an upper bound of G. Taking r sufficiently long, then G(r) - G(O) >::::J O. Thus in both cases,

dJ:',

TV

+ 2T = 0;

if r --;

00

which is the well-known virial theorem and -~ W is called virial of Claussius. The time scale over which the process of virialisation takes place is called the virial time scale of the system. This theorem has wide applications in Stellar dynamics and forms the basis for statistical calculations of stellar clusters. Ex 1.9.1. Consider a system in which the total forces acting on the particles consist of conservat'ive forces PI and frictional forces Ii proportional to the velocity. Show that for such a system the virial theorem holds in the form T = -~ 2: P:r t , providing t

the motion reaches a steady state and is not allowed to die down as a result of the frictional forces.

SOLUTION: Here the forces are the sum of non-frictional forces (conservative) PI and the frictional forces ft, proportional to the velocity. Thus the virial W = 2: Pir t t

depends only on the conservative forces PI and have no contribution of it- Further it is assumed that the motion of the system must not be allowed to die down as a result of the frictional forces. The KE T of the system is T

1

1

2 = -2 '" L.J m'v t t = 2 '" L.J m'v'v' t t t i

t

Here given that the frictional force

Ii

=

CVi; C

= some constant and

Vi

= velocity.

1.10. EXERCISE

35

From Newton's fundamental equations of motion, we have,

Pi = F: + I! = F: + CV! => 2:P!r! = 2: F:ri + C 2: VITi =>

2: Pir! = 2: FIT! + 2: V!T!; C

Taking time average

To maintain the motion, energy will constantly involved so that all time averages of the quantity which is a total time derivative of a bounded function zero, we have,

1.10

Exercise

Ex 1.10.1. State Newton's laws of motion. What are inertial frames? Give examples of inertial and non-inertial frames. Ex 1.10.2. Define angular momentum of a system of particles. Prove that the rate of change of linear momentum of a system is equal to the mass x acceleration of the mass particle at the centroid. Ex 1.10.3. Prove that, the total angular momentum of a system about a fixed origin o is (i) the total angular momentum of the mass particle at the centroid about 0, and (ii) the angular momentum of the system about the centroid for motion related to the centroid. Ex 1.10.4. Show that the angular momentum is conserved in motion under central force. Ex 1.10.5. Prove that if the centre of mass of a system is at rest or moves parallel to its position vector, the angular momentum of the system of particles is independent of of the choice of the reference point. Ex 1.10.6. Find the period of oscillation for the potential V(x)

= -Vosech2 ax.

Ans: T= ~Jj. Ex 1.10.7. Prove that the centre of mass of a system of particles moves as if the external force on the system on the constituting particles were concentrated at the centre of mass. Ex 1.10.8. The KE of a system is the sum of (i) the KE of the mass particle at the centroid and (ii) the KE of the system for motion relative to the centroid.

36


Ex 1.10.9. The rate of change of angular momentum of a system ahout a fixed point is equal to the SU'll of the moments of the external forces about the same points. Ex 1.10.10. Show that the KE of a system of particles is equal to the sum of the KE of the motion of the centre of mass and the KE of motion of the particles with respect the centre of mass. Ex 1.10.11. A rocket of total mass 1.11 x 105 with fuel 8.70 x 104 Kg is to be launched vertically. It is designed to consume fuel at a constant rate 820Kg/ s. What should be the minimum exhaust speed for lift off at launch? Vr = 1.327km/ s Ex 1.10.12. A two stage rocket is to be built capable of accelerating a lOOK 9 payload to a velocity of 6000m/ s in free flight in empty space. If the exhaust velocity of the fuel used be 1500m/ s and an empty rocket (without fuel or payload) weights 10 percent as much as the fuel it can carry, then find the optimum choice of masses for the two s.tages so that the total take-off weight is minimum. Show that a single 36, 196kg, 1769kg. stage rocket can never attain such a speed.

Chapter 2

Generalised Co-ordinates A particular co-ordinate system chosen from the symmetry of the physical system helps to simplify the problem. For example, when the equation of motions are expressed through vectors, these equations do not depend on any specific co-ordinate system and the physical significance of the equation is also clearly seen. Such a dependence on the co-ordinate system, however, is undesirable and we should be able to write the equations of motion Without any specific reference to the co-ordinate system used. Here we shall define 'neralised co-ordinates which are independent of the co-ordinate system and may not be the usual spatial co-ordinates.

2.1

Constraints

We know motion of a system can not proceed arbitrarily and is restricted to occur alollg some specified path, or on a surface (plane or curved) arbitrarily oriented in space. Any restriction (geometrical conditions) which must be satisfied by the coordinates, on the freedom of motion of a dynamical system, in some way, is known as constraint and the force responsible is called constraint force or force of constraint. (i) A constraint is analytically written in the form

f(7 k, -t k, t) = 0 where 7

k,

(2.1)

-t k are position vector and velocity of the kth particle respectively,

t is time. This type of constraint is called differential or kinematical constraint.

(ii) A system without any constraint is called free system.

-t

(iii) A finite or geometric constraint is free from k and can be written in the form f(7 k, t) = O. If the finite constraint does not contain the time t explicitly, i.e., = 0, then the constraint is called stationary, otherwise it is non-stationary.

¥t

n

(iv) The general form of the constraint can be written as

k=l

lk' D are functions of t and

.

L lk 7 k + D

-t k, lk are not simultaneously zero. 37

= 0, where

CHAPTER 2. GENERALISED CO-ORDINATES

38

Physically, constraint motion is realized by the forces which arise when the object in motion, is in contact with the constraining surfaces or curves. They reduce the number of co-ordinates needed to simplify the configuration of a system and therefore simplify the mathematical description of the problem. A motion along a specified path is the simple example of the constraint motion. In a rigid body, the constraints on the motion of the particle are the distance Tij remain unchanged. The constraints may be classified as follows, for all dynamical systems, according to different types of constraints and forces to which the dynamical system may be subjected.

2.1.1

Holonomic Constraints

If the constraints involved in a system be such that the condition of constraints can be expressed as equations connecting the co-ordinates of the particles ( and possibly the time) having the form

(2.2) then the constraints are said to be holonomic and the system is called connected holonomic system. In this equation, j denotes the jth constraint. We know, the distance between any two particles with position vectors Ti and Tj of a rigid body remains unchanged, Le., (Ti - TJ )2 = Cfj. This equation of constraint is an example of a holonomic constraint of rigid body. It may be pointed out that, sometimes ." the constraints on a system are specified by a restriction on the velocities of the particles, rather than the co-ordinates and still the constraints are holonomic. This is true due to the fact that the velocity equation is integrable with respect to time to yield a resultant relation of the form (2.2), which is free from velocities. Ex 2.1.1. Prove that the constraint qlql holonomic.

+ q2q2 + qlq2 + q2ql = k,

k

= constant,

is

SOLUTION: The given constraint can be written as 1d

2

2

d

2dt [ql + q2J + dt (qlq2) = k => [q~ + q~J + 2qlq2 = 2kt + c => [ql + q2]2 = 2kt + c. Hence the given constraint qlql + q2q2 + qlq2 + q2ql = k is holonomic. 2.1.2

Non-Holonomic Constraints

Constraints that can not be expressed in the type (2.2) is called non-holonomic constraints. It may be in the form of an inequality. Some of the constraints of a system may be expressed as (2.3)

For example, the walls of a gas container constitute a non-holonomic constraint. The constraint of a particle placed on the surface of a sphere, under the effect of

2.1. CONSTRAINTS

39

gravity, which can be expressed as r2 - a 2 2: 0; a = radius of the sphere, is also non-holonomic. There are, however, dynamical systems in which equations among velocities cannot be integrated to give the relation between the co-ordinates. Such dynamical systems with non-integrable differential constraint are then taken as nonholonomic constraints.

2.1.3

Scleronomic Constraints

If the constraints are explicitly independent of time, i.e., finite stationary constraint, then it is called scleronomic. In other words, a scleronomic system is one which has only 'fixed' constraints. For example, the constraint in the case of a pendulum with fixed support is scleronomic constraint.

2.1.4

Rheonomic Constraints

If the constraints contain time as an explicit variable, i.e., non-stationary, then it is called rheonomous. In other words, a rheonomic system has 'moving' constraints. For example, consider, a simple pendulum whose length changes with time, its constraint expression is time dependent and this assigned motion is rheonomic. Similarly, if the radius of the sphere is changing with time.

2.1.5

Bilateral Constraints

When the equations of constraint are in . . he form of equalities, then it is called bilateral constraints. The general form of the bilateral constraint can be written as

(2.4) At any point on the constraint surface both the forward and backward motions are possible. In the bilateral constraint of motion on the surface, the force of constraint may be either direction normal to the surface.

2.1.6

Unilateral Constraints

When some of the constraints are expressed in the form of inequalities, then it is called unilateral constraints. At some points no forward motion is possible. The general form of the unilateral constraint can be written as : (2.5)

In the unilateral constraint of motion, the force of constraint acts only one direction which is the outward normal to the surface.

40

2.1. 7


Conservative Constraints

When the total mechanical energy of the system is conserved, while performing the constraint motion, is called conservative constraint. In this case constraint forces do not do any work. A conservative system is subjected to potential type of forces. For example, gravitational force field is conservative.

2.1.8

Dissipative Constraints

When the total mechanical energy of the system is not conserved, and the constraint forces do not work, then it is called dissipative constraint. In this case constraint forces do not do any work. For example, Lorentz force (force on a charged particle moving in an electromagnetic field) is a conservative force, because, the potential energy (V) of the charged particle in an electromagnetic field is velocity dependent. Result 2.1.1. When we are to solve mechanical problems, the constraints introduce the following two types of difficulties 1. Non-independence of Co-ordinates: Since the co-ordinates qi of the particles of a system are connected by the equations of constraints, they are no longer all independent. Hence the equations of motion for the particles of the system -t

F(ext)

+ ""' ~

-t

F .. tJ -

d-tp. dt

-_.

J()#i) -t

-t

cannot be solved independent of all others, where F(ext) is an external force, F ij is the internal force on the ith particle due to the lh particle, Pi is the linear momentum of the ith particle. 2. Forces of Constraints : Constraints on a system are always associated with the forces exerted on the system by which they impose restrictions on the motion of the system. Such forces are called the forces of constraints. They are among the unknowns of the problem and cannot be specified directly and must be obtained from the solution we seek. The problems involving holonomic constraints allow elimination of dependent coordinates (by introducing proper set of generalised co-ordinates) and hence a formal solution. So we shall overcome the first difficulty, we assume that any constraint, if present, is holonomic. We shall overcome the second difficulty by formulating the laws of mechanics in such a way that the forces of constraints disappear and in that case a problem can be solved with only known applied forces. This is done by using the fact that "the virtual work done by the forces of constraint is zero" .

2.2

Generalised Co-ordinates

The cartesian co-ordinate system is not only possible system of co-ordinates required to define the position of mathematical system. In fact, the choice of the co-ordinates is perfectly arbitrary and is determined on the basis of the problem concerned. Now,

2.2. GENERALISED CO-ORDINATES

41

we may think a wide variety of possible co-ordinate system, any set of parameters which give an unambiguous representation of the configuration of the system of coordinates in more general sense. The minimum number of independent co-ordinates required to completely specify the configuration of the dynamical system at any given time is called the generalised co-ordinates of the system. The generalised co-ordinates may be any parameters, they are not necessarily positional ( i.e., rectangular cartesian, curvilinear, etc) coordinates in the conventional orthogonal system, dimensionless parameters, angles, axes, moments or their combinations or any set of suitable parameters. Following are some examples of generalised co-ordinates

Ex 2.2.1. A dynamical system be a motion of a simple pendulum (of length l) oscillating in a vertical plane; the suitable generalised co-ordinate is (), the angular displacement from the vertical line through point of suspension. Similarly, for a spherical pendulum, the usual polar angle () and ¢> are suitable generalised coordinates. Ex 2.2.2. A particle on the surface of the sphere; generalised co-ordinates are (), ¢> where (), ¢> are the polar co-ordinates on the ~urface. Ex 2.2.3. A rod lying on a plane surface; generalised co-ordinates are x, y, () where (x, y) are the co-ordinates of one end of the rod and () is the angle between x-axis and the rod. While choosing a suitable set of generalised co-ordinates in a mathematical problem, the following points are to be noted: (i) The values of the selected generalised co-ordinates will specify the configuration of the system completely. (ii) They can be varied arbitrarily and independently without violating the constraints which may be imposed on the system. (iii) There is no uniqueness in the choice of generalised co-ordinates, the different sets so chosen will be functionally related. Then our choice should fall on a set of co-ordinates that will give us a reasonable mathematical simplification of the problem. Unlike, cartesian co-ordinates, generalised co-ordinates belonging to any particular particle are not divided into groups of three that can be associated together to form vectors. They are associated with a system as a whole, so that the configuration of the system can be described by a minimum number of independent co-ordinates. Number of independent variables being minimized, constraints are automatically obeyed. So one should not introduce the forces of constraints in treatment with generalised co-ordinates.

CHAPTER 2. GENERALISED CD-ORDINATES

42

2.2.1

Notation for GC

A set of n generalised co-ordinates are represented by ql, q2, ... , qn or simply qj j (j = 1,2, ... , n). When we are to describe a specific problem, the symbols ql, q2,···, qn correspond to co-ordinates that we choose to describe the motion. For example, (i) When a particle moves in a plane, it may be described by cartesian co-ordinates x, y, z, then ql = x, q2 = y, q3 = z. (ii) When the problem involves some spherical symmetry, it is suitable to use spherical co-ordinates ql = r = (x2 +y2 +z2)1/2j q2 = () = cot- 1 (x2+=2)1/2 j q3 =

4> = tan-l~. with greater advantage.

2.2.2

Transformation Equations

Consider, a system of N -point masses each with position co-ordinates -r'i for the ith particle. Suppose there are k constraints among the 3N position co-ordinates so the number of generalised co-ordinates will be reduced accordingly. Let the number of independent co-ordinates n = 3N - k, which are known as generalised co-ordinates and are denoted by ql, q2, .. . , qn. If the old co-ordinates -r'i contain the set of mathematical equations (2.6) expressing the old co-ordinates -r'i; i = 1,2, ... , N as function of generalised coordinates ql, q2, ... , qn and time t are called transformation equations for the set of T7i variables to the qj set. Also the generalised co-ordinates qi can be expressed as the function of old co-ordinates T7 1, -r' 2, ... , -r' Nand t by the equation qi

=

qi ( -+ r 1, -+ r 2, ... , r+ N, ) t .

The necessary and sufficient condition that the transformation form a set of coordinates qj to the set T7 i is effective is that the Jacobian determinant J of the equation (2.6) be different from zero at all points. For the solution of a I!lechanical problem, first we set up the equation of motion in terms of generalised co-ordinates qj and then integrate the simple equations of motion to obtain time dependence. Then the function -r'(t) can be obtained from (2.6).

2.2.3

Configuration Space

The span of the set of all independent co-ordinates qij i = 1,2, ... , n in n-dimensional Euclidean space ~n is known as configuration space of the dynamical system. Any point in the configuration space describes all constitutes of the system. The point is called a system point. At any given time t, the system is located at some point in its configuration space, and as the system moves, its co-ordinates will change and the system point in the configuration space will describe a curve giving the trijectory of

2.3. GENERALISED DISPLACEMENT

43

the system. Thus, in case of a system of N -particles, the 3N dimensional space in the absence of constraints or n dimensional space, when the constraints are present, is called the configuration space.

2.2.4

Degrees of Freedom

The minimum number of independent variables, required to describe the configuration uniquely of a given dynamical system which are compatible with the given constraints is called the degrees of freedom of the system. If the dynamical system of N particles is connected by k independent constraints then, the degrees of freedom will n = 3N - k in three dimensional space. (i) Holonomic systems are characterized by the fact that, the number of generalised co-ordinates require to specify the configuration of the system is equal to the number of degrees of freedom. (ii) For non-holonomic system, the number of generalised co-ordinates require is greater than the number of degrees of freedom of the dynamical system. These coordinates are called quasi-genemlised co-ordinates. The number of degrees of freedom is also defined as the number of independent ways, a mechanical system can execute, without violating any constraints imposed on it. For example, let us consider, the motion of a particle, constrained to move on the surface of a sphere of radius a. The centre of the sphere coincides with the origin of the cartesian co-ordinate system, then x 2 + y2 + z2 = a 2. Let us introduce ql = ~,q2 = ~, then q3 = Jl - q? - q~, so that q3 is not an independent co-ordinate. So number of degrees of freedom = 2.

2.3

Generalised Displacement

When a body is moved from one position to another, the change position is called a displacement of the body. Consider, a dynamical system of N particles of masses mi(i = 1,2, ... , N), having position vectors 7i with respect to some origin 0, in which constraints are also present. In the N system constraints, let the number of generalized co-ordinates be n = 3N -k. Then for the ith particle, the transformation equation 7i = 7 i(qt, q2, .. ·, qnj t)j i = 1,2, ... , N gives,

6-+ r i

n

a-+

ri = "LJ" a.

j=l

6qj

a-+ ri + 7ft

6tj

.

t

= 1,2, ... ,N.

(2.7)

%

Any real displacement d7 is always associated with the change in time dt. A virtual displacement 67 i is assumed to take place at fixed instant t, so variation with respect to time is absent, i.e., 8t = o. Hence from equation (2.7) the virtual displacement 67 is given by (2.8)

44

CHAPTER 2. GENERALISED CO-ORDINATES n

fJqJ

=L

ai:?fJr\; detlaijl =j; O.

)=1

8qj are called the generalised displacements for the ith particle. The equation (2.8)

gives the virtual displacement of the ith particle of the system in terms of generalised displacements fJqj. Since qj'S need not have the dimension of length, so generalised displacements fJqj's do not necessarily have the dimensions of linear displacement. But, because of equation (2.8), ~l fJqj must have always the dimensions of linear displacement. If qj is the angle co-ordinate, then fJqj is the angular displacement.

2.4

Generalised Velocity

Here we obtain an expression for the velocity of a particle in terms of the generalised co-ordinates. Velocity may be described in terms of time derivative qj of the generalized co-ordinate qj, is called generalized velocity. The N holonomic, time-dependent constraint T'i = T'i(q1, q2, .. ·, qn; t); i = 1,2, ... , N gives, --+

d ri dt

n!l--+!l

~Urt

- - = L..J - J=l

aqJ

u%

-at

!l--+

Uri +-= at

n!l--+

~uri.

L..J - - qj . aq)' )=1

!l--+

Uri +at

(2.9)

a:;t,

appears only when there are constraints depending where f3iJ = ~~' and -;y = on time or in some cases where it is convenient to introduce moving co-ordinate --+ axes. For the scleromic constraint (-;y = 0), the generalised velocity qj are given by

q] =

n

.

L:: 13;/ T'i.

It is easy to express velocities in generalised co-ordinates because

.1=1

from the equation (2.9), we see that, the velocity components of the ith particle is the linear functions of the generalized n velocity components qj no matter how the generalised co-ordinates are defined. Since q] need not have the dimensions of length, qj do not necessarily have the dimensions of linear velocity. But because of equation (2.9), ~~i qj must always have the dimension of linear velocity. Deduction 2.4.1. Velocity in cartesian co-ordinates: In cartesian co-ordinates, the generalised co-ordinates are q1 = X, q2 = y, q3 = z. The orthogonal unit vectors are i,J, k. Any arbitrary displacement dT' of a point may be resolved into three mutually perpendicular displacements: dxi, dYJ and dzk. Thus, dT' = dxi

=>

-+ V

d--+ r

= -

dt

+ dYJ + dZk .": = x~ + Yl + zk .~

.A

2.5. GENERALIZED ACCELERATION

45

Deduction 2.4.2. Velocity in spherical polar co-ordinates: In spherical polar co-ordinates, the generalised co-ordinates are ql = r, q2 = (), q3 = ¢. The orthogonal unit vectors are r, 0, ¢. Any arbitrary displacement d--;;7 of a point may be resolved into three mutually perpendicular displacements : drr, rdOO and r sin ¢d¢¢. Thus, ~

~ ::::} V

2.5

+ rdf)f) + r sin ¢d¢¢ A

d r = drr

=

d--;;7 dt

A

.

A

•

A

= 'H + rOf) + r sin ¢¢¢

Generalized Acceleration

Consider a system of N particles with --;;7i as the position vector for the ith particle. For acceleration, differentiate the expression of (2.9) with respect to t, we get

(2.10)

Thus we see that, the cartesian components of acceleration are not linear functions of compon'ents of generalised accelerations ijj alone, but depend quadratically and linearly on the generalised velocity components qj.

2.6

Generalized Kinetic Energy

We consider a dynamical system in which the ith particle has co-ordinate -;;;"'i and mass mi' Let us assume constraints are rhenomic. Now we write down an expression for kinetic energy in terms of generalized velocities. The kinetic energy T of a system of N free particles for a generally rhenomic system is given by ,

46


n

=

n

LL

n

O!jkqjqk

+L

j=lk=l

where O!·k = ! 3

(3jqj

+ "y = T(2) + T(l) + T(O)

(2.11)

j=l

f m·~r-i ~T\.

2 i:::;:l

1

qj

q,,'

(3. = 3

f m·~r-i

i=l

1

qj

alL

t'

"y

=!

f

2 i:::;:l

m·(aJii)2 '

are definite

functions of the ~'s and t and hence of q and t, called the coefficients of inertia. Thus the KE in terms of generalised velocities has three distinct terms : contains terms homogeneous quadratic in generalised velocities and this fact is indicated by the superscripts 2 in T(2). This form is known as canonical form. Since the matrix [O!jkJ is symmetric and positive definite, so T(2) is nonnegative. If this term is free from cross terms, i.e., ~~i ~:' = 0; j =fi k, the generalised co-ordinate system in qk's is referred to as an orthogonal system.

(i)

T(2)

(ii)

T(l)

(iii)

T(O)

a-+

-+

is a linear combination of velocity qj and vanish when ~ = 0, i.e., when ~ i does not contain the time explicitly. This situation occurs when moving co-ordinate system are not involved. Also T(l) is either positive or negative. a-+

is independent of generalized velocities. It will vanish when ~ = From the expression we see that T(O) is non-negative.

-+

o.

Thus the KE is in general made up of a term that is independent of the generalised velocities, a term linear, and a term that is quadratic in them. When the constraints are sc1eronomous, i.e., the transformation equation (2.6) is independent of time, then a;;. = 0, 'i = 1,2, ... , N. Hence (3j = 0 and "y = 0 but O!jk =fi 0 so that T(O) = T(l) = 0 and T = T(2). In this case T(2) ~ a hom~eneous quadratic form of generalized velocities or a canonical form. But ~;.i and ~;ki, by defillition, represent -+

-+

J

two base vectors b j and b k at ~i. When the constraints are scleronomous, then n

from the expression KE

T

=

n

E E O!jkqjqk of the system, we have, j=lk=l

47

2.6. GENERALIZED KINETIC ENERGY =

n

n

n

n

n

n

j=l

k=l

;=1

k=l

k=l

j=l

L L Ojkt5jlqk + L L Ojk t5klQj = L OlA:Qk + L OjlQ;

naT

nn

nn

or, Lril[j7" = LLOlkrikql 1=1 q, 1=1 k=l n

+ LLOjlqjql. 1=1 ;=1

n

= 2 L L olkQkQ, = 2T.

--

1=1 k=l

In orthogonal (q) system, b;. b k = Ojj -::/: k and consequently, 0jk = 0 when j n

In this case, T

=/: k.

= !m E Ojj Q1j=l

Deduction 2.6.1. KE in cartesian co-ordinates: For the orthogonal cartesian co-ordinate system, the generalised co-ordinates are q1 = X, q2 = y, q3 = z. The transformation equation is -:t = xi + YJ + zk which does not conta.in time explicitly, so that ~ = O. The system being orthogonal, so

87 87

A

87 87

A

A

A

87 87

A

A

8x' 8y =i.j =OJ 8y' 8z =j.k=Oj 8z' 8x =k.i=O.

Therefore, the KE is given by

T =

1

3

" Ojj qj·2 = 2m. '~

·2 Ollq1

+ 022q2·2 + 033q3·2

j=l

1 a-:t a7 1 1 = 2m ax . 8x = 2mi .i = 2m . A

where,

all

Similarly, 022 = 033 =

A

~m.

Thus the KE of a particle of mass m is T = !m[x2 + iP + z2].

Deduction 2.6.2. KE in polar co-ordinates: In plane polar co-ordinate (r,6) system, we have x = rcos6,y = rsin6, where (x,y) is the orthogonal cartesian co-ordinate system. Hence, Therefore, the KE is given by 2

1 m '~ " Ojj qj ·2 = oUq1 ·2 · 2 Ql = r, Q2 T = 2 + 022Q2j

=6

j=l

where, au = Similarly, 022

8Y2 21m [( 8x2 8r) + (ar) 1=

1 2 .2 1 2m[cos 6 + sm 6] = 2m .

= ~m[( ::)2 + (:: )2] = ~m[r2 sin2 0 + r2 cos2 OJ = ~mr2.

Thus, the KE of a particle of mass m is T = !m[f2 + r 292J. Similarly in spherical polar co-ordinate system, the KE is T = !m[f2 + r 292 + r2 sin2 6~2J.


48

Ex 2.6.1. KE of double pendulum: A double pendulum in a uniform gravitational field consists of two bobs of masses ml and m2 at ends of two weightless rods of length II and b. The second pendulum i~ ~uspended from the first bob of the first pendlllum, which is fixed to a rigid support. Let us assume that the pendulums move in the xy-plane only. Since the pendulum is constraint to move in the vertical plane, it has two degrees of freedom. The generalised co-ordinates selected are the angles ql = (h, q2 = fh made

h~O

x

y

Figure 2.1: Double pendulum by the two rods respectively, (which are actually angular displacements) with the vertical. Taking 0 as origin, let the co-ordinates of PI and P2 be (Xl, Yl) and (X2, Y2) respectively. Now from geometry, Xl

=

h sint'h;

Yl = h cosfh

X2 = h sin 01 + l2 sin 02; Y2 = h cos 01 + l2 cos 02· Thus the KE can be written as T

1~

1'2

1

..

1

·2

= 2 L..J CXjk iwik = 2CX(hOl fh + 2[CX0102 + CX0201](h02 + 2CX020202 ),k

OXI 2 0Yl 2 OX2 2 0Y2 2 whereCXOI0l=m1[(001) +(00 ) ]+m 2[(001) +(001)] 1 = ml [l~ cos 2 01 + l~ sin2 01] + m2[(h cos 01)2 + (-ll sinOl)2] = (ml + m2)l~. OXI OXI 0Yl 0Yl OX2 OX2 0Y2 OY2 CX0102 = ml [00 1 002 + 001 002] + m2 [00 1 002 + 001 002] = m1 x 0 + m2[it cos 0112 cos 02 + (-it sin 01)( -l2 sin 02)] = md1l2coS(01 - (h) = CY0201· OXI 2 0Yl 2 OX2 2

0Y2 2 CX0202=ml[(002) +(00 ) ]+m 2[(00) +(002)] 2 = m1 x 0 + m2[(l2cOS02)2 + (-l2sin02)2] = m2l~. Thus the kinetic energy T for the double pendulum is given by

1 T = 2[ml

2'2

+ m2]l101 + m2hl2 COS(OI -

..

1

2·2

O2)01 02 + 2m2l202 .

49

2.7. GENERALISED MOMENTUM

Deduction 2.6.3. If we are to use the plane polar co-ordinates (r,O) as generalised co-ordinates, then the KE (T) is T = ~m[r2 + r 2(P]. The associated generalised momenta are given by Pr

aT . = ar = mr =

linear momentum; Po

aT . = aiJ = mr 2 0 =

angular momentum.

Ex 2.6.2. Find the KE for a free particle in paraboloidal co-ordinates ((, ry, ¢). SOLUTION: Using the relation between cartesian co-ordinate (x, y, z) and parabolic co-ordinate system ((, ''7, ¢), we get

x =

~ i;

~ry cos ¢;

y =

~ry sin ¢;

= ~ry cos ¢ + ~iJ cos ¢

y = ~ry sin ¢

- ~ry sin ¢¢

+ (iJ sin ¢ + f.ry cos ¢¢;

i = ~~ - ryiJ

Thus the KE T for a free particle is T

= !m[i;2 + y2 + i 2] 2

= ~m[(~ry cos ¢ + ~iJ cos ¢ - ~ry sin ¢¢)2 + (~ry sin ¢

+ ~'r,sin¢ + ~rycos¢¢)2 + (~~ - ryiJ)2] = !m[~2(ry2 + e) + iJ2(ry2 + e) + ery2¢2] 2

= ~m[(ry2 + e)(iJ2 + ~2) + ery2¢2].

2.7

Generalised Momentum

The momentum associated with generalized co-ordinate qk is similarly defined and is called the generalized momentum Pk associated with the co-ordinate qk as

aT

Pk = aqk = mkq.. k,

T

1·2

= K .E = -2 "m;J7· ~ , ~

(2.12)

so that the generalised components of momentum can readily be obtained from the generalised expression of the KE of the system. Differentiate the equation (2.11) with respect to qk, we get

a---tri a---trio ",,1 mi = ~ ~ 2 T·a N

Pk

11.

qj

i=l j=l

q)

qk

N

a---t a---t

" ri ri + ~ mia·fit· i=l

(2.13)

qk

The first term is a linear function of generalised velocities. Last term of (2.13) will again be absent if the generalized system is stationary. It is a linear function of generalized velocities. Pk need not always have dimensions of linear momentum. The product of any generalised momentum and the associated/conjugate co-ordinate must always have the dimension of angular momentum or of the dynamic variable 'action'.


50

Deduction 2.7.1. Let us consider a particle, defined by (r, e) generalised coordinates in which Fr, Fo are the components of force. Then the generalised force associated with rand e co-ordinates are respectively as

where F = FrT + FoB and 7 = rT. Here Qr is the component of force in r direction and Qo is the torque acting on the particle to increase e.

2.8

Generalised Force

Consider a N particle system with ith particle having the position co-ordinates 7 i, there are k number of holonomic constraints, then n = 3N - k is the number of generalised co-ordinates ql, q2, ... , qn.Here we consider, the amount of work done 6W by the force L F i on the system in terms of arbitrary small displacement 67 i i

of the dynamical system. Then the work done for arbitrary small displacemet:lt is

(2.14) Qj are called the components of the generalised force associated with the co-ordinates qj and possibly on time t and also determines the work done in a virtual displacement in which qj alone changes. Also, qj need not have dimensions of length, generalised, force Qj need not have dimensions of force, but Q j 6qj must have always dimensions of work. For this reason the generalised force need not always have the dimensions force. As Qj is the coefficient of 6qk in (2.10), so its name generalised force. We will make use of generalised force to obtain Lagrange's equation of motion using D' Alembert's principle. If qj be an angle, the 6qj measures the angular rotation of certain part of the system about certain axis, Qj would be torque corresponding to that angular co-ordinate qj. If the system is conservative, then, if V be the potential energy of the system in any configuration space, we have 6W = -6V. In this case, V = V(ql, q2,.··, qn). Therefore

If the dynamical system be holonomic, dql, dq2,"" dqn are independent and so we get Qk = - g~ k = 1,2, ... , n.

;

2.9. GENERALIZED POTENTIAL

2.9

51

Generalized Potential

If the forces acting on the system are derivable from scalar potential energy function V of r+ i or qj but not 7 i or qj, the work done by the force on the system is an arbitrary displacement dr+ i of the system as N 8V 8V 8V &W = -&T = - 2:(-dXi + -&Yi + -dZi) i=l 8Xi 8Yi 8 Zi N

8V = 2:(8V 8xi 8qk i=l 8 Xi 8qk

(2.15)

+ 8V 8Yi + 8V 8zi) aYi 8qk

aZi aqk

N 8-+ r = - 2:(Fi. 8 i) - Qk.

i=l

qk

Thus, the generalized forces are given by the potential energy function Qj = -~, in this sense, the definition of Qj as generalized force is a natural one. When the system is not conservative, we define the generalized force Qj associated with a co-ordinate qj by

8U Qj = - 8qj

d 8U

+ dt(8qj)

where the generalised potential energy function U = U(qk, qk, t), which is a function of position co-ordinates q and generalised velocities, may be called a 'velocity dependent potential' or simply 'generalized potential' since it gives rise to generalized force Qj. We can come across such a velocity dependent potential when we consider the electromagnetic forces on moving charges. Ex 2.9.1. The Lorentz force experienced by a charged particle in an electromagnetic field is given by -+

aA -+ -+ -+ F=e[E+v xB; E=-"V4>- at; B="VxA.

-+

-+

-+

-+

-+

-+

-+

-+

-+-+

Here e is the charge, 4> = 4>( r ,t), A = A ( r ,t) are the scalar and vector electromagnetic potential respectively. In such a case, the velocity dependent potential U is given by -+-+

U = e4> - e( v . A). As the electromagnetic force is gyroscopic in nature, so it can be incorporated in the suitably generalised potential U as defined above. Ex 2.9.2. Let us consider the motion of a particle subject to Weber's electrodynamic law of at.tract.ion t.o a fixed point, t.he force per unit mass acting on t.he particle being F = ..!.[1- f2 - 2rf] r2

c2

52


where T is the distance of the particle from the centre of force. In this case, the velocity dependent potential function U(r, r, t) is given by U = ~[1 + ~l. This is another example of velocity dependent potential.

2.10

Principle of Virtual Work

The word virtual means possible or permissible. Here we shall discuss one important feature of constraint forces.

2.10.1

Virtual Displacement

Any imaginary displacement which is consistent with the constraint relation at a given instant is called a virtual displacement. The system is subjected to an virtual (infinitesimal) displacement with the forces and constraints imposed on the system at the given instant t. If the transformation equation be (2.6), the virtual displacement 8r\ is linked with the virtual displacement 8qj by (2.16)

as there is no passage of time in a virtual displacement. This change in the configuration of the system is not associated with a change in time t, i.e., the displacement takes place instantaneously.

2.10.2

Virtual Work

The work done by any force on a particle due to its virtual displacement is called virtual work. If the sum of virtual works done by reactions caused by constraint be zero, then the constraints are called ideal constraint.

2.10.3

Principle of Virtual Work

Let us consider a dynamical system of N -particles which is in static equilibrium and in which time-dependent holonomic constraints are also present. Then the total --t --t force on each particle vanishes, i.e., F i = O. Now let us give a virtual displacement to the system which refers to a change in the configuration of the system as a result of any infinitesimal change of the co-ordinates 8-:t i consistent with forces and constraints imposed on the system at the given instant. Then the virtual work of the iLl! particle is F .. 8-:t. and so, the total virtual work done on such system due to the arbitrary virtual displacement must identically vanish 8W

= """' ~ --t F i.8 --t r • = o.

(2.17)

2.10. PRINCIPLE OF VIRTUAL WORK Let us divide the force

F

t

53

into the applied force F~a) and the force of constraint

f 1.,.l.e., -Fi = -w Fi + -f i, then

We now consider the system for which the virtual work of tbe forces of constraints is zero (i.e., L ft.8-:ti = 0). An example of such a system can be had in mind if t

we consider that the forces of constraints being perpendicular to the surface while virtual displacement tangential to it, then virtual work done by forces of constraints will be zero. Thus the condition of static equilibrium reduces to (2.18)

This condition for equilibrium of a system is known as the principle of virtual work and which is applicable to the statics and indeed to mechanics of a system of particles. Thus the principle states that, a system of particles is in equilibrium, if the total virtual work done by the applied forces of constraint is zero.

2.10.4

D' Alembert's Principle

D' Alembert's principle is based on the principle of virtual work. To interpret the static equilibrium of the system, D'Alembert adopted an idea of a reversed force. He conceived that a system will remain in equilibrium under the action of a force . ( which is the linear equal to the actual force F i plus reversed effective force Iii momentum of the particle according to Newton's law of motion). The equation of motion of the ith particle in a dynamical system is

-

Itt

appears as an effective force called the reversed force of inertia on the where particle, supplementing the already existing externally applied force F i . Thus the principle of virtual work takes the form ith

L)Ft -lti)' 8-:ti = 0 or,

L)F(a) t

or, I:(F~a)

-

~) Pi .

8-r i + I:.f i· 8-:ti = 0

-lti)' 8-:ti = O.


54

It is called D'Alembert's principle in which forces of constraints disappear from the problem. Since forces of constraints no longer appear, D'Alembert's principle may be expressed as

(2.19)

Fi

where means the applied force on the ith. particle of the system. Thus the equation (2.19) implies that the sum of the virtual work done by the external force and the reverse effective force is zero, which is known as D'Alemberl's principle. Since the virtual displacement 8-;;:\ are not.independent of each other, but are related by the constraints, so the coefficients of 8~ i can not be set equal to zero and as a result, D' Alembert's principle can not be used to obtain the equations of motion for the particles of the system. Thus our required task to transform the equation (2.19)into an expression involving virtual displacements of independent generalised co-ordinates. Ex 2.10.1. Plane pendulum:

Let us take the plane pendulum problem, where the generalised co-ordinate is chosen as O. So the work done by the applied force is (-mg sin 0) (180), where 80 is the virtual displacement that 0 undergoes. Also, the acceleration of the bob is Ie. Thus from D'Alembert's principle, we have,

(-mgsinO)(IMJ) - mle(180) = 0 9 9 '* 0.. = -(T)sinO ~ -TO;

80

-I 0,0 small.

This is the usual equation of the SHM. Next suppose that, the length of the string changes with time. HereJ;he only change in the above equation is that the pendulum

s

0

mg

Figure 2.2: Simple pendulum bob has a component of acceleration (Ie + 2iiJ) in the 0 direction. Hence the work done by the inertial force is m[le + 2iiJ]180. So the modified form of the equation is

(-mg sin 0)(180) - m[le + 2iO]180 =}

d

=0

2'

dt[ml 0] = -mglsinO.

The LHS is the rate of change of the angular momentum of the bob about the point of support which, in the absence of gravity, remains constant even when the length of the pendulum is changing with time.

2.11. EXERCISE

55

Ex 2.10.2. A bead of mass, Tn is constrained to move on a rod in xy-plane. The rod i8 made to rotate with constant angular velocity w in the xy-plane. The reaction of the rod on the bead is R, the force of constraint. Prove that the virtual work done by R is zero. SOLUTION: Let the rod OA be moving with constant angular velocity w around O. Let P(r,O) be the position of the bead at any instant t, then

x = r cos (); y = rsin(}; () = wt. If the bead of mass Tn is confined with the rod, then x = r cos wt, y = r sin wt. Hence for arbitrary displacement, 8s, in which bead stays with the rod and time varies, Le., 8s is a real displacement, then 8s = (ox, 8y), where

8x = 6r cos wt - rw&t sin wt &y = &ysinwt + rwcoswt&t.

In the real displacement, &s = ac. Therefore, we easily see R&s =I- 0, as R has component along ac. Hence for real displacement force of consistent does do work. Consider now the virtual displacement, in which time-variation is not taking place, Le., &t = 0. In this case, &x

= &rcoswt;

&y

= &rsinwt.

In this virtual displacement case &s is along &r therefore, R.8s = 0, as R is orthogonal to rod, and r is the radius vector along the rod. Hence for virtual displacement, the virtual work done by forces of constraint is zero.

Exercise

2.11

Ex 2.11.1. What are generalised co-ordinates of a dynamical system? Define holonomic and non-holonomic dynamical system. Write short note with examples on unilateral and bilateral constraints.

Ex 2.11.2. For a scleronomous dynamical system of n dJ., prove that the KE is a homogeneous quadratic function of generalised velocities. Ex 2.11.3. For a scleronomous holonomic dynamical system of n dJ., moving under conservative force field, prove that KE +PE =constant.

Ex 2.11.4. For a scleronomous holonomic dynamical system of n dJ., prove that 'It

L qk k=l

zt -

L = constant, is the integrals of motion, where qi'S are the n generalised

"

co-ordinates, L= Lagrangian. Ex 2.11.5. A body subject to gravity moves on a smooth sphere along a vertical great circle. Find its angular acceleration assuming that the great circle is at rest in an inertial frame.

56


Ex 2.11.6. A body is sliding down an inclined plane which is moving horizontally with constant velocity. Find the position of the body as the function of time. Ex 2.11.7. Show that for a hoop on a plane, the constraint of rolling without slipping i:::; not holonomic. Write down the constraint in differential form.

Chapter 3

Variational Principle Calculus of variations is one of the most important mathematical tools of great scientific significance used in Mathematicians, Physics, Engineers. The generalization of the elementary theory of maxima ana minima of function of a single or several variables is the origin of the calculus of variations. The basic object of calculus of variation is to explore for finding extrema or stationary values of a quantity whose values are determined by one or more functions, known as functional. The domain of a functional is the set of admissible functions, rather than a region of a co-ordinate space. For example, (i) A functional is the length L between two given points (xo, YO) and (Xl, Y1) on the curve y = y(x) which is Xl

L[y(x)J

=

J[l + (~~)2P/2dx Xo

where we are to determine the curve y = y(x). (ii) The functional is the area A of a surface bounded by a given curve C as A is determined by the choice of the surface

-JJ

az 2 +(ay)J az 2 1/2 dxdy [l+(ax)

A[z(x,y)J-

D

where D is the projection of the area bounded by the curve C on the xy-plane. Several physical laws can be deducted from concise mathematical principles to the effect that a certain functional in a given process attains a extrema. Here Xl

we are to find the stationary values of the functionals like

J F(x, y(x), y'(x))dx, Xo

Xl

Xl

J F(x, y(x), y'(x), . .. ,y(n) (x))dx, J F(x, Y1 (X), ... ,Yn(x), yi (x), . .. , y~(x))dx and Xo

57

CHAPTER 3. VARIATIONAL PRINCIPLE

58

J J F(x, y, z(x, y),

Zr,

zy)dx in which the function F b given, and the functions

D

y(x), yt(x), ... , Yn(:r), z(x, y) are the arguments of the functionals. Eor example, moment of inertia, the co-ordinates of the CG of a curve of surface are functionals, as their values are determined by the choice of curve or surface.

3.1

Properties of Functionals

A variable quantity I[y(x)] depend on a function y(x) is a functional if to each function y(x) belonging to a certain class of functions, there is a definite value of I. Exoperty 3.1.1. [Variation] : Let y(x) and Yl(X) be two functions belonging to a certain class. Then the difference 81/ of the argument y(x), defined by 8y = y(.r.) - Y1 (x) is called increment or variation. Property 3.1.2. [Closeness ]: The function y(x) and Yl(X) are closed, if IY(x) Yl(x)1 is small for all x for which y(x) and Yl(X) are defined. WIlen it occurs, we say, y(x) is closed to Yl(X) in the sense of zero-order proximity. However, with this definition, the functional

J X2

I[y(x)]

=

F[x, y(x), y'(x)]dx,

Xl

which occurs in many applications, is seldom continuous due to the presence of y'(x) in the argument of I. This necessities the extension of the concept of closeness of the curves y = y(x) and y = JJ1(x) such that both Iy(x) - Y1(x)1 and Iy'(x) - y~(x)1 are small for all values of x for which y( x) and Yl (x) are close in the sense of first order proximity. In general the curves Y = y(x) a-ad Y Yl(X) are close of the nth order proximity, if Iy(x) - Yl(x)l. IY'(x) - y~ (x)I ... , Iy(n)(x) - yin) (x) I are small for values of x for which these functions are defined.

=

Property 3.1.3. [Continuity]: If a small change in y(x) results in a small change in the functional I[y(x)], then the functional I[y(x)] is said to be continuous. This definition is, however, somewhat imprecise since we have not specified what we mean by 'a small change in y(x)'. The functional I[y(x)] is said to be continuous at Y = yo(x), in the sense of kth order proximity if given any positive number E, :l a 8> 0 such that,

II[y(x)]- I[yo(x)]1
O

+ lin

f3[y(x) , ll'£5yj/ll'l max l£5y/ a

a->O

= L[y(x), 5yJ = M;

as L is linear.

Thus, the variation of the functional I[y(x)J is given as

a

all'I[y(x)

+ ll'£5yj

at a

= O.

Property 3.1.6. [Optimal value of the functionalj: A functional I[y(x)J attains a maximum on a curve y = yo(x) if the values of I on any curve close to y = yo(x) for which 61 = I[y(x)J - I[Yo(x)J ~ O. Similarly, if 61 = I[y(x)J - I[yo(x)J ~ 0, the minimum of I[y(x)J attained on the curve y = yo(x). Theorem 3.1.1. If a functional I[y(x)J attains its optimal value on y = yo(x) such that the domain of definition belong to a certain class of functions, then at y = yo(x),£5I = O. The calculus of variations explores methods that permit finding maximum or minimum values of the functionals. Such problems are known as variable problems. Several physical laws can be deducted from concise mathematical principles to the dfect that a certain functional in a given process attains a optimal. In mechanics, we have the principle of least action, the principle of conservation of linear momentum, Fermat's principle in optics, principle of Castigliano in the theory of elasticity are some examples of variational problems.

60

3.2

CHAPTER 3. VARIATIONAL PRINCIPLE

Euler's Equation

Let us assume that J(x, y, y') is thrice differentiable. Then the necessary condition for the extremum of the functional b

I[y(x)] = j J(x, y(x), y'(x))dx; y(a) = Yl, y(b) = Y2

(3.1)

a

where Yl and Y2 are specified at the fixed boundary points a and b, is that its variation must vanish, assuming that the admissible curves on which an extremum is achieved, admits of continuous first order derivatives. PROOF: Let y = y(x) be the curve which extremizes the functional I[y(x)] in (3.1) such that y(x) is twice differentiable and satisfies the boundary conditions y(a) = Yl, y(b) = Y2. Let y = y(x) be an admissible curve close to y(x) such that both y(x) and y(x) can be included in a one-parameter family of curves

y(x, a) = y(x)

+ a[y(x) -

y(x)].

(3.2)

For a = 0, y = y(x) and for a = 1, y = y(x). The difference y(x) - y(x) is the variation 6y of the function y(x). Now on the curves of the family (3.2), the functional (3.1) reduces to a function of a, say 'ljJ(a). But the extremizing curve corresponds to a = 0 so that 'ljJ(a) is extremized for a = 0, i.e., [~]Q=o = 0, where b

'ljJ(a) = I[y(x)

+ a6y] =

j J(x, y + a6y, y' + a6y')dx a

b

= j J(x, y(x, a), y'(x, a))dx Q.

b

'* 'ljJ'(a)

= j [Jy(x, y(x, a), y'(x, a))8y + Jyl(X, y(x, a), y'(x, a))8y']dx a

b

'* ./[Jy(x, y(x), y'(x))8y + Jyl(X, y(x), y'(x))6y']dx = 0 a b

b

'* j[Jy(x, y(x), y'(X))6y]dx + j a

Jyl d~ (8y)dx = 0; as d~ linear

a b

b

[J.y (x, y(x), y'(x))8y]dx + [Jy8y]~ - j [:x Jyl ]8ydX = 0

'* j a

a

b

'* j a

[Jy(x, y(x), y'(x)) -

d~Jyl]8YdX =

0 as [8Y]a

= 0 = [8Y]b'

(3.3)

3.2. EULER'S EQUATION

61

In view of assumptions made on J(x,y(x),y'(x)) and the extremizing curve y(x), it follow that the integrand in (3.3) on the curve y(x) is a given continuous function, while 8y is an arbitrary continuous function subject to [8Y]a = 0 = [8Y]b. Thus we must have

Jy(x, y(x), y'(x)) -

d~JYI =

0

(3.4)

on the extremizing curve. This equation is known as Euler's equation. If J(x, y, y') does not contain x explicitly so Jxy' = O. Using d~Jyl = Jxy' + JyylY' + JylylY" from (3.4), we get (3.5) which is the second order equation in y(x). The two arbitrary constants appearing in the solution are determined from the boundary conditions yea) = Y1, y(b) = Y2. Now we have the following discussions: (i) The existence of the solution (3.5) can not be taken for granted and even if solution exists, it may not unique. However in many problems, the existence of a solution is evident from the geometrical or physical nature of the problem. Hence in such cases, if the solution is unique, then this solution gives the solution gives the solution of the variational problem. (ii) When the function J in (3.1) depends on y and y' only, Euler's equation becomes Jy - JyylY' - Jylyly" = O. Now,

-d [J - Y'f] yl dx

= fyy' + fy'Y " - Y"fyl - Y'2fyy' - f ylyly'" Y = y'[Jy - y' Jyyl - JylyIY"] = 0 =} J - y'Jyl = C1; where C 1 is a constant.

(iii) Invariance of Euler's equation: Suppose that a functional I[y(x)]

=

b

J F(x, y, y')dx is transformed by replacing the independent variable or by a sia

multaneous replacement of the Jesired function and the independent variable. Then, the extremals of the functional are found from Euler equation for the transformed integrand. Herein lies the principle of the invariance of Euler's equation. Let x

J

=

x( u, v), y

F(x, y, y')dx

= =

=

J J

y( u, v) and

IXu Xv I =I- 0, then Yu Yv

F[x(u, v), y(u, v), Yu m2, mi will move downward with acceleration m 1 +-m 2 g. Obviously, mi = m2 indicates ml m2 equilibrium and then either mi and m2 remain at rest or move with uniform speed in opposite direction.

84

CHAPTER 4. LAGRANGIAN MECHANICS

.Ex 4.1.4. Linear Harmonic Oscillator A body of mass m is connected to a spring of negligible mass and rests on a frictionless horizontal floor. The spring has a spring constant k. When the body is displaced from its normal equilibrium position by x, the potentIal energy of the spring becomes

J X

V=-

J

--+ --+ F.dr

=-

1 2 (-kx)dx=2kx.

o

The mass m being springless, i.e., rigid, it does not contribute to V. The spring being assumed massless it also does not contribute to the KE of the system. Therefore, KE of the system is the KE of the body which is given by T = !mx 2 . For x as the generalised co-ordinate, the Lagrangian is given by

1 2

1 2 2 8L aL. -;:} = -kx; -8' = mx. .2

L = T - V = -mx - -kx =}

x

IIX

Therefore, Lagrange's equation for one dimensional oscillator is

mx+kx

= O.

Thus the motion executed by the mass m attached to the spring is S.H.M. and the period of oscillatory motion is T = 211" JIfI. Ex 4.1.5. A particle of mass m is projected with initial velocity u at an angle Q with the horizontal. the resistance of the air may be neglected. Use Lagrange's equations to describe the motion of pmjectile. Let a particle of mass AI is projected from the fixed point 0, with a velocityu at an angle Q with the horizonal. Let P(x, y) be the position of the particle at any time t. Let us consider the co-ordinates x and y as the generalised co-ordinates. At P. the kinetic and potential energies are given by SOLUTION:

T

= ~m(i:2 + Ii); V = mgy =} L = ~m(i:2 + Ii) - mgy.

Since x and yare generalised co-ordinates, the Lagrange's equations of motion for the projectile are

~(f)L) _ 8 Lx = 0', dt ax

=}

a

=}

aL

- ay = 0;

= 0; dd (my) + mg = 0 t x = 0; jj + g = O.

d (mx) - 0

dZ

d (aL) dt ay

4.1. LAGRANGE'S EQUATION OF MOTION

85

Now the initial conditions are t = 0; x = 0; i: = ucosa; Applying the initial conditions, the solution is given by X

= ( U cos a ) t;

=

y

.) ( 7L sm a t -

12

- gt =? Y = xtana -

2

y

=

0;

iJ = usina.

9 2 x. 2 2 2u cos a

e,

Deduction 4.1.3. Spherical Polar Co-ordinate System: Let P(r, ¢) be the position of a particle at any time t of mass m in spherical polar co-ordinate System (r, e, ¢). Let F r , Fe, F be the components of force acting on the particle in the increasing direction of r, e and ¢ respectively. Here the dynamical system consists of a single particle of mass m and generalised co-ordinates of the system are r, e, ¢. 1£ Q.r, Qe, Q be the generalised force components, then

8W = Qr 8r + Qe8e + Q¢8¢ and 8W = Frdr + rFede + rsineF¢d¢ =? Qr = Fr; Qe = rFe; Q¢ = rsineF¢. At any time t, the Kinetic Energy T of the system is given by T =

~rn['r2 + '1'2(;2 + T2¢2 sin 2 e].

Now, using the Lagrange's equations of motion we have

d [aa~] - aaT r dt '1' d aT

-d [-.]t

ae

d aT

-d [-.]t a¢

= Qr

aT -a e=

Qe

+ rJ} sin2 e) = Fr

=?

dd (mr) - m(rfp t

=?

-d (mr e) - mr ¢ sin e cos e = rFe t

d

2·

2 ·2

aT d 2 2 . -a = Q¢ =? -d (mr sin e) = rF¢sme ¢ t

or, m(f - riP - r¢2 sin2 e)

= Fr

m[! (r 2iJ - r2¢2 sin e cos e)] = rFe

d

.

2

m dt (r¢sin e)

= rF¢ sine.

These are the equations of motion of a particle of mass m in spherical polar coordinates from Lagrange's equation. Ex 4.1.6. Spherical Pendulum In a spherical pendulum, bob moves through the entire space about the point of suspension. The bob of such pendulum moves on the surface of the sphere whose radius is equal to the length of the pendulum. The position of the bob is located The distance x of the bob from the origin will be the radius of the sphere on which the particle or bob moves. The force in this case will not be central, but is constant in the vertical direction. The kinetic energy of the pendulum is T = ~mI2(iJ2+sin2 e¢2)


86 z

~------y

p

:Figure 4.2: Spherical Pendulum and by spherical co-ordinate 9, e/>. the potential energy of the bob due to gravity relative to the horizontal plane z = 0 is V = mgl cos 9. Hence the Lagrangian function L for the dynamical system is given by

L= T- V =

aL aiJ

= mr

~ml2(92 + sin2 (J~2) -

mgl cos

e

2 '2 ej aL ae == mr sin ecos ee/> + mgr sin e

2'

a~ = mr2 sin2 e~j ae/>

a~ = o. ae/>

The equations of motion corresponding to the co-ordinates 9 and e/> are

d aL aL 2" 2 '2 dt ( ao ) - a(} = 0 => mr e - mr sin 9 cos 9e/> L d (a~) _ aa = 0 => mr2dd (sin2 e~) = O. dt ae/> e/> t

mgr sin 9 = 0

We shall discuss the nature of the solution in the next chapter.

Deduction 4.1.4. Non-Conservative System: If potentials are velocity dependent, called generalised potentials, then though the system is not conservative, yet the above form of Lagrange's equation can be retained provided Qj, the components of generalised force, are obtained from a function U(qj, qj) such that

au

Qj = - aqj

d au

(4.9)

+ dt ( aqj ).

Hence from equation (4.6) we get

d aT aT au d au dt ( aq'J) - aq·.1 = - aqJ + dt ( aq'J) or,

~[~(T dt aqj d aL

or, -d (-a' ) t qj

U)]-

aL

-a qJ

~(T aqj

= 0

U) = 0 (4.10)


87

which is the same form of the equation (4.8) in which L = T - U is the Lagrangian. U is known as the generalised velocity dependent potential and we across such a potential when we consider the electromagnetic forces on moving charges.

Ex 4.1.7. Velocity-Dependent Potential of Electromagnetic Field Here we derive the Lagrange's equation of motion for the non-conservative system. Consider a charged particle of mass m and charge q is moving in electromagnetic field. The Maxwell's equations are --+ a B --+ V'xE+-=O' --+

at

--+

--+

--+

--+

' a--+ E

V'.E = cO

--+--+

= J-loj + J-lOC07ft; A

V' x B

--+

P

--+--+

--+

V'.B

= O.

--+

The quantity cO E = D and B = J-loH are respectively called electric displacement --+ and the magnetic induction respectively. From this equation, we observe that V' x --+ --+ E I- 0, so E can not be expressed as a gradient of a scalar potential function. But, --+ --+ --+ --+ --+ --+ --+ as V'. B = 0, B can be expressed as B = V' x A, where A is called a magnetic vector potential. --+

--+

V' x E

+

a--+ B at

--+

=

--+

--+

+

=> V' x [E

--+

--+

0 => V' x E

a--+A

at]

=

a

--+

+ at (V'

--+

--+

0 => E

--+

x A)

a--+A

=

+ 7ft = -

--+

0

--+

V'

where is the scalar potential functions. Now the Lorentz electromagnetic field --+ force F on a charged particle (in cgs Gaussian system) having charge q and moving in an electromagnetic field is given by --+ --+ 1 --+ --+ F = q[E + -(V x B)]; c = Velocity of light

c

a--+

1 A 1 --+ --+ --+ or, F = q [- V' - ~ at - ~ (V x V' x A)]. --+

--+

--+

--+

in which the magnetic force %(V x B) is gyroscopic in nature. The power associated --+

--+ --+

with these forces F is zero, i.e., F. V = O. To write more convenient form, let us write the x component of various terms on right hand side. Now,

(V x V x

A)x

= Vy(aA y _ aAx) _ Vz(a.Ax _ aAz) =

ax Vx aA x ax

a

az B.T + 11: aAy + Vz aA z _ Vx aA x _ 11: aAy _ v~ aA x y ax ax ax y ay ~ az ay

--+ --+

= ax ( V . A) Al

dAx _ aA x so, dt - at

+

v

V. V' Ax

aAx x ax

dAx _ aA x at

=> dt

--+ --+

+

11: aAx y ay

= --+v ~A .v

x

+

v z

aA x az


88

Now, as the velocity V does not contain x explicitly, we have ~ = o. Also as %~7: = 0, we have (V.A) = Ax. Thus, consider the generalised co-ordinate x co~ponents of the va;ious terms, the Lorentz's force is given by

dt

PI;

a 1 aAx = q[- ax - 7;Tt a

1 -.

+ 7;(V

1 aAx

-; x

v

-;

x A)x]

1 a -; -;

1-; -;

1-;-;

= q[- ax - 7;Tt + 7; v.v Ax + 7;ax(V.A) - 7; V. 6A x a

1 dAx 7;Tt 1-;-; 1d = q [- ax ( - 7; V . A) - 7; dt {8V (V. A ) }] x 1-;-; 1d 1-;-; = q[--, ( - - V.A) - --{-( - - V.A)}]; as ax c edt avx c au d au = - ax + dt(DV)· 1-; -;

= q[- ax ( - 7; V .A) -

a

a -;-; .

a

a

= (x,y,z,t)

-;

where the generalised potential energy is U = q - ~ A .11. This shows that, when a charged particle moves in electromagnetic field, the force acting on the particle is derivable from a potential dependent on velocity of the particle. Therefore the force in generalised co-ordinates is given by

Such a particle system in electromagnetic field is called non-conservative one and the equations of motions in generalised co-ordinates is given by d DT

dt(~) -

a8TqJ =Qj

uqJ d 8T aT au or, dt ( 8qj) - 8qj = - 8qj

.:!.- (8(T -

or

, dt

8q]

d au

+ dt ( 8qj )

U)) _ 8(T - U) 8qJ

d 8L t qj

=0

8L

or'-d(-a.)--a =0. qj

Thus the Lagrangian for the charged particle moving in an Electromagnetic field is, L

=T

- U

=T

- q

q-; -;

+ -c A . v

which shows that the form of Lagrange's equation remains same even if the system is non-conservative, where U is the generalised velocity and time dependent potential.

Deduction 4.1.5. Explicit form of Lagrange Equation :We shall now show how the second derivatives of the co-ordinates with respect to the time can be

4.2.

DISSIPATIVE SYSTEM

89

found explicitly from Lagrange equation. Let the configuration of the dynamical system considered be specified by co-ordinates qI, q2, . .. , qn, we shall suppose that the configuration can be completely specified in terms of these co-ordinates alone, without t, so that the kinetic energy of the system is a homogeneous quadratic function of IiI, 1i2, .. , , tin. Also from the previous section, we see that, this is always the case when the constraints are independent of time, but not in general when the constraints have forced motions. Suppose the Kinetic Energy is

where akl = alk, and where the coefficients akl are known functions of q1. q2,· .. , qn' The Lagrangian equations of motion for the system are

d aT dt

aT (-a') - -a. = Qj; j qJ

= 1,2, ... ,n.

qJ

j = 1,2, ... , n

= 1,2, ... , n

j

where

e

(4.11)

~) is the Christoffel's symbol of second kind and is defined by the relation

J

•

These equations, being linear in the accelerations, can be solved for the quantities = laijl be the determinant and Ars be the minor of ars in 6.. Multiply the n equations of (4.11) by Ali, A 2i , ... , Ani respectively, and add them we get iis. Let 6.

These n equations of iiI, ii2, . .. , iin which are give explicitly as a functions of (iI, tI2, ... , tin; ql, q2, ... ,qn can be regarded as explicit form of Lagrange equation of motion.

4.2

Dissipative system

When the forces acting on the dynamical system are derivable from a potential, Lagrange's equation of motion is given by (4.6), where L contains only conservative forces and Qj includes the forces that are not derivable from a potential. The forces


90

Qj are dissipative, i.e .. to energy of the system is not conserved. This type of force occurs in the situation when frictional forces are present, and that the frictional force is proportional to the velocity of the particle. Thus the x component of such force may be written as

where kx is the :1' component of the frictional force per unit velocity. Frictional forces of this type may be derived in terms of a function'S, known as Rayleigh's dissipative function, and can be written as

which covers all the particles of the dynamical system. Since the frictional forces ---> -+ -+ -+ F f can be derived in terms of the function'S, we have, F f = \1 v'S where \1 v = ~ -aa + 3aaVv + kaaVz is t.he velocit.y• differential operator. For physical int.erpretation lJ r of Rayleign!s dissipative function'S, let us calculate the work done by the system against friction, which is given by dWf

-+ -+-+ = - ---> F f.d r = - F f. v dt = [kxv; + kyv~ + kzv~ldt

dW:r _ 2 2 2 _ => dt - kxvx + kyv y + kzv z - 2'S. Thus the rate of energy of dissipation due to friction is equal to twice Rayleigh's dissipation function. The component of generalised force Qj arising as a result of force of friction is given by n

~-+

a-+r i

n

~-+

a-+r i

Qj = Lt F if'--a-: = - Lt \1 v'S'--a-: ! % ! qJ _

~ -+

0.
! _

a'S

- - ~ V' v'S· aqj - - aqj .

Hence the Lagrange's equation of motion becomes

Therefore, to derive the equations of motion, if the frictional forces are acting on the syst.em, we lllUst be specified two scalar functions - the Lagrangian L and the Rayleigh's dissipative function'S. Ex 4.2.1. Viscous force F retarding a sphere in a fluid of co-efficient of viscosity 'TJ -+ is given by Stokes law F = -6'Jr'f]'r 11, where 11 is the terminal velocity of the sphere. Find thf Rayleigh dissipation function.

4.3. NON-HOLONOMIC SYSTEM

91

SOLUTION: Let z be measured from the ground. Let Fz be the dissipative force, then Fz = -kzv z ; kz = 67rT7J. Therefore, the Rayleigh's dissipative function;} is given by

1

2

;} = -'2kz1Jz; k z =

67rT7J

where V z = i is the velocity along z axis. Let p be the density of the sphere and M be the mass of the sphere, then M = ~ 7rT3 p. The potential energy (V) is V = - M 9 z, and so the Lagrangian (L) is L = ~1\1v~ - Algz. Thus the lagrange's equation of motion is given by

-d [oL - - -oL J + -a;} = 0 ::::} Mooz dt

4.3

oi

8z

oi

67r7JTZ.

+ M 9 = 0.

Non-Holonomic System

Let q1. q2, ... , qn be the generalised co-ordinates of a non-holonomic bilateral dynamical systems having N number of particles and k be the number of holonomic constraints. When the constraints acting on the system are non-holonomic, the generalised co-ordinates are not independent of each other. Let for such dynamical system have k' non-holonomic constraint relations, with degrees of freedom = n, so the number of quasi-generalised co-ordinates required is n+k' where n = 3N -~~-k'. Let the differentials dqi of the generalised co-ordinates are connected by m( < 11.) relations of constraints as n

I: alkdqk + alt dt = OJ

I

= 1,2, ... ,m.

(4.13)

k=l

where alk's are functions of qk's and t. These equations are the equations of the constraints for a non-holonomic system. The virtual displacements tSqk at any instant t satisfy the following equations of constraints n

I: alk

tSqk = OJ I = 1,2, ... ,m

(4.14)

k=l

as Hamilton's principle does not involve variation of time. These equations can be used to reduce the number of virtual displacements to only that of the independent ones. In the case of non-holonomic system, the virtual displacements tSqk's are not all independent, but they are connected by m relations (4.14), so that (n - m) of 6qk's are independent. To eliminate these extra dependent virtual displacements we shall use Lagrange's method of undetermined multipliers. Multiply (4.14)by m constants ),1, ),2,· ·.,),m respectively, sum over I and finally integrate it we get (4.15)


92

where the unknowns Ak'S are called Lagrange multipliers which may be, in general, functions of time. Now the Hamilton's principle for conservative system we get

L

=T - V =

Lagrangian

(4.16)

All the qk's are not independent but are connected by m equations (4.14). However, the first (n - m) of these co-ordinates may be chosen independently, the last m ones being fixed by eqnations (4.14). Since AI'S can be chosen arbitrary so let us choose Ak'S (k = 1,2, ... ,m) such that 8L

d

8L

'Tn

-8qk - -(-. ) + '" Al dt 8qk ~

alk

= 0;

k

= n-m+ 1, ... ,n

(4.17)

1=1

which are the nature of the equations of motion for the last m of the qk variables. Having found these Ak'S from equation (4.17), we can write equation (4.16) as

(4.18)

where £5'11. £5'12, ... ,£5qn-m are independent. Thus the expression in the third bracket must vanish, therefore

a8Lqk - dtd (88Lqk. ) + L Al 'TIL

alk

= 0;

k

= 1,2, ... , n

- m

E4.19)

1=1

Combination of equations (4.18),(4.19) are then finally the complete set of Lagrange's equations for non-holonomic system. Although there are (n - m) + m = n equations in all, we have (n + m) unknowns, namely n generalised co-ordinates qk's and m constants AI. Thus the additional equations are given by the constraints (4.13)as n

L alk

qk

+ alt = 0;

l

= 1,2, ... , m.

(4.20)

k=l

Therefore equations (4.19) and (4.20)together constitute (n + m) equations for (n + m) unknowns. Physical significance of A :Suppose, we want to remove the constraints on the system, but instead. applied external forces Q~ in such a manner as to keep the motion of the system unchanged. The equations of motion would likewise remain the

4.3. NON-HOLONOMIC SYSTEM

93

same. Clearly, these extra applied forces must be equal to the forces of constraint, for they are the forces applied to the system so as to satisfy the condition of constraint. Under the influence of these forces Q~ the equations of motion are

But these must be identical with the equations

d 8L 8L -d (-a' ) - -a t

qk

Hence we can identify

qk

=

L

),lalk;

l

k

= 1,2, ... , n.

L ),lalk with Q~, the generalised forces of the constraints.

In

l

this type of problems, we really do not eliminate the force of constraints from the formulation and they are obtained as apart of the answer. Result 4.3.1. The above equations(4.19) include holonomic constraints as a special case. Let the generalised co-ordinates ql, q2, ... ,qn be connected by the m relations fl(qI, q2,"" qn, t) = 0;

1 = 1,2, ... ,m

where m < n. Differentiating, we get 1 = 1,2, ... ,m

which is of the form of equation (4.20). Thus the coefficients are given by alk = ~t~; au = 1 • Thus the Lagrange's multiplier method can also be used for holonomic constraints when (i) it is inconvenient to reduce all the generalised co-ordinates qk's are independent or (ii) we might wish to obtain the forces of constraints.

1t

Ex 4.3.1. Particle In a Sphere Let a particle of mass m, move on the surface of a frictionless surface of a smooth sphere of radius l, under the action of gravity. Let the origin of the co-ordinates be at the centre of the sphere and let the z axis be vertically upwards. Here co-ordinates of any point on the sphere will be (r, 0, ¢) of which r is constant as the particle is constrained to move on the surface of the sphere of radius r = l. The equation of constraint is r - I = 0, where r is the radial distance of the particle. Thus the coefficient in the equation of constraint are, a r = 1, ao = 0 = ar/J. Let us suppose that the particle is initially at rest and let it slide down along the surface. The particle will obviously move in a vertical plane which we shall take for convenience ¢ = O. The Lagrangian for the particle is,

L=

~[r2 + r 2lP]- mgz = ~12(P -

mg1' cos O.


94 Using the results r

= 1, r = f = 0,

-mliP

the equations of motion are

+ mg cos (J =

Aj ml 2 (j - mgl sin (J = O.

The undetermined multiplier A is dependent on (J, in general. Also, differentiating the first equation containing A with respect to time t, and using (j = Tsin (J from the second equation, we get, ... . -2ml(}8 - mg sin (J(J

'* -3mgsin8 =

dA' d(Je

=

dA· d88

'* A(e) =

3mg cos (J

+c

where c is the constant of integration, determined by the condition 8 = 0, A = mg, the force constraint at the top of the sphere. Thus A(8)

= 3mgcos 8 -

2mg ~ 0

as the particle will move on the surface as long as the surface pushes it outward. The equality sign holds, when cos(Jc = ~, i.e., at an angle 8c = cos- 1 ~ the particle flies off the surface.

Ex 4.3.2. Cylinder Rolling on Inclined Plane Consider a cylinder rolling without slipping on a inclined plane of length l, which makes an angle 0: with the horizontal. We wish to find the acceleration of the cylinder and the frictional force of constraint. Let the initial position of the cylinder of radius

o x

Q

B~------'------'A

'Figure 4.3: Cylinder rolling, without slipping r be 0 and roll down the plane along the line of greatest slope without slipping. Let P he the position when it has moved through a distance x down the inclined plane. Let C be the centre of the cylinder and LQCP = (J. Here the generalised co-ordinates are x and (J and they are connected by a relation x = r(J. Then, the equation of the constraint is rd() = dx, as the friction is sufficient to prevent the slipping. The constraint force is the force of friction. Since there is only one equation of constraint, only one Lagrangian multiplier ,is required. Now, the coefficients in the equation of the constraints are aT = Tj ax = -1. Due to the translation and rotational motion, the kinetic energy is made up by the two components. Thus the

4.4. CYCLIC CO-ORDINATES

95

kinetic and potential energy of the cylinder is 1

T= -mi:

2

1'2 1 2 1 2 '2 + -I() = -mi: + -mr () 2 2 4

2 V = mgh = mg(l- x) sina =?

~mi:2 + lmr 2(P -

L= T - V =

mg(l- x) sina.

The Lagrange's equation for () is given by

~(a~) _ aL = a A

a() a() ~mr20 - rA =

()

dt

=?

=?

2 I" "2mx

-

O· as rO = x ' Imx ". /\'=0 =?' /\ = "2

The Lagrange's equation for x is given by

d aL

aL

*

mgSina~mx = 0

dt (ai:) - ax = axA * mx - mg sin a + A = 0

mx "

=? x

2.

= 3gs111a =?

()"

2 . = 3rgsllla.

Hence the frictional force of constraint A is img sin a.

4.4

Cyclic Co-ordinates

If the Lagrangian L of a dynamical system does not contain a co-ordinate explicitly, then that co-ordinate is called an ignorable or cyclic co-ordinate. Thus, some of the co-ordinates ql, q2, ... , qk are not exactly contained in L, although the corresponding velocities ql, q2, ... , qk are not so centained. The Lagrangian L is written as a function of qj and q]. If. anyone co-ordinate say qj {j~. an ignorable ?o-~~dina1;e, then ~ = 0, where may not~e zero. We have seen that the motIOn of a conservative, holonomic dynamical system with n degrees of freedom, for which the co-ordinates are ql, q2, .. . , qn and the kinetic potential L, is determined by the differential equations

gt

d 8L 8L -d (-8' ) - -8 = 0 t

qj

qj

=?

d 8L 8L -d (-8' ) = 0 =? -8' t

qj

qj

= Pj =

Constant = /3j.

This Pl is known as the conjugate momentum for q]. So, th«:: Lagrangian of a dynamical system does not contain a co-ordinate qj, then the corresponding conjugate momentum is conserved. The presence of ignorable co-ordinates is the most frequently occurring reason for the solubility of particular problems by quadratures.

96

4.4.1


Systems with ignorable

co~ordinates

Let a dynamical system be represented by n generalised co-ordinates ql, q2, ... , qn' Suppose ql, q2, .. ·, qs(s < n) be cyclic. We define a function R, the Routhian as s

R

= R(ql, ... ,qn, Pl,· .. , Ps, tis+1, ... ,tin, t) = LPitii - L. i=1

n 8L n 8L 8L Now, dR= Lqidpi + LPid,qi - L ~dqi - L ~dqi - Ft dt s

s

~1

=1.

~1

~

~1

~

n 8L· 8L n = Ltitdpi - L ~dtii - a;dt - LPidqi t=1 t=s+1 ~ i=1 n 8R n 8R s 8R 8R dt . Also, dR = L ~dqi + L ~dtii + L a.dpi + t=1 qt t=s+1 qt i=1 Pt s

m

(4.21) (4.22)

The infinitesimal quantities occurring on the above expressions are arbitrary and independent. Comparing, (4.21) and (4.22), we have

8R . 8R . i = 1,2, ... , s -8- = -Pi; -8- = q.i; qi Pi 8R 8L 8R 8L - - ; - - - - 8 ; i=s+1,s+2, ... ,n. 8tii 8qi qi

(4.23) (4.24)

Now the Lagrange's equation of motion gives,

(4.25) in which the Routhian function has replaced the Lagrangian function. Now, the Routhian R is a function only of the variables qs+1, qs+2, .. . , qn; til, ti2, . .. , tin, and the constants {31, {32, ... ,{3s. Thus this is a new Lagrangian system of equations, which we can regard as defining a new dynamical problem with only (n - k) degrees of freedom, the new co-ordinates being qs+1, qs+2, ... , qn, and the new Lagrangian being R. When the variables qs+1, qs+2, ... , qn have been obtained in terms of t by solving this new dynamical problem, the remainder of the original co-ordinates namely ql, Q2, ... ,qs can be obtained from the equations qj

=-

J;~

dt; j

= 1,2, ... ,s.

Thus, a dynamical problem with n degrees of freedom, which has k ignorable coordinates, can be reduced to dynamical problem which has only (n - s) degrees of freedom. This process is called ignoration of co-ordinates.


97

Ex 4.4.1. Kepler's Problem for planetary motion: Let us now apply Routhian procedure of ignoration of co-ordinates to the Kepler's planetary problem. In plane polar co-ordinates (r, (J), the Lagrangian function L of such system is given by

where V (r) = - ~ is the potential. Here the Lagrangian L does not contain explicitly, so (J is ignorable co-ordinate and so Po

oL = mr 2'(J = constant = [( = -. of}

(J

say)

where [ is the constant magnitude of the angular momentum. Also, it follows that :it 2 0) = O. But we know that 2 0 is just the areal velocity- the area swept out by the radius vector per unit time. This says that, the well-known Kepler's second law of planetary motion: the radius vector sweeps out equal areas in equal times. We now set up the Routhian function as,

(!r

!r

r

Note that R contains rand

only. Now, the equation of motion is given by

.!!:..(oR) _ oR

=0

dt oqi Oqi d. '2 =? - (mr) - mr(J dt

d

=?

+ -r2k = 0 [2

-d [mr2 t

+ 2V + -2] =0 mr

which is the equation of a particle moving under central force. Also, mr2+2V + ~~2 = 2T + 2V, so that the constant of the above equation is nothing but the two times of total energy E. Its first integral gives the conservation of total energy E as 1,2

-mr

2

[2 e E + -+V = onstant = mr2

J r

=? t -

-

ro

dr [2(E - V - 6)]1/2 m

2mr

where ro is the initial value of 'r. This equation gives t as a function of l' and the constants E, [ and roo However, it may be inverted, at least formally, to give r as a

98


function of t and the constants. Once the solution of r is obtained, the solution 0 follows immediately as t

T

o- 00 = 1 j'

=

dt

mr2

o

j' mr2 . /1. [E - V - L ] ldr

.

Ym

TO

2r1lT'
0 - 0' = cos- 1 (

u -

. 12mE

Y!2

du m2k2 _ (u _ mk)2]

--y:r

= [cos- 1 (

u. /2mE

V !2

j2

j2

+ m2k2

--y:r

)]U uo

!!f# I ).

+ --y:r m 2k2

The quantity 0' can be determined using the initial condition. Thus we can identify four constants of integration I.E, ro, 00 . Also, the above equation can be written as U

=

mk

r

2mE

m 2k 2

+ ------z:r- + [ 4 cos( 0 -

l2/mk

, 0)

2El2 1 + mk2 cos(O - 0').

or'-r- = 1 +

Now, the general equation of a conic with one focus at the origin is ~ = 1 + ccos(O - 0'), where e is the eccentricity of the conic section. Comparing we get,

VI

e= + ~S· The nature of the orbit depends on the magnitude of e according to the following scheme: Values of E Values of e Nature of Orbit Hyperbola E>O e>1 Parabola e=1 E=O Ellipse E cos

-1

J

dq2

y'hq~ + /2

+

Constant

QlvfJh -1 Q2vfh [ ~] - cos [. r;:;;; J = Constant 1 - /1 v/1

= 11' -

/(

say )

I I # , #,

=> cos- aql - cos- bQ2 = /; a = :::} cos- 1 [aQlbQ2 => 1 - b2q22 - a 2 Ql2 :::} a 2qr + b2qi

-; b = /1

-

/1

V(1-

+ aQf)(l - bQi)] = 11' - /. + a 2b2Ql2Q22 = ( - cos / - a bQl)Q22 + 2abQlQ2 cos/ = sin2 /.

Thus the problem is completely solved and the relation between ql and q2 is given by a2qr + /)2qi + 2abQlq2 cos/ = sin2 /, where a = ~; b = ~, / are constants of Y-::rJ. Y"il integration. Ex 4.4.5. A particle moves in a plane under the action of two Newtonian centres of attraction at the points (c, 0) and (-c, 0), the attractions being ~ and f.z respectively, where r, r' being the distances from (c, 0) and (-c, 0) respectively. Show that the problem is of Liouville's type and hence solve it. SOLUTION: Here the KE (T) of the particle whose rectangular co-ordinates are (x, y) is T = ~mv2 = r;: [::i:2 + 1?J and its potential energy (V) is J-L'

y'(x + c)2

+ y2


103

Let us take the new variables ql, q2 defined by the quantities r' r = ql - q2. Therefore the potential energy is given by

(qf - c2 )(c2 (qf - c2 )(c2

+ qlq2)2 + [qlql(C2 -

-

q~)(qlq2

-

c2(qf - c2 )(c2 - q~) q~)[qfq~ + qfq~] + qfqf(c2 - q~) c2 (qf - c2 )(c2 - q~)

=

ql

+ q2

and

q~) - q2q2(qf - c 2 )j2

+ q~q~(qf -

c2 )

,c

=}

T

=

c2(qf - q~)[(qf - c2)q~ + 2 - q~)qf] c2(qf - c2 )(c2 - q~) 1 '2 '2 -2 (qf - q~) [ 2 ql 2 + 2 q2 2]' c - q2 ql - c

Thus the given problem is the Liouville's class of Lagrangian. Comparing with equation (4.27), w~ get, Ul(ql) = qf, U2(q2) = -q~; Vl(ql) = ~,V2(q2) = h; Wl(ql) = ql C C q2 -(fL + fL')ql, W2(q2) = -(fL - fL')q2. Therefore,

~u2qf = hUl(ql) -

Wl(ql) +"11

=}

"21 (2 ql -

·2 = h ql2 + ( fL + fL ') ql q22) ql

. '1ar Iy, SImI

"21 (2 ql -

·2 q22) q2

= - I1q22 + ( fL -

+ "11

fL ') q2

+ "12·

Dividing, the the equations and integrating, we get,

The reader may simplify this problem as similar of the previous problem.

104


Ex 4.4.6. Show that the dynamical system for which 2T = r1r2(fr ~. + can be expressed of Liouville's types. Hence solve it.

/2

+ f~)

and V

=

Hints: Put r1 = f]1 +f]2 and r2 = f]1-f]2· Then T = (f]r-f]~)(qr+q~); and V = ~. ql q2 Comparing with equation (4.27), we get, 11.1 = f]r,11.2 = -q~,v1 = 1,v2 = 1,W1 2q1. W2 = O. The solution is of the type,

=

. -l(ql - l/h) -sm . h- 1 ( -q2- ) ="'f. sm

v'Il

a

4.5

Conservation Theorems

Lagrange equations of motion are second order differential equations. To solve them completely for a system having k degrees of freedom, k such differential equations have to be integrated 2k times and likewise 2k constants of integration will be involved. These 2k constants of integration can be determined, if k initial values of C/k and qJ,; are known. It is not always possible to integrate every equation of motion. However, sufficient information about the physical nature of the motion of the system can be obtained without solving these equations completely through the constants of motion. The conservation theorems for a system in motion provide the knowledge of the constants of motion. As we shall see, the conservation theorems are the relations in the form of the first order differential equations. As such equations can be readily solved, the solution of the mechanical problem by employing conservation theorems becomes much easier. Before discussing Lagrangian formulation of conservation theorems, let us first discuHs about cyclic co-ordinates.

4.5.1

Conservation theorem of energy

Consider a conservative system, then the potential energy V is a function of generalised co-ordinates only, so that aay = 0 and ~ exist. Thus, . qk aqk

av = ~ (T _ V) = 8qk

8qk =

8T _ 8V 8qk 8qk

8 (~1 mkqk .2) a~ "2 qk

k

-

0

. = mkqk·

Also, let the constraints do not change with time so that the co-ordinate transformation equation is r?i = r?i(ql, q2, ... ,qn). Its Lagrangian L is a function of f]1, ... , f]n; q1, ... ,qn; t. If the Lagrangian does not contain the time explicitly, then

or,

L dL

=

L(ql, q2,· .. ,'qn; q1, Q2, . .. , qn) ~ 8L. ~ 8L ..

di= ~aqk+ ~a-qk k

qk

k

qk

d ( 8L ) . ~ 8L.. =~ ~ dt a- qk + ~ a-qk; k

qk

k

qk

U' L . smg agrange equatIOn

4.5. CONSERVATION THEOREMS

105

' " d (8L . )

= L...J dt ~qk k

Lk

~(8~ qk dt 8qk

'

qk

L) = 0 =>

~(8~ qk dt 8qk

L) = 0

d

=> dt (mkqkAk - L) = 0 => mkq~ - L = Constant = .1. The constant of motion .1, is one of the first integrals of Illotion and is, called .1acobi's integral of the system, and it represents the total energy of the system. Since !mkq~ = T and L = T - V, we have 2T - (T - V) = constant

=> T + V

= constant.

Therefore, if L of a conservative system does not contain time explicitly, then the total energy of the system is conserved. It is the conservation theorem of energy.

4.5.2

Conservation theorem of linear momentum

Let qk be the generalised co-ordinate and dqk represents the translation of the dynamical system in some given direction. Lagrange's equation for the motion of a dynamical system are given by

Let us consider the system is conservative. Then, its potential energy V is function of the generalised co-ordinates qk only so that g~ = 0 and ~ exists. Since the KE T depends upon velocities (which involve time derivative of displacement) so T of the dynamical system is independent of qk i.e., ~ = 0 and also unaffected by a shift of origin. Thus the Lagrange's equation of motion becomes, d

8T

8V

.

8V

-(-. ) + -8qk = 0 => Pk = -8qk - = Qk dt 8qk where Qk represents the components of the generalised force along the direction of translation of qk and Pk is the component of linear momentum along the same direction. Now, we interpret the equation Pk = Qk. The generalised force is Qk = - t aa- represents the unit vector along the direction of translation "~ F i. aTqki. Here, ~ aqk ~

of generalised co-ordinate qk. If we denote it as

n,

then

-

the component of the total force along the direction of n

-

along the direction of translation of the co-ordinate qk.


106

~-.

~

= L-J P t· n

= -. p.n ~

i

so that Pk is the component of the total linear momentum along the direction of translation of the co-ordinate qk. Thus the equation Pk = Q k represents the equation of linear motion of the system in generalised co-ordinates. Now suppose that the generalised co-ordinate qk is cyclic. Then,

Qk =

a=v .Pk = --a 0 qk =}

0 =>- Pk

=

constant.

Thus, if a given component of the total applied force on the system is zero, then the corresponding component of the linear momentum is conserved. It is the conservation theorem of linear momentum.

4.5.3

Conservation theorem of angular momentum

To ptove the conservation theorem of angular momentum, consider that the cyclic

(,)

N

o

o Figure 4.4: Angular momentum

-F"""-

co-ordinate qk is such that dqk corresponds to the rotation of the system of particles about some axis. Since T cannot contain' qk and V is independent of qk, it can be again proved that (as before) Pk = Qk. Since dqk corresponds to rotation of the system about some axis, we shall show that now the generalised momentum Pk represents the component of the angular momentum and the generalised force Qk represents the component of torque along the direction of rotation of the generalised


107

co-ordinate qk. Th~gure shows th~ictorial representation of the rotation of the position vector OP = ~i(qk) to OQ = ~i(qk + dqk) through an infinitesimal rotation dqk, so that PQ = d~i and LPNQ = dqk. Keeping the magnitude of vector constant, we find

-

Id"7 i l = =>

(NP) x dqk

= TisinfJdqk

. (J I d~il dqk = Tism .

n be the unit vector along ON. It follows that dd7qk is perpendicular to both ~i and n and is given by ~;:i = n x ~i' Now, the generalised force Qk is given by Let

!

=

L n."1\; r\ =

torque acting on the ith particle about the axis of rotation

i A,,",~

= n. L..J

T i

A~

= n.

T

i

where T = is the total torque. It shows that Qk is the component of the total torque acting on the system along the unit vector n, i.e., the axis of rotation. Again, from the relation Pk = ltr-, we get, oqk

~ = fI,. ""'~ L..J Li = n.L ~

where L is the total angul~mentum. Thus Pk is the component of the total angular momentum of the system along the unit vectors n, Le., the axis of rotation.Therefore we say that the equation Pk = Qk is the equation for .rotational motion of the system of generalised co-ordinates. Now, suppose that the generalised co-ordinate qk is cyclic, so, Qk =

_aOVqk = 0 => Pk = 0 => Pk =

Constant.

Thus, if a given component of the total torque on the system is zero, the corresponding component of the angular momentum is conserved. It is the conservation theorem of angular momentum.


108

Ex 4.5.1. A particle of mass m moves on a plane in the field of force given by -+ F = - kr- cos (H, where k is a constant and f is the radial unit vector. Will the angular momentum conserved? SOLUTION: Let P(r, 0) be the position (in polar co-ordinates) of the particle of mass m in the given plane. The KE of the given particle is given by

Since the given force is only in the radial direction, so Qo = O. Thus from the Lagrange's equation of motion, in terms of the ignorable co-ordinate 0 is d aT dt [ao d

~ dt [mr

aT

- ao] = Qo

2'

2'

0] = 0 ~ mr () = constant.

Now the quantity mr 2 0 is the angular momentum of the particle of mass m about the origin and so is conserved. Also the Lagrange's equation of motion in generalised co-ordinate r, where Qr = Fr = -kr cos (), is

aT d aT dt [ar - ar] ~

mr - mr0

2

= Qr + kr cos () = 0

which is the differential equation of the orbit of the particle.

Theorem 4.5.1. If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations, then L' = L + clj; (ql, ... , qn, t) also satisfies Lagrange's equations where F is arbitrary, but differentiable, function of its arguments. PROOF: Here we are to show that Lagrange's equation of motion does not change, if we add to the Lagrangian a total time derivative of any arbitrary function F(ql' ... , qn, t). Let L(qk, tik, t) be the Lagrangian of a system and F(qk' t) be arbitrary differentiable function. Then L + clj; is also satisfies Lagrange's equation of motion, for

!!:..[8(d~/dt) _ ~(dF) dt

8qk

= !!:..[-t(8F

8qi dt

dt 8qk 8t

+ 8F til)]- ~[8F + 8F tid 8ql

=~(8F)+~(8F)iIl_ 8t 8qk

8qi 8t 2

8ql 2

8 F _ 8 F iII 8qkat 8qk8ql

8ql 8qk 4 (.) 8F 8qi =OasF.,Fqj so-8' =Oand- =0. qj 8 qj

The arbitrary function F(qk' t) is called the Gauge function for the Lagrangian. The transformation L' = L + of the Lagrangian, which keeps the dynamical equations of motion unchanged, is known as Gauge transformations. This introduces

!Jft


109

an arbitrariness in the form of the Lagrangian that can preserve the same form of the equation of motion written in terms of L. So if L = L(qk' qk, t) can be produce a set of equations of motion through Lagrange equation of motion, then any other Lagrangian L' = L + dF~tt) formed from any arbitrary choice of F(qk' t) can be plugged in the same Lagrange equations of motion replacing the original L.. The explicit forms of the equations of motion in qk and t would not be different for tb.ese two different Lagrangians. However, the generalised momenta and Jacobi integral corresponding to L' = L + dF / dt are

aL'

aL'

a aF

aF

aF

P~=-a' ·· a qk qk =-a' qk +-a' qk [-at+-a·qzj=Pk+ql I • aL' I . aF aF aF J = qk-. - L =.J + qk- - - = J - - . aqk aqk at at

Therefore, under Lagrange Gauge transformation, the canonical momenta and the Jacobi are also changed. The momenta change due to explicit spatial variation of the Gauge function F and the energy-like Jacobi integral changes due to the explicit time variation of F. In this process the new gyroscopic potential energy and its ordinary energy is also modified.

4.5.4

Condition of invariance

Equations of motions are said to be invariant when these are same for the old and new variables corresponding to some suitable transformations of co-ordinates and velocities. Such a transformations are called invariance transformations. Let us define a inverse point transformation q: = q~(ql' q2,.' ., qn; t) with q~ forming the Lagrangian L'(q', q', t) which is

L'(q', q', t) = L[q(q', t), q(q', q', t), t] which is the condition of invariance for the transformation defined by q~ = q~(ql' q2, ... , qn; t). For example, same equations of motion are obtained if L' differs from L by a total time derivative, say ~f, where F = F(q, t), i.e.,

L'( qI , q. /,t) = L( qI , q",t)

dF (ql,···, qn, t ) . + dt

At the infinitesimal level, keeping first order terms in 8q and 8q, we have,

L(q, q, t) = L(q + 8q, q + 8q, t) . = L(q, q, t)

n aL

+ :t (8F) aL.

d

ql

t

+ I)-a, 8qi + -a', 8qi] + -d (8F) i=l

ql


110 n

8L

d 8L

d 8L

d

d 8L dt 8q·t

d n 8L dt L...J 8q·t i=l

=> "[-5q' - -(-)5q·] + -(5F) = 0 L...J a· t + -(-5q·) dt a·· t dt 8·. 1. dt ~1 ~ ~. ~ 8L L...J 8q·t i=l n

=> " [ - - -(-. )]5qi+-["-. 5qi+5F] =0 d

=> dt

8L_ [L: ~oqi + 5F] = 0; qt n

by Lagrange equation

i=l

=>

t

~~ 5qi + 5F =

Constant in motion.

qt

i=l

Thus associated with an infinitesimal invariance transformation there is a constant of motion. Also the constancy of angular momentum from the rotational invariance of n

the Lagrangian may be readily derived from

2: g~ 5qi + 5F = i=l q,

Constant in motion,

for the general setting of the physical system with n particles. Ex 4.5.2. The point of support of a plane pendulum of mass m and length I is moving along horizontal axis with constant velocity V. Write down the Lagrange's equation. Let at an instant t, the co-ordinate of the point of support of the plane pendulum be (0, Yo), where x axis is taking vertical. Let P(x, y) be the position of the mass m, which makes an angle 8 with the vertical. Thus, SOLUTION:

x

= Leos 8, y = Yo + l sin 8; yO = V t.

Thus the KE and PE are given by,

T

= ~m[x2 + :z?] = ~m[1202 + V2 + 2Vlcos80];

V

= mgl(l- cos 8).

Using Lagrangian L = T - V, the Lagrange's equation of motion for the generalised co-ordinate 8 is given by,

d 8L dt ( 80 ) For

=

8L

..

g.

DO => 8 + I sm 8 = o.

s~all 8, the above equation becomes jj + r8 = 0, the time period is T = 27rJI.

This is the same result as finds in simple pendulum. The term mVI cos 80 did not contribute because, /t(VI sin 8) = ftF(t). Hence one can add a term which is a total time derivative to the Lagrangian L. Ex 4.5.3. Discuss the problem of damped oscillator described by the Lagrangian L = e"Yt [~mq2 - ~ kq2];" m, k, positive constants by considering the different cases. SOLUTION:

For the Lagrangian, L = e"Yt[~mq2 - ~kq2], the equation of motion is

!!...(8~) _ dt 8q

8L oq

= 0 => !!...(e"Ytmq + e"Ytkq) = 0 dt .

k 0 => q.. +,q. + -q =

m


111

which is the differential equation for the damped oscillator. Let us take the trial solution of the form q = eQt for the linear second order differential equation, we have the auxiliary the equation

Three cases may arise

(i) Let

~
(psin2 () = C2 where O2 is a integration constant determined from the initial condition () = ~ and ¢ = J2ngjl. Thus C 2 = J2ngjl and consequently ¢sin 2 () = J2ngjl. Using the Lagrange's () equation A.(U) - aT = oW = 0 we have

!

dt ae

Be

7JO

'

(2ml20) - 2ml2¢2 sin () cos ()

=

-mgl sin ()

=> 210 - 2l2¢2 sin () cos () = -g sin () => 210 - 4ngcot(}cosec2(} = -g sin () => 210 2 + 4ngcot 2B = 2g cos B + 03; Integrating where the integration constant 0 3 is determined by the initial condition () position) and 0 = O. Thus C3 = 0 and therefore 2102 + 4ngcot2() = 2g cos (). The rod will fall down till iJ position () is given by,

=

O. Also as cos ()

=

2n cos 2 B - cos Bsin 2

or, cos or, B = Since

0 comes out to be positive,

()

=

~ (horizonta

0 gives the initial position, the

=0

+ 2ncos(} -1 = 0 cos- 1 [Jn 2 + 1 - n].

2

()

hence from this position, the rod will rise again.

Ex 4.5.5. A bead slides without friction on a wire which is rotating with angular velocity w in the force free space. Write down the Lagrange's equation of motion and find the radial velocity and acceleration.

SOLUTION: Let us consider the polar co-ordinate (r, ()) so that x = r cos B, y = r' sin B. As the wire is rotating with constant angular velocity, we have iJ = w(given) so that () = wt. Now the KE and PE are respectively T

= ~m[f2 + r 2iJ2];

L

=T

=>

- V

V

= mgy = mgr sin ()

= ~m[f2 + r 2iJ2]_ mgr sin 0

The Lagrange's equation of motion for rand () are given by

~(aL) _ aL _ dt af

=> f =

'2

rB - gsinB;

ar -

d 2' dt(mr 0)

= 0,

o.

~(aL) _ aL -

, dt ao 2'

or,mr ()

=

aB -

o.

,

constant.


114

These two equations give the required equations of motion. For the force free space, the kinetic energy (T) is conserved, so that

= ~m[r2 + r 2w2] => r =

T

r = riJ2

Also,

[r 2w2 +

constant]~.

= rw 2 -

- 9 sin e => f

9 sin wt.

Ex 4.5.6. A bead slides on a wire in the form of a cycloid described by the equations x = a( 0 - sin 0),11 = 0.(1 + cos 0); 0 ::; 0 ::; 271", where a is the radius of the generating (;l,rde.

Neglecting fnctwn between bead and w'ire .find the equation of mot'ion.

Let P(x, y) be the position of the bead of mass m at any time t on the cycloid x = a( e - sin e), y = a(l + cos e); 0 ::; e ::; 271". The KE of the bead is SOLUTION:

T =

~m[x2 + I?]

=

~m[a2(1

2 2 2 '2 = ma [1 - cose]O .

- cos 0)2

+ a2 sin2 e]iJ2

Taking x-axis as the reference level, the potential energy of the bead is given by

v = mgy = mga(l -

cos e),

where y is measured upwards, opposite to mg. Thus the Lagrangian of the motion can be written as L

=T

- V

= ma 2[1- cose]02

- mga(l- cos e)

in which we see that 0 is the only generalised co-ordinate. Thus the Lagrange's equation of motion for the generalised co-ordinate e, i.e., [~~] ~~ = 0 gives

1t

-

d [2ma2iJ(1 - cos e)] - [ma 2iJ2 sin e + mga sin e] = 0 d,f .. 1 '2 9 => (1 - cosO)e + -2 sinee - - sine = 0 2a '2

or jj ,

+ ~ cot ~ 2

2

-

JL cot ~ = 0 2a

2

'

which is the desired differential equation of motion. Further let, U = cos(~), then 2 2 Thus the differential equation du - _12 sin(f!.)iJ and (llj'Id u - _12 sin(f!.)jj - 14 cos(f!.)iJ dO2 2 2 . of motion becomes, d 2 ,u d02

9

=-

4a u.

This differential equation represents a simple harmonic motion. The solution of this equation gives,


115

where C1 and C2 are constants. This shows that cos(~) returns to its original value

/f-,

after a time of 271" which is the required period of one oscillation between cusps of the wire. As this period is same as that of the simple pendulum of length 4a, hence this system is also referred to as Cycloidal pendulum. Ex 4.5.7. Consider a dynamical system with T = !(qr + q~), V = f(q1 - q2), where f is a given function. By choosing suitable new co-ordinates Q1 and Q2, reduce the problem of determining the motion to the evaluation of an integral involving the function f. Determine q1, q2 as functions of t if f (x) = x 2 . SOLUTION: Let us choose the new co-ordinate Q1 and Q2 as Q1 = q1 - q2, and Q2 = q1 + q2· Therefore, the old co-ordinates in terms of new co-ordinates are q1 = !(Q1 + Q2) and q2 = !(Q2 - Q1). The KE and PE in terms of new variables Q1 and Q2 are

and so t~e Lagrangian L becomes L for Q2 gives,

=

HQr+Q~)- f(Q1). The Lagrange's equation

where A and B are constants of integration. The Lagrange's equation for Q1 gives,

Now, given that, f(x)

= x 2 , so the above equation gives,

Thus we get, two equations, q1 these equations give

+ q2 =

At + B and q1 - q2

=

ke it . The solution of

116

4.6


Exercise

Ex 4.6.1. Find the Lagrangian of the Sun-Earth system.

GATE -1997

Ex 4.6.2. The Lagrangian for the Kepler's problem is L = ![f 2 + r 2e2] + ~; f.L > 0 where Cr, (7) denote the polar co-ordinates and the mass of the particle is unity. The angular momentum of the particle about the centre of attraction is a constant. GAT E - 2001 Ex 4.6.3. A rod of mass m and length l is free to move in any direction on a frictionless inclined plane. The angle of inclination of the plane with respect to ground is O. How many degrees of freedom are there for the rod? Write down the equations of motion for the rod. What are the conserved quantities for the motion of the rod? Ex 4.6.4. The lagrangian L for a motion of a particle of unit mass is L = !(i;+y2 + i 2 ) - V + i:A + yB + iC; where V, A, B, C are functions of x, y, z. Show that the equations of motion are x = -~~ +y[~~ - ~:l-i[~~ - ~~] and two similar equations. Ex 4.6.5. Deduce the Lagrange's equations of motion of a dynamical system of n degrees of freedom specified by n generalised co-ordinates qk{k = 1,2, ... , n) and moving under a conservative field of forces. Ex 4.6.6. A bead is able to slide along a smooth wire in the shape of a parabola z = CT2. The parabola is rotating with constant angular velocity w about its vertical symmetry-axis. Write down the equation of motion of the bead. Ex 4.6.7. Use Lagrange's equations to find an expression for the angular acceleration of a pendulum bob moving in ~ vertical plane. Also find the tension in the string which supports the bob. Ex 4.6.8. A particle of mass m, moves on a frictionless horizontal plane {xy)plane. It is attached to one end of each of two identical massless springs A and B. Each spring is of upstretched length l and spring constant k.' The other ends of A and B are fixed at (0, l) and (~, ~) respectively. The equilibrium position of the particle is at (0,0). Obtain the Lagrangian of the particle. Assuming that x « l, y « l, simplify L, retaining terms up to order x 2 ,y2 and xy. GATE -1991

Ex 4.6.9. An artificial satellite moves in the gravitational field of the earth of mass M. Regarding the satellite as a point particle of mass m, obtain the Lagrange's equations of motion for the satellite. Hence find expressions for the canonical momenta Pr , Po, Pcp. 'Which of these is constant in time. GATE - 1992 Ex 4.6.10. A thin, uniform rod of mass m and length R slides with its ends on a smooth vertical circle of radius R. Set up the Lagrangian of the system. Find the period of small oscillations in terms of given quantities. GATE - 1994 Ex 4.6.11. A thin uniform rod of mass M and length l is oscillating in a vertical plane about a fixed horizontal axis through one of its ends. Write Lagrange's equation of motion of the rod and determine its time period for small oscillation. GATE - 1995

4.6. EXERCISE

117

Ex 4.6.12. A rod of mass m and length l is free to move in any direction on a frictionless inclined plane. The angle of inclination of the plane with respect to ground is (). How many degrees of freedom are there for the rod? Write down the Lagrange's equation of motion. What are the conserved quantities for the motion of the rod? GATE -1997 Ex 4.6.13. Consider the lagrangian L = !q log(q2) - )..q + qf(q); where f(q) an arbitrary function and ).. > O. Write down the Lagrange's equation of motion. calculate the Hamiltonian of the system. Is total energy of a constant of motion. Does the Lagrangian transform covariantly under the transformation q --+ aq for a a real constant. GATE - 2001 Ex 4.6.14. A particle of mass m is constrained to move on the plane curve xy = c(c> 0) under gravity (y-axis vertical). Find the Lagrange's equation of motion for the particle. GATE - 2001 Ex 4.6.15. A particle of mass m is constraint to move on the surface of a sphere of radius R. The sphere is resting on the ground. Set up the Lagrangian for the particle. What are the constants of motion for the particle. GATE - 2000 Ex 4.6.16. Consider the motion of the particle with Lagrangian L = ~(:i:+y2 + z2) as viewed from the rotating co-ordinate system x' = x cos () + y sin (), y' = - x sin () + y cos (), z' = z where () = ()(t) is some function of time. Find the new Lagrangian L' in terms of those co-ordinates. Ex 4.6.17. Prove that a dynamical system having n dJ. and k( < n) ignorable co-ordinates can be reduced to one (n - k) rJ. Ex 4.6.18. Define cyclic co-ordinates. If all the co-ordinates of a dynamical system are cyclic, prove that the co-ordinates can be found by integration. In particular, if the system is scleronomous, show that the co-ordinates are linear functions of time. '2

Ex 4.6.19. The kinetic potential of a dynamical system is L = aq;!+b +!q~+2q~+cq21 where a, b, c are given constants. Find the q2 in terms of elliptic functions. Ex 4.6.20. The kinetic potential of a dynamical system is L = ~f(q2)qr+h~-1P(q2). Show that, the relation between the variables ql and q2< is given by a differential equation (~) ~ = 0, where q~ = ~, and where L' is defined by the equation,

d:l

-

1

L' = [2h - 21P(Q2)] 2 [f(q2)

1

+ q~]2.


Chapter 5

Hamiltonian Mechanics The study of classical mechanics based on Lagrangian dynamics is not only simpler but also advantageous in solving complex problems. The study of classical theory of fields is extremely suitable via, the approach. For a holonornie dynamical system, described by a set of n generalised co-ordinates, the Lagrange's equations of motion are the second order differential equations. Here we shall derive an alternative formula of dynamics in terms of Hamiltonian, consists of two first order differential equations for a system with n degrees of freedom, in generalised co-ordinates and momenta. We shall assume that, the constraints are holonomic and the forces are monogenic, i.e., they are derivable from potentials which depend either on position or are velocity dependent. This Hamilton formulation is the basic foundation of statistical and quantum mechanics, field theories, that are outside classical mechanics, Hamilton-Jacobi theory for complete classical solution of dynamical problems, that is within classical mechanics.

5.1

Phase Space

Let us consider a system of n particles. The configuration space is not adequate to represent the history of the system. If only 3n momenta are used as axes, we will get the momenta space. Any point in the momenta space will describe the state of motion of the whole system and the locus of the point is called Hodogmph. The space traced by the dynamical system during its motion, formed by 6n coordinates; 3n generalised co-ordinates ql, q2,"" q3n and 3n momentum components PI, P2, ... ,P3n is called phase space. The extended phase space or state space is (2n+ 1) dimensional space where one more dimension is added to the phase space to include the parameter time. We know, only three initial values (only of co-ordinates) for the particle and hence specification of the path followed by the system is not possible. Hence there are infinite number of possible paths through any point in the configuration space. The specification of a point on the path in the phase space need six initial values for each particle, so that the state of the system is defined completely. Therefore, there is less arbitrariness about the path in phase space as

119

CHAPTER 5. HAMILTONIAN MECHANICS

120

compared to the path in configuration space. Thus, there is unique path in phase space, which refers to actual dynamical path.

5.2

Hamiltonian Function

Let Pk = 8!- be the conjugate momentum corresponding to the generalised co• I' . 0 f varIa . bl ordinate Pk.8qk The pair (qk,Pk) is called the canomca conjugate paIr es. Consider a Lagrangian L = L(qk, qk, t), which is a function of positiWl'-co-ordinates qk, the generalised velocities qk and time t of a dynamical system. Then, dL -dt,

~ 8L .

=~ k

8L .. 8L aqk qk + ~ ~ ;y-qk + at k qk '

~ dt d ( 8L ) . ~ ~qk 8L .. 8L =~ ;y- qk + ~ + at; k

qk

k

d 8L.

qk

8L ..

d or, dt

8L . [2:k ;y-qk qk

L]

~ d

8L

= 2:[dt (-8' )qk + ~qk] + at k qk qk

b L 'E t' y agrange s qua IOn

=

8L.

8L

~ dt (~qk) + at k qk

8L

= -at

d

or, dt[2:Pkqk - L] =

8L

-at;

Pk

=

8L 8qk

k

or, dd H(q,p, t)

t

=-

L 88 ;

H

t

= 2:Pkqk -

L.

k

The function H = H(q,p, t) = LPkqk - L is called precisely Hamiltonian or Hamilk

ton's function. Normally, the Hamiltonian for each problem must be constructed via Lagrangian formulation. Thus H consists of two quantities :

(i) L Pkqk is the part that is homogeneous in qk in the second degree and k

(ii) L is a part of Lagrangian independent of generalised velocities. The quantities qk and Pk form a set of independent variables in Hamiltonian formalism, whereas qk and qk form the independent set in Lagrangian formulation. We shall use the relation H(q,p, t) = LPkqk - L to obtain Hamilton's equations of motion. k

It is at once implied from the above relation that if L does not contain the time t explicitly, then H is reduced to a constant in time. For the dynamics of microscopic systems like atoms, molecules etc., the forces involved are known definite forces exerted by microscopic particles on one another and can be taken as the applied forces on the system. Consequently, the concept of constraints lose its meaning. Thus, for microscopic systems, Hamiltonian dynamics enjoys wide application. 2x 2 Ex 5.2.1. For an anharmonic oscillator the Lagrangian L(x, ±) = !±2 ax

3

+ f3x±2.

!w

Find the corresponding Hamiltonian.

5.2. HAMILTONIAN FUNCTION

121

=

From the given Lagrangian L(x, x) get the conjugate momentum as SOLUTION:

oL P = ox

.

=x+

2{3" xx => x

!x 2 - !w 2x 2 - ax 3

+ (3xx 2, we

P

= 1 + 2{3x'

Therefore the Hamiltonian H is given by H(x,p) = xx - L(x, x) = xx -

~x2 + ~w2x2 + ax3 -

px = 1 + 2{3x

p2 2(1 + 2{3x)

2px - p2 - 2(1 + 2{3x)

+ 2w

1

2 2

x

+

(3xx 2

1 223 2w x + ax

+ ax

3

.

Ex 5.2.2. Weber's electrodynamic law of attraction: Let us consider the motion of a particle subject to Weber's electrodynamic law of attraction to a fixed point, the force per unit mass acting on the particle being

where r is the distance of the particle from the centre of force. In this case, the velocity dependent potential function V(r,r,t) is given by V = ~[1 + ~J. Here we construct the Hamiltonian H for velocity dependent potentials. Now the Lagrangian for the system is

The conjugate momenta corresponding to the generalised co-ordinate r is given by pr

oL 2r => r. = Pr [m = n-:= mr. - -2 u1' 're

2

-2

re

J- 1 .

Thus, the one dimension Hamiltonian H for the velocity dependent potential is

1 ·2 + -1 [1 +r 2J Prr. - -mr 2 r c2 2 -1 1 2 2 -2 1 [ p ; 2_2 = pr2[m - -J - -mp [m - -J + -r 1 + -(m - -) J. rc2 2 r rc2 c2 rc2 1 1 '2 Now, T + V = -2mi·2 + -[1 + r 2J r c 1 2 2 2 1 2 2J = -mPr[m + -[1 + -c2 m - ) - ¥= H. 2 rc2 r rc2 H

= Prr. - L =

-r

Hence the given system is not conservative.

p; (

122


Deduction 5.2.1. We have seen that for a conservative, scleronomic system the n

quantity

2: qi g!-q, i=1

L is a constant and that if the KE is a homogeneous quadratic

function of the velocities, the total energy is constant as well, i.e,

Therefore the Hamiltonian H can also be interpreted as generalised energy. Deduction 5.2.2. For the kinetic energy function homogeneous in the second degree of the generalised velocities as given by T = ~ 2: 2: o,ijqilj, the Lagrangian L can i

j

be written as L

=~

L L aijqiqj - V(q) i

J

where q = q(qI, q2,' .. , qn) and aij = aji. In matrix notation it can be written as L =

~qT Aq -

V(q).

Let aij1 =1= (aij) -1, in general, be the elements of A-I, then from Lagrange's equation of motion we have, n n n 8 8V d ~ .1 1~ ~ aij .. - [ ~ak'q' = - ~~--q'q' - dt .1=1 J J 2 i=1 .1=1 aqk t J 8qk

8L ::::::} Pj = a'. =

%

L aji(q)qj; n

inverting and qi = aijlpj

i=1

. 1 (-1 ) 8 a ij (-1 ) av ::::::} Pk = -2 ais Ps -8 ajmPm - - a qk qk = ~pT A-I aA A- 1p _ av 2 aqk 8qk 1 1 8A8V = __ pT _ _ p _ _ . 2 aqk 8qk

Thus the Hamiltonian H can be defined as H(q,p) -

aH·

DH

aqk ;

apk'

qk =

=

~pT A-lp

+V

with Pk =

Ex 5.2.3. Write down Hamiltonian in spherical polar co-ordinates for the Lagrangian . 2 L = ~m r+ + ~; J.L constant.

123

5.3. HAMILTON'S EQUATION OF MOTION SOLUTION: In spherical polar co-ordinates the Lagrangian is given by

I. TA f..L = -q q+-

2

r

where the matrices A and A-I are given by

A

=

m 0 0 ) 0 mr2 0 ; ( o 0 mr2 sin2 ()

A-I

=

(

m 1. 0 0 ~

o

0

0

)

0

mr

I

.

mr 2 sin 2 8

Thus the Hamiltonian H becomes

;:pr)

(

5.3

f..L

- ~

Hamilton's Equation of Motion

We have already introduced the Hamiltonian function H, which is related to the Lagrangian L by H(q,p, t) = EPkqk - L(q, q, t). The Hamiltonian H of a dynamical k

system is function of generalised co-ordinates qk, generalised momenta Pk and the time t. Here we want to describe the motion of the system in terms of equation of motion involving the Hamiltonian H, by transforming the set of variables (qk, qk) to a new set (qk,Pk). So,


124

Comparing the coefficients of dPk, dqk, dt, we obtain,

. qk

aH

= apk;

. -Pk

aH

aL - at

= aqk;

=

aH at'

(5.1)

These 2n set of first order differential equations of motion (5.1), are known as H amilton's canonical Equations of motion or simply Hamilton's equation, due to William Hamilton[1835]. The 2n constants of integration can be evaluated in terms of initial conditions. Equations (5.1) predict that generalised momenta and generalised co-ordinates are dynamically equivalent sets of variables because their roles can be trivially interchanged just by making a change of sign. The Hamiltonian methods are not particularly superior to Lagrangian techniques for the direct solution of the mechanical problems. Rather, the usefulness of the Hamiltonian viewpoint lies in providing a framework for theoretical extensions of many areas of Physics. This gives an elegant and most useful method in the theory of classical mechanics. Ex 5.3.1. Generalised Harmonic Oscillator: For the generalised harmonic oscillator, the Lagrangian L is given by,

L = -1,2 q - -1 [X - -y2] q2 - Y. -qq 2Z 2 Z Z where X, Y, Z are time dependent. The conjugate momenta is given by

aL 1. Y P=aq=zq-zq; =*q=pZ+Yq. Thus the Hamiltonian H is given by, H = pq. - L 1.2

= Z1 [.q -

Y] .

1'2

q q - 2Z q

+"21 [X -

0

zy2] q2 + Y. zqq

y2 2

1

= 2Z q + "2 [X - Z]q 1

= 2Z[PZ + Yq]

2

1

+ "2 [X -

y2 2

1

ZJq = "2[Zp

2

2

+ 2Ypq+Xq J.

5.3. HAMILTON'S EQUATION OF MOTION

125

The Hamilton's canonical equation of motion gives,

. 8H . 8H p=--=-(Yp+Xq); q=-=(Zp+Yq] 8q 8p

~ (~) = (-J -:) (~) ~ x(t) = Ax(t) where x(t)

=

(~)

and A

=

(-J -:). The eigenvalues of

A are given by,

1-YZ- A Y-~A = 0 1

~ A = ±JY2 - XZ = A1,A2' The linearly independent eigenvectors corresponding to the distinct eigenvalues AI: A2 are give by (Y + Al Z)T and (Y + A2 z)T respectively. Thus the solution is given by x(t)

= C! (Y ~ AI) eA1t + c2 (Y ~ A2)

eA2t.

Deduction 5.3.1. Physical significance of H : Let us consider the Hamiltonian H = H(q,p, t). Now, the total time derivative of H is given by,

dH

"

dt = ~ k =dH ~ dt

aH . aH Pk + at aaHqk qk. + "~k a:Pk

"~Pkqk .. k

8L

. . -8L +" ~qkPk k

at

an

= -at = 8t; from (5.1).

If the Lagrangian of a system does not involve time t, then ~f

dd~

= 0

~H

= 0 and consequently

= LPkqk - L = Constant = E.

Thus, quite generally for the generalised co-ordinates and its canonical conjugate the Hamiltonian function H(qk,Pk) may be interpreted as the energy of the system. This gives a physical meaning of the Hamilton's function. Therefore, if the time t is cyclic in L, then t is also cyclic in the Hamiltonian H and so the Hamiltonian H is constant in motion. In this case H = E is the energy integral for conservative systems, for which H = T + V = E, otherwise in general H # H(t) and L # L(t), there will exist a constant of motion, called the Jacobi '8 integral given by H = constant = J

which need not be identical with the actual energy E.

126


Deduction 5.3.2. Now let us consider the conservative system. For a conservative system, potential energy V is a function of generalised co-ordinates qk only and is independent of the generalised velocities, i.e., aay = 0 and the conjugate momentum qk .IS gIven . by Pk = ~. aT [lqk

Deduction 5.3.3. If the Hamiltonian H does not involve a particular co-ordinate qk, then from (5.1) we get, 8H . = 0 =? Pk = 0 =? Pk = constant . 8qk

Such co-ordinate qk is called cyclic or ignorable co-ordinate. Thus, a cyclic coordinate in the Lagrangian will be absent in the Hamiltonian H. Conversely, if a generalised co-ordinate does not occur in H, the conjugate momentum is conserved. Ex 5.3.2. The Hamiltonian of a dynamical system is given by H = qp2 - qp + bp, where b is a constant. Solve the problem. SOLUTION: The Hamiltonian of the given dynamical system is H = qp2 - qp + bp. Hamilton canonical equatkms of motion give . 8H d'p = 8H q=-an 8p 8q =? q = (2p - 1)q + band _

p=

p2 - p.

From the last equation, we have,

dp p2 _ P

= -dt =?

1

[p -

1 P _ 1ldp = dt

log -p- = t + k; integrating p-1 e t +k 1 t +k =? P = et +k _ 1 = 2"[1 + coth(-2-)]' =?

From the first equation, we have,

. (t+k) q = qcoth -2-

+ b =?

=? qe-[J coth(~)dtJ =?

=

dq (t+k) dt - qcoth -2b

t+k = b qcosech 2(-2-)

J J

e-[J coth(~)dtJdt + c

t+k cosech2(-2-)dt

t+ k = -2bcoth(-2-) =?

=b

+c

+c

t+k t+k t+k q = cSinh 2(-2-) - 2bcosh(-2-)sinh(-2-)'


127

Ex 5.3.3. The Hamiltonian 'of a dynamical system of two degrees of freedom is given by H = qlPI - q2P2 - aqf + bq~, where a, b are a constants. (i) Solve the problem. (ii) Also show that the quantities H = P1 ~:q1 and F2 = qlq2 are constants of the motion. Are there any other independent algebraic constants of the motion'? SOLUTION: (i)" The Hamiltonian of the given dynamical system is H = qlPI - q2P2 aqf + bq~. Hamilton canonical equations of motion give

.

aH

.

aH

i=1,2.

qi=-a andpi=--a; Pi

=}

qi

liI = ql; q2 = -q2; PI = -(pI - 2aql); P2 = P2 - 2bq2·

The first canonical equation gives,

-dql = dt =} ql = et+ c1 • ql

Similarly, q2 we get

= e- HC2 , where CI

=}

and

C2

-dPI + PI = dt

2ae t+ C1

Ple t

e2HC1 dt

= 2a

J

are constants. Also using PI

=

-(pI - 2aql),

+ C2 =} PI = 2aeH C! + eC2 - t .

Similarly, using h = P2 - 2bq2, we get P2 = -2bte t +C2 + eHC3 • Thus the expressions for ql, q2, PI, P2 give the solution of the dynamical problem of two degrees of freedom. (ii) Using the Hamilton's canonical equation of motion, we have ~[PI - aql] = q2(PI - aliI) - Q2(PI - aql) q2 q~ dt

-

=}

FI

PI - aql

+ PI -

aql

q2

= -PI

+ 2aql -

= PI -

aql

q2

=

aql q2

. ; as q2 = -q2

+ PI -

aql = 0

constant.

Similarly, it can be shown that P2-q1bq2 is also constant of motion. Also

=

_..!!.

Q2 q2

_P1-:Q1 = PI - 2aql q2 q2 ql

Thus FI and F2 are independent integrals of motion.

i= o.

~~~~--~~~~

128


Ex 5.3.4. A Hamiltonian of one degree of freedom has the form H = ~ - bqpe- at + b;q2e- ot [a + be- nt ] + where a,b,a and k are constants. Find a Lagrangian corresponding to this Hamiltonian.

kf

SOLUTION:

From the Hamilton's canonical equation of motion, we have,

.

8H 2p ab = - bqe- at + _(2p)e- at [a 8p 2a 2 = E - bqe- nt + abpe-at[a + be-nt]

q= -

+ be-at]

a

+ bqe- ot = p[-1 + abe-at(a + be- nt )] a q + bqe- ot q + bqe- nt =*p= =~-~~ + abe-at(a + be-ot) D =* q

where D = ~ + abe-ot(a Hamiltonian, we get,

+ be-ot).

From the relation between Lagrangian and

q= 8H 8p q[q + bqe- nt ] p2 -at kq2 = D - ?:D + bqpe - 2 nt ot q2 + bqqe- _ 1 [q + bqe- ]2 + bqpDe- ot

L=pq-H;

_

•

2

kq2

-

D 2 q2 _ b2q2e-2ot _ 2bqqe- at + 2bqpe- at D kq2 2D 2 ii + 2bqe- at [q + bqe- at ] - b2q2e-2od kq2 = - - ; pD = q+bqe- at 2D 2 q2 + b2q2e-2at + 2bqqe-at kq2

2q2

+ 2bqqe- at _

2D

(q

+ bqe- at )2 2D

kq2 -

1

2 - '2

2 (q + bqe- at )2 ~ + abe-at(a + be-at)

which is the required Lagrangian for the given Hamiltonian.

Deduction 5.3.4. Cylindrical Co-ordinates: Here we are to use the cylindrical co-ordinates (r, e, z), which are taken as generalised co-ordinates, we are to write the Hamilton's equations of motion for a particle of mass m moving in a force field of potential V(r,e,z). Using the transformation equations x = rcose; y = r sin e; z = z between cartesian and cylindrical co-ordinates we get T

= ~m[±2 + ii + z2]

L

=T

2

- V

=

~m[r2 + r 2iP + z2] 2

= ~m[1~2 + r 2iP + z2]_ VCr, e, z)

129


which does not contain the time explicitly. The conjugate momenta are then becomes

=

Pr

m a'r

= mr;

Po

=

m aiJ

= mr

2'

e; pz

=

m aZ

.

= mz.

where Pr is the component of linear momentum along the line through the origin in xy-plane, pz is the third component of linear momentum of the particle, PIb is the third component of angular momentum of the particle. Since L does not contain the time explicitly, the Hamiltonian H is given by H

= T + V = ~m[f2 + r 2iJ2 + z2J + V(r, e, z) 1

P;

= -2 m [-2 m

p~

P;

+ 22 + 2"J + V(r, e, z) mr m

= -2 [p~ + P~r + P;J + V(r, e, z). m 1

We also calculate H from the relation H = Pr'r + poiJ + pzz - L, and get the same result, Thus the Hamilton's canonical equation of motion gives · aH P~ av . aH PI' p r = - - = -3- - - · r = - = ar mr ar ' apr m · __ aH __ av. iJ _ aH _ ~ PO ae ae ' - apo - mr2 · aH av Pz=--=--; az az

. aH pz z=-=-. apz m

When V is not a function of e, then Po = 0, i.e., Po = constant. If Po = 0 and V is independent of r, then Pr is a constant. Potential functions that are independent of e are said to be possesses cylindrical symmetry. Deduction 5.3.5. Spherical Co-ordinates: Here we are to use the spherical polar co-ordinates (1", () , ¢ ), which are taken as generalised co-ordinates, we are to write the Hamilton's equations of motion for a particle of mass m moving in a force field of potential V(r, e, ¢). Using the transformation equations x = r sin e cos ¢; y = r sin esin ¢; z = r cos e between cartesian and spherical polar co-ordinates we get, T

= ~m[x2 + 'Ii + z2J = ~m[f2 + 1'21;2 + r2 sin2e~2J 2

=> L = T - V =

2

~m[f2 + r 2iJ2 + r2 sin2e~2J -

V(r, e, ¢).

Since the Lagrangian L does not contain the time explicitly, the conjugate momenta are then becomes aL . aL 2. aL 2. 2 . Pr = af = mr; Po = aiJ = m1' e; PIb = a~ = mr sm e¢.


130 The Hamiltonian H is given by

= Pri· + poe + P

t2

[pkoqkn~ -

j(Pk

t2

+ ~~)Oqk + j(qk - ~~)OPk =

tl t2

0

tl t2

=> - j(Pk +

~~)Oqk + j(qk - ~~)OPk =

tl

0 at tl,t2.

tl

Since the system is holonomic and now described in the phase space, qk and Pk'S are all independent and Oqk's and 0Pk'S are arbitrary at all points of the path. Thus the above integral vanish, only if .

8H

0

Pk+- = ; 8 qk which are the Hamilton's canonical equations of motion. Ex 5.4.1. Using Hamilton's principle write down the equations of motion in spherical polar co-ordinates.


138

SOLUTION: In spherical polar co-ordinates the KE T can be written as T = !m[i'2 + + 'r 2 sin2 O¢2l. Let the potential energy V = V(T, 0,

Deduction 5.4.3. Here we are to deduce the Newton's second law of motion from Hamilton's principle. Let us suppose that, a particle of mass M at the position -;;7 ---+ is moving under the action of force F. Let -;;7 be its position vector at any instant. The kinetic energy of the particle is given by .

T

=

1 1~12 -m r

2

=

1 (~~) -m r. r

2

1·· .. ::} 8T = 2[-;;7.8-;;7 + 8-;;7.-;;7]

..

= m-;;7.8-;;7.

Also, small variation in potential energy may be expressed as 8V Now, a,ccording to Hamilton's variational principle,

J -:t t2

[m .8-:t +

tl

F.8-;;7]dt =

0

=-

---+

F .8-;;7

= 8W.

139

5.5. ACTION PRINCIPLES

::}

Im~.c5~I~~

Jm7·.c5~dt+ F.c5~dt = t2

-

0

tl t2

::::} -

r .c5 r dt + -F.c5 r dt = 0 as c5r =0 j m ...;tl t2

::} j[F - m7l.c5~dt = o. tl

- -

Again, as c5~ vanishes at end points, the above equation is true for every virtual .. = 0, which displacement c5~ and hence the integrand must vanish, i.e., F - m~ is Newton's second law of motion.

5.5

Action Principles

P. M. L. Maupertuis (1740) asserted that all activities of nature are performed with a least possible expenditure of 'action', which is defined as the product of mass t2

velocity and distance. In mechanics, the quantity A =

J L,Pkqkdt

is known as k action. A variation of the path in configuration space is always considered to be infinitesimal and the variation in the generalised co-ordinates qk is always taken at the same instant. There are certain physical situations in which moving boundaries are relevant. This is in contrast to the case of Hamilton's principle where the varied path shares with the actual path the same and points which are fixed. tl

5.5.1

Action as function of co-ordinates

We have defined the action as A = J Ldt; L = Lagrangian, so that ~1 = L, where the integral is taken along a path between ql and q2 at t = tl and t = t2 respectively, so ql = q(tl), q2 = q(t2). Of the many paths that can be found between these points only one path is the actual one for which A is minimum. For the actual path also, the Lagrange's equation of motion (~t) - ~~ = 0 is satisfied. Let us now consider another aspect of A, when we regard A as a quantity characterizing the motion along the actual path, and compare the values of A for paths having a common beginning at ql, but passing through different points at t = t2, i.e., end point variation at t = t2 different from 0 but c5q = 0 at t = tl. Thus,

fit

140

CHAPTER 5. HAMILTONIAN MECHANICS = LPk8qk;

taking 8q for 8q(t2)

k

8A

8L

=> 8qk = Pk; Pk = 8qk' Now, L ,

= =

dA dt

=

8A at

+L

8A dqk k 8qk 8t

8A ' " 8A qk 8A ' " . at + L.J a . = at + L.JPkqk k qk k

= LPkqk -

H(q,p, t)

k

=> dA = LPkdqk - H(q,p, t)dt k

= LPk(t2)dqk(t2) -

H2 dt 2 - LPk(tl)dqk(tl) + H1dtl

k

k

if the end variations at t = tl and t = t2 are not zero. Thus equation shows whatever the external forces be present in the system, the RHS is a perfect differential.

5.5.2

Principle of Least Action

The principle of least action is of fundamental importance in classical mechanics. An important variational principle associated with Hamiltonian formulation is the principle of least action. In 8-variation process (i) the varied path in configuration space always terminated at endpoints representing the system configuration at the time tl and t2 as the correct path, for the derivation of Hamilton's principle. (ii) the varied path return to the same endpoints in configuration space, to derive the Lagrangian equation of motion.

In this general type of variation, known as 6-variation of the path of a system, we allow to vary the position co-ordinates as well as time. The principle of least action t2

for conservative system is then expressed as 6

J E Pkqkdt = 0 on the actual path tl

k

as compared with some neighboring paths, where 6 is the variation in A with time. PROOF: Let a configuration of dynamical conservative system be represented by n generalised co-ordinates ql, q2,.'" qn. Let us consider, a family of possible varied paths defined by the functions

where ex is an infinitesimal parameter and for the correct path it is zero. The continuously differentiable function 1}k do not necessarily have to vanish at the endpoints,


141

either the original or the varied. We are to evaluate the 6- variation of the action integral as ~

J

6

Ldt =

tl

~+6~

J

~

L(o:)dt -

tl +6tl

J

L(O)dt

tl

where L(o:) means the integral is evaluated along the varied path and L(O), to the actual path of the motion. Since the variation includes the change in time for every path, i.e., t is a function of 0: and therefore, each point qk will be function of 0: and t, i.e., qk = qk(O:, t), so that

as change in t occurs as a result of 6-variation, so, j!do: = 6t. From the above expression, we see that, the change in generalised velocities is not same as the time derivative of the change in the generalised co-ordinates. Thus a point P on the actual path now goes over to pIon the path where qk -+ q~ = qk + 6qk and 6qk = 0 at the end points of the path. Now the 6-variation in the Lagrangian L = L(qk, qk, t) is given by

Let 6-variation consists of two paths : (i) The first arises from the change in the limits of the integrand. For first order approximation, this part is simply the integrand on the actual path times the difference in the limits in time. ~ ..

(ii) The second is caused by the change in the integrand on the varied path, but now between the same time limits as the original integral. Thus the 6-variation of the action integral can be written as t2

6

J

Ldt = 6I(t2) - 6I(tl);

tl

142

CHAPTER 5. HAMILTDNIAN MECHANICS

= [8I(t2) + j(t2)6t2J - [8I(tl) + j(tl)6tlJ = [8In~ + L6t2 - L6tl; as j = L t2

=6

J

Ldt + L6tl~~·

t1

The 'variation of the first integral can be carried out through a parameterization of the varied path, as like Hamilton's principle except, the variation qk does not vanish at the end points. Thus

t2

=> 6

J

Ldt

=

t1

[L6t -

L:Pkqk6tJ~~. k

In this equation 6qk refers to the variation in qk at the original endpoint times tl and t2. To obtain the principle of lest action we restrict our further considerations by three important qualifications : (i) Only systems are considered for which L, and so H, are not explicit functions of time, and consequently H is conserved.

(ii) The variation is such that H is conserved on the varied path as well as on the actual path: (iii) The varied paths are constrained requiring that 6qk vanish at the endpoints but not 6t.


143 t2

Under the above prescriptions, taking the action A

= J L,Pktlkdt,

we get,

tl k

t2

t2

A = j LPkizkdt = j[L + Hjdt; as H is conserved tl

tl

k

t2

= j Ldt + H(t2 - tl) tl t2

=> 6A = 6 j Ldt + H 6(t2 - tI) = [L6t - LPktlk6t + H 6tn~ 4

k

= [(H

+ L - LPkizk6t)6tj~~ = 0 asH = LPktlk k

L

k

t2

=> 6A = 6 j LPkizkdt = 0 tl

k

which is the principle of least action. In recent times it is more customary to refer to the integral in Hamilton's principle as the action. Thus the physical trajectory is the one for which the action is stationary. The end points are fixed in both the variational principles. In Hamilton's principle, the time (t2 - tl) is prescribed ( fixed as the body moves from one point to another ) in the configuration, while in the principle of the least action, there is no such restriction on the time (t2 - tl), but the total energy between the end points is prescribed. This is the basic difference between Hamilton's principle and principle of least action.

5.5.3

Second Form of Least Action

Let the co-ordinate transformation equations of a dynamical system do not contain the time explicitly, then L = T - V and H = T + V, so that,

L Pktlk = H

+L =

T

+ V + (T -

V) = 2T.

k

In this case the principle of least action takes another form t2

6 j 2Tdt = 0 => 6A = 0 tl

144 where A


=

t2

I2Tdt. Now we are to derive the parametric form of the principle. Now, tl t2

A

t2

J = .l =

V2TV2Tdt

J

=

y2(H -

tl

V)VL m(~:)2dt

tl

t2

V)VL m(:~ )2d)';

).

J

dql dq2

dqn

where 1(ql, q2, ... ,qn; qL ... ,q~.); function. Thus,

q~

J2(H -

= arbitrary parameter

h P2

=

1(ql' Q2,· .. , qn; d)' , d)' , ... , d)' )d)'

PI

81

d

=

~; k

= 1,2, ... n independent of t, is a

81

8qk - d)' (8q~) = 0; k = 1,2, ... ,n

which is the the required parametric form. If the system does not involve any external force, then T is conserved, and then t2

6.

J

dt = 0

tl

'* 6.(t2 -

tl) = O.


145

This conditions leads to Fermat's principle in geometrical optics, "a light ray travels between two points along such a path that time taken is the least." Ex 5.5.1. Brachistochrone Problem:

Brachistochrone Problem is concerned with finding a plane curve that joins two points such that the time required for a particle falling under gravity along it, from an upper point to the lower, is a minimum. Let x axis be horizontal and z axis vertically upwards. P and Q are initial and final positions of the particle respectively. If the velocity and height at P is (v, z) and the same at Q is (Vq, Zq), then the total energies at P and Q are respectively given by,

Since it is a conservative force field, by conservation of energy, we have Tp i.e., 1 2 2mv

Now, the velocity v

=

=

TQ,

1 2 + mgz = 2mVq + mgZq 2 =} v = V~ + 2g(Zq - z).

~: for an element of arc ds of PQ in time dt. Hence

p

t

= f[v; + 2g(zq - z)r 1/ 2ds Q Zo

=

viI + x'2 ,dx ---;======d x . fZ Jv~ + 2g(zq - z) Z; = dz'

as ds 2 = dx 2 + dz'2

Zo

= f L(x, z, x')dz;

, dx x =-. dz

Z

The equation has the form of the action with x playing the role of 'generalised coordinate', Z that of 'time' and the integrand that of the 'Lagrangian L', so L = vi ~ Z ). Since L(x, z, x') does not contain x explicitly, so x is ignorable. Also v q +2g

Z'J-

as L does not contain time t explicitly, Brachistochrone is the path of minimum time obtainable from (/::::,. is replaced by J as t ignorable), Zo

' d aL aL J f L(x, z, x )dz = 0 =} dz (ax') - ax = 0 z =}

aL ax' = constant

= c( say)

146

CHAPTER 5. HAMILTONIAN MECHANICS X, ~

VI + Xl\/V~ + 2g(zq l+x'2 X'2

~

x'2

=

1

2 2

2

1

2+2gZq

2

1!q

29

dx=j

j

~ x = a + b[0 2

[v~

x'2

+ 2g(zq - z)] c2[v~ + 2g(zq - z)]

where a = -c Vig~2gC Zq and b = integration since we have,

1

= 1 + - = - - : : - - - - - - - 0 - - : -2

c2[v~ 1-

=c

z)

sin 0]

•

+ 2g(zq -

_ b- Z - a+Z

_ -

z)]c

[dX]2 dz

Setting, Z = _[a;b]_ [at b] cosO facilitates

l+cosO[a+bsinO]dO I-cosO 2

+A =

4 1 2 [0 - sin 0] gc

+ A.

Thus we get the curve for minimum time, the Brachistochrone, in parametric form with x and z appearing as functions of 0 and this represents the equation of the cycloid.

5.5.4

Maupertui's Principle

Let we are interested only in the motion but not the trajectory, we can write a restricted form of variational principle. Let us take the situation in which the initial and final positions are fixed and a given initial time, however we allow the variation in final time. Hence

If further we are interested in real motion in which energy is conserved, then H = E and so b"A + EJi = O. Also, if the Lagrangian L does not contain time t explicitly, then, Pk = = Thus writing the action for real motion, we have for A,

:t gr·

t

A = j[LPkQ.k - H]dt· to

k

tl

= j LPkQ.kdt - E(t - to); to

as H = E

k tl

= Ao - E(t - to); Ao = j LPkQ.kdt to

b"A = b"Ao - Eb"t b"Ao = b"A + EJi = 0

~

~

k

147


The quantity Ao is called Hamilton's characteristic action or observed action. This equation satisfies least action principle for all paths for which energy is conserved. Let the co-ordinate transformation equations do not contain the time explicitly, then the KE of the dynamical system is a homogeneous function of velocities. Thus

2T = 2[E - V(q)] = LLajkq/lk J

'* dt =

k

2[E - V(q)]

=/

[LLajkdqjdqk]2[E - V(q)]. k

j

Thus the principle of least action takes the form c5Ao

[L L ajkdqjdqk]2[E - V(q)] =

= c5 /

o.

k

j

For a single particle, the principle of least action takes the form t2

~/

2Tdt

= ~/

m

d(3

= O.

1I

Further, for such a system, the Hamiltonian H = T + V(q), so that the special form (known as Jacobi's form )of the least action principle becomes

~

J

J2[H - V(q)] d(3

= O.

For one particle system, the principle of least action can be expressed in terms of arc length ~ f J2m[H - V]ds = O. Ex 5.5.2. Kepler's Problem: For the planetary problem, the KE T and PE V are respectively given by T = ~m[r2 + r2(j2] and V = -~. Since the system is conservative, the total energy in Kepler's problem is given by H

'* H

= T + V = !m[i·2 + r2(j2] - ~ = E 2

- V

r

= !m[r2 + r2(j2] = E +~. 2

r

148


The Jacobi's form of principle of least action gives ~ J y'2[H - V] df3 = 0, so that

~

J

2m(E +~) y'm 2(r 2d0 2 + dr 2)

or, ~

J

k 2m(E + -;:-)[1"2

or, ~

dr

+ (dO)2]dO = 0

J(

I) 1 dr f r, r dO = OJ r = dO·

+ ~)[r2 + r /2 ] do not contain the co-ordinate e explicitly, using same pattern as Lagrangian [2: g~ qk - L = H], we get, k qk

Now, f(r, r') so,

=0

=

J2m(E

af 1 - f ( r,r ') = lj l = constant ar,r r-----2m(E + klr) 12 2 m (E + -k)[ r 2 + r 12] -- l --';:;---;:-....:...r r2 + r /2 r 12 2mr 2 2 l2 =} r = -l2-[Er + kr - 2m]

or,

o

=}

J

J

o

~

r

dO = _l_ J2m

dr r[Er 2 + kr -

2

_I P/2 2m

We shall discuss it in the next chapter. Ex 5.5.3. A particle of unit mass is projected so that its total energy is h in a field of force whose potential energy is ¢( r) at a distance r from the origin. Deduce the differential equation of the path zs c2[r2 + (~~)2] = r 4 [h - ¢(r)], where c is a constant. SOLUTION: Given that, the force field have the potential energy V = ¢(r), r is the distance from the origin and the total energy is h, so by the energy equation, •

H

Now, the

~,ction

= T +V

=}

T = h - ¢( r).

A is given by,

J =] =J t1

A

= h

=

2Tdt

to

J Sl

=

y'2[h - ¢(r)]ds

So

SO

Sl

y'2[h - ¢(r)][l + r 20/2]dr

so

J Sl

=

f(r, O')dr( say),

So

149

5.6. SYMMETRY, CONSERVATION LAWS

where ()' = ~~ and f(r, ()I) = )2[h - d()'

=

constant

r2()1

df

= dO = 0 In

= y 2C( say)

=> V2[h - 2[h - C 2[r2 + (dr )2] = 1.4[h - H = constant

=> E

= T +V =H =

constant .

It is the law of conservation of energy. Thus homogeneity of flow of time leads to the law of conservation of energy.

Ex 5.6.1. Spherical Pendulum:

A spherical pendulum is essentially a simple pendulum which is free to move through the entire space about the point of suspension. The bob of the pendulum can swing on the surface of the sphere in any direction whose radius is equal to the length of the pendulum. It is obvious that spherical polar co-ordinates are most suitable to locate the position of the bob of the pendulum. Moreover, since 1 is constant. the generalised co-ordinates are 0 and ¢. As the mass Tn traces out a sphere of constant length l. the KE of the pendulum is given by

T

= !m[r2 + r 2iP + r2 sin2 O¢2] 2

= !ml 2[(P + sin2 0¢2]. 2

Potential energy V of the bob due to gravity is V = -mgl cos 0; relative to the horizontal plane z = O. Hence, the Lagrangian function L for the system is thus given by

Then, the Lagrange's equations corresponding to

e and ¢ are given by

.!!:.-(ml2iJ) - ml 2 sinOcosO¢2 -mglsinO dt ., '2 g => 0 = sin 0 cos O¢ + I sin 0 and

:t

(mZ2 sin 2 O¢)

= O.

=0

5.6. SYMMETRY, CONSERVATION LAWS

153

The second equation implies that the angular momentum in the vertical direction, z, i.e., direction is a constant of motion. Hence the corresponding momentum is conserved. Thus

= ml 2 sin2 O¢ = constant = . A ~ = ml 2 sin2 () . PcP

A(say)

Substituting the value of in terms of PcP, the one dimensional equation of motion in 0 turns out to be 2"

ml 0 = mgl sin 0 + '2

A2

2

P~ cos 0 2' 3

ml sm

2g

e

~ 0 = - m2l2cot () + TCOS(}

+B

where B is a constant. We can find the total energy of the pendulum. Since ~f the total energy E is given by

·8L E = "~ qjpj - L = ()-.

. .1

8(}

= 0,

·8L - L + -.

8

= ~m12e2 + ~ml2 sin 2 (}¢2 + mgl cos () 2 1

= '2ml

1

2 '2

e + '2 ml

= '2ml2e2 ~

2 1 1

2. 2

sm

A2

em 2l4 sin4 () + mgl cos ()

A2

+ '2 ml2 cosec2() + mgl cos () = T + Ve

dE . 2'" A2 2 . = ml (}() - --2 cosec (}cot() - mgl sin 00 dt ml

-

= o.

Hence it is a constant of motion. Further Ve = ~ ~ cosec20 + mgl cos 0 is known as the effective potential energy and it depends only on O. When the pendulum is restricted to move only in one plane, say = 0 plane, then the equation of motion becomes .. g () = I sinO which is the familiar equation occurring in the case of simple pendulum with the difference that () here is measured with respect to the vertically upward direction.

5.6.4

Noether's Theorem

In classical mechanics, Noether's theorem occupies an important position. This theorem states that " if a symmetry is found to exist in a dynamical problem then there is a corresponding constant of motion." Discussion : We see that symmetry properties of the Lagrangian or Hamiltonian


154

imply the existence of conserved quantities. Thu~, if the Lagrangian L does not explicitly a particular co-ordinate of displacement, then the corresponding canonical momentum is conserved. The absence of explicit dependence on the co-ordinate means the Lagrangian L is unaffected by the transformation that alters the value of the co-ordinate; it is said to be invariant, or symmetric under the given transformation. Similarly, invariance of Lagrangian under time displacement implies conservation of energy. The formal description of the connection between invariance or symmetry properties and conserved quantities is contained in Noether's theorem. This theorem enables us to obtain first integrals of systems of Euler's equations for furictionals admitting of some invariance properties.

5.7

Conclusion

The Hamiltonian formulation has several advantages over that of Lagrangian, as (i) Hamiltonian formulation provides deeper insight into the behavior of a dynamical system. (ii) Hamilton's formulation proves superior in the indirect integration ofthe equations of motion by means of suitable transformation.

(iii) The transformation used in Hamiltonian mechanics are known as canonical transformation, which is characterized by a single function, namely, the generating function. If a generating function can be found, any dynamical problem can be reduced to one of the differentiations and elimination.

5.8

Exercise

Ex 5.8.1. For the Lagrangian L

= !q2 - qq + q2,

find the Hamiltonian.

Ex 5.8.2. Show that the constant of motion associated with the infinitesimal transformation for the spatial displacement leads to the constant of motion of the linear momentum. Ex 5.8.3. Set up Hamilton's equations of motion for the Lagrangian L(q, q, t) = mlq2 sin 2wt + qqw sin 2wt + q2w2]/2. Ex 5.8.4. For the Lagrangian L = qt+q~+qlq2-qt-q~, find the conjugate momenta and obtain the Hamiltonian. Write down the Hamiltonian's canonical equations and solve them. Ex 5.8.5. Write down Hamilton's equations in spherical polar co-ordinates for the Lagrangian' L

. 2

= !mT7 + 7;

J.L constant.

5.8. EXERCISE

155

Ex 5.8.6. Consider the Lagrangian L = ~qlog((?) - )..q + qf(q); where f(q) an arbitrary function and)" > O. Write down the Lagrange's equation of motion. calculate the Hamiltonian of the system. Is total energy of a constant of motion. Does the Lagrangian transform covariantly under the transformation q -+ aq for a a real constant. GATE - 2001 Ex 5.8.7. For the Lagrangian L

=L

fk(qk)q~ - LVk(qk), obtain the Lagrangian

k

k

and Hamiltonian equation of motion. Solve the Lagrange's equation. Ex 5.8.8. Define phase space. Starting with a Lagrangian L = L(q, q, t), obtain the Hamilton's canonical equations of motion aaH = qk and aaH = -'Pk. Pk qk' Ex 5.8.9. State Hamilton's principle. Deduce it from D'Alemberts principle. Ex 5.8.10. Solve the simple pendulum problem by using Hamilton's canonical equations of motion. Ex 5.8.11. Write Hamilton's canonical equations of motion for the compound pendulum. Ex 5.8.12. Using Hamilton's canonical equations find the equations of motion of a p;rojectile in space. Ex 5.8.13. A particle moves under the influence of gravity on the frictionless inner surface of a cone x 2 + y2 = c2 z2. Obtain the equations of motion. Ex 5.8.14. A heavy bead of mass rn is freely movable on a smooth circular wire pf radius a which is made to rotate about a vertical diameter with a spin w. Prove that, the action will be 62

A

=

rna 2

J[ rna2~ + 2ga cosO +

w 2 sin 2

Ol~dO

61

where H is the Hamiltonian and with downward vertical.

e is the angle made by the radius through the bead

Ex 5.8.15. Show that the shortest distance between two points on the surface of the sphere is a great circle. Ex 5.8.16. Show that the area of the surface of revolution of a curve y = y(x) is X2

21l"

J yy'l + yl2dx.

Hence show that for this to be minimum, the curve must be a

Xl

catenary. Ex 5.8.17. A particle of unit mass moves along x axis under a constant force .f starting from the rest at the origin at time t = O. If T and V are KE and P E of to

the particle, respectively. Calculate J(T - V)dt also calculate for the varied path o in which the position of the particle is given by x = !ft2 + c.ft(t - to), where c is a constant. Show that the result is in agreement with Hamilton's principle.


Chapter 6

Transformation Theory We have seen that the integration of a dynamical system which is solvable by quadratures can generally be effected by transforming it into another dynamical system with fewer degrees of freedom. The way of canonical transformation to obtain solution of a dynamical problem is to transform old set of co-ordinates into new set of co-ordinates that are cyclic. In this way, the new equations of motion can be integrated much easily to give a solution. In this cha~ter, we shall discuss several bracket notations and canonical transformation which underlies this procedure, and, indeed, underlies the solution of all dynamical systems. In the next chapter, we will consider Hamilton-Jacibi theory of equation of motion which is another aspect of canonical transformation theory. At first, we introduce some Bracket not.at.ions to deal with the transformation theory.

6.1

Poisson Brackets

Given any two dynamical variables u(q,p, t) and v(q,p, t), of any dynamical system, the Poisson brackets of two functions u, v with respect to the canonical canonical variables (q, p) is defined as

(6.1) Here (q,p) refers to the set (qk,Pk, t) of independent variables pertaining to a holonomic system with the number of degrees of freedom 11, with respect to which poisson bracket is evaluated. The poisson brackets (PB) are found to be very useful tool in quantum mechanics and field theory. In this notation the equations of motion can be written in a symmetrical form. PBs can be a useful tool for finding out some of the hidden constants of motion. Ex 6.1.1. If ql and q2 are generalised co-ordinates and Pl and P2 are corresponding generalised moments. Find [X, YJ where X = q~+q~ andY = 2Pl+P2.GATE -1991

157

CHAPTER 6. TRANSFORMATION THEORY

158

SOLUTION: Let X bracket, we get [X,

=

q~

+ q~

and Y

Yl = (ax DY

_

=

2Pl

Using the definition of Poisson

ax DY ) + (DX

Dql DPI BPI Dql X 2 - 0) + (2q2

= (2ql 6.1.1

+ P2.

X

DY _ DX DY ) Dq2 DP2 DP2 Dq2 1 - 0) = 4ql + 2q2.

Properties of PB

In this section, we shall derive some of the properties of poisson brackets. Let 11.(q,p,t),v(q,p,t) and w(q,p,t) are assumed to be any three dynamical variables pertaining to a holonomic system with degrees of freedom n, whose generalised coordinates and momenta are denoted by (q, p). The algebraic properties of PB are therefore of considerable interest. Property 6.1.1. The PB of any two dynamical variables is non-commutative, i.e., [u, vlq,p

=

-[v, ulq,p,

PROOF: Using the definition of PB, we have [u, vlq,p

= L( DDu k

=-

qk

DDv - DDu DDv) Pk Pk qk

"" uu Dv Du Dv L:(DPk Dqk - Dqk DPk)

= -[v,ulq,p'

Hence the PBs are anti-commutative. In particular, when, v = u, then [u, ulq,p = -[u, ul'l,p = 0, i.e., the PB of a function with itself is identically zero. Property 6.1.2. If c is a constant independent of (q,p, t) then [cu, vlq,p

= c[u, vlq,p =

[u, cVlq,p'

PROOF: Using the definition of PB, we have

Similarly, [u, CVlq,p

= c[u, vlq,p'

This is called homogeneous property of PBs.

Property 6.1.3. The PBs also satisfy the distributive property with respect to addition, i.e., [u + v, wlq,p = [u, wlq,p + [v, wlq,p'

6.1. POISSON BRACKETS

159

PROOF: Using the definition of PB,

[u -I- v, wlq,p

a -I- v)."!l ow a ow = "" L) "!l(u - "!l(u -I- v)."!l} k

uqk

UPk

UPk

·uqk

=L{ou.oW _ ou.ow}-I-L{ov.OW _ ov.ow} k oqk OPk OPk oqk k oqk OPk OPk oqk = [u, wlq,p -I- [v, wlq,p. Also,[u-l-v-l-w, xlq,p = [u, xlq,p-l-[v, xlq,p-l-[w, xlq,p. In general, when a, b are constants then [au -I- bv, wlq,p = a[u, wlq,p -I- b[v, wlq,p, which is the linear property of PBs. Ex 6.1.2. If ql and q2 are generalised co-ordinates and PI and P2 are corresponding generalised moments. Find [X, Yl where X = qr -I- q~ and Y = 2PI -I- P2.

SOLUTION: Using the definition of Poisson bracket, we get

[X, Yl = [q~ -I- q~, 2PI -I- P2l = [q~,2PIl -I- [q~,2pIl -I- [q~,p2l -I- [q~,p2l = 2qI -I- 2qI -I- q2 -I- q2 = 4qI -I- 2q2.

Property 6.1.4. [uv, wlq,p = [u, wlq,pv -I- u[v, wlq,p, for three dynamical variables u,v and w. PROOF: For dynamical variables, u, v, w, we have

This is the elementary property of differentiation of PB. Also,

[uvw, xlq,p = [yw, xlq,p; y = uv = y[w, xjq,p -I- [y, xjq,pW = uv[w, xlq,p -I- [uv, xlq,pw.

Property 6.1.5. If c is a constant, then poisson bracket of c with any dynamical variable u is given by [e, ul = O. PROOF: Since e is a constant, we have aac qk dynamical variable is given by

[e, ulq,p

"" oc au

= 0 = aacPk .

= L....,,["!l"!l k uqk UPk

oc au

"!l"!ll

UPk uqk

Now, the PB of e with any

= o.

160


Property 6.1.6. Let u, v, w be three dynamical quantities, having continuous second order derivatives, then the double PB relation holds

[u, [v, wjq,pjq,p + lv, [w, ujq,pjq,p + [w, [u, vjq,pjq,p = 0 i.e., the sum of the cyclic permutation of the double PB of the three quantities u, v, w is zero. PROOF: There is no simple algebra to prove the identity of PBs. We shall prove this by introducing some new variables. The definition of PB can be written as ~ au a au a [u, vjq,p = 6 (aq ap. - ap aqk)v k=l k k k 2n av

=Du v =

where

~i

Laia~i

i=l

represents a set of (qk,Pk) variables and Du is an operator defined by

2n

Du

= L: 0i a~.· .=1

2n

Similarly, we can define Dv

= ?: f3j J=l

at. Now

where we have used variables qi and Pi and the coefficients of as constants Aj., Bj respectively. First let w = Pj, then

g, ~ are expressed

[u[v, pjll - [v, [u, pjll = Aj av au => Aj = [u, -a j - [v, -a j qj

qj

av au a =[u'-a l+[-a ,v1=-a [u,v1· qj

Similarly, taking w

= qj

[u, [v, W1q,p1q,p

we have B j

qj

qj

= -k[u, v1.

+ [v, [w, u1q,p1q,p =

Hence

~awa

awa 6( -a. -a. [u, v1- -a. a-:-[u, v]) i

P. q]

[u, vll => [u, [v, W1q,p1q,p + [v, [w, u1q,p1q,p + [w, [u, v1q,p1q,p =

q] p]

= -[w,

o.

6.1. POISSON BRACKETS

161

This is an important identity known as Jacobi '8 identity satisfied by Poisson brackets. This identity says that the bracket product is non associative. This property of PBs is very important in defining the nature of the PBs. Let VI, V2, .. . , Vn be a set of dynamical quantities (all functions of q, p, t) and let f(vl, V2, ... , vn ) be differentiable functions of VI, V2, ... , Vn . Then

of [u, f(vl, V2,···, v1£ )] = ~[u, VI] UVI

of

of

+ ~[u, V2] + ... + n-[u, v n ]. UV2 UV

n Since the PB have the properties anti-commutative, Linearity and Jacobi's identity, the PB defines a particular type of non-associative algebra, called Lie algebra. Property 6.1.7. If [u, v]q,p be the poisson bracket of u and derivative of any PB relation is 9t[u, v]q,p = [~~]q,p + [u, ~~]q,p. PROOF: Using the the definition of PB, we have,

a ot [u, v]q,p

V

then the partial

~oa,u

=~

ov ou ov ot (oqk OPk - OPk Oqk)

= I)~(ou) ov + ou ~(OV) _ ~(ou) ov _ OU ~(01J)] k oqk at OPk oqk OPk ot OPk ot oqk OPk oqk ot =

L[~(ou) ov _ ~(ou)~] k oqk ot OPk

6.1.2

OPk

at oqk

+ I) ou ~(OV) _ ~(ou)~ k oqk OPk ot

OPk ot oqk

Basic PBs

The PBs constructed out of the canonical co-ordinates as [qi, qj] and momenta [Pi, pj] themselves are called elementary PBs. These are called fundamental PBs as with the help of these, one can calculate all PBs of daily use. Now by definition of PB, we have

_ _] _ ~(Oqi oqj _ oqi Oqj) [q~, q] q,p - ~ 0 0 0 0 . k qk Pk Pk qk Since q~( or qj) is not a function of Pk, we have

Oqi UPk

~

oqj UPk

= ~ = 0 =}

[qi, qj]q,p

= 0 = [Pi,Pj]q,p'

~(Oqi Opj

Also, [qi,Pj ]q,p = ~ k

oqi OPj) 8 8- 8 8 qk Pk Pk qk

= L

Oqi OPj; as oqi = 0 = oq] k oqk OPk OPk OPk

=

L dikdjk = dij. k

Thus for the elementary basic or fundamental PBs we have,

[qi, qj]q,p

= 0 = [Pi,Pj]q,p

and [qi,Pj]q,p

= dij = -[Pj, qi]q,p.

(6.2)

162


The invariance of the fundamental PBs is always equivalent to the symplectic condition for the canonical transformation. Result 6.1.1. As qi = qi(t) and Pi = Pi(t) are explicit functions of the time t, it is possible to invert these relations. Thus we can write the dynamical variable t as t = t(CJi, Pi) and

6.2

Lagrange Bracket

For any two independent dynamical variables u(q,p, t) and v(q,p, t) pertaining to a dynamical system with the number of degrees of freedom = n, the Lagrange's bracket of (u, v) with respect to the basis (q, p) is defined in the analogous way of Poisson bracket as

OPk _ OPk oqk} (u, v )Q,l1 = ~{Oqk ~ OU ov ou ov .

(6.3)

k=l Here (q, p) refers to the set (qk, Pk, t) of independent variables pertaining to a holonomic system with the number of degrees of freedom n with respect to which Lagrange bracket is evaluated. Property 6.2.1. Lagrange bracket is antisymmetric, i.e., (u, v)q,p = -(v, u)q,p. PROOF: Using the definition of Lagrange bracket we get

(u, v)

q,p

= ~ {a qk apk _ apk aqk } ~

k=l

au av

au av

= _ I)aPk aqk _ aqk aPk} k=l au av

au ov

= -(v,u)q,p" Property 6.2.2. The fundamental brackets have the same value in both the bracket systems. PROOF: Here we are to show that (qi, qj) = 0; (Pi,Pj) = 0; (qi,Pj) = &ij. Since the co-ordinates (q, p) are independent in nature, the relations hold from the general expression of Lagrange bracket. Since q's and P's are independent, we have

Using the definition of Lagrange bracket, we get

'(qi,qj)= t{a qk apk _ aqk aPk }. k=l aq.i aqj aqj aqi

163

6.2. LAGRANGE BRACKET Similarly, (pi, pj)

= O. Further (% pj) =

t

=

t

{Oqk OPk _ OPk Oqk} k=l Oqi OPj OPj Oqi

k=l

~qk ~Pk = qt PJ

t

8ki8kj

= 8ij .

k=l

Property 6.2.3. Lagrange and PBs stand in some kind of inverse relationship to each other, but the precise form of the relation is complicated. When the coordinates are not canonical, then the receiprocal character of the two brackets manifests itself the relation 2n (6.4) L[U/,Ui](U/,Uj) = lSij, 1=1 where [Ul, Ui] and (u/, Uj) are the PB and LB respectively. PROOF: Let us suppose that there are 2n independent functions U1, U2, ... , 'U2n of the variables (qi,Pi); i = 1,2, ... ,n, so that ql,q2, ... ,qn;Pl,P2'''',Pn may be regarded as functions of Ul, U2, ... , U2n. From definitions of PB and LB we have

I)t(

I)Ul,Ui](U/,Uj) = OUI OUi _ OUI OUi) f)Oqk OPk _ oqk OPk 1=1 l=l m=l oqm 0Pm 0Pm oqm k=l OUI OUj OUj OUI

=

-_

n

f

{ t (OUI OUi _ OUl OUi) t(Oqk OPk _ oqk OPk)} /=1 m=l oqm 0Pm 0Pm oqm k=l OUI OUj OUj OUl ~

~(OUi 0Pm ~

1=1 0Pm OUj

OUi Oqm) _ + oqm OUj

OUi ( ) _ OUi _ ~ .. qm, Pm - UtJ' OUj OUj

The properties of the LB and PB parallel to each other in many aspects but LBs do not obey Jacobi's identity. So LBs do not obey the 'product' operation in the Lie algebra.

6.2.1

Angular Momentum and PB relations

The canonically conjugate momentum corresponding to any angle variable represents a component of angular momentum. Nevertheless, the total angular momentum of a system is best defined in known of cartesian co-ordinates and momenta of the individual particles. The angular momentum L can be expressed in terms of linear momentum P and co-ordinate q as ~

L

= q x P;

i.e., Li = f.ijk qj Pk

where the alternating tensor f.ijk is given by f.ijk

=

+1 when i,j,k are in even permutation ofl,2,3 0 when any two of the indices i,j, k are equal { -1 when 'i, j, k are in odd permutation ofl, 2, 3

(6.5)


164

Property 6.2.4. Let us consider a one-particle system with the ith cartesian component of its total angular momentum vector is Li = €ijkXjPk. Now the PB of Li with x J is [Li' Xj]

= [€ikIXkP/, x J ] = €ikl[XkPI, Xj] = -€ikI X k c5jl - €ikjXk = €ijkXk

Property 6.2.5. For the angular momentum [L'i, Lj]

Li

and

Lj,

we have,

= [€inkXnPk, €jlmXIPm] = €inkEjlm[XnPk, XIPm] = EnkiEjlm{XnPk[X n , Pm]

+ XnPm[Pk, xd}

= EnkiEjlm {c5nm X lPk - c5kl X n Pm}

= EnkiEjlnXlPk - EnkiEjknXnPm = EnmiEjlnXlPk - ElkiEjkmXlPm = (EnmiE.1ln - EkiIEkm.i)XlPm = (c5mj c5il - c5ml c5ij - c5im c5l j + c5ij c5lm )XIPm

= (c5jm c5il Similarly, [Lj, nents, we get

Lkl

=

Li;

[Lk' Lil

[L~:, Lyl

= Lz;

c5im c5jl )XlPm

=

Lj.

= EkijEklmXlPm = EijkLk.

When expressed in terms of x, y, z compo-

= Ly = [Lx, Lxl = [Ly, Ly] = [Lz, L z ] = O. [Ly, Lzl

= Lx;

[Lz, Lxl

Thus if we have any two components of L as constants of motion, then the third component and hence L as a whole must be constant in motion. So, if any two components of the angular momentum are constant, the total angular momentum is conserved. If a and b be any two vectors then we have the following PB relation [a.L, b.Ll

= (a x b).L.

Using this result we can generalised the above results. Thus if a and b represent any two unit vectors along which the components of L are separately conserved, the component of L perpendicular to the plane formed by a and b is also conserved. Similarly, we get the relation [L2, Lil = 0, for i = 1,2,3. Property 6.2.6. We know, [Pi,P.il = 0, but [Li' L j ] = EijkLk =F o. Hence if one component of the angular momentum along a fixed direction is taken as the canonical momentum, the two perpendicular components cannot be simultaneously the canonical momenta. But [Li' L j ] =F 0 shows that the scalar magnitude of the angular momentum and anyone of its components can be simultaneously canonical.

6.3. POINT TRANSFORMATION

165

Property 6.2.7. We know, the Hamilton's canonical equations can be written as [p, Hjq,p = P and [q, Hjq,p = q. Now the PB of Li with H is

[Li' Hlq,p = [EtJkqjPk, Hl = Etjk[qjPk, Hl = Eijkqj{ qj[Pk, Hl + Pk[H, qJl}

= Eijk{ qjPk - pk F1{q, Q) = -2- cotQ = -2- q cotQ => Q = tan-1{../km q ). p

Since Fl does not contain time t explicitly, the Hamiltonian is unaffected by the transformation. Hence, the new momenta is given by

P

aFt

mwq2

2

= - aQ = -2-cosec Q =

1

rm

2

p2

2V k{kq + m)

~ ../2mwP.

=> q = J2P sinQ => f{P) mw

Therefore, the Hamiltonian H in terms of the new variables (Q, P) becomes

H

1 = _[p2 + m 2w2q2]

2m 1 2P = -[mwq2.mwcot2Q+-sin 2 Q] 2m mw

= P[wcos 2 Q +

+

mw

sin2 Q]

= wP =

~p,

V~

which is quite simple form. Since the Hamiltonian H is cyclic in Q, hence, the conjugate momentum P is a constant, i.e., P

H E = - = - = a{

constant,) say w w where E is the constant energy. The equation of motion for co-ordinate Q reduces to the simple form

.

Q

aH = -aP = w => Q = wt + 8

where q. = -m Dqk

eE

=> q = -P t + qo and P = eEt + Po; qo = q( t = 0), Po = p( t = 0) m t

=> q = -[eEt + pol + qo and P = eEt + Po m

eE

2

Po

=> q = - t + - t + qo and p = eEt + Po. m

m

For the free-particle Hamiltonian for mass m, the transformed Hamiltonian K(Q, P) assumes the form K(Q, P) = [>2. The new co-ordinate and momenta (Q, P) must m,


182

be satisfied the Hamilton's canonical equations of motion

. Qk .

8K

= 8Pk

. and Pk

oK

= - 8Q k

=?

P . Q = - and P = 0

=?

p Q = - t + Qo and P = Po·

'm

m.

Eliminating t, between the equations, we get cErn

q=~[p(Q-Qo)]

or, q = eEm[

Q - Qo

P

2

]

2

[IT"

po ,m

+ mlp(Q-Qo)]+qo;p=eE p(Q-Qo)]+Po

+ Q + (qO

- Qo)];p

m

= eEl P (Q - Qo)] + Po·

Thus the required canonical transformation is given by (qO Q2

q = eEm p2

6.7

= Qo = 0; Po = Po = P)

Q

+ Q;p = meE P'

Condition for Canonicality

Let us take time as independent parameter. The conditions of canonicality of a given phase space transformation are seem to be evaluated at constant t, that indicates the purely geometrical nature of the canonical transformations that take place in terms of the phase space co-ordinates alone. We shall now obtain the explicit analytical expression for conditions for which a transformation to be canonical. The condition for canonicality must be such that as to preserve the form of Hamilton's canonical equations of motion.

6.7.1

Exact Differential Form

Let we are to show that perfect differential as a condition for a transformation to be canonical. Let the generating function F which transforms variables (q, p) -+ (Q, P), when the time is held fixed, then

For a transformation to be canonical, the equation (6.13) must be satisfied. Thus """ ~Pkqk - H(q, P, t) k

k

"". =? """ ~Pkqk - " ~PkQk k

. - K(Q, P, t) + dt dF = """ ~ PkQk dF as K = H = dt

k

=? I:[PkdqJ,. - PkdQk] = dF k

6.7. CONDITION FOR CANONICALITY

183

where dqk and dQk are the differential changes in the old and new co-ordinates in real time dt. Thus if the transformation (qk, Pk) - t (Qk, Pk) be a contact transformation, n

then the expression

L: [PrdQT -

Prdqr 1 would be an exact differential and this result

1'=1

can be obtained by using any generating function. Since the transformations in phase space between two sets qk's and Qk'S are essentially geometrical, we can look upon dqk and dQh: as the elements of the static geometrical curves rather than evolving paths, so far as the testing of the canonical condition is concerned. Ex 6.7.1. Calculate pdq - PdQ for the transformation Q = log(1 + .;qcosp); P = 2(1 + .;q cos p).;q sin p. Show that zt is canonical and find the generating function. SOLUTION: The given transformation Q = log(I+.;qcosp); P = 2(1+.;qcosp).;qsinp is canonical if the expression pdq - PdQ is an exact differential. Now, pdq - PdQ

cosp - 2qsinpdp

.

= pdp - 2(1 + y'qcosp)y'qsmp x 2 yq in(1 in ) + yqcosp = pdq + 2q sin 2 pdp - cos p sin pdq

= d(pq)

1 - d[2"(sin2p)q] = a perfect differential.

Since the expression pdq - PdQ is an exact differential, the given transformation is canonical. Now the generating function is given by Q sin 2p -1 e - 1 F1(q, Q) = pq - -2-. q = qcos /q-

F:.,,(p, (.2)

-

(e

Q

./

- l)y q - (e Q -1)2

sin2p Fl - pq = --2-· Q

=

= - sinpcosp

(e Q - 1)2 2 = -tanp(eQ _1)2. cos p

But neither q nor p can be expressed explicitly in P, as a result other two forms of generating function may not be written. However, as

~~ a at> a = IJI = ap

1, the

8q 8p

transformation are invertible, and 4(e Q q

=

-

1)2(e Q + 1)2 + p2 4(e Q + 1)2 ; cosp

Ex 6.7.2. Show that the transformation Q ical. Find generating function.

=

=

2(e Q - 1)(eQ + 1) Jp2 + 4(e Q - 1)2(e Q + 1)2

y'2(je t cos p; P

=

y'2(je- t sin p is canon-

SOLUTION: The given transformation Q = y'2(jc t cosp; P = y'2(jc- t sinp is canonical if the expression PdQ - pdq is an exact differential. Now, PdQ - pdq

= y2qc- t sinp[(2q)-1/2c t cos pdq = sin p[cos pdq - 2qsinpdp] - pdq =

81.

~[-qsm2p - pq]dq

uq 2

y2qc t sin pdp] - pdq

81 + ~[-qsin2p up 2

pq]dp

= dF.


184

Since it is an exact differential, the given transformation is canonical and the generating function is F ( q, p) = ~ q sin 2p - pq. From the transformation equations Q = J2qe i cos p; P = y'2ije- t sinp, we have Q . cos P = rrr;:. t => S111 p y2qe

=

Rf,,2 1-

--2t·

2qe

-

So, P(q, Q. t) = qsinpcosp - pq - Q - ei

~J

V2"

1

2t

Q2 _ -1 .j2qe 2qe2i qtan Q

-

Q'

.

Ex 6.7.3. Show that the transformation Q = logei~p)j P = qcotp is canonical. Find generating function. SOLUTION: The given transformation Q = 10gCi~P)j P expression pdq - PdQ is an exact differential. Now, pdq - PdQ

= pdq -

8

qcotp is canonical if the

oQ oQ P[ oq dq + op dp]

dq = pdq - qcotp[ - q = 8q [qp

=

+ cotpdpj 8

+ qcotpjdq + 8p [qp + qcotpjdp = dF.

Since it is an exact differential, the given transformation is canonical and the generating function is F(q,p) = qp + qc;otp. From the transformation equations, we have

6.7.2

Poisson Bracket Form

PBs can be used to. test whether a given transformation is canonical. Now, we shall give another form to the conditions that a transformation (q,p) --+ (Q, P) may be a contact transformation. The fundamental PB as [Qi,n]q,p = ,",(8Q 2 8P) _ 8Qi BPj 8qk OPk 8Pk 8qk

7

= 2:(8 Q i k iJQi = &Qj Similarly, [Qi, Qjjq,p

=0=

8qk 8qk 8Qj

+ 8Qi

= 6i,j =

[QI,' PjjQ,p.

)

8 Pk ) 8Pk 8Qj

[Qi, QjjQ,P and [Pl , Pj1q,p

=0=

[Pi, PjjQ,p.


185

The PBs such as [Qi, P j ] = Jij , [P,;, P j ] = [Q~, Ql] = 0 are elementary useful for testing the canonically of any given phase space transformation. Ex 6.7.4. Let (q,p) be canonical variables and let Q = Q(q,p), P = P(q,p) be transformation equation from (q,p) to (Q, P). Show that the transformation will be canonical if [Q, P]q,p = 1. Hence show that Q = q cos fJ - !u;; sin fJ; P = mwq sin fJ + p cos fJ is canonical. SOLUTION: Let the transformation Q = Q(q, p), P = P(q,p) be canonical. Then the expression pdq - Pdq is an exact differential. Also,

pdq - Pdq = pdq -

P[~Q dq + 8 Q dp] 8p

oq

= [p _ p8 Q ]dq _ p8Q dp.

8q

8p

The condition for perfect differential is

~[p _ p8 Q ] = ~[_p8Q]

8p 8q 8q 8p =? 1- 8P 8Q _ p~(8Q) = _ 8P 8Q _ p~(8Q) 8p 8q 8p 8q 8q 8p 8q 8p 8P8Q 8P8Q =? --8 -8 + -8 -8 = 1 =? [Q,P]qp = 1 p q q p , which is the required condition for the transformation to be canonical. Conversely, let [Q, P]q,p = 1 i.e., Poisson bracket of Q and P is given as 1. But the Jacobean of transformation .J from (q,p) to (Q, P), where Q = Q(q,p), P = P(q,p) is defined as J

=

-8(Q P)

'

8(q,p)

£9. £9.

= jJp fjp 8q

ap

= 8P8Q _ 8P8Q

8q 8p

Dp 8q

= [Q,P]

q,p

= 1.

Thus the given transformation is canonical. Now

[Q, P]q-p

8P 8Q

8P 8Q

sinO.

= -8 -8 - -8 -8 = cosf) + -.mwsmf) = 1. qp pq mw

Hence the given transformation is canonical. Ex 6.7.5. Under what condition will the transformation P would be canonical, where a, b, c, d are constants?

= ap + bq, Q = cp + dq

SOLUTION: It would be canonical if [Q, P] = 1, when it is given that [q,p] :::;: 1. To see this, we calculate the PB of [Q, Pl. Then,

[Q, P]

= [cp + dq; ap + bq] = cb[p, q] + da[q, p] = ad -

be.

For [Q, P] to be unity we must have ad - bc = 1.

186

6.7.3


Lagrange Bracket Form

Here we shall give another form to the conditions that the transformation (q, p) -+ (Q, P) may be contact transformation. Let the transformation Qi = Qi(q,P, t); Pi = p/,(q,p, t); i = 1,2, ... , n be canonical. Then Pic5qi - Pi 8Q, = an perfect differential.

Since the old variables (qi,Pi) may be regarded as functions of new variables Qi, P~; the necessary and sufficient condition for a transformation to be canonical is that

for a fixed value of time. The conditions for perfect differential gives

However, Qi and Pi are independent variables, so that these conditions reduces to

6.7.4

Bilinear Covariant Form

Let (qI, q2, ... ,qn) be any set of n variables and consider a differential form ( known as Pfaff's expression) in the variables (qI, q2, ... ,qn) as

where Q1, Q2, . .. , Qn are the functions of (qI, q2, ... ,qn)' Let us write

where 8 is the symbol of an independent set of increments. Therefore 6e 1 - (If}2

= 6[Ql d ql + Q2 dq2 + ... + Qndqnl

- d[Q1 8q1 + Q2 c5q2 + ... + Qnc5qnl = c5Ql d ql + ... + c5Qn d qn + Ql c5(dql) + ... + Qn c5(dqll) - dQl c5ql - ... - dQn c5qn - Ql d(c5qn) - ... - Qnd(c5qn).


187

S"ince the variations d and .; Px = -mw Y sin>. + P x cos A, Py = -mwX sin>. + ~Qx cos >., k = mw 2 , we get

!f!

K(Px, Py, X, Y) =

2~ [( -mwY sin A + PTcos A)2 + (-mwX sin>. + P.T cos A)2]

+ ~2" [(X cos A + 2

= ntw (X2

2

6.8

y P sin A)2

+ (y,cos A +

'InW

x P sin >.)2] nLW

p2) + y2) + _1_(p2 2m x + y.

Properties of CT

In this section we shall discuss some properties of canonical transformations. Property 6.8.1. From the transformation equations, in generating functions, diffcnmtiatillg Ollce more the first two relatiolls of each of sets of t.he t.ransformation equations, we get

where 'i; j = 1,2, ... , n. These symmetrical formed equations are to be satisfied by all univalent canonical transformations, which are totally independent of the generating functions or the Hamiltonian H. These sets of equations together are known as the direct conditions for a canonical transformations. Thus given a set of canonical transformations either in the form of point transformations or in the form of transformation of configuration space, they must satisfy all the above sets of equations. Property 6.8.2. Any finite volume in the phase space before and after the transformation remains canonically invariant. This is known as Liouville's Theorem.

190


Let ql, q2,···, qn and PI,P2,'" ,Pn be a set of generalised co-ordinates and their corresponding conjugate momenta respectively to characterize a given state of a physical system. Then the corresponding phase-space (q, p) is 2n-dimensional space with ql, q2, ... ,qn and PI, P2, ... ,Pn denoting the 2n-number axes system. The elementary volume elements dVI , dV2 in this phase space (q,p) and (Q, P) are respectively given by PROOF:

dVI = dql.dq2 ... dqn.dPI.dp2 ... dpn dV2 = dQI·dQ2· .. dQn.dPl·dP2···dPn. We are to show that, under canonical transformation this elementary volume is an invariant. Let us consider an univalent canonical transformations carries the set (q,p) ~ (Q,P) as Qk = Qk(q,P,t); Pk = Pdq,p,t);k = 1,2, ... ,n. Then the Jacobian of transformation is defined as

.7= 8(Q,P) = 8(q,p)

Hence the Jacobian determinant of the univalent canonical transformations is unity. Now the change in the volume element in two different bases (q,p), (Q, P) is related with the Jacobian as

dql.dq2 .. ·dqn. dPI.dP2 .. ·dpn

=

IJldQI.dQ2 ... dQn.dPI.dP2 ... dPn

= dQI.dQ2 ... dQn.dPl.dP2 ... dPn

which shows that the phase space volume is invariant through canonical transformations. This is generally valid when the phase space is a constructed in cartesian fashion. Ex 6.8.1. Show that the cartesian to polar co-ordinate transformation in phase space given by (Xl, X2, X3, PI, P2, P3) ~ (1', e, , PI, P2, P3) is not canonical. SOLUTION:

The transformation equations from cartesian (Xl, X2, X3) to polar (T, e, or 00 o ~~~ or 00 o sin () cos ¢ r cos () cos ¢ -r sin () sin ¢ sin () sin ¢ r cos () sin ¢ r sin () cos ¢ cos () -r sin () 0 ~~~

= r2 sin () i=

1.

Since the Jacobian of the transformation is not unity so the given transformation is not canonical. Now, let us take a transformation (x!, X2, X3,P!'P2,P3) (r, (), ¢, Pr, Po, PP=-o' =mx=>x=-. x m Thus, the Hamiltonian H of the system is given by, p2 H=px-L=-. 2m

Therefore,

Thus, H is a constant of the motion. Also, given thp.t, ,f. =: x - ~, so, ,

dF of dt = at + [F,H] = of + (oF.oH

ot p

ox op

P

t

_ of.oH) op ox

= -+ 1.+ -.0 = O. m m m

•


200

Thus, F is a constant of the motion. Thus [H, F] is a constant of the motion. Now,

uH uF aH uF [H,F]=-a '-a --a '-a x P P x

= o.(-~) - ~.1 = -~. m

m

Since Hand F are constant of motion, so is

m

0:.

Now,

of = _~ = [H,F].

at

Tn

Hence the result is verified. Ex 6.9.4. Write Hamilton's equation in matrix form and obtain the sympletic condition for a canonical transformation. SOLUTION: Hamilton's canonical equations are prescribed in a phase space. Let "7 be the column matrix of order 271, x 1, defined by "7 = (ql, q2, ... , qn;PI,P2,'" ,Pn)T. Let J be the antisymmetric matrix of order 271, x 271, given by,

=

J

(~~)

where Q is the null matrix of order 71, and I is the identity matrix. Hamilton's canonical equation can be written as

.

"7=J

aH

ary ·

Therefore,

(6.24)

Let us consider the canonical transformation, where time does not appear and is given by Pi = Pi(qbPk), Qi = Qi(qk,Pk), i.e., a transformation from 2n variables (ql,q2, ... ,qn;PI,P2, ... ,Pn) and (QI,Q2, ... ,Qn;PI ,P2, ... ,Pn) = ((say). For the restricted canonical transformation, ( = (("7) and clearly,

So, in matrix notation, we have,

.

aH

(=M,,=MJ-

ary

where NI is a 271, x 271, matrix, whose (i, j)th element is given by ~. This matrix 111 is called Jacobean matrix of transformation. Now, by the inverse transformation H can be considered as a function of ( and in fact

6.10. INFINITESIMAL CONTACT TRANSFORMATION

201

Now, for restricted canonical transformation, the Hamiltonian H is same as the new Hamiltonian K and therefore K = H. Since the Hamilton's canonical equations of motion for the ( system is

(= J~~. For a restricted canonical transformation H MJM

T

=K

and hence

= J.

(6.25)

This is known as sympletic condition for the canonical transformation and the matrix AI satisfying the condition is called sympletic matrix.

6.10

Infinitesimal Contact 'fransformation

If a canonical transformation is introduced very near the identity transformation, then it is known as infinitesimal canonical transformation or infinitesimal contact transformation. As in case of infinitesimal rotations such a transformation is one in which the new co-ordinates differ from the old by i.lfinitesimal amounts. Let the transformation equation, when we retain term correct to first order terms, can be written as (6.26) where 6qk and 6Pk are the infinitesimal changes in the co-ordinates and momenta. Thus to find a transformation mean the identity transformation, the generating function F2 (q, P) will also differ only by an infinitesimal amount and is given by

F2(q, P) = qkPk

+ EG(q,p, t)

where E is the some infinitesimal parameter of the transformation, indcpcndent of qk, Pk and G is arbitrary differentiable function of its (2n + 1) arguments. The transformation equations become are

Pk

aG

aG

= Pk + E a - ; Qk = qk + Eap qk

=> 6Pk = Pk - Pk =

k

aG -E aqk ;

6qk

aG

= Qk - qk = E aPk .

These equations gives the infinitesimal changes in the conjugate variables which are generated by arbitrary function G. Since the terms under consideration is of the first order in E, so Pk - Pk is infinitesimal. Thus, we can replace Pk by Pk and G(q, P) ~ G(q,p), which gives (6.27) The diffen~nce involved ill this replacemcnt is of second order of smallncss. The new generating function G(q,p) is called infinitesimal generator of the infinitesimal canonical transformation.


202

..

Deduction 6.10.1. Here we are to derive if G is the infinitesimal generator of the canonical tnt~sformation, then the change of Hamiltonian is given by the Poisson bra,cket o(H with G. The change in the value of the Hamiltonian H = H(q,p) as a result of substitution of Q for q and P for p at instant t and t + dt, in one dimension nndc£ the infil¥tesimRI transformation it; 8H oH_ fJH = -fJp + -oq 8p oq 8H 8G 8H 8e = --;:;}€!=} + -8 €-8 = €[H, Gl q p -D = :c. Py


204

Hence the infinitesimal generating function G becomes,

G

= XPy -

Y]Jx

=

Lz

= (r x p) z

so that fix infinitesimal three-dimensional rotation the infiuitesimal generator is the angular momentum with de in the present case as the infinitesimal parameter L Since the z axis is chosen arbitrary, generally the generating function corresponding to an infinitesimal rotation about an axis -t

G= L.'n where n is the unit vector along the direction of the infinitesimal rotation vector. Thus, angular momentum is the generator of the infinitesimal rotational motion of the system. Hence if Hamiltonian is invariant under infinitesimal rotation, the angular momentum is conserved.

6.11

Finite Transformations

The fillit(~ transformation could be looked on as a sum of an infinite succession of infinitesimal canonical transformations, each corresponding to an infinitesimal displacement along the continuous trajectory in phase space. By integrating the expression for the infinitesimal displacements, it is possible to obtain the finite transformation. Note that, each point on the trajectory in the phase space corresponds to a particular value of the parameter a, say, starting from the initial system configuration say a = o. Let 1L = 1L(a) be some continuous function of system configuration along the trajectory where 1L0 = 1L(0). The infinitesimal change of 1L on the trajectory is given by

61L = [1L, G]q,pda

=?

du dO'

= [u, G].

By integrating this differential equation, we can get u( a) and the effect of finite canonical transformation. Also, by the repeated application of the above equation, taking [u, G], flu, G], G], ... , etc as a function of the configuration.

d2 u -d 2 0:

=

d3 u da 3

= do([[u, G], G])

d d1L

-Z (-Z ) (0:

d

(a

=

d

-z [u,G] = [[u,G],G] (a = [[[u, G], G], G], ... etc.

Thus the formal solution can be obtained by expanding u(a) in a Taylor's series as

6.12.

LIOUVILLE'S THEOREM

205

where the zero subscript meaning that the value of the Poisson bracket taken at the initial point a = o. Let us consider a particular case, G = II and the parameter is the time t. Then the equation of motion for u is ~~ = [u, II] and the formal solution for u(t) is given by

u(t)

t3

t2

= uo + t[u, II]o + 2! flu, II], II]o + 3! [flu, II], II], H]o + ...

where the zero subscript refers the initial conditions at t

= o.

Ex 6.11.1. Obtain the motion in time of the linear harmonic oscillator by means

of the formal solution for the Poisson bracket version of the equation of motion. Assume that, at time t = 0, the initial values are xo and PO. SOLUTION: The Hamiltonian H for a one dimensional harmonic oscillator is H(q, p) = + ~x2 where x = the position co-ordinate, m = mass and k is the force constant. From the definition of Poisson Bracket, we have

Fm

[x H] = ax aH _ ax aH , ax ap ap ax [[x,H],H]

= 1 E- _ 0 x (kx) = E.m

m

= [E-,II] = ~[p,H] = ~{apaH _ apaH} m

TIL

TIL

1

p

k

m

m

m

Dx Dp

Dp ax

= -{O- - l(kx)} = --x. [[[x, H],H], H]

k

k

m

'Tn

k ax aH m Dx Dp

ax oH Dp ax

= [--x,H] = --[x,H] = - - { - - - - - } =

k p - - { l - - O(kx)} m

m

=

kp --2

m

=

constant.

Since [[ [x, H] , H], H] is constant, the higher order bracket vanishes identically. The formal solution for the Poisson Bracket version of the equation of motion is given by

x(t) = Xo

t2

t3

+ t[x, H]o + 2! flu, H], H]o + 3! [flu, H], H], H]o + ...

Po kxo t 2 kpo t 3 =·-z:o+-t- - - - - 2- . m m 2! m 3! This is the familiar elementary solution to the problem.

6.12

Liouville's Theorem

This is fundamental theorem of statistical mechanics. We know, at some instant of time, every possible dynamical state of a system in classical mechanics can be uniquely represented by a point in the phase space. This point in the phase space

206


is known as the image point of the system. Also, this point will not be stationary bur will move along a definite trijectory determined from the equation of motion q] = PJ and pj = - q) where H is the Hamiltonian of the system. Now, if we consider many mechanical systems with slightly different q's and p's, all of them can be represented by a set of distinct points in the phase space, occupying an infinitesimal volume in it. In course of time, the system points move along different paths in the phase space due to different initial conditions. As a result, the density of systems in the neighborhood of some given systems in phase space changes with time. Liouville's theorem states that "the density of systems in the neighborhood of some given dynamical system in phase space remains constant in time." PROOF: Case 1:0ne degree of freedom: We shall first prove the theorem with one degree of freedom. Let p be the density of a dynamical system with one degree of freedom, in the neighborhood of a given system of point in two-dimensional phase

gH

gil,

p

C '~ A Ii B D

L -_ _ _ _ _ _ _ _ _ _ _

q

Figure 6 1..\rea element in phase space

space described by q and p co-ordinates. When we are to evaluate the density p at a fixed point in the phase space, it may still vary with time. Since p = p( q, p, t), the total time derivative is given by dp dt

as

(j)' .

~H, PJ

up).

-

=

ap at

ap .

ap .

+ aq q + apP =

ap at

~H, where the Poisson bracket

uq)

.

+ [p, Hj; [p, H] arises from the implicit

dependence and !fit from the explicit dependence. Let us consider the area element dqdp of two dimensional phase space. At time t = 0, the boundary of the area is formed by some surface of neighboring system points. As the time progresses, the different shapes of infinitesimal areas, defined by system points will move in phase space. Let the speed with which the representative points enter the element ABeD through AD be q. Then, the number of representative points entering the element through AD per unit time will be pqdp. Similarly, the number of points leaving the element through BC will be pqdp + gq (pq)dqdp. Thus the number of representative points that remain in the element ABC D are found to be pqdp - pqdp - :q (pq)dqdp = - :q (pq)dqdp.

Similarly, considering the number of representative points extering through AB and leaving through BC, we find that the number of points thnt remains in the element

6.12.

LIOUVILLE'S THEOREM

207

ABC D are found to be - gp (pp )dqdp. Hence the increase in the representative points per unit time in the typical element as -[gq (pq)

+ ~ (pp)]dqdp.

But the rate of increase of points per unit time in the above typical element id ?!j[dqdp. Therefore, '::: =

dp or, -d t

-[:q aq

(pq)

+ ~ (pp)]

ap

+ p( -aq + -ap ) = O.

(6.28)

Now, we know the motion of a system point in time is simply the evolution of a canonical transformation generated by Hamiltonian. Therefore, the co-ordinates (q, p) at time t2 are related to the variables at time tl by a particular canonical transformation. The change in the infinitesimal area element about the system point over the time interval is given by same canonical transformation. Recalling t 's canOl1lca . 1 equat'IOns qj . = op]' 8H· H amI' Ion Pj = - 8H' oqj' I.e., ~ dq + £i!. up = 0 , tlIe equat'Ion (6.28) becomes,

ap =}

at

+ [p, H]

= 0

dp =}

dt = 0

which is the Liouville '8 theorem. Case 2:n degrees of freedom: Let p be the density of a dynamical system with n degrees of freedom, in the neighborhood of a given system of p,oint in phase space. Through the phase space, there will be an implicit an explicit dependence of time because the 2n co-ordinates (qk,Pk) ofthe system vary with time. Since p is function of (q,p, t), the total time derivative is given by

dp

dt

=

ap

at

~ ap .

ap.

+ L.Ja-qk + a k=l

qk

Pk

Pk ] =

8p

at

+ [p, H]

where the Poisson bracket [p, H] arises from the implicit dependence and !Jff from the explicit dependence. Consider an infinitesimal volume of a typical element as

surrounded a given system point in phase space. At time t = 0, the boundary of the volume is formed by some surface of neighboring system points. As the time progresses, the different shapes of iufinitesimal volumes, defined by system points will move in phase space. The transformation in the infinitesimal volume is a canonical transformation which is a composition of an infinite sequence of infinitesimal canonical transformations. Now Poincare '8 Theorem states that the volume of any

208


arbitrary region in the phase space in canonically invariant, by the virtue of the fact that its Jacobian is unity. Thus, the size of the volume element about the system point cannot vary with time. Thus,

So both the number of systems in infinitesimal region dM and the volume dV are constants, and so the density p = ~t! must also be constant in time. This shows that the numerical density satisfies the equation = 0, which is the Liouville's theorem. From this theorem, we have the following discussions :

*

(i) The above theorem can be written alternatively by the partial differential equation + [p, H] = o. In statistical arguments it is useful to employ, instant of the numerical density p, a probability density (J obtain by dividing p by the total number of dynamical system considered.

!fit

(ii) When the number of systems dM in a given state is constant in time, then the systems is in statical equilibrium. Hence the density p of system points at an instant in phase space does not change in time. Thus the statistical equilibrium condition can be written as [p, H] = o. The characteristic of the essence will be determined by the choice of the function p. Conversely, we can insure equilibrium by choosing the phase density p to be function of the constants of motion of the system, for then its PB with H would vanish. (iii) For such a conservative system, p is a function of the energy E, i.e., stationary and the statistical equilibrium condition is automatically satisfied. So a region of a phase space containing the particles evolves as a whole but the density of the pha.'le particle remain unchanged. (iv) Liouville's theorem can not be applied to the dissipative system for which the constant energy surface containing the trajectories in the phase space keeps on shrinking in volume due to dissipation.

6.13

Exercise

Ex 6.13.1. Calculate the commutators [x, L z ] and [x 2 , L z ].

GATE -1996

Ex 6.13.2. A particle of mass m is subjected to a potential V(x) = kx; k > O. If H is the Hamiltonian of the particle, calculate [H, V(x)]. GATE - 2001 Ex 6.13.3. Find the relation between rotation and angular momentum through the PB relations.

6.13. EXERCISE

209

Ex 6.13.4. For what value of a and j3, the transformation Q qCt sin j3p is canonical?

=

qCt cos j3p and P

=

Ex 6.13.5. The transformation Q = y1qcos2p,P = y1qsin2p is canonical. Show that the generating function is Fl = ¥cos -1 ( ~) - ~ q - Q2. Find also F2 , F3, F4.

J

Ex 6.13.6. Find out the canonical transformation in which the generating function F2 is given by F2(q, P) = qcot- 1 ~ + P. Ex 6.13.7. Using fundamental PB, find the values of a and j3, for which Q = P = qf3 sin ap represents a canonical transformation. Find the generating function F3 (Q, P) for this case. qf3 cos ap,

Ex 6.13.8. Prove that, the form of the Lagrange's equations is invariant under a point transformations. Ex 6.13.9. Prove that, the transformation Q canonical, where a is any constant.

= tan":'I(~n, P = ~aq2[1 + Ct~:2] is

Ex 6.13.10. Show that the Hamiltonian for the simple harmonic oscillator, namely H = ~ + !mw 2 x 2 , can be written in the form H = waa*, where a = vfTI¥(x+~) and a* its conjugate. Evaluate the PBs, [a, a*], [a, H], [a*, H]. Ex 6.13.11. A particle of mass m is acted upon by a constant force F. Using the techniques of PB show that x and p are in the forms x = Xo + ~} + !!fit 2 and p = Po + Ft. Ex 6.13.12. Find the position x and momentum p at time t for the motion of a particle given by the Hamiltonian H = ~ + !mw 2x 2. Ans: x = Xo cos wt + ..E.!L sin wt; p = -mwxo sin wt + Po cos wt. m,w


Chapter 7

Hamilton-Jacobi Theory We have seen that if the Hamiltonian H is conserved, we can solve the problem by the transforming to new canoilical co-ordinates which would all cyclic. HamiltonJacobi equation provides a general method to find the trajectories of the particles with Lagrangian and Hamilton's canonical equation. Mechanical problems are easily solved by the general procedure named canonical transformations. There are two ways of effecting such transformations. (i) If the Hamiltonian H is conserved, we can solve the mechanical problem by transforming old set of co-ordinates into new set of co-ordinates which would be ,all cyclic, as the equation in the new form are easily integrable. Consequently all momenta are constants. (ii) Alternative way to obtain the solution the problem is to seek canonical transformation from co-ordinates and momenta (q, p) at time t, to a new set of constant quantities, which may be the known 2n initial values (qO, po) at t = O. Let the relation between the old (initial values) and new canonical variables for a mechanical problem be

q = q(qo,po, t); p

= p(qO,PO, t)

(7.1)

which is actually the desired solutions of the problems as the function of initial values and time. This is most general procedure and is applicable at least in principle, to a case when the Hamiltonian involves time.

7 .1

Hamilton-Jacobi Equation

The equations of motion in terms of the transformed Hamiltonian K and the new K(Q, P) and the old Hamiltonian H(q,p) are

. Qi

=

oK·

oK

BPi;

BQi ; K

P.l = -

211

of = H + at

CHAPTER 7. HAMILTON-JACOBI THEORY

212

where F is the generating function of the transformation. As the new variables Qi and Pi are constant co-ordinates so, . Qi

.

= 0 = Pt

BK BPi

=? -

BK OQi

= 0 = -,-.

If the transformed Hamiltonian function K to be zero, then the system will be said to be transformed into the equilibrium problem. Thus if the transformed Hamiltonian K will be zero, then the generating function F satisfies the equation

H(q,p, t)

+

of

Bt = O.

(7.2)

Now the generating function F is a function of old co-ordinates q1., the new constant momenta Pi and the time t, so we take the generating function of the form F2 = F 2 (q, P, t). For such a generating function, the equation of transformation is Pi = HF2~'l,TJ,t) , so that the equation (7.2) assumes the form q,

BF2

oF2

H(q1, . .. qn; ~ •... ,-;-; t) uQ1 uqn

BF2

+~ = O. uqi

(7.3)

This first order partial differential equation in (n + 1) variables q1, q2, ... , qn, t of the desired unknown function F2, is known as the time dependent Hamilton-Jacobi equation for a holonomic system having n dJ. Like the Lagrangian and Hamilton's canonical equation, Hamilton-Jacobi equation also provides a general method to find the trajectories of the particles.

7.2

Hamilton principal function

We know, every first order partial differential equation like (7.3), has a solution, depending on an arbitrary function; such a solution is called general integral of the equation. The general solution of equation (7.3) is called Hamilton principal function and is denoted by S, which will merely provide dependence on old-coordinates and time as 5 = S(Q1,Q2, ... ,qn,t). Every first order PDE has a solution, termed as general integral, depending on an arbitrary function. But general integral of HJ equation is not very useful for mechanical problems, so we need to know the complete integral of such equations which contains many arbitrary constants as there are independent variables. Since the PDE (7.3) in (n + 1) independent variables qI, £]2,···, qn, t, its complete solution must involve (n + 1) independent constants of integration aI, 02, ... a n +1. However, S does not appear directly in PDE (7.3), only partial derivatives with respect to Q and t appear in (7.3). Further S is a solution of (7.3), (S+k) must also be a solution, where k is some constant. Hence out of (n+1) constants of integration mentioned above. one is an additive constant attached to S. But as this additive constant will have no effect in a generating function, we can take S to be involving only n constants aI, a2, .. . , an, and the complete solution of (7.3)is, therefore, of the form

(7.4)

7.2.

HAMILTON PRINCIPAL FUNCTION

213

in which none of the n constants aI, a2,. " ,an (may depend on time) is additive and the constant A appears as an additive constant. Thus S is a function of n co-ordinates qi, n constants at and time t.

7.2.1

Connection with canonical transformation

Let a solution containing n arbitrary constants in addition to the additive constant, be known. We now, investigate how the solution, given by (7.4), is related to the solution of the equation of motion variables qk(t); k = 1,2, ... n. To find this relation we employ canonical transformations. Consider the canonical transformation (q,p) ~ (Q, P) via the generating function F 2 (ql, q2,.'" qn, PI"'" PrI , t), where we can take liberally the n constants aI, a2, ... , an of integration to be the new momenta at = Pi; (i = 1,2, ... , n). Since qk's are old co-ordinates and ak's are new moInenta, so S(ql, .. . , qn, 0'1, Ct2,· .. , CY n , t) corresponds to F 2(ql, ... , 'In, PI, . .. , Pn , t), i.e., F 2(r./1, ... ,f]n,Pl, ... ,Pn ,t) = S('ll, ... ,f]n,O'l, 0'2, ... ,an ,t). In order that the transformation be canonical we must have (i) First, let us write the transformation equation as

which are n complete equations and at the time t relating a's in terms of initial values q and P, as

=

0 these give n values

It does not indeed satisfy the definition of momenta in terms of Hamilton's principal functions.

(ii) Further, the second transformation equation which provide the new constants co-ordinates appear as (7.6) which are n-equations and at time t = 0 give the n values of (3i in terms of the known initial values of qt as qi = qi(fJi, at, t), and so Pi = Pt(fJi, ai, t). This set of equations defines a set of new co-ordinates Qi in terms of qt, ai, t. (iii) The relation between the new Hamiltonian K(Q, P, t)

aF2

= H + at

J(

= H

and old H as

as + at =

0

(7.7)

The equations (7.5) and (7.7) clearly identify ('i)S to be Hamilton's principal function (ii) the desired complete integral of the HJ equations of motion. As the transformed

214


= 0, the corresponding Hamilton's equations of motion . oK . oK Qt = o~. = 0; Pi = - OQi = 0 =* Qi = canst. = f3i and P = canst. = ai

Hamiltonian K(Q, P, t)

t

where D.! are the constants of integration in the HJ equation. Also Qi = f3, = aas a, is the new results of the identification of S with the generating function F2, i.e., F2 = S. fJt are called the second integrands of motion. Thus the set (f3i, D t ) is canonically related to the set (lji, Pi) through a generating function the same as Hamilton's principal function. Thus the 71, co-ordinates qk's can be expressed in terms of time t and 2n constants ak, fJk. This gives the solutions of the equation of motion for the q's. Also, since S satisfies Hamilton's equation, the Hamiltonian function of the new system is zero, and consequently the equations of the system are da t dt

df3t

"

= 0; ill = 0; z =

1,2, ... ,71,

so that (aI, a2,···, an, (31, f32, .. . , f3,,) are constant throughout the motion. It follows that, if S denotes a complete integral of Hamilton's partial differential equation, containing 71, arbitrary constants (1:1, 0'2, ... , an, the the equations {3,

=

aS

-~; U(l:i

Pi

=

aS

!:l';i uljt

=

1,2, ... ,71,

constitute the solution of the dynamical problem, since they express the variables (fJl, .... ljn, P1, P2, ... ,]In) in terms of t and 271, arbitrary constants (Dl, D2, ... ,O:n, (31, /h, .. . , (3n)· In this way the solution of any dynamical system with n degrees of freedom is made to depend on the solution of a single partial differential equation of the first order in (n + 1) independent variables.

7.2.2

Significance of S

Hamilton's principle function S is the generator of the canonical transformation to constant canonical co-ordinates and momenta. The choice of the new momenta (l:i, to a certain extend, is arbitrary, so one can choose any n quantities "Ii which are independent functions of the constants of integration O:t, i.e., "It = "It(a1,··· an). According to this choice, the Hamilton's principal function is S(qi, "Ii, t) and the rest derivations remains unaltered. In order to understand the physical significance of Hamilton's principal function S, let us write the total time derivative of S as

d~ dt

=

L. t

aS q.t + oqi

= LPtqt =* S =

J

H

L. t

aS C¥i oat

=L

Ldt + constant

+ aS at

7.2.


215

where g;i, = Pi ai = 0, as a, are constants and H + ~~ = 0. Thus the Hamilton's principal function S differers from the indefinite time integral of the Lagrangian by a constant term. If qt and p, are known functions of time t, then the integration is performed and the problem is solved but practically it has no use. Thus the same integral when indefinite form shapes the Hamilton's principle which results in the Holution of mechanical problems, while the indefinite form it shapes the Hamilton's principal function that provides an alternative way of solving the problem.

Theorem 7.2.1. Let S(q,p, t) be any complete integral of the Hamilton-Jacobi equation ~~ + H (q, ~~, t) = 0, then each motion with associated momenta given by ~~ = P = - (3 is a natural motion satisfying the canonical equations. and

g;

PROOF:

The Hamilton-Jacobi equation for a dynamical system is

as at

+ H(q,p =

oS

oq' t)

= 0.

(7.8)

Let S = S(q, a, t) + A be the complete integral of the HJ equation (7.8), which contains (2n + 2) independent quantities ql, ... , qn, al, ... , an, t, A. Differentiating equation (7.8) partially with respect to ak and qk respectively, we get

a2 s oakat a 2s and - aqk ot

11.

a2 s

oH

+ ~ OPr oakaqr = aH

+-

oqk

°

02S 2: -aH = 0. apr aqkaqr 11.

T

(7.9) (7.10)

r=l

g;

Further, we are given = -(3, which represents n equations. Now, differentiating this equation with respect to t, we get

a2s

11.

a2s

ataak

r=l

aqroak

- - + 2:

qr =0.

(7.11)

This equation represents n linear equations and can be solved to get ql, q2, ... , tin. It may be noted that, equations (7.9) and (7.11) are precisely of the same form ulI , ... , aoH . Thus we have the first and so equation (7.9) can be solved to get auH ,a PI P2 Pn Hamilton's canonical equation qk = aaH. Also we note that, aasq = p. Differentiating Pk this equation partially with respect to t, we obtain 02S ata qk

11.

o'1.s

+ 2: r=l aqr a qk tir = Pk·

(7.12)

Comparing equations (7.10) and (7.12), we have at once, the second canonical equation Pk = - aaH . Thus the two equations constitute HamiJton's canonical equations qk and hence each motion described by the characteristic function S is a natural motion. This is known as Jacobi's theorem, follows from Hamilton's equation of motion in terms of the transformed quantities, with the fact that K = Constant = E and it will help us to determine the natural motion for the given dynamical system.

216


Ex 7.2.1. Simple Harmonic Oscillator The form of the Hamiltonian H of the one dimensional harmonic oscillator as 2 = + ~ kq2, which is independent of time, k = force constant. Put p = ~~, where S = Hamilton principal function, in the equation of H, we have I-I (q, p)

tm

H(q,p)

= 0,

Now the new Hamiltonian K

1 as 2 1 2 = -(-) + -kq

aq

2m

2

for that requirement, the HJ equation becomes

as

K=H+-=O

1 as 2 =:;. - ( - ) 2m

aq +

at

1 2 -kq 2

+ -as at = O.

Since the only term that involves an explicit dependence of S on it is the last term, so the solution can be written in the form of separated variable as

S(q, 0', t) = W(q, 0')

-

at

where 0' is the constant of integration. With this choice of solution the time can be eliminated and so

where c is an additive constant, which will not affect the transformation. The constant 0' is thus to be identified with the total energy E. We, however, do not need the value of S, but g~. Now, the solution for the co-ordinates q arises out of the transformation equation

1m] J

as - aO' -

Vk

fi

sin- 1

(3 -

=:;. (3

+t =

=:;. q

=

I¥-

dq _ t 2f: _ q2

q{f

sin{ {f(f3 +

tn =

I¥-

sinw(f3 +

tn

7.2.


217

jF;,.

where w = This is a usual solution for a one-dimensional harmonic oscillator problem. Finally the solution for the momenta p arises out of the transformation equation p

=

as aq

=

aw aq

=

V2ma -

mkq2

= V2macosw(f3 + t).

Now we are to evaluate the constants (a, 13) in terms of initial values (qO, po) at time t = 0. Let the particle is at rest at t = 0, i.e., Po = 0, but it is displaced from its equilibrium position by an amount qo. Then

which is the initial total energy of the system. Thus qo

=

V% k

=?

q = qo sin w (13

+ t).

Since at t = 0, q = qO, we see that 13 = 0, and thus for particular initial condition, the solution becomes q = qo sin wt. Thus Hamilton's principal function S is the generator of the contact transformation transferring harmonic oscillator with a canonical momentum a = H, the total energy and a co-ordinate 13 which vanishes initially at t = 0. Thus Hamilton's principle function can be written as S(q, a, t) = Vmk j[2; -

q2]~dq 1

=Vmk

j

[q5-q 2J2dq-

= kq5 j[cos 2 w(f3 + t) -

at

kq2 20t

~]dt;

q

=

qo sinw(f3 + t).

Also the Lagrangian becomes

which shows that S is the time integral of the Lagrangian. Ex 7.2.2. Show that the function S = ~w [q2 + a 2 ] cot wt - mwqa cosecwt is a solution of the HJ equation for H principal function for the Harmonic Oscillator with H = 2;n [p2 + m 2w 2q2]. Prove that this function generates a correct solution to the motion of the Harmonic Oscillator in time.


218 SOLUTION:

Hamilton-Jacobi equation satisfied by Hamilton's principal function is 8S H (q, 8q)

1 8S 2 or, 2m [( 8q) For the given S

=

~ [q2

+

8S 8t

+m

+ a 2]cot wt -

=0

2 2 2]

w q

8S

+ at = o.

mwqa cosecwt, we have

2 2 2] - 1 [(8S)2 +mwq 2m 8q

8S +8t

1 ·mw 2q2 mw 2 x m 2w2(qcotwt - acosecwt)2 + - __ [q2(1 2m 2 2 + a 2cosec2wt - 2qacosecwtcotwt] 1 1 1 = 2mw2(qcotwt - acosecwt)2 + 2mw2q2 - 2mw2q2

=-

+ cot2wt)

1 - 2mw2(qcotwt - acosecwt)2 = O.

Thus given S satisfies Hamilton-Jacobi equation. The transformation equations give 8S

Q = f3 = 8a = mwacotwt - mwqcosecwt =}

q=

mwacotwt - f3 mwcosecwt

= acoswt -

f3 . smwt, 'mw

which is familiar solution for a harmonic oscillator. Also, 8S

p = 8q = mw(qcotwt - acosecwt)

= mwcotwt[ar.oswt cos 2 wt

= mwa[. smwt = -mwasinwt -

L sinwt]- mwacosecwt m,w

1 - . - ] - f3 cos wt smwt f3coswt = mq.

Thus the given S generates correct solution to the motion of one dimensional harmonic oscillator.

7.3

Hamilton's characteristic function

We see that, if the Hamiltonian H does not contain the time t explicitly, the Hamiltonian principle function S should be separated into two parts - one involving coordinate q only and the other involving time f, only. Hence in this case also the integration of the HJ equation is possible and HJ equation for S can be written as (7.13)

7.3. HA.MILTON'S CHARACTERISTIC FUNCTION

219

The first term of (7.13) is concerned only with the dependence of S on the qk's and the second term involves only the dependence of t. Hence the solution for S can be written in the form

This substitution gives the reduced time independent differential equation as

as

H(qk; - ) = aqk

a1 =

constant.

One of the constants a1 of the integration appearing in S, so this equal to the constant value of H. Although, when H is constant, the time independent function W appears here as a part of principal function S, it can be shown that W separately generates its own canonical transformation that has properties very much different from that generated by principal function S. Consider a canonical transformation in which the new momenta are all constants of motion ak and where a1 in particular is the constant of motion H. Let the generating function for this transformation be W(q, P), then

Since W is independent of time, so the old K and new Hamiltonian H are equal, i.e., J( = O:t. The function W = W(qk, O!k) is known as Hamilton's characteristic function, which generates a canonical transformation in which all the new co-ordinates are cyclic. The canonical equations for Pk and Qk are

Ther~fore,

Q1 is the only co-ordinate which is not a constant in motion. Q1 is related with time and we have an instance of the conjugate relationship between the time t as a co-ordinate and the Hamiltonian as its conjugate momentum.


220

7.3.1

Significance of W

The total time derivative of the Hamilton's characteristic function W = W(qk, Uk) is given by,

=

action of the system.

Ex 7.3.1. A particle of mass m projected with a velocity u at an angle () .with xn,xis. Write down the HJ equation for the motion. Find the complete integral and the equation of the trijectory. SOLUTION: Let P(x, y) be the position of a particle of mass m at any time t. The KE (T) and PE (V) are given by

T = .!m(i;2 + ii)j V = mgy 2 The Lagrangian and hence the conjugate momenta Px and Py are given by 1

L = T - V = 2m(x2 Px

=

8L ai;

.

= mXj

Py

+ ii) =

aL ay

mgy.

.

= my.

We use the HJ technique for solving such mechanical system. The Hamiltonian H is given by H (Px,Py,y )

1 + (2 1 m (x.2 +y.2) +mgy= 2m =2 Px+P2) y +mgy.

Since H does not contain X explicitly, so Px [2

H(py, y)

=

2m

=

Constant

= 1 (say).

p2

+ 2:'n + mgy.

Let S be the action and put Py = ~;, then the HJ equation is

as

H(ay'Y)

as + at

= 0

l2 1 as 2 as ::::} -2m + -(-) + mgy+ - = 0 2m ay at

Hence

7.3. HAMILTON'S CHARACTERISTIC FUNCTION

221

Since H is constant of motion, so the solution can be written in the form S(y, a, t) = W(y) - at, where a is a constant of integration. Hence ~~ = -a and so the HJ equation becomes H - a = 0, i.e., H = a. Thus,

l2 2m

1 oW + _(_)2 +mgy = 2m oy

oW oy

= J2ma - 2m 2gy - l2

=> => -

=> W(y, a) =

a

J

J2ma - 2m2gy -[2dy + c

2V2

2ma -l2 ~ - Y]2 + C 2 3 2m g 2V2 2ma _l2 3 --3-mj9[ 2m 2g - y]"2 - at + c

= --mj9[ => S(y, a, t) =

where c is the additive constant and this gives the complete integral for S. For the general solution of the equation of motion, we put {3=

oS

oa

=> (3 + t = m

oW _ t

=

J

oa

•

dy J2ma - 2m 2gy - [2

1 2 2 1 = --[2ma -l - 2m gy]"2

mg a l2 g 2 => y = mg - 2m 2g - 2({3+t) .

Let the particle be initially projected from 0 with velocity u at an angle () with the x-axis. Then Velocity at t = 0 along x axis = u cos () Velocity at t = 0 along y axis = u sin (). and put Px = 1 given 1 = mu cos (). Also the constant a is evaluated from H at t = 0 as H = !mu 2 = a. Therefore {3

= -t +

J

dy Ju 2 sin 2 ()

-

2gy

= -t - ~ 9

viu 2 sin

2

() -

2gy.

When t = O,y = 0, then {3 = -~usin(),x = ucos().t. Hence the equation becomes t -

~ucos () = _~vlu2 sin2 () 9

2gy

9

. ()]2 or, [-X- - -Usm ucos () 9

= -1

g2

[u 2 sm . 2 () - 2 gy ]

which is the equation of the trijectory and this represents a parabola.


222

7.4

Separation of Variables

Under suitable conditions, the variables in the HJ equation can be separated. Consider systems in which the Hamiltonian is one of the constants of motion, which may not necessarily be the total energy. In such a case, we need consider only the contact transformation generated by the Hamilton's characteristic function W and its corresponding HJ equation. A co-ordinate qi occurring in the HJ equation is said to be separable, if the Hamilton's principal function can be written in the form

W

= L Wi(qi, 0:1,···, O:n, t) i

which will split the HJ equation in n equations

Hi(qi,

aW -a qi t

j 0:1, ... , O:n)

= O:i·

The constants O:i are referred to as the separation constants. Each of the equations involves only one of the co-ordinates qi and first order partial derivative of Wi with respect to qt. These equations are solved for a first order differential equation ~ and the integrand with respect to qi to get W. Now if the Hamiltonian H does not explicitly contain time, one can linearly decouple time from the rest of 8 and write

Under this assumption, the HJ equation with this trial solution becomes

Since the RHS and the LHS are now functions of totally different variables, they can be equal for all values of these variables only if the LHS and the RHS are separately equal to constant independent of all the above variables. Let us write

a82 -8 = t

-0:1,

8W then H(qi, -8 ) qi

= 0:1

where 0:1 the constant value of Hamiltonian, is the constant of separation. The first gives the solution 82 = -O:lt and the second can be viewed as the time independent HJ equation with W to be identified as Hamilton characteristic function. In any case, the solution to the time dependent HJ equation for conservative dynamical systems, thus has the form, apart from a trial additive constant term

Now, let all the co-ordinates except q1 are cyclic. Since q1 in non-cyclic coordinate, the conjugate momentum PI is constant, and the solution for W be of the form n

W

= LWi(qi,Pi ). i=l

7.4. SEPARATION OF VARIABLES

223

Since all the coordinates except q1 are cyclic, we have

aWi -a = qi

p

i

= O:i =}

UT

Hi

. ....L 1 = O:iqij~;.

Thus the HJ equation becomes

H(ql,

aWl -a ,0:2,"" q1

O:n)

= 0:1

which is the first order ordinary differential equation for WI in the independent variable q1 and hence can be solved easily. Thus for a single cyclic co-ordinate the completely specify function W can be written in the separated form as n

W

= W 1(q1, 0:) +

L

O:iqi·

i=2

Ex 7.4.1. Hydrogen Atom Problem Here we consider the well known Hydrogen atom problem or Kepler problem. This problem, in general, consists in finding the trajectory of a charged particle under the action of central force in an inverse square field. Let a particle of mass m with charge e is moving about the stationary nucleus with positive charge ,e, where , = atomic number. Thus the field of central potential is given by

,2e V(r) = - - = r

k

k = ,2e > O.

--j

r

In three-dimension spherical polar co-ordinate (r, (), ¢), the Hamiltonian His,

Since H does not contain ¢ explicitly, so ¢ is cyclic co-ordinate and so, PcP = mr2 sin 2 ()¢ = Lz = Constant.

This shows that the motion is confined to a two dimensional xy-plane and so the Hamiltonian H in two dimensional polar variables is

1 .2 H = -m[r 2

+ r 2 ()'2 ] -

k r

- =

1

2

-[p 2m r

+ P~ -J --kr r2

where Pr = mr and Po = mr 2 iJ are the conjugate momentum along the rand () directions respectively. Further, the last equation shows that H does not contain () explicitly, so () is cyclic and so Po = constant = I (say), which also characterized by for a central force motion the angular momentum is conserved. Therefore

1 H = -[p~ 2m 1

[2

k

+ -J --. r2 r

CllAPTER

224

r

HAMILTON-JACOBI THEORY

As the system is conservative, the Hamiltonian H represents the total constant energy E of the system and hence 1 _[p2

2m

r

Z2 + _] __k = E

r2

r

Z2

=> Pr = [2m(E - 2" r

k

1

+ -)]2. r

Now H does not involve the time T explicitly, so we can separate the variables. Hence our Hamilton's principal function S has the spatial part lV which is to be taken as the generating function for the transformation. Hence the Hamilton's characteristic function can be written as

W(r, 0) = Wr(r)

+ Wo(O)

fr

= 8;:r j Po = ~~ = 8':0°. Also Po = 8':0° = Z(constant) so that Wo = l() + q(constant). Hence the HJ equation can be written as

where Pr =

aW Pr = -r = [2m (E. - -l2

ar

=> Wr (-r) = "hm

z2

J

(E - 2" r

r2 k

+ -kr )] !.2. 1

+ -r )2" d-r + c

Therefore the generating function for the transformation is

S = W(r, (}) = Wr(r)

+ Wo((})

=l()+v'2m

J

Z2 r2

k r

1

(E--+-)2"dr+c

where c = q + C2 is the constant of integration. Using this expression for Hamilton's characteristic function the transformation equations become t + i31 = ~ = ~~ and

which is the equation of the conic with eccentricity


225

The path is ellipse, parabola, hyperbola according as e < 1, e = 1, e > 1, i.e., according as E < 0, E = 0, E > 0. In Hydrogen-atom problem, we are interested in elliptic path. If the length of major axis is 2a then, 2 _ 2Z 2 /k _

1- e2

a -

-

-

k E'

When the trijectory is ellipse, the energy can be expressed as k 2a

,2£

E=--=-where 2a

=

2a

length of the major axis of the ellipse.

Ex 7.4.2. A particle of mass m is constrained to move on a smooth parabolic wire, x 2 = by under the action of gravity. Write down the Hamilton-Jacobi equation for the system and find the general solution of the equation of motion. Let P(x, y) be the position of the particle of mass m at any time t. The KE (T) and PE (V) of the particle moving on the parabola are SOLUTION:

2

1 [.2 .2] 1 4x . 2 mg 2 T= 2 mx +y = 2m[I+t;2]x; V=mgy= bX'

4X 2

1

'2

mg

2

=> L = T - V = 2 m [1 + t;2]X - b X . Hence Px

aL

4x2 .

= ax = m[1 + t;2]x.

Therefore, the Hamiltonian (H) is given by

H = Px x. - L 1

1

p2

= _

[1

4x2]

= 2m + b2

x

2m 1 + 4x 2

·2

x

mg X 2

+b

+mgx2

IT

Let S be action and Px = ~~. Since H is constant in motion, put S and ~7 = -a. Hence the HJ equation becomes

as H ( ax ' x)

as

= W(x, a) -

aw

+ at = 0 => H ( ax ,x) = a 1 1 oW 2 2 => -2 4 2 (-a ) + mgx = ml+ x x

a

V

aw ax

=> W(x, a)

=

2

4x = [2m(1 + -)(a - mgx 2 )Jfrac12 2

b 4x2 1 [2m(I+t;2)(a-mgx2)]"2dx

J

=> S(x, a, t) = W(x, 0') -

at

=

J

[2m(1

4x2

+ b2 )(0' -

1

mgx 2)]"2dx - at.

at


226

For the general solution, we put f3

=

g~

=

~~

-

t so that,

Ex 7.4.3. A particle of unit mass moves in a field given by the potential V = ~ Fz, where k and F are constants. Prove that HJ equation is completely separable in parabolic co-ordinates (C rJ, ¢) given in terms of cylindrical polar co-ordinates (p, ¢, z) by the following relations z = ~(e2 - rJ2), P = erJ, ¢ = 'P. SOLUTION: Here we use Pilrabolic cylindrical co-ordinates, where the transformation equations are,

x = pcos¢ = erJcos¢; y = psin¢ = erJsin¢; z = r

~(e - rJ2)

= vx2 + y2 + z2 = JerJ 2 + 4 ~(e2 - rJ2)22 = ~(e +· rJ2)

In this co-ordinate system, the KE in terms of e, rJ and 1 variables is given by, - -1 [x·2 T -2

+ y·2 + z.2J

= ; [(irJ cos ¢ + ~r,cos ¢ - erJ sin ¢¢)2 + (irJsin¢ + er,sin¢

+ ~rJcos¢¢)2 + (~~ - rJ1])2J = ~' [(e 2 + rJ2)(e + r,2) + erJ2¢2]. In terms of

~,17

and ¢ variables, V

k

= -:; - Fz =

e 2k+ rJ2 2

F

2

2

"2(e - rJ ).

Therefore, the Lagrangian and so the momenta are given by,

The Hamiltonian for the system is


227

Since H is free from explicit time dependence, the system is conservative, so the HJ equation is H = constant = E, E being total energy of the system. Hamilton-Jacobi equation satisfied by Hamilton's characteristic function W can be written as

Since cP is cyclic co-ordinate so, P O. Find the energy in terms of action J. Show that the period is independent of amplitude. SOLUTION: The Lagrangian and so the corresponding momenta are given by 1 .2 1 2 A L=-mq --[kq +-] 2 2 q2 8L . . p =}p=-. =mq=}q=-. 8q m

The Hamiltonian H of the system is H

.

p2

1

A

2

= pq - L = -2m + -2 [kq + -]. q2

Since H is free from explicit time dependence, the system is conservative, so the HJ equation is H = constant = E, E being total energy of the system. Thus,

p2

H = 2m =}

P=

1

).

+ _[kq2 + -] = 2 q2 1

E ).

± 2m[E - _(kq2 + -)] 2 q2

vmk 2 2 1 = ±-[(,B - q )(q - a)]2, q

where a, ,B are the roots of the equation,

2;; x -

x2

-

i

say,

= 0, i.e., say a = E-~

7.5. ACTION-ANGLE VARIABLES and f3

=

237

E+q=u, where f3 > a > O. The action variable J is given by,

7r /2

J J

= v'mk[

7r /2

{a(l

+ cos 20) + f3(l -

cos 20) }dO

+ 2( a + f3)

o

7r/2

2

+ af3-f3

o

a

J

dO

0 2

sec 0dO /f3 201; put, ~

+tan

2· 2

= a cos 0 + f3sm ' . 0

Now the period of oscillation is given by

so that it is independent of amplitude. Also the angle vfl,riable is, w = vt + c, .where c is a constant. Hamilton-Jacobi equation satisfied by Hamilton's characteristic function W can be written as H = constant = E, E being total energy of the system. Thus,


238

Now, w = a~~j,J) gives q in terms of wand J. Thus the co-ordinate q is given by a function of t as,

E

= [k +

q

VE2 - >'k . k sm(//t + c)]1/2.

Ex 7.5.5. A particle of mass m is constrained to move under the action of gravity in the vertical xz- plane on a smooth cycloid curve given perimetrically by, x = 2

a( e + sin e), z = a( e - cos e). Show that a suitable Hamiltonian is H = 4ma2(i~cos e) + mga(l - cos e). Use action-angle variables to show that the frequency of oscillation of the particle is independent of its amplitude with lei < 1r. SOLUTION: The equation of the given cycloid is x The KE T of the moving particle is given by

= a(e +

sine), z

= a(e -

cos e).

1 T = _m(±2 + z2) 2

= ~mZ2 [(1 +

cos e)2 + sin2 e](j2 = mZ2(1 + cos e)(j2.

= mgz = mgl(l - cos e). Thus the Lagrangian is given by L = mZ2(1 + cosO)(j2 - mgZ(l - cosO).

The PE V is V

=> Po =

8L

80 = 2ml

2

.. Pe (1 + cose)e => 0 = 2m12(1 + cosO)'

Now, the Hamiltonian of the system is,

Pe2

.

= poO - L = 4m12(1 + cosO)

H

+ mgl(l- cos e).

Since the system is conservative, so the energy equation gives H = E, where E is the total energy of the system. Thus, H = E => Pe = ± 4m -=Z2-;'"(l-+-co-s-=-e:-:-)[E=----m-g--=Z-;-(1---cl)-s-=-O:-:')1

Jr:-

= ±4mZJ9icos ~Ja2 -

sin

2

~j

2 a = 2!9Z > O.

Thus the action variable J is defined by,

J

=

f

=

8va 2 2 x 8mZJ9i[ 2·

PedO

=

2 x 4mZ J9i

J ~J cos

a2 - sin

2

~ j where sin ~ = a

-0

~2 => E =

X

4~

The frequency is

4mly'9ia'"

IfJ 1/

=

82

a2 8 (J +"2 sin-l(~)l~aj sin 2 =

~ 2 x 4mly'9i"

= H.

WJ = 4~ If

X

2!91

8

~ 4"E~

Chapter 8

Reference Frames The solution of many physical problems in Physics, Applied Mathematics can be readily obtained by introducing the moving frames of reference, i.e., non-inertial frames of reference, for example, while discussing the motion of the particles near earth's surface, separation of liquids of different density in a centrifuge, the analysis of the motion of the rigid body, the experiences of an astronaut in an orbiting satellite, etc. Usually for such problems, we are to consider two frames - the inertial frame and the moving frame. The frame which is at absolute rest or in uniform motion, is called an inertial frame. Relative to a fixed frame of reference, a moving frame can posses a translation or rotational velocity or both. If the frame of reference moving with constant velocity relative to a fixed frame, then it is called an inertial frame of reference. If a frame of reference has some acceleration or retardation relative to a fixed frame, then it is called non-inertial frame of reference. Basically, earth 1S itself a non-inertial frame and any frame associated with earth in noninertial. Introduction of non-inertial systems also enables us to explore some of the conceptual difficulties of classical mechanics, such as properties of space, validity of Newton's law and meaning of inertia.

8.1

Translation motions

Let us consider two co-ordinate systems 0' (x' y' z') fixed in space and O{ xy z) fixed to a certain body whose orientation in space are observed to remain fixed. O{xyz) is moving with a translational velocity with respect to the frame of reference O'{x'y' z') so that the axes are parallel. The physical principle involved are not affected by such a particular simple choice of translation. At time t = 0, the origins 0 and 0' of the two co-ordinate systems will coincide. At a certain time t, let R be the instantaneous position vector of the point 0 with respect to the point 0', i.e., 00' = R. A single observer, who sees the two co-ordinate systems, is able to describe the motion of a particle with respect to the either of the reference frames. Let the position vectors of the same point P with respect to the base points 0 and 0' be -:;;+ and -:;;+, in the ~

~

239

~

CHAPTER 8. REFERENCE FRAMES

240

p

~-------------y

)C-------------y' x'

Figure 8.1: Translation motion

two frames respectively, then ---+

r" = R + r' ~,

~

~

::::} 11' =

~,~

+ rand r = 11' + V; and 0;' =

::::} r = R

~

+r 0; + a: R

(8.1)

where 11", 0;1' mean the velocity.,and .. acceleration of a with respect to a'. Thus the velocity and acceleration (r' ,-:t)' of a particle at point P measured in the ~~

fixed system is the vector SUlll of velocity and acceleration (R, R) of the moving frame and the velocity and acceleration (-t, ~) of the particle at the point P in the ---+ moving frame. Let F be the total external force acting on the particle P. Then the equations of motion of the particle at the point P in the fixed frame and moving frame of reference are .. ,

---+

~

---+

mr' = ..

.. ,

and m -:t = m r'

-

F ~---+

m R = F - m R = F ef f

(8.2)

---+

where F eff is the effective force acting on the particle at the point P. The equation {8.2)has the same form as Newton's equation of motion in an inertial frame, where ---+

---+

~

~

F is replaced by F - m R. The quantity -m R is called pseudo or fictitious force, which arises out of the acceleration of the reference frame and is not due to particle interactions. This pseudo force sharply acts in direction opposite to the acceleration of the frame. Hence, an acceleration frame is non-inertial. From equation (8.2), we have the following discussions : ~

(i) If a moving frame of reference has an acceleration R, the effective force acting ~

on the particle at point P is smaller than the actual force by an amount m R , ~

due to a result of the acceleration m R of the moving frame. ~

(ii) If 0 moves with constant velocity relative to a', then R = 0 and the Newton's equation of motion will hold in both the systems. Thus all co-ordinate system

241

8.2. ROTATING FRAMES OF REFERENCE

which move with uniform velocity relative to one another will be inertial if any one of them is inertial. This is known as principle of Newtonian relativity. (iii) There is in fact no dynamical way to distinguish one inertial system from another.

Ex 8.1.1. A pendulum in a Lift Let us consider a pendulum of length l and bob mass m within a lift, which is moving The pendulum is oscillatory in a vertical plane with upward with an acceleration -t small angular amplitude (). Let T be the tension of the string with respect to an observer in the lift, the total force acting on the bob is

a.

-t

T

+m

-t

9 -

-t

m

a

-t

A

A

= T - mgk - mak -t = T -m(g+a)k A

where k is a unit vector in the vertically upward direction. This equation at once suggests that the pendulum oscillates in a gravitational field which is effectively increased from -mgk to -m(g + a)k. Now the radial and tangential components of the gravitational force is meg + a) cos () and m(g + a) sin () respectively. The radial components of the forces supply the necessary centripetal acceleration to keep the bob moving on the circular arc of radius l. Thus, the equation of motion is, d2 s

m dt 2

d 2 ()

= -m(g + a) sin() (g

+a

.

=> dt 2 = - - l - sm();

as s

= l()

2

=> ddt () _ (g l +-a ()., when () is small (:-:;- 40). -2 This shows that the motion is simple harmonic, and the period of oscillation is Tl = 27f 9 +1 a. • Similarly, when the lift descends with a downward acceleration a( < g),

V

V

then the period of the simple pendulum becomes T2 = 27f g~a' When the begins to fall freely, then a = 9 and T2 - t 00. This shows that the pendulum stop oscillating.

8.2

Rotating frames of reference

The transformation from an inertial system to a rotating system is fundamentally different from that to a translating system. The system translating uniformly with respect to an inertial system is also inertial) whereas a uniformly rotating system is intrinsically non-inertial and involves several fictitious forces including one which is velocity dependent. Let us consider a reference frame 0' (x' y' Zl) fixed in space and another frame O(xyz) fixed in t.he body. Their origin and base vectors of the two systems coincide at time t = 0, but at any time t, they may not necessarily coincide. When the body rotates with angular velocity c;J about some instantaneous axis

242

CHAPTER 8. REFERENCE FRAMES z' z

p

y

~~---------------y'

x' x

Figure 8.2: Rotdional motion passing through 0, frame O(xyz) will rotate with respect to the frame O(x'y'z') with angular velocity W. The frame O(x'Y' z') is the rotating frame of reference. Let the unit vectors i',]', /';' as the fixed unit vectors forming a rectangular cartesian triad in the primed system O(x'y'z') and the unit vectors i,l,/.; in the unprimed system O(xyz) which are changing their directions along with the rotating axes. The position vector of a particle at P can be written as ~

= x'i' + y']' + z' /';'; = xi + Yl + z/.;;

space set of axes body set of axes.

Now the transformation equations from unprimed system to primed system are ,

X

=

~

~,

r.2

~ ~,

~

~,

~,

= n.2 + YJ.2 + zk.2 A

, A. A., A Y = T.J = X2.J + YJ.J + zk.J , A, A, A, A A, Z = T.k = xz.k + YJ.k + zk.k ~~,

~

~

Le.,

~,

~

(~:) = (!f. z'

~,

~

nH) (~) .

i.k' j .k' k.k,'

(8.3)

z

where the dot products i.i',l.i', .. .etc. on the right hand side of the above equation are simply the cosines of the angle between the corresponding axes. Similarly we get the equations for inverse transformations.

8.2.1

Rate of change of arbitrary vector

Here we are to find the transformation rule relating the time derivatives of an arbitrary vector G = G (t) in the inertial and rotating systems. Physically, the vector G is the same in both the frames, i.e., G f = G r , the suffices f and r are used for fixed and rotating frames. The components of G will be different in the two systems because the unit vectors in both the frames are pointing differently. Thus, ~

~

~

~

~

~

~

Gf

,..

....

....

= G~i' + G~j' + G~k'

A A A =G1 i+G2j+G3 k = Gr· ~


243

---t

The time derivative of G will differ in the above two co-ordinate systems: Since the ---t rate of variation of G is measured by an observer in the fixed frame so the time ---?

derivative of G f is denoted by constants, so ~:

d,--40

Z.

f.

In the primed syste~ as the unit vectors are

= ~ = ~' = 0 and so the time derivative in the primed system is ---t

d' G f = dG~ i' dG; "./ dG~ k' dt dt + dt J + dt . ---t

,~

is the rate of change of G with respect to the rotating frame because in this £, £., dk' d'G ---t case d~t = Tt = dt = o. Also is called rate of growth of G. Similarly, the time ---t derivative of G r with respect to an observer in the rotating frame in which vectors i,3, k are treated as constant unit vectors is ' d :;; f

Tt

---t

dG r _ dG 1i dG 2 ". dG 3 k dt - dt + dt J + dt . In the rotating system, however, the unit vectors are changing in directions. Thus ---t the time derivative of G with respect to 0' frame in unprimed components is

*,

The remaining terms ~, ~~ arise as a result of the rotation of the system. Since the unit vectors i, k are moving (changing directions) with time with respect to 0', they can not be treated as constant vectors in 0' frame. is the velocity of

3,

that point on OX' whose position vector is primed components.

8.2.2

i.

*

Similarly, we can also find

d~r in

Velocity and Acceleration

Let us assume that, the origin of the fixed and rotating frames are the same, but their axes may not be parallel. Let -:t be the position vector of a point with respect to the base point O. The body is rotating with angular velocity C;; about ~he axis passing through the origin. When a vector changes, then in both the frames ---t

---t

---t

dr=rsin(Jdn=}dr =dn x r ---t

---t

dr dn ---t ---t ---t ---t =}v=-=-xr=wxr. dt dt

(8.4)

The quantity C;; x -:t is called rate of transport. This expression is the linear velocity of a particle having position vector -:t and rotating with angular velocity C;;. In


244

dX

this expression, 7 is any position vector rotating in a body and is the time derivative of 7. So the expression can be used for particular cases to unit vectors i, 3, k. Applying this to the unit vectors i, 3, k in a system rotating with angular velocity of rotation TJ, we have,

di dt

~ d3

-t

= w x z, dt =

~ dk x J, dt

-t

W

=

-t

W

A

x k.

Hence the rate of change of vector 7, when the unprimed system is rotating with an angular velocity TJ with respect to primed system is given by

d7 dt

d

A

A

A

= dt(xi+yj+zk) dx ~ dy ~ = dt Z + dt J + dx dy = dt i + dt j +

dz dt k dz dt k

dx ~ dy = dt Z + dt J

dz

A

A

A.

di

A

A

d]

dk

+ x dt + y dt + z dt -t

A

+ x( w

A

x i)

-t

-t

+ y( w

-t

A

x j)

+ z( w

A

x k)

-t

+ dt k + w

x r

d7, d7 r -t -t =?--=--+wxr dt dt

(8.5)

dJ;

where r represents the velocity of the rotating system and the term TJ x 7 arises as the result of rotation of the system. The relation can be used to obtain expressions for velocity of the particle situated at the point P. In operator symbol this equation can be written as d'

d

-dt = -+ dt

-t

(8.6)

wX

it

where is the time derivative in the rotating co-ordinate system. This relation may be used whenever we wish to discuss the motion of a particles, relative to a rotating co-ordinate system. In particular, if we operate it in TJ, we get, d-t d-t d,-t W W -t-t W -t-t-t (8.7) dt = dt + W x W = dt as W x W = O. This shows that the angular acceleration rotating frames. Now the acceleration is

d,27 d' d'7 d dt2 = dt (dt) = (dt d27 dt d2-t r

-t

+W

-2-

-t

x r

+ -Wt x

d7 dt

will be the same in both fixed and

[d7 x) dt

-t

+W

-tJ

x r

d7 - t -t -t + W x (w x r) dt d-t d-t -t r -t -t -t) W -t = -d 2 + 2w x -d + W x (W x r + x r. t t ~ -t -t dr -t, -t -t -t -t -t) dw -t a =a+2wx dt +wx{wxr + dt xr.

=

+ -dTJ dt

iJ

+ -Wt x

(8.8)


245

dJ!

The term W x (w x ~) + X ~ is called acceleration of the transport. The expression is the acceleration of the particle situated at the point P, relative to the rotating frame where, (i)

(ii)

d;X = rt is the acceleration as measured in the unprimed system. dX = velocity relative to the rotating axes. 2w x dX is called Coriolis acceleration. This acceleration is present only when the particle is moving in the unprimed system with velocity 7. It arises from the motion of the particle in the non-inertial frame.

(iii)

w

(w

x ~ = velocity due to rotation of the axes. Wx x~) is called Centripetal acceleration. This term reduces to w 2 r if w.~ = O. This acceleration of the particle situated at point P is directed towards the axis of rotation and is perpendicular to it. Its magnitude 2

Iw x (w x ~)I = w 2 r sin () = ~() rsm where v = wr sin () is the speed of the particle when it rotates in a circle of radius r sin (). (iv)

dJ!

x ~ = angular acceleration of the particle due to the acceleration of the rotating axes. It is known as Euler acceleration. When W is constant, it vanishes.

The equation (8.8) is sometimes referred to as Coriolis theorem. Thus we see that, in a rotating non-inertial frame, two pseudo forces will appear, one of which is called coriolis force and other is centrifugal force. The action of the first one is conditional and is not effective for the rest particle with respect to rotating frame, where as the second force always acts in rotating frame. Deduction 8.2.1. Velocity and acceleration in plane polar co-ordinates: Let P(r, ()) be the position of the particle at any time t, i = unit vector in the radial direction (r increasing), ] = unit vector in the cross-radial direction (() increasing) and let i x 3 = k. Now with the rotation of particle P, i,3, k form a rotating set at -+ .A ---+-+A 0, rotating with the angular velocity w = ()k. Now, OP = r i. So, -+ V

d

-+':

~':

-+

= dt ( r ~) = r ~ + w ~"

= r i

-+':

x ( r ~)

." --+..... -:..,." --+.,.. + ()k( r i) = r i + ( r ())j.

Hence the velocity components in radial and cross-radial directions are respectively ~ -+' 1° and r (). Also, -+

a

d1J

d ~A

-+' A

= dt = dt [ r i + ( r ())j]

= 7i + (70 + ~(j)] + 70] - ~02i = [7 - ~02]i + [270 + ~(j]).


246

Hence the acceleration components in radial and cross-radial directions are respectively -:P - -;;7iP and 270 + -;;70.

8.2.3

Equation of motion

In a inertial frames of reference, the Newton's second law of motion is valid. Let us assume that Newton's law of motion holds in the primed system. If we are to consider the motion of a particle of mass m, then its equations of motion from the point of view of an observer who is at rest with respect to primed system as ---+/

ma

= ---+ F

d2 -;;7 ---+ d-;;7 ...... ---+ ---+ dc;J ---+ ---+ or,m-2-+2mw x-+mw x(w x r)+m- x r =F dt dt dt d2 -;;7 ---+ ---+ d-;;7 ---+ ---+ ---+ dc;J---+ ::::} m dt 2 = F - 2mw x dt - mw x (w x r) - m dt x r

(8.9)

---+

where F is the force act on mass m. This relation shows that Newton's equation ---+ of motion retains its form in a rotating frame if to the applied force (F) we add a number of fictitious forces. Thus, the fictitious forces acting on the particle in the rotating frame are ---+

(i) The real force F

= m ddt->f , it 2

is as usual term as Newton's second law.

(ii) The force -m c;J x (c;J x -;;7) arising as a result of the rotation of co-ordinate axes. The term -mc;J x (c;J x -;;7) is called the centrifugal acceleration, in the general sense, for nonuniform in rotation which always points radially outwards. This

N ~------:,.:P,----

o

cox(coxr)

rXm

Figure 8.3: Centrifugal force acceleration will be forced to be zero at the pO,les where c;J is parallel to -;;7, . and has the maximum value at the equator. It's magnitude is mw2 p, where p = sin e is the distance of the point, from the axis of rotation. (iii) The coriolis force -2mc;J x d:;; is an velocity dependent force arising as a result of the motion of the particle in the rotating system. Like the dissipative forces it depends on the velocity. It is obviously perpendicular to the plane containing c;J and When a particle is fixed with respect to the rotating .... -> ---+ co-ordinate system, dd; = 0 and the coriolis force is zero.

d:;;.

8.3. LAGRANGIAN AND HAMILTONIAN

247

(iv) The force mTJ x (TJ x 7). For a uniformly rotating system, TJ is constant and hence TJ x (TJ x 7 is zero. Thus the acceleration (11 referred to a uniformly rotating system is ---t ---t 1

a

=

-F

m

---t

-

---t

(---t

---t)

w x w x r

- 2w Xd -r . ---t

dt

The centrifugal and coriolis force are arising because of use of non-inertial frame. Thus non-inertial forces are often helpful in dealing with the problems concerned ---t with rotational motion. Now the effective force F ef f is given by ---t

F eff

= ---t F -

d7 - mw ---t (---t ---t dTJ---t x w x r) - m - x r. dt dt

---t

2mw x -

---t

=F +

non-inertial forces.

The distinction between a true force and a fictitious force is that while the former is due to interaction between the particles, the later is due to acceleration of the reference frame. Thus, for a satellite moving around the earth in the fixed system, the only real force acting on it is the gravitational force that produces the centripetal acceleration.

8.3

Lagrangian and Hamiltonian

Since the Lagrangian L is a scalar quantity, we still use the same form as L = T - V in this frame also. The Lagrangian of the system has to be evaluated in the inertial frame only and then should be expressed in terms of quantities defined in the rotating frame. For a single particle, the PE is V = V(171) and the KE T is given by, T =

1

~2

2m r

=

1

---t

---t

---t

2

2m[ v + w x r 1

= ~ml"if12 + ~mltt x TJI2 + m"if.(TJ x tt) where "if is the velocity of rotating frames. Therefore, the Lagrangian L is

L

1 ---t 2 1 ---t ---t 2 ---t ---t ---t' () = T - V = 2ml v I + 2ml r x w I + m v .( w x r) -'- V r 8L '* 8"if = m v + m( w ---t

---t---t

x r) = Pr

where Pr is the conjugate momenta corresponding to the co-ordinate 7. Thus the true momentum of the particle is not just m 1/ but m[1/ + TJ x ttl which is the momentum observed in the fixed frame. Also,

8L

---+

---t

87 = m v x w

---t) ---+ + m( ---t w x r x w -

---+

V V.


248

The Lagrange's equation of motion gives,

~(8~)_ 8~=O

--=-t ='?mv=

dt 811 8r 2mvxw (--+ --+) +mu.,xr (--+ --+) xW-v. --+ MV

The first term 2m(11 x w) of the RHS of the above expression is called Coriolis force and the second term m(w x ~) x w is known as centrifugal force. The Hamiltonian H of the system is H = --+ P r' --=-t r - I

8.4

J

2 2 1 1--+1 1 1--+ = 2m v - 2m r x --+1 w + V(--+) r.

Motion of rotating earth

The earth rotates about its north-south axis from west to east with an angular velocity 0.7292 x 1O-4 ra dj sec. When we ordinarily think of co-ordinate system, we generally take the laboratory system, i.e., one fixed to the earth, in which our necessary equipments are at rest. In fact, this system is rotating with the earth, so that it becomes a non-inertial system and so the fictitious forces appear. To find the effect of rotation, we need a fixed frame. We may imagine one such with its origin 0 at the centre of earth, its z axis north-ward along the earth's rotation axis and x, y axes at right angle to this. The origin 0' of the other co-ordinate system is fixed to the earth at the place of observation. It's z' axis is the upward vertical at 0', x' axis is eastward along the line of latitude and the y' axis is northward along the line of longitude. Let a particle of mass m be situated at P near or on the surface --+ of the earth and suppose some applied force F acts on it in addition to its weight Let ~, be the position vector of the particle at P with respect to 0 and with --+ respect to 0' is ~. Let 00' = R, R = radius of the earth. Then,

mg.

--+ ~'=~+ R

='?

--+ d~' d~ --+ --+ dR ='?--=-+w x r +dt dt dt d2--+ d--+ d 2~ r r --+ (--+ --+) --+ r ---;]l2 = dt2 + w x w x r + 2 w x dt

+

d--+ w--+ dt x r --+

+

d2--+R dt 2

where the rotating system has a translation as well. Also, R is a vector of constant magnitude and rotates with 0' frame with an angular velocity W about 0 Z axis.

iii

--+

--+

Thus, dt = w x Rand 2 --+ --+ --+ --+ d R = ~(dR) = dw x x dR dt 2 dt dt dt + dt --+ (--+ --+) iasWIS --+. fi xe. d =wx wxr

11 w

8.4. MOTION OF ROTATING EARTH

249

(;J x (w x 7) is directed towards the rotation axis of the earth and perpendicular to it. In the inertial frame, the equation of motion of the particle will be

d2 7'

m---;]i2 =

-+-+

m

9

+F

d 2-+

T -+ -+ -+ (-+ => m-= m 9 + F - mw x w

X

-+ -+ (-+ = m -+ 9 + F - mw x w

X

~

d-+

-+

-+ T) - 2mw x -

-+

-+

'r

&

2-+

d R - m-~

d7

T) - 2m w x dt'

This is the equation of motion on the surface of the earth rotating with constant angular velocity It shows that if the form of Newton's law (valid in inertial frame) is to be preserved in the rotating laboratory frame (no~-inertial frame).

w.

Deduction 8.4.1. Effect of g: Let earth be a sphere of radius T and centre O. ~hoose the co-ordinate frame OXYZ, such that Z-axis coincides with the axis of rotation. Then, X~ -plane represents the equitorial plane of the earth. The earth

z OJ N ___- r - _

c

Fe

~'-------:r--Y

s Figure 8.4: Effect of g

w

rotates about Z axis with uniform angular velocity from west to east, so that = wk. Consider a particle of mass m placed at a point P on the surface of earth, where the geographical latitude is A. The position vector of the particle is given by 7 = r cos A} + r sin Ak. If r is the unit vector along OP, then f = cos A} + sin Ak. The particle at point P is acted upon by the following forces:

w

(i) The real force equal to the weight mg of the body along PO, i.e.,

mg = mg(-r) = -mg[coSA}

+ sin Ak].


250 ~

(ii) The fictitious force Fe is given by ~

~A

A

A

Fe = -m w k x [wk x (r cos Aj

= -mwrk:

+ r sin Ak)] A

+ sin AD] = mw 2r cos A].

x [cosA(-i)

~

The central force Fe acts along CP, i.e., radially outwards along the circular path traversed by the particle on the surface of the earth due to rotational motion. If m 9' be the effective weight of the body, then, ~,

~

m9 = m9

*

~

+ Fe =

-mg( cos Aj + sin Ak) A

2

~,

A

A

+ mw

2

A

r cos Aj

A

9 =-[gCOS)..-w rcosAJj-gsinAk.

Therefore, the magnitude of the effective acceleration due to gravity ,is given by, g' = J(g cos A - w2rcosA)2 + (gsinA)2 2w 2r w4r2 = g[l - - - cos2 A + - - cos 2 AJ1 / 2. 9 g2 Now,'r

= 6400km = 6.4 w2r

9 2

x 1Q6m ;w 21T

= 24x~~x60rad8-1 2

= [24 x 60 x 60 J x

6.4

and 9

= 9.8ms- 2, so that

6 X 10 9.8 = 0.003436«

Since w r « 1, the third factor containing ~ 9 9 fore, the above equation reduces to

=

(w2r? 9

1.

can be neglected. There-

When the particle lies at equator, i.e., A = 00 , geq = 9 - w2r and when the particle lies at poles, i.e., A = 900 , gpo = g. From these results, it follows that due to the rotation of earth, thee effect on the acceleration due to gravity is zero at poles and it goes on increasing as one moves from the poles to the equator, and becomes maximum at the equator. Ex 8.4.1. How fast than its present speed of rotation, earth should start rotating, so that the effective acceleration due to gravity at the equator will become zero? SOLUTION: Due to rotation of the earth, the effective acceleration due to gravity at latitude 'A is 9' = 9 - w 2 r cos2 A, where w is present speed of rotation of earth and r is its radius. At equator, A = 00 , we have, geq = 9 - w2r. Suppose that Wo is the speed of rotation at which, if the earth rotates, the effective acceleration due to gravity at equator will become zero. Therefore, if w = wo, geq = O. Thus


251 24X~~X60rads-l and

Now, radius .of the earth, r = 6400km = 6.4 x 10 6 m; w g = 9.8ms- 2 , so that w 2r

9

=

27r 2 6.4 x 10 6 [24 x 60 x 60] x 9.8

=

1 291

~

w2

J = 291 ~

Wo

~

17w.

Thus, if the earth starts rotating with angular speed of about 17 times its present value, then the acceleration due to gravity at equator will become zero. Deduction 8.4.2. Effect of coriolis force on a particle moving on the surface of earth : Here we are now interested about the effect of coriolis force on a

T y

IS Figure 8.5: Local co-ordinate system

particle moving on the surface of earth surface. Let earth of radius r and centre o rotates about its axis of rotation from west to east with uniform velocity TJ. Consider a mass m lying at point P in the northern hemisphere where latitude is ).. Consider that a co-ordinate frame P XY Z is attached to the surface of earth at point P, such that its Z axis, i.e., PZ coincides with the vertical line OP, Y axis, i.e., PY is directed towards north of the earth, so that X axis, i.e., P X is directed towards east. Let the particle moves with velocity 11 in the XY plane, then 11 = vxi + vyJ. The angular velocity TJ lies in Y Z plane makes an angle ). with the Y axis, so TJ = w cos).) + w sin ),k. Due to the earth rotation, the coriolis force acting on the particle is given by -+

-+-+

i

j

k

Fe = -2mw x v = -2m 0 wcos). wsin). Vx Vy 0 = 2mwsin).[vyi - Vx)J

+ 2mwcos).(vxk).

•


252

The acceleration due to the coriolis force is given by ---+

a

1 ---+ m

~

~

~

= -Fe = 2wsin>,[vyi - vxjl + 2wcos>.(vxk)

a:

a:

w here H = 2w sin>. [v yi - vxYl and v = 2w cos>. (Vx k) are the horizontal and the vertical components of the acceleration due to the coriolis force.

a:

(i) When the particle is moving along X axis, then Vy = 0 and H = -2w sin >'vxY, i.e., the acceleration of the particle will be along negative Y axis. Therefore, a particle moving in northern hemisphere along X axis will be deflected towards negative Y axis. Similarly, it can be shown that in case the particle is moving in southern hemisphere along X axis it will be deflected towards left.

a:

(ii) When the particle is moving along Y axis, i.e., Vx = 0, then H = 2w sin >.vyi, i.e., the acceleration of the particle will be along positive X axis. Therefore, a particle moving northern hemisphere along Y axis will be deflected towards positive X axis. Similarly, it can be shown that in case the particle is moving in southern hemisphere along Y axis, it will deflected towards left. Thus the maximum magnitude of the coriolis acceleration is at the north pole, or, the south pole and is ac

= 2w sin ~v = 2 x

7.29

X

1O-5 v

where v is the velocity in the horizontal plane. Thus we conclude that although the magnitude of coriolis acceleration is small, it plays an important role in many phenomenon in the earth. It is take into consideration the effects of the coriolis acceleration in the flight of missiles, the velocity and the time of flight which are considerably large. Hence the velocity 11 of the moving particle in the horizontal plane on earth surface, we can immediately find out the angle of deflection for particle motion along the duration t as, distance travelled in time t in the deflected direction a = distance travelled in time t in the direction of projection

_ ~(2wsin>'v)t2 -"'------- = w sin >.t. vt Thqs, on the north pole (>. = 900 ), a = wt, which is maximum and for t = 1808, then a = 7 x 1O-4 rad = 0.040 , which is found to be quite small, but it assumes considerably important in guided missiles.

Ex 8.4.2. Deflection of a freely falling body

253


Let us assume that the angular velocity of the earth's rotation is small so that the variation in 9 can be neglected. In this situation, we are to find the horizontal deflection from the vertical of a freely falling particle (neglecting effects of wind. viscosity) caused by the coriolis force near the earth's surface, falling in the earth's gravitational field. Let P be the position of the particle (Fig.8.5) of mass m, which is falling freely from rest under gravity towards the earth from a height h( P A = h) in the northern hemisphere. Although the coriolis force will give rise to small velocity components in the horizontal plane, we can neglect them in comparison with V z • Thus the velocity 11 at P is given by

11 = gt = -gtk = (0,0, -gt). OPZ is along z-axis which is in vertically upward direction and makes an angle () = 900 - ). with ON. The particle is falling along t.he z-axis freely under gravity. o is the centre of the earth and P is the origin of the xyz system. OB is the polar axis of rotation of the earth with angular velocity (;J. P B is parallel to ON, then LNOZ = (J and the angular velocity is along P B. Since the particle velocity is almost vertical, the angular velocity vector (;J lies in north-south vertical plane, i.e., YZ plane and makes angle.(J with y-axis, so (;J = (O,wcos(J,wsin(J). The component of (;J along the vertical direction is w cos (J and dhected upward in the northern hemisphere and downward in the southern hemisphere. Hence, a body falling freely in the Northern hemisphere will be deflected to the east. PT is the tangent at P and is along the north direction denoting y-axis. The line P R 1S along x-axis towards eastern direction. The particle is acted upon by the following forces: (i) The real force equal to weight m 9 of the body along PC, i.e., m 9

=

mg(-k) = -mgk. (ii) The fictitious force F curialis = -2m (;J x 11. To find the coriolis force on the particle, let us evaluate the deflecting force 2m d;J: x (;J is in the east-west direction as

--

i

j

vxw=x

iJ

k

Z ow cos (J w sin (J

i 0

3

k

0 -gt = wgt cos (Ji. ow cos (J w sin (J

-

Therefore the coriolis force on the particle is F curialis = 2mwgt COS (Ji~ and so the effective force on the particle is given by

F

Effective

- -

= F Real + F Coriolis = -mgk~ + 2mwgt cos (Ji~

Hence the path of the particle will be deflected towards the right in the northern hemisphere and left in the southern hemisphere due to coriolis acceleration. The


254

equation of motion is given by ~

= 2m ~ v

m r =}

= my =

mx

~

"

x w - mgk

i

=2 x

°

J

y

z

-mgk

wcosOwsinO 2m(ywsinO - zw cos O)j East ward -2mxw sin OJ

North ward

mz = 2mxw cos 0 - mg. Vertically upward dy .. 2 . 0 dz 0 =} x = dt w sm - dt w cos =}

d .. dt x

=

d2 y . d2 z 2 dt 2 w sm 0 - dt 2 W cos 0

= 2w sin O[-2xw sin 0] = -4w 2 [x - :w cosO]. Let p =

2w cos O[2xwcos 0 - g]

x - fw cos 0, then this equation becomes, 2

ddt p 2 =}

x

+ 4w 2P =

°

=}

P = A cos 2wt + B' sm 2wt

= JL cosO + A cos 2wt + B sin 2wt.

2w The body falls under the action of gravity, so the velocity will be almost opposite to z-axis. The small correction on deflection of V z . produced by the coriolis force is quite small and hence the value of the force have negligible components along x and y direction. As 9 is directed along z direction the components of acceleration of the particle at time t = are

°

= 0, y = 0, Z =

x

-g

=}

x

= 0, y = 0, Z = -gt

as the particle is falling along z- axis. Using these initial conditions, we get x = :w sinO(1- cos2wt).

The x- component equation changes. It is evident from this result that the particle falls following a spiral path about the vertical. Again integrating the above equation, with respect to t, we get the coriolis displacement Xc toward east, as Xc

J

dx =

2~ cos 0

o =} Xc

t

J

(1 - cos 2wt)dt

0

9 sin2wt = -2w cos O[t - 2w ] 9 1 (2wt)3 = 2w cosO[t - 2w {2wt - -3-!-

1

= '3Wgt3 cos e.

+ ... }]


255

Hence if h is the height from which the the particie falling from rest, the time. of falling T is T = ~, follows from the equation z = -g. Hence the deflection of a particle towards the east when it is dropped from rest at a height h, at latitude () in the Northern hemisphere will be deflected eastward by a distance Xl

1 2h 2V2 = -w cos ()g{_)3/2 = --w cos () 3 9 3

fJ3 -

9

where we neglect the resistance of air. This is the eastward effect on a freely falling body, due to the effect of coriolis force. For example, if the particle be dropped from a height of 100 m from rest at latitude e = 45°, it will be deflected by about 1.55 x 1O- 2 m toward east in northern hemisphere. The coriolis deflection in a vertical fall is greater at the equator e = 0, where 1/ is perpendicular to (;J and it is absent at the poles e = I' where 1/ is parallel to (;J.

Ex 8.4.3. A particle is projected vertically upward with a velocity V from a point 0 on the earth's surface. Prove that when it returns to the horizontal plane through 0 it will have a westerly deviation 4V3~80SA, ), being the latitude of the place 0 and w the earth's angular velocity of rotation. SOLUTION: Let P{x, y, z) be the position of the particle at any time t, then the equation of motion of the particle relative to the rotating earth,

x - 2wysin), = 0; ii + 2w[xsin), + zcos),] = 0; z - 2wycos), = -g where), is the latitude of the point P{x, y, z). The given initial conditions are x(o) = y(o) = z(O) = 0; x(O) = y(O) = 0; z{O) = V. Integrating the first equation and using initial condition we get

x - 2wy sin), = constant

= 0 => x = 2wy sin ),.

Integrating the third equation and using the initial condition, we get

z - 2wy cos),

= -gt +

Substituting the values of x and

constant => z - 2wy cos), = -gt + V.

z in the second equation and neglecting w2 , we get

ii + 2w[sin )'(2w sin),) + cos )'(2wy cos), + V => ii + 2w{V - gt) cos), = 0 => y + 2w{Vt - ~gt2) cos), = constant = 0 1

- gt)]

=0

1

=> y + 2w{ '2 V t 2 - '6 gt 3 ) cos), = constant = O. Substituting for

y in the expression for x and neglecting w2 we obtain, x

= 2wsin)'[2wcoS)'{~Vt2 - ~gt3)] = 0 =>

X

= constant = O.

256


Substituting for

y in the expression for z and neglecting w 2 we obtain,

z= V

- gt

=}

z - Vt+

~gt2 = 2

constant

= O.

Let at· time t = T the particle returns to the horizontal plane, then

1 2V VT - 29T2 = 0 =} T =

g'

At time T =

2:,

the deviation is given by

y

= -2wcOSA[~ 4V3

_

2 g2

f!. 8V3] = _ 4V3wCOSA 3g2

6 g3

Since this is negative, the particle suffers a westerly deviation of amount 4V3~l)s>,.

Ex 8.4.4. A particle is fired at latitude .X with velocity V directed towards the west and at an angle 0 with the horizontal. Prove that the time taken to reach the maximum is ~ sin 0 - 2~f cos Asin 0 cos O.

SOLUTION: Let P(x, y, z) be the position of the particle at any time t, then the equation of motion of the particle relative to the rotating earth,

x-

2wy sin A = OJ ii + 2w[x sin A + Z cos A]

= OJ Z -

2wycosA

= -g

where A is the latitude of the point P(x, y, z). The given initial conditions are x(O) = y(O) = z(O) = 0 x(O) = OJ y(O) = -V cos OJ z(O) = V sin O. ~ntegrating the first equation and using initial condition we get x - 2wy sin A = constant

= 0 =} x = 2wy sin A.

Integrating the third equation and using the initial condition, we get Z - 2wy cos A = - gt + constant =}

z - 2wycos A =

Let at time t = T the particle reach the

-gt + V sinO.

m~imum

hight and then Z = 0,

2wy cos A - gT + V sin 0 = 0 V. 2wy =} T = - sm 0 + -- cos A.

9

9

Substituting the values of x and Z in the second equation and neglecting w2 , we get

ii + 2w cos A[gt + V =}

sinO] = 0

Y+ 2w [V sin Ot + ~ gt 2] cos A = constant

=}.y

+ 2w[~ V sinOt 2 + ~gt3] cos A =

= - V cos 0

-V cosOt.


257

Neglecting the term containing w2 , we get the time as T

V.

2wy

= -smO + --COSA 9

9

= V sin 0 + 2w cos A[ - 2w( ~ V sin Ot2 + ~ gt3 ) cos A - V cos Ot] 9

2

9

6

= V sinO- 2w cos AV cos O.t 9 9 V . 0 2wV2 = - SIn - - - 29

COS

\. sm 0 COS 0

A

9

Ex 8.4.5. Foucault's pendulum

If a freely suspended simple pendulum with a heavy bob and a long suspension, is

~--~~-------y

x

m

Figure 8.6: Foucault's pendulum set swinging at the north pole in the given plane in space, then its linear momentum perpendicular to the plane is zero, and it will continue to swing for a long time in the invariable plane while the earth rotates beneath it. J.L. Foucault [1851] established that, the plane of oscillation of a pendulum is caused by the coriolis force to precess about the vertical axis through the point of suspension. To analyze the motion of Foucault's pendulum, the following assumptions are made (i) The displacement of the bob from the equilibrium position is very small, since the suspension is very large, the motion of the bob will be practically confined in a horizontal plane. As such, z becomes constant and negligible and so z, Z becomes negligible, i.e., Z « :i; or iJ and Z ~ o.

(ii) When the displacement of the bob is very small and suspension is very long, the tension of the suspension becomes practically vertically upward and nearly equal to the weight of the bob. In this pendulum, the size of the bob is of no consequence. Here we discuss the effect of coriolis force, on the oscillation of the simple pendulum. The pendulum is under the action of the following forces :


258

(i) The true force on the pendulum is the downward force of gravity. Thus the real force, i.e., the weight m 9 of the bob is acting vertically downwards and

mg

= mg(-k) = -mgk. ~

(ii) The real force equal to the tension T in the string. When the displacement of the bob is very small, in case of a pendulum of very large suspension, the tension T in the suspension will almost vertically upwards, i.e., T is nearly equal to mg. ~

~

~

- - -

(iii) The fictitious force F Coriolis = - 2m W x 11, where v = xi + iJj + ik is the velocity of the bob at any instant. The angular velocity vector W lies in yz-plane and makes angle () with the y-axis, so W = (O,wcos(),wsin()). Therefore, the coriolis force on the bob is given by ~

F Coriolts

~

= 2m v

~

x w

i

=

j

k

. z o w cos () w sin ()

. 2m x

iJ

= 2mw[( - i cos () + iJ sin ())i -

xsin ()] + xcos ()k].

Now, as coriolis force is always at right angles to W the effective component in the horizontal plane will be due to w sin ()k, where () is the geographical latitude of the place. Hence the x and y component of coriolis components are 2mw sin ()iJ and -2mw sin ()x, which acts perpendicular to the plane of oscillation of the pendulum and is solely responsible for rotation of it. ~

(iv) The fictitious force F Centrifugal = -m Wx (w x 7), where 7 is position vector of the bob with respect to the origin of the co-ordinate frame. The angular velocity of the earth is very small. Since the magnitude of the centrifugal force is proportional to w 2 , it can be neglected.

.

~

(v) The fictitious force F Transverse = -m W X 7. Since the angular velocity of the earth is constant, F Transverse = O. Hence the effective force on the bob of Foucault's pendulum is given by ~

~

~

~

~

FT=FR+Fc m -it = (m 9 + T)

+ 2mw[( -i cos () + iJ sin())i - x sin ()] + x cos Ok].

Assume that the displacement of the bob from the equilibrium position is very small. Since the suspension is very large, the motion of the bob will be practically horizontal, so that z will be constant negligible and so i, Z will be zero. Now the motion of the pendulum is confined to the horizontal xy-plane, so we shall consider only the horizontal component of coriolis force. The equation of motion for the pendulum is ~ mr

= 2m

d7~~ dt x w - V'V

~~~ x w - V'V

= 2m v


259

where V = !k2m(x2 + y2) and k 2 ~ sr-; l =the effective length of swing. Hence the equations of motion in terms of cartesian components becomes,

+ k2x = ij + k 2 y =

x

2w(iJ sin () - i cos ()) ~ 2iJw sin () -2xw sin ().

From this, we see that the equations carry the contributions from the coriolis acceleration in addition to the usual ones of the harmonic oscillator. In order to solve the equations, multiply the second equation by i = P and then add with the first equation we get

(x + iij) - 2w sin ()(iJ - ix) + k 2 (x + iy) = 0 or, ~ + i20e + k 2 = 0; 0 = wsin()

e

=> e=

Ae-i[n+~ ]t

+ Be-i[n-~

=> x + iy = [Ae-i~ t + Bei~ = e-iflt[(A + B) cos Jk2

]t

t]e- iflt

+ 0 2 t + i(A -

B) sin Jk2

+ 0 2 t]

= e-iflt(X + iY) where X = (A + B) cos "/P + 0 2 t and Y = (A - B) sin ../k2 + 0 2 t are solutions when coriolis force is absent and (A!~)2 + (A~~)2 = 1, which is the equation of the ellipse. Thus, due to the rotation of the earth, the plane of oscillation of the bob traces an elliptical path. The factor e- iflt tells that the plane of the oscillation of the bob is rotating with an angular velocity 0 = w sin () in the direction opposite to the direction of rotation of the earth. The continuous change in the plane of oscillation is known as precession. Now we have the following discussions (i) The period of complete rotati.on of the plane of oscillation of the Foucault's pendulum is given by

T _ 211" _

211"

_ 24h sin()

- n - wsin() -

where () is the geographical latitude of the place. At the equator, () = 00 and hence

T=

w

211" =00 sin 00

which shows that, at the equator, the plane of vibration of the Foucault's pendulum does not rotate at all.

(ii) At latitude (), the period of rotation of the plane of oscillation of a Foucault's pendulum is T = wsm 2?," o' Now, 211" w = 24h = 1 day, the period of the earth, so


260

that T = si!oday. If n be the angular velocity of the plane of vibration of Foucault's pendulum, then 21f

21f

.

n = -T = 1/.sm a = 21fsm(} rad/day. At the pole, A = 90 0 , so that

npo1e =

21f sin 90 0 = 21f rad/ day

which is universally true. At pole, the plane of vibration of a Foucault's pendulum rotates through 21f radian in a day. Thus, once an ideal pendulum is set swinging in a given vertical place, it will observed to suffer deflection a precession, in the course of 24hrs, the plane of oscillation will be turned through an angle 21f sin (). (iii) At the north and south pole,

e = 90

0

and 270 0 respectively and

n = ±w.

Thus Foucault's pendulum freely swings in an invariable plane while the earth rotates beneath it. The precessional motion of a Foucault's pendulurr. is then not perceptible to an ~bserver on the north and south pole, but an observer at any other place on the earth, if he were able to watch the pendulum at the pole, could see the plane of oscillation precessing and completing the rotation once a day. Thus Foucault's pendulum is used to demonstrate the rotation of the earth.

Ex 8.4.6. Oceanographic Phenomena The coriolis force -2mTJ x U arising from the earth's rotation, plays a significant role in oceanographic phenomena involving displacements of masses of matter over long distances, such as the river flow on the surface of the earth. Basically, rivers flow approximately in a horizontal plane for which a slight downward slope in the direction of flow is required to maintain the speed of the flow. When the river flows, there is a small component of gravitational force in the direction of flow (forced direction). In sllch a situation, the coriolis force acts across the river flow, which tends to change the direction of flow in the direction of the force. Let the direction of the flow to be the x axis, the transverse horizontal axis to the left of the flow direction as the y axis and the local vertical as the z axis. Let the direction of the flow at a place having geographical latitude A, make an angle ¢ with respect to the geographical north direction. Then the angular velocity TJ of the earth can be expressed as

TJ

= w[sin Ak + cos A cos ¢i

The velocity U of the river flow is U (/ c

= 2u

x TJ

=

= vi

-

cos>. sin ¢)].

and so, the coriolis acceleration is

-2wv[sin >.) + cos>. cos ¢k].

Hence the transverse component is - 2vw sin A, which is independent on the direction (¢) of the river flow but depends on the latitude (A) of the place. Thus the river /


261

deviation occurs to the right of the river flow in the northern hemisphere and to the left of flow in the southern hemisphere. This other effect of the coriolis force also tends to slightly raise the right banks of the rivers in the northern hemisphere than the left banks. The opposite is the case for southern hemisphere. Ex 8.4.7. Change of Meteorological Phenomena (a) Cyclones: When a low pressure zone in atmosphere is generated ay any particular place of the earth, pressure gradient will set up and the pressure gradient

High pressure

Low !Jt'eSRure

Figure 8.7: Cyclones

flows occur towards the low pressure zone for equilibrium of pressm·e. Due to the horizontal pressure gradient the long range motions of air masses take place, which we perceive as winds. In the absence of coriolis force, the direction of the wind velocity occurs perpendicular to the isobars (the equi-pressure lines). On the northern hemisphere, coriolis force acts on the wind to deviate its direction towards its rig aIHI this process is continuous so that the wind flows spirally towards the centre of the low pressure in an anticlockwise direction. The deflection to the right continues, until the wind motion becomes parallel to the isobars. The wind, then continues parallel to the isobars, circulating in anticlockwise direction about the centre of low pressure. Such circulating winds give rise to whirlwinds and cyClones. This cyclone activity is limited to a smaller area on the earth. Since the coriolis force acts perpendicular to the trajectory, it will provide the centripetal force. Let R be the radius of the wind trajectory, the centripetal acceleration is v2

v

R

2w

-=2wv~R=-.

III the southern hemisphere, the coriolis force will deflect the wind motion towards left and so in a whirlwind or a cyclone set up in the southern hemisphere, the wind whirls in clockwise direction. At equator, >. = 0°, so the horizontal component of


262

acceleration due to coriolis force is zero and so no cyclones are found at the equator. (b) Trade Winds and Tropical Winds : If the moderate pressure gradients form over large distances, the wind flow is still from high pressure zones to a low pressure zones, but due to deflection by coriolis force ultimately becomes parallel to the low pressure isobar and the resulting coriolis force just balances the force due to

Low pressure

High 'pressure

Figure 8.8: Trade winds pressure gradient. The wind flow the continuous along isobar around the centre of the pressure. If such a flow occurs in tropical zones, these winds are called tropical winds. The same effect is responsible for the direction of trade winds. The heating of the earths surface near the equator causes the air to rise, and to fill up the vacuum cooler air from north rushes to these areas. Because of the effect of coriolis force, the wind does not flow directly north to south, but deviates towards the west in the northern hemisphere, this giving rise to north-eastly trade winds. In the southern hemisphere, the same effect causes the south-east trade winds.

Ex 8.4.8. Larmor Precession: Larmor frequency has a great importance in atomic and nuclear physics. According to classical electromagnetic theory, electrons revolve around the nucleus in discrete orbits under the influence of an central attractive electrostatic forces exerted on then by the nucleus. Let in a magnetic field, a particle of charge q is moving in an orbit round a fixed point of charge q'. The magnetic force of the moving charged particle is F = q 7 x B, where B is the magnetic density. This frame has the same form as the coriolis force. The magnetic moment AI and the angular momentum L of a system of moving changes, relative to a particular origin are given by

- - -

- -

-

-q---M=2c[r x v];

-

L=m[r x v].

Here both M and L have similar form. Including electric force and magnetic force B, for such a bound system, the equation of motion of the particle of mass" m can

to#f,


263

be written as ~ qq' ~ q ~ ~ 'mr =--er +-['I' x BJ '1'2 e d11 qq' ~ q ~ ~ or m - = --e + - [v x BJ 'dt '1'2 r e

where Er is the unit vector along T7, 11 is the velocity, i.e., absolute time-derivative of the position vector T7. Let the co-ordinate system is rotated about the origin with angular velocity TJ. Then using the expression, 11 = -:t + TJ x T7, we have

d11~ dt = 'I' + 2[ ~~J~~~J w x 'I' + w X [w x 'I' • Thus the equation of motion of the charge in the rotating system is ~ ~J ~ [~ ~J '1'+ 2[~ wxr+wxwxr

qq' ~ q ~ ~ ~ =---2er+-[r + w x '1'] mer me

=-

qq' 2 Er mer

~ X

B

+ (-2 q )2(iJ x (iJ x T7)J; me

~

q~

w =---B. 2me ~

Since the magnetic force that we have eliminated in linear in B, whereas the magnetic term (~)2 [B x (B x T7) ] is quadratic in B, it is possible in the limit of weak magnetic fields to neglect this term. Let us assume that the magnetic field B is weak, such that ~

~

~

~

i.e., it is equivalent to requiring that TJ be much less than any frequency associat.ed with the motion in absence of the magnetic field. Then with this weak field approximation, the equation of motion becomes ~ 'I'

qq' ~ = - mer2 - - er

which is the equation of motion of a particle in an inverse square-law field. Therefore, in a rotating frame, the orbit is precessing ellipse, precessing with angular velocity TJ . Since w2 < < w5, the axis of the ellipse will precess only through a small angle in each revolution. This effect is known as Larmor effect and for the classical gyromaganetic ratio r = ~, the precession angular velocity TJ = ~ B is known as the Larmor frequency. For electrons-2 is negative, and the Larmor precession is anticlockwise around the direction of B. Hence we conclude that the effect of magnetic field on the electron in its orbit round the nuclear remains essentially the same except for a slow precession of its orbit around the field direction. The effect results in an ~


264

observable shift in the ~ectrallines of atom in presence of a magnetic field, known as Zeeman effect. Let 0 be the angular momentum, then ~

dO q~ [~ ~J - = - r X vxB. dt c ~

~

Let us choose B k B and the orbit is to be circular. Thus in the weak field is rotating around B with Larmor approximation, dd~ = - 29,~Jk X "!::1], i.e:,

n

~

frequency with the orbital plane precessing around B. Larmor frequency is found as the rate at which the magnetic moment vector precesses about the magnetic field vector. Larmor's theorem can be expanded to include a collection of charged particles, all with the same charge-to-mass ratio, acted on by mutual electrostatic forces and an external central force directed toward a common centre. This generalization is possible because these forces depend only on the inner particle distances, and the distances of the partiCles from the attracting centre of the external force which must coincide with the origin.

8.5

Exercises

Ex 8.5.1. Let an S' frame be rotating with respect to a fixed frame S having the same origin with the angular velocity V = 2ti - t 2) + (2t + 4)/::, where t is the time. The position vector of a typical particle at an instant t as observed in S' frame is T' = (t 2 + l)i - 6t) + 4t 3 k. Find the acceleration of transport and the coriolis acceleration at time t = 1. Ex 8.5.2. If a particle is dropped from rest at height h, show that any time t the position of the particle is x = 0, y = lwg cos ).,t3, z = h- ~gt2. Find also the equation of the trajectory. Ex 8.5.3. An object is thrown downward with initial speed Vo. Prove that after time t, it is deflected east of the vertical by an amount wvot2 cos () + ~wgt3 cos ().

Chapter 9

Rigid Body Motion By a rigid body we mean, a set of dynamical discrete mass points subject to the holonomic constraints where the distances between all pair~ of constituent points remains invariance throughout the motion and whose orientation does not change under any external force. However, in the strict sense, no material body of the world is perfectly rigid in nature. Before proceeding to discuss the problems in the dynamics of rigid· bodies (which can be solved by quadratures), it is convenient to introduce a number of constants, which can be assigned to a rigid body, depending on its constitution.

9.1

Moments, Products of Inertia

The moment of inertia determine the dynamical behavior of a body. Let a~y rigid body be considered, and let the particles of which it is constituted by typified by a particle of mass mk situated at a point, whose position vector is r+ k = Xki+YkJ + zkk. The scalar quantity

(9.1)

3

where n = either i or or k, is known as the moment of inertia about an fixed axis. Let us introduce the nine symbols lxx, l xy , l xz , l yx , l yy , l yz , l zx , l zy , lzz as n

lxx

= L mk(Y~ + z~) = A k=l

n

lyy

n

= L mk(Y~ + z~) = B;

lzz

= L mk(x~ + Y~) = C

k=l

k=l

n

lxy

=L

n

mkxkYk

= lyx = H;

lyz

k=l

k=l

n

lzx

=L

= L mkYkZk = jzy = F

mkxkzk

= lxz = G

k=l

265

266

CHAPTER 9. RIGID BODY MOTION

where the summation extended over all particles of the system. The three scalar physical quantities lxx, I yy , I zz are called moment of inertia about x, y, z axes respectively and the remaining six scalar physical quantities I xy , I xz , lyx, I yz , I zx , lzy, are called products of inertia of the body associated with corresponding planes. Each scalar component has the dimension [Ml L2]. If in a body, the matter is continuously distributed and the density function is p = p(?7), then the moments and products of inertia can be written as lxx

=

J

p(?7)(r2 - x 2)dT; lxy = -

J

p(?7)xydT

where the integration is performed over the entire volume of the body.

9.1.1

Axes theorem for moment of inertia

The moment of inertia for a rigid body about any axis parallel or perpendicular to the axis passing through centre of gravity, can be obtained by two axes theorem, which are (i) Perpendicular axis theorem. (ii) Parallel axis theorem. (i) Perpendicular axis theorem: For a two dimensional laminated body, if Ix, ly, lz be the respective moment of inertia for rotation about x, y and z axis respectively, then perpendicular axis theorem states that,

+ ly; 1:1; = ly + lz; ly = Ix + lz; lz = Ix

when the body exists in xy plane when the body exists in yz plane when the body exists in xz plane.

This means that the sum of the moments of inertia of a lamina about two mutually perpendicular axes in its plane is equal to its moment of inertia about a perpendicular axis passing through the point of intersection. For a three dimensional body, if Ix, ly, lz be the respective principal moment of inertia for rotation about x, y and z axis respectively, then perpendicular axis theorem states that, Ix

+ ly + Iz = 2 L

m(Op)2

when P is imaginary point at which total mass is supposed to be concentrated, and o is arbitrary chosen point. This theorem is applicable to laminar bodies, i.e., when the particles may be considered as confined to a plane. (ii) Parallel axis theorem: This theorem states that if AB and CD be two parallel axes such that the former axis passes through the centre of gravity of the body then for moment of inertia lAB and ICD about AB and CD axis respectively, ICD

= lAB + Ma 2

where M = mass of the body, a is the normal separation of two parallel axis lAB and ICD. Thus the moment of inertia (1) of a body about any axis is the sum of its

9.1. MOMENTS, PRODUCTS OF INERTIA

267

moment of inertia about parallel axis through the centre of mass and the product of mass of the body by the square of the distance between the two axes. This is known as parallel axes theorem.

9.1.2

Radius of gyration

The dimension of moment of inertia are [M L 2 J. SO the moment of inertia of a body can be equated to product of mass M = E mi of the body and square of some i

appropriate length K. This means that, we can always write

This quantity K in this equation is called the radius of gyration of the given body about the given axis. If we had placed a particle of mass equal to that of the body at a distance K from the given axis, the moment of inertia of the particle would be equal to that of the body. So far as rotation about the given axis is concerned, the body behaves like particle so placed. The concept of radius of gyration helps us to replace a body with extension by a particle.

9.1.3

MI of some symmetrical bodies

1. Uniform rod of length 2a and mass M : (i) About a line through the middle point and perpendicular to the rod = iMa 2 . (ii) About a line through one end and perpendicular to the rod = 1Ma2 . 2. Rectangular plate of sides 2a, 2b and mass M: (i) About a line through the centre and parallel to the side 2a is = 1 M b2 . (ii) About a line through the centre and perpendicular to the plate= iM(a 2 +b2 ). 3. Rectangular parallelopiped of sides 2a, 2b and 2c: About a line through the centre and parallel to the side 2a is = iM(b2 + c2 ). 4. Circular ring of radius a and mass M: (i) About its diameter is = ~ M a 2 • (ii) About a line through the centre and perpendicular to its plane = Ma 2 . 5. Circular plate of radius a and mass M: (i) About its diameter is = ~Ma2. ' (ii) About a line through the centre and perpendicular to its plane = ~Ma2. 6. Hollow sphere of radius a and mass M: About its diameter = ~Ma2. 7. Solid sphere of radius a and mass M: About its diameter = ~M0,2. 8. Elliptic disc of axes 2a and 2b and mass M: (i) About the axis 2a = IMb 2 • (ii) About the axis 2b = ) M 0,2. (ii) About a line through the centre and perpendicular to its plane = ~M(a2+b2).

268

9.2


Independent Co-ordinates

The position of a rigid body is fixed as soon as three non-collinear points in the body are fixed. Any arbitrary point in the body can be localized with reference to these three basic co-ordinates in some convenient co-ordinate system. Thus a rigid body with n particles can almost have 3n co-ordinates. But, for a rigid body, as r 1.j = Cij, where r 1.) is the distance between ith and yth particle and ct ) are constants, the actual number of degrees of freedom can not be obtained by the 3n - ~n(n -1) relations. Thus to find a point in the rigid body, it is necessary to specify its distances to all other points in the body, one needs only the distances at any three ot.her non-colli~lear points. Hence to fix 3-points AI, A 2 , A3 in a body we need nine co-ordinates. But the three reference points are themselves not independent, there are in fact three equations of rigid constraints imposed on them

which reduce the number of independent degrees of freedom to six. Three independent co-ordinates out of six will be required to locate a point on the instantaneous axis of rotation in the rigid body, which is underlying a translation motion relative to some fixed or inertial frame of reference. Two independent co-ordinates are required to locate the axis of rotation passing through the above mentioned point. Lastly, the orientation of the body can be specified in terms of an angle. Therefore, to discuss the motion of a rigid body, we shall use two co-ordinate system (i) Inertial frame, fixed in space and (ii) Non-inertial or Body co-ordinate system, fixed in body. Thus to specify the configuration of a rigid body in space, we need six independent generalised co-ordinates : 3 degrees of freedom represent the translational motion (each particle of the rigid body is the same direction and of equal magnitude) and another 3 degrees of freedom represent the rotational motion (the body rotates about some point itself) of the body.

9.2.1

Generalised Co-ordinates

Let a rigid body be free to rotate about a fixed point. To specify the orientation of the rigid body the Lagrangian formulation needs three independent parameters. When these generalised co-ordinates are known, one can write a Lagrangian for the system and obtain Lagrange's equation of motion. The most common and useful set of parameters are Eulerian angles. In the Eulerian angles, we have a set of three co-ordinates, defined rather unsymmetrically it is true, yet suitable for use as a generalised co-ordinates describing the orientation of the rigid body.

9.2.2

Eulerian angles

So far, we referred the motion of a rigid body to a body frame of axes of which are coincident with the principal axes of the body. These axes rotate along with

269

9.2. INDEPENDENT CO-ORDINATES z

y

x

Figure 9.1

the body. The position of a rigid body, one of whose points is fixed, is completely determined by three co-ordinates, namely, the three Eulerian angles (), Lx = Wx

n

L mk(rk -

L mkXkYk -

xk) - Wy

k=l

Lz

=

=

L

-W,l;

Wz

L mkYkZk

k=l

n

Ly

n

k=l

n

mkXkYk

n

+ Wy L

mk(rk - Yk) - Wz

k=l

k=l

n

n

n

k=l

k=l

k=l

L mkYkZk k=l

L mkXkzk - Wy L mkYkZk + Wz L mk(r~ -

-Wx

z~)

which contains the linear function of all components of (J'. Let us introduce the symbols lxx, lxy, lxz. lyx, lyy, lyz, lzx, lzy, lzz for the coefficients of w x , w y , Wz as n

lxx =

L mk(r~ -

n

x~)

k=l

=L

n

mk(Yk

+ Zk);

lzz

k=l

= L mk(x~ + Yk) k=l

n

lxy

=-

+ z~)

k=l

n

lyy

mk(Y~

= L

L

n

mkxkYk

= lyx;

k=l

lyz

=-

L

mkYkzk

= lzy

k=l

n

lzx = - L'mkxkzk = lxz· k=l

The quantities lxx, lyy, lzz are called moment of inertia about x, y, Z axes respectively and lxy, l~·z, lyx, lyz, lzx, lzy, are called products of inertia associated with corresponding planes. Thus, the angular momenta can be written as

The symmetric second order tensor I which operates on angular velocity (J' is known as moment of inertia tensor, with three diagonal elements, principal moment of inertia where as, the off diagonal elements are product of inertia. In tensor notation we can write, lij

=L

m[ r20ij - rirj]

where Oij is the Kronecker delta. The coefficients lxx, ... can always be calculated if the distribution of the particles about the axis is known.

9.5

Kinetic Energy

The kinetic energy of a rigid body which is in motion, can be calculated in the same way as angular momentum. The kinetic energy of motion of the rigid body is a


274

scalar quantity and is defined by 1~ 2 1~ ~ 12 (9.2) T=2"~mkVk=2"~mklvk k=l k=l where the summation is taken to all the particles of the body and Vk is the linear velocity of the kth particle whose position vector is 7 k with respect to the fixed point o. The expressions for the inertial velocity 11k must come from the general equation 11k = 11 + 01 x 7 ki 11k = velocity of mk relative to the inertial frame, 11 is the velocity of translation of the body relative to the inertial frame and so, 2T

n

n

k=l

k=l

= I: mkv~ = I: mk 11k. 11 k n

= Lmk(11 + 01 x 7 k ).(11 + 01 x 7 k ) k=l n

11

= L mkl 11 12 k=l

+ 11. I: ffik(W

n

x

k=l

n

= Mu 2

+ I: m}.; 7

n

k (11 x

01) +

k=l = Mu = Mu

2

2

+ (11

+ (11

I: mk(w x 7 k).(W x 7 k)

7 k) +

k=l

I: mk(w x 7 k ).(w x 7 k ) k=l

x 01).

x 01).

n

n

k=l

k=l

I: mk 7k + I: mk(w x 7 k ).(w x 7 k ) n

n

k=l

k=l

I: mk 7 k + w. I: mk[(7 k.7 k)W n

(01.7 k)7 k]

n

1 M u 2 + 2" 1 (u ~x~ 2 (~~ )2] ~ T = 2" w). '"' ~ mk ~ T k + "21 '"' ~ mk [2 W Tk W. T k k=l k=l = T(t) + T(m) + T(r) where, T(t) = translational kinetic energy of the whole body at the given instant. T(m) = is the term having mixtures of rotational and translational velocities of individual particles and can therefore be termed as the mixed term. It simply vanishes if either the centre of mass is chosen to be origin of reference frame or the whole body translates in a direction parallel to the axis of rotation of the body AND T( r) is the rotational kinetic energy of the body. Also the rotational KE can be written as n

2TCr)

= ~.[7 k

x (~ x

7 k)] = ~. L mk[7 k k=l

x (~ x

7 k)]

9.5. KINETIC E)reRGY

275

~.,

, '-.. where we can easily put the expression for KE in simpler form if we assume in much that the centre of ljass of the rigid body is at origin of the body. Using the principal axes transformdHoll, we write Iij = Ii8ij and in this case,

.'

which involve the principal moments of inertia Ix, I y , Iz along x, y, z directions. Thus the products of inertia are all zero and the moments of inertia Ix, I y , Iz are the non zero elements. This can be achieved, as Iij is symmetric, by choosing the axes in the body such that the products of inertia are zero. These axes are called the principal axes of inertia. In many practical problems, we are handling the rigid bodies with regular shapes, which helps us to locate the principal axes, (i) For each point of a rigid body, there exists a set of principal axes and the corresponding moment of inertia and the principal inertia. If two of the three principal components Ix, I y , Iz of the moment of inertia about the principal axes are equal, i.e., Ix = Iy -1= Iz then the body is called symmetrical top. In general, bodies for which Ix = Iy -1= Iz behave like homogeneous bodies of revolution about an axis of symmetry. For example, let us consider a right circular cylinder, the axis of the cylinder is taken as z axis. The symmetry of the cylinder shows that Ix = I y. (ii) When all the components of moment of inertia about principal axes are equal,

i.e., Ix

= Iy = Iz

then the body is called spherical top. Such equality may also be observed for bodies of an absolutely arbitrary shape when their mass is properly distributed. For a body with central symmetry such as homogeneous sphere or cube etc. are the examples of spherical top. (iii) When all the components of MI about principal axes are not equal, i.e.,

for all points of the body, then the body is called asymmetric top. (iv) When for a body Ix = I y , but I z = 0, then the body is called rotator ,Le., it has two degrees of rotational symmetry. When we are to e~press the KE T in terms of Eulerian co-ordinates, then T will depend upon angulat velocities ~, 8, -¢ and co-ordinates (},,,p. Since the KE term does not contain 4> explicitly, it will be ignorable co-ordinate, if it is absent in the PE also.

276

9.6

CHAPTER 9. RI6ID BODY MOTION

'"

Euler's Theorem

)

To describe the motion of 8, rigid body completely, we must specify two independent motions of rigid body: (i) a translation motion (ii) a rotational motion. Euler's theorem is one of the basic theorem to confirm the description of the motion of a rigid body. Theorem 9.6.1. A general displacement of a rigid body with one point of which is fixed, is a rotation about some axts through the fixed point. Deduction 9.6.1. The most general displacement of a rigid body is a translation with a rotation about some axis. This is known as Chasle's theorem.

9.7

Euler's Equation of Motion

Let the motion of rigid body with one point 0 of which is fixed,(chosen as our base ---t point) will take place under the action of torque N acting on it. We take principal axes at the body at 0 as a set of moving axes about which the components of angular velocity of the body are W x , w y , W z • Newton's approach leads to a set of equations known as Euler's equation of motion for such a rigid body. Thus the rotational motion of the rigid body in the inerti,al frame of reference attached to 0 is

(9.3) The time derivative in a fixed frame of reference is related to the time derivative of a vector in a rotating frame of reference rotates with an instantaneous angular velocity (;] by the following relation

(9.4) ---t

Here the angular momentum L about origin and the instantaneous angular velocity (;] are quantities that refer to a fixed inertial frame, known as 'body frame of reference'. Hence the equation (9.3) can be written in terms of body axes is

(9.5) which is the appropriate form of the Newtonian equation of motion relative to body axes. Let W x , w y , W z be the components of angular velocity (;] along the principal axes, then for such choice of principal axes, ---t

~

~

~

~

~

~

L = hwxi + 12wyj + 13wzk

---t

=?

dL dt

= hwxi + 12wy .i + Iswzk

(9.6)

9.7. EULER'S EQUATION OF MOTION

277

where the body base vectors i,], k are constants in time with respect to the body co-ordinate system. Now, components of moments of inertia are, however, constant with respect to the body frame of reference, and hence, --+

--+

dL) ( dt T

[dL] dt f

=

[d --+] dt (I w ) f

=

....:...

(9.7)

= I w.

Also, -;;; is same in the fixed and body frames of reference. Hence equation (9.5) can be written as ....:...

--+

--+

--+

(9.8)

Iw+wxL=N

from which the angular momentum will be easily calculated, if we orient the axes of the body frame of reference such that they coincide with the principal axes of the body. Thus the x component of (9.8) is obtained as Nl

=

--+ ";

N.2

•

--+--+

= hwx - (w

x L)x

= hwx - wyL z - wzLy = hwx - (I2 - h)wywz .

Similarly for others. Thus the equations of motion of the rigid body becomes

h~x - (h - h)wyw z :: Nl } 12wy - (h - h)wzwx - N2 hw z - (II - 12 )wx wy = N3

(9.9)

which are Euler's equations for the motion of the rigid body moving with one point of it is fixed. The components of torque and of angular velocity in equations (9.9) are taken along the axes of the body frame of reference, which coincide with the principal axes at the origin. The solution of equations (9.9) will enable us to understand how angular velocity of a rigid body changes with time with respect to the principal axes under the cation of the known torque. Deduction 9.7.1. Rate of work done by the torque: If the torque acting on a rigid body is zero, according to the law of conservation of angular momentum, the angular momentum of the body is constant of motion. Thus, --+

--+

--+

--+

--+

--+

wxL=O::::}wx(Iw)=O. --+

--+

Hence the angular momentum L is parallel to the angular velocity TJ, i.e., L = ITJ, where I is the moment of inertia taken about one of the principal axis. As the torquefree motion of the rigid body ITJ is parallel to TJ, the angular velocity TJ is directed along the principal axis of the body. Hence --+ --+

w .N =

--+

dTJ

w.I dt;

--+ --+

= ~~(TJ.ITJ) = dT 2 dt

--+

w .w x L = 0

dt

278


where the differential takes place either in the fixed or in the body frame of reference and T = ~ (w .JLJ) is the kinetic energy. This is the relation between the rate at which work is done by the torque and the rate of change of kinetic energy with respect to time.

Deduction 9.7.2. Rotation of a free rigid body: The total KE is just the rotational KE if the rigid body is rotating about a fixed point. Let a rigid body --t --t rotating freely, i.e., in the absence of any external torque (N = 0). Hence from equation (9.5) we get, --t

dL --t --t --t Tt+wxL=O --t

dL L. dt

--t

=}

=}

d

1

=}

L2

+

--t

--t

=0 --t

dt(2L2) =0;

--t

--t

L.( w xL)

L=ILI;

= ILI2 = constant.

--t

--t--t

Hence, when w is not parallel to L, and N = 0, then --t

dL dt

--t

+w

--t

x L

#

--t

2

--t

0 and L = 1 L

--t

--t

2 1

= constant.

--t

This shows that, L is fixed in space (as N = 0), but with respect to the body --t frame, only the direction of L appears to change without changing its magnitude. Also, --t

dL w.Tt

--t

--t + --t w.(w x

--t)

L =0

--t

dL dT w '-d = 0 =} -d = 0 =} T = constant. ,t ,t

--t

=}

dli

Thus for freely rotating rigid bodies LJ and are perpendicular to each other and the rotational kinetic energy T is a constant in motion. Hence when a rigid body is rotating about a fixed point, then L2 and T are the two constants of motion.

9.8

Force/Torque-free motion

Let us assume that a rigid body is rotating about a fixed point freely, i.e., the motion of the body about a fixed point not subject to any external forces or torques acting --t --t on it. In the absence of any torques we have (N = 0) i.e., Nl = N2 = N3 = 0 and hence the Euler's equations becomes

9.8. FORCE/TORQUE-FREE MOTION

279

The solution of this equation (9.10) will give the required rotational motion. But two integrals of great importance can immediately be obtained. MUltiplying equations of (9.10) by wx,wy,w z respectively and adding, we get,

=}

hwxWy h w; + hw~

+ hW2W2 + hW3W3 = 0 + hw; = constant = 2T;

(9.10)

Integrating

where T is the KE of the body. Multiplying equations of (9.10) by hwx , hwy, hwz respectively and adding and integrating, we get,

But we have already seen that the components of W, i.e., the angular 'momentum of the body about the principal axes through the base point 0 are hwx , I 2w y , hwz . Therefore, (9.11) Thus there are two constant of motion H2 and T for rigid body motion, rotating freely about a fixed point, in absences of any external forces. Since there are no external force or torque acting of the system, the KE and the total angular momentum vector must be constant in time, and so they are two integrals or constant of motion. Again, if l, m, n be the direction cosines of the resultant of W = (wx, wy , wz ), then

J[2 + m 2 + n 2 V(hw x )2 + (I2Wy)2 + (hw z )2

1. H'

This gives the direction of the invariable lines. From the equations (9.10) and (9.11), we might be able to eliminate only two of the three unknowns wx,wy,wz ' But the solution of the equations in the remaining unknown can only in general be expressed in terms of elliptic functions. Using the identity, w; + w; + w; = 1, we have trom Cramer's rule,

1

h h 2T

h 12 h

Ii Ij H2

Ii Ii Ij

1 1 1

111 2 Al

2T(I2

=

+ h) -

H2

hl3

A~ = 2T(I3 + h) - H2 hh \2 _ "3 -

2T(h

..

+ 12 ) h/2

H

2

280


Differentiating the identity w;

+ w; + w; = 1, we get

where A = w 2 and c is the constant of integration. Thus using initial conditions it is possible to integrate the (9.9) completely in terms of elliptic functions. But this -PTOcess is very laborious.

Deduction 9.8.1. Integration of equation (9.9) under certain conditions : When there are no external forces the Euler's dynamical equation is given by the equation (9.9). We have seen that, the general solution of Euler's dynamical equation involves Jacobean of three fundamental relative function. However there are two cases in which the equation can be integrated without using any relative function. The first one when two of the principal moments are equal and the second, if the motion so started that H2 = 2hT. Case I: Let us take a rigid body symmetrical about z principal axis so that the two principal moments of inertia are equal, i.e., II = I 2 . Hence the equations take the form (9.12) The third equation shows that, the component of the angular velocity W z along the z axis is constant, i.e., W z = constant = n(say). Differentiating, the first two equations of (9.12), with respect to t, we get

hw x = (h - I3)W y W z ~ I{wx = -(h - h)2w y w; .. (h h- h)2 2 ~ Wx = Wz Wx = -0 Wx

i/

where 0 2 = It 3 W z = constant. The differential equation represents the equation of SHM. The solution is Wx = acos(Ot + b), where a and b are constants. Now,

ThiR shows that, the angular velocity components Wx and Wy change in such a manner that the vector c;J p = wxi + WyJ has a constant magnitude a and rotate uniformly about the z axis of the body with angular frequency O. The sense of 0 is the same as that of W z if h > h and opposite to that of W z if h < I3. For finding the position

9.S. FORCE/TORQUE-FREE MOTION of the body in space at any time t, when It equation,

= 12wy = hw z =

11 Wx

281

= 12 ,

we get from Euler's dynamical

= H sin () sin 'l/J O2 = H sin () cos 'l/J 0 3 = H cos ().

resolve part of c;J about 0 1 resolve part of c;J about resolve part of c;J about

Dividing the first two equations, we get,

tan'l/J

Itw x = Wx = -1- = -cot(Dt + b) lWy

~ 'l/J

Also for Ix

= Iy , iJ

=

1r

"2

Wy

+ (Dt + b).

from Euler's first kinematical equation of motion gives

= Wx cos'l/J - Wy sin 'l/J 1r 1r = acos(Dt + b) cos["2 + (Dt + b)] + asin(Dt + b) sin["2 + (Dt + b)] = 0 ~ () = constant.

Also, () = cos

-1

hw z

as hwz = H cos ().

(H);

Also from Euler's second kinematical equation of motion gives

;p = Wx sin 'l/J + Wy cos 'l/J =

H sin () sin 'l/J . nl. H sin () cos 'l/J nl. sm'f/ + cos 'f/

It

It

= Hsin(} ~

¢ = Ht.

It

It

Case II: Let H2 = 2I2T and It > 12 > 13 . If H2 (9.10) and (9.11), we get

= 2I2T,

then using equations

(9.13) Now, by Euler's second equation, we have

(It - h)(I2 - h) [2T _ I 2] It/3 2Wy

hwy = =f ~ Wy

= =fJL(n 2 - w;)

where JL2 = (II-I~~~~2-13) > 0 as It > this equation is Wy

h > hand n =

= ±ntanh(nJLt).

I¥i = i;·

The integral of

282


Thus, when t (9.13), we get

Wx

-+ 00,

= ±n

tanh(nJ-lt)

-+

1, so that, JWyJ

h(h - h) h (h - 13 sech(nJ-lt);

Wz

= ±n

-+

n

=

%as t

-+ 00.

Now from

12(h - h) 1 ( sech(nJ-lt). 3h-h

As t -+ 00, w x , W z -+ 00. Consequently, the axis ofrotation ultimately coincides with the mean axis of 12 axis or the mean axis of the momental ellipsoid. The angular velocity about this axis, tends to the value but the instantaneous axis never actually coincides with it. The equation of the double sign can be effected by noting the initial values of w x , w y , W z .

%,

9.8.1

Poinsot's geometrical representation

Let a rigid body is rotating freely through a fixed point 0, about an instantaneous axis 01 so that the KE T is constant in motion. However, it is possible to describe the geometrical representation of motion of rigid body, known as Poinsot's construction. based on the integrals of motion without solving the problem completely. For a proper choice of the axes in the body, the products of inertia may be made vanish, for this the quadratic surface is (9.14) Momental ellipsoid: axis at 0 be

Let the equation of the momental ellipsoid to the principal

(9.15) where !vI is the total mass of the body and c is sublinear factor referred to the -.. principal axis of the body at O. Thus, if we consider any radius vector OP through z

~~-------+~--y

x

Figure 9.4: Momental Ellipsoid

o meeting the ellipsoid P. Momentum inertia of the body about OP is inversely proportional to the square of 0 P. Let the instantaneous axes of the rotation 01

9.B. FORCE/TORQUE-FREE MOTION

283

meets the ellipsoid at P(xo, Yo, zo), where OP be l, m, n, then

= r.

Let the direction cosines of OP

Xo = I = Wx ,. Yo = m = Wy., Zo = n = r w r W r rwx rwy 1"Wz *Xo=-; Yo=-; ZO=-· W

W

,

wz. W

W

Thus, from equation (9.15), we have, T 2 1lW x2 + .l2Wy

2

+ 12 3W z =

4

2

4w M c'

2

2r

{IT

* (2T).r =Mc.w *w=rv~

which shows that, the resultant angular velocity is proportional through the radius vector. The equation of the tangent plane to the momental ellipsoid is (9.16) Let A, fL"

be the direction cosines of the normal to the tangent plane, then

This shows that, this normal is parallel to the invariable line and hence tangent is parallel to the invariable plane. Hence the equation of the tangent plane can be written as AX + fLY

+

,Z = p.

(9.17)

Now since equations (9.16) and (9.17) represent the same plane, we have

Since p is constant, the tangent plane to the ellipsoid at Q is fixed in direction and at a fixed distance from 0 and hence the tangent plane is fixed in space. Hence it follows that" The motion of the body is such that the momental ellipsoid rolls on the invariable plane with its centre fixed at 0 with angular velocity which is proportional to the radius vector of the ellipsoid in the direction of the instantaneous axis. The radius vector to the point of contact is the instantaneous axis of rotation." The momental ellipsoid stands for the space of the instantaneous angular velocity. Higher the amount of rotation KE the bigger·is the size of this inertial ellipsoid. Ellipsoid of inertia: When a rigid body rotates about some fixed point 0, the


284

instantaneous axis of rotation always passes 'through 0 but its direction changes with time. The MI of the body about these different axes through 0 will, in general, be different. These different values can be represented geometrically in a very elegant way if we define a vector X in the form -+

X

-+

w

= wVi =

-+

w m;

-+

w= Iwl·

Using this notation, equation (9.14) becomes

F(X) =

XT1X =

1

which belongs the angular velocity space in which every points corresponds to a set of values of w x , w y , W z . Also, the expression F(X) is always finite and never zero but

, Figure 9.5: Inertia ellipsoid positive and so it is a closed surface. This equation of quadratic surface is elliptic in nature in w-space and is known as inertia ellipsoid relative to the point 0, which is the centre of the ellipsoid. The axes of the inertia ellipsoid lie along the principal axes of the body which are fixed in the body. This equation looks like the equation of the ellipsoid of inertia provided we identify with X. Now

k

Vx F = 21-+X =

(2 L. m = Vf

2I(;]

Thus (;] will always move such that the corresponding normal to the inertia ellipsoid is in the direction of the angular momentum, where the distance from the centre of the inertial ellipsoid to any point on the surface is ~. Let us draw a tangent plane to the surface of the inertial ellipsoid at the point P. The perpendicular distance of this plane from the origin of the inertial ellipsoid is given by

ON

=

-+

-+

0 P cos (); () = angle between wand L -+

_H

rnrr.

-+-+

w.~ =

m 1(;]11 L I

-+2T

IL 1m

= v~T;

IL I

T=KE


285

---+

FDr all time, the KE T and L are constants of motion, and the tangent plane is so always a fixed distance ON from the origin 0, of the ellipsoid. A line through the ---+ point. in the fixed directioll of L is called invariable line. Also, the normal to the ---+ plane has the fixed direction, i.e., along the direction of L, so the tangent plane is called invariable or fixed plane. Hence we conclude that, the inertia ellipsoid of a rigid body under free rotation always touches the invariable plane, the perpendicular to which drawn from the origin points always toward the fixed direction of the angular momentum of the body.

9.8.2

Polhode and Herpolhode

The motion of a rigid body about a fixed point under no forces, may be represented by the rolling of an Poinsot's ellipsoid fixed in the body upon an invariable plane. The invariable plane is fixed in space. The point of contact of the ellipsoid and the invariable plane lies on the instantaneous axis of rotation and so there can be no slipping at the point. Hence the ellipsoid rolls on the invariable plane. Now we have the following facts ---+

(i) Both the invariable plane and the angular momentum L are fixed in space. (ii) The invariable plane has to meet tangentially the inertial ellipsoid at some or other point and this point of constant determines the direction of the instan- . taneous angular velocity c;J of the body. (iii) As c;J changes with time, this point of contact on the surface of the inertial ellipsoid ought to change with time. We know, for all time, the invariable plane has to remain fixed in space and so the point of contact can change only if the inertial ellipsoid itself changes its orientation with time. Also as the direction of c;J changes continuously with time, the force free motion of the rigid body as being such that the inertia ellipse rolls over the invariable plane, without slipping. As the ellipsoid of inertia rolls without slipping over the invariable plane the point of contact between the ellipsoid of inertia and the invariable plane indicating the direction of instantaneous c;J with respect to the body as well as the fixed plane traces out simultaneously two curves. The curve traced out on the inertia ellipsoid by the point of contact P between the inertia ellipsoid and the invariable plane is called polhode and similar curve on the invariable plane is called herpolhode. The equations of any polhode may be found from the fact that it is the locus of a point on the momental ellipsoid such that the tangent plane at the point is at a constant distance from the centre. If p be the distance from the centre of (9.15), to the tangent plane at (x, y, z), then (9.18)

286


The equations, (9.15) and (9.18) represent, for different values of Hand T, the polhode on the ellipsoid (9.15). Eliminating Mc4 from these two equations, we get,

which describes a cone in the body, called polhode cone. This cone will be real if, and only if, p lies between the greatest and the least semi-axis of the ellipsoid (9.15), or if ~; lies between the greatest and the least of the principal moments h, h, h. The projections of the polhode on the co-ordinate planes, found from (9.15) and (9.18), are

If h > h > h, then the first and the last equations represent ellipses and the middle represents hyperbola, so that the polhodes are closed curves surrounding the axes of greatest and least moment. Now we consider the polhode cone for different values H2 of 2T' (i) If ~; = h, the cone is the line y = 0, Z = 0; i.e., the axis of greatest moment is the instantaneous axis, and the distance of the plane from the centre does not permit of contact at any point but the en ~; > 12 > h, then the cone cuts the plane x = in imaginary lines and the planes y = 0, Z = in real lines, so that in this case the cone surrounds the axis of greatest moment, and therefore so does the polhode.

°

(iii) If ~; = h, the mean moment, the cone is two planes passing through the axis of mean moment, cutting the ellipsoid in two ellipsoid in two ellipses which separate the polhodes round the axis of moment h from the polhodes round the axis of moment h.

°

°

(iv) If 12 > ~~ > C, the cone cuts the planes x = 0, y = in real lines and Z = in imaginary lines, so that the cone surrounds the axis of least moment and so does the polhode. (v) If ~; = h, the cone is the line x = 0, y = 0; i.e., the axis of least moment is the instantaneous axis and only possible for it.

Though the invariable line is fixed in space, as the body moves the line traces a cone in the body and is called the invariable cone.


9.8.3

287

Body and Space cone

Let us consider a torque free motion of symmetric body having its principal moments of inertia It = 12 . described by Euler's equation of motion, such that, Poinsot ellipsoid is an ellipsoid of revolution. The components of (J, i.e., WI, W2, W3 are along the principal axes of the ellipsoid. The KE T = ~ I: 1iw; and the magnitude of the ---'> angular momentum 1L 12 = L2 = I: l]w; arc constants of motion. Let us define k = [31/1; h > It, then the torque free Euler's equation becomes

+ kW2W3 = 0; ~ WI + k2WIW~ = 0; WI

W2 - kWIW3 W2

= 0;

+ k2W2W~ =

= O. W = O.

W3

0;

The solution of the third equation gives W3 =constant = n, say, which indicates the component of the angular velocity in the direction of the inertial symmetry axis of

z

L

OJ

z Body cone

o

o

Figure 9.G: Body and Space cone

the ellipsoid does not change with time. The first two equations give

WI

+ k 2 n 2 wI

+ iW2 = ~ w~ + w~ =

~ WI

2

~w =

= 0; W2

Ae(kntHO)i;

A,

+ iW2)(WI w~ + w~ + w~ = (WI

+ k 2 n 2w2 =

eo

0 constant andi =

iW2) = A2 2 A2 n =

+

V-I

constant.

This shows that the angular velocity vector (J rotates about the body z axis describes a cone with the vertex at origin. Let ab be the angle between (J and the symmetry axis, then W3

cos a b = 1---'>1] ~ tanab W

A = -. W3

Both A and W3 being constants, ab is fixed in magnitude. As ab remains constant with respect to an observer fixed to the body, the instantaneous axis (J rotates about symmetry axis (also called body axis) in the form of a cone about the inertial

288


symmetry axis of the body of the constant semi angle O:b with the inertial symmetry axis of the ellipsoid. This cone is called body cone. Also, A = Iwl sin O:b and W3 = Iwl cos O:b. Thus the polhode on the ellipsoid is clearly a circle about the symmetry axis, and the herpohode on the invariable plane is likewise a circle. With respect to the body co-ordinate system, the angular velocity vector W is also constant in magnitude and it traces out the body cone with respect to the rigid body, whose intersection with the inertia ellipsoid is the polhode. Accordingly, an observer fixed in the space axes sees W move on the surface of a space cone whose intersection with the invariable plane is a hepolhode. Now let us see how an observer fixed in the spase would view this motion of W. In a fixed inertial frame as N = 0, so L has a fixed direction (also magnitude) in space. Therefore, it is easier to locate W with respect to L. Also, ~

~

~

~

~~

2T = w. L

~~

=

constant =*

w.L

IwllLI

= cos O:s = ~

constant ~

where o:~ is the constant angle between the two vectors wand L. Thus the vector --> W rotates about the vector L in a cone with semi-vertical angle O:s given by, cosO: s =

This means that, the instantaneous axis W makes a constant angle O:s with the. axis of L, i.e., moves on the surface of a cone with the direction of L as axis and apex at the centre of mass. This is called space cone. The instantaneous axis of rotation is the line of contact between the space cone and the body cone at any instant. This axis in the body being instantaneously at rest, the body cone rolls without slipping around the space cone. Now ~

~

Since h = 12, so this' shows that the space cone lies inside the body cone when n > 0, i.e., 13 > h and outside when n < 0, i.e., h < h.

9.8.4

Binet Ellipsoid

The Poinsot construction shows how the angular velocity W moves, but gives no information as to how the momentum L vector appears to move in the body system of axis. Here we describe geometrically, the path of L as seen by an observer in the ~

~


289 -+

principal axes system. With respect to the set of axes each of the components of L would involve only the corresponding components of W as

Ll

= hWI; L2 = hW2; L3 = I3w3.

1

2

2

2

Li

L~

L~

=> T = KE = 2[It Wl + hW2 + h W3J = 2It + 2h + 2h

(9.19)

As T is constant, equation (9.19) represents an ellipsoid fixed in the body axes, -+ known an Binet ellipsoid. Now the angular momentum vector L must lie on the sphere

(9.20) follows from the conservation law of angular momentum. For a given initial condition, KE and angular momentum - the path of L is a body given by the intersection of the ellipsoid (9.19) with the sphere (9.20). Also equation (9.19) can be written as

Li L~ L~ 1 -2+ - 2+ - = a a a2 123

where ai = .j2TIi ; i (9.20) it follows that

= 1,2,3 are the semimajor axis of the ellipsoid.

From equation

a; < L2; aj = smallest semi major axis

> L2; aJ = largest semi major axis Let the principal moments of inertia are labelled as h < h < h, then

J2Th < L < J2Th. Hence the momentum sphere of radius L always intersects the KE ellipsoid. Ex 9.8.1. A body moves under no force about a point 0, the principal moments of inertia at 0 being 3A,5A and 6A. Initially, the angular velocity of the body has components WI = n, W2 = 0, W3 = n about the principal axes. Show that at any later time, W2 = ~ tan( ~) and ultimately the body rotates about the mean axis. SOLUTION: Here given that It = 3A, h = 5A and h = 6A. Euler's equation of motion of the rigid body under no forces, with one point of it is fixed is given by,

hWl - (12 - I 3 )w2 w3 = 0; hW2 - (13 - It)W3Wl hW3 - (It - h)WIW2

= 0; =0

=> 3Wl = -W2W3; 5W2 = 3W3Wl; 3w = -WI W2. From the last two equations, we get

9WIWI

+ 5W2W2 = 0 => 9wr + 5w~

= a(say).

290


Initially, WI = n, W2 = 0, so a = 9n 2. Using the first and last equations, we have

Initially, when WI = n, W3 = n, we get b = 0, so that w~ = w~. From the equation 5W2 = 3W3WI, of the Euler's equation, we have,

Using the relation, 9w~

+ 5w~

= 9n 2, we get,

Thus at any time t, the angular velocity components are obtained, which satisfy the given initial conditions. Further, when t -- oo,Le., after very long time period of motion, tanh[fsl -- 1, so W2 -- ~. Using the relation 9w~ + 5w~ = 9n 2 , we get WI = and W3 = 0. Ultimately, body keeps rotating about the mean axis with a limiting angular velocity ~. Now we can find the direction ratios of the invariable

°

line. The direction ratios of the line are

hlP, 12J12 , 13113 , where H

is given by,

Thus, the direction ratios are ]gsech(fs)' tanh(fs)' Jssech(fs)' At t = 0, these are ]g,0, As t -- 00, these become 0,1,0, Le., invariable line, Le., axis of rotation coincides with mean principal axis.

Js.

9.9

Motion of Symmetrical Top

Let us consider the motion of a symmetrical top (like child's top and gyroscope) about the axis of symmetry as an application of the methods of rigid dynamics. The top is fixed at its lower tip 0 and is acted upon by only the gravitational force mg acting at G downward, where OG = l. G is located on the symmetric axis z'. The symmetry axis is one of the principal axes and will be chosen as z' -axis of the co-ordinate system fixed in the body, so h = 12 . As one point is stationary so

9.9. MOTION OF SYMMETRICAL TOP

291

z

z'

~~---------------y

x' Line of nodes

Figure 9.7: Symmetrical top

the description of the motion of symmetric top is completely specified by the three convenient set of generalised co-ordinates as Euler angles : () : it gives the inclination of the z axis from the vertical. cp : it measures the azimuth of the top from the vertical. 'ljJ : The rotation angle of the top about its own z axis. To obtain the solution of the motion of the top, we use Lagrangian procedure. The angular velocities along individual axes we have the components of W as

Wx' = ¢ sin () sin 'ljJ + iJ cos '1jJ Wyl = ¢ sin () cos 'ljJ - iJ sin 'ljJ WZI

= ¢ cos () +

;po

Let h, h, h be the moment of inertia about ox', oy', oz' respectively, where h = I 2 . The KE of the symmetrical top can be written in the form as

T. =

1 2 "2h[Wxl

1

2

+ Wy/] + "2hwz1

=

~h

=

~h[¢2sin2()+iJ2] + ~h(¢cos()+;P)2

[(¢sin () sin 'ljJ + iJ cos 'ljJ)2 + (¢sin () cos'ljJ - iJ sin 'ljJ)2] +

~h(¢cos () +

;P)2

which is the expression for T in terms of Eulerian co-ordinates and it depends upon only one generalised co-ordinate (). Let us take the horizontal plane through 0 is as the zero level, then the potential energy of the top is V = M gl cos (), in terms of Euler's angles, where 9 is the constant vector for the acceleration of gravity. The Harpiltonian H of the symmetric top is 1 '2 2 '2 1· . 2 H = T + V = "2 h [cp sin () + () ] + "2h(cpcos() + 'ljJ) + Mglcos(). Since H does not contain time t explicitly, H = E =Energy and hence it is conserved, Now the Lagrangian function L of the conservative system is given by 1 '2 2 '2 1, , 2 L = T - V = "2 h [cp sin () + () ] + "2h(cpcos() + 'ljJ) - Mglcos(),

(9,21)

292


Taking, e, ¢, 'ljJ as generalised co-ordinates, where ¢ and 'ljJ are ignorable co-ordinates, the Lagrange's equation of motions are (i)dd

(8~)

t 8¢

dd [h¢sin2 e+hcose(¢cose+¢)] =0

=

o=}

=}

h("j; + ¢cos e) = constant = ha(say).

t 2 2 =} h ¢ sin e + hJ; cos e + h¢ cos e = constant = h b( say) d 8L d·· (ii)-d (-.) = 0 =} -d [h(lj, + ¢cose)] = 0 t 8'1/) t

The motion due to change in e is called nutation and that due to change in ¢ is called precession. The general motion of the top about the fixed point 0 is the combination of nutation and precession.

9.9.1

Integrals of motion

Now from (9.21), we see that the Lagrangian function L does not contain ¢ and 'IjJ explicitly, hence they are cyclic co-ordinates. Hence the total energy E and two canonical conjugate momentum p¢ and P'I/J are constants of motion. Only three additional quadratures are needed to solve the problem, and they can easily be obtained from these three first integrals without directly using Lagrange's equation. Hence the two integrals of motion are

8L . 2 . . 2 8¢ 8L .. P'I/J = - . = h['ljJ + ¢cose] = I3Wz = ha; a = constant

P¢ = - . = h¢sin e + h'ljJ cos e + h¢cos e = hb; b = constant

8'ljJ

.

=}

13'1/) . =} ¢

.

= 11 0, =

.

2

.

I3¢COS () and h¢sin e + I3'IjJ cos f} b - a cos e ni. h - b - a cos e cos e . 'f/ = -a 2 sin ( ) ' h sin 2 e

.

+ h¢cos

2

()

= hb (9.22)

from which rb and 'ljJ are known if e is known. Here the two constants of motion P¢ and Pw are expressed in terms of new constants a and b. Another integral is in the above E = T + V = H. Now, W z is constant in time and the term

t;a

{ta

2 equation is also constant. Hence E is also another constant of the motion and so the energy equation can be written as

E

,

=E

If

2

1

'2

2

'2

1·

.

2

If

2

- - a = -h[¢ sin e + e ] + -I3(¢cose + 'IjJ) + MglcosO - - a 2h 2 2 2h 1 b - a cos e 2 . 2 '2 1 h 2 If 2 =-h[( .2 ) sm O+O]+-h[-a] - - a +MglcosO 2 sm e 2 h 2h 1 '2 1 (b-acosO)2 = -2hO + -2h sm . 2 + Mgl cosO 0 1 '2 1 (b - acose)2 = -2he + V'(O); V'(e) = -2 h . 2 + Mgl cos e (9.23) . sm

e


293

which becomes an one-dimensional problem in () co-ordinate, and V' (()) is the effective potential as a function of () alone. The problem of motion of the symmetric top is solved in principle, when we know the dependence of () on time. Hence from (9.23) we get

iJ = dO = dt

~[E'(()) h

! JX = VI;! = -VI;! = _! =-!

'* t(()) =

V'(O)]

d()

[E'(()) - V'(())] sin ()d()

y'2E' sin 2 0 - h (b - a cos ())2 - 2Mgl sin2 () cos 0 du ; y'2E'(1 - u 2) - (b - au)2 - 2Mglu(1 - u 2) du

. a

y'a(l - u 2) - (b - au)2 - (3u(l - u 2)'

=

2E' (3

h '

u

= cos()

=

2M gl

h

du 2 )(1 - u )(a - (3u) - (b - au)2·

The method of solution involves elliptic integrals, but the general nature of the motion can be discussed without performing the integration. Knowing ()(t) ¢ and 'I/J can be evaluated from (9.22) by quadratures. Now, we consider the variation of effective potential energy function V'(()) with the co-ordinate O. The effective potential energy function V' (()) assumes infinite values in 0 < 0 < 7r and V' (()) ~ 00 for () = 0, 7r. Now, 8V'(()) _ M 1 . () J [b-acos()] -(b-acosO)cos() + asin 2 () !)() - 9 sm + 1 . 0' x . 2 () u sm sm . () J (b-acosO)(a-bcosO) = - M 9 1sm + 1 . 3 sm 0 8V'(()) 0 8() > ; for () ~ 7r and a =1= b

< 0;

for ()

~

0 and a =1= b.

Hence a~Jf}) must vanish in 0 ::; () ::; For extreme value of V' (()), we have

7r,

where the minimum value occurs for V'(()).

8V'(()) _ -M 1 . () A.J a - bcos() _ . 8() 9 sm + 'f' 1 sin 0 - 0,

'* -l'vIgl sin() + sm.¢() [h(~ + ¢cos()) -

as

A. _ b - acos() sin2 ()

'f' -

(h¢sin 2 () + h~cos() + h¢cos 2 ()) cosO]

=0

'* - Mgl sin2 0 + ¢[h( ~ + ¢ cos 0) - h ¢ sin2 () cos () - h cos2 ()(~) + ¢ cos ())] = 0 '* -1vIgl sin2 () + ¢[h( ~ + ¢ cos ()) sin2 () - h ¢ sin2 () cos ()] = (j '* ¢2 [h - h] cos () - ~h¢ + M gl = 0 (9.24)


294

If the top has a particular value of energy E~ = E, () motion is bounded motion, confined between the two values ()l and ()2, which are the roots of the equation ,

1

El = M gl cos 0 + - h [ 2

b-acos() 2 . () J. sm

These two points represent turning points and they coincide if the symmetric top has the energy E' = E~, which corresponds to the minimum value of V'(O) for fixed () = 00, then the symmetric top has the constant angular velocity given by ¢ = b-~ cos 2 eo about the vertical axis. 8m 00

9.9.2

Precession without Nutation

The variation in the angle () is referred to as the nutation of the symmetry axis of the symmetric top and is an up and down motion of the symmetry axis. We have seen that, if the top has energy E' = E~, then the motion of the top is restricted between two values of 0, say ()I, 02, known as turning points, where () is the root of the equation E~ = MglcosO + ~h[b-s~~~SeJ2. Thus the axis of symmetry performs nutational motion confined between values ()l and ()2. If the axis of symmetry z' -axis I

I I I I

V(8)

I I I

I I --------- ----------------------------------- - --------r----I

I

:

:

: E'l

I I I I I I I I I I I I



!

: E'z

---------+---------------

i

I

---------------+---------.. ----

I I I I I I I

I I I I I I I

O~----~I----------~--------~I

I I I I I I I

_____'~_____.

°2

1t

Figure 9.8: Precession with nutation is inclined at an angle ()o, the top possesses uniformly about the vertical z-axis, with the precessional velocity ¢o = b-~1o:eo. From equation (9.24), it follows that, for a sm 0 given

eo, there are two angular frequencies of precision ¢* given by . ¢*

[hJ ± -'-V______________ ft#,-J;2 - 4(h - h)coseMgl _ J = ___ 2(h - h) cos 0

=

h

2(h - h) cos 00

[1±


=?

=?

295

b - acos()o 13 sin2 eo = 2(h - Is) cos eo [1 ± b - a cos ()o

sin2 ()o

Is

= 2(11-3COSO I ) () [1 ±

1 - 4Mgl(

h - Is cos eo [2

) - .-

3

'ljJ2

.

]'IjJ (9.25)

The +ve sign of radial precession will be fast while that with negative sign is called a slow precession. Also, ;p* is real if 1 - 4M 1(h - h ) co~ ()o > 0

Ij

9

'2

[2 =?

cos ()o ::; I

1 -

'ljJ2-

3I

'IjJ

.

4Mgl'

3

1["

for ()o
O. Hence j(oo) > O,j(-l) = -(b+a)2 < O,j(l) = -(b-a)2 < 0 and j ((0) < O. When '/1, = ± 1, the roots correspond to a vertical top which is unusual case. Hence at least one of the roots of the cubic must satisfy '/1, > 1, a region that does not correspond to real angles. Physical motion is possible when f(u.) is positive in 1J, E (-1,1), i.e., e E (0,11"). The curve of the effective potential energy indicates that, there are two real roots '/1,1 = cos el, '/1,2 = cos e2 , lies between -1 and +1 and the third root on consideration of equation (9.26), 'U3 = cos e3 > 1 and is not physical interest, Initially, when the top is spinning fast, nutation motion is absent and cos el = cos e2 = cos eo. Also, when iJ = 0, initially, the initial value cos e1 of cos e satisfies equation (9.26), During nutation, the precessional velocity is J; = b~~1(~o, from which it is clear that ¢ changes with e. If Ibl < lal, i.e., Ip¢1 < Ip'l,bl, let us define an angle e* as

cose* With this value of

e > e*,

=!!.; Ibl < lal. a

e, the precessional velocity is given by . ¢

For any

= P¢ P'I,b

=

a -'-2-(cOS e* S111

e

we have, cos e* > cos e and

-

cos e).

¢ has

the same sign of a and that of

9.9. MOTION OF SYMMETRICAL TOP·

297

z

Figure 9.!O: Turning angle

If however, () < ()*, ;p has a sign opposite to W z'. Also, dV:JB) IB=Bo < O. Hence ()* corresponds to the angle satisfying ()* < ()o. Thus for the two angles ()1 and ()2, if fh > 0*, then ;p has the same sign as W z' throughout the nutational motion, whereas if ()1 < ()* < ()2, ;p has opposit.e directions as shown in t.he figures, which show the loci of the axis of the top on a unit sphere. Now we consider the case of symmetric top start.ed with a spin angular velocity W z ' = at· the given angle () = ()1. It is clear that initially iJ = 0 and ;p = o. The constants of motion are

W z'.

;p

In this case, where

()*

= ()1 the motion of the symmetric top is in the following figure z

Figure 9.11: Ocillation of spin axis· nutation

298


In this case, the effective potential energy function is given by V'((J)

= Mgl cos (J + ~h[hwzl COS;l ~ ~WZI cos (J]2 Ism

2

_ ~1' 1 (J lVJ 9 cos

-

+ ~1 2 1 X

_ l§w;1 [2M g l 2h l§w;1 _ l§w;1 [ (J

- - - --cos -

2

I

1

a cos

+

(J

1§ 2 [cos (J1 - cos (J]2 12 WZI . (J 1 sm cos (J)2] sin (J

+ (COS(J1 -

(cos (h - cos (J )2]. _ 2M glh . (J , a [2 2 sm 3Wz1

The angle 01 is one of the roots of the equation V'

l§w;1 [

given by

0 + (cos Olsm!7 .- cos 0 )2] = M 91cos 01 = al§w;1 21 cos 1 1 + a cos 0 sin2 0 = a cos 01 sin2 0 cos 0)2 + 0' sin2 O( cos 0 - cos Ol) = 0 0

-1- a cos 2 1 or, (cos Ol - cos 0)2

or,( cos Ol -

= E',

II

or,(cos Ol - cos 0) [cos Ol - cos 0 - a sin2 OJ

=0

or,( cos Ol - cos O)[a cos 0 - cos 0 + (cos Ol - a)] = 0 1 or, cosO = COSOl; 2a[1 ± }1- 4a(cosOl - a)] 2

or, cos 0 = cos Ol, cos O~, cos O~ where cos Of = 2~ [1 + VI - 40' cos (h + 40'2] > 1, which has no physical value and cos O~ = cos (J2. Thus, when a spinning top initially at rest is released at an angle (h, it falls slightly under the action of gravity and gains in the precessional and the nutational motion. The axis of the top reaches the limiting angle 02 where it has maximum precessional velocity and zero nutational velocity. It describes successive cusps with the turning points on the surface 0 = Ol. The precessional velocity at o = (h has the value ).. _ b 'f'-

a, cos 0 2

sin 0

_

_ 2M gl

-wz1a---. hWzl

Thus the axis of the top undergoes precession very slowly ( as ¢ = 0) and nutation at very large velocity at the smaller angle 0 = Ol. The situation is exactly opposite at the larger angle 0 = (}2.

9.9.4

Steady motion

The motion of the symmetrical top is said to be steady, if 0 = constant = a, say. We know, one of the constants of motion is

..

'I/J

h

+ ¢cos () = h a =

constant

= k(say).


299

This means that the total angular velocity k about the axis of symmetry remains constant, i.e., -J; = constant = n(say). From Lagrange's 0 equation, we have

he - h¢2 sinO cos 0 + I 3(;P + ¢cos O)¢sin 0 - Mgl sinO = 0 ::::} -hn 2 sin a cos a + hknsina - Mglsina = 0 ::::} h n 2 cos (Y - hkn + M gl = 0; a i= 0 ::::} n

=

hk ± JIjk 2 - 4h cosaMgl

2h

= nI, n2·

This shows that there is a pair of real and distinct zeros of n, provided

I§k 2 - 4h cosaMgl > O. Thus, it)!' a sufficiently great spin k about the axis of symmetry the top can execute a steady motion 0 = a about the vertical under two possible precessional angular velocities nI, n2·

.9.9.5

Stability of steady motion

The motion of the symmetrical top is said to be steady, if 0 = constant = a, say. Here we are to discuss the stability of steady motion of the top when the axis is not vertical. From first integrals of motion, we have,

., + ¢cosO

'IjJ

II

= ha=

h¢sin2 0 + I3n cos 0

= hb =

= n(say)

constant constant

= k(say).

Since 0 = constant = a, we have from the second equation, ¢ The Lagrange's 0 equation of motion gives,

=

constant

= w(say).

he - h¢2 sin 0 cos () + h[;P + ¢cos (;I]¢sin () - Mgl sinO = 0 he - h¢2 sin 0 cos 0 + I3n¢sinO - Mgl sin (;I = 0 he + F(O) = 0 where, F( (;I) = - h ¢2 sin (;I cos (;I + 13n¢ sin (;I - M gl sin (;I

() I [k - I3ncosO] . (;I M l . 0 . 0 = - I 1 [k - I3ncosO]2 . 2 sm cos + 3 n sm 9 sm . 2 hsm 0

-- -

cosO [k - I 3ncos 0]2 h sin 3 ()

hsm 0

+ -I3n[k-I3ncosO] . h sm 0

Ml' 9 sm 0.

Now, we calculate P'(()), which is given by,

= [k - hn cos Oj2 + I§n 2 sin2 () - 4Mglh sin2 Ocos() F(a) = -h¢2 sin a cos a + I3n¢sina - Mgl sin a = -Ie = O·, as 0 = constant = a

h sin2 ()P'(O) Also,

l\1gl ::::} J,n = hwcosa + --. w

3h sin()cosOP(O).


300

Suppose that the axis of the top is disturbed slightly from its original position, then e = a + e, where e is very small quantity. Thus,

= P(a + e) = P(a) + ep'(a) = ep'(a); as F(a) = 0 =? hi + f.P'(a) = 0 p(e)

=?

a? I§n2 4M 1 ]C 0 I 1

~.

5M "3¢ sin •

2

~

+ 2M "3 n cos fJ = 5

fJ

+ 2ncosfJ = 2n => ¢ =

2

=> 5¢sin

~

()

5M

.

4n

2

25 A

~

"3 2n

5(1 + cosfJ)

Putting the value of A and using this value of (p, form second equation we get, ·2

2

=>

·2 5a fJ

=>

5afJ

A¢ sin fJ

where we use n

=

·2

JTfi.

As,

5a02

2n + C*n cos fJ = 5A

+

4n2a(1 - cos fJ) 5(1 + cosfJ)

+

6

fJ

gcos

4n 2 a

1 - cos fJ

+ 6g cos fJ = --[1fJ] 5 1 + cos 8n 2a cos fJ

12g cos fJ

5(1 + cos fJ)

1 + cos fJ'

0 is in the direction of the fJ decreasing sense, we have,

= 6g cos fJ[

2

-1]

=

l+cosfJ

=>

0= _

f69

V~

Vf69 ~ o

6gcosfJ(1 - cosfJ) l+cosfJ

cos fJ(l - cos fJ) l+cosfJ

J =- J t

=>

4n 2a

+ -5-

t

1 + cos () dO; cos fJ(l - cos fJ)

dt

Put x 2 = cos fJ

211" cos- 1 (x 2 )

=2

J

1 ~ x 2 dx

= 2 tanh -1 x

1

=>

6 g t = 2 tanh-l [Vcos fJ] => cosfJ = tanh 2 [tJ 3g J. f69 lOa

V~ 5

Ex 9.9.2. The principal moments of inertia of a body at the centre of mass are A, 3A, 6A. The body is so rotated that its angular velocities about the axes are 3n, 2n, n respectively. If in the subsequent motion under no force WI, W2, W3 denote the angular velocities about principal axes at that time t, show that WI = 3W3 = ~sechu and W2 = 3ntanhu, where u = 3nt + ~ log~ 5. Hints: Here the given principal moments are A, 3A and 6A, so the Euler's equation of motion for no force becomes


302

with the initial condition WI = W2 = 3n, W3 = n at t = O. These results gives, 1 3n

t = -tanh

-1

W2 lr,:: (-) - -10gv5. 3n 3n

Now using the relations tanh-I(~) = ~ log ~~~ and u = 3nt required result.

+ ~ loge 5,

we get the

Ex 9.9.3. A solid cube is in motion about an angular point which is fixed. If there are no external forces and WI, W2, W3 are the angular velocities about the edges through the .fi:uod point, prove tha.t WI + w2 + W3 = constant and also WI + + wj = constant.

wi

Hints: Let 0 be the fixed angular point and G be the CG of the cube. The line OG and two lines through 0 perpendicular to OG form a set of principal axis, for which the DCs are (~, ~, ~), (~, - ~, 0), (~, ~, - ~). Obviously the angular velocities about OG and the other principal axes respectively are Wx = WI +~+W3 ,Wy = 2w3 Also I = 2ma 2 l = llm(J2 I = llma 2 Then uSl'ng Euler's wI- W 2 W V2 ' z = WI +W2,j6' , 1 3 ,-2 3' 3 3' equation with no forces and 12 = h, gives the result. Ex 9.9.4. If a rectangular parallelopiped with its edge!; 2a, 2a, 2b rotates about its centre of gravity under no forces. Prove that, its angular velocity about one principal axis ~s constant and about the other axis it is periodic, the period about the first axis is (a 2 + b2) : (b 2 - a2) . Hints: As the body moves under no forces, we use the Euler's equations of motion. Let m be the mass of the parallelopiped, then 11 = W(a 2 +b2) = 12,13 = ~(a2+a2). From the last equation, we get W3 = 0 => W3 = constant = n say.

Thus, W3 = n implies that the angular velocity about one principal axis is constant and is given by TI = ~. Using the other two equations we get,

which shows that WI is periodic with period T2 = ~(~;i!~). Find

R.

Ex 9.9.5. A rectangular plate spins with constant velocity W about a diagonal. Find the couple which must act on the plate in order to maintain this motion. Hints: Let 0 be the centre of the lamina. The unit vectors along the principal axes at 0 are i, k, where k is normal to the plane of the plate and they are fixed in body. Also, It = k mb2 ,12 = kma2,h = W(a 2 +b2) and WI =wcosa,W2 =wsina and W3 = 0, where tan a = ~. Using Euler's equation, we get NI = 0, N2 = 0, N3 = 1 2 a 2 -b 2 "3 mw ab a2 + F ·

3,


303

Ex 9.9.6. A rigid body which is symmetric about an axis has one point fixed on this axis. Discuss the rotational motion of the body, assuming that these are no forces acting on the body, other than the reaction force at the fixed point. Hints: Let us assume that, the axis of symmetry coincides with one of the principal axis, say the direction of k and A = B. Using Euler's equation with no force, we get .. W2

°

+ £"'12 W2 = ; H

£"'12

H

C- A 2 = -A--W3'

This equation can be solved by using the initial condition t = 0, W2 = 0, so W2 = bsin(Ot),wl = bsin(Ot). Hence w2 = constant, which shows that the angular velocity is constant in magnitude and the axis of W precesses round the k axis with frequency 2~' Ex 9.9.7. A circular disc of radius a and mass m supported on a point at its centre. It is set spinning with angular velocity Wo about a line making an angle a with the normal to the disc. Find the angular velocity of the disc at any subsequent time. Hints: We choose unit vector k along normal to the plane of the disc at the centre o of the disc. i and 3 vectors on the plane of the disc are chosen suitably to form a right-handed system. We assume that, the initial angular velocity vector w lies on the k - 3 plane. Let at subsequent time w = WI i + W2) + W3k and initially, WI = 0, W2 = Wo sin a, W3 = wo cos a. For the circular disc, A = B = tma2 and C = !ma 2 . Use Euler's equation of no force to obtain WI = -Wo sin'a sin(wo cos a.t), W2 = Wo sin a. cos(wo cos a.t) and W3 = Wo cos a. Ex 9.9.8. A uniform circular disc of mass m be initially at rest but is free to move in any manner, is suddenly set in motion by an impulse applied to a point A of its circumference. The initial velocity of A has components U along the radius of the circle through A, V along the tangent to the circle at A and W along a third perpendicular. Prove that T, the initial KE of the disc is given by 2T = m[U 2 + iV2 + ~W21· Hints: Let C be the centre of the disc. With C as origin, we consider a set of rectangular axes CXYZ. These are principal axes of the disc and A = B = C = Let w= wli + W2) + W3k be the angular velocity of the disc at A(a, 0, 0), then the velocity components f) = VI i + 1)2) + V3k of C will be given by,

mt,

m;2.

i ) k ("11)c = (l1)A

+Wx 7

= (U, V, W)

+

WI

a

W2 W3

°0

then, the velocity of the eM of the body VI = U, V2 = V + aW3, 713 = W - aW2. The angular momentum about C = (AWl, BW2, CW3), so that the momentum about A is (AWl)i + (BW2 - amv3)) + (CW3 + amv2)k. Since the disc was rest before impulse, using conservation law of momentum about A,


304

Using these equations, find

9.10

aW2, aW3

and then use the formula

Exercise

Ex 9.10.1. Find the moment of inertia tensor of a uniform square plate of side a and mass m, taking the origin at one corner and the x and y axes along the edges. What are the values of the principal moments of inertia? CATE - 1994 Ex 9.10.2. Three particles of mass m each are situated at the corners of an equilateral triangle of side a. Consider the inertia tensor about point 0, the midpoint of base BC. Using symmetry arguments, suggest a set of principal axes. Evaluate the principal moments. CATE -1999 Ex 9.10.3. A body moves under no force about a point 0, the principal moments of inertia at 0 being 6A, 3A and A. Initially, the angular velocity of the body has components WI = n, W2 = 0, W3 = 3n about the principal axes. Show that at any later time, W2 = -vgntanh( vgnt) and ultimately the body rotates about the mean axis. Ex 9.10.4. An elliptic disc ~ + ~ = 1 at rest, is put into motion by making the points (a, 0, 0) and (0, b, 0) move with velocity (0,0, U) and (0,0, V) respectively. v. Find the resulting motion and prove the velocity of the centre is

Ut

Ex 9.10.5. A rigid body is rotating about a fixed point at which A, A, C are principal moments of inertia, under a couple ->'w x , ->'w y , ->'w z , where w x , w y , W z are the angular velocity components under the principal axis. Prove that, at any time t,

where

(J'

=

n(CiA); E,

n, a are arbitrary constants.

Ex 9.10.6. A top an axes of symmetry OC, where G is the CM and it spins with the end 0 on a rough horizontal table. The mass of the top is M and its M I about OC and any axis through 0 perpendicular to OG are C and A respectively. Initially, OC is vertical and the top is set spinning with spin n about its axis. It is then slightly displaced. If the subsequent motion e is the angle OG makes with the vertical and ¢; is the angular velocity about the vertical, show that, A¢ sin2 e = Cn(l - cos e) and A( ¢2 sin 28 + (;i2) = 2M gh(l - cos 8), where OC = h. Ex 9.10.7. A body rotates under no forces about its one point, the principal moments at this point are 4k, 3k and k. Initially angular velocity of the body has components WI = n, W2 = 0 and W3 = nv'2 about the principal axes. Find the angular velocity components at any time t and direction ratios of the invariable line.

Chapter 10

Central Force Field In classical mechanics, 'central force' plays an important role, as the knowledge about central force is important for dealing with several inverse square force, specially, in case of planetary motion. Newton's gravitational force, Ven'dar Waal's force between atoms and molecules in atomic structure, are examples of central forces. This plays also an important role in spacecraft dynamics. Here we shall discuss the characteristics and several applications of central forces.

10.1

Central force

If a force (attractive or repulsive) acting on a particle is always directed from or towards a fixed centre along the line joining the origin to the particle and magnitude of which is a function only the distance from the fixed point, then it is called central force. The fixed point is called centre of force. Both the gravitational force between two masses and the electrostatic force between the two stationary point charges are central' forces. Mathematically, it is written as -+

F

= F(r)r

(10.1)

where r is unit vector in the direction of the force along the line joining the centre of force and the particle. The central forces have the following properties, which are independent of the form of the force law : (i) The force may be repulsive or attractive, which is directed towards or away from the centre of force. Gravitational forces between two mass particles, electrostatic force between two unlike charges, etc. are the examples of attractive forces.

(ii) Central forces, which do not depend an time explicitly, are conservative forces, both have a natural occurance in classical mechanics. (iii) Since central forces are conservative forces, they can be expressed as the neg-+ ative gradient of corresponding potential energy V(r), i.e., F = -~~ r. 305

306

CHAPTER 10. CENTRAL FORCE FIELD

(iv) When a body moves under the action of a central force, its energy and angular momentum are conserved. (v) When a body moves under the action of a central force, its motion is confined to a plane, known as orbital plane. In this orbital plane, the motion lies and it is the normal to the angular momentum. (vi) Only radial dependence in F(r) implies that the isotropy of spacve is preserved by F(r) about the origin. The idea about central force is, in general, applicable in discussion on tides, dynamical manipulations of the orbits of spaceships in the space age, the geometry of the orbits of planets, etc. The motion of a particle under a central force field can be classified as 1. Bounded motion: In such type of motion, the distance between two bodies never exceeds a finite limit. For example, the motion of the planet's around the sun. 2. Unbounded motion: In such type of motion, the distance between two bodies is infinite, initially and finally. For example, scattering of alpha particles by nuclei of a gold foil as the Rutherford experiments. Apply the Lagrange's equation of motion, we shall discuss the problem of two bodies moving under the influence of a mutual central force. Also, under mutual interaction, the two bodies move in such a way that their centre of mass remain fixed in space. Here we take this centre of mass remains fixed in space. Here we take this centre of mass as the centre of force. This is an important part of theoretical physics as this type of motion performed by the planets around the sun, by satellites around the earth, by two charged particles with respect to each other.

10.1.1

Angular momentum

Let us consider, the motion of a particle in a plane under central force. So, the torque acting on it is ---t

T

= ---t r

---t

x F

= F(r)( ---t r

---t

x r)

= ---t O.

---t

Since we have for effective angular momentum L, ---t

---t

T

=

dL

---t

dt = 0 =} L = constant.

So far particle motion in a plane under conservative central force, its angular momentum will remain conserved.

10.2. CENTRE OF MASS

10.2

307

Centre of Mass

Let us consider a-system of n-particles having masses ml, m2,"" m n . Suppose that the particles constituting the system are situated at points in space, whose position vectors are r\, r' 2, ... , r' n respectively. Then the position vector of the centre of mass of the system is given by --t

R =

where M =

L

Li

mi T'i --==----

Lmi

= --M

(10.2)

mi is the total mass of the system. The centre of mass of a system

i

have the following properties (i) The total linear momentum of the system is equal to the product of the total mass of the system and the velocity of the center of mass. Form the equation (10.2), we get --t

MR

~

--t

= L..tmiTi i

(10.3)

--t

~

--t

~

where V = R and v i = T i are the velocities of the centre of mass and of the ith particle of the system respectively. Hence the result fo1l9wS from (10.3). --t

(ii) If there is no force act on the system, i.e., F = 0, then,

L mi 11 --t

:::::} V

2

= a constant vector = a constant vector.

Thus, if no external force acts on the system, then the centre of mass moves with constant velocity and the total linear momentum of the system must be conserved. (iii) The product of the total mass of the system and the acceleration of the centre of mall is equal to the sum of the external forces acting on the individual particles of the system. For this, from (10.3), we get

MV

= Lmitfi i

Ma: = Lmiit = Lmia:i = LFi i

(10.4)

308

CHAPTER 10. CENTRAL FORCE FIELD ~

~

~

~

where a = V and a i = v i are the accelerations of the centre of mass and of the ith particle respectively. Thus, when a system of particles mov~s under the action of external forces acting on the particles of the system, the centre of mass moves as through it were a particle of mass equal to that of the whole system with all the external forces acting directly on it. (iv) Centre of mass of a system of the two particles divides the distance between the two particles in the inverse ratio of their masses.

10.3

Two-Body Problem

Here we shall consider the motion of the two particles which are free to move in space under the influence of mutual forces of attraction or repulsion, acting in the line joining the particles and dependent on their distance from each other. Here we shall assume that, the masses involved are spherically symmetric, homogeneous and Newton's laws hold. Let us consider the interaction of two particles under a mutually attractive force characterised by a potential energy V of the system, which depends only on the position of particles relative to each other, between the two

o Figure 10.1. Two-body'problem particles or their relative velocity. Let --;;71 and --;;72 be the position vectors of the monogenic system two particles of masses m1 and m2 respectively. The KE T and PE V can be written as: 1 ~2 1 ~2 T = 2m1 T 1 + 2m2 T 2j V = V (1--;;7 1 - --;;721) = V (T) where the masses m1 and m2 separated by distance for centre of gravity, then ~

R

~

=

~

1

7n1 ~

~

T.

Let R be the position vector

~

+ 'fn2 (mIT 1 + m2 T 2)

=:';>T1=R+

7n1

m1 +m2

~

~

~

T;T2=R-

(10.5) m1 ~ r. m1 +m2

It is always possible to reduce the problem of the motion of two bodies so that of an equivalent one body problem in a central force field of mutual interaction, by

10.3. TWO-BODY PROBLEM

309

introducing the concept of reduced mass. Such a reduction of two body problem to an equivalent one body problem is found to be very convenient. Now the Lagrangian for such system is given by

Let ]7~ and ]7~ be the position vectors of the masses mI, m2 with respect to the centre of mass, where ]7 = ]71 - ]72 = ]7~ - ]7~, then ~I

'"""

L.J mk r k

~I m2 ~I =~ 0 =} r 1 = - - T 2

m1 m2 ~ ~I m1 ~ =}r1= rir2=r. m1 +m2 m1 +m2 k

~I

Thus the kinetic energy is given by ~

.

T( R, J7)

= KE

eM + KE of motion about the eM

of motion of the

1 ~2 = 2(m1 +m2)R +T1 1 ~2 1 . 12 1 . 12 = 2(m1 + m2) R + 2m1]71 + 2m2r2 1 ~2 1 m1 m 2 .~2 =-2(m1 +m2)R +-2 r m1+ m 2

1

~2

1

~2

= -M R + -J-L r 2 2

where M = m1 + m2 = total mass of the system and J-L =

m 1+m 2

ml

m,2

= (_1_ + ..1..-)-1 m,l m,2 ~

is known as reduced mass of the system. In terms of two variables Rand Lagrangian L becomes ~

~

T,

the

.

L = T(R, J7) - VCr)

=

1 ~2 -M R

2

1 ~2

+ -J-L T 2

-

VCr). ~2

Since V is a function of relative co-ordinates, the first term ~ M R

represents the

~

free motion of the centre of mass. As the three components of R does not appear in L, so they are cyclic and consequently the centre of mass is either at rest or moving uniformly. Lagrange's equations of motion in two variables Rand r will be d(8L) 8L 0 ( z.) dt fiR - 8R = => tt(MR) = 0 =} R = constant which shows that the velocity with which the eM moves in a straight line with uniform velocity. .. ) d (8L) 8L 0 ( Z'/, dlN-8r= => dtd ( J-Lr.) + 8V ar = 0 =} J-Lr.. = - 8V ar = f() r


310

which represents the equation of motion for the system under consideration at a distance T. This equation shew that, the motion of m2 relative to ml is the same as if mi were fixed and m2 were attracted to ml with the force derived from the potential energy mlml +m2 V. Thus, if the two free particles move in space under any law of mutual attraction, the tangents to their paths meet an arbitrary fixed plane in two points, the line joining which passes through a fixed point. Hence the problem of two-body motion has an exact solution. Dropping the first term, the Lagrangian becomes L

=

1 2

~2

-/1 T - V(T)

(10.6)

which is the Lagrangian, if we had a fixed center of force with a single particle at a distance 7 from it having mass /1 = (2+ _1 )-1. Thus under the central ml m2 force the motion of two bodies about their center of mass can always be reduced to an equivalent one-body problem. Now we have the following discussions about the Lagrangian L in equation 10.6. (i) In this Lagrangian, 7 is the position vector of the mass m2 with respect to the mass mI. Hence it represents the Lagrangian of mass /1 is situated at the position of mass m2 and its position vector with respect to mass mi is 7. (ii) In order to appreciate the importance of the reduction of a two body problem to a one body problem, consider that mass ml is very very large as compared to mass m2, i.e., ml > m2. The reduced mass of the system is given as 1

1

1

1

/1

ml

m2

m2

-=-+-~-.

This shows that the reduced mass of the system will be practically the same as the mass of the heavier body. This is used in Bohr's theory of the Hydrogen atom, Satellite motion around the earth, Planetary motion around the sun etc. Further let us suppose that the centre of mass are equal, m2 mlTI = m2T2 ::::} TI = - T 2 mi ::::} Tl ~

0 as

ml » m 2 ·

Thus the centre of mass of the system of two masses m 1 and m2 coincides with the heavier mass mI. For example, for studing the motion of artificial satellites, ballistic missiles or space problems orbiting the earth or the motion of a planet about the sun, the mass of the orbiting body is very much less than that of the central body. Such a reduction of a two body problem to an one-body problem is very convenient and is very often used in classical as well as quantum mechanics. The motion of an electron around the nucleus, before the advent of quantum mechanics, was studied as two body problem in classical mechanics.

10.3. TWO-BODY PROBLEM

311

Deduction 10.3.1. From Newton's laws of motion, we have,

where (! 1 and (! 2 are the constant vectors. If M = pair, then the centre of mass of the system is

ml

+ m2, the total mass of the

which shows that, the centre of mass of the system moves in a straight line in space. Ex 10.3.1. Calculate the changes in the values of energy and angular momentum, when a two body system interacting through gravitational force is reduced to an equivalent one body case. SOLUTION: Let us consider a system of two bodies of masses ml and m2. Let at any instant, -:t 1 and -:t 2 be their position vectors and 111 and 112 are their respective --t --t velocities. If R 1 and V 2 be the position vector and the velocity of the centre of mass respectively, then --t

--t

--t--t

Ii = ml r 1 + m2 r 2 j V = ml v 1 + m2 v 2 ml

+m2

ml

+m2

Now, the reduced mass J..L of the system, is J..L = :::'~2 and let the reduced mass J..L is located at the position of the body of mass m2 and -:t is the position vector of the reduced mass w.r.t. mass mI. Then, --t

--t

--t

...!.t

...!.t

...!.t

r=r2- r l=?r=r2- r l --t --t --t =?V=V2- V I,

where 11 = 7 is velocity of the mass m2 or the reduced mass w.r.t. mass mI. Let be the distance between the two bodies. Then the total energy E and E' of the system of two masses and reduced mass are given by, r

Therefore, the change E - E' ·in the energy, is given by,

312


which is the KE of the centre of the mass system. Therefore, on reducing a two body system to a one body case, the total energy of the system decreases by an amount equal to the KE of the centre of mass. Let us now calculate the corresponding change in the angular momentum. The total angular momentum of the system of two bodies is given by, ~L

~

= r 1

~

~

~

x ml v 1 + r 2 x m2 v 2

and the angular momentum of the reduced mass is given by

The change in the angular momentum is given by ~

~,

~

~

~

~]

ml v 1 + r 2 X m2 v 2 mlm2 [~ ~ ---+ ~ ~ ~ ~ ~ 1 rlXV1+r2XV2-rlXV2-r2XVl ml+m2 1 [~ ~ ~ ~ ~ ~ ~ ~l - - - ml r 1 x ml v 1 + m2 r 2 x m2 v 2 + ml r 1 x m2 v 2 + m2 r 2 x ml v 1 ml + '!n2 m 1 T 1 + m2 T 2] [m 1 v 1 + m2 v 2] = (rnl + m2 ) [ x ml +m2 ml +m2

L - L = [r

1 X

~

= (ml

~

~

+ m2) R x V

~

~

=R

~

~

~

x (ml + m2) V ,

which is the angular momentum of the centre of mass about point O. Therefore, on reducing two body system to a one body case, the angular momentum of the system decreaSes by an amount equal to the angular momentum of the centre of mass.


10.4

313

Lagrange's equation of motion

Let the origin be centre of force and a particle of reduced mass J.t at a distance r, the separation distance between the two particles. Let us describe the position of the hypothetical particle in spherical polar co-ordinates (r, e, ¢), then x = Tsinecos¢; y = l' sine sin ¢; Z = TCOSe X = rsin ecos ¢ + r cos ecos r sin esin ¢¢

¢a -

y = rsin esin ¢ + l' cos esin ¢O + l' sin ecos ¢¢

z = rcos e- l' sin eo KE=

~/j,[±2 + y2 + z2]

=

~J.t[r2 + r 2iP + 1'2 sin
dA 2

dt

=

~r2d() 2

dt

=

~h 2

dA

=> dt = Constant. Thus the conservation of the angular momentum is equivalent to saying the areal velocity of the particle in a central force field is constant. Hence the rate at which t+d/

o ----'-------11--Figure 10.2: Areal velocity the area is swept out by the radius vector is constant, which is Kepler's second law of planetary motion. Hence the conservation of areal velocity is not only restricted to an inverse square law of force but is a general property of central force.

10.4.1

Expression for total energy

From the Lagrange first equation of motion 10.7, we have the second order differential equation in r as

.

l J.Lr2

()= -

315

lOA. LAGRANGE'S EQUATION OF MOTION

This first order differential equation in r is known as energy integral and it solve for the motion. Here! f..Lf2 represents the KE, 2~:2 the centrifugal potential energy (energy due to rotational motion) while V(r) represents the gravitational potential energy of the system. Since the forces are conservative, we get another first integral of motion, which is actually the total energy E of the system. Thus

-

E

12

1

= 2f..Lf2 + 2f..Lr2 + V(r)

(10.8)

= ~f..L[f2 + r 202] + V(r)

= T

+V

= Constant

which shows that the conservation of energy E can be derived directly from the equation of motion.

10.4.2

Potential energy

By orbit, we mean a scheduled path of any object moving under a central force. The equation of the orbit in the differential form is found to be a great use. The one dimensional equation of motion of the particle moving under a central force is

f..LT

.

dV dr

= f..LrrP - -

=-

[2

f..Lr 3

+ F(r) = Fe

(10.9)

where the additional term ~ is known as centrifugal force. This equation of motion in r, with iJ expressed in terms of [, involves only r and its derivatives. An equivalent statement can be obtained from the conservation theorem of energy. Equation (10.9) can be looked upon as a one-dimensional equation of motion in r with the effective potential energy Ve. As the centrifugal force does not arise out of interaction between two particles, so it is not strictly a force and is known as fictitious force or pseudo or false force. It arises due to a accelerated motion of the particle in the orbit. The effective force Fe and the the effective or fictitious potential energy Ve are given by

Fe

[2

= F(r) + -f..Lr3 ; Ve(r) = -

J

Fe dr

[2

= V(r) + -f..Lr 22'

When the particle situated at 00, Le' r --+ 00, then the attractive force is asymp2 totically vanishes. Also, the terms 2!r 2 in Ve(r) is the potential energy associated with centrifugal force. Now the total energy for such a system is

E

= ~f..Lf2 + Ve(r).

t

This equation for energy suggests that the radial KE f..Lf2 and the effective PE for the radial motion in Ve(r). Again, Ve(r) consists of two parts, namely V(r), which is the actual PE and ~ is the centrifugal PE for the radial motion. If the motion is bounded in r, then f vanishes at the extreme values of r say 7' = rmin, r = rmax. Both of these bounds must exist for a bounded motion, Hence

!

E

= Ve(r)

at r

= rmin, rmax

and E - Ve(r)

= ~f..Lf2 > 0, "ir.

316

10.4.3


Integrals of motion

To solve the equations of motion, a total four integrations are needed, as there are two variables rand o. The first two integrals are J.Lr 2iJ = l, and the equation (10.8), the remaining two integration can be accomplished in a variety of ways. For simplicity, from the equation (10.8) we get,

1

[2

2

"2.'1.r + 2J.Lr2 + V(r) 2

r=

=}

= E [2

1

[-(E - V(r) - -)J2 J.L 2J.Lr2

J r

t

=}

=

dr

(10.10)

1

[~(E - V(r) - ~12 )J2

ro

~

~r

where r(t = 0) = TO. From this we calculate r = r(t), the constants of integration are E, l, roo Using this value of t, we can calculate 0 as

[J 0= t

. de e = -d

t

=

[ -2 J.Lr

=}

J.L

o

dt r t

~()

+ eo

where e(t = 0) = 00. Also the equation of the orbit in the integral form as,

de

iJ

= ...,.dr = t

dr

l -2

2

1

J.Lr [~(E - V(r) - 2~r2)]2

+ -[J r

=}

e

= eo

J.L ro

dr 2 1 r2[~(E - V(r) - 6)J2 ~ 2~r

(10.11) .

which is. the equation of the orbit in integral form. When V(r) is given these integrations are easily performed consisting of four constants E, [, ro, eo. The above two expressions for t and 0 are the two remaining integrations. But practically, the integrals are usually quite unmanageable and in any specific problem it is often more convenient to perform the integration in some other fashion.

Ex 10.4.1. Kepler's problem: For all central force problems, the inverse square law plays an important role. The differential equation of the orbit of a particle subjected to an attractive force F(r) = -f?z, Le., potential force V(r) = -~, is

J.LT = F(r)

[2

k

[2

+ -J.Lr 3 = -r2 -+- 3 J.Lr

where k is a constant signifying the strength of the attractive force be,tween sun and planet, J.L denotes the reduced mass of sun and the planet. Putting 'It = ~, the


317

differential equation of the orbit is d2u J1k . d2y d0 2 + u = ~.e., => d0 2

=> y = u 1

r' + y = 0; rJ1k = Acos(O - (

y=u-

J1k

r

0)

J1k

=> -r = -l2 + Acos(O -

(0)

l2 j J1k

l2 A (10.12) cos(O - (0) l' kJ1 where 00 and A are constants of integration. The value of the cQnstant 00 can be obtained as a value of 0 when y has the maximum value. 0 is called the true anomaly or the angle between the perihelon. The words aphelion and perihelion are generally used instead of apocentre and pericentre when the centre of force is supposed to be the sun. Equation (10.12) gives the orbit of the body moving under the effect of inverse square force field. Comparing this with the standard equation of the general conic section with the origin as focus as = 1 + ecos(O - a), we say that the orbit is a conic section with semi latus rectum L = and eccentricity e = l~:. L j e is the distance of the focus from the directrix. The parameters e and L determine the shape of the conic. In order to know which of the conic sections can possibly represent a planarity orbit, it is necessary to obtain a relation between the orbital eccentricity (e) and the energy (E) of the planet. To calculate eccentricity (e) in terms of the energy (E) and angular momentum l, we have

=> - - = 1 + -

f

0-00

J =- J v'

ljr2dr J2J1(E _ V -

=

2!:2) = -

du ; A A + Bu + Cu 2

=> l2 j J1k = 1 + r

k:

J

J

= 2J1E

rl

+ ~u _

B

= 2J1k, C = -1

2

z2 '

·du

2El2 1 + J1k 2 cos(O -

u2

l2

(0);

where the quantity 00 is a constant of integration determined by the initial conditions, but not necessarily the same as the initial angle 00 can now be identified as one of the turning angles of the orbit. Also, three of the four constants of the integration appear in the equation of orbit. In effect, the fourth constant locates the initial position of the particle on the orbit. Note that, at the ends of the transverse axis of the conic, i.e., apses r = O. The turning points are given by an apsidal quadratic in l' whose roots are say 'T' = 1'1, 'T'), for which cos a = ±1 and they are given by 1

J1k

-1'1 = -l2 +A;

(10.13)

318


Now the turning points are given by 1

ILk

/l,k

2El2

=+ l2TI l2

1 + ILk 2

.

The value of r, thus obtained correspond to those at the ends of the transverse axis of the conic. Comparing these with (10.13), we get

As E, l, IL and k are known the values of A and e are known. As the eccentricity is a function of E, for possible values of E, we get different types of orbit as : Values of E E> 0 E=O vemin < E < 0

1 k2

,

E=~mm=-2~

Values of e Nature of motion Nature of Orbit Non-parabolic Hyperbola e> 1 Bound and Non-parabolic Parabola e=l Ellipse Parabolic O<e 0 at r = ro gIven by

d~(r)

dr

= O.


324

Let the equation of the orbit under any central force f{r) is given by

d2 u J-l d(}2 +u=-[2u2F{u) where u

uo,

2

d u dJj'I

=

~ and f{~) 0 . Hence

=

= F(u).

Now for a circle of radius r

= ro,

i.e., u

=

r~

=

Hence the equation reduces to

For a small displacement along a radial direction, we take u = u + 0: where 0: is a small perturbation to u, i.e., 0: « Uo. Then the differential equation of the orbit becomes

d2u d(}2

u5 F{uo+o:) F{uo)· (uo + 0:)2 Uo 0:. 2 = --F() .F{uo + 0:)[1 + - j Uo Uo Uo , 20: = --p() [F{uo) + o:F (uo) + ... ][1- Uo Uo

+ Uo + a = -

+ ...l.

Neglecting the squares and higher powers of Uo and keeping the angular momentum constant we get,

d2u, d(}2

+ Uo + 0: = d2u d(}2

F'{uo) -Uo + 0:[2 - F(uo) uo]. F'(uo)

= -0:[3 - F(uo) uo] = -m

2 ll'.

The general solution of the differential equation is 0:

= C1 cos m(} + C2 sin m(}j m > 0 = qcoshm(} + c2sinhm(}j m 2 < 0 = q(} + C2j m = O.

,

The first solution is interesting but the other solution 0: increase indefinitely with continuously increase of (). The approximation to the path is

u

= un + Cl cosm(} + C2 sin m(} =

Uo

+ Acos(m(} + B).

10.5. GENERAL EQUATION OF THE ORBIT

325

Thus for such small deviations from the circularity conditions, the particle execute simple harmonic motion in u about uo, where A is the amplitude, that depends on the deviations of the energy from the circular orbit and m 2 = [3 - ~(~~? uo]. Hence the orbit will be elliptic close to the circular orbit. Hence a circular orbit of radius r = TO = is stable if m > 0, i.e.,

;0

F'(uo) F( uo )

3 uo

< - = 3ro

d

=> [du log F(u)]u=uo < 3ro

J3 -

with the apsidal angle 7r/ ~(~~?uo. Under this condition, the maximum and minimum values of u are a + A and a - A. Hence the new path will be ellipse with semi-axes given by the reciprocal of a - A and a + A. Ex 10.5.2. A system is in motion under no force and its K E is ~ (A0 2+ B J>2), where A, B are functions of () alone. Show that a steady motion is possible in which () = a, and J> = w, a constant, provided that in the position () = a, B is a maximum or a minimum and further that this state of motion is stable if B is a maximum and unstable if B is a minimum.

SOLUTION: The KE and PE in motion of the particle are T = ~(Ae2 V = 0 and so the Lagrangian L is given by

L

+ BJ>2)

and

= T - V = ~(Ae2 + BJ>2).

As A, B are functions of () alone, so 4> is ignorable. Therefore,

aL

..

{3

aJ> = constant = (3(say) => B4> = (3 => 4> = B(())

The Routhian function R is given by, R

=L

. - (34>

1'2

'2

= "2(A() + B4>

. ) - (34>

1

'2

1· {32

= "2 A () -"2 B(())"

The equation of motion gives,

d aR dt ( ae)

aR

- a() = 0 => ..

d· {32 , d() (A()) - 2[B(())]2 B (())

1"2

(32

=0

,

=> A(())() + "2A (())() - 2[B(())]2 B (()) = o. Steady motion is possible at () = a is possible, if jj = O. From the above equation, we see that, B'(()) = 0 for () = a is possible if B(()) is either maximum or minimum. In that position 4>

(3

= B(a) =

constant

= w.

326


Now, to examine, the stability of the motion, take, () compared to unity. Thus, ..

1

1'2

..

1

I

fJ2

I

A(())e + '2A (())e - 2[B(())J2 B (()) '2

A(a + O~ + '2A (a + ~)~

=

a

+ €,

where ~ is small

=0

fJ2

- 2[B(a + ~)J2 B

I

(a +~) = O.

Expanding in Taylors series and neglecting the square and higher order terms, we get,

Since A(a) > 0, it follows that the motion is stable if B"(a) < 0 and unstable, if B"(O') > O.

10.6

Motion under arbitrary potential field

Here solution for a specific force field, we use only the equations of motion and conservation laws, without performing the integration. To discuss the motion of a particle in an arbitrary potential field, consider a system of known energy and angular momentum, the magnitude and direction of the velocity of the particle can be written as

r=

J~[EVel = J![E- V 11 m

_l2_l. 2 2mr

The conservation of angular momentum gives the angular velocity iJ and this together

l'

Figure 10.3: Curve of arbitrary potential field

10.6. MOTION UNDER ARBITRARY POTENTIAL FIELD

327

with r gives both the magnitude and direction of r. In the figure, we draw an arbitrary potential energy continuous curve. At the points A, B we have r = 0 and they are correspond to the intersection of Ve = E = constant, and on the potential energy curve. Such points where the radial velocity r = 0 are defined as turning points. At the intermediate points between A and B, we have

which is the KE of the particle at that point. Now we discuss the motion of the particle at the different range of the value of r. Case 1: Let r < rl. In that region the PE Ye > E, the total energy, so that r will be imaginary, as KE is '-'ve. Hence the region is forbidden for the particle. Even if E < Ve , the particle always has some probability of penetrating the potential barrier in quantum mechanical description. This is known as turning effect. Case 2 : Let l' E h,1"2], in this region, E > Ye. The points l' = 1'1 and r = 1·2 are the turning points. The motion of the particle is, therefore, oscillatory in the potential wall and the particle WIll be confined to the region. The particle does

Figure 10.4: Bounded motion not possess enough KE to cross over the potential barriers on either side. The orbit of the particle may not be closed and might be bounded in the region between two circles of radii rl and r2. rl rv r2 is called the apsidal distances. In quantum mechanical description, the particle does not possess any arbitrary energy E in the potential well, but the energy will have discrete values. Case 3: Let r E (r2, r3). Here also Ye > E and like case 1, the region is forbidden for the particle. Case 4: Let r 2 r3. In that region, r = r3 is the only turning point. A particle initially at an infinite distance from the center of force is able to approach up to r = 1'3 where it gets rebound and proceeds unhindered to infinity ag tin. This phenomenon is known as scattering. Thus two regions [1'1,1'2] and r 2 '13 are respectively the bounded and unbounded motions. With the help of the arbitrary potential energy curve, we are to understand the nature of the orbits.


328

~~----------~-------.x

Figure 10.5: Scattering

10.7

Kepler's Law

Kepler's law of planetary motion were (i) Each planet moves round the sun in an elliptic orbit, with sun situated one of the foci. (ii) The radius vector drawn from the sun to the planets sweeps out equal ares in equal time. (iii) The square of the time period of the planet around the sun is directly proportional to the cube of the semi major axis of the elliptical orbit of the planet.

Deduction 10.7.1. Kepler's Law from Newton's law: For the verification of the third law, let a and b be the major axes of the orbit. Then the area of the orbit is nab and the rate of describing area ~h. Hence the periodic time T, i.e., the time of one revolution along the orbit, is

nab 2nub - h/2 - .Jiil

T------

=}

T2

4n 2

a

p,

3" = -

-

2nab 2n 3 --a2 p,b 2 /a - VIi

J

= a constant

=} T

2

aa 3

Thm; we have shown that, all the three laws of Kepler can be verified from Newton's law of gravitation.

Deduction 10.7.2. Newton's law from Kepler's Law: Here we are to deduce the Newton's law of gravitation from Kepler's law. Second law of Kepler states that It, is constant, hence T2iJ is constant. But we know that the expression for cross-radial acceleration is ~fh,C,.2iJ) and hence it vanishes. There is no cross radial force, i.e., the force is only radial. Hence the force is directed towards the sun. Again, from the first law, we know that the path is an ellipse, whose equation may be written as 1

l

;;: = 1 + e cos e =} u = y (1 + e cos e) d2 u =}

e

de 2 = -y cos e.

329

10.8. UNBOUNDED MOTION Hence, for the central orbit, the force F per unit mass is given by,

F

= h2

2 2[d u U d()2

Again, for an elliptic orbit, we have = constant. Therefore, we have a

+U T

=

1= h

2 2 u =

l

h

l

2

~

r2'

2~ab, and from the law of Kepler, we get

5-

constant

h2 1

=

=> - =

(27rab)2 47r 2 b2 47r 2 ,2 3 = -,-2- = h2 1 1 a 1 a 2 47r = constant for all points constant

= J1( say)

J1

=> F= 2"' r Thus, for each planet the force per unit mass (i.e., acceleration) is towards the sun and varies inversely as the square of the distance from the sun. Hence Newton's law of gravitation " every particle in this universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely as the square of the distance between them", follows from Kepler's law. .

10.8

Unbounded Motion

The scattering or collisions experiments constitute an important source of information regarding the nature of the interaction between the particles. We assume that, the velocities of the particles are very small as compared to the velocity of light, i.e., the treatment is non-relativistic. The discussion of collision between two bodies is found to be very simple and symmetric, if we describe the effects of collision in two co-ordinate systems : (i) Laboratory system: In general, the collision between two particles is oblique collision where both moving particles approaches to each other in a non-linear manner with respect to their line of impact. After collision, they move away from each other in the same fashion. So, in general case, collision takes place in a plane and obviously, the direct or linear collision is a special case of oblique collision in which particles moves along their line of impact before and after collisions. Experiments are often carried out in which one of the particles is rest in the laboratory and the other approaches it and collision takes place. Such a set up in which a particle collides with another particle is at rest is called laboratory frame of reference. (ii) Centre of mass system: The discussion of collision between two particles becomes very simple and symmetric for motions of particles before and after collision which are taken such that, the centre of mass of the system be rest throughout the collision. Such set up is thus called centre of mass (eM) frame, and, obviously, in this system, the collision between two particles is treated as they have equal and opposite momenta initially.

330

10.9


Scattering

Below, we are discussing three types of collision, in which the law of conservation of linear momentum and angular momentum holds.

10.9.1

Elastic Scattering

The collision between two particles is said to be elastic in nature, if the linear

-> X' ______ , _ _ II,,,-!_

... ______ , _ _ ___

In!

Before collision

Figure 10.6: Elastic Scattering momentum and kinetic energy of the system of the two particles is conserved and the masses of the scattered and recoil target particles are same as those of the incident and target particles respectively. Let, two particles A and B have masses ml and m2 respectively. A is moving with velocity ttl (called incident particle) collide against the particle B moving with velocity tt 2 (called the target). After collision, the particle A moves with velocity 1/1 (called the scattered particle) in the direction making an angle a with the incident direction and the particle B moves with velocity 1/2 (called the recoil target) in the direction making an angle j3 with the incident direction. Using the law of conservation of linear momentum, we have

and the conservation of kinetic energy of the system,

In such a process, the internal nature, or the structure of the particles does not change and without taking into consideration, the internal energies at all, we can apply the law of conservation of energy.

Elastic Scattering in Laboratory SystE'm Let us consider :the incident particle of mass ml moving with velocity ttl along x-direction collides against the target particle of mass m2 at rest tt 2 = 'o. Suppose, after the collision, the scattered particle moves with a velocity 1/1 making an angle

10.9. SCATTERING

331 y

1."

Figure 10.7: Elastic Scattering in L~boratory System

a with the direction of the incident particle, while the target particle recoils with velocity 7J 2 in the direction making an angle (3 with the incident direction. According to the law of conservation of linear momentum, we have ---t

---t

---t

---t

ml U I + m2 0 = ml v I + m2 v 2 =? ml UI i = ml (VI cos ai + vI sin a]) = (ml VI cos a =?

mIVI cos a

+ m2( V2 cos (3i - V2 sin (3]) + m2v2 cos (3)i + (ml VI sin a - m2v2 sin (3)3

+ m2v2 cos (3 = mlUlj

mIVI

sin a -

m2v2

sin (3

= O.

These equations describe the conservation of linear momentum of the system along x and y direction respectively. Now, using the law of conservation of KE of the system, we have, 12121212 "2mIUI

+ "2m20 = "2mIVI + "2 m2v2

=?

1 2 1 2 1 2 "2mIUI = "2mIVI + "2 m2v2 '

The above mentioned three equations contain four unknowns VI, V2, a and (3, but the values of these four unknowns can not be found from the above three equations, either the angle a or (3 is measured experimentally. Deduction 10.9.1. Let ml VI

cos a

=? (UI -

in this case the above equations become

+ V2 cos (3 = Ulj VI sina - V2 sin (3 = 0 and u~ = v~ + v~ VI cos a)2 + (VI sina)2 = (V2 cos (3)2 + (V2 sin (3)2 =? U~ + V~ - 2UI VI cos a = V~ = U~ - V~ =?

=? (UI

= m2 = m,

VI

= UI cosaj

cos a) sin a =

(UI

V2

= UI sina

cos a) sin (3

=?

a

+ (3 = ~.

Thts shows that, if in the lab system, incident particle collides with the target particle of equal masses at rest, the two particles move at right angles to each other after the elastic collision.


332 Elastic Scattering in

eM

System

Let us consider the incident particle of mass ml moving with velocity 1/ 1 with respect to the observer ('If the lab system against the target particle of mass m2 at

--> --> =1 1 - V 0

x' ---It;- - _ "

nn_no

_

V

~ n

G

In!

_ _ _ _- -

tn2

Figure 10.8: Elastic Scattering in eM System -->

rest u

-->

2

-->

= o. Let V be the velocity of the eM of the system of the particles, then -->

ml u

1

+ m2 --> 0 =

(ml

--> + m2) --> V =} V =

m1 ml+ m 2

-->

u

1·

(10.15)

From the equation (10.15), it follows that the eM moves in the direction of the incident particle. Let 1/~ and 1/~ be the velocities of the two particles in the eM system. Then according to Galilean transformations, we have (10.16) (10.17) From these equations (10.16) and (10.17), it follows that the two particles move towards each other in the eM system. Figure 1 represents the velocities of the incident particle, target particle and the centre of mass in the eM system. Let 111 ->

v'!

--"-- --------- X

Figure 10.9 and 112 be the velocities of the two particles with respect to the observer of the lab system. If 11~ and 11~ be the respective velocities of the two particles in eM system, then according to Galilean transformations, --> I

v

1 =

-->

v

1 -

-->

--> I

V and v

2 =

-->

V 2 -

-->

V.

10.9. SCATTERING

333

Using the law of conservation of linear momentum in CM system, we have ml

ml m 2 ::::}

~/

ml +m2

~/

+ m2 u 2 =

~ U 1

ml m 2 - . ml +m2

u

1

~/

~/

+ m2 --+/ v2 ~ ~/ --+/ u 2 = ml v 1 + m2 v 2

ml

v

1

ml~/

::::} V 2 = - - V l ' m2

This shows that, after collision, the two particles in CM system move in opposite directions. Further, the equations suggest as to how the velocity vectors v~ and v~ of the particles after collisions are oriented in the xy-plane. If we draw the velocity --+ vectors VI, 112 after collision in LAB system and V, then the velocity vectors v~ and v~ was shown in figure 10.8. As in CM system the particles move towards each other before collision, and move away from each other after the collision, the collision between two particles in CM system may be depicted as shown in figure 10.9.

10.9.2

In-elastic scattering

The collision between two particles is said to be inelastic in nature, if the linear momentum of the system is conserved (KE is not conserved) and the masses of the scattered and the recoil target particles are same as those of the incident and target particles respectively. An inelastic collision, the KE of the system decreases, i.e., 12121212

rm1'U1

+ 2m2u2 >

2m1v1

+ 2 m2v2 '

The decreases in KE of the particles appears in some other form of energy such as heat, sound, etc. This loss of KE must be taken into account while using the law of conservation of energy. When changed particles interact, they may, due to acceleration acquired as a result of the force of interaction, radiate energy in the form of electro-magnetic waves. Most of the collisions that occur in nature are Inelastic.

10.9.3

Reaction

In some interactions, the masses of the particles may change or new particles may be created or some particles may coalesce into one. Charges are known to take place in the internal structure of the particles, or the total number of particles in the final state may be different from that in the initial state. The collision between two particles is said to be reaction, if the linear momentum of the system is conserved, but the masses of the scattered and the recoil target particles are entirely different from those of the incident and target particles. In case of reaction between the two particles, the scattered and the recoil target particles are better known as outgoing particle and the residual particle respectively. In a reaction, the total KE of the system may be less or greater than the final KE of the system, i.e., 1212

2m1u1

1212

+ 2m2u2 >, < 2m1v1 + 2 m2v2 '


334

10.10

Direct /linear collisions

Two smooth spheres of masses ml and m2 moving with respective velocities UI and Let e be the coefficient of restitution between them and VI, V2 be the velocities of the two spheres just after impact in the direction of UI and U2 respectively. Then the velocity of approach is ttl -11'2 and the velocity of separation is V2 - VI. By, Newton's experimental law of impact, we have, U2 in the same direction impinge directly.

(10.18) Again, since the impulsive action and reaction is along the line of centres, by the principle of conservation of momentum, we have, (10.19) From these equations, we get the velocities impact, as

10.10.1

VI

and

V2

of the two spheres just after

Loss of KE

The loss of KE t:.T, due to impact is given by

t:.T

1

= 2[(ml'u~ + m2'u~) 2(

-

1 ) [(ml + ml+m2

(mlv~

+ m2v~)1

m2)(mlu~ + m2u~) -

(ml +

m2)(mlv~ + m2v~)1

1 2 2

) [(mIUI + m2 U2) + mlm2(UI - U2) 1 ml+m2 - 2( mlm2 ) ( 1 - e 2)( UI - U2 )2 ; 0::; e ::; 1. ml+m2 = 2(

Now we have the following conclusions : (i) For perfectly inelastic collision of bodies, e = 0, and so and mlm2 ( )2 t:.T = 2( ) UI - U2 ml+m2

VI

= ml~!t:~U2 = V2

and this two particles will move together after collisions. (ii) For perfectly inelastic collision of bodies, e = 0, and so m2) + 2m2tt2 u2(m2 - ml) + 2mlUI ; V2 = -'"-----'-'----ml +m2 ml +m2 and so t:.T = 0, i.e., for elastic collision, particles will move separately along the same line of impact after collision without any loss in KE. and this two particles will move together after collisions. VI

= uI(ml -

10.10. DIRECT/LINEAR COLLISIONS

10.10.2

335

Characteristic of direct collision

We see that, for direct collision, between two bodies moving in the same direction before and after collision along their line of impact. The velocities after collision are

°

and the loss of KE for such collision is 6.T = 2f;~~2) (1 - e 2 )(-ul - U2)2 j ~ e ~ l. With these results, we can have the following c aracteristics for direct impact, (i) For ml = m2, VI = U2, V2 = UI, for e = 1, i.e., for elastic collision between two bodies having same mass, the velocity exchange takes place between them. (ii) For elastic collision, e = 1,6.T collision between two particles. (iii) For elastic collision, e initially, then,

= OJ, i.e., no loss of KE occurs for elastic

= 1, if U2 = 0, i.e., if the second particle is at rest

For, ml = m2, VI = 0, V2 = UI, i.e., the first body is stopped and the second one takes off with the velocity which the first one originally had. Both the momentum and KE of the first are completely transferred to second. For m2 » ml,VI ~ -UI and V2 = 0, i.e., when a light body collides with a much heavier body at rest, the velocity of the light body is approximately reversed and the heavier body remains approximately at rest. For m2 « ml, VI ~ UI and U2 ~ 2ul, i.e., when heavy body collide elastically which remains practically unchanged, but the light body rebounds approximately with twice the velocity of the heavy body. Deduction 10.10.1. Maximum energy transfer due to head on elastic collision : Consider a particle of mass ml which collides with another mass m2, initially at rest. For such head of collision, if the initial velocity UI of mass ml changes to the velocity VI, then for U2 = 0, we have VI

=

ml-m2 UI· ml+m2

The fractional loss of the KE of mass 6.T T

ml

for such collision is


336

where x

=

~. ml

The transfer in energy will be maximum if 1

2

'2m2V1

=0*

4x (1 + x)2

=1*

.

:1:

= 1; z.e., m1 = m2

which is the condition maximum energy transfer for head on collision.

10.11

Oblique collisions

Two smooth spheres of masses m1 and m2 moving with respective velocities U1 and at angles a1 and a2 with their line of centres impinge directly. Let e be the coefficient of restitution between them and VI, V2 be the velocities of the two spheres just after impact in the direction making angles (h and (h with the line of impact. Since the spheres are smooth, the impulsive action and reaction will be along the line of centres and, therefore, there will be no change in velocity components in the direction perpendicular to the line of impact. Resolving the velocities perpendicular to the line of impact, we get, U2

VI

sin 81 = U1 sin a1; V2 sin 82

= U2 sin a2·

Again, the velocity of approach is U1 cos a1 - U2 cos a2 and the velocity of separation is V2 cos a2 - VI cos a1. By, Newton's experimental law of impact, we have, (10.20) Again, since the impulsive action and reaction is along the line of impact, by the principle of conservation of momentum, we have, (10.21) From these equations, we get the velocities VI and V2 of the two spheres just after impact, as VI

cos 81 =

V2cOS 82=

10.11.1

U1 'U2

cos a1 (m1 - em2) + m2 u 2 cos a2(1 + e) cosa2(m2 -

m1+ m 2 em1) + m1 u 1 cosa1(1

+ e)

ml+m2

; .

Loss of KE

The loss of KE 6T, due to oblique impact is given by 6T

= ~[(mlu~ + m2u~)

-

(mlv~ + m2v~)]

= 'I[ 2 mlul2( cos 2 al +

. 2al) + m2 u 22( cos 2 a2 + sm . 2a2) sm 2 2 - {m1v~(cos2 81 + sin 81) + m2v~(cos2 (h + sin 82)}]

= '12 [m1 Ul2 cos 2 al + (

1

2 ml +m2

) [(ml

2 cos2 a2 m2 u 2

- {ml VI2 cos2 81 + m2 v 22 cos 2 82}]

+m2)(mlu~cos2a1 +m2u~cos2a2)

10.12. EXERCISES

337

- (ml +m2)(mlV~Cos2(h +m2v~cos202)]

2(

1 Tnl +m2

) [(mlUICOS a l +m2U2COSa2) 2 +mlm2(UICOSal-U2COSa2) 2

- {(ml vI cos fh ml m 2

(

2 ml + m2

10.12

+ m2v2 cos O2)2 + mlm2(V2 cos (h - VI cos 01 )2] 2

2

)(1-e )(Ulcosal-U2cosa2); O~e~1.

Exercises

Ex 10.12.1. In elliptic motion under the inverse square law, the time occupied in describing any arc depends only on the major axis, the sum of the distances from the centre of force to the initial and final points, and the length of the chord joining these points, so that if these three elements are given, the time is determinate, whatever be the form of the ellipse-prove this. This is known as Lambert's theorem. Ex 10.12.2. Describe thR nature of orbits for a particle in motion in an inverse square law force field. Ex 10.12.3. Describe the nature of orbits for a isotropic harmonic oscillator. Ex 10.12.4. The total energy expression of a body moving under the effect of a central force is given by E = ~mr2 + 2;~2 + U. Discuss the centrifugal potential energy of the body. Ex 10.12.5. A particle moving under the action a central force describes a spiral path r = e bO . Find the law of force.

-:!&

Ex 10.12.6. A particle of mass m moves under the action of a central force directed towards a fixed point. Show that the velocity of the particle at any point in its orbit is V = [b(e 2 - 1) + !~P/2, where e is the eccentricity of the orbit and L is the angular momentum of the particle.


Chapter 11

Theory of Oscillations In everyday life we come across various things that move. The motion of physical systems can be classified into two categories - translation and vibration motion. If the position of a body varies linearly with time its motion is translatory. A motion that repeats itself in equal intervals of time is called periodic motion. If a body in periodic motion moves back and forth over the same path, its motion is called vibratory or oscillatory. Oscillations are the most general form of motion of dynamic systems about the equilibrium position. For small derivations from the equilibrium position, the oscillations are harmonic and acquire a special significance. In this chapter, we shall study three types of problems : (i) First, we shall study the simplest and smoothest type of free oscillatory motion of undamped physical systems. (ii) Secondly, we consider the forced and damped oscillation. (iii) Thirdly, we shall discuss, the theory of free oscillations of small amplitude about the position of stable equilibrium that only the fundamental frequencies are excited.

11.1

Equilibrium

While dealing with oscillations executed by various systems, we shall frequently come across the condition of eqUilibrium. For any periodic motion, there exists a position of a particle in which the net force acting on the particle is zero. This is the equilibrium position.

11.1.1

Stable equilibrium

If a slight displacement of the system from its position of equilibrium results only. in small bounded motion about the point of equilibrium, then the system is said to be stable equilibrium. At the extreme point xo at which the potential energy = function V is minimum, if the particle is displaced, then the force F( x) = -

%:

339

CHAPTER 11. THEORY OF OSCILLATIONS /

340

slope of the potential energy curve will tend to return it and it will oscillate about eqUilibrium point, performing bound motion. These points are actually points of stable equilibrium. For example, the bob of a simple pendulum in its equilibrium state satisfies the stability criteria.

11.1.2

Unstable equilibrium

If a slight displacement of the system from its equilibrium results in bounded motion, then the system is said to be in an unstable equilibrium. If the particle is displaced sightly from the point of equilibrium Xo then it will be acted upon by a force F(x) = ~~ which will tend to push the particle away from equilibrium point, when released. For this reason such points are called the points of unstable equilibrium. The egg standing on end is an example of unstable eqUilibrium. Hence the equilibrium is unstable if the extreme value of the PE V is a maximum.

Ex 11.1.1. Simple pendulum: Taking the zero of the potential energy at the bottom of the swing, the potential energy of the bob of mass m of the pendulum at a position of angular deflection o is V(O) = mgl(l - cosO), where I is the length of the pendulum and 9 is the acceleration due to gravity. The equilibrium position of the pendulum is given by,

:~ = mgl sin 0 = 0 => 0 = 0, 7r. Now, ~:¥ = mgl cos 0 is positive at 0 = 0 and negative at 0 = 7r. Therefore, 0 = 0 is the position of stable equilibrium at which the b01tcan hang for an indefinite period, which is not at all possible at the position of uri"'Lable equilibrium at 0 = 7r. The pendulum can oscillate only about the position of stable equilibrium.

11.2 11.2.1

Oscillatory problems Simple pendulum

Simple pendulum is an example of a dynamical system. It consists of a heavy particle of mass m called the bob attached to one end of a weightless string of length l, which oscillates back and forth in a vertical plane, while the other end of the string is attached to a fixed point O. Taking the fixed point 0 as the origin of the co-ordinate system, let the typical position of the particle at time t be given by (x, y) or (r,O) in polar co-ordinates. Here, the particle moves under the action of two forces, namely the gravIty and the tension T of the string. The tangential and normal equations of motion are d2 s m d0 2

or,

=

v2 -mgsinO; mT

zii = -g sin 0;

v2 mT

=T

=T

- mg cos 0

- mgcosO, as s

= Ie

11.2. OSCILLATORY PROBLEMS If () is small, then sin ()

~

341

() and the first equation reduces to .. l()

..

g.

= -g() => () = -1 SIll()

which is a simple harmonic motion of period

T, = 27r/fr = 27r/f If () = a(very small) when iJ = 0, then () at any instant is given by () To solve the general solution for the motion exactly, we have

iJ.

=

a cos

j¥t ..

()

j d(();) = -y j sin()d() o

a '2

2g

=> () = T[cos() - cosal· Since () decreases with t, we have

.

. /2g

() = -Y T[cOS () =>

f29T -

YT

cosal

o

-

-

j ·Jcos() - cos a d()

~ v'2

-

a

Q

j 0

d()

·!sin2 g

V

_

2

sin2!!.

2

where T is the time to reach A. To integrate the right hand side, let us put sin ~ sin ~ sin (AI - gn) L L ~kaklaim = O.

(11.17)

k=li=l

of.

If Al

A::n, then (AI - A::n) is non zero in general, and hence n

n

L

L

~kaklaim = OJ

of. m.

l

k=li=l

Now, we consider Al - A~

=

n

0, i.e., Al or n

terminate. We are to show that

n

wl are real and 2: 2: ~kaklaim is indek=ll=l

n

L: 2: ~kaklaim of. 0, for l = m.

For this purpose

k=ll=l

separate akl into real and imaginary components as

n

With this separated value ok akl the expression

n

L: 2: ~kaklarm can be written as k=ll=l

L k

LTjkakla;m = L

LTjk(Qkl

k

j

=L

+ i{3kl)(Qjm -

i{3jm)

j

LTjk [QklQjl

k

+ {3kl{3jz] + i L k

j

= LLTjk[QkIQJI + {3kl{3jz]j k

as Tjk

LTjk [{3kI Qjl - Qkl{3jz] j

= T kj .

j

2: 2: TJkaklajm > 0 and this sum vanish only if the system is not k j Using this result we also conclude that the KE T = 2: 2: TjkT]kT]j is also

This shows that, in motion.

k

j

positive and

Hence the eigenfrequencies are all real and the motion for a given w 2 will be completely o~cillatory about the position of stable equilibrium. Also, if wl of. w~, then L k

LTjkaklaJm J

= OJ

l

of. m

(11.18)


351

wl

which is called the orthogonal property. Also when the values of are substituted in equation (11.14), we shall determine the ratios of akl. This indeterminacy can be removed by requiring the additional condition L

LTJkaklajm

k

= 1;

l

=m

(11.19)

J

to be satisfied by ak's, i.e., akl are normalized. The equation (11.18) and (11.19) can be written as the orthognormality conditions LLTjkaklajm = 81m , j

k

i.e., aFTa m = 81m ; in matrix notation, where 8kl is the Kronecker delta and al is the vector corresponding to frequency WI. The eigenvectors a r so defined are said to form an orthonormal set. They are orthogonal in the sense of the condition of equation (11.18) and they are normalised in accordance with the condition of equat,ion (11.19).

11.4.2

Normal co-ordinates, Frequencies of Vibration

We see that, the equations of motion will be satisfied by an oscillatory solution of the form (11.12) not merely for one frequency but in general for a set of n frequencies Wk. So a complete solution of the equations of motion therefore involves a superposition of oscillations with all the allowed frequencies. Thus if the system is displaced slightly from equilibrium and released the system performs small oscillations about the equilibrium with the frequencies WI, W2, ... , W n . The solutions of the characteristic equation (11.14) are therefore often designed as the frequencies of the vibration or resonant frequencies of the system. The normal co-ordinates of a system are the parameters in terms of which the equations of motion of the system can be written each equation contains only one dependent variable. If a solution of equations of motion involve only one single frequency, the co-ordinate appearing in it will be called the normal co-ordinate. Thus, the generalised co-ordinates, each one of them executing oscillations of one single frequency are called normal frequency. The general solution of the equations of motion for T/j may now be written as n

T/j

=L

Ckajke-iwkt; i

= V-I

k=1

where for each resonant frequency Wk, there being a complex scale factor Ck' The solution for each co-ordinate is, in general, a sum of simple harmonic oscillations in all the frequencies Wk satisfying the characteristic equation. Unless it happens that all of the frequencies Wk are commensurable, i.e., rational fractions of each other, T/J" never repeats its initial value and is therefore not itself a periodic function of time. From algebra, "when two homogeneous quadratic functions of arbitrary number of variables, one of which is positive for all values of the variables, then by a real

352

CHAPTER 11. THEORY OF OSCILLATIONS

transformation we may clear both expressions of the terms containing products of the variables and at the same time change the coefficients of the square terms in one of the expressions so that each is equal to unity." However it is always possible to find a linear transform of the co-ordinates 1]j to a new set of generalised co-ordinates, that are all simple periodic functions of time, i.e., the KE and PE are expressed only the new generalised co-ordinates in simple way. Let us now define the new set of co-.g.rdinates (J related to the original co-ordinates TIJ by linear relations as

where alj's are constants. From this equation, it is observed that (k = Cke-iwkt are the quantities that would undergo oscillations with only use of frequency Wk. Then the KE reduces to the simpler form in the new co-ordinates as

The PE reduces to the simpler form in the new co-ordinates as V =

1

2L j

=

1

2L r

=

L ~k1]j17k

1

=

2L

k

J

L V)k [L ajr(r] [L aks(s] k

s

r

L[L L V)ka.7raks](r(s s

k

j

~ LLw;[LLTjkajraks](r(s = ~ LW;(;' r

s

k

J

r

It should be remembered that the off-diagonal elements of T and V rise to the coupling between the particles. Thus, when the normal co-ordinates are used, the KE and PE are become simultaneously diagonal and hence this will uncouple the co-ordinates and the problem will become completely separable into the independent motions of the normal co-ordinates. In terms of the new variables, the Lagrangian L can be written as

The Lagrange's equation of motion gives

L[e + W~(k] = 0 k

which are a set of n independent equations. The solution of the single equation is (k = Cke-iwkt. Thus, each of the new co-ordinates is thus a simply periodic function


353

involving only one of the resonant frequencies. These quantity ('s are called the normal co-ordinates or principal co-ordinates of the vibrating system. Each normal co-ordinate corresponds to a vibration of the system with only one frequency, and these component oscillations are spoken of as the normal modes of vibration, Le., the system is vibrating in a normal mode. Further physical properties of all normal modes that when the system is vibrating in a normal mode is simply periodic, passing twice through the equilibrium position in every complete oscillation, and the velocity of every particle of the system vanishes at the same instant twice in every complete oscillation. All of the particles in each mode vibrate with the same frequency and with the same phase, the relative amplitudes being determined by the matrix elements ajk. The complete motion is then build up out of the sum of the normal modes weighted with amplitude and phase factors contained in the Ck'S. Ex 11.4.1. Two Coupled Pendulums Here we shall discuss the motion of two oscillating systems in which the motion of one system can influence that of the other. The two systems are then said to be

o 0' --~~--------------------~------~y

"--____...+-____--'l B X2

mg

Figure 11.1: Two coupled pendulums coupled to each other and the resulting oscillations are called coupled oscillations. The system consists of two identical simple pendulums, each having length 1, a bob of mass m coupled to each other by means of a massless horizontal spring of force constant /-t. The separation between the two bobs is such that the spring is relaxed in the equilibrium position. The system is disturbed slightly from the equilibrium position as shown in the figure and relaxed. At any time t, let Xl, X2 represent the horizontal displacements and YI, Y2 be the vertical displacements of the two bobs from their equilibrium position, such that they move in the same vertical plane. We shall consider, the small horizontal oscillation about the position of equilibrium. The spring is stretched or compressed depending on whether Xl > X2 or Xl < X2. Let us choose (h and e2 as generalised co-ordinates, we get from figure, Xl

YI

= 1sin el; X2 = 1sin e2 = 1(1 - cos el ); Y2 = l(1

- cos e2 ).


354

The kinetic energy T of the two bobs can be written as

. = e. T Te· ,

T

(ml2 0 ) 0 ml 2

=

((h)

·e = (h' ' X2-Xl

I

The potential energy due to extension of spring is

J-lxdx

=

~(X2 - Xl)2 and so

o the PE V of the two bobs measured from their equilibrium position is V

1

= mg(Yl + Y2) + "2J-l(x2 = mgl[(l - COSOl)

- Xl)

+ (1 -

2

cos ( 2)]

+ ~J-ll2(sin02 -

sinOl)2

oi ( 8~ 1 1 2( . . 2 2 -···)+ 1-1+ 2 -",) +"2J-ll sm82-sm8l) 1 2 2 1 2 2 ~ "2 mgl (8 l + ( 2 ) + "2J-ll (8 2 - ( 1 ) [

=mg l(1-1+

=

~(mgl + J-ll2) (0; + O~) -

= ~(Ol' (2) (mgl + 2J-ll2 -J-ll

2

2 J-ll 0l02

(81)

2 -J-ll 2 ) mgl + J-ll

=

82

eTve.

Now the characteristic equation for determination of normal frequency of vibration IV - w 2 TI = 0, which gives

w 2 is

1

=> mgl + /1,[2 => (mg[ WI

=

- w

- J-ll2

+ J-l[2 -

2 2 ml mgl

w2m[2)2

1-

- J-ll2 0 - w 2ml2 -

+ J-ll2

= J-l2[4

If, JT+ ~ Jwi + ~, W2

=

=

are the normal frequencies of vibration. Let the normal co-ordinates 1]1,7}2 corresponding to WI, W2 respectively, then the equations of motion in terms of normal co-ordinates as

WI, W2

The angular frequency WI of the oscillation is the same as that of either pendulum oscillating in isolation. This case arise as both pendulums are always in phase and the spring is at its natural length throughout the motion. N ow we shall proceed


355

to express 1]1,1]2 in terms of generalised co-ordinates{h, rh. Obviously, we are to calculate the eigenvectors aI, a2 corresponding to the normal frequencies WI, w2 of vibration. (i) For WI = let al = (an, a21f, then

j¥,

J

=?

(!D2 _:z;2) (:~~) = (~) =?

an = a21 = a.

Using the normalization condition, we get

JEt + ~

=

(ii) For the angular frequency W2 (aI2' a22f, then

=

(0)

-a22

= (3.

-JLl~ -JLl~) (a I2 ) ( -JLl -JLl a22 =?

U~ing the normalization, we get (3

=

a12

=

=

vwt

+~ >

WI, let a2

0

lvkn. Hence, the normal co-ordinates

1]l, 1]2

corresponds to WI, W2 respectively are given by

~*r ) ((hfh) = (l1m v'2m v'2m 1

=? 1]1

1

(1]1) 1]2

= lJ2mUh + O2); 1]2 = lJ2m(01

- 02).

The two pendulums are oscillating harmonically at the same frequency W2 and are always out of phase. The spring is either extended or compressed during the motion and the coupling is effective. Also, it is clear that the two vectors are orthogonal to each other. The co-ordinates 771 and 7]2 are the desired co-ordinates and they are referred to as the normal co-ordinates of the system because equations of motion in them contain a single frequency of normal modes of oscillation. Now, we shall interpret modes of vibration. (i) To discuss the mode associated with the normal co-ordinate 1]1 and hence with WI, we put 7]2 equal to zero. Thus, 01 = O2 , which means two pendulum oscillate in phase, the frequency of motion is WI = Obviously, the spring constant JL is not occurring in the expression of WI, the two pendulum can vibrate as if they are independent. Hence there is no scratching or compressing of spring

j¥.


356

at any time during the motion justifying the inphase condition. The normal mode of oscillation is called a symmetric mode and the particles are oscillating always in phase. (ii) To discuss the normal mode associated with the normal co-ordinate TJ2 and hence with W2, we put TJI equal to zero. Now, fh = -(h, so they are out of phase which is not hard to understand, because the frequency associated with

V

this mode W2 = 7+ ~ involves spring constant /1. The spring is compressed and elongated, and exerts a restoring force. Hence restoring force due to spring will come into action. These oscillations are called mode of oscillations since the particles always move out of phase with frequency W2. We see that, the normal co-ordinates T}1 and TJ2 are independent of each other, as the system can oscillate with one frequency only while the other is suppressed.

Ex 11.4.2. Linear Triatomic Molecule: Here we obtain the resonant frequencies and normal co-ordinates of vibration of a linear symmetrical triatomic molecule, and in particular C02-molecule. The configuration of such molecule is given in the following figure: All three atoms are

m M m o~----~~--~Or--~~~----o

Figure 11.2: Linear Triatomic Molecule on one straight line. The oxygen atoms are co.nnected to the carbon atom by springs. For simplicity we shall consider only vibration along the line of the molecule, and the actual complicated interatomic potential will be approximated by two springs of spring constant /1 joining the three atoms. Let TJl, TJ2, TJ3 be the displacements of Oxygen, Carbon, Oxygen atoms from their equilibrium positions. Let m be the mass of Oxygen and A1 be the mass of Carbon atoms. Let the atoms interact harmonically, then the potential energy as a function of co-ordinates TJI, TJ2, TJ3 is V =? 2V

=

1

2,/1(TJ2 - TJI)

2

1

+ 2,/1(TJ3 -

= /1[TJ~ + 2rJ~ + TJ5 = (T}1, TJ2, TJ3)

TJ2)

2

- 2TJ1TJ2 - 2TJ2TJ3J

(~/1 ;: _0/1) (~~) = TJTVTJ· ° -/1 /1 TJ3

Now the KE of such linear triatomic molecule is T =?

2T

1 [.2 TJI

= 2,

+ TJ3.2] + 2,1 M TJ3.2

= (1h, il2, il3)

(~ ~ ~ )

°°

m

(ill) .

~~

·T

= TJ

.

TTJ·


357

Now the characteristic equation for determination of frequency w2 is i.e.,

IV -

w 2 rl

= 0,

J.L - w2m J.L 0 J.L 2J.L-Mw 2 J.L =0 2 o J.L J.L - mw 2 2 w [J.L - mw ][J.L{ M + 2m) - M mw 2] = 0 WI WI

= 0,W2 =

(Ii

/ J.L

2m

V-;;;"w3 = VmY + M]'

= 0 refers to translatory motion but not oscillatory motion of the atoms and the

rest two to the oscillatory motion. Now we are to find the eigenvectors aI, a2, a3 corresponding to WI, W2, W3. (i) For WI = 0, let al = (an, a21. a31)T, then the components aij are determined for WI = 0 by the equations,

j

=}

~J.L) (:~~) (~)

(!J.L ; : = o -J.L /1, a31 0 =} an = a21 = a31 = a(say).

This of course, is exactly what was expected from the translational nature of the motion. In this situation, the displacements of all the atoms are in the same direction

and in equal amount as shown in the figure. Using the normalization condition to fixes the values of aij, we get, 2

m{an

2 2 + a31) + M a21 =

1 =} a

=

1 J2m+M

~:===::;=;;

(ii) For W2 = ~, let a2 = (aI2' a22, a32)T, then L(\tij - w~Tij)aj

= Q; i = 1,2,3

j

=}

(~J.L 2J.L __J.L!:f!: ~J.L ) ( :~~) = ( ~ ) o

=}

-J.L

0

a22 = 0; al2

a32

0

= -a32 = (3(say).

In this situation, the central atom does not take part in motion and the end atoms oscillate with actual amplitude but opposite in phase as shown in the figure. Using


358

the normalization, we get the numerical value of the quantities as 2

m(a12

2 2 + a32) + Ma22 = 1 =? f3 =

1 y2m

!7'L:'

In this case, the centre of the atom is at rest, while the two outer ones vibrate exactly out of phase, because they must in order to converse linear momentum. (iii) For W3

= ...j~[1 + ~],

let a3

L)~'J

= (a13,a23,a33)T,

-

w~1ij)aj

then

= Qi i = 1,2,3

J

=?

(-2o -

-~1

';lJ.L J.L - J.L m -J.L

=? a13

-

0

fJ

)

-2!!Jjf

= a33 = ')'i

(a13) a23 a33

a23

rrL

= -2 M

(0) 0 0

')'(say).

This shows that, the end terms vibrate in phase with equal amplitude whereas the ~

,,-

Figure 11.5:

central atom vibrates in opposite phase with a different amplitude as shown in the figure. Any general longitudinal vibration of the molecule that does not involve a rigid transformation will be some linear combination of the normal modes W2 and W3. The amplitudes of the normal modes and their phases relative to each other will be determined by the initial conditions. Using the normalization, we get

Since in the third case, the molecule is asymmetrically stretched, oscillating dipole moment associated with the motion and corresponding absorption based will appear in the infarct region. Hence the normal co-ordinates (1, (2, (3 associated with normal frequencies WI, W2, W3 respectively as 1]1) 1]2 ( 1]3

=

M ]1/2) 1 1 [2m(M+2m) v'2m+M v'2m 1 0 [ M ]1/2 v'2m+M - 2m(M+2m) ( 1 1 [ M ]1/2 v'2m+M - v'2m 2m(M+2m)

Hence the relations complete the discussions of the linear triatomic molecule. Ex 11.4.3. Oscillation of a Particle on a String


359

Figure 11.6: Oscillation of a Particle on a String

Here we consider a light string of length L = (n + 1)1 with n equal masses m placed along it at regular equal spacing interval 1. The string is stretched to a force F. We shall consider small transverse or longitudinal oscillations of the particles. Let the transverse displacements be Xi from the equilibrium positions of the ith particle, where they are taken as generalised co-ordinates. Thus the KE of the system of particles is

The equilibrium length betweeni h and U+1)th particles is l. But when the particles are displaced, the changed length between lh and U+ l)th particles is 01 and hence

Now, F is the tension, hence in the change ol = (xJ~Xj? the work dOlle = F.ol. Since the displacement Xj are transverse, the PE comes from the tension of the string. Hence the PE is V

= F.ol =

F

2l [x~

+ (X2

- Xl)2

+ ... + (xn -

F n+l

= 2l

L:)Xj-l - Xj)2;

XO

= Xn+1 = o.

j=l

Thus, the Lagrangian L of the system of particle is

X n _l)2

+ x~l

360


and the Lagrange's equation of motion for Xj are

d aL dt ( ax) J

=>

aL

- axJ_=

mx-J -

=> Xj

0

F(x_ l J- 1

2 = Wo(Xj-l

-

2x-J

- 2Xj

+ x-+!) = J

0

+ Xj+!) = OJ

2

Wo

F = -. ml

Thus the equation of motion for the particle clearly shows that it depends on displacements Xj-l and Xj+! of its neighboring particles. We seek solution

Xj

= ajeiwtj

i

= F1

then we get the following set of secular equations (2W5 - w2)al - W5a2 = 0

-W5al -W5an-l

+ (2w5 + (2w5 -

w2)a2 - W5a3 w2)an = O.

The condition that an's have solution for w2 is 2W5 - w2 -w5 o· .. -Wo2 2Wo2 - W2 -W 2 '"

=0

o o

O

o o =0

o

O· .. 2w5 - w2 -w5 0 -w5 2w5 - w2

0 0

o

where Dn is the determinant of order n. Expanding the determinant with respect to the first row, we get Dn - (2w5 - w2)Dn_l + w6Dn-2 = 0 2

=> D~ - (2 - W2)D~_1 + D~-2 = OJ Dn = w5nD~.

Wo To solve the linear homogeneous difference equation, put

D~

w2 ,sin{n + 1)0 - (2 - '2) sin nO:+ sin(n - 1)0

Wo

=>

;

= sin(n + 1)0, then

=0

w2 Wo

2 - '2 = 2cosO

=> w2 = w5(2 - 2 cos 0)

= 4w5 sin2 ~

=> Dn = W5 n sin(n + 1)0. Now the zeros of Dn are given by sin{n + 1)0

= 0 => 0 = ~j n+l

k

= 1,2, ... , n.

Hence the eigenvalues, i.e., the normal modes are given by w~ = 4w5 sin2 2(!~1); k = 1,_2, ... ,no


11.4.3

361

Oscillation under constraint

Here we shall consider the effect produced on the periods of normal vibration of a dynamical system about a configuration of stable equilibrium when the number of degrees of freedom of the system is diminished by the i~troduction of an additional constraint. Suppose that, the original system is specified in terms of its normal co-ordinates (1]1, 1]2,. " ,1]71.), so that the KE and PE of the constraint system have the form _

T -

[.2.2 .2]. 21 1]1 + 1]2 + ... + 1]71.'

_ 1[ 2 2

V -

2 2 2 2J. 2 A11]1 + A21]2 + ... + An1]n ,

and the additional constraint be expressed by the equation

Since 1]1, 1]2, ... ,1]71. are small, we can expand the function 1 in ascending powers of 1]1, 1]2,··.,1]71. and retain only the first terms of the expansion. As the relation 1(1]1, 1]2, .. . ,1]71.) = 0 is to be satisfied in the position of equilibrium in which the co-ordinates vanish, this constraint may be expanded and can be express by the equation, A 1'1]1

+ A 21]2 + ... + An1]n =

0

where AI, A 2, ... , An are constants. As the equilibrium configuration is supposed to be compatible with the constraint, there will be no constant term. By means of this equation, we can eliminate qn, we get T

1 [1]1' 2 2 + ... + 1]n-1 ·2 1 (A' ' )2] =2 +' 1]2 + A2 11]1 + A' 21]2 + ... + A n-11]n-1 71.

V

=

~[Ai1]~ + A~1]~ + '"

+ A~-l1]~-l + ~~ (A11]1 + A21]2 + '" + An-11]n-1)2J. 71.

The Lagrange's equation of the constrained system are therefore the (n-l) equations .

!!.. ( aT ) + 8V + p 81 dt 81]7'

.. or, 1]r

81]r

\2 1 (A 11]1 .. + ... + A ) + .l\r1]r + A r [ A2 n-1 7" J71.-1

71.

+ or, i]r where, p

= 0

81]r

A2 A~ (A11]1 n

+ A21]2 + ... + An- 11]n-dJ =

+ A;1]r + Ji,Ar = 0; =

~271. (A1ih + '"

iin A~1]n =-An-A ' n

r

= 1,2, ... , n -

0

1

+ An- 1i]n-1) + ~~ (A11]1 + '" + An- 11]n-1) n


362

Hence the equations of motion of the constraint system can be written in the form of the n equations as ilr

+ )..~TJr + J.LAr = 0;

T

= 1,2, ... , n

where J.L is undetermined. Now, consider a normal mode of vibration of the modified system, defined by the equations TJr = {r cos )..t, for T = 1, 2, ... ,n; and J.L = v cos )..t. Substituting in the equations of motion, we have

Substituting the values of Qr'S given by these equations in the equation Al Ql ... AnQn = 0, we get an equation in )..2 as

This is of degree n - 1 in

)..2,

+

and by writing it in the form

we see that if, )..1 < )..2 < '" < )..n, F()..I) is positive, F()..2) is negative and so on. So this equation in )..2 has (n - 1) roots, which from the form of the equation are evidently interspaced between the quantities )..i, )..~, ... , )..;, i.e., the periods under constraint separate the free periods. The quantities corresponding to these roots are the periods of the normal modes of vibration of the constrained system, and it therefore follows that the (n - 1) periods of normal vibration of the constraint system are spaced between the n periods of the original system.

2;

11.4.4

Stationary property of the normal modes

Let frictionless constraints be imposed until the system retains only one degree of freedom. Here we shall consider the effect of adding constraints to a dynamical system to such an extend that only one degree of freedom is left to the system. Let (TJl, TJ2, ... , TJn) be the normal co-ordinates of the original system. Since the constraints may be represented by linear equations between the normal co-ordinates, so they may put in the form

where Tl, T2, . .. , Tn are constants and ( is the new variable which may be taken as defining the configuration of the constrained system at time t. So the KE and PE of the original system have the form


363

27r are I·t speno . d s 0 f normal·b A1' 27r A2 ' ... , An VI rat·Ion. Then t h e KE an d PE 0 f t h e so 27r constraint system have the form

T

=

1[ 2 2" rl

Hence the equation

+ ... + r n2] be the inclinations to the vertical of the radius through the fixed point and the radius through the ring. Then the KE and PE are given by

2T

= ma2[3()2 + 20Jy + Jy2]; 2V = mga[20 2 + ¢2].

Here we use the stationary property of the normal modes. Suppose that, we impose a constraint represented by the relation ¢ = rO. Then the KE and PE becomes

2T = ma2[3 + 21' + r2]()2; 2V = mga[2 + r2]02. The Lagrange's equation of motion for 0 gives d 8T dt ( => ma 2[3

av

ao )+ BO = 0

+ 21' + 1'2]8 + mga(1 + r2)O = .. 9 2 + r2 =>0+0=0 a 3 + 21' + ,. 2

0


366

which represents oscillations with period value of the function p2 are given by dp2 = 0

dr

2

=}

PI

Thus the periods of oscillations are

2;, where p2 = ~ 3';t~r2'

=} 7'

= -2 1

2g

2

'

= -; P2 = a 27r Pl

The stationary

9

-2 . a

= 27f VG. a and 2 2g

27r P2

= 27f V!2Q. 9

Ex 11.4.6. A light string stretched to tension P has its ends fixed at a distance 91 apart. Masses 2m, m are attached to the string at distances l,51 respectively from one of its fixed ends. Find the normal modes of transverse vibration, and show that

the periods are

47f~, 47f1fj.

Let ql, q2 denote the small displacements of the particles of masses 2m, m at time t, gravity is neglected. We may neglect vibrations in the tension due to the slight extension of the string as they would only introduce terms of higher order in ql, q2 than those which we retain. Thus the KE of the system is given by SOLUTION:

m

2m

Figure 11.7

2T

·2 = 2mql·2 + mq2'

Let QI: Q2 denote forces which would keep the particles at rest in the displaced position, then

The infinitesimal increment in PE, negative of which will be equal to work done for infinitesimal displacements in the masses, can be written as


367

Also for such conservative system the PE can be obtained as

The Lagrange's equation are then

These equations might have been written down at once as the equations of motion of the two particles. To solve them, we put, ql = Ml cos(pt + c), q2 = M2 cos(pt + c) and we get p

_M:p2

+ Blm (5Ml

- M2)

and - M2p2

+ B,m P (4M2

- 2Ml)

==>

_p2 + 5A 2A

2 =

==> p

z

A _p2

+ 4A;

=0 =0 P

A=

SZ;;;

2 9A ± VBIA - 4.1BA2 = 6A 3A = 3P 3P 2 ' 4 l m ' Blm'

Thus the periods of vibrations are Tl =

;7 = 41r/fji and T2 =

;~

= 41rJf/s.

Ex 11.4.7. A light string of length 3l is stretched horizontally between two fixed points, gravity is neglected. Masses 15m, 7m are attached to the points of trisection. The tension in equilibrium is Aml. The particle of mass 15m is drawn aside a distance a, the other remaining undisplaced, and both are simultaneously released. Prove that in the subsequent motion the displacement of the particle of mass 7m is 15 [!3i. 26 a cos 35 t -

V

cos

Vr;"3 t 1.

Let ql, q2 denote the small displacements of the particles of masses 15m,7m at time t respectively. The KE of the system is given by

SOLUTION:

2T = 15mq~

+ 7mq~.

Let Ql: Q2 denote forces which would keep the particles at rest in the displaced

7m 15 m

Figure 11.8

368


position, then ql Ql = >..ml [T

+ ql -l

q2 - ql Q2 = >..ml [ l

q2]

=

q2] +T =

(

>..m 2ql - q2

)

) >..m{2q2 - ql .

For such conservative system the PE is given by

The Lagrange's equation are then

These equations might have been written down at once as the equations of motion of the two particles. To solve them, we put, ql = MICOS{pt+e),Q2 = M2COS{pt+e) and we get

=:}

{2>.. - 15p2)Ml - 7M2 = 0; ->..M1 + {2>.. -7p2)M2 = 0 (2X - 15p2)(2).. - 7p2) - >..2 = 0 3>" >..

=:}

p~

= -35

and p~

= -. 3

Substituting the two values of p2 in succession gives two sets of corresponding values for !vII, 1\12 which may be write, Mu M21 M12 M22 -7- = -5- = 01; -1- = -3 = 02·

Thus the complete solution of the differential equations in

Ql,

q2 can be written as

ql = 701 COS(plt + e1) + 02 COS(P2t + e2) q2 = 501 COS{p1t + e1) - 302 COS{p2t + e2) where aI, 0'2, e1, e2 are arbitrary constants, to be determined by using initial conditions. The initial conditions are q1 = a, q2 = 0, til = 0, ti2 = o. The vanishing of the velocities shows that e1 = 0 = C2, then we have a

= 701 =:}

+ 02; 0 = 501 - 302 3 5 01 = 26 a ; 02 = 26 a .

Hence the displacement of the second particle of mass 7m is


369

Ex 11.4.8. A uniform bar of length 2a is hung from a fixed point by a string of length b fastened to one end of the bar. Show that, when the system makes small normal oscillations in a vertical plane, the length l of the equivalent simple pendulum is the root of the quadratic [2 - (1a + b)l + ~ab = O. Let AB be the string of length b and Be be the rod of length 2a, where centre of gravity at G(x, y). The string AB makes an angle 0 with the vertical and the rod Be makes an angle ¢ with the vertical. Then x = bcosO + acos¢, y = bsin 0 + a sin q;. The KE of the rod Be is given by SOLUTION:

T=

~m[x2 + 1?1 + ~Iw2 2 2 1

.

.

2

•

• 2

1

(2a?'2

+ asin¢¢) + (bcosOO + acos¢¢) 1+ 2'm 12 ¢

= 2'm[(bsinOO

= ~m[b2e2 + ~a2¢2 + 2abe¢cos(0 - ¢)l 1

4

2 '2

.. + 2abO1;cos(0 - dt = dt - v; dt' = dt; dt' = dt ,

I

=> U x = U x -- v; u y => 11' = 11 - V'

,

= uy; U z =

(12.5)

Uz

where 11 = (u x , u y , u z ) is the velocity of the particle with respect to the observer of the frame 8 gnd 11' = (u~, u~, u~) is of the frame 8'. The set of transformation equations (12.5) is the classical velocity addition formula in Newtonian Mechanics. Also from equation (12.5), we conclude that velocity of a particle is not a Galilean invariant. Also, the inverse Galilean velocity transformations are

12.4.4

Invariance of acceleration

Differentiating the transformation set of equations (12.5), w.r.t time t, we get du~

dt =>

dux du~

=

du~

dt; dt dux du~ dt ' dt'

=

du y d11.~ du z dt ; dt = dt du y du~ dt ' dt'

, du z ast=t dt .

-=-'- =-'-= -

dt'

Since the components of the acceleration of the particle measured in the two inertial frames are independent of the uniform relative velocity of the two frames, acceleration of the particle is invariant under Galilean transformation.

12.4.5

Invariance of Newton's Law

In fact, Newton's first law of motion, whether a given frame of reference is an inertial frame of reference or not. Since Newton's first law of motion always holds good in an inertial frame of reference, it will also hold in two inertial frames of reference moving with uniform velocity relative to each other. We know, that the measurement of acceleration of a particle is independent of the uniform relative motion of the two inertial frames, i.e., 71' = 71. Let m be the mass of the particle. In Newtonian mechanics, mass is a constant quantity and is independent of the motion of the observer. Multiplying both sides of the above equation with m, we have, ~,

ma ~

~

~

~

= ma =>

~,

F

=

~

F,

~

where F' = m a' and F = m a. Thus, the observers of the two inertial frames of reference moving with uniform velocity with respect to each other measure the force on the particle to be the same. Therefore, equations of motion (F = m 71) of a ~

CHAPTER 12. RELATIVISTIC MECHANICS

378

particle is invariant under Galilean transformation. In other words, Newton's second law of motion is Galilean invariant. Since measurement of force is not affected by the uniform relative motion between the two inertial frames of reference, Newton's third law of motion F 12 = - F 21 must also hold in the two inertial frames moving with uniform velocity relative to each other. Hence Newton's laws of motion are invariant under Galilean transformations. ~

12.4.6

~

Conservation law of linear momentum

Consider two particles of masses m1 and m2 moving with velocities 111 and 112 respectively in an inertial frame 8 at rest. Suppose that, the two particles collide and after the collision, they move with velocities 1/ 1 and 1/2 respectively. According to the law of conservation of linear momentum in frame 8, we have,

Let us now view this collision between the two particles from another inertial frame of reference 8' moving with a uniform velocity 1/ with respect to the frame 8. Suppose that the observer of frame 8', the velocities of the two particles before and after collision appear as 11~, 11~ and 1/~, 1/~ respectively. According to inverse Galilean transformations for velocities,

Thus the law of conservation becomes, m1(U~

+ 11) + m2(11~ + 1/) =? 'rn 1 'U

= ml(1/~

+ 1/) + m2(11~ + 1/)

1 + m2 u 2 = ml v 1 + m2 v 2·

~,

~,

~,

~,

It is the law of conservation of linear momentum in frame 8'. It has the same form in

the frames 8 and 8' moving with a uniform velocity relative to each other. Hence, the law of conservation of linear momentum is invariant under Galilean transformation.

12.4.7

Conservation law of Energy

Consider two particles of masses m1 and m2 moving with velocities 111 and 112 respectively in an inertial frame 8 at rest. Suppose that, the two particles collide and after the collision, they move with velocities 1/1 and 1/2 respectively. If F is the amount of energy that appears as heat or in some other form, then according to the law of conservation of energy in frame 8,

1 ~2 1 ~2 1 ~2 1 ~2 2m1 u 1 + 2m2 u 2 = 2m1 v 1 + 2m2 v 2 + F. Let us now view this collision between the two particles from another inertial frame of reference 8' moving with a uniform velocity 1/ with respect to the frame 8. Suppose that the observer of frame 8', the velocities of the two particles before

12.4.

GALILEAN INVARIANCE

379

and after collision appear as 1ti,.1t~ and 1ii, 1i~ respectively. According to inverse Galilean transformations for velocities, ........ ........, ................ ........ , ................ ........ , ................ ........, ........ Ul=ul+v; U2=u2+ v ; Vl=vl+v; V2=v2+ v . According to the law of conservation of linear momentum in frame 8, we have, ........

, + m2 ........,2 = ml ........v ,1 + m2 ........v ,2

ml U 1

=? [mll1i

U

+ m21t~].11 =

[mll1i

+ m211~].11

Also, the law of conservation of energy becomes,

1

2m1

(........ , U1

) ( ........ , ........ ) 1 ( ........ , ........ ) ( ........ , ........ ) + ........ V . U 1 + V + 2m2 '!L 2 + V . U 2 + v

= 1 =? [2m1

........ '2 U1

1

2ml

( ........ , ........ ) ( ........ , v 1 + v . V1

) 1 ( ........ , ........ ) ( ........ , ........ ) F + ........ V + 2m2 v 2 + V . V 2 + V +

1 ........ '2] [ . . . . , ........ '] . . . . + 2m2 U 2 + ml U 1 + m2 U 2 . V 1 ........ '2 1 ........ '2] [ ........ , . . . . ']........ F = [2m1 v 1 + 2m2 v 2 + ml v 1 + m2 v 2 . V +

1

=? 2m1

........ '2 U1

1 ........12 + 2m2 U2 =

1

2m1

........ '2 v1

1 ........ '2 + 2m2 v 2 + F.

It is the law of conservation of energy in frame 8'. It has the same form in the frame 8 and 8' moving with a uniform velocity relative to each other. Hence, the law of conservation of energy is invariant under Galilean transformation. 82~

82~

a2~

1 a2~

Ex 12.4.1. Show that, the electromagnetic wave equation ~ + 8? + Fz"2" - ? ift"T is not invariant under Galilean transformation, where 1> is scalar potential. SOLUTION:

=0

The Galilean transformation from unprimed to primed frame to refer-

ence are

x' = x - vt; and y' = y; z' = z; t' = t 8x' 8y' 8z' 8t' 8y' 8x' 8z' 8t' = 1 and = - = - =0· - = 1 and = - = - =0

fu

~

Bz =

fu

fu

fu

'~

~

~

~

~

1 and 8z = 8z = 8z = 0;

~

~

8t = -v;

~

~

at = at = 0;

~

~

8t = 1.

Here the electromagnetic wave equation in frame S is 2

2

2

2

8 1> 8 1> 8 1> -2- -c12-88t-1>2 0 2 8y2+ 8z 8x+ - . In frame 8', the scalar potential 1> is function of x', y', z' and t', i.e., 1> = 1>( x', y', Thus,

z', t').

380


a2 dJ For this, Now, we calculate iJi'T'

Thus, the electromagnetic wave equation in frame 8' becomes,

a2¢ a2¢ v 2 a2¢ 1 a2¢ 2v a2¢ ax,2 ay,2 3 z'2 c ax'2 c at'2 c ax' at' v 2 a2¢ a2¢ a2¢ 1· a2¢ 2v a2¢ =} [1- - j - + - + - - - - + - - - =0 c2 ax'2 ay,2 az'2 c2 at'2 c2 ax' at' . fJ2¢

-+-+---- - -2+ - -2 - = 0 2

Thus, the two equations do not have the same form in the two inertial frames of reference moving with uniform velocity v relative to each other. Hence, the electromagnetic wave equation is not invariant under Galilean transformations.

12.5

Special Theory of Relativity

In 1905, the special theory of relativity which deals with the problems involving the inertial frames of reference is developed by Einstein. He realized that a relativity principle existed both for mechanics and electromagnetism. This theory is the outcome of the analysis of the physical consequences that would result if no uniform frame of reference is in existence. The special theory of relativity is based on the following two postulates: (i) All the basic laws of Physics which can be expressed in the form of equations have the same form in all the inertial frames of reference moving with an uniform velocity with respect to one another. (ii) The measured speed of light in free space has always the same value c for all the observers irrespective of the relative motion of the source of light and the observer. The first postulate bears the fact that, there is no existence of an universal frame. If we reformulate the different laws of Physics for observers in a relative constant motion with respect to one another, we obtain other set of transformations in place of Galilean Transformation, under which the laws would be invariant. If there is no difference, then this difference enable us to determine which objects are 'stationary' in space and which objects are 'moving'. Practically, an universal frame of reference is not available and we are not able such differences in physical laws. The second postulate follows directly from the negative results of MichelsonMorley experiment, i.e., the speed of light in free space is constant irrespective of the motion of the source or the observer. Since the speed of the light does not depend upon the motion of the earth, there is no path difference and hence no

12.6. LORENTZ TRANSFORMATION

381

fringe shift is expected. Further, the velocity of light is not invariant under Galilean Transformations, they need to be discarded in view of the second postulates also. These postulates of the special theory of relativity do not appear so radical at first sight, but lead to very revolutionary ideas. The concepts of the invariance of mass, length interval and time interval and their mutual independence are neither invariant nor independent of one another in special theory of relativity.

12.6

Lorentz Transformation

In 1904, H.A. Lorentz derived a set of transformations for space and time coordinates of an event in two inertial frames moving with a uniform velocity relative to each other. However, Einstein obtained these transformations quite independently on the basis on the two postulates of the special theory of relativity. These equations are known as Lorentz-Einstein transformations or simply Lorentz transformations. This tra'1sformation should confirm to the following general requirements: (i) The transformation must be linear, i.e., any single event in one inertial frame must transform to a single event in another frame, with a single set of coordinates. (ii) The postulate of the equivalence of all interval frames requires that the direct and inverse transformations should be symmetrical with respect to each other. (iii) In case, there are no relative motion between the two frames of reference, Lorentz transformations should be reduce to identity transformations. (iv) In case, the velocity of moving frame is very small as compared to the velocity of light, Lorentz transformation should reduce to Galilean transformations. (v) The relativistic law of addition of velocities as derived from the Lorentz transformations should leave the velocity of light as invariant. Let us consider two inertial frames of reference Sand S' having their respective axes parallel to each other. The frame S' is moving with a relative constant velocity v along x-direction. According to Galilean transformation, the velocity transformations are

which violate the postulates of the special theory of relativity. The first postulates of the special theory of relativity implies that a uniform linear motion is one frame of reference must appear as the same in the other inertial frame also and likewise a linear relationship among x, y, z and t must go over a linear relationship among x' , y', Zl and t'. Let us assume that x and x' may be connected by the relation

x'

= k(x -

vt)

(12.6)

382


where k is a constant of proportionality, which does not depend upon x, t but may • be a function of v. When k = 1 and v «< c, then x' = x - vt which is known to be correct in Newtonian Mechanics. Further as the motion of the frame S' is only along x-direction, we have y'

=y

and z'

= z.

(12.7)

In the special theory of relativity, space and time are independent and they form a space time continumn. Therefore time co-ordinates t and t' may be assumed to be connected by the relation t'

= pt + qx

(12.8)

where p and q are some constants. To determine the constants k, p, q we use the second postulate of special theory of relativity. Let at t = t' = 0, the origins 0 and 0' of the frames coincide with each other and at that instant, a light signal is emitted in the space from the origin 0 of the frame S. With the passage of time, this light signal will grow into a.spherical wavefront of ever increasing radius for the observers of the two frames, with the origin of that frame as the centre of the spherical wavefront. Let the velocity of light in frame S is c and that in S' is c' = c - v. If according to the observer of the frame S, the spherical wavefront passes through the point P(x,]}, z) in space at time t, then (12.9) Let the point P in the frame S be observed as P'(x', y', z') at a time t' in frame S'. Since according to second postulate of the special theory of relativity, the speed of the light is same in all the inertial frames, i.e., x,2 + y,2 + z,2 = (ct')2 2 2 =} k (x - vt)2 + y2 + z2 = c (pt + qx)2 2 2 2 2 =} (k _ c q2)x + y2 + z2 _ 2(vk2 + c pq)xt

= (C 2p2 -

k 2v 2 )t2.

Since (12.1O)is identical with (12.9), we get k 2 _ c2q2 = 1; vk 2 + c2pq = 0; c2p2 _ k 2v 2 = c2 2 2 2 =} v[l + c q J + c pq = 0; _v 2[c 2q2 + 1J + c2p2 = c2 1- p2 =} vpq + p2 = 1 =} q = - vp 1p2 2 2 =} vk + c p. _ _ = 0 vp 2 2 c c2 vp 2 =} P = c2 _ v 2 and so k = c2 _ v 2 ; q = - c2 v/c2 q = - 731=_=(;=v'if'2/;=c2irr)

-V

(12.10)

12.6. LORENTZ TRANSFORMATION

383

Hence, putting these values in (12.6) to (12.8), we obtain,

x' =

x - vt . y' VI - (v 2Jc2) '

y' z'

=,

z' t'

=,

2

=

t - (vx) Jc (v 2Jc 2)

Vl-

(12.11)

These set of equations are called Lorentz transformations. From this set of equations, one can obtain x, y, z and t in terms of x', y', z' and t' as

x

=

x' + vt' , -y'r=I=_=(~v:; 9 x 10 6m = 336 => 6m = 9 x 1016 kg =

Ex 12.10.4. Calculate the increase of mass of500g of metal of specific heat 0.15cal/g/oc when it heated through 900° . SOLUTION: From the given data, we have,m = 500g; s = 0.15cal/g/oC; 6T = 900°C. Thus the heat absorbed by the metal, is given by,

6E

= ms6T = 500 x 0.15

x 900cal

= 500 x 0.15 x 900 x 4.2J = 2.835 x 105 J. If the heat energy absorbed by the metal appears as increase in mass 6m, then,

6E

= 6mc2 = 6m x (3 x 2.835 x 105

108)2J

=> 6m = (3 x 108 )2 = 3.15 x 10

-12

kg = 3.15 x 10

-9

g.

Ex 12.10.5. A proton has a total relativistic energy 9000MeV. Find its KE and its speed, assume that the rest mass of proton is 1.6 x 1O- 27 kg.


400

From the given data, we have, mo = 1.6 x 1O- 27 kg; E 9000 x 1.6 X 10- 13 J. Thus the KE T of the proton is given by,

=

SOLUTION:

T

=E-

moc2

= [9000 x = 12.96 X

1.6 x 10- 13

-

1.6

X

10- 27 x (3

mo

. 11

--;====;~::;;: =}

V

\11 - v2 /c 2

Jl _ v2/c2 =}

12.11

v2 1- 2 c

108 )2]J

10- 10 J.

If m be the relativistic mass of the proton, then using m

=}

X

9000MeV =

- V

2/ 2 _ C

= 1.6 X 10-

moc E

= J I-v m0 2' 2 Ie

2

---

27

x (3 X 10 9000 x 1.6 X 10- 13

=

0.01

=} V

we get,

8

)2 =

0.1

= 0.995c.

Momentum and energy

Consider a body of rest mass mo moving with a velocity 11. Then its relativistic mass is m = J m02 2' The relativistic linear momentum of the body is, given by, I-v Ie -+

-+

p=mu=

mO-+

Jl- v2 /c 2

u

(12.21)

which depends upon the velocity 11 of the body. If 11 = (u x , U y , u z ) and p = (px, Py, Pz), then we can find the components of relativistic linear momentum of the body. Now, according to Einstein's mass-energy relation, the body of mass m is equivalent to energy E, given by

(12.22) This is called relativistic energy relation. From equation (12.22), it follows that the energy of the body may be positive or negative. The physical meaning to the negative energy by predicting the existence of antiparticles was described by P.A.Dirac. Deduction 12.11.1. We know, the relativistic energy of a particle of rest mass mo

12.12. TRANSFORMATION OF ENERGY AND MOMENTUM and momentum pis E

=

401

J p2c2 + m6c4. Also,

E = moc2 + T

=}

2 Vp 2c2 + m6c4 = moc + T p2c2 = T2 + 2moc2T

=}

p=

=}

J~: + 2moT.

This is the relation between the momentum and KE.

Ex 12.11.1. Calculate the linear momentum of a 25GeV proton, assuming the rest mass energy of the proton is IGeV. SOLUTION: The given quantities are moc2 energy of the proton is given by,

E Using the relation E

pc = =}

12.12

P=

IGeV; T

=

25GeV. The relativistic

= moc2 + T = (1 + 25)Gev = 26GeV.

= ±Jp2c2 + m5c4,

V

E2 - m6c4

=

the relativistic energy of the proton is,

J26 2 - 12

25.9S x 1.6 x lO lD 8 kgm/s 3 x 10 .

= 25.9SGeV =

= 1.3S6 x

10

-17

25.9S x 1.6 x 1010 J kgm/ s.

Transformation of energy and momentum

Consider two inertial frames of reference 8 and S' having their respective axes parallel and the frame 8' is moving with a constant velocity v with respect to frame 8 along X - direction. Consider that a particle of rest mass mo is moving with a velocity u with respect to the observer of frame 8. If U x , u y and U z be the components of velocity of the particle, then

(12.23) To the observer of the frame 8, the mass of the particle will appear as m

= J l-u rn P • 2 /c 2

Suppose that to the observer of the frame 8', the particle appears to move with velocity u'. If u~, u;, and u~ be the components of velocity of the particle, then

(12.24) To the observer of the frame 8', the mass of the particle will appear as m' rnp According to velocity transformation law, we have J1-'u /2 /c 2 • ,

Ux

=

'U x

1

'V

v

- CIux

, -

;

uy -

u Y Jl 1

- v 2 /c 2 . v

- CIux

,

2

U'

z

u z ' __ /1 - v / c = _--'v ---'_ 1-.3i.-u c~ x

2

=

CHAPTER 12, RELATIVISTIC MECHANICS

402

The X - component of the relativistic momentum of the particle in frame 8' is

",

= rnu x =

Px

rna

ux-v

' VI - u,2/ c2'1 v - ~ux

To express the X - component of momentum in frame 8' in terms of the X - component of momentum in frame 8, let us first evaluate the factor u,2 / c2 , From ' th e re1a t lon, 11,'2 = U '2 U z ,we h ave, x + U '2 y +'2

VI -

=?

11,,2

1- -

c2

(u x - v)2 + (u 2 - u;)(1 - v 2/c2) = 1 - ..:....-:=------.:....-::-:.....:.....-~~_:::__-!.-.:... 2 2

c (1 - vu x /c )2 2 (c - v ) - (u; + u 2 - u;)(1 - v 2/c2) c2 (1 - vu x /c 2)2 2 2 (1 - u /c )(1 - v 2/c 2) = c2(1 - vux /c2)2 2

Thus, the X - component of the relativistic momentum of the particle in frame 8' is given by

It is how the X - component of momentum of the particle in frame 8 transforms to X-component of the momentum in frame 8', Now, the Y - component of the momentum of the particle in frame 8 f is given by,

12.12. TRANSFORMATION OF ENERGY AND MOMENTUM

403

Similarly, p~ = pz. Therefore, the transformation equations for momentum components in unprimed frame 8 to primed frame 8' are ,

Px

=

PT -

./1

V

vE/c2 2/2;

-v c

,

,

Py=Py;

Pz=pz·

(12.25)

It follows that just as in case of Lorentz space-time transformations, the time t joins the space co-ordinates x, y, z as the fourth co-ordinate; in momentum transformations, the energy E combines with momentum components Px,Py and pz as the fourth co-ordinate. Let us now obtain the transformation equation for energy from unprimed to primed frame. If E' be the energy of the particle in frame 8', then 2 E' = m'c2 = mo c2 = mo(l- vu x /c ) c2 Jl - u'2/c2 Jl - u 2/c 2 Jl _ v 2/c 2 ' _ mc2 - muxv _ E - vPx . as Px = mu x , E = mc2. (12.26) - Jl- v2/c 2 - Jl- v2/c 2 ' The equations (12.25) and (12.26) are the momentum- energy transformations from unprimed frame 8 to primed frame 8' moving with constant velocity v relative to 8. The inverse momentum- energy transformations can be obtained by changing v to -v, primed quantities to unprimed and vice-versa, i.e., _

p~+vE/c2 .

,_

,_

Px - Jl-v 2 /c 2 ' Py - Py,PZ E _ E'+'Vp~

-

pz

} .

(12.27)

~h-v2/c2

-

Ex 12.12.1. 8how that, the quantity [P2 - ~] is invariant under Lorentz transformations. SOLUTION: According to momentum energy transformations, we have , Px - v E / c2 " , E - vP~ x P = Jl-v 2/c 2 ; PY=PlI,pz=pzandE = Jl-v 2/c 2 Using these relations, the quantity [p'2 E'2 p'2 _ _ c2

Ff)-] in 8' frame becomes,

E'2 = p'2x + p'2y + p'2z---;}2 2

]2 2 2 [E - vp~ ]2 = rPx - vE/C +p +p . }1- v 2/c 2 y z }1- v 2/c 2 1 2] 2 2 E 2 v 2") - 2" (E - vpx) + Py + Pz C c 1 2 E2 1)2 2 E2 2 2 1 -v 2/ c2l(px - 2c - "2(Px 2] + Py + pz c C

1

1

2/ C2[(Px V

2 E2 2 2 =Px - 2 +Py +pz c E'2 E2 :::} p'2 _ _ = p2 c2 -~.

404


Thus, the quantity [P2 - ~] is invariant under Lorentz transformations.

12.13

Relativistic force

---> ---> The momentum p and the force F are connected by the Newton's second law of motion. In the non-relativistic case, we have, ---> d Ii --> ---> F = Tt; p =mv ---> =d - (---> 'rn'u)=rna;

(J = constant. dt In fact, for the relativity theory, we have to use the relations (12.25) for momentum, and so we have, 2

--->

d ---> d11 ---> d(3 v 1/2 F = dt(m(3v) =m(3 dt +mv dt; (3=(1- c2 )- .

ft

The second term m11 is in the direction of the velocity 11. Also, the first term contains d]{, which may be in any direction. So we resolve it into a component perpendicular to the velocity and a component parallel to the velocity. Writing the perpendicular component as the perpendicular component of the acceleration, i.e., d1i (lIth -_ ---> a1., we have,

---> d11 ---> F 1. = m(3( dt h = m(3 a1..

(12.28)

Now, we calculate the component of force parallel to the velocity. Now, 2 2 d(3 = ~(1 _ v )-1/2 = _1_(1 _ v )-3/2~(v2) dt dt c2 2c 2 c2 dt where v 2 = 11112. If () be the angle between the vectors 11 and 0;, we have

~( 2) dt v

= ~(1--->12) = 2---> d-V = 2--->.--> dt v v . dt v a = 2111110;1 cos () = 2vl0;1 cos ().

Thus, 10;1 cos () is the component of the acceleration in the direction of the velocity, i.e., 10;1 cos() = alI-. so that ft(v 2 ) = 2van· Thus, d11 ---> d(3 v v 2 3/2 Fll = m(3- + m v - = m(3an + mv-(l -)an dt dt c2 c2 2 2 2 2 = m[(l - ~ )-1/2 + ~(1 - ~2 )-3/2]an = m(l - ~ )-3/2 an . (12.29) 2 2 2 c c c c Equations (12.28) and (12.29) give respectively the components of force, perpendicular and parallel to the direction of motion. The mass m(l - ~) -3/2 is called

congitudinal mass" where as m(l - ~ )-1/2 is called the tr~nsverse mass. The difference between these two masses implies that at high speed (Le., in the relativistic region) the acceleration vector is not parallel to the force vector.

12.14. EQUATION OF MOTION

12.14

405

Equation of motion

-

Suppose that, a particle moves along a trajectory. We denote the unit vectors normal and tangents to the trajectory ny nand f. The total force F acting on the particle can be decomposed into normal and tangential components as

Each component of the force produces an acceleration in the appropriate direction, determined by the inertia of the body in the direction. Since the normal acceleration is '~ and R is the radius of curvature of the trajectory, and v is the particle's velocity and the tangential acceleration is ~~. Thus the force can be written as -F

=

mo v2 mo dv n+ I 'T 2 2 2 2 Jl - v /c R (1- v /c )3 2 dt mo df mo dv Jl - v 2/(;2 v dt + (1 - v 2/c 2)3/2 dt'T A

A

A

=

d movf dt [Jl _ v2/c21

which is the relativistic equation of motion of the particle and it is a generalisation of Newton's equation of motion. It can be written in a more convenient form as dp dt

= Fj

p

= mV,m =

mo J l - v 2 /c2

(12.30)

The quantity m is called the relativistic mass, or simply mass, mo is the rest mass.

12.14.1

Relativistic rocket

In the relativistic case, a mass M must be considered relativistic, i.e. M

= J 1-v M02 2' Ie

where Mo is the variable rest mass of a rocket. The rest mass of the rocket decreases during its motion. Taking this into account, we can write, (12.31) If necessary, we can easily take into account the external forces acting on the rocket. As a result, we arrive at the relation

(12.32)

The difference 'U - v is not the velocity of ejection of gases relative to the rocket because we must use use formula for the velocity addition in the relativistic case.

406

12.15


Inelastic collision

Let us consider two inertial frames of reference Sand S' having their respective axes parallel and the frame S' moving with a constant velocity v along x-direction w.r.t. frame S. Suppose that two identical bodies A and B, each of mass m, are moving in frame S' with velocities u' and -u' respectively along x'- axis. Further, that the two bodies collide inelasti'cally, so that after collision they stick together to form a single body. In the frame S', the combined mass of the two bodies will be at rest and let it be equal to Mo. When viewed in the frame S, suppose that the two bodies A and B appear to be of masses ml and mao Further, suppose that in the frame S, the two bodies appear to move with velocities Ul and U2 along x- axis before collision. According to the velocity transformations, we have, u' +v . - 1 + vu'lc2 '

Ul -

and

U2

-U' +v = -----.,,-2 1 + v( -u')lc

-U'+v 1-vu'lc2 '

Let us assume that the frame S' is moving with velocity u' w.r.t. frame S. Then, setting v = u' in expressions for Ul and U2, we have, Ul

=

+ u' 2u' = . 2 1 + u' u'1c 1 + u'2 1c 2 ' U'

and

U2 =

-U' + u' 1 -u'21 c2 =

o.

As such, the body B of mass m appear at rest and the combined mass of the two bodies will appear to be moving with velocity v = u' in frame S. Suppose that in the frame S, the combined mass of the two bodies appears as M, then according to mass variation formula, M = J M02 2' If mo is the rest mass of the either of the l-u'

Ie

two bodies A and B, then we have, m1 =

rno

rno

"-t==;;r:;=::;:;:

2

J1- [1+~%le2]2/c2 mo[l + u'2/c2] mo[l + u'2/c2] J[l + u'2/c2J2 - 4u'2/c2 1 - u'2/c2 J1- uVc

From the law of conservation of linear momentum in frame S, we have,

Since the factor J1 - u r2 1c2 < 1, from the above equation, it follows that, Mo > 2mo, i.e., during an inelastic collision, the mass of the system increases. The increase

12.16.

407

MINKOWSKI SPACE

in mass is given by,

6m

= Mo - 2mo = = 2mo[

Vl-

2mo u,2/ c2

Vl-

- 2mo

1 -1] . u,2/ c2

The energy equivalent of the increase in mass during the inelastic collision,

It may again be pointed out that the increase in mass in an inelastic collision is due to materialization of the kinetic energy possessed by the two colliding bodies before collision. The KE possessed by the two bodies before collision,

Now, the mass of the either body w.r.t. the observer of the frame 8', is m rnQ and so Jl-u /2 /c 2 ' T

= 2[

mo 2 - mo]c u,2/ c2

VI -

= 2mo[

VI -

1 2 - l]c , u,2/ c2

which is same as the expression for 6E. In other words, energy equivalent of the increase in mass of the system is equal to the total KE of the two bodies before collision.

12.16

MiIikowski space

In special theory of relativity, the concept of space and time are independent of each other. The length of a radius vector in three dimensional geometry, i.e., x 2+y2+z2 is not invariant under Lorentz transformation. However, the quantity x 2+y2+z2_c2t 2 is invariant under Lorentz transformations, i.e.,

The above equation may be written as, (12.33)

where x', y', z' and ict' are the four coordinates in the frame 8', while x, 1], z and ict are those in frame 8. Just as the radius vector in the three dimensional space has three mutually perpendicular components x, y and z, the space components x, y, z and time components ct may be treated as the components of the radius vector in the four dimensional space, called Minkowski space or space time continuum. Since


408

it is not possible to draw four mutually perpendicular axes through a point in space, the fourth component ct has been multiplied with i so as to remind that this fourth dimension is imaginary. Thus, in Minkowski space, the fourth-radius vector having the co-ordinates (x, y, z, ict) may be represented as --+

...

r = xi

..

,..

+ yj + zk + ctw,

(12.34)

where w is the unit vector along the fourth imaginary axis. The four coordinates x, y, z and ict define a vector in the four-space. Such a vector is often termed a four vector. The representation of space-time can be extended to give a geometrical interpretation of the relativity of simultaneity, length contraction, time dilation etc. The fact that L2 is invariant suggests that a Lorentz transformation can be merely regarded as a rotation of coordinate axes x, y, z and ict in space time. Introduction of four-vectors in the four space has another advantage. When physical laws are expressed in terms of four-vectors then the law takes the same form in any frame moving with a uniform velocity with another frame. Thus, the invariance of the physical law is automatically proved when expressed in four vectors. The Lorentz transformation (12.11) for x and t co-ordinates are x'

=

t - (vx)/c 2

x - vt I \1"1 _ (v2 / c2)' t

= -vr=I=_=(O=v~2/:;::;c2~)

which can be written in the form, I ct - (v/c)x ct = -Vr=I=-=(~v:;;=2/~c2~)

I x - (v/c).ct or, x = -vr=I=_=(-Fv2~/:=;c2~)

I

or. x

=

X -

{3w

\1"1- {32;

I

w

=

W -

{3x

v

Jl- (32;{3 = ~,w =

ct.

(12.35)

Note that, the Lorentz transformations for x and t co-ordinates are not symmetrical. But introducing ct as the fourth dimension, the Lorentz transformations assume the form of the equations (12.35), which become just symmetrical to each other. Minkowski referred to space-time as the world and a point in space-time as the \ -world po-into The motion of a particle in space-time can be represented by a curve called wprld line, which gives the loci of space-time points corresponding to the motion. The world line of a particle must lie in its light cones. Now we are to apply this concept to understand some implications of the theory of relativity. We draw only x and waxes of the frame of reference and ignore the y and z axes, because y' = y and Zl = Z. The tangent to the world line at any point is always inclined at an angle 0« 45°). It is because, the slope of the tangent to the world line at any point, m

dx

Idx

'v

= tan () = -dw = -edt - = -. c ,

Since v 0, then it is called time like interval. If (dr)2 = 0, then the corresponding interval is called a light-like interval.

12.16.2

Four velocity ~

Having defined a four-dimensional length vector XI-" we can now define a fourvelocity. The rate of change of the position vector dX I' of the particle with respect ~

to Its invariant proper time dT, i.e., the four vector quantity

d~"

is called the four

~

velocity and is denoted by V Ii' Tnus,

Thus the four-velocity is the particle proper time rate of change of its four position vector. It should be noted that the three dimensional velocity 11 = (~, ~, ~) of

410


the particle has magnitude v which is the same as that occurring in factor . ~, 2 V 1-8

since we are finding the proper time rate of change positIOn vector of the particle. For the velocity four vector, we find d Xi)2 1 _ (32 ~ dt 3

V2

J.l

= _1_[,",(

_

c

2J =

2

2

v - c 1 _ (32

=_ 2 0 C

< .

t=l

Since squares of the world velocity is less than zero, it IS tIme like and has a constant magnitude. Thus, V~ is invariant under Lorentz transformation. The four acceleration can be defined as the particle's proper time rate of change of the four velocity. Thus, the four acceleration A J.l is given by, ~

~

2~

-; dV J.l d XJ.l d [ 1 -; J AJ.l=-d-=-d 2 =-d ~(v,zc). T T T VI - (32 Thus, the first three components A J ; j = 1,2,3 of the four acceleration are non zero when 7J is not constant and the fourth component A4 is zero.

12.16.3

Four momentum

In Newtonian mechanics, the momentum of a particle is defined as the product of the mass and the velocity of the particle. We are to extend the same definition in relativistic mechanics, on the basis of the following assumptions: (i) Since time t is not Lorentz invariant, it should be replaced by proper time

T.

(ii) The mass of the particle is truly the characteristic of the particle and is independent of the velocity. For the purpose, we take mass of the particle measured in a frame in which it is at rest. Such a frame is called rest frame or the proper frame of the particle and the mass so measured is called rest frame or the proper frame of the particle and the mass so measured is called rest mass or the proper mass of the particle and is denoted by rna. Then the four momentum is defined by, ~

P = rna dX J.l = ( L

I

dT

~

rna v , irnac );"V' Jl _ (32 Jl - (32

= (dX1 , dX2 . dX3). dt

dt

dt

-;

The first three components of four momentum P dimensional momentum. Thus,

and the fourth component of momentum is P4

=

IL

are the components of the three

~

~

~. Ab:" P J.l" P J.l 2

V 1-13

=

_m,2r2

< O.

12.16.

12.16.4

MINKOWSKI SPACE

411

Four force

In. Newtonian mechanics, the generalised equation of motion is given by,

which is not invariant under Lorentz transformation. Its relativistic generalization should be a four-vector equation, in special part of which reduce to the above equation in the limit as fJ - t O. Now we have the following assumptions: (i) Since time t is not Lorentz invariant, it should be replaced by proper time r. (ii) The mass m can be taken as an invariant property of the particle. (iii) In place of the velocity

Vi,

world velocity vJ1. should be substituted.

(iv) Force Fi should he replaced by some four vector KJ1.' called the Minkowski force. Thus taking into account all these changes, the equation of motion is given by

-t

The first three components of four force K J1. are the components of the three dimensional force. Thus,

-t

The time like part K4 of the four vector K J1. can be obtained as ~

--->

K 11.' V p.

d

-t

--->

= V Jl.' dr (m V I') d

mc2

= - dr (-2-) = 0; V·

=> K

J

J

yh -

(J2

+ K4

d 'm - t

= dr ( 2" V

as

mc2

2

-t

J1.'

V J1.)

is constant

zc

JI - fJ2 = 0 'l.moc

VI - f32


412

12.16.5

Kinetic energy

In the classical case, the KE T is given by, ~~

dT i dT F. v = - =? K4 = ---;===.:::-dt c {32dt

VI -

Also, from the fourth component of Minkowski force, we write dId ( - dr ' {32 dt dT 1 d mic f3 2dt = f32 dt ( (32) K4

i

=?

-

~ VI -

- ( mv4)

VI -

VI -

dT

=?

dt =

mic ) {32

VI -

VI -

d ( "me? ) T dt f32 =?

VI -

=

mc2

VI _ f32 + To

(12.37)

where To is the constant of integration. Also, equation (12.37) can be written as T

= (1- (32)-1/2mc2 + To = (1

1 v2

+ "2 c2 )mc2 + To;

For v

< < c.

Hence T will be classical KE if To = -me? Generally, we take, To = 0, so that K4 = t(i~). Using this relation, the fourth component of momentum is P4 = ~T. To express KE in terms of four momentum, we use the Lorentz invariant of four momentum. Now,

12.17

The relativistic Lagrangian

Our prior discussion of Lagrangian and Hamiltonian dynamics is correct only in the non relativistic limit. The extension to a relativistically correct formulation is straight forward, however. For a single particle moving in a velocity-independent potential, we defined the generalised momentum as Pk = ~t. We shall adopt this definition and require that relativistic Lagrangian, when differentiated with respect to qk, must yield the components of the relativistic expression for the momentum . {)L mqk, = -{)'; qk

,= Vl- f32.

12.18. EXERCISE

413

Since this imposed requirement involves only the velocity-dependent part of the Lagrangian, we would expect only this part (the KE) to differ from the non relativistic Lagrangian. If we write the relativistic Lagrangian as

then, we have, mc2 - - - , Integrating ,

,

where the constant of integration can be ignored. Thus, the relativistic Lagrangian becomes,

Note that, this expression is not equal to T - V, as in the non relativistic case. The equations of motion are found by operating the above Lagrangian with the Euler-Lagrange equations in the usual way. The Hamiltonian is given by, H= Li]kPk-L= L k

= mc2,

k

m i]2

r:;--;;? . k +mc2 yl-{32+V }1 - {32

+ V = T + Eo + V = E.

Therefore, the relativistic Hamiltonian is equal to the total energy, including the rest energy.

12.18

Exercise

Ex 12.18.1. Show that, the expression dx 2 + dy2 Lorentz transformations.

+ dz 2 -

c2dt 2 is invariant under

Ex 12.18.2. Prove that, the four dimensional volume element dxdydzdt is invariant under Lorentz transformation. Ex 12.18.3. Show that, under the Lorentz transformation, the wave equation for the propagation of the electromagnetic potential ,,2<jJ - ~~ = 0, where 6,2 = ~~ is the D'Alembert operator remains invariant.

,,2 _

Ex 12.18.4. Find the relativistic expression for KE of a particle, whose rest mass is mo, moving with velocity 11. Obtain the relation E2 = p 2c2 + m5c4. Show also that, in the limit of low velocities the usual expression for KE can be obtained from the relativistic expression of it. Ex 12.18.5. Two electrons move towards each other. The speed of each being O.gc in a Galilean frame of reference. What is their speed relative to each other?

414


Ex 12.18.6. What should be the speed of a particle of rest mass mo in order that its relativistic momentum is mac. Find its total relativistic energy. v'2moc2. Ex 12.18.7. A rod 1m long is moving along its length with a velocity of O.6c. Calculate its length as it appears to (i) an observer on the earth (ii) moving with the rod itself. Ans : O.8m, 1m Ex 12.18.8. The earth revolves around the sun with an orbital speed of 30Km/s. Due to this motion, the diameter of the earth will appear shortened to an observer sitting on the sun. Find the percentage contraction in the diameter of the earth. Ans : 0.5 x 10- 8 . Ex 12.18.9. A rod of proper length L rests in the xy plane inclined at an angle e to the x axis. What does the observer moving with speed v along the x' axis. Find its length and the angle of inclination. Ex 12.18.10. A rod of length l, always parallel to the x-axis of frame 8, moves with a speed 'lJ, in the y direction with respect to s-frame. The ends of the rod are at (0, nt) and (I, ut) in the xy-plane of 8. Another frame 8' is moving with respect to 8 along their common xx' axes with speed v so that the usual Lorentz transformations are valid. Find the tangent of the angle which the rod makes with the x' - axis as observed in the frame 8'. GATE - 1996

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