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~
fi
Unlike the classical case, the particle therefoce cannot be at rest in the bottom of the potential well, since this would cause both <x> and
to vanish. It must necessarily "vibrate a little". A rough estimate of the energy is given by
E
2
=~ 2m
+ tmw2<x>2
We want to minimize this subject to the constraint (5.22). Replacing
and minimizing the resultin~ expression with respect to variation in produces the minimum
~/<x>
by <x>
Except for the factor !, which we cannot account for by such a primitive argument, this is the same as the zero-point energy (5.21).
Once we have the propagator at our disposal we also control the dynamical evolution of the Schrodinger wave function.
As in the
free-particle case we can now find an exact and especially simple solution to the Schrodinger equation. the normalized wave function
At the time
t=O
we consider
177
(5.23)
1jJ(x,O)
=
'rmw
y'~
I
mw 2 exp[ - 2fl(x-a) ]
This corresponds to a Gaussian probability distribution centered at the point
x=a.
At a later time
t
the wave function will, accord-
ing to (2.18) evolve into IK(y;tIX;O)1jJ(X,O)dx
1jJty,t)
4/iii;; 2/ lItJJ V ~ V21TlflSinwt
=
I
illtJJ 2 2 exp[2li.Sinwt[(y +X )Coswt - 2yx} -
2 2li.(x-a)] dx
IlliJ
As usual this is just a Gaussian integral. Using the identity
illtJJ 2 2 2fISinwt [(y +x )Coswt - 2yxJ -
lItJJ 2 m(x-a)
lItJJie iwt . t 2 2 1W (aSinwt - iy)] - ~[y 2li.Sinwt [x - iethe integration can be immediately performed, and the wave function at the time t reduces to
(5.24)
~(y,t)
· · WK exp[- ~(y-aCoswt) 2 ]exp[- l~t]exp{_ ~[2aySinwt -
~
=
V
2
~ Sin2wt]}
Apart from a complicated phase factor this is of the same form as (5.23).
Consequently the corresponding probability distribution,
P(x,t)
11jJ(x,t) 12 =
/§I
exp[ -
~w
(x - a Coswt)2]
is still a Gaussian distribution, but this time it is centered at x = a Coswt This is highly interesting because this means that the wave packet
oscillates forth and back following exactly the same path as the classical particle! If
a=O
the probability distribution reduces to the stationary
distribution
corresponding to a classical particle sitting in the bottom of the potential well. state.
function for (5.25)
Thus it is closely related to the classical ground
It should then not come as a great surprise that the wave a=O, i.e.
178
I
represents the quantum mechanical ground state. the simple time dependence
exp[- ~ Eotl. with
equal to the ground state energy
!hw.
Notice that it has Eo
precisely being
It must therefore in fact
be the eigenfunction of the Hamiltonian with the eigenvalue
!fiw.
Exermse 5.2.1 Introduction: Consider a free particle, where the corresponding wave fUnction at the time t=O is given by ~ x2 1jJ(x,O) = ~2 exp[-1j"(?"l Problem: a) Show that its wave fUnction at a later time t is given by
(Unlike a particle which is at »rest« in the bottom of a potential well, this wave packet thus spreads out in time!) b) Show that the corresponding probability amplitude in 'momenttun space« is given by
i2 an
~(k,t) =
2(t)
HIt 2 2 exp[ - k a (t)l exp[8ma2cr2(t)lexp[
and as a consequence
<x> =
'!\ "2
Also the rema~nlng eigenfUnctions can easily be extracted from the propagator. With the third characterization of the propagator in mind we decompose it as follows K(y;Tlx;O) = e- iwT / 2
mw 2' T exp{ mw 2'wT [~(y2+x2)(1+e-2iwT)_ 2xye- iwT l} 11fl(1-e- lW ) 11(1-e- 1 )
vi
If we introduce the variable K(y;Tlx;O)
z=e-
iwT
this can be rewritten as follows
r-mw= z-~ exp[- ~(X2+y2)l~ exp{ J..
(2
2) 2
_ ~[ x +Y1_~2-
This should be compared with
,'"'
E ~~ (y) exp[_i E Tl = z-· E ~ (x)~ (y)zn
n=O n
n
fJ. n
n=O jn
n
If we put
~n(x)
= ~ exp[-
~x2J(2nn! )-!Hn(~)
and introduce the rescaled variables
u=~ it follows that (5.26)
and
Hn(u)Hn(v)
v=~, has the generating formula ( 1 -z 2
,
)-.
exp
2 [uvz - (u2 +v 2) Z 2 1 1_Z2
2
xyzl}
179
I
Since the left hand side is a Tay~or series in z we can actually find 2-n H (u)H (v) by differentiating the right hand side n times and thereafter put z o. I~ fol~ows trivially that H (u) is a polynomial of degree n in u. In fact it is a Hermite polynomial and thg above formula is the so-called Mehler's formula. We leave the details as an exercise:
Worked exercise 5.2.2 Introduction: mula
The Hermite polynomials Hn(x)
z
are defined by the generating for-
n
n=o Problem:
a) Deduce Rodriques' formula 2 d n _x2 Hn(x) = (_On eX e n dx b) Show that e
-x
2
1
liT f
_t,2
e
+ 2it,x dt,
and consequently H (x) = (_2i)n e x2 n
liT
Jt,n e-t,2 + 2iE,x dE,
c) Prove Mehler's formula
(5.28)
L
n=o
=
[l-z2j
-! exp [ 2xyz -
(2
2) z2
x 2+ Y - z
Using Mehler's formula we can in fact recalculate the propagator. must then determine the eigenvalues En and the eigenfunctions ~n(x) Hamiltonian: 2 2 h d 2 2 (- + jlllL1l X )$ (x) = E ~ (x) 2m dx2 n n n
First we for the
(This eigenvalue problem is treated in any standard text on quantum mechanics!). Once we have the eigenvalues and the eigenfunctions at our disposal we can finally explicitly perform the summation in (2.70) by means of Mehler's formula. Notice that the phase ambiguity is still present. The Taylor series in Mehler's formula has convergence radius 1, due t~.&¥e poles at z = ±1. We are especially interested in the behaviour at z = e ~ • We are thus actually working directly on the boundary of the convergence domain! This boundary, i.e. the circle Izl= 1, is decomposed into two disjoint arcs by the poles z = ± lJ Furthermore the poles correspond precisely to the caustics wT = nTI. At two times t and t , h separated by a caustic, we have thus no direct relation between the pftases.
180
I
Let us finally tackle the problem of the phase ambiguity in the propagator for the harmonic oscillator. Below the first caustic we know that the exact propagator is given by K( ~,·TI Xa'·0) -- e-i1T/4 v'ZnMInwT rrrr;;- exp {imu 2 2) COSW:r - Z~Xa]} mSinwT [( ~+Xa
(5.17)
When
T
=~
1T T <w
this reduces to the particular simple expression 1T.
K(~;Zwlxa'O)
=e
-i1T/4 !iii;
vin% exp[-
.1Illl
lh~xal
From the group property (2.25) and the translational invariance in time we know that
K(Xb;~IXa;O)
= =
!K(Xb;!wIX;O)K(X;z:IXa;O)dX i1T Z e- / i1TJexp[- i~(Xb+Xa)xl~dx
which by (5.5) reduces to
This takes care of the first caustic. We now proceed in the same way with the second caustic 1T K(xbl : lx a ;O) = J K(x b ; ~IX;O)K(X: ~IXaIO)dX
Continuing in this way we finally obtain the propagator corresponding to an arbitrary caustic: (5.29 ) But the propagator on a caustic serves as the initial condition for the propagator on the subsequent segment, i.e. we must demand lim T ...
!!!
+
K(xb;T!xa;O)
=e
-in.~
n o[x b - (-1) Xa1
w
Consider the expression e
- i 1T /4 / mw { imw [( 2 2) T 1} V21T~ISinwTlexp ZnSinwT Xb+Xa Cosw - ZXbXa
I
181
(Notice that it differs from Feyruman's expression (5.17) by an inclusion of the absolut value of SinwT under the square root). If wT
=
nrr
0 a e +'" 2 iAx dx Tn . 1T Je -~'4 e when A < a
r
jU 1m
=/¥ =
Now, going back to the action of the Fourier path, (5.37)
i
exp lii S 1
we observe that the analogue of
2 2}
2 (k 2 7T an --:2 - w ) k=1 T
N-1 = exp {imT 4n I A
is negative when
k < wT/1T, and
positive when k > WT/1T. Consequently there are Ent[wT/1Tl ti ve terms" (where Ent [x 1 denotes the entire part of x). multiple integral is therefore actually given by
"negaThe
For the free-particle propagator this ambiguity does not occur. corrected formula for the propagator thus comes out as follows:
The
(5.38)
But that is precisely the Feynman-Soriau formula! Now that we have seen how the limiting procedure based upon Fourier paths work, let us return for a moment to the problem of the integration measure. If you want to calculate a path-integral directly, i.e. you do not calculate a ratio any longer, you must necessarily incorporate an integration measure in the limiting procedure. Feynman suggested that one could use the same integration measure as for the piece-wise linear path. This assumption leads to the formula:
I
188
x(T)=O
exp{i}[~(dX)2_ TIO 2 dt
(5.39)
J x(O)=O
r;;- N J
lim N ...
V(x)]dt} D[x(t)]
(V21iThi) a>
J . N-1 k t ... exp{i' S[ L akSirr.f-]}dx ... dx N- 1 k=1
lim N ...
a>
But here the right hand side diverges as you can easily see, when you try to calculate it in the case of a free particle, cf. exercise 5.3.1 below:
Exercise 5.3.1 Introduction: Consider the matrix A involved in the linear transformation (5.34), i.e. 'k Sin [,T I ] j,k 1, ... ,N-l Ajk N
a) Show that
N-1
AXT
N
I
=2
Det LA.]=
and
[~]-Z
b) Show that the series involved in the limiting procedure (5.39) is given by lim N-""
1
[N]
r (N)
TI
N-1~ 21TihT
and use Stirlings' formula for the r-function to verify that it diverges. In fact, by considering the case of the free particle propagator, it is not difficult to construct the correct integration measure, which (when you integrate over the Fourier components) turns out to be given by:
XjT) =0 i exp{[ S[x(t)]}
D[x(t)]
=
1T N-1
Nl!~ (vrz)
rm
r(N)[V~]
NJ
r
. N-1
1
.
k1Tt]} N-1
_jexp{~[k:1akSln.y- d
x(O)=O (As a consistency check you can u~ this formula to rederive the propagator of the harmonic oscillator). So now you 'see why we prefer to neglect the integration measure: Every time we introduce a new limiting procedure, i.e .• a new denumerable complete set of paths, we would have to introduce a new integration measure!
5,4
ILLUSTRATIVE EXAMPLE: THE TIME-DEPENDENT OSCILLATOR
As another very important example of an exact calculation of a propagator we now look at the quadratic Lagrangian: L
dx 2 = tm(at)
- tmW(t)x
2
a
189
I
Since it is quadratic we can as usual expand around a classical solution, so that we need only bother about the calculation of the following path-integral x(tb)~O
J
t imfb dx 2
exp{211 [(CIt)
x(t }~O
ta
a
Let us first rewrite the action in a more suitable form.
Performing
a partial integration we get tb S [x (t)]
=- ~ J [ ta
Here we have neglected the x(t a )
tions
= x(t b ) =
O.
boundar~
terms due to the boundary condi-
Inserting this we see immediately that
the above path-integral is actually an infinite dimensional generalization of the usual Gaussian integral since it now takes the form:
x(tb)=O
J
t
im b dZ exp{- 2fJ. fx (t) [dt Z
+
W(t)]x(t)dt}D[X(t)]
ta
x(ta)=O
To'compute it we ought therefore to diagonalize the Hermitian operator -
d
2
dt
+ WIt)
2
At the moment we shall however proceed a little different.
By a
beautiful transformation of variables we can change the action to the free particle action!
Let
f
be a solution the second order
differential equation
d2 {-2 + W(t)} fIt) dt
(5.40)
i.e.
f
=
a
belongs to the kernel of the above differential operator.
The solution can be chosen almost completely arbitrarily within the two-dimensional solution space. that
f
The only thing we will assume is
does not vanish at the initial endpoint: f (t a ) " O.
(Notice that
fIt)
is not an admissible path, since it breaks the
boundary conditions!).
Using
f
we then construct the following
linear transformation, where x(t) is replaced by tha path y(t):
r
190 t
(5.41 )
x(t)
f (t)
J
~ f (s)
ds
ta Here we assume that the transformed function y(t)also satisfies the boundary condition y(t ) = 0 a Differentiating (5.41) we obtain t
x' (t) = f' (t)
J Yf'(~sl
ds + y' (t)
f' (t)
"""f'(t) x(t) + y' (t)
ta so that the inverse transformation is given by t
(5.42)
y(t) = x(t) -
I
f' (s) x (s) ds
fTs)
ta We can now show that the above transformation has the desired effect. Using that t
x"
f"
(t)
J Lill f(s)
(t)
d
s
+ ft (t)f (t) fIt
+ y" (t)
we obtain:
{;i$
t
+ W(t)}x(t) ={f"(t) + W(t)f(t)}
I[~'(~S)]dS
+
f'n~r(t)
+y"(t)
ta But here the first term vanishes on account of (5.40).
Consequently
the action reduces to:
~
S[x(t)] =
-1 J
t
[F(t)f' (t)y' (t) + F(t)f(t)y"(t)] dt
with
F(t)
= J L...!& f(s) ds
ta Performing a partial
integrat~on
on the second term this can be
further rearranged as: tb
s[x(t)
- 1 [X(t)Y'(t) Jt
1
a
But here the boundary terms vanish due to the boundary conditions satisfied by
x(t).
Thus we precisely end up with the free-particle
action in terms of the transformed path
y(t)!
191
I
There is only one complication associated with the above transformation, and that is the boundary condition associated with the final end-point
tb'
The boundary conditions satisfied by
x(t)
are transformed into the following conditions on the transformed path
y(t): tb
J
~dS
o
ta The second boundary condition is non-local and therefore not easy to handle directly.
We shall therefore introduce another trick!
the identity, Ii (x(t )) = -,}; b cf.
J exp
Using
{-ia x(t ) }da
b
(5.5), we can formally introduce an integration over the final
end-point: x( t ) arbitrary b
x(1)=O
2~
J exp{iS[X(t)]}D[X(t)]
Jf~eXP[-iaX(~)]eXP{kS[x(t)]}da D[x(t)]
x(ta)~O
x(ta)=O
This is because the integration over
a
now produces a Ii-function
which picks up the correct boundary condition!
(Notice that if we
attempt to calculate the path integral by a limiting procedure we must now also integrate over the final end-point
x ). N
Changing
variables we then get
Here the infinite dimensional generalization of the Jacobi-determinant is independent of linear.
y(t)
because the transformation (5.41) is
The remaining integral is Gaussian.
"Completing the square"
the whole formula therefore reduces to ~
1
OX = #ed liy ]
J
exp{ -
-'" with
y(tb ) arbitrary b dt im bdy 2 f (\)f -z-}da exp{crifCat) dt}D[y(t)] t f (t) t a yet )=0 a a t
ill Z 2
2iif
t
yet)
~
yet) -
~f(~)ff~) ta
•
J
I
192 At this point we get a pleasant surprise: the a-integration!
We can actually carry out
Furthermore the remaining path-integral is with-
in our reach, since it only involves the free particle propagator. In fact we get:
y(1)
arbitraIb
Jexp{~ {(~~)2dt}D[y(t)] y(ta)=O
== f -0>
Ko (x;1b IO ;ta )dx
a
==
lz C1b 7Tih
0>.
-t )
a
LeXP[~
2
(1, ~tidx ==
This should hardly come as a surprise since, by construction, the above path-integral represents the probability amplitude for finding the particle anywhere at the time
tb.
The total path integral thus
collapses into the simple expression:
- W(t)x 2]dt}D[X(t)]
(5.43)
It remains to calculate the Jacobiant! using a very naive approach.
We shall calculate it
(The following argument is only includ-
ed for illustrative purposes, it is certainly not a rigorous procedure!).
As in the approximation procedure for path-integrals we
discretize the linear transformation by introducing a time-slicing. The paths
x(t)
and
y(t)
are then replaced by the multidimensional
points
The linear transformation (5.42) can then be approximated by T
- N
n
f' (t k )
(x k " x k - 1 )
k~ 1 ~ ---'2"---:":"--:""
(This is actually the delicate pOint, since the discrete approximation of the integral is by no means unique, and the Jacobi-determinant turns out to be very sensitive to the choice of the approximation procedure).
Okay, so the Jacobi matrix has now been replaced
by a lower diagonal matrix. from the diagonal, i.e.
The determinant thus comes exclusively
I
Taking the limit
N
we then find:
-T
N
lim N +
exp[log n (1 k= 1
a>
£'(t k ) T
N
lim N +
exp[ E log(1 - ! £( t )
N) 1
k
k= 1
a>
tb
f'J11 exp[-! T(t)dtl
fta
Consequently
Inserting this into
(5.43)
our formula for the path-integral fi-
nally boils down to the following remarkable simple result:
The path-integraZ corresponding to the quadratic action (5.44)
t t m b dx 2 2 m b d2 S[x(t)l == 2 f[(dt) - W(t)x ldt == - 2 fx(t) [dt2 ta ta
+
W(t)lx(t)dt
is given by x(~)=O
(5.45)
f
exp{~[x(t)l}D[x(t)l
x(ta)=O where
f(t) is an almost azobitrary soZution to the differentiaZ equation 2
{~2
+
W(t)}£(t)
the onZy constraint being that
=
0
f(t a )
~
O.
Notice that the above formula in fact includes the free-particle propagator (with W(t)=O) and the harmonic oscillator (with W(t)~2). You can easily check that
(5.45)
in these two cases by putting
reproduces our previous findings
f(t):1, respectively
f(t)=Cosw(t-t ). a
194
5.5
I
PATH-INTEGRALS AND DETERMINANTS Now that we have the formula for the propagator corresponding to
a quadratic Lagrangian at our disposal, we will look at it from a somewhat different point of view. tb Sex (t)
;:J
1
x(t)
d
Since the action is quadratic,
2
{~+
W(t)} x(t)dt
dt
we can "diagonalize" it.
For this purpose we consider the Hermitian
operator d
2
- -:-2 - WIt) dt which acts upon the space of paths,
x(t), all satisfying the bound-
x(t ) = x(t ) = O. b a normalized eigenfunctions ¢n(t):
It possesses a complete set of
ary conditions:
A given path,
x (t),
can now be approximated by a linear combination tb
N xN(t)
I
with
an¢n (t)
n=1
an
f
¢n(t)x(t)dt
ta
Notice that the corresponding action of the approximative path reduces to
The summation over approximative path's can therefore immediately be carried out since it is just a product of ordinary Gaussian integrals:
I. ·f exp{~s In the limit where
[XN (t)
N
+
=
1 }d3., ... da N the approximative paths fill out the whole
space of paths and the path-integral is therefore essentially given by
By analogy with the finite-dimensional case we define the determinant of the Hermitian operator eigenvalues.
_a t2
- WIt)
to be the infinite product of
Of course the determinant will in general be highly
195
I
divergent but we can "regularize" it in the usual way by calculating the ratio of two determinants.
From the above calculation we then
learn the following important lesson:
The path integral corresponding to a quadratic Lagrangian is essentially given by the determinant of the associated differential Operator, i.e·X(s,)=o ~
f
(5.46)
exp{~fX(t)[-:tr2-
x(t )=0 a
MDet[-~2- W(t)J}-~
W(t)]x(t)dt}D[x(t)] =
ta
hlhere the right hand side should actually be interpreted as a limiting procedure (i.e. it is a short hand version of the follOhling expression: N
(5.47)
lim
_1
L',(N)[ IT;')
N ->- co
2
fexp[~ l: Ana~l(vj~~/dNa N
lim L',(N)f.-
=
n=1
N ->- co
n"'1
Notice that we have included a factor
(12 ~ft f in
11
the integration
over the generalized Fourier components, i.e. the proper integration
measure for evaluating the determinant is given by
~
(5.48)
k=1
[/2:Hi da k ]
Apart from that we still need a further integration measure
L',(N)
since the determinant is only proportional to the path-integral.
The
above characterization of the path-integral is the one used by e.g. Coleman [1977]
.
Since we already know how to compute the path integral we can now extract a relation for the determinant.
To avoid divergence problems
we calculate the ratio of two determinants.
According to (5.45) it
is given by
Det[-a~- W(t)]
(5.49)
Det[-a~- V(t)]
In this formula it is presupposed that at
tao
fW
and
fV
does not vanish
It can however be simplified considerably by going to the
singular limit, where
fW
and
fV
does vanish at
tal
Let us de-
I
196 note by
{- a~ -
fO
the unique solution to the differential equation,
W
W(t)}f(t)
=
0,
which satisfies the boundary conditions
2.. fO (t ) dt w a Similarly we denote by
f~
= 1
the solution which satisfies the boundary
conditions:
o In the above identity (5.48), we can then put fO
f = fO + d 1
f1
W+ E W
V
V
V
It follows that and Finally the limit of the integral tb
f
dt ta [f (t)]2
w
fO at tao But since almost all W the contribution to the integral comes from an "infinitesimal neigh-
diverges, due to the vanishing of
ta (in the limit where
bourhood" of
s+O), it follows that i t di-
verges like
Consequently lim
s
+
0
tr, f -w
dt
{
t
f2 (t) }
a Thus the identity
I{
(5.49)
ftr, e
dt (t)}
,V a collapses into the extremely simple detert
minantal relation: Det[ -a~-W(t) ]
(5.50)
DetH~-V(t) ]
If, eg., we put V(t)=Q
(and consequently
Det[-a~-w(t)l
DetH~l
f~(t)=t-ta) it follows that
197
I
This can be used to calculate a propagator (corresponding to a quadratic Lagrangian) relative to the free-particle prop~gator Ko: . (5.51)
K(~;tblxa;ta)
K(O;tbIO;t a ) exp{~[xcll}
==
(
1
Det{-d~-W(t)}]-2 KO(O;tbIO;t ) Det{-d~} a
-I
m
2TIiflf~ ( tb )
exp{~s[xCl]}
E.g. in the case of the harmonic oscillator we put W(t)=w 2 and consequently 1 fW(t); ;;Finw( t-t ). In this way we recover the by now well-known result (5.11). a /
Rema:r>k: Incidenti.lly the
investigation of determinants corresponding
to linear operators has a long tradition in mathematics.
E.g. the
basic determinantal relation (5.50) has been known at least since the twenties*).
This is very fortituous, since in our derivation some
dirt has been swept under the rug.
The passage from quotients of
path-integrals to quotients of determinants only works if the integration measure
n(N)
introduced in (5.47)
of the potential function
WIt).
is actually independent
Since the result we have deduced,
i.e. the determinantal relation (5.50) is known to be correct, we have thus now justified this assumption. In fact the determinantal relation (5.50) beautiful reasoning which emphasizes its basic who are acquainted with more advanced analysis argument: Consider the expressions 2 Det(- a g(A)
can be proven by a very general and position. For the benefit of those we include the main line of the
WIt) - A) t 2 Det(- a - V(t) - A) t
and h (A)
is important for the following that A is treated as a complex variable. can be proven quite generally that a differential operator of the form,
It
- a2t
- WIt)
has a discrete spectrum of real eigenvalues 1.. 1 ,1.. 2 ' ••• , An' •••
*) J.H. Van Vleck,
Proc.Nat.Akad.Sci.
li,
178 (1928).
It
198
I
which are bounded below and which tend to infinity as n~. Each of these eigenvalues has multiplicity one, i.e. the associate eigenspace is one-dimensional. When A coincides with one of these eigenvalues, say A = An' it follows that the "shifted" operator 2
{-at - wit) - An}
has the eigenvalue zero, so that its determinant vanishes. As a consequence the function g(A) becomes a meromorphic function with simple zerOs at the eigenvalues A~ and simple poles at the eigenvalues A~. Consider next the function to the differential equation
f~+A(t).
{- at2
-
By definition it is the unique solution
w(t)
-
A}f (t) = 0
which satisfies the boundary conditions
flta) It fo llows that
=0
is an eigenvalue of the operator
WIt)} if and only if
o
f IW+A) It b ) = 0
and when that happens f(W+A)(t) is in fact the associated eigenfunction (except for a normalization factor). As a consequence the function heAl V becomes a meromorphic function with simple zeros at AW and simple poles at A k The quotient function n' g (A) /h IA) is thus an analytic function without zeroes and poles! (Thus it has a behaviour similar to e.g. exp[A]). Furthermore, from general properties of determinants (respectively solutions to differential equations) it can be shown that g(A) (respectively heAl) tends to 1 as A tends to infinity except along the real axis. The same then holds for the quotient, which as a consequence must be a bounded analytic function. But then a famous criterium of Liouville guarantees that it is constant, i.e.
Specializing to
5,6
A=O
**=
1
we precisely recover the determinantal relation (5.50).
THE BOHR-SOMMERFELD QUANTIZATION RULE
we shall now encounter the important problem of computing quantum corrections to classical quantities. Especially we shall consider quantum corrections to the classical energy. In the case of a onedimensional particle we know that it can be found by studying the trace of the propagator. Inserting the path-integral expression for the propagator, this trace can be reexpressed as
I
199
x(T)=x
(5.52)
j J eX~t~S[X(t)]}
G(T)
-00
x(O)=x
D[X(t)]dx o
o
== J exp{~S[X(t)]} D[X(t)] x(O)=x(T) where we consequently sum over all path's which return to the same pOint.
This expression is known as the path-aum-traae integral.
There are now in principle two different approximation procedures available for the evaluation of this path-cum-trace integral. first method is the hleak-aoupling approximation.
The
Let us assume that
the potential has the general shape indicated on fig. 58.
To calcu-
late the low-lying energy states we notice that near the bottom of the potential well, we can approximate the potential by V(x)
RJ
W"
(0)x
2
I
iV"(O)x 2
E
:
E"
E3
E2 El
Eo
X-axis Fig. 58
Thus the problem is essentially reduced to the calculation of the path-cum-trace integral for the harmonic oscillator! already solved, cf.
This we have
(5.19-20) and the low-lying energy states are
consequently given by (5.53)
with
v"(O)
mw
2
The second method is the so-called WKB-approximation which can be used to find the high-lying energy states. non-perturbative method.
It is thus essentially a
It leads to the so-called Bohr-Sommerfeld
quantization rule, which in its original form states that
200
f pdq
(5.54) where
q
= n
I
• h
is the position coordinate,
we integrate along a periodic orbit.
p
the conjugate momentum and
Notice that the left hand side
is the area bounded by the closed orbit, cf. fig. 59, so that the quantization rule states that the area enclosed in phase-space has to be an integral multiple of
h.
The original WKB method which was
based upon the construction of approximative solutions to the schrodinger equation, will not be discussed here.
But before we jump out
in the path integral version of the WKB method it will be useful to take a closer look on classical mechanics and the original derivation of the quantization rule (5.54).
p-axis Phase-space diagram.
--+--"'------""'t'---f-"'--..... q-axis
ILLUSTRATIVE EXAMPLE:
Fig. 59
THE BOHR-SOMMERFELD QUANTIZATION RULE
Let us first collect a few useful results concerning the action. Suppose
(xa;t a )
and
(xb;t b )
are given space-time points.
Then
S(Xb;tb[xa;t a ) will denote the action along the classical path connecting the two space-time points.
The
res,ult~ng
function of the initial and final
space-time point is known as Hamilton's principal funation.
(If
there is more than one classical path connecting the space-time points (xa;t a ) and (xb;t b ) it will be a multivalued function). All the partial derivatives of Hamilton's principal function have direct physical significance:
as
(5.55) Here
~ Pa
as
Pa
ata
E
as
at b
is the conjugate momentum at the initial point
the conjugate momentum at
xb
and
served along the classical path).
E
E
xa ' P b is the energy (which is con-
201
I
Proof: A change, 6x b , in the Epatlal position of the final space-time point will cause a change, 6x(t), in the classical path. The corresponding change in the action is given by tb
J [~~
6S =
6x(t) + aL Qx(t) J dt ax
ta
~
tb CaL
J
x
t
-~~J dt "aX
6x(t)dt + [aL "aX
6x(t) ]
a But here the integrand vanishes due to the equations of motion and we end up with: 6S
= a~(tb)QX(tb) ax
It
-
aL ) 6X b = -;-(t b
aL(t )6x(t ) • a a ax
Pb 6X b
ax
follows that 6S 6X b
Pb Similarly we may consider a change,
6t , in the temporal position of the b final space-time pOint. Notice first that
The corresponding change in the action is also slightly more complicated,
[~ 6x(t) ax
+
~ Qx(t)Jdt ax
ta where the first term is caused by the change in the upper limit of
As before we can now make a partial inte-
the integration domain.
gration leaving us with the formula 6S
=
aL L(t b )6t b + -;-(t )6x(t b ) b dX
=
L(t b )6t b -
aL' ) J 6tb = [ L(t ) - -;(tb)x(t b
ax
b
But the energy is precisely given by
aL
•
~(tb)x(tb)6tb
dX
202 E =
p~
I
aL
- L
a~
X- L
so that we end up with the identity
-
6S =
as
Le.
Eot b
3t
-
~
E
b
Notice that the above considerations were in fact anticipated in the discussion of the Einstein-de Broglie rules, cf. the discussion in section 2.5. After these general considerations we now return to the discussion of a one-dimensional particle in a potential well like the one sketched on fig.
59.
In the general case there will exist a one
parameter family of periodic orbits whi9h we can label by the fundamental period
T: x = xT(t)
Each periodic orbit will furthermore be characterized by its energy E, which thus becomes a function of the period
T, i.e.
E
= E(T).
From (5.55) we learn that dS = _ E
(5.56 ) where
OT S(T)
is the action of the periodic orbit
xT(t).
We can now
introduce the Legendre transformation of the action function
S(T).
It is defined by the relation: dS W(E) = S(T) - dTT=S(T) + E'T where
T
should be considered a function of
E.
(Notice the simi-
larity with the passage from the Lagrangian to the Hamiltonian).
In
analogy with (5.56) we then get: dW _ dS dT
dT
dE - OT OE + T + E • dE
(5.57)
= -
dT dT E dE + T + E CiE = T
Finally we can get back the actfon function by a Legendre trans formation of
W(E),
(5.58)
where
S(T) E
= W(E)
- T'E
= W(E)
dW - dE • E
should now be considered a function of
T.
Notice that
the Legendre transformation is in fact given by the phase integral in (5.54) T
W(E)
i.e.
S(T) + E'T
f [m(£E) o
T
2
- V(x)]dt +
2
f [m(~~} o
+ V(x}]dt
203
T
(5.59) since
I o
WIE) dx = m at
p
quantity
W(E)
I
2 m(dx) dt
dt
for this simple type of theory.
It is thus the
which we are going to quantizel
From (5.57) we now get,
where
dE
= -T1
w
is the cyclic frequence of the periodic orbit.
dw = ~ dw 2~
w
with
2~ =T
Following
Bohr we then assume that only a discrete subset of the periodic orbits are actually allowed and that the transition from one such periodic orbit to the next results in the emission of a quantum with energy
nw, where
w
is the frequency of the classical orbit.
The
last part of the assumption is Bohr's famous correspondence principle.
It follows that
W can only take a discrete set of values and
that the difference between two neighbouring values is given by
= n.
2~
Thus the quantization rule for W(E ) = n
where
c
2~(n+c)ft
=
2~·n
W can be stated as +
2~ftc
is a constant which we cannot determine from the corre-
spondence principle.
You can look upon this formula as the first
two terms in a perturbation expansion for powers of
1/n.
W(En)
where we expand in
Bohr and Sommerfeld simply assumed it was zero, but
using the WKB-method it was actually found to be ,.
Thus the correct
quantization rule for the above case is given by (5.60)
As an example of its application we shall as usual consider the harmonic oscillator. features:
This example turns out to have two remarkable
1) The assumption of a one-parameter family of periodic
solutions labelled by the period consequence the action function monic oscillator.
T
breaks down completely.
SIT)
As a
cannot be defined for a har-
2) The quantization rule (5.60) is exact.
Since the action function cannot be defined, we shall define W(E)
through the phase integral (5.59).
A general periodic orbit
is given by x(t)= ACoswt +
BSinwt
The associated energy is consequently
E
=
mW 2 (A2 + B2)
while the phase integral turns out to be
I
Thus W(E) = 2rr E w
and the Bohr-Sommerfeld quantization rule reduces to the well-known result (5.20)
o
Now that we understand the Bohr-Sommerfeld quantization rule we return to the path-cum-trace integral (5.52). This time we are going to expand around a classical solution x(t) = xcI (t) + nIt) The potential energy is expanded to second order: V[x(t)]
2
V[Xcl(t)] + V'[xcl(t)]n(t) + Pl"[x c1 (t)]n (t) Inserting this, and using the classical equation of motion, the action then decomposes as follows Rl
T
S[X(t)]
Rl
S[Xcl(t)] +
b{I(~)2-!V"[XCl(t)]n2(t)}dt d2
T
1
+ 1bn(t){-~- ffiV"[Xcl(t)]}n(t)dt
Consequently the path-cum-trace integral reduces to (5.61)
G(T)
Rl
7
l:
-00
x
n(T)=O
exp{~[xcl(t)]} cl
J expHKn{-~- ~Vff[xc1(t)]}ndt}D[n(t)]dxo
n(O)=O
where xcl(O) = xcl(T) =xo' But the remaining path integral we know precisely how to handle. According to (5.45) it is given by
j __m_-J""-Td-t 2rrifJ.f(O)£(T)
with
£2 (t) j
o In fact we can relate f to the classical path: solves the Newtonian equations of motion: 2
d XcI m~
- V'[x Cl (t)]
Differenting once mOre, we thus find 3 d x
m
~ dt
= -
V"[x lIt)] C
The classical path
205
I
Consequently we can put
Thereby the path-cum-trace integral reduces to (5.62 )
T dt
dx
IlxdP o
0
where we sum over all classical paths satisfying the constraint xcl(O) = xcl(T) = xo.
Despite its complicated structure it is
essentially a one-dimensional integral of the type +00
I
exp{if(x)} g(x)dx
Such an integral can be calculated approximatively by using the Thus we look for a point
stationary phase approximation.
the phase is stationary, i.e.
f' (x ) = O. o
Xo
where
We can then expand
around this point: fIx) "'" f (x o ) + ,f" (x o ) (x-x o ) g(x) Rl g(x ) + g' (x o ) (x-x o ) o
2
In this approximation the integral then reduces to a Gaussian integral: +co
(5.63)
f
exp{if(x)}g(x)dx
Rl
g(xo)eXp{if(Xo)}vI£,,(:~)
Using this on the path-cum-trace integral (5.62) we see that we need only include contributions from the classical paths which produce a stationary phase, i.e.
o = xa ao
But according to (5.55) this gives
o i.e. the momenta at
t=O
and
P b - Pa t=T
are also identical.
Thus the
stationary phase approximation selects for us only the purely periodic solutions! od
T.
They can be parametrized by their fundamental peri-
Corresponding to a given
T
we should therefore only in-
clude the contributions from the periodic orbits:
206
I
x T/ 3 (t) r •.. rXT/n (t) I •••
XT / 2 (t)
Before we actually apply the stationary phase approximation we can therefore restrict ourselves to a summation over these periodic orbits:
N::>tice that the parametrization of the periodic orbit is only determined up t, i.e. we can choose the starting point quite arbitrarily on the closed path. p-axis Each of the x-values between the turning points x 1 and x 2 occurs furthermore twice on the path, cf. fig. 60. We are now ready to perform the x -integration. The x x 0 ~~____________-+____+-__-+~2~.. action of the periodic orbits does X-axis not depend upon the choice of the starting point, and neither does the expression to a translation in
T
Fig. 60
J
o
dt (~) 2
Thus the only contribution to the integral comes from x2 +00 dx o 2 dt = ~ I~T(O) I x 1 Cit
J
f
J
i
n
Having taking care of the xo-integration we must then compute the integral over the reciprocal square of the velocity. Here we get ~2
T
Jo
•
dt
(x T / n )
2
2n
J
3
(2[E(T/n) - V(xB-~) dx m
The same integral can however be obtained by a different reasoning! Consider the phase integral W(E), which is given by x 2
WeE)
= 2n
J /2[E
- vex) 1 dx
x1 for the periodic orbit in question. respect to E we obtain:
Differentiating twice with
207
I
-3/2
dT dE
(2[E - V(x)])
dx
consequently dt
T
Jo
•
(x T / n )
_ m3/2 dT dE
2
Putting all this together, the path-cum-trace integral thus finally reduces to: G(T)
(5.64)
Rl
1.
I
_1_
m I2nifl n=1
-[T]n nT~dEI dT
exp{i nS - } ft
T T~
n
This was the hard part of the calculation! energy levels.
Now we can extract the
As usual this is done by gOing to the transformed
path-cum-trace integral and looking for the poles. path-cum-trace integral is given by, cf. i
G(E)
(Notice that
E
G(T) exp{fi·ET} dE
now both represent an integration variable and the
energy of the periodic orbit. latter as
The transformed
(5.9)
Ecl I).
To avoid confusion we shall write the
Inserting the above expression for
G(T)
we
obtain
G(E)
i
Rl
1
iiiKv'21Til'i
""
n~,Jo
Here it will be preferable making an exchange of variables, i.e.
~
~
= Tin,
is the fundamental period of the periodic orbit in question:
G(E)
Rl
. 1 ~v'21Ti~
""""f· ~ dT E exp{1f(ET+S[T])}T/I~Iv'n
n=1
T
o
The T-integration is then performed by use of the stationary phase approximation.
A stationary phase requires
a o = 1T(EI
+ S[T])
Thus for a given value of
E
=E
as
- ~
=E
- Ecl
we get the main contribution from the
periodic orbit which precisely has the energy (5.63) the stationary phase approximation now
E! give~:
According to
208
G(E)
Rl
~ jZTTift
2- _,_
mI'lv'ZTTifl n='
n
I
_,_
fd2S cr:rr
Using that dE
and
dT
W(E)
+ ET (E)
S (T)
the formula finally reduces to
(5.65)
G(E)
Rl
~
T(E)
~
exp{%nW(E)}
n='
i mt
T
(E)
This clearly has poles when (5.66) and thus we have precisely recovered the old Bohr-Sommerfeld quantization condition (without the half-integral correction term).
The
reason we missed the half-integral correction term is, however, very simply:
It is due to the fact that we have not taken into account
the phase ambiguity of the path integral.
precisely as for the
harmonic oscillator we should pick up phase corrections. pression (5.62) is completely analogous to (5.38).
The ex-
The additional
phases then come fran the singularities in the integrand
T
-,
Uo f2 ~:) ] These singularities correspond to the zero's of
f(t)
=
x(t), i.e.
In the present calculation the x and 1 are thus analogous to the caustics. Each time we pass
to the turning pOints turning
~ts
a turning pOint we therefore phase-factor
e
iTT/2
points in the orbit
, cf.
e~pect
.
(5.29).
xT/n
that we pick up an additional
Since there are
2n
turning
we consequently gain an additional phase
e- iTTn = (_1)n which should be included in the formula for the propagator.
The
transformed path-cum-trace integral (5.65) is thereby changed into
(5.67)
G(E)
Rl
~
T(E)
~
n=1
(-1)nexp {%nW(E)}
209
I
This causes a shift in the poles, whichare now given by (5.68)
and that is precisely the correct quantization rule according to the original WKB-calculation. Notice too that near a pole we get the expansion
Using the approximation 1 + exp4i W(E)}
Rl
it follows that
1 + exp{k W(E
ll
G(E)
)
}exp{~ T(E) (E-En )} = 1 - {1
+
~ T(E) (E-En )}
behaves like G(E)
Rl
1 E--=-E n
when
E
Rl
En'
This should be compared with (5.10) and it clearly
shows that the approximative path-cum-trace integral has the correct asymptotic behaviour close to a pole. we have been working hard to derive a result which, as we have seen, can in fact almost be deduced directly from the correspondance principle!
When we are going to use the path-integral technique in
quantizing field theories we will actually have to work still harder Before we enter into these dreadful technicalities I want to give a few examples of almost trivial applications of the preceding machinery. The first example is concerned with the free particle.
To find
the energy levels we enclose the free particle in a box of length L
and use periodic boundary conditions.
Thus we have effectively
L.
replaced the line by a closed curve of length effect that the energy spectrum is discretized. uous free-particle spectrum back by letting
L
This has the We get the contin-
go to infinity.
The free particle can now execute periodic motions by going around and around the closed circumference:
The periodic orbit
given by xT(t) =
Lt T
It has the following energy and action: m dx 2 _ mL 2 E(T) = Z(dE) SIT)
-;;Z
ET
x
T
is
210
I
Consequently W(E) ; SIT) + ET ; LI2mE Since there are no turning pOints in this problem the quantization , rule is given by (5.66), i.e. Ll2mEn
21Tnfl
i.e. (5.69 )
These are the correct energy levels in the discrete version.
To see
this we notice that the free particle Hamiltonian is given by
~
fi2
d2
H=----
2m dx2
Since the SchI1.5dinger wave function must be periodic, the eigenfunctions are given by 21Tnx] o/n(x) = exp [ ± 1. -Land the corresponding eigenvalues precisely reproduce the above result (5.69)
The second example is concerned with a field theory.
In the
discussion of the sine-Gordon model we found an interesting family of periodic solutions:
The bions (or breathers).
In a slightly
changed notation, where we emphasize the cyclic frequency are given by (cf. ~(x,t)
(5.70)
= ~4
w, they
(4.50): n Sinwt ] with Arctan [ Cosh(nw~
n
/].l_w 2
= ---w--
and
O<w
Here.the parameter So
is given by
(5.110) .Thus Co can be computed once
we have specified the potential. follows that
$0
Since
Xo
is an even function.
is an odd function, it
Consequently we may summa-
rize its asymptotic behavior as follows: (5.120)
I
CO e
$0(1:)
Co
-W1:
e W1:
as
T
as
1"..... T ...,..
~
+00
The second order differential equation (5.118) has two independent solutions. $0
The second one can be constructed explicitly from
as follows:
Put 1: dS
Jo $~
(5.121 )
(Notice that $0
has no nodes and that
$0
since the integral is 1:-dependent).
(S) ~o
identity we now get 1: I
~o (1:)
$~(1:)
~~ (1:)
~~ (1:)
dS
J0 $~ (S)
1
+~
and 1: dS
J $~ (S) 0
They imply the following two identities: (5.122)
W[$o;~o]
d1/J
= $0
erro -
~o
is not proportional to
Differentiating the above
d$o
2
~ (O,O, ... ,O,t)
They are called the coordinate lines. By differentiation we now obtain the tangent vectors
..
(6.6)
e
n
They are called the canonical frame vectOl's. To investigate 1
dA dt
-n
x
,Jt
a
2
x
them a little closer we will
1
write out the parametrization of the coordinate lines in
U
their full glory:
Adt)
FIG. 91
~l(t'O'.'.'O)l = . , .•. ,
[
·N et> (t,O, ... ,a)
By differentiation we get
[
~~ (0, ••• ,a ,t) et> (0, ...
,a ,t)
l.
271
...e
II
a
theY'l1lore, when XCt=XCt(A) extremizes the funational I, then Ct ax ax S is constant along the geodEsic, so that we can idEntify gCtS(x(A)) 7i."A F the parameter A with the arc-length s.
Proof:
Consider the Lagrangian Ct dxCt L(x iax-) = l.;g~V(X(A»
dx~ dx v ax- ~
It leads to the Euler-Lagran0e equations
Le.
but they are identical to the sreodesic e0uations if we can show pro·· perty (2). This is done in the following way: Ct dxo Let XCt=XCt(A) extremize the functional I. Observe that L(x iax-) Ct dx is homogenous in the variable ax- of degree 2. Thus it satisfies the Euler relation 2L
dxCt We must show that L(XCt(A);ax-)iS constant alonCT XCt=XCt(A). Consequently we consider the total derivative: dL dA
USing the Euler-Lagrage equations this is rearranged as dL 2 d)..
where we have taken advantage of the Euler relation
• But this im-
plies immediately that ~~ = O,i.e. A must be proportional to the arclength s.
D
II
298
Observe that when you have found the geodesic equations you can directly extract the Christoffel fields as the coefficients on the rigth hand side. By an abuse of notation the above functional is often denoted as
Exercise: 8. 7 • 2 Problem: Consider the two-sphere 8 2 with the line element ( cf. (6.26) ), ds 2 = dS 2 + 8in2Sd~ (1) Determine the geodesic equations and show that the meridian ~ is a geodesic. (2) Compute the Christoffel field.
= ~o
Worked exercise: 6.7.3 In Minkowski space the line element is given by dT2 = _ dt 2 + dr 2 + r2(de 2 + 8in28d~2)
Probl~m:
when it is expressed in terms of spherical coordinates, cf. (6.39). Compute the aSBociated ChriBtoffel field.
Okay, you might say: This is very nice. We can work in arbitrary coordinates if we are willing to pay the price of fictitious forces and non-trivial metric coefficients and all that, but why should we do it? After all, we have at our disposal a nice family of coordiante systems, the inertial frames, which gives a simple description of physiscs! There is however a subtle reason for working in arbitrary coordinates which was pointed out by Einstein. Fictitious and gravitational forces have a strange property in
co~on:
If you release a test particle in a gravitational field its acceleration is independent of its nass. Two test particles with different masses will follow each other in the gravitational field. This is known as Galilei's principle and it has been tested in modern tines with an extremely high preCision! ROTATING FRAME OF REFERENCE
INERTIAL FRAME OF REFERENCE
x
i
(THE PARTICLES ARE
BO'l'Ii AT REST)
FIG.
112
(THE PARTICLES APPEAR TO BE ACCELERATED)
299
II
But the fictitious forces have the same property: The acceleration of the test particle depends only on the field strength -
rVas '
and
the field strength itself only depends on the metric coefficients. Hence if we release free particles, then their accelerations relative to an arbitrary observer, say a rotating frame of reference, .rill be independent of their masses. Esc:>ecially if
~le
release two free partic-'
les close to each other, then they will stay close to each other! Okay, you might say, I accept that they have this property in common, but have they anything else in cOl'!1I!lon? E'or instance I can transform the fictitious forces completely away by choosing a suitable coordinate system. But you cannot transform away the gravitational field! It is true that I cannot transform away the gravitational field -completely. But I can transform it away locally! Consider a gravitational field, say that of the earth. Now let us release a small box so that it is freely falling in the gravitational field. Inside the box we put an observer with a standard rod and a standard clock.
EINSTEIN BOX
Thus he can assign
coordinates to all the events inside his box (See fig.113) But now the miracle happens. If he releases partic-
OBSERVER
les in his box, they will be freely falling too. Therefore they will move with a constant velocity relative to his coordinate system, and
lLJ
they will act as free particles relative to his coordinate system! Consequently we have succeeded in trans-
SURFACE OF THE EARTH
FIG. 113
COORDINATE SYSTEM IN BOX
forming the gravitational field away inside the box by choosing a suitable coordinate system! When we say that we have transformed away the gravitational field inside the box, this is not the whole truth. Actually if we were very carefull we would find that if we released two test particles as shown on figure 114 they would slowly approach each other. Consequently there is still an attractive force between the two test particles. This force is extremely weak and is known as the Tidal forl;J2. The Tidal force owes its existence to the inhomogenities of the gravitational field, and it is a second order effect, so we may loosely say, that we can
300
II
transform away the gravitational field to first order inside the box! EINSTEIN BOX
So there is a good reason why we might be interested in studying fictitious forces:
It might teach us something about gr(]1)i tationa l forces!
SURFACE OF
Fig. 114
THE EARTH
+ 6.8
CevICTORS The next concept we want to introduce on our manifold M is a co-
vector:
Definition 13 A c ovector
~p
at the point P is a Unear map
~p: Tp(MJ ....
R
Here we will use a notation introduced by Dirac. Given a tangent vector it by
!vp>'
vp
we denote
The symbol
I>
<w I~ > +P P
is
called a keto Similarly a covector ~p is denoted = dt
313
II
* is a linear map Tp(M) ~ R, since the bracket is bilinear. Therefore vp
...
generates a unique tensor of rank 1, which we also denote vp'
Exercise 6.9.1
.
...
...
Problem: Conslder a tangent vector vp ....Show that the components of vp as a tangent vector and the components of vp as a tensor of rank 1 are iaentical.
Ylhen we have tensors of an arbitrary t:'pe at our disposal w" can ge-
neralize the contractions. Let, for instance, Tp be a tensor of type
(1,2). It is then characterized by its components
n with respect to some coordinates (x1, ... ,x ). Here the index a transforms contravariantly and the indices band c transform
covariantly.
But then we may contract the indices a and b obtalltingthe quantity:
As we sum over the index a, this quantity is characterized by only one index
The important point is now that the quantity Sc transforms covariantly! This follows from the computation j a S Ta av Ti ax axk (Toc (21 ac axi C!l jk aya a~/
a j where we have used that (~) and (~) are reciprocal matrices. But axl. aya if Sc transforms covariantly, we may regard it as the components of a cotencor Sp' We say that Sp is generated from Tp by contraction in the first two variables. This is obviously a general rule: Whenever you
contract an upper index with a Zower index, the resuZting quantity transforms as the components of a tensor. (where the degree of course is lowered by two!). The only trouble with contractions is that they are almost impossible to write in a coordinate free manner! As contractions play an important role in the applications, we will therefore write down many equations involving tensors using component notation.
314
II
In the following table l'/e have summarized the most important properties of mixed tensors:
A mixed tensor Tp of type termined by its components i
Components
T
l • .. i
k
.
. = T (dx
JI'''~
i
l
(k,~)
is completely de'
; ... ; dX1k;~. ; ... ;~. )
P
JI
J~
with respect to a coordinate system. The indices i l , ... ,i transform contr8:variantly , and the indices k jl, •.. ,j~ transform covariantly.
If Sp andTp are mixed tensors of the same type and A 1S a real scalar, then you can form the mixed tensors Sp+Tp and ATp of the same type (k,~). Furthermore these mixed tensors are characterized by the components (k,~),
Linear structure
(6.71)
Tensor product
If Sp and Tp are mixed tensors Of type (kl'~I) and (k 2 ,P-2), you can form the mixed tensor S1' ~ Tp of type (k l + k2 '~l +P-2). The tensor product is characterized by the components
(6.72)
S
il ..• i k
il ... i 1. T JI"·Jt
1
k
2
jl·"j~2
If Sp is a mixed tensor of type (k,~), you can form a mixed tensor Tp of type (k-l ,~-1) by contracting a contravariant and a covariant index. The contraction is characterized by the components
Contraction
(6.73)
In what follows we are going to deal a lot with tensor fieZds.
To con-
struct a tensor field T of rank 3 we attach to each point P in our man nifold a tensor Tp of rank 3! Let us introduce coordinates (xl, ... ,x ) n on M. Then the tensor at the point P(xl, •.. ,x ) is characterized by its components T
abc
(x 1 , ... ,x n )
315
II
\·7e say that the tensor field T is a smooth field if the components Tabc(X', ... ,xn) are smooth functions of the coordinates. If nothing else is stated we will always assume the tensor fields to be smooth.
The unit-tensor field.
Illustrative example:
Let M be an arbitrary manifold. Then we construct a mixed tensor of type (1,1) in the following way. At each point P we consider the bilinear map
The corresponding components are i .... To( dx ie.) • J
'" < ax
i
" l e.> J
i.e. the Kronecker-delta! It is a remarkable fact that the components of this tensor have the same values inall coordinate systems. As the components are constant throughout the manifold, they obviously depend smoothly upon the underlying coordinates! We therefore conclude:
The Kronecker-delta oi. are the components of a smooth tensor field on M, called J
the unit tensor field of type (1,1).
Exercise 6.9.2 Problem: Show that the Christoffel fields r
V
as
=
l.
..,g
v].! [a/>:].l(:t
a/>:S].!
g a aS ]
ax
ax].!
--S+-a --ax
are not the components of a mixed tensor field. (Hint: Show that they do not transform homogenously under a coordinate transformation. )
Exercise 6.9.3 n Problem: Consider a point P on our manifold. To each coordinate system (x', ... ,x ) around P we attach a quantity T·lJk with two upper indices and one lower index. Apriori the upper indices need not transform contravariantly, and the lower index need not transform covariantly. Show the following:
If the quantity U~ given.by Q, " RU k - T1-J S
-
k
ij
are the components of a mixed tensor of type (1,1) .Ulhenever S1.ij are the eomponents of a mixed tensor of type (l j 2), then p1-J k are the components of a mixed tensor of type (2,1)
316
II
The method outlined in exercise 6.9.3 is very useful when you want to show
that a given quantity transforms like a tensor. Clearly it
can be generalized to mixed tensors of arbitrary type. Suppose now that we have attached a
metria g to our manifold M. Then
we have previously shown (Section 6.8) how to identify tangent vectors and covectors using the bijective linear map
I:T;(M) - Tp(M)
,
generated by the metric. If we write it out in components it is given by
Exercise 6.9.4 Problem: Show that I(
axk )
is characterized by the components gki
11/e can now in a similar way identify all tensor spaces of the same rank. For simplicity we sketch the idea using tensors of rank 2. Let T be a
cotensor of rank 2 characterized by the
covariant com-
ponents T ij . Then we identify T with the following tensor of rank 2: II (T)
(~;4)d~f.T(I(~) ;In two. o.f the indices must always co.inside (since an index a can i o.nly take the values l, •.• ,n), whence F aj ••• a vanishes! So. in the k case o.f k-fo.rms we do. not have an infinite family o.f tenso.r spaces! We will use the convention that co.tensors of rank 1 are co.unted as l-fo.rms. Of ~o.urse, it has no. meaning to. say that a co.vecto.r is skewsymmetric, but it is useful to. include them amo.ng the fo.rms. In a similar way it is useful to. treat scalars as O-fo.rms. Co.nsequently the who.le family o.f fo.rms lo.o.ks as fo.llo.ws:
*
n
n+l
A~ (M) =R;A~ (M) =Tp (M) lJ\.~ (M) l ••• lAp (M) lAp
n+2
(M) ={O} lAp
(M) ={O}; •••
Wo.rking with o.rdinary co.tenso.rs we have previo.usly intro.duced the tensor pro.duct: If F and G are arbitrary co.tensors then the tenso.r pro.duct F €II G def ined by
is a tria
co.tenso.r o.f rank k+m. But if we restrict o.urselves to. skewsymmeco.tenso.rs this co.mpo.sitio.n is no. lo.nger relevant because F ~ G
is no.t necessarily skewsymmetric.(If yo.u interchange V. and 1.
say no.thing about what happens to. F(Vji ••• ;vk)G(Uji •.. therefo.re try to. mOdify this compo.sitio.n:
;rtm».
rt.J
yo.u can
We will
If Waj •••• a is a quantity with indices aj, ••• ,a k then we can co.nk struct a skewsymmetric quantity in the fo.llo.wing way (7.1)
w[ aj ••••• a 1 k
1 = k!
!(_l)TI w TI TI
(all ••• TI (a ) k
329 where we sum over all permutations
TI
II
of the indices al, ... ,ak,and
(-If
is the sign of the permutation, i.e. (-1)
TI
(-1) TI
= =
+1
if
11
is an even permutation
-1
if
11
is an odd permutation
For instance we get W[ab]
1 = 2![Wab
- wba]
and W[abc]
=
1 3! [w abc + wbca + wcab - wacb - wcba - wbac ]
The quantity w[al .... ~] is called the skewsymmetrization of wal ••• ak . ] is completely skewsymmetric, and that if ~ wal ....• ak is born skewsymmetric, then
Observe that w[a
I····
1 (This is, of course, the reason why we have included the factor k!') If we introduce the abbreviation
+l if (b l ... bk) is an even permutation of (al"
(7.2)
sgn[bl .•. b k al ... ak
]= { -1
.ak)
if (bl ... bk) is an odd permutation of(al ... ak)
o otherwise
we can write down the skewsymmetrization as an explicit summation W = 1 sgn[bl ... b k ] W [al" .ak ] k! al'" The vector ; is characterized by the foZlowing three properties: -+ ... -+ (1) n is orthogonal to each of the tangent vectors u 1 , .•. ,un - 1 . (2) The length ~f (3)
n
is-+equal to the "area" of the "paralleLZogram" spanned by u , ... ,u _ , n 1-+ 1-+ -+ The n-tuple (n,u 1 , ... ,u _ 1 ) is positively oriented. n
Proof: ... First we deduce a useful formula. Let v be an arbitrary tangent vector. Then bl b r: un-Iva -+ -+ ...) (7.36) !I(n;~) = nava .gE b bU... = £(ViUl;···;un_l a 1'" n-l
II
352
The rest now follows easily:
= £(~.;~ i •.• ;~ ) which vanishes automatically since t (1) g(;i~.l 1. 1. 1 n-l is skew symmetric. (2) Consider the normalized vector ti/litiU . It is a unit vector orthogonal to ~ , •.. ,~ . Therefore the volume of the parallelepiped 1 -+ n:'l -+ -+ spanned by (n IUnll),u , ••• ,u reduces to the area of the paral4.1;- 1 lellogram spanned by u1, .•. ,U _ ' Consequently we get that n 1 -+- ..• iUn_l] -+-+- ... ;u -+- - ) = g(n; -+- -+Area[uli .:.( -+nl 11-+-11 n ;u1; nl II-+-I nl) n 1
r +. IInll
-+ -+
g(ni n )
=
-+
IInll
(3) We immediately get that £(ti;~ i ••• ;~ 1
n-l
)
g(ti,ti)
-+
Lemma 3 motivates that we define n -+
-+
Ul, ..• ,U _ n l -+
> 0
0
to be the cross product of
and we write it in the usual manner -+
-+
n = Ul x • ,.xu _ n l
Exercise
7.4.4
Problem: Consider the Euclidian space ~+1. (a) Let (~'~l'" .,~ ) be an arbitrary (n+1)-tuple. Show that the volume of the parallelepiped gpanned by (~';l""'; ) is given by the familiar formula n
(bl Let Nfl be an orientable n-diemensional manifold in Rn+1. At each point in M we select a positive oriented orthonormal frame (til""'~ ) in the corresponding tangent space. Show that the unit normal vector ¥ield -+
-+
-+
n=u1x ...x U n
is a globally smooth normal vectorfield on M (Compare with exercise 7.4.2)
7.5
THE DUAL MAP As another important application of the Levi-Civita form we will
use it to construct the dual map *, which allows us to identify forms of different rank. This gives a greater flexibility in the manipulations of various physical quantities.
353
II
Let F be a k-form with the components considering the
We start by components
Then we contract this tensor with the Levi-Civita form and obtain !,/g(x) e: b b Fbl ••. bk [PositivelY orientated] k. al···an-k 1'" k coordinate system! 1 is included to make life easier! where k! As e:al"'~ is skewsymmetric we have thus produced a form of deThis form is called the dual form and it is denoted *F gree n-k * or sometimes F
Definition 8 Let
*)
be a k-form on an orientable manifold
F
Then the dual form nents
al'"
a
with a metric
g.
is the (n-k)-form characterized by the compo-
*F
~!/g(X) e:al"'~_k (7.38)(*F)
Mn
bl •.. b k Fbl •.• bk
(~~~~i metric)
n-k
~!/-g(x) e:
al···an _ k b1···b k F
bl' .. bk (Minkowski metric)
with respect to a positively oriented coordinate system. Exercise 7.5.1 Problem: (a)
Let ~ be a scalarfield on a Riemannian manifold and let be positive oriented coordinates. Show that
(xl, ...• xn)
*~ = ~;gdxlA ... Adxn
(7.39) (b )
Show that E
(Euclidean metric) (Minkowski metric)
= *1
Exercise 7.5.2 Problem: Show that the contra variant components of *F are obtained by contracting the components of F with the contra· variant components of 1:. i.e. _1_ e: a l · , • ~-k b l ,·· bk F
(7.41)
(*F)
al···a -k
k!/g
n
{
- 1
k!/-g
*)
(Euclidean metric)
b l .. ,bk
e:
a 1 ·, .an - k b l ·· ,bk
F
b " .b k l
(Minkowski metri c )
This definition differs slightly from the one generally adopted by mathematicians. See for instance Goldberg [1962] or de Rham [1955].
354
II
In this way we have constructed a map from
hk(M)
to
An-k(M)
It is called the dual map or the Hodge-duality. By construction it is obviously linear
*(AF) = A(*F)
*(F+G) = *F + *G
(7.42)
But a more interesting property is the following one: Theorem 5
**T **T
(7.43)
*
So up to a sign
* :
=
(_I)k(n-k)T
(Euclidean metric)
T
(Minkowski metric)
= -(-l)k (n-k)
is it's own inverse. Especially the dual map,
Ak(M) ~ An-k(M), is an isomorphism.
Proof: For simplicity we only consider the case of a Minkowski metric. Let F be an arbitrary k-form. Then *F has the components
1. ,I-g k!
E:
a l ·· '~-k b l ·· .bk
Fbl' .. b k
We want to compute the components of **F The first thing we must do is raising the indices of *F, i.e. determine the contravariant components of *F. According to exercise 7.5.2 they are given by the formula
Then we shall contract the Levi-Civita form with factor! )
(6)
(**F)
c l ·· .c k
= -(n-k)! -I ; : g E: c
*F (remembering the statistical (*F)al ·· .an- k
l ·· .ck a l ·· .an- k
1
k!(n-k)!
E:
cl,,·c k al·"a n _ k
E:
a l ·• .an - k b l ·· .bk
F
bl".b k
The rest of the proof is a combinatorial argument! First we observe that
But
is skew-symmetric and therefore
Sgn [bcl ... bc k ] F = k! F 1,,· k bl" .bk c l " .c k and if we insert that into (0) we are through.
0
From theorem 3 and 5 you learn the following important rule whenever you deal with the operators in the exterior algebra:
355
II
"Be wise - appZy them twice" We shall evaluate the sign associated with the double map
**
so
often that it pays to summarize them in the following table
(7.44)
Euclidean metric k even
**F
n even
k odd -F
F
n odd
(7.45)
F
F
Observe that the dual form,
**F
Minkowski metric k even
k odd
n even
-F
F
n odd
-F
-F
*F ,involves the Levi-Civita form. Thus
it depends on the orientation of our manifold. If we exchange the orientation with the opposite one, then the dual form is replaced by the opposite form
F -F
*F ~ '= For this reason physicists also call the dual form
*F
a pseudo-tensor.
&eT'Cise 7.5.3
Introduction: Let MP be an orientable Riemannian manifold. A differential form F is called seZf dual i f it satisfies *F = F and anti self dual if Problem:
Show that self dual and anti -self dual forms can only exist in Riemannian spaces of dimension 4, 8, 12, 16, .,.
Exercise 7.5.4
Let MP be an orientable Riemannian manifol~ and A a+l-form+on M. Then A is equivalent to a tangent vector a. Let (ul""'~-l) be an arbitrary in-l)-tuPle. Show that the volume of the parallelepiped spanned by (iIl , ... '~n-l,a) is given by -+ -+ -+ -+ -+ -+ -+ -+ Vol[ul;···;un_l;al = *A(ul;···;un _ l ) = ai u l x ..• x un - l ) This gives a geometric characterization of the dual form *A'
Problem:
Exercise 7.5.5 Problem:
a)
Show that al
*(dx b)
A ••• Adx
ak
_ Itg ) -
(n-k)!
E..
ll" .In-k
h .. ·jk
g
jlal jkak i l i n- k ... g dx A . . • Adx
Consider spherical coordinates in Euclidean space dual map is given by *dr
= r 2 SinedeA
It converts a I-form A into the 2-form
dA with components
V x t.
But- this 2-form obviously represents the curl
Finally d converts a 2-form into a 3-form. If the 2-form represents a vector field, i.e. we consider the 2-form *A. then the 3-form d*A represents a scalar. It is this scalar we want to examine. As d*A is characterized by the components ai(*A)jk + aj(*A>ki + ak(*A)ij the 123-component is given by
Dualizing this we find that
*A
represents the scalar:
*(d*A)
=~ rg
a.(lgAi) l.
If you remenber that I:i = 1 in ~art~sian coordinates, you immediately see that d*A generalizes the divergence g. a. Notice however that we may express the divergence more directly by means of the co-differential 5. If A is a I-form we know that
-&A
1 rg
= ....,...
.
a.l. (/gAl.)
(Compare (7.64». This is in agreement with the result above since according to -5 .. *d*.
(7.59) we have
Exercnse 7.7.2 Problem: (a) Let F be a 2-f~rm r!pre~enting the vector a. Show that 5F rerepresents the curl V x a. (b) Let G be a 3-form representing the scalar field ,. Show that 5G represents the gradient ~
V,.
We know that the exterior derivative
d has some simple properties
rJ.2 = d(F" G) = d F"G+
(-1)
0
deg F
F"dG
Almost all of the identities involving grad, div and curl are special cases of these two rules. see the scheme two pages from thisl
371 (7.71)
II
The exterior derivative in R3.
O-form:
l-form:
4>
d4> : (a14>, d24>, 3 3 4»
l-form:
2-form:
A: (AI, A2, A3 )
dA,:
...al p
0
OJ bD
rl
al
... .....0 ...OJ
2-form:
+'
~
Ii
*A:
~,
A3 0
AZ _A1
-A'j
"Z A3- d3AZ
xxx
0
0
- - --------------------Conventional Vector analysis
xxx
0
i (d*A)IZ3 '" "i (y'gA )
AI
3-form: G : (GhZ3= G
"1 ~ -"zA1
xxx ",A,-3 A, 3-form:
4-form (Zero !)
dG
= 0
-------- ------------- -------------------------------Exterior algebra
Covariant formulas
Gradient: VtP
...al
i>o
0: 0
..... +' ....."
Divergence: 'i/ • a ~
~
A
a itP
dtP
- SA
n1 ai(lgAi )
Curl:
vx;: Laplacian: lltP
• k
dA
.rg E:ijkaJ A 1
- &H
.
~d i Vial.,)
The preceding discussion of the dual map, the scalar product, the wedge product, and the exterior derivative should convince you that exterior algebra is capable of doing almost anything you can do in conventional vector analysis. But exterior algebra is not just another way of saying the same thing. It is a much more powerful machinery for at least two reasons I 1. 2.
It works in any coordinate system, i.e. it is a covariant formalism. It works in any number of dimensions.
372
(A) l.
d2 = 0
(A)
If 'We apply
to a scalar field
'We get
.p.
d(d.p) = 0
(*) ~
II
As d.p represents the gradient of .p, the second curl . Thus (*) general ize s the rule
N implies tRat
n
P E V () n
Thus we conclude that tion! i
P
is an interior point if
n
an
This property of
an
an
in. But
a>J
N and that is a contradic-
an
an
is a com-
has property 1. Therefore you
are interior points, it follows that
boundary. Consequently
FOr a
~
itself is a regular domain! But as all the point in the
regular domain
Theorem 2
n
has important consequences. As
pact submanifold all the points in see that
n
(Cartans regu~ar
a(an)
an
has no
is empty:
~emma)
domain
n
the boundary
an
is also a regular doma-
has vanishing boundary
a (an)
o.
We shall include a point as a zero-dimensional regular domain, although it really falls outside the scope of our definitions. We do not assign any boundary to a pOint, so that the rule
aan = 0
still holds
in this very special case! As this point we had better look at some specific examples all using
M = R3
•
400
II
Examp l~ 1: If n is the closed unit ball, n
= {xl
then
intO
<xix> ~ I}, is the open unit ball,
= {xl
intn
<xix> < I},
0
and the boundary of
is the
two-sphere,
=
an
i.e.
{xl <xix>
= S2.
ao
= I}
,
Observe that
s2
is itself a two-dimensional regular domain without boundary
aan
=
0. Fig. 141
Examp Ze 2: If
n
is the northern hemisphere
82,
of
E R31 <xix>
n = {x then
intO
= 1, x
3 > oJ,
is given by,
intn = {x E R31 <xix> = 1, x 3 > and the boundary of n is the
oJ,
unit circle, an
= {x
Le.
1, x 3
E R31 <xix>
an
=
OJ,
sl ,
sl is itself a onedimensional regular domain without
where
boundary:
d(dF) = o.
a(an)
=
0
reminds us of the
They are, in fact, very intimately connected.
a differential form
cLosed form. the property
Fig. 142
aan = 0.
Finally we observe that the rule rule
ao=sl
F
had the property
dF
=0
In a similar way we will call a regular domain an
o ,
a closed domain.
If
, we called it a 0
with
This is in agreement with
the familiar expressions "a closed curve" and "a closed surface". At this point you should begin to have a feeling for the regular domains which are going to be the domains of the integration! There is, however, still a problem which we will have to face, and that is
the orientability of
n.
401
II
If we return to our intuitive discussion of how to integrate a twoform on a surface, it is clear that if we interchange the coordinates Al
and
then we also change the sign of the integral because ~i
-+i
~ F(e l ,e 2 )E
2
x
Fig. 143
But, if we interchange
Al
A2 , this simply means that we
and
construct a new coordinate system with the opposite orientation! This clearly shows that the integral depends on the orientation. Consider now a regular domain
0
in a manifold
M
Then int 0
is a k-dimensional submanifold which can be covered by adapted coordinate systems.
provided
int n
We say that the r~gular domain 0 is orientable is an orientc.ble manifold. (Compare with the
discussion in Section 7.4). The interesting point is that whenever
o
is an orientable regular domain, so is
orientation chosen on orientation on int 0
ao
Now let
int 0
aO!
In fact, the
induces in a canonical manner an
Tb explain this we choose an orientation on (~,U)
ao,
be an adapted coordinate system on
then: (xl, .•• , xk, ..• , xn) (xl, .•• ,xk,o, ••• ,0) xl < (0,x 2 , .•• ,xk,o, .,0)
°
serves as local coordinates on
M
serves as local coordinates on
int 0
serves as local coordinates on
ao
The local coordinates generate a positive orientation on ao if and onZy if the loaaZ coordinates (x l ,x 2 , •.. ,x k ,0, ••• ,0) generate a positive orientation on int Q. We have tried to exemplify this on This makes i t reasonable to define:
(0,x 2 , ••• ,x k ,0, ••• ,0)
fig.
144a-c.
402
II
k
Orientation of intll induced by a coordinate system.
Fig.
11111':--1,,--, 1I
l44a
by a coordinate k = 2
12 has a »hole«! Observe that the :s)inner« boundary
and the »outer« boundary get opposite orientations.
Fig. l44b
k dO
=
=1
,
I
I ,
I
III
Orientation of intO induced by a coordinate system.
~
P UQ
Q(+)
PH A point is negatively oriented when we move away, positively oriented when we move towards the point.
Observe, that in the case where
Fig. 144c
dO
reduces to a finite set of
points, we will have to construct a special convention, as you cannot introduce coordinate systems on a single point!
403
II
8,3 THE INTEGRAL OF DIFFERENTIAL FORMS With these preparations we are ready to define integrals of forms. , So let n be an orientable k-dimensional regular domain in Mn and let F be a differential form of degree k defined on Mn We assume for simplicity that
int
n
is a simpZe
manifold which can be
covered by a single coordinate system (not necessarily adapted!). We denote the coordinates of
int
n
by
(Al, ••• ,Ak) and those of
M
by
(xl •••• ,xk ••••• xn).
R
F
Fig. 145
The coordinates tated.
(Al •••. ,Ak)
are assumed to be positively orienk (Al, ••• ,A ) in U corresponds a set of
To each point
canonical frame vectors ~
~
el,···,e k As
F
number:
has degree
k, it maps (~l""'~k)
F(el, ••• ,~k).
ordinary function on
into a single real
. cons~der
-+
~
Thus we may F(el, •••• e ) as an k U • and therefore we can define the integral
as (8.5)
Jn F
def.
=
f u
-+
~
1
F(el, ••• ,ek)dA ••• dA
k
We may rearrange this formula by introduci~g the canonical frame ~s -t
~l'
on M.
-t
••. , ~n
404
II
(8.6)
Inserting this we get
Therefore we finally obtain: Definition 3
n
Let
=Mn
metrized
be an orientable k-dimensional regular domain paracoordinates (A1 , •• • ,)..k).
by
form of degree
k
(xl, .... xn) on
respect to coordinates integral of
\J
''i n J,
be a differential
F
F
with
al • • •ak
Then we define the
M
through the formula
F
-+ F(e l ,···, e k ) dAl ... dAk ~
fu
F
Let
characterized by the components
(8.7
Ju
F
a r .. a
ax
al
ax
~
k
al
~
dAl ... d).k
The formula (8.5) clearly shows that the integral is independent of the coordinate system (xl, ••• ,x n ) on
M
but we still have to
show that it is independent of the coordinate system ().l, ••• ,).k) on
n . Therefore we consider two sets of coordinates (AI, ••• ,).k) and (~l, ... ,~k) which both cover int n and which specify the same
int
orientation on positive.
-+
-+
F(el;···:e k ) (2)
n ,
int
i.e. the Jacobiant
i dA 1
= F(~i
(1)1 aliI
(2)
But F is a k-form and Therefore we conclude F (~.
; ••• ; ;;.
(1)
(1)
~l
~k
)
=
E.
1
(2)
~k) (2)
i dA k k
i (l)k aJ.1
.
1 · · · lk
F(E!l;···: (1)
-+
, .... , e
i l ,··· ,i k
Inserting this we get F(tl ;···;
Det
= F (e-+
is always
i (1)1
~
; ••• ; e. )
(If
i
i
aliI
k
~ ... dA
can only take the values
F (t ; ••• ; ~k) l
(1)
(a~~) d).J
First we observe that
(1)
all
k
1, ••• ,k.
405
II
Using the transformation rule for Riemann integrals we then obtain
~
I
e k ) Det (Olli) ---j I dA l •••k dA OA
(2)
Det(dll~) J
But
is assumed to be positive and therefore we can forget
OA
about the absolute value.
f F(el ;···; ek )
(8.8)
U2
(2)
n ,
d]Jl .•• dll
k
(2)
and this shows that int
We then finally obtain
IF n
=
f F(el ;···; ek ) Ul
(1)
dAl ••• dAk
(1)
is independent of the coordinate system on
as long as it generates a positive orientation!
up to this point we have only considered the situation where int Q is a simple manifold, i.e. it can be covered by a single coordinate system. We now turn to the general situation where we have to use an atlas (cj>i,Ui\EI consisting of overlapping coordinate systems, which together cover integral of a differential form over n.
IF
int
n.
We want to construct the
n
Now there is one case where the answer is obvious: Suppose there exists a special coordinate system (cj> o,Uol so that the restriction of F to n vanishes outside cj>o(U )' Then of course, we define o the integral as
fn Iu F(~ l·'···;~ F =
kldAl ••• dAk
o
Fig. 146 F ;,t 0 Next we want to consider the general case of an arbitrary differential form F • This requires, however, the introduction of a oechnical but important machinery: Recall that an open covering of a topological space subsets (OiliEr that covers M, i.e. M
=
M is a family of open
U O.
iEr
1
(Open coverings play an important role in the characterization of aompaat spaces. A topological space M is compact if an arbitrary open covering of M can be exhausted to a finite covering) • Then we consider a family of functions (WiliEr on M,
1/\ :
M~
R
We sfI¥ that such a family is ZceaUy finit6, when at an arbitrary point P E M only a finite number of the functions Wi have a non-zero value. Observe that this property guarantees tllat the sum
406
II
l: 1/J.
iEI
1
is well-defined in a completely trivial sense, even if the index set I is uncountable. (At each point P the sum reduces effectively to a finite sum). Consider once more a family of functions (1/Ji)iEI on our toplological space M. We say that it is a partition of unity if it is a locally finite family of non-negative functions such that l: 1/J. = 1 iEI 1 {i.e. for any point
P we have
L1/J.{P)
i
= 1)
1
Now we can filially introduce paracornpaat spaces. As you might expect,they are characterized by a suitable property of their open coverings: A topoboeioaZ spaae M is paraaompaat if we to an a:t'bi trary open covering {O i \EI Clan find a partition s~ah that 1/J i vanishes outside 0i' (We say that such a partition of unity is subordinate to the open covering. Observe that the partition of unity and the open covering have the same index set I). To get used to the concepts involved, you should try to solve the following useful exercise:
of unity {1/Ji)iEI
Exeraise 8.3.1 Introduction:
Let (
4> (P)
P
i.e. it just reproduces the value of the function
4>
This is the
most trivial and most uninteresting case. A scalar field
4>
can also be represented as the n-form
(7.39)
M
But then it can be integrated over n-dimensional domains including itself.
Usually such a domain n can be parametrized by the coordinates (xl, ..• ,x n ) on M itself and the integral reduces to (8.18) where
In
U
*
4>
J
n
4>
;,g
dx
l
A ••• A dx
n
Iu
is the coordinate domain corresponding to
4>(X);,g dxl ••• dx n.
n
This shows
that we can generalize the ordinary Riemann integral in Euclidian space l n I 4>(X)dX ••• dX , n which is valid only for Cartesian coordinates, into the covariant expression
which is valid in arbitrary coordinates.
All you have to do is to
replace the Cartesian volume element with the covariant volume element ;,g dxl ••• dx n • When we integrate a scalar field 4> on a manifold M
414
g
with metric
II
we always use the covariant integral (8.18).
JQ
=
$
As an application we can choose
JQIcJ dxl ••. dx
£
then represents the volume of
Q
1.
The integral
n
since
E
= ;.g d
xlA .•• Ad xn
is the
volume element in the tangent space, (compare the discussion in Section 7.4) (8.19)
Vol [Q]
=
J ;.g dxl .•. dxn Q
Here
Q
is restricted to bean n-dimensional domain, but the formula
(8.19) can easily be generalized to cover the "volume" of arbitrary k-dimensional domains. i.e. the length of a curve, the area of a twodimensional surface etc.
To see how this can be done we consider an
orientable k-dimensional regular domain fold way
The metric
M Q
on
Q
in our n-dimensional mani-
induces a metric on
M
becomes a k-dimensional manifold with metric.
be a parametrization of
and in this Let p,l, •••• Ak )
The intrinsic coordinates (Al •..• ,Ak)
Q. -+-
-+-
generates a canonical frame (el •••• ,e ) in each of the tangent spaces k (compare figure 148). The induced metric is characterized by the intrinsic components
g(~i;~j) The k-dimensional volume of
Q
is therefore given by
)
(8.20)
Volk[Q]
=
JI Q
1
-+--+-
Det[g(e.;e.)] dA •.• dA ~
k
J
This can also be reexpressed in terms of the extrinsic components of
g:
The volume formula is then rearranged as
JQ/ Det [g ab
(8.21)
a
b
ox ---j ox ] dA 1 ••• dA k ---. oA~ 01..
E:x:ercise 8.4.2 Consider a curve r connecting two points P and Q Show that reproduces the usual formula (6.40) for the length of a curve.
Problem:
Next we consider vector fields. represent
t..
by a one-form
A.
dimensional domain, i.e. a curve
r
with the parameter
Let
t..
be a vector field.
VOll[r]
We can
It can then be integrated over a one-
r
If we parametrizes this curve
A, (cf. figure 151), the integral reduces to
415
(8.22)
A2 r J A=J
r
I
< A
AI
Observe that in a
~ > dA
=
JA2
II
d~i
Ao ---d~ dA
"I
l.
1\
Euclidian
R
space
Q ->-
->-
dr dA
A·-
p
and the integral (8.22) therefore generalizes the usual line integral
12 .. Fig .151
We shall also refer to Suppose
A
fA
as a tine-integraL
r
$
is generated by a scalar field
i.e.
d$ ,
A
then Stokes' theorem gives
Jr d$ The boundary
ar
= Jar $
consists of the points
negative oriented and
Q
P
Q, where
and
is positive oriented.
P
is
By convention we
therefore put
Jar
$
= $ (Q)
-
$ (P)
Stokes' theorem consequently produces the formula (8.23)
which generalizes the usual theorem of line-integrals
(sec. l.l}It
does not, however, use the full strength of Stokes' theorem. If we use that
Jr
d$
=J
~
A i 2 dx AI axl. dA
dA
=
J"2 AI
~
dA
1\
we irrmediately see that it is a simple reformulation of the fundamental theorem of Calculus. Worked ezercise 8.4.3 Problem:
We use the notation of exercise 7.6.9. Especially manifold in R2, which we identifY with C, (a) Let r be a curve in M Show that
M is a two-dimensional
Jrh(z)dz = frh(z)dz where the integral on the right hand side is the usual complex line-integral.
416
II
(b) Consider an analytic (holomorphic) function curve connecting A and B . Show that
h(z).
Let
r
be a smooth
Ih'(z)dz = h(B) - h(A)
r
(c) Consider a two-dimensional regular domain Q bounded by the closed curve r . Let h(z) be an analytic (holomorphic) function in M. Prove the following version of Cauchy's integral theorem: fh(z)dz=O
r
A
A vector field
can also be represented by a (n-l)-form
Then it can be integrated over a (n-l)-dimensional domain. dimensional surface
Q
is usually called a hypersurface.
If
an orientable hypersurface then it allows a normal vector field
i.e. at each point
P
to the tangent space
on
Q
the normal vector
Tp(Q).
Let us parametrize the surface Q with the parameters (AI, ••• ,A n - l ). They generate the canonical frame vectors -+-+(e l , ••• ,e n _ l ). Furthermore we let ti denote the normal
...
ti(P)
Q
n
is I
is orthogonal
r
...
vector n = elx ••• xen_l (Compare the discussion in
Fig. 152
section 7.4). According to exercise 7.5.4
*A (-+-el,···,e-+-) ~ = A' n_l
*A.
A (n-l)-
we have
(-+-+-) elx ••• xen_l
The integral of *A therefore reduces to
fQ* A -- fu*
(8.24)
Thus
I* A Q
-+-) ,1 d n-l -A (-+el,···,e n _ l dA ••• A
fuA7,-+-n d,lA •.• dA,n-l
simply represents the fZuz of the vector field
the hypersurface
Q
A
through
For this reason we refer to the integral of
* A
as a fluz-integral. Having introduced flux-integrals we can now explain why theorem 4 is called Stokes' theorem. If S is a surface in R3 bounded by the curve r theorem 4 gives us fs dA = Ir A The right hand side is the line integral of A along
r.
We also have
d A = *(ilxA) (Compare the discussion in section 7.7). The left hand side therefore represents the flux of the curl ilxA. Consequently theorem 4 generalizes the classical Stokes' theorem from the conventional vector calculus, cf. (1.16).
417
n
Let
be an n-dimensional regular domain bounded by the hyper
an
surface
II
A
If
divergence of
-+-
A .
is a vactorfield, then
&A
-
represents the
We can integrate this divergence over
and get
the volume integral
According to the corollary to Stokes' theorem this can be rearranged as
=
(8.25)
(_l)n-l J
on
*
A
up to a sign we have therefore shown that the integraL of the divergence is equal to the flux through the Qoundary. This is the gene-
ralization of Gauss' theorem to an arbitrary manifold.
We can also
work out the corresponding covariant coordinate expression.
Using
(8.24) and (7.64) we get
(8.18) ,
a. (/g Ai)dnx
(8.26)
Jn
~
where the normalvector (8.27)
n.~
...
n
= Ig
= JAin. on
~
dAl •.. dA n - l
is characterized by the covariant components
e:. .
~Jl'"
j
n-l
We can also reproduce the integral theorems from the two-dimensional vector calculus. Consider for instance a region n bounded by a smooth curve r (see figure 141). Then it is well-known that the area of n is given by the line-integral
(8.2 8) Area (Here to
;
*;
'f
-+- -+- . [n) = ~ r*roar
r
is the vector orthogonal
and with the same length. Remember
also that
x
~*;oAt represents the area of
the triangle spanned by ; and 1+6;). It is convenient to write out this formula in Cartesian coordinates
(8.29) Area
[n)
= ~Jxdy-ydx
Fig. 153
To deduce this formula we use that dx. A dy is the "volume form" in R2. Consequently Area [n] i.e. Area [n) An application of Stokes' theorem now gives
Area [n)
= ~Jrxdy
- ydx
which we immediately recognize as being equivalent to (8.29).
418
II
We conclude this section with a discussion of simple forms, where we can give a naive interpretation of the integral. For simplicity we consider the threedimensional manifold M = R3. If we choose a coordinate system on M, then the coordinates (Xl, x 2 , x 3 ) generates several simple forms. for instance: dxl.
dx l
A
dx 2
•
dx l
A
dx 2
A
dx 3
•
dxl.
Consider first the basic one-form Q. (See figure 154).
Let
r be a smooth curve from P to
p
a Fig. 154
Ir
We want to interpret the integral [a;a+ll. [a+l;a+2l, ••.
xl
We divide [a,bl in unit intervals
Then the integral is roughly given by:
Jrdxl The coordinate
dx l
"" L
< dxll
i
"tli
>
generates a stratification characterized by the surfaces: xl = -1. xl
O. xl = +1, .•.
According to our analysis in section 7.2 , the number
l
can be inter-
preted as the number of surfaces intersected by the curve segment corresponding to the i'th unit interval. Therefore the total integral can be naively interpreted as the number of surfae~8 intersected by r. This can be demonstrated rigorously using Stokes' theorem. It immediately gives us
JIl
dxl
= J xl = xl(Q) or
- xl(p)
and the number xl(Q) - xl(P) cle~ly represents the number of surfaces intersected by r. Observe that if r is a ciosed curve, then each surface is intersected twice from opposite directions. Therefore
fr
dx
l
=0
This is also in accordance with Stokes' theorem
Next we consider the basic form dx l A dx 2 . (See figure 155).
Let
Il
be a smooth surface in
M.
419
A2
II
U
AI
Fig. 155
x
1
We want to interpret the integral
III We divide
U into unit squares.
I
.bel A
dx'
Then the integral is roughly given by:
dxl
dx 2 "" 1: "-A ....1
A
i
Il
A
dx2 ( +e i l'. +e i 2 )
The coordinates (x l ,x2 )
generate a honeycomb structure in M According to our I 2( +i +i ) analysis in section 7.2 the number dx A dx e ; e can be interpreted as the 2 l total number of tubes intersected by the surface segment corresponding to the i'th square. Therefore the integral can be naively interpreted as the number of tubes intersected by Il. This can also be justified more rigorously by the following argument: For simplicity we assume that we can use (x l ,x2 ) as adapted coordinates on Il, i.e. Il is parametrized on the form:
x3
= x 3 (x l ,x2 ).
III dxl
A dx
2
=
Then we get
Iu dxl ax2
U is an open subset of the xl -x2-Plane. Tubes in M correspond to unit cells in the x l -x2-plane. The number of tubes intersected by Il is equal to the number of unit cells contained in U. On the other hand f ax l ax2 U is equal to the area of U and this number also represents the number of unit cells contained in U. If Il is a closed surface, then
Here the coordinate domain
In dxl
A
dx
2
=
a
because each tube is intersected twice with opposite orientations. This illustrates Stokes' theorem for the basic form
dx l A dx 2 ,:
o
II
420 Exercise 8.4.4 Problem:
Let
Q
= S!
be the northern hemisphere of the unitsphere in
that the "number" of tubes intersected by
S!
is
R3.
Show
Compare this with exercise
Tl.
8.3.3. Exercise 8 . .4.5 Problem:
2 Consider the basic three-form dxl A dx A dx3 The coordinate functions xl,x2 and x 3 generate a cell-structure in M. Let Q be a
3-dimensional regular domain in
f
M.
Show that
2 l 3 dx A dx A dx
Q
can be interpreted as the number of cells contained in
8.5
THE HILBERT PRODUCT OF TWO DIFFERENTIAL FORMS Consider two differential forms
We have previously (sec. 7.5) T
Q.
and
U, cf. 1
T
U
and
of the same degree
k
introduced the scalar product between
(7.52-53):
(Tlu)= k: T
i l ··· i k
~*(*T
U..
~l·· ·~k
A
U)
This is the relevant scalar when you look at a specific point
P,
i.e.
when you want to introduce a metric in the finite dimensional vector space
k Ap(M).
Globally the differential forms of degree dimensional vector space
Ak(M)
k
span out an infinite
and we would like to associate an
inner product with this space.
To motivate it we consider real-valued smooth functions on a closed interval [a;b].
Here we have the natural inner product: < fig>
.fbf(X) g(x)dx , a
=f
*f A g
[a,b]
In analogy with this we consider the n-form (7.50)
*T
1
A U = -- T
i l ·· .i k
U
il ••• i k
k:
,
If it is integrable, we define the inner product in the following way: def. 1 i l ·· .ik 1 n A U = --k' T U . . Ig dx ••• dx (8.30) < T I u >
f
M •
~l·· ·~k
We shall refer to this inner product as the Hilbert product. compact, then the Hilbert product is always well-defined. If compact, the integral is not necessarily convergent.
If M is M is not
II
421
It is, however, always well defined if one of the differential forms has compact support. Consider a Riemannian manifold. with compact support.
Let Ak(M) denote the vector space of k-forms o The Hilbert product defines a positive definite metric in
this infinite dimensional vector space.
Thus Ak(M) o complete it and obtain a conventional Hilbert space
is a pre-Hilbert space. We can
~(M) called the Hilbert space of square-integra?le k-forms •. An element of the form 1 l.l l.k T = k' T. • (x) d x A ••• A d x • l.l· .. l.k
~(M) is on
with measurable components
is convergent. For a pseudo-Riemannian manifold things are slightly different. Here the scalar product (Tlu) is indefinite and therefore the Hilbert product is indefinite too. If we complete
Ak(M) o
we therefore get a Hilbert space with indefinite metric.
In the following we shall always assume that all the differential forms we are considering have a well defined Hilbert product.
Observe
first that up to a sign the dual map is a unitary operator, i.e. it preserves the inner product between two differential forms:
Theorem ? (a) The dua'l map * is a unitary operator on a Riemannian manifo'ld: = < T I u > (b) The dual map * is an anti-unitary operator on a manifold with Minkowski metric: = -< T I u > Proof: This follows immediately from the corresponding local property (Theorem 9, section 7.5): < *T
I
*u > =
t
(*T
I
±
*u)'
Then we finally arrive at a most important relationship between the exterior derivative -form, U
d
and the codifferential
5.
a k-form and consider the inner product:
=
I*U
A dT
M
This can be rearranged using a "partial integration":
Let
T
be a
(k-l)
422 d(*UAT)
=
II
d*U A T + (_l)n-k *U A dT
From exercise (7.6.1) we know that
d * U
=
(_l)n-k+l * IS U •
Furthermore
JM d(*u either because
M
A
T) =
JaM
*U
A
T = 0
is compact without boundary or because
"vanishes sufficiently fast at infinity".
J * U A dT = J *
(8.31)
& U A T
i .e .
Thus we see that due to our sign convention, 6 operator of
d.
*U A T
Therefore we obtain: < UI d T > =
is simply the adjoint
It should be emphasized that this holds both for
Riemannian manifolds and manifolds with a Minkowski metric. This has important consequences for the Laplacian operator.
M be a Riemannian manifold.
Let
Using (8.31) we can rearrange the
Laplacian as
...
where
&'
denotes the adjoint operator.
B~rmitian operator.
positivQ k-forms.
Consequently
-A
is a
The above argument applies to arbitrary
In the special case of scalarfields it is well-known from
elementary quantum mechanics, where the Hamiltonian, H
=-
i'J2
2mA ,is a
positive Hermitian operator reflecting the positivity of energy! We can also use (8.31) to deduce an important property of harmonic forms on a compact manifold.
Theorem 8 Let
M
be a compact Riemannian manifold.
A k-form
T
is harmonic
i f and only if it is primitively harmonic, i.e. A T
=0
iff;d T
= &T =
0
Proof: Let us first observe that = + = <ST 1ST> + If
T
is harmonic, i.e. AT
0, then the left-hand-side vanishes
automatically. But the right-hand-side consists of two non-negative terms:
Hence they must both vanish: < & T I & T > = = 0
II
423
But as the inner product is positive definite, this implies
&T
=d
T
0
=0
Consider for instance a zero-form, Ii
q,
vanishes automatically and
i. e. a smooth function, _m2] From this we immediately deduce
o
dS =- de: I e:=O
428
II
( H ere we have used that 1jJ vanishes on the boundary to throwaway the boundary term coming from the partial integration). But this is only consistent if ~ satisfies the equation (8.40) -Sd~ = m2~ , which is the geometrical form of the Klein-Gordon equation: This is a one-form A , and the action is based The Max~eZZ fieZd: upon the Lagrangian density (3.50) which leads to a covariant action given by
This is rearranged as (8.41)
Then we perform a variation,
A where
S(g)
-~
-~
From
A+E:U
is a one-form, which vanishes on the boundary of
u
£:2
- € - :f
this we immediately get dS o = cr---= - = -
€I€=o
( Here we have used that U vanishes the boundary of n to throw away the boundary term coming from the partial integration). But this is only consistent if (8.42 )
Le.
-OF
o
which is nothing but the Maxwell equations! Exercise 8.6.4 Problem:a)Consider the massive vector field by
A
Show that the action (3.51) is given
(8.43) b)Perform the variation tion
A ~ A+£:U
and deduce the following equations of mo-
(8.44) c)Show that they are equivalent to
(8.45)
cA= m2 A
-6A=O
We may summarize the preceding discussion in the following scheme:
429
FIELD (8.39)
~
(8.41)
Maxwell
( 8.42)
A
(8.43)
Massive vector
(8.44)
Equation of motion
Action
Klein-Gordon
(8.40)
II
S
= f~~(*d~Ad~)-~m2(*~A~)
S
=
S
=J
-Is, ~ (*dA) AdA
-6d~ = m2~
-odA
-~(*dAAdA)-~m2(*AAA) -odA
n
=
0
= m2 A
A
8.7 INTEGRAL CALCULUS AND ELECTROMAGNETISM As an other example of how to apply the integral calculus we will use it re-express in a geometrical form various electromagnetic quantities like the electric and magnetic flux through a surface and the electric charge contained in a 3-dimensional regular domain. We start out peacefully in 3-space to get some feeling for the new formalism. Remember that tially
B
E
is a I-form, but
B
is a 2-form. Essen-
is the dual of the conventional magnetic field. Now let
n
be a 3-dimensional regular domain. From the discussion of fluxintegrals (8.24) we get B
The magnetic flux through the closed surface
an .
(8.47)
Jan*E
The electric flux through the closed surface
an .
(8.48)
f;p
The electric charge contained in
(8.46)
fan
n
We can now use the integral calculus to deduce some wellknown elementary properties: Example 1 If n ~ke
c~ntains
no singularities, then the magnetic flux
an
closed surface
vanishes.
This follows from an application of Stoke's theorem:
~ (due to (7.76».
J an B = JndB =
0
0
~
through
430 Example 2
II
(Gauss' theorem)
The electric flux through the closed surface electric charge contained in
an
is
n.
This follows from an application of theorem 5 (Corollary to Stokes' theorem). From the Maxwell equation ;
o
x
[the charge]
=
(7.78)we get
£lJ *p 0
n
=
-J n*QE
Example 5:
This time we consider a static situation. Let face with boundary to
S
be a sur-
r . Then the flux of current through S is equal
£oc 2 times the circulation of the magnetic field along Observe first that the Maxwell equa-
r.
tion (7.79) reduces to &B
=
1 £oc2 J
for a static configuration. The proposition then follows from an application of theorem 5 (Corollary to Stokes' theorem) :
Fig. 156 We conclude the discussion of electromagnetism putation of two important integrals:
in 3-space with the explicit com-
a) Consider the spherically symmetric monopole field. Let 52 be the closed surface of a sphere with radius r . Then polar coordinates (r,8,~) are adapted to the sphere, and we can choose (8,~) to parametrize it! We Can now compute the magnetic flux through the closed surface 5 First we observe that the monopole field B is gi ven by (cf. (7. 89) ) : B
= t,;- 5in8d8AiAp
Then we immediately get: 2Tf
(8.49)
Tf
J52B = tL ~Of 8=0J 5in8d8d~ = g Tf
Fig. 157
431
II
b) This time we consider the magnetic field around a wire with current j. Let r be a circle around the wire with radius p. Then the cylinder coordinates (p,~,z) are adapted to the circle, and we can choose ~ to parametrize the circle. We can now evaluate the line integral of the magnetic field along the closed curve r. For this purpose we must find the dual form *B which is the one-form representing the magnetic field. This has been done previously (cf. section 7.8) *B -
-L
- 21TE: o c 2
z
dip -t
J
Thus we get •
2'TT
fr *B = 21T~ c 2 fd~ oo
x
•
=
E7 n
Fig. 158
0
Then we proceed to consider electromagnetism in Minkowski space, i.e. 4-dimensional space-time. Here it is more complicated to express suitable quantities, so we shall adopt the following terminology: Suppose we have chosen an inertial frame S. Let (xO,X 1 ,X 2 ,X 3 ) denote
suppressed~
One dimension
the corresponding inertial coordinates. We say that the three-dimensional submanifold
F,; is a space slice if it is a subset on the form F,; = {XEMlxo=tO}
i.e.
F,;
consists of all the spatial to . Ob-
points at a specific time
serve that the spatial coordinates (X 1 ,X 2 ,X 3 ) are adapted to F,; The fundamental quantities describing the properties of the electromagnetic field are the field strengths A , and the current
J . Now let
n
Fig. 159 F
and *F
,the Maxwell field
be a 3-dimensional regular domain
contained in a space slice relative to the inertial frame
S • Then we
can form the following integrals which we want to interpret:
Ian F
Ian*F
Observe first that the restriction of
fn*J
and dxo
to
n
vanishes. Conse-
quently the integrals involve only the space-components of the integrands. But (8.
50~
F
and
F
=
*F
are decomposed as
[*]
and
*F
[*J
432 So the restriction of (respectively striction to
F
II
(respectively
*F) to
an
is given by
B
*E). Similarly the dual current has the following reintn
*JI~
..
=
(*J)
123
dX ' Adx 2 Adx 3
= -JOdX ' Adx 2 Adx'
We can now generalize the results obtained in (8.46)-(8.48) to the following Lemma 1 Let
n
be a J-dimensional regular domain contained in a space slice.
Then (8.51)
JanF
r
(8.52)
Jan
*F
-In*J
(8. [, J)
The magnetic flux through The electric flux through
=
an an
The electric charge contained in
n
Exercise 8.7.1 Introduction: Let n be a. 3-dimensional regular domain obtained in a space slice and let F be smooth throughout n. Problem: Use the 4-dimensional integral formalism to re-examine the following well known results: a) The magnetic flux through an is zero. b) The electric flux through an is equal to the electric charge contained in n (As usual we have put EO = c = 1 ).
ILLUSTRATIVE EXAHPLE:
MAGNETIC STRINGS IN A SUPERCONDUCTOR
We have earlier been discussing some of the features of superconductivity, especially the flux quantization (see section 2.12). Recall that in the superconducting state of a metal, the electrons will generate Cooper pairs. These Cooper pairs act as bosons and we can'therefore characterize the superconduc~ing state by a macroscopic wave function
~
called the order parameter of the superconducting metal. The square of the order parameter,
1~12,
represents the density of the Cooper pairs.
We want now to study equilibrium states in a superconductor. Consider a static configuration ~(~), where ~(~) is a slowly varying spatial function. In the Ginzburg-Landau theory one assumes that the static energy density is given on the form: (8.54)
H
433
II
Here a is a temperature dependent constant (8.55)
a
=
aCT)
=
a ~c T
(with a positiv)
c
where Tc is the socalled critical temperature. The constant y is just inserted to normalize H to be zero at its global minima. The equilibrium states are found by minimizing the static energy. This leads to the Ginzburg-Landau equation (8.56)
Now consider the potential (8.57)
It has the well-known shape shown on figure 160. Above Tc the vacuum
T>T
c
TO
if
x:TW-WaIT Ua8 (x)dA
1
2,
dA d x
Within the framework of distributions we therefore have
(9.28) S
We summarize the main properties of the singular form
in the
following lemma Lemma 1 (Dirac's lemma) The singular form 1)
S
has the following properties:
It vanishes outside the string.
(9.29 )
2)
dS
(9.30)
3)
fns
=
*K
=-gM
n
for any cZosed surface
surrounding the mono-
pole. Proof:
(1)
If
x
lies outside the sheet
o'(x-X(A 1 ,A 2 »
(2)
To check the relation _1_
a
s
(Fg a8)
;=ga
L,
then the
o-function
automatically vanish. (9.2~
we
u~e
that it is equivalent to
= KS
(chain-rule)
(Stoke's theorem) = (3)
Finally we must
CCIlIpUte
rounding the monopole.
th= flux through a
closed surface sur-
But a closed surface
n
the monopole is the boundary of a regular domain
surrounding
W con-
481
II
taining the monopole. Consequently we get using (9.29) Ins =
Worked Problem:
laws
Iw ds
=
=
-lw*K
=-gM
D
e~ercise
Remark:
9.2.2 Prove (9.30)by an explicit computation of the integral.
We can also reformulate (9.29 ) as
5*S = - K Within the framework of distributions the boundary operation coincides with the codifferential and we therefore get &*S = -g~l:
= -gMal: = ~rM
= - K
F.
Using Dirac I s lenuna we can now "cure the diseases" of
It was
generated by a magnetic monopole and an electrically charged particle and therefore had the properties dF=-*K
(9.24)
InF=gM
We now choose an arbitrary string extending from the monopole to infinity. Associated with this string we have a weak form
S
with the
properties:
dS
(9.29,30)
= *K
Consequently we see that
F + S
is exact, although singular, and we
therefore can find a global gauge potential A, The gauge potential A
which generate
F + S •
will, of course, be singular too. It will be singu-
lar at the position of the monopole, reflecting the singularity of and it will be singular at the string, reflecting the singularity of S • Formally
S
represents a concentra-
ted magnetic flux flowing towards the monopole. Hence we may formally interpret
F + S
as a concentrated magne-
tic flux flowing towards the monopole position along the string and then spreading out to produce the monopole field. However, it should be emphasized that
S
has no physical meaning. The
position of the string can be chosen completely arbitrarily and the introduction of
S
is a purely formal
trick, which cures the diseases of
F!
Fig. 176
F ,
482
II
9.3 DIRAC'S LAGRANGIAN PRINCIPLE FOR MAGNETIC MONOPOLES We are now in a position where we can state the Lagrangian principle of
Dirac~)Dirac's
(9.32)
S
action consists of three pieces:
= SpARTICLES
+ SINTERACTION + SpIELD
where SpARTICLES
SINTERACTION
SpIELD
q
Jr
A dxpa dA
a dA
q
=
Notice that the interaction term only contains a coupling between the electrically charged particle and the field! However, if we look a little closer at
SpIELD
we see that it contains information about the
monopole trajectory because (9.33)
P
~v
=aA ~
v
-aA-S v ~ ~v
Hence when you vary the trajectory of the monopole, you will have to vary the sheet which terminates on the trajectory. But that will force S~v to vary, and thus SpIELD contains the coupling between the monopole and the field. That the above action in fact gives the expected
equations is the contents of the following famous theorem: Theorem 1 (Dirac's theorem) All equations of motion for a system consisting of monopoles, electrically charged particles and the electromagnetic field can be derived from Dirac's action, provided you respect Dirac's veto, i.e. netic strings are never
allo~ed
the mag-
to cross the worldline of an electri-
cally charged particle. Proof: The proof is long and technically complicated and you may skip it in a first reading. First we list the equations of motion which we are going to derive d2 x a dx ~ dx v a dx S e + ra e e ] _ qP 6--e(1) ( 3) me [ --;rrzdF =-* K ~ v liT dT
---err -
(2)
d*F =-* J
(4)
Then we list the dynamical variables to be varied:
*) The theory of magnetic poles, Phys. Rev. 74 (1948) 817
483 (a) (b)
(c)
Xea(A) xma CA) All (x)
II
Trajectory of electrically charged particle. Trajectory of monopole. Gauge potential.
Observe, that the string coordinates X(A I ,A 2 ) are not considered as dynamical variables. They are fictitious coordinates and shOuld be completely eliminated in the end of the calculations. Okay! Let us go to work: Eq. (1) is not a dynamical equation. It is a purely kinematical equation which is built into the model from the beginning: F + S = dA but by construction we have:
dF + dS = dS = *K
dEIE=o
dF
A A + EB Il Il Il fni=gJa(X)Ba(x)d'x
Eq. (2). Performing the variation
dS I
0
~
-dS
we get:
which leads to the desired equation of motion:
X
Eq. (3) follows from the variation of the trajectory Il(x) . Performing the variation XIl(A) ~ XIl(A) + E~(A) we get from a previous Ealculation (exercise 8.6.3) :
dS I dEl E=O
r
2
-a) dA ([aAa_S]dXCX + A 3L. BY dA adA ax aAa dx B a dxa ~S - 8 (IT' Y - ax8 (IT' ya]dA q J[aA ax dAS)dXCX -B - - - - - Y dA qJC) dXa dA q
Al
From which we conclude: v 8 dx ll dx \ d2 B me g a8 ( ~ + r Ilv+at+13
(9.37)
Let us fix the time-scale so that at
t=o
the particle is as close as
possible at the origin, i.e. r(O) = d(= distance of closest approach) Then eq.
(9.37) reduces to
(9.38) This has an important consequence. If the speed
vio , then the par-
ticle comes from infinity and returns to infinity. Thus there are no bound states.
[The case
v=O
where the electric charge stay at rest is
a very special situation. In fact, it is unstable: The slightest perturbation and the electric charged particle will move to infinity! It is therefore irrelevant when we investigate the scattering of charged particles in monopole fields.] Then we consider the orbital angular momentum
~:
t =
rxmv
Since the monopole field is spherically symmetric we expect that there should be a "total angular momentum" which is conserved. First we observe that the orbital angular momentum is not conserved:
d ......
. . ...... dv
...... r
rv-(;.v)r r2
dt(rxmv) = vxmv+rxrnat = KrX(vxr') = K where
rr
r =
is the unit vector pointing towards the charged particle.
To rearrange the right hand side we observe that d
A
d ( ... ,
... ... dr
rv-r -dt
d~ = dt ~) = ----r~~
so that we finally obtain:
490
d ~~) 'dt(rXmv
II
dr Kdt
But then we have found the following constant of motion: ...
~
j
(9.39)
rxmV-Kr
This means that the total angular momentum of the system must be iden-
j . The second term,-K~, is, of course, nothing but the
tified with
angular momentum of the electromagnetic field. Step 4:
t
nents
~
Since we have decomposed and
Kr
into two orthogonal compo-
J
we immediately get
(9.40)
But
is a constant, and
K
is conserved. Consequently
J
served too. So, although the orbital angular momentum
t
t
is con-
is not con-
served, we find that its size is conserved! This has a curious consequence: Consider the system at t(-oo)
t=-oo
= mv(-oo)b
and
and
t=O , then we get t(O)
mv(O)d
Particle Distance of closest approach Impact parameter (distance of closest approach for a free particle)
Tangent at infinity
Fig. 179
where
b
is the impaat parameter. But
t
and
v
are conserved. Thus
we finally obtain: b
(9.41)
=
d
Observe, that if we point directly towards the monopole, i.e.
b=O,
we will hit it! But this is a very exceptional situation: If we point directly towards the monopole, the electriCally charged particle Will
IlOve
freely because ...
~
vxB
0
But the slightest perturbation and the particle will react to the force from the monopole field and no longer hit the monopole. Consequently this is an unstable situation and we shall neglect it.
491 ~:
tion (9.39) with
r
we get
(9.42) But
J
II
Finally we observe that multiplying both sides of the equa-
i.e. is conserved and
K
+
~
Cos(J,r)
-K
=J
is a constant. Thus the angle between
~
J
and
is conserved and therefore the partiale is aonstrained to mave on a aone with j as axis:
g
Worked exeraise 9.4.2 da Problem: Compute the differential scattering cross section dn for the scattering of electrically charged particles in a monopole fielu.
9.5
QUANTIZATION OF THE ANGULAR MOMENTUM As we have verified, the magnetic strength
gM
of the monopole en-
ters into the expression (9.39) for the angular momentum of a charged particle in a monopole field. When we quantize the motion of the charged particle we know that the angular momentum becomes quantized, and we expect this to liminate the possible values of the possible values of of angular momentum:
K A
K
•
To determine
we must therefore determine the operators A
A
J l ,J2 ,J 3 . For a charged particle moving in a magnetic field we have previous-
ly determined the Hamiltonian (Cf. eq.
H=
(9.43 where
A
(2.65)):
JL(-i~V-qA)2 2m
is the gauge potential generating the magnetic field. Con-
sequently we must first find
~gauge ~tliU
~
A
producing the monopole
492
field. But here we run into the wellknown trouble that no globally defined gauge potential can produce l:he monopole field, i.e. A will necessarily contain singularities. However, as we have seen in section 9.2, we may concentrate the singularities on the z-axis. We can still make a choice: The singularities form a string which can be either infinite or semi-infinite. As we shall see, the actual choice of the string has consequences on the quantum mechanical level and we shall therefore work out both cases. The semi-infinite string was originally introduced by Dirac, and we shall speak about the Diraa formalism. The infinite string has been especially advocated by Schwinger and we shall speak about the Sahwinger formalism. Let us collect the appropriate formulas in the following table: (9.44)
Sahwinger formalism:
'-/
A
(9.45)
Diraa formalism:
- ~cose,,1P
A
41T
zy _ --Y--_r(r:z) qA
= K
=
K
[
r(r-z)
o
Observe that the Hamiltonian (9.43) becomes a singular operator, but there is nothing to do about this for the moment. We will have to live, for some time with the singularities along the z-axis and we will have accept similar singularities in the wave function w(r,t). Next we construct the operator corresponding to angular momentum. On the classical level we know that ~ ~ ~ 7 j = ~ rxmv - Kr =; r x (p-qA) Kr (9.46) This corresponds to the operator (9.41) ~ r x (-ifiV-qA) - Kr = -ifirxV - ( rxqA + Kr)
*0
A
Consequently we have found the following candidates for the angular momentum operators:
493
Schwinger formalism:
(9.48)
J
2
31 =-ifl(Y~ ay az - z~)
- x~) =-ifl(Z~ ax dz
Kxl:y2
32
1\
J
Dirac formalism:
KX 2 +y2
l =-ifl(y,}Z
1\
(9.49)
- Z~) ay -
1\
J
II
3 =-ifl(
xr
=-in( z-1ax
- x~) dZ
~3 =-ifl(, x~ "y - Ya"x)
x"~ - y"~)
K-L
r-z
- K~
r-z
+ K
Before we continue the discussion we recall some general aspects of the angular momentum. According to Dirac the components of the angular momentum are always represented by three Hermitian operators ~1'~2'~3 satisfying Dirac's commutation relation: (9.50)
If the system is spherically symmetric they must furthermore commute with the Hamiltonian, i.e. [ft;~i]
(9.51)
=
0
This guarantees the conservation of the angular momentum, since according to the quantum mechanical analogue of (2.61) we have ifl
d~i
dt
=
[ft;~.] ~
= 0
In many problems with spherical symmetry we can use the operators of orbital angular momentum (9.52)
which trivially satisfy Dirac's commutation rules (9.50). But we cannot use them as angular momentum operators in this particular problem because they do not commute with the Hamiltonian (9.43). On the contrary the above operators (9.48-49) not only satisfies Dirac's commutation rules (9.50), but they commute with the Hamiltonian (9.43) as well. The verification of this is left as an exercise: Exeraise 9.5.1 Problem: (a) Let
f(X)~ be a differential operator. Show that: [f(X)a"x;g(x)]
= f(X)¥X
i.e. the commutator is a multiplication operator. (Hint: Let both sides operate on a testfunction ~(x)). (b) Show that the operators of angular momentum «9.48-49) satisfy the commutation rules with (c) Show that the operators of angular momentum «9.48-49) satisfy the commutation rules (9.50) and (9.51) as required by Dirac.
494
II
We pr=eed to investigate the sp:ctrum of the angular m:mentum. Using the commutation relations (9.50) one can determine the possible eigenvalues for the angular momentum operators.
(For details see e.g. Schiff [196S]).
First we introduce the square of the total angular momentum: A
It follows that
J2
we can diagonalize states (9.55b)
A
A
A
J2 = J~ + J~ + J~
(9.54)
commutes with J2
and
J
J
1
,J2
and
J
3
,
and consequently
simultaneously. If we label the eigen-
3
we get:
1jJjm
=
(9.55b)
=
rnfI1jJ.
Jm
j (j+l)f'1.21jJ.
Jm
the possible eigenvalues are given by
Furthermore it (9.56)
m
-j,-j+l, .•• ,j-l,j
j
O,t,1,t,2,f,···
Finally it is preferable to introduce the raising and lowering operators (9.57) They satisfy the commutation rules (9.58) and as a consequence the operators J+ and J genfunctions with a fixed j.
fl/j (j+l)
m(m+l) 1}!.
= fi/j(j+l)
- m(m-l) 1jJ.
=
(9.59)
connect the different ei-
Jm Jm
In the particular example of a charged particle in a monopole field we can now use the explicit form (9.50-51) for the angular momentum operator to determine which of the possible eigenvalues (9.56) that are actually realized. Recall that classically (9.40) j2 = 12 + K2 This equation has a quantprn mechanical analogue We have decomposed the quantum operators in the following way:
~
=~
x
(-iflV-qA) -
K; = t - Kr
But then we can square this operator relation:
Worked ea:erais!il,. 9. 5. 2
Problem: Let Show
L b.lO. the operator
'r, = ~x(-iI'lV-q.Jt)
thatt~e operat~r4Products
L'r
both vanishes.
and
r'r,
495
According to exercise 9.5.2
II
both operator products
automatically vanishes. The following operator relation has therefore been established:
~2
(9.60)
Here
K
~2
/I
j2
is a C-nurnber and
r?
and
therefore have the same
eigenfunction~:
= [j(j+l)h 2-K 2 l1jl.
r?1jI.
)~
)m/l = L2+L2+L2 /I
/I
is a positive operator, becau/lse But 1 2 3 are Hermitian operators. Thus the eigenvalues of L2 and we conclude
i.e.
(9.61)
j(j+l) .::
/I
L ,L 1
/I
2
and
L 3
are positive
2 (..:-K) 11
This is our main result because this shows that in a monopole field j=O
is not allowed. Consequently the minimal value of
j
is non-tri-
vial and the system therefore possesses an intrinsia spin! To determine the exact spectrum, i.e. the allowed values of m
eigen~unctions
together with Xhe
expressions for
j2
and
J
3
j
and
Yjm , we must now use the explicit
It is convenient to use spherical coor-
•
dinates. After a long but trivial computation you obtain the following explicit expressions:
Schwinger formalism:
J~2
-
_~2r __l__ JL(s;ne~)
-
11
LSine ae
~
3e
(9.62) J
3
Dirac 3:2
formalism: -
l'l2[
I a ( . eJ,. +_1 32-J_2iflK___I ___ JL+2lC2 ___ 1 ___ Sine ae s~n-ae) SinTGW . I-Cose 3IP . l-Cose
(9.63)
-ifl JL +
alP
K
The Schwinger formalism is the most complicated one. But let us start with it to get a feeling for the general machinery. We look for eigenfunctions of the form gate the eigenvalue equation (9.55a) i.e.
J31j1. flml/J. 3)m :lm -ifl a\llY jm (0,1P) = mflY
jm
(0,\Il)
yjm(e,IP)
• First we investi-
496
II
But this shows that we can factorize y, in the following way )m IP (9.64) Y (0,1P) = Pjm(COs0)e i m jm Since Yjm (0,1P) ,must be a smooth function we demand that m is an integer, so that e 1mIP is periodic with the period 2n You should observe that
e imlP
is still singular at the z-axis where it is discon-
tinuous. (If we approach the z-axis along the line ~O but if we approach the z-axis along the line IP-~ then
then eimlP=l eimlP=_l)
Usually we get rid of this singularity by demanding
O=Pjm(Cos0) on Yjm itself will be smooth. In our models, however, we have singularities on the z-axis from the beginning since the z-axis, because then
...
the gauge potential A
itself is singular on the z- axis. We shall there-
fore neglect the problem. Because
m
can only take integer values we
see that onl.y bostmia states are possibl.e. Then we must use the second eigenvalue equation the expression (9.64) and introducing (9.65)
x=Cos0
m2+2nix+ (K) 2 l-x 2
2 [ - :x [ (1-x );x ] +
(9.55b) . Inserting
we obtain: ] j(j+l) Pjm(x)
-
o
We should then look for regular normalizable solutions of this equation on the interval
[-1,+11. It can be shown that eq. (9.65)
regular normalizable solutions if and only if
m
both integers or both half-integers. Furthermore
and j
has
K
Ii are either is constrained
through the relation
)'
(9.66)
~
hl 1'1
in agreement with the previously obtained result (9.61).
Wopked exepaise 9.5.3 PrOblem: Consider the differential e~uation (9.65) on [-1,1]. Determine under what circumstances it has regular normalizable solutions and show that such solutions are given by -Hm+!'5.)
PJ'm(x)
~ (l+x)
= N.·(l-x) Jm
In the Schwinger formalism integer valued. Consequently
K
-Hm-!'5.)
j-m
~ ~(l-x)
(j +!'5.)
dxJ-m
(j _f.)
~ (l+x)
we know from the beginning that
~]
m
is
must be integer valued tOQ' To summa-
rize we have obtained the following results:
Theopem 2 The Sahwingep
fopmal.ism ,
l.eads to a bosonia speatpum: K
m')'K
ape al.l. integeps
Fupthepmope IKI is the intpinsia spin of the state. The eigenfunations ape given by the fopmul.a
497 (9.64)
II
Pjm(COS0)eimq)
where
(9.67 )
Njm(l-x)
-l:;(m+~) 11
(l+x)
-l:;(m-~) l1d j-m[
j-m (I-x)
(j+~) fl
(J"-~l fl
(l+x)
dx Then we briefly discuss the Dirac formalism
. Here the spectrum is
more easily obtained. From the eigenvalue equation
(9.55a) we get
-i ~Y" alP Jm (0,n) ,0/ = (m-~)Y" fl Jm (0,n) ,0/ But this shows that we can factorize
Yom as follows: )
(9.68) Since
K
i(m--)IP
Yjm (0,1P) = Pjm(COs0)e ei(m-fi)1P
fl
has to be periodic we conclude that
ger valued. But from the general theory, we know that
is inte(m-K) m is either
half-integer valued or integer valued! We have now two possibilities: Either
(j,m'K)
are half-integer valued, or
(j,m'K)
are all integer
valued. Thus we have the possibility of a fermionic spectrum! To check that all the listed combinations are admissible we must investigate the second eigenvalue equation (9.55b). Inserting the expression (9.68)
and introducing
tion (9.69)
[
_
But that is
~[ l-x 2 ~] ax ( ) ax exactl~
x=Cos0
we obtain the differential equa-
2 K (K)2
+ m +2ll1fix+.fi" I-Xi
] -j (j+l) Pjm(x) = 0
the same equation as the one we analyzed in the
Schwinger formalism .! (See eq. (9.65) ). But there we found that K there existed regular normalizable solutions, provided m and fi
were both half-integers or both integers! Thus they are all admissible and we conclude Theorem
:3
The Diraa
formalism
has both a fermionia and a bosonia speatrum:
are all integers or all half-integers Furthermore
IKI
is the intrinsia spin of the state. The eigenfunations
are given by the formula: (9.68)
where (9.67 )
PJ"m(x) = N" (I-x) Jm
-l;(m+K)
(l+x)
-l:;(m-K) d j - m [ (j+K) (j-K)] --"- (I-x) (l+x) dxJ-m
II
498
9.6
THE GAUGE TRANSFORMATION AS A UNITARY TRANSFORMATION As we have seen, the choice of the gauge potential
A
has consequences
on the quantum mechanical level. This may be somewhat surprising: Usually the choice of a gauge potentialis unique up to a gauge transformation and gauge transformations leave physics unchanged. When discussing monopole fields you should, however, be careful! The crucial point is that here different choices of the gauge potential need not be related through a global gauge transformation. E.g. the spherically symmetric monopole field has been represented by the two gauge potentials:
(9.44-45)
A2 =-*(COS0+l) dIP
and
Formally we have
A2 = Al - *dlll but
III
is not smooth throughout space time. It makes a jump somewhere
between
0
and
2rr
and therefore
Al
and
A2
are not related
through a global gauge transformation. On the quantum mechanical level ,things behave a little different. Here gauge transformations are represented by unitary transformations, as we will now explain: Suppose the state of a quantum mechanical system is represented by ~(r,t)
the Schrodinger wave functions
and the various physical quan-
PI,P2,P3,
tities are represented by Hermitian operators
etc. If
U
is a unitary operator then the same state can be represented by the transformed wave function
~ '(r,t) = U~(r"t) and the various physical quantities by the transformed operators
p;
PI' = Ul? IU- I
=
UP 2 ij- I
The transformation: (9.70)
is called a unitary transformation and it leaves all matrix elements invariant ~
~
~ xi + a i
(a) Show that a dilatation is a conformal map with the conformal factor
n 2 (x) =
1.
2
.
(b) Show that an inversion is a conformal map with the con-
formal factor
1
2
---
(x) = <xlx>2 (Strictly speaking we must restrict ourselves to the manifold M = R1 {xl<xlx> ,. O} since the inversion breaks down on the "cone" <xix> = 0).
n
.>+,
(c) Show that the. transformation
C = ITr is given by i x1+a 1 <xlx> y 1+2+<xlx> (which is strictly speaking only well-defined on the manifold RP+~{xI1+2+ <xlx> # O}). Show furthermore that it is a conformal map with the conformal factor
n2 (x)
=
(1 + The transformation C transformation. (Hint: conformal maps f and g formal factor given by
2 + <xlx»2 is called a special conformal Show that the composite of two is again conformal with the con-
534
II
Wor'ked exercise 10. 3 . 2 Problem: Show that the stereographic projection from the sphere onto the plane rr:S2'{N}~ R2, is an orientation reversing conformal map with the conformal factor
=~ 2
n2(8,(j))
Sin
where
e
2
is the polar angle.
Okay, by now you should feel comfortable apout isometries and conformal maps. We proceed to investigate various concepts which can be derived from the metric. Let Mn be a manifold with metric
Then we have previously in-
g
troduced an equivalence relation between tensors of various types.
(See
section 6.9) In coordinates this corresponds to the raising and lowering of indices using the components of the metric tensor. Thus the cotensor .
'k
T~j = g~ Tki etc. Now when we use a diffeomorphism to transport tensors from one manifold to another we should be careful. Suppose f:(Mn,gl) ~ (N n ,g2) Tij
is equivalent to the mixed tensor with components
is a diffeomorphism and that
T
then there is no reason why sors on
N
and
f.T
and
g2)
(with respect to
•
T'
M
are equivalent tensors on
f*T'
should be equivalent ten-
But we know that
f*
commutes
with tensor products and contractions. If for instance
we therefore get i
(f*T') j Consequently we conclude
Lemma 4 Suppose f:(Mn,glJ + (N n ,g2J is a diffeomorphism. If T and T' are equivalent tensors on M with respect to gl, then f*T and f*T'
are equivalent on
N
Thus we see that unless
with respect to f
f*gl
is an isometry it will not respect the
equivalence relations induced by the initial metrics M and N.
gl
and
g2
on
This observation is of vital importance in physics. Consider a scalar field
~
(M,g)
and let
be the Euclidian space
R3
(with the
standard metric) representing the physical space. The energy density is then given by
H
=
~ai~
ai
~
But here we have used the equivalence relation induced by the From the field itself we can only construct the
covector
d~
metric~
with the
535 components
di~'
So when we use the contravariant
II
di~
components
it is implicitly understood in this expression that we have used the
metric components to raise the index. Thus it would be more correct to write out the energy-density as
= ~gijd.~d.~ l. J
H
This is a very common situation in physics: Indices are contracted using the metric. Now observe that the metric is a fixed geometrical quantity. It is physically measurable, e.g. the arclength of a curve in a physical space can be measured with great accuracy in the laboratory. We are not free to exchange this metric if we want to compare the predictions of the theory with the experimental results. Suppose now that we have been given a diffeomorphism of space into itself f : R3 ~ R3 • Then we can investigate the transformed field configuration
~'
=
f*~
.
E.g. we can compare the energy densities of the original and the transformed field configurations at corresponding points and But they are only identical if
f
is an isometry since the metric is
fixed. This distinguishes the isometries from a physical point of view: They leave various physical quantities invariant, i.e. they act as symmetry transformations.
We have previously discussed time-like geodesics on a manifold with Minkowski metric.
(See sec. 6.7). We will now extend the concept of a
geodesic. Motivated by the discussion in section 6.7 we define Definition 9 A geodesic on a manifold paramatrized by
x~
d2x~
(6.49)
The parameter
M
= X~(A) ~ +
with a Minkowski metric
g
is a curve
which satisfies the geodesic equation
r~
dx~ dx 8 = 0
~i3 dA
dA
A involved in the geodesic equation is called an
~
fine parameter.
Observe that the affine parameter is only determined up to an affine parameter shift
536
A = as+b
II
(aiO)
i.e. a geodesic satisfies the same equation (6.49) with respect to the new parameter
s
.
Consider a point at
• Then
P
around
P.
vp
~p,
P
To see this we introduce coordinates
In these coordinates P is represented by coordinates ai A geodesic through P .with tangent vector
~P
and
-+
there is exactZy one geodesic which passes through
and has tangent vector
x~
M and a non-trivial tangent vector
in
P
by
then has to satisfy the geodesic equation d2xi
-a-v-
+
r
i
dx
jk
j
dxk
"d."I""d."I"
0
with the boundary conditions and But this second order differential equation has a unique solution by the well-known uniqueness and existence theorem for ordinary differential equations. If we perform an affine parameter shift
A
=
as+b
(afO)
the tangent vector placed by
a~p
-+
vp
is re-
so we have ac-
tually shown that to each point P
and each direction at
P
correspond exactly one geodesic. Worked exercise 10.3.3 Problem: (a) Let r be a curve on M parametrized by xCJ. = x(l(s). Show that is a geodesic ir and only if it satisfies an equation of the rorm
(10.25)
A(s)
a::
(b) Show that the corresponding arrine parameter (10.26)
A
=
J:
r
A is given by
expU:2A(SI)dSl ]dS 2
(up to a linear change or the parameter)
The affine parameter is not only characterized by the simplest possible form of the geodesic equation. It is also distinguished by the following property
537
II
Theorem 9 Let vector
A
be an affine par·ameter on a geodesic. Then the tangent
~ = ~~
has constant length, i. e.
(1(v(>.) ,V(>.))
is constant
along the curve.
Proof:
r
d dx el dx i3 , dALgai3CDl CDlJ
dx ll dX" , CDl d>' J
= 2g ai3
Here we can exchange
due to the identity
a dx [ By ldx" gaBCDl ~g dllg"y CDl
=
a dX [ B ]dX" gaSCDl r II" CDl
Thus we get a a 2 i3 d [ dx dXS] dX [d x d'\ gai3CDl CDl = 2g aB CDl CfP" +
r
S
ll dx dX"] II" CDl CDl
and this last expression vanishes automatically for a geodesic.
0
So the affine parameter is a natural parameter on a geodesic! We can now divide the geodesics on (a)
Time-~ike
time-like so that (b)
M into three classes:
geodesics, where all the tangent vectors are
(The affine parameter can then be normalized dx el dxS gaSCDl CD\ = -1 ).
Nul~-geodesics,
where all the tangent vectors are null-.
vectors. (c)
Spaae-~ike
geodesics, where all the tangent vectors are
space-like. so that
(The affine parameter can then be normalized dxel dx S gaSCD\ CD\ = +1 )
We can then show Theorem 10 (1)
Isometries map geodesics into geodesics and preserve the affine parameter.
(2)
(This is valid fo-::' Riemannian manifolds too).
Conformal maps preserve null-geodesics. If parameter on the null geodesic affine parameter on
f(r),
r,
where
then
n2
,\
is an affine
fn2('\)dA
is an
is the conformal factor.
Proof: The first proposition is almost trivial since isometries preserve arc lengths and since geodesics are in general characterized as extrernizing the arclength. Special care should however be paid to nullgeodesics, but here the result will follow from the second proposition.
538
II
To prove (2) it suffices to consider a single manifold consider a rescaling of the metric
g
= g.
n2 g
+
Now if
M and to r
is a
null-geodesic then it especially satisfies the geodesic equation
d~~
(6.49)
W
+
ga~
+
r~
dx~ dx~
~e dI" dI"
=a
Under a rescaling r~~e
the Christoffel field
r~~e
g~e
= r~~e
=
n2(x)ga~
given by (6.47) is changed into
+ o~~aelnn + o~~a~lnn - g~ea~lnn
Thus in the rescaled metric we get
d2x~ -~ dx~ dx il _[d~~ ~ dx~ dX B ] dx~ d ~. dx~ dx il ~+r ~~dI"dI"- "(IT2+r ~ildI"dI" +2dI"d"lnn-[a lnnlg~~dI"dI" Here the first term vanishes because
r
is a geodesic, and the third
term vanishes because the tangent vectors are null-vectors. Thus we get
But according to exercise 10.3.3 this shows that
r
is a null-geodesic
with respect to the rescaled metric. However it gets the new affine parameter given by
~(,,)
=
J"exp o
[Js 2dds (lnn 2 )ds 0
l
]dS 2
o
=
-------
In many applications it is preferable to control the set of all possible global isometries: Consider a manifold
M with metric
and suppose
= (+)z Z + <ealeb>(!)z Z = <ealeb> Since the embedding is an isometry we can simply identify
RP
x
Rq
with this particular section M, which henceforth will be denoted q M(R P x R ). The section M(R P x Rq ) will ultimately be replaced by another section of the null-cone
K, but before we proceed with the construc-
tion we must take a closer look at the conformal structure of the nullcone
K
A
K is a null vector
generator of ~
=
It generates a line R,+
w
(u,i",v) ~~
K,
on
= \(u,t,v)
(\(+00
called a characteristic line. A given characteristic line !~ will intersect the section M(R P x Rq ) at most once, and there are characteristic lines which do not intersect M(R P x Rq ) at all. They correspond to the lines which are parallel to the hyperplane
~ = (u,t,v), where
they are generated by null-vectors are precisely the null-vectors where
t2
u-v = 1, i.e. u=v.
But these
= O.
Thus there is a one-toone correspondence between characteristic lines missing M(R P x Rq ) and points on the null-cone through the origin in the original pseudoRP x Rq (cf. fig. 204 a). Consequently the exception-
Cartesian space
al lines represents points on the null-cone at infinity! Consider now two local sections N2 on the null-cone K. N1 and Suppose furthermore that the characteristic lines intersect N1 and
N2
at most once.
Then we have a natural map 11
:
N1
~
N2
543
II
Fig. 204b
Fig. 204a
obtained by projection along the eharacteristic lines (fig. 204b). The basic observation is the following one: Lemma 5 The projection along characteristic lines,
IT:
Nl
~
N , 2
is a
conformal map. Proof: . d uce new coord · ·In R'+p x Rq+' lnates , . We nee d two radial It is preferable to lntro variables r"r Z and p+q homogeneous variables a' , ... ,eP,.p , ••. .pq: Z
r,
rZ Z
z Z u + (x') +
v
z
+ (y' )
z +
...
...
+ (xP ) + (yq)
z
,a P
xp/u
.p' = y'/v, ••. ,¢q
yq/v
8'
Z
x'/u, ••.
(As usual we have troubles with the homogeneous coordinates, which break down when u=O or v=O. Since we are only interested in a local result, we will simply assume that u=O and v=O does not intersect N, or N ). The null-cone K is then Z characterized by the equation If we put the common value equal to r we can introduce the following intrinsic coordinates, (r,e', .•. ,eP,.p', ••• ,.pq) on the null-cone K. A characteristic line is then given by a fixed set of the homogeneous coordinates. Furthermore the local sections can be parametrized as follows: r = f
(e ' , .•. , eP , ¢ 1 , •• • ,.pq)
r = g(e' , •••
,a P ,¢' , .•• ,¢q)
We can therefore use (e', ••• ,eP ,¢' , ••. ,.pq) as intrinsic coordinates on N, and HZ. With this choice of coordinates the projection map IT is simply represented by the identity map!
544
II
We must now determine the various mrtrics inyolved in the game. Consider first q the complete pseudo-Cartesian space R +P x R + with the canonical metric de
2
= -
du
2
1 2 p 2 1 2 q 2 + dv2 - (dx) - ••• - (dx) + (dy) + ••• + (dy )
In the new coordinates this is reexpressed as
K, where
The null-cone ds
2
IK
=
dS~
HZ
dS~
1
2
.
now gets the induced metric: .
kl
and
N1
f2(8,<j»[ =
= r 2 = r,
r [ - e ij (6)d8~d8J + d
Finally the two sections N1
r
C
ij
k
I
(<j»d<j> d<j> ]
are equipped with the induced metrics
N2
j (e)de i d6 + dkl(<j>ld<j>kd<j>l]
g2(6,8)[ - C (e)d8 i d8 ij
j
+ dkl(<j»d<j>kd<j>l]
Consequently
i.e.
Exercise 10.4.1 q RP x R as the interIntroduction: In the above discussion we have embedded 1 using the bijective section M1 of the null-cone K and the hyperplane u-v map +1 <j>1 (z1) = ( 2 q We could equally well have embedded RP x R as the intersection COlle Jl and the hyperplane u+v = 1 using the bijective map 1 + 1 - $2 (z2) = ( 2 z2 2) Problem:
of the null-
Show that the projection along the characteristic lines, 1T:
M1 '" M2
'
corresponds to an inversion in ' RP x Rq •
Having obtained this len'ma we can now apply it to project q M(R P x R ) into a suitable section of K (analogous to the sphere in the Euclidean case). This new section should then include the "nullcone at infinity", i.e. each characteristic line should intersect it exactly once. Unfortunately we run into a slight technical problem: In general it is not possible to find a single section, which is intersected exactly once by each characteristic line. We shall therefore adopt the following strategy:
Let r and 1 Cartesian space
r
545 II denote the radial variables in the complete pseudo2 1 p R + x Rq + 1 :
2 =u2 12 + ••• + (x p2 + (x) )
r 1
Denote by N the intersection of the null-cone the hypersphere
Clearly straint
N is a submanifold defined by the following equations of conr
1
~
Topologically N is therefore a product of the two unit spheres sP Rq + 1 • Consequently N is a hyper-torus: in R1+ p and sq in
N
sP
x
sq
The hyper-torus sP x sq is a nice section on K , but each characteristic line will actually intersect it twice in antipodal points (see fig. 205). Consequently each point in the originaZ pseudo-Cartesian space
sP
x
RP
x
Rq
is represented by a pair of antipodaZ points on
sq.
Fig. 205
q If follows from lemma 5 that the projection map, ~: M(R P x R ) ~ sP x sq, is a conformal map. From the pOint of view of investigating conformal transformations we can therefore equally as well work on sP x sq, except that we must restrict ourselves to transformations which maps pairs of antipodal points into pairs of antipodal points. q In coordinates the projection from sP x sq to RP x R is given by ~(u,z]l,v)
u-v
II
546 consequently the pOints where
u
and
v
coincide are "sent to infin-
ity". But according to our previous analysis these points are in a one-to-one correspondence with the null-cone in RP x Rq . Thus sP x sq is obtained from RP x Rq by adding a "cone at infinity". Because sP x sq is a compact subset of R1 + p x Rq + 1 it is often q referred to as the confopmal compactification of RP x R • q Consider once more the Euclidean space R . In this case the conformal compactq ification reduces to soxS , but the zepa-dimensional sphepe SO only consists of two points, u = ±l. In this case (and only in this case!) the conformal compactification therefore breaks up into two disconnected components {+l} x sq and {-1} x sq Thus we need not double count the points in the conformal compactification because we can cimply throwaway the component {-l} x sq ! But then it is superfluous to enlarge the space with f time-like coordinate, i.e. we simply enlarge Rq to the q Euclidean space R +, and use the unit sphere sq in this enlarged space as the conformal compactification. It is now easy to show that the point at infinity corresponds to the north pole and that the projection along the characteristic lines corresponds to the stereographic projection (cf. fig. 206). Thus we have a nice simplified picture in the Euclidean case.
Conformal compactification of the line.
Fig. 206
RP
Using the conformal compactification of the pseudo-Cartesian space q it is easy to construct conformal transformations on RP x Rq • x R
Consider the matrix group nal matrices operating on
O(1+p; q+1) consisting of pseudo-orthogo1 R + p x Rq + 1 . Each matrix in O(1+Piq+1)
generates a conformal transformation on
sP
x
sq
in the following way:
Notice first that a pseudo-orthogonal matrix S preserves the inner product in R1 +p x Rq + 1 • Especially it maps the null-cone into itself.
Consequently it maps the hyper-torus
caZZy onto a new subset,
S[Sp
x
sql, of
K.
sP
x
sq
K
isometpi-
To get back to the hyper-
torus we then project along the characteristic lines!
In this way we
547 II have constructed a mapping of the hypertorus into itself which we denote by
1[(S']:
Alternatively we can describe 1[[8] in the following way: Each pair sP x sq generates a unique charac{p , -p}, on of antipodal pOints, teristic line
S
formation line
l'.S(P) •
and ~p
x
sq.
.9- p on the null-cone K The pseudo-orthogonal transmaps this characteristic line into another characteristic
The image of
P
is then the intersection between
According to lemma 5 the combined transformation
.9- S (P)
1[[S]
is now a conformal transformation of hyper-torus into itself. From the construction follows immediately some basic properties of 1[[S].
First it maps a pair of antipodal pOints into a pair of anti-
podal pOints, i.e. it can also be considered a conformal transformation RP x Rq . Next the assignment of a conformal transformation to
on
each pseudo-orthogonal transformation constitutes a representation of the pseudo-orthogonal group, i.e. = = 1[[ S2S1]
-
= 1[[ S2]
Notice, however, that the correspondence between pseudo-orthogonal transformations in O(1+p;q+1) and conformal transformations in q q RP x R is not one-to-one. This is because a point in RP x R responds to a pair of antipodal points on
sP
x
sq.
cor-
A pseudo-ortho-
gonal matrix which interchanges antipodal points will therefore generate the identity. matrix:
-
T.
There is precisely one such pseudo-orthogonal
It follows that each conformal transformation is in fact
generated by a pair of pseudo-orthogonal matrices
{S, -S}.
This is
the price we have to pay when we want to represent conformal transformations by matrices! The conformal transformations generated from
O(1+p,q+1) evidently
constitute a group, known as the conformaZ group and denoted by In anaZogy with the conformaL compactifioation of now represent each conformaL transformation in "antipodaL" matrices in
RP
C(p,qJ
x
Rq
C(p,q).
we can
by a pair of
O(l+p,q+l).
Let us investigate the structure of the conformal group a little closer.
548
II
Suppose that the pseudo-orthogonal transformation S actually preserves the additional coordinates (u,v), i.e. that it is on the form
o 5 o
(10.30a)
s E O(p,q)
Then the corresponding transformation on ZCl
~
~ '"
(u; (u-v) zCl ;v)
RPxR q
RP
Rq
x
reduces to
(u; (u-v) s (3z (3 IV) CI
sP x sq
sP x sq
Consequently the conformal group
C.(p,g)
contains the group of
origin preserving isometries 0 (p,g). In the remaining investigation it suffices to consider pseudo-ortho-
gonal matrices of the form
* S
[:
I
*
:]
They will be divided into four general types
+ --2-
(10.30b)
s
C].!
(I)
-2-'-
(10.30c)
Ii (A)
-sinhA
o
c\}
1 + --2-
c
- -r-
coshA
1 -
2 -2-
I
-ev
- c" 1
c\}
].!
;
-2-
I
1
c\l
1
-
c" 1
-2-
Each of these types constitute a representation of a particular subgroup of e(p,g), i.e.
The first type, (10.29)
The second type, (10.31)
The third type, (10.32)
S(I), represents the inversion ].! z
~
z].!
T(c].!), represents the group Of translations z]'!
~
z].!+ c].!
D(A), represents the group of diZatations z].! ~ eAz ll
549
Finally the fourth type, formal translations
II
C(c~), represents the group of special con-
(10.33)
Let uS check the last of these statements just to illustrate the principle: A point z~ in RP x Rq is represented by the point [u,iu-v)z~;v] on the hypertorus sP x sq. By the pseudo-orthogonal transformation C(c~) this is mapped into the point [u ' ; z'~;v'] where (u+v) u + (u-v ) + ---2--
u'
z,1l = (u-v)z~ + c ll (u+v) v'
RP
In
x
=
v - (u-v ) - ---2-- (u+v)
Rq this corresponds to the point z,~ ~,
(u-v)z~ + c~(u+v) (u-v) + 2(u-v) + (u+v) z~ + cU~
u-v
+ 2 + u+v
u-v
[u; (u-v)z~; v]
It remains to determine the factor (u+v)/(u-v). Since on the hyper-torus, it follows that u2 + (u_v)2 x 2 = v 2 + (u_v)2y2
is a point
From this relation we especially get u 2 _ v2 = (u-vt{y2-x 2 ) i.e.
u2 _v 2
----2
(u-v) Inserting this the induced mapping from z~ ~
u+v
=-;:;::v RP x Rq into itself finally reduces to
z~ + c~ 1 + 2 +
We can also summarize the above findings in the following way:
RP
Pseudo-orthogonal transformations in D1+P x Dq+l:
conformal transformations in
Transformations preserving both and v
Pseu~o-or~ho&onal ~ransformations: ,"S" E O(p,q) • z ~ S SZ
u
zCl
x
zCl + c Ci
Transformations preserving
u-v
Translations:
Transformations preserving
u+v
Special conformal transformations: CI zCl + cCl z ~ 1 + 2 +
Transformations in the u-v-plane
Dilatations:
Reflection
Inversion
in the u-z-hyper-plane
z CI _
e Az (l z~
zCI ~ .
q
R
550
II
By now you should feel comfortable about the general structure of the conformal group
C(p,q).
It has the following simple characteri-
zation:
Lemma 6 On a pseudo-Cartesian space
RP
x
Rq
the conformal group,
C(p,qJ,
is the smaZZest group containing the isometry group and the inversion. Proof: First we make the trivial observation that we can generate special conformal transformations using only translations and the inversion. This is due to the identity
(cf. exercise 10.3.1). dilatations.
It remains to show that we can also generate
But that follows from the identity
a 1 = _c = a = a = _c a D(1+k. Consequently p(z) itself must be a polynomial
o
10.7 WINDING NUMBERS AS an application of the preceding machinery we now specialize to
the case where
M
same dimension
n
and
N
Let
that almost all points in
are compact orientable manifolds of the f : ~ ~ Nn be a smooth map. Then we know
N
are regular values (Sard's theorem). We
can now show
Lemma 11 Then either Let Q be a regular value in N. or it consists of a finite number of points.
f-I(Q)
is empty
Proof: The proof is somewhat technical and may be skipped witbout loss of continuity. If ~1(Q) is non-empty we consider a point P in it. Then f* maps Tp(M) isomcrpbically onto T (N) and there exists open enighbourhoods U,V around P and Q, so that f ~ps U diffeomorphically onto V (compare the discussion in section 10.1). Especially P is an isolated point in the pre image ~1(Q) . On tbe other band, ~1(Q) is a closed subset of M. As M is compact, r-1(Q) must itself be compact. If f-I(Q) was infinite, it would then contain an accumulation point, which by definition is not isolated. (Eacb neighbourhood of the accumulation point contains other points from the preimage. ) Therefore t-1(Q) can at most be finite. 0
563
II
Let us now introduce coordinates which generate positive orientations on M and N. Let Q: (yl, .•. ,yn) and P : (Xl, .•. ,xn ) be chosen so that Q
=
f(P)
•
lies in the pre image of the regular value
p
Since
f.
maps
V
of
Q.
U
of
Q, ,
i.e. we
Furthermore
f
diffeomorphically onto a neighbourhood
p
If the Jacobiant is positive, then
the orientation and if it is negative then orientation. Since the preimage f-l(Q) number of points, we can now define: Definition
TQ(M)
Det[dyi/dxj]IP' is not zero~
know that the Jacobiant, maps a neighbourhood
isomorphically onto
Tp(M)
f
U
f: U
~
V
~
V
preserves
reverses the
can at most contain a finite
11
The degree of a smooth map
n
f
M
~ Nn
at the regu Zar vaZue
Q
is the integer (10,54)
Deg(fiQ)
E
p.Ef-l(Q)
j Sgn I dyi /dX I I P ,
~
~
Here
sgn[ dy i/ 3xj]
±l
according to whether
ses the orientation. Cons'equently tive number of times
Q
If we let
Deg(fjQ)
=
O.
denote the number of points in
p
f-l(Q)
denote the number of points in
q
Jacobiant, then
preserves or rever-
simply counts the effecis is covered by the map f . I f C l (Q)
empty, it is understood that Jacobiant and
f*
Deg (f j Q)
p
the compact manifold sional example.
and
N
q
need not be constants as
f
Q
with positive with negative ranges through
This is easily seen already from a I-dimen-
E(1;amp tel: Consider the following map, f : S I ~ S I
presented
f-l(Q)
,
where we have re-
by a periodic function:
N
p
q
p-q
1
0
1
2
1
1
3
2
1
2
1
1
1
0
1
211
+ -;------------------~2-11----~
M Fig. 210
564
II
But the following deep theorem holds, which we shall not attempt to prove: Lemma ]0
(Brouwer's lemma)
The degree of a smooth map. values in
f
M
~
N.
is the same for all regular
N.
This common value is then simply denoted the effective number of times The integer
Deg(f)
N
Deg(f)
is covered by
and it measures
M by the map
f.
is therefore referred to as the winding number-,
since i t counts how many times
M
is wound around
N
In the lite-
rature it is also referred to as the Brouwer degree. Lemma If
11 f: M -
N
o.
fails to be surjective. then the winding number is
Proof: If
f-l(Q)
f
fails to be surjective there exist a point is empty. Such a point
point with
Q
Q
so that
is therefore trivially a regular
0
Deg(fiQ) = 0
At this point we take a look at some illuminating examples: Example 2:
As an example in dimension
n=l
we look at
M = N = Sl
If we introduce polar coordinates, we may consider the map
f
.
given
by: f(cp)
Cos ncp [ Sin ncp
i.e.
This is obviously a smooth map with winding number Example J:
In
n=2
dimensions we look at
:
[:::::::] f3(8,CP)
n
M = N = S2.
duce polar coordinates, we may consider the map f(8,cp)
cp' = ncp .
f
If we intro-
given by:
Sin8 Cos nCP] Sin8 Sin ncp i.e.
(8' ,cp' )=(8,ncp).
Cos8
This is clearly a smooth map outside the poles and it maps the first sphere
n
circle
8
N •
times around the second sphere. In fact, it maps each little 8
0
in
M n
times on the corresponding little circle in
565
If
f
II
was a smooth map,
it would therefore have winding number
n.
fortunately
is not
necessarily
f
Un-
smooth~
In
general it is singular at the poles. This cannot be seen from the above coordinate expression since the polar coordinates themselves break down at the northpole and southpole. This is a subtle point, so let us investigate it in some detail. We may introduce smooth coordinates at the northpole, say the standard coordinates
x
and
y
They are related to the polar coordi-
nates in the following way:
If
f
x
SinS Coscp
y
SinS SinCP
Arcsin~
S
i
Arctg
is smooth, then the partial derivatives af ax
will depend smoothly upon (x,y) . sic coordinates of
-+
-+
and
ex
af ay
and
e
For simplicity we compute the extrinthat are tangent vectors to the second
y
sphere. Using that:
~l
dS ax
dCP ax
-1
ax
ay
as
dCP ay
2Y 2Y
acp ax
1
r
l-
acp
as
Coscp CosS
Sincp CosS
Sincp SinS
Coscp SinS
we easily find the following extrinsic components of
and
Cos'INlNET As an exemplification of the machinery built up in this chapter
we will now look at a famous model from solid state physics. We have previously studied superconductivity ( see section 2.12 and 8.7). This time we will concentrate on a ferromagnet. A single atom in the ferromagnet may be considered a small magnet with a magnetic moment proportional to the spin. At high temperature the interaction)energy between the local magnets is very small compared with the thenreU energy. As a consequence the direction of the local magnets will be randomly distributed due to
~l
vibrations.
FOr sufficiently low temperature, how-
ever, this interaction between the local magnets becomes dominant and the local magnets tend to lign up. We thus get an ordered state characterized by an order parameter, which we may choose to be the direction of the local spin vector. Thus the order parameter in a ferromagnet is a unit vector. If we introduce Cartesian coordinates it can be represented by a triple of scalar fields
[$1 (X);$2(X) ;$3(X)] subject to the constraint
$a$a
=
1.
CONTINUUM LIMIT $Cx)
H
=p
=
[$IC~);$2(~);$3C~)]
I V$a. v$a
d2x
R2 We are going to study equilibriun configurations in a ferromagnet. Consider a static configuration $(~), where $(~) is a slowly varying spatial function. In the Heisenberg model for ferromagnetism one assumes that the static energy functional is given by (10.63)
corresponding to a coupling "between nearest neighbours". To obtain the field equations for the equilibrium states we must vary the fields $a. But here we must pay due attention to the constraint $a$a
= 1.
The order parameter
$
is not a linear vector field, i.e. we
are not allowed to form superpositions!
571
II
The field equations for the equilibrium configurations are given by
1
(10.64a)
or equivalently
1
(10.64b)
Proof: We incorporate the constraint by the method of Lagrange multipliers. Consider therefore the modified energy functional
~[~a(X);A(X)]
=
~
+
Now all the fields,~a(x) and A(X), must be varied independently of each other. Performing the variations A(X)
~
A(X) +
£~(x)
and
we obtain the "displaced" energy functional
~[£]
=
~[O]
+
£J
+
£2~
+ £2DegQ. Thus we have finally
explicitly construced all spin waves with a negative winding number.*) Similarly a spin configuration with a positive winding number corresponds to an anti-holomorphic map on the sphere, i.e. it is on the form: (10.79b)
w(z) =
~m
with
P,Q
polynomials, DegP>DegQ
It would also be nice to find a simple formula relating the winding number to the polynomials P and Q. That is easy enough: The map w=P/Q is a smooth map and according to Sard's theorem there are plenty of regular values. Let Wo be a regular value. Then the preimage consists of all solutions to the equation P(z) - woQ(z) =
o.
As DegP>DegQ it has Degp distinct solutions. map preserves the orientation. Thus this means that
$
w
FUrthermore a holomorphic
has winding number
DegP
and
has winding number -DegP.
Remark:
Rather than relating spin waves to conformal maps one can relate them directly to holomorphic (or anti-holomorphic) functions. Using a stereographic projection IT from the unit sphere in field space to the complex plane we can project the order parameter down to a single complex field given by ~2
1.\1
W
= -~l_iji3
+ i-'l'-I.\l_iji3
In the same way "he tangen" vector a,$a on the unit sphere is projected down into the tangent vector diw in the complex pl~e, i.e. d$a ~* dw Since IT is conformal it preserves the right angles. Furthermore it reverses the orientation so that ~* -id.w #d$a ]!.* -idw i.e. 1
Similarly it follows from the linearity of Aa
e: i } }
1f*
~
e:, .d.W 1J J
IT*
i.e.
that
*d$a JJ..* *dw
*) A.A. Belavin and A.l1. POlyakov,"lIetastable states of two-dinensional isotropic ferromagnets", JETP Lett. 22 (1975) 245.
577
II
Putting all this together we therefore see that the double self-duality equation (10.78) is projected down to the ksual self-duality equation for a holomorphic function *dw = -idw
Okay, this concludes our discussion of spin waves, i.e. the ground states for the various sectors. We might still ask if there are other finite energy solutions to the full field equations (10.64) Apriori there could be local minima lying somewhat above the ground state, or there could be "saddle points". But Woo*) has investigated this problem and using complex analysis he has shown that the answer is negative: Any finite energy solution to the seaond order equation (10.64) automatiaally solves the first order equation (10.78) Worked exeraise 10.8.4 Problem: a) Show that a spin configuration represented by the complex valued function w(z) has the energy density 2 {ow H = (l+ww)Z
(10.80)
oZ
b)
c) ( 10.82)
~2W
OZdZ
=
~
oW
ozJ
oZ
zw
illi: ~
and
oZ dZ
(1+WW)o;2~
=
2w
~ ~
Consider the following two complex-valued functions
awaw
fez)
d)
+
Show that the corresponding equations of motion are given by (l+ww)
( 10.81)
aw
oZ-
=
awllw
oZ oZ (l+wwP
and
g(z)
=
az oZ (l+ww)Z
Show that fez) is holomorphic and that g(z) is anti-holomorphic provided w(z) solves the equation of motion Show that if f has no poles, then w itself must either be holomorphic or anti-holomorphic.
In the above exercise we have almost proven that a solution to the full field equation (10.64),which has finite energy, is automatically represented by a holomorphic or an anti-holomorphic function. There is however one possible loop-hole: The quantity
f (z)
=
oW oW oz
az
(l+wwF
might have a pole, or equivalently the energy density might have Consider, e.g., the non-admissible solution w(z)
*) G.
\\00,
=
zls
"Pseudoparticle configurations in two-dimensional ferromagnets",
J. Math. Phys. 18 (1977) 1264
a"pole~
578
II
(It is non-admissible because it is multi-valued!) Neglecting the branch cut for a moment we notice that it produces the energy density
1 H = 2Izl(l+lzl)2
Consequently the energy density has a "pole" at
z=O . But it is still
integrable (as you can easily see if you introduce polar coordinates). Thus we cannot exclude the possibility that
f(z)
might have a pole.
One must then investigate the solution to the first order equation (10.82) very carefully to exclude that possibility too, i.e. near a pole of like
f(z)
w(z) =
we must show that
w(z)
is necessarily multi-valued
z~. FOr further details you should consult the original
paper of Woo. Exeraise 10.8.5 Introduction: Let F be an ordinary function: F:R1R and let w(z) be a complex valued field characterized by the energy density H =
F(wW){~ ~ az az
+
~ az ~} aZ
In analogy with the Heisenberg ferromagnet we will only be interested in static configurations with a finite energy. Problem: a) Show that the equations of moti~n are giv~ El 2 F d_W =-F'w ~ ~ and F ~ =-F'w ~ ~ azaz dZ dZ azaz dZ az b) Show that
fez) = F(ww) ~ az ~ az
is a holomorphic function provided w(z) solves the equations of motion. c) Specialize to a static massless complex Klein-Gordon field in (2+1) space-time dimensions and try to characterize the static solutions with finite energy. (There are none except for the trivial solutions w(z) = constant!) Hint: Show that it corresponds to the case F=l.
10.9
THE EXCEPTIONAL f4-MODEL
As another interesting example we will consider a model in two spacedimensions, which on the one hand is related to the Ginzburg-Landau model for superconductivity, on the other to the Abelian Higg's model (cf. the discussion in the sections 8.7-8.8). It will be based on the static energy functional:
(10.83)
D
579 ,cf.
(B.60) and (B.BB).
II
For the moment we leave the potential unspe-
cified except that we assume that it is gauge invariant, i.e. it is
,
a function of 1~12
and furthermore that it only vanishes at the non-
zero value I~I= ~o' The field equations for a static equilibrium configuration are given by
au
(10.B4a)
-
(10.B4b)
D*D~
2~2~[
- &B
i~[~D~ - ~D~J 2
cf. the worked exercise B.7.3. As usual we consider only finite energy configurations. This leads to the boundary conditions: (10.B5)
lim p -+
=
B
0
D~
lim p +
co
=
lim
0
..,
p +
co
As in the case of ordinary superconductivity we have furthermore flux quantization, i.e. (10.B6)
"-model possesses the Bogomolny decomposition (10.92)
H
= !\lB+/2U[cjl)EW
+
~1*Dcjl+iD$1I2 : n1lcjl~
We therefore conclude as usual:
(with
II
1\2= < I> )
581 (a)
The energy in the sector
H ~ Inl1f
+
!a ClL d dX a ~ + a(d cj» dA(~ ~ a Cly dX
A!). obtain:
7aA(y(x»))
dnx
603 +
L[~ (I-g(y(x))
+ I-g(x)
II
~ Ifxl ]}dnx
Here we need some identities which we leave to you as an exercise:
Worked exercise 11.4.1 Problem: Remember that a)
Let
I~I
V
denotes the components of the characteristic vector field.
denote the Jacobiant. Show that
~Iffl =~ (~) = aIJ alJ dA ox dA axlJ
(11 . 38)
b)
a
i.e. it suffices to differentiate the trace of the Jacobi matrix, which in turn gives the divergence of the characteristic vectorfield. Show that
(11.39)
Here it is instructive to divide the terms into two groups. Those containing V and those containing a
o
= fU{A + B}/-g dnx
B separately ~ divergences. Consider first A: aL d~a A=a;p~+~alJ~ a IJ a Using the covariant equations of motion (8.35) A is immediately reduced to
We want to rewrite
A
~d
oL d~a
( 11.46)
A =
----±.T-g
a- (.T-g _ _ oL_.~) IJ a(dlJ~a) dA
Then we turn our attention to the second group: B = - aca o~ ) IJ a
av~a
0 a IJ
V
+ La alJ + L----±- a (.T-g)alJ + a'dL (avgall)a v IJ .T-g IJ gall
[o(~~~a) °v~a + olJvL ]alJa +{L I~g v
a).T-g) +
a!~1l
avgall}aV
But here you recognize an old fellRw in the first paranthesis. It is nothing but the canonical energy momentum tensor ~ IJ (Cf. (3.17)). The second term is due to the fact that we work in a covariant fo¥nallsm where .T-g need not degenerate to a constant. We need still a useful identity:
Worked exercise 11.4.2 Problem: Show that the fOllowing identity holds, provided of motion: ( 11.42)
~a
solves the equations
604
II
Using (11.42) We can finally rearrange the B-term as: (11.43)
--La (;::g¥llaV )
B=
I-g
v
Il
Okay, it was a long tour de force, but now A and B has been rearranged as divergences and we obtain the identity which we have looked forward to find: (11.44) As
0
aL ~ = dS dA = Ju[ 1 a {/-g(~ d" I-g Il lJ"'a
--=
+ aV Tv ll ) ~ I-g dnx O
~
U is arbitrary, this is consistent only if the integrand vanishes.
This is the main theorem! To apply it we must gain some familiarity with the current, especially the last part. Let us try to calculate d<PaA/d"
explicitly. We consider first the case where
<Pa
is a collec-
tion of scalar fields. Then by definition (f,,)* we get the characteristic vector field: v
a V = dy = b~<xlx> - 2x~ dA1A=o Furthermore we know from exercise 10.3.1 that Q A
= (1
0A
is given by
+ 2A + A2 <xlx»-1
Consequently dQ A = _ 2 dAIA=o Inserting this the Noether current (11.120) reduces to
~2 1
- 2x n- A P \l
- 2x
4-model 582 625 Volume expansion Volume form 343-44,351 Volume form on a sphere 567 414 Volume of a regular domain Vortex-string in a superconductor 78,441
Wave equation 557 Weak coupling approximation 199 Weak k-form 458 Wedge product in 3-space 6 Wedge product of differential forms 330 Weierstrass' theorem 561 Wick rotation 212 Winding number 564 WKB-approximation 204-209 World-line 284 World-sheet 442 Wronskian 241-42 Yukawa-potential Zero-mode in a non-linear field theory Zero mode in quantum mechanics
97 158 238
