m
u
--Eca
m'
>-u
am >~ ..c ~) E ... c. Q)
..ccaCl)
... u-c
c-- c::::: m __ -W--I ca E "'l--= cae..: .!iUL...
C
-= - -
=-..c
-=...a -=C-c
0
.1::
."
•
.-= Q
C.
c.caQ)
/'
•
,
P
Contents
Pion Limited, 207 Brondesbury Park, London NW2 SJN ,.",
Authors Preface
© 1971 Pion Limited All rights reserved. No part of this book may be reproduced in any form by photostat microfilm or any other means without written permission from the publishers. SBN 85086 023 7
·,1,·, [
( >1
+
[J
J 3
The laws of thermodynamics (P. T.Landsberg)
v Mathematical preliminaries " Quasistatic changes • The First Law The Second Law
"Simple ideal fluids
Joule - Thomson effect
v Thermodynamic cycles v Chemical thermodynamics 'V The Third Law
"' Phase changes
" Thermal and mechanical stability
"oj
v
0 K
(
\
( ... -..~
Set on IBM 72 Composers by Pion Limited, London.
Printed in Great Britain by J.W.Arrowsmith Limited, Bristol.
v
vii
I
I
4
6
9
II
21
23
28
33
34
37
2 Statistical theory of information and of ensembles (P. T.Landsberg) VEntropy maximisation: ensembles
v Partition functions in general
vEnt ropy maximisation: probability distributions "Most probable distribution method
.Some general principles
44
44
48
51
53
60
3 Statistical mechanics of ideal systems (P. T.Landsberg) v Maxwell distribution
~ Classical statistical mechanics
"' Virial theorem v Oscillators and phonons
or The ideal quantum gas
",Constant pressure ensembles v Radiative emission and absorption
63
63
67
72
74
80
92
93
4 Ideal classical gases of polyatomic molecules (C.l. Wormald) -The translational partition function "Thermodynamic properties and the theory of fluctuations v The classical rotational partition function "The quantum mechanical rotational partition function A convenient formula for the high temperature rotational partition function Thermodynamic properties arising from simple harmonic mode" of vibration Corrections to the rigid rotator-harmonic oscillator model Contributions to the therm9dynamic properties arising from low lying
electronic energy levels Calculation of the thermodynamic properties of HCI from spectroscopic data Thermodynamic properties of ethane
98
98
100
103
105
108
110
,((3>
118
121
127
Contents
Contents
ii
5 Ideal relativistic classical and quantum gases (P. T.Landsberg)
132
6 Non-electrolyte liquids and solutions (A.J.E.Cruickshank)
Cell theories of the liquid state
Equation of state treatment of liquids
Binary solutions
140 142 154 168
7 Phase stability, co-existence, and criticality (A .i.E. Cruickshank) Singulary systems Binary systems
193 193
~
..
iii
23 Time dependence of fluctuations: correlation functions, power spectra, Wiener-Khintchine relations (C W.McCombie)
472
24 Nyquist's theorem and its generalisations (C W.McCombie)
489
25 Onsager relations (C W.McCombie)
502
.../ 26 Stochastic methods: master equation and Fokker-Planck equation (I. Oppenheim, K.E.Shuler, G." Weiss)
511
214
27 Ergodic theory, H-theorems, recurrence problems (D. ter Haar)
531
8 Surfaces (/.M.Haynes)
The Gibbs model of a surface
230 230
28 Variational principles and minimum entropy production (S.Simons) Macroscopic principles Electron flow problems
9 The imperfect classical gas (P.CHemmer)
The equation of state
The virial expansion
Pair distribution function. Virial theorem
246 246 254 261
549 549 557
10 The imperfect quantum gas (D. fer Haar)
The equation of state
Second quantisation formalism
v 11 Phase transitions (D. fer Haar) Einstein condensation of a perfect boson gas Vapour condensation Hard sphere gas
270 270 278 282 289
293 303
13 Green function methods (D. fer Haar)
Mathematical preliminaries
General formalism
The Kubo formula
The Heisenberg ferromagnet
319 319 320 324
14 The plasma (D.ter Haar)
333
15 Negative temperatures and population inversion (U.M. Titulaer)
Dynamic polarization
A model of laser action
341 346 348
16 Recombination rate theory in semiconductors (/.S.Blakemore)
350
17 Transport in gases (D.J.Gri{{iths)
378
323
18 Transport in metals (/.M.Honig)
401
19 Transport in semiconductors (/.M.Honig)
432
\(20 Fluctuations of energy and number of particles (C W.McCombie)
448
v21 Fluctuations of general classical mechanical variables (C W.McCombie)
457
Fluctuations of thennodynamic variables: constant pressure systems, isolated systems (C W.McCombie)
~
282
12 Cooperative phenomena (D. fer Haar)
oJ 22
Author index Subject index
465
\
r
563 565
I Authors
.)
•
J .S.Blakemore
Department of Physics, Florida Atlantic University, Boca Raton, Florida
A.J .B.Cruickshank
School of Chemistry, University of Bristol, Bristol
D.J .Griffiths
Department of Physics, University of Exeter, Exeter
J.M.Haynes
School of Chemistry, University of Bristol, Bristol
P.C.Hemmer
Institutt for Teoretisk fysikk, NTH, Trondheim
J.M.Honig
Department of Chemistry, Purdue University, Lafayette, Indiana
P.T.Landsberg
Department of Applied Mathematics and Mathematical Physics, University College, Cardiff
C.W.McCombie
J.J. Thompson Physical Laboratory,
University of Reading, Reading
I.Oppenheim
Department of Chemistry, Massachusetts Institute of Technology, Cambridge, Massachusetts
K.E.shuler
Department of Chemistry, University of California, San Diego, California
S.Simons
Queen Mary College, London
D.ter Haar
Department of Theoretical Physics, University of Oxford, Oxford
U.M.Titulaer
Institut voor Theoretische FYsica der Rijksuniversiteit, Utrecht
G.H.Weiss
National Institute of Health, Bethesda. Maryland
C.J.Wormald
School of Chemistry, University of Bristol• Bristol v
I Preface
1
Problems and solutions! Their production is an annual ritual-feared and unpopular-for the university teacher. Nor are examination questions particularly liked by students. Yet, take away the examination aura, and the technique appears in a new light. The problem-and-solution style of writing presents to the author new difficulties and constraints, and thus offers him a novel challenge. Perhaps it is like teasing a sculptor with a new and promising type of stone, or a poet with a new rhythm. Viewed in this light, the potential author has the opportunities which go with a new medium: for example, a novel way of arranging important results and of setting them up to be seen more clearly than is possible in a uniformly flowing and elegant exposition. As to the reader, he finds before him a series of hurdles. They are easy enough at first to bewitch him, beguile him and persuade him to join into the fun-until he gradually participates with the author in a neo-Socratic dialogue. The existence of these opportunities must surely be one of the considerations which explain how it was possible to find so select and experienced a group of persons to help in the construction of this book. Each author contributed in his own special area of research with the result that the book presents a penetrating view of statistical physics and its uses which, as regards the width of its sweep is, I suspect, beyond any one of today's experts. Thus, in spite of the age of our subject and the love and care which has been bestowed on it by successive generations, this book presents in some sense a new unit. I have planned its outline and have discussed various points with the authors, but have imposed only a limited uniformity of style. I have also attempted to ensure that overlap of material is not extensive, that cross references are adequate, and that the early problems in each chapter are reasonably easy. In this way I wanted the reader of each chapter to feel drawn into a fascinating world which opens up for him once he realises he can actually derive results for himself. Here then is a book for teachers, undergraduates or graduates who want to know what can be done with reasonably simple models in thermodynamics and statistical physics. It is also suitable for self-study. It should appeal to a wide range of numerate readers: mathematicians, physicists, chemists, engi neers and perhaps even economists and biologists. The foundation of the general theory of statistical physics is involved and difficult, and its discussion takes a great deal of time in a lecture course. One can turn to this book for ideas and for inspiration vii
viii
Preface
if one wishes to pass on to areas of application while condensing the time spent on the foundations. This procedure will appeal to those who feel, as I do, that the foundations of the subject are best discussed briefly at first, and then again later from time to time as the effectiveness and the power of the methods are being appreciated. It will come as no surprise to the experienced teachers that the need for care in the setting of a problem can stimulate original work. There are new presentations in various places (e.g. in Problems 1.24 to 1.26, 1.29-1.31, and 3.19 to 3.22), and in several sections of this book previously unpublished ideas will be found. For example, in Section 6 some looseness of logic inherent in previous work has been eliminated, through discussions between the author and the editor, and there are new approaches to equations of state, to the law of corresponding states and to the relationship between reduced equations of state (a) for chain molecules and (b) for their components. Problems 17.7 to 17.9 have also some novelty in the use made there of the random walk approach. There are several other places where recent research work has here been incorporated in book form for the first time (for example, in Problem 5.5). In conclusion, I wish to thank all contributors and the publishers for the cooperation which made this venture possible.
,
P. T.Landsberg
.,
r
1 The laws of thermodynamics P.T.LANDSBERG (University College, Cardiff)
MAJHEMATICAL PRELIMINARIES
VI.I If each of the three variables A, B, C is a differentiable function of the other two, regarded as independent, prove that (a)
(~;) (~~t G~t C
= -1,
(~~)B = I/G~t· Solution
Let the functional dependence of A on Band C be expressed by f(A, B, C) = 0, then
(a~) B,CdA + (:;) A,CdB + (:~) A.BdC
= If A is constant this becomes
(:;) A'CG~) A= - (:~) A,B' i.e.
G~) A= -(:~t.B/(:;) A,C' Similarly
(:~t = (a~)B,cl(:~t'B' (~~) C= -(:;) A,C I(a~) B,C' Multiplication of these three equations yields
(~~)C(~~t (:~t
=-1.
O.
Chapter 1
2
1.1
Interchange of A and C in the second of these equations yields
(~~)
=
B
1/ G~t .
1.2 (a) Integrations over the following two paths in a plane are to be performed: (i) the straight lines (XI, Yl) ~ (x 2 , yd ~ (X2, Y2); (ii) the straight lines (Xl,Yl) ~ (X 1 ,Y2) ~ (X2,Y2)' P(xl>yIl, Q(X2,Y2) are two points, andx 1 =1= X2,Yl =1= Y2' The differen tial forms to be integrated are
1.2
The laws of thermodynamics
Solution
(a) We have
i i
(dx+dy)
(I)
(dx+dy) = (Y2 -Yl)+(X2 -xd·
These two line integrals have the same value, namely u(Q)-u(P) = (X2+Y2)-(Xl +Yl)'
On the other hand
r dv
du
0)
where u
=X
and
=
f
f
dv
(I)
du
= u(Q)-u(P),
(il)
=1=
f
(ii)
~(x~ -Xi)+X2(Y2 -Yl),
J(i)
f
dv == x(dx+dy).
f
= (X2 -X 1 )+(Y2 -YI),
(ii)
du == dx+dy, Show that
3
= X 1 (Y2 -yd+t(x~ -xi).
dv
J(ii)
Since the two results are unequal, there does not exist any function vex, y) of which the differential form dv can be considered as an exact differential, for otherwise both integrals would yield v(Q)-v(P). In the present case av, although inexact, has an integrating factor, i.e. a function g(x, y) exists which converts dv to an exact differential
dv,
dF = gdv .
and discuss the result. [We shall denote differential forms with this property by av instead of dv, and call them inexact, while du is an exact differentiaL In relations of the type du(x,Y, ... ) g(x,Y, ...)crv(x,y, ... ), g(x,y, ... ) will be called integrating factors.] (b)If dF = X(x,y)dx+ Vex,
If one integrating factor exists, then there exists an infinity of them, and a simple example is furnished in the present case by
is an exact differential, show that
so that
(~~t = (~~t· (c) Pfaffian forms have the general form n
dv or av
=
i
L A/(x = I
1> x 2 , ..· xn)dxi'
(dv may be exact or inexact). Show that for n = 2, if the Xj are single-valued, continuous and dif ferentiable functions, av has always an integrating factor provided X 2 is non-zero in the domain of variation considered. 3 need not have an integrat (d) Verify that a Pfaffian form with n ing factor by considering dv = X dy + k dz, where k is a non-zero constant.
g
= 1/x,
dv
du .
(b) For an exact differential dF
ax y dx+ (aF) ay x dy == Xdx+ Ydy, ( aF)
=
aF (ax) axay = ay 2
x
(ay) ax y'
(c) Let
crf
= Xdx+ Ydy
,
where X, Yare continuous differentiable and single-valued functions of the independent variables x, y. This restriction on X and Y together with Y =1= 0 means that the equation dy X -=- dx Y has a solution of the form F(x,
=
C,
where C is a constant.
i.e. dF
=
(~~)y dx+
C;t
dy
0,
Chapter 1
4 Comparing coefficients in df
= 0 and
dF
±(~~t = ~(~~)x Hence gdf
1.2
(a) From the first two expressions for dQ an equation for dt is found:
== g(x,y).
(C p -Cv)dt
(
ax
y,z
= 0,
x
k
(aF)
az
g'
= Iv dv -I p dp
.
Substitution for dt in the second expression for oQ
= gXdx+gYdy = dF
(aF) ay x,Z
5
Solution
= 0,
and dfhas the integrating factor g(x, y). (d) Suppose dv(x, y, z) == xdy+kdz = g(x, y, z)dF. Then aF)
The laws of thermodynamics
1.4
g
x,y
It follows from part (b) that
aF 1 1 (a g ) aXay=0=g-g2 ax y,; aF = ° = -g2k (aaxg ) axaz 2
_
IvCp
(
dQ - C p -C dv + lp v
IpCp ) C -C dp. p v
Comparison with the last form for oQ yields the required result. (b) The first equation under (a) yields this result at once. (c) C v is the heat required per unit rise of empirical temperature at constant volume, also called the heat capacity at constant volume; Ip is the heat required per unit rise of pressure at constant temperature, some times called the latent heat of pressure increase. The other coefficients can be described analogously.
2
1.4 Let
y,z'
a~
x(~)
ayaz = -g2
az
k(~)
-g2 ay
x,y =
cxp
==
K t
QUASISTATIC CHANGES(I)
1.3 For a fluid and other simple materials any three of pressure p, volume v, and empirical temperature t are possible variables. They are connected by an equation of state so that only two of the three variables are independent. An increment of heat added quasistatically may then be expressed in the alternative ways = Cvdt+lvdv = Cpdt+lpdp
= mvdv+mpdp,
where the coefficients are themselves functions and are characteristic of the fluid. The term empirical temperature refers to an arbitrary scale and is used to distinguish it from the absolute temperature, denoted by T. Prove that the following relations hold: (a)
_ IvCp __ IpC v mv '!!.E.. _ . mv - Cp -C v ' mp Cp -C v ' 1v + Ip - 1,
(b)
= Cp -C v (aatp ) v = _ Cp l-Cpv' (av) at p Iv·
(c) Express in words the physical meanings of the coefficients in the expressions for dQ.
\
_.l(av) ap V
t
be the isothermal compressibility of a fluid. The notation of Problem 1.3 will be used. (a) Show that the Griineisen ratio r == cxpv/KTCv of the material satis fies
r = v(~~)jcv.
(b) Show that the ratio of heat capacities 'Y == Cp/Cv satisfies
(~~))(~~)t' - 0 = G~)j G~)p' 'Y~l = (~~))(~~)v' 'Y
=
where ( )a denotes a quantity evaluated under quasisfatic adiabatic con ditions, i.e. for dQ = 0. (c) If Ka is the adiabatic compressibility, prove that
Kt
K (I) Changes consisting of a continuum of equilibrium states.
~(~~ \
be the coefficient of volume expansion at constant pressure, and let
These equations cannot be satisfied by a finite function g(x, y, z).
oQ
==
x,;
a
= 'Y.
Chapter 1
6
(b) Prove that the Griineisen ratio of Problem 1.4 is v
(d) Show from a consideration of d(lnv) that
( ~) ap
= t
_(aKat
t)
7
The laws of
1.6
1.4
r
. p
v
I(~~)o
mp
(c) Find the most general equation of state of a fluid whose Griineisen ratio is independent of pressure.
Solution
(a) We have
r
-vG;)p
=
Solution
v(~)v
where Problem 1.1 (a) has been used. (b) From the first equation in Problem 1.3:
v
ap )
( av
a
_ Cpl mp - Cvlp
[from Problem 1.3(a) 1
G;)a
Iv '
(~~)a
lp ,
(~~)t
Iv Ip '
p
where x
=0
_
y
= pdv+xdy
Tor p. If oW were exact we would have p
) (a ay v
=
(ax) av
= 0 y
•
This yields 1 0 if y is chosen to be p, and hence oW is inexact. (b) We start with oQ = dU+pdv = Cvdt+l"dv whence
We now take the ratios of terms in the first column to corresponding terms in the second column. The first pair yields "I at once, The second pair yields, in conjunction with Problem l.3(a),
.'
Co = Ca~)v (~~)v(~)v' =
"I-I'
The third pair yields "1/("1-1). (c) This follows from (b). (d) We observe that t and p are the independent variables needed. The form of the equation to be proved suggests that the procedure of Problem 1.2(b) may be involved. Fortunately we also have the hint to consider d(lnv). Hence dOnv)
oW
'G~)p = mv a) ( at v
_Cvmv __ Cv~ Cp Iv CpCp-Cv
I
(a) This is obvious since dU = oQ -ow. If dW were an exact differential, one could integrate to find Q = U+ W+constant, in contra diction with the inexact nature of oQ. Alternatively, note that we have two independent variables, so that one could write
= ~dV = ~ [(~~)p dt+(:;)tdP].
This is OI.p dt-Ktdp, and we can indeed apply Problem 1.2(b). THE FIRST LAW
1.5 The first law states that an internal energy function U exists such that for a fluid or similar material we have, in addition to the first equa tion of Problem 1.3, oQ = dU+pdv, where p is the pressure, and p dv is the mechanical work done by the system. fa) Show that the increment of work dW pdv is inexact
Substitute in Problem 1.4(a) to find
r = v(~~)) C
v vI(~~)v' =
(c) Integrate the result of (b) to find
1
U = r(v)[pv+f(v)],
where rand f can be functions of volume, and fjr is a constant of the integration with respect to p. The equation of state of a solid is some times taken in this form: pv = r(v)U(v, T)
1.6 Show that (a)
(~~)v'
mp = mv
(~~)p +p,
=
(~) av
p
=
(amap
v
1.
) v
8
Chapter 1
1.6
(b)
(~~)p - (~~)v (~~\ (~~t·
(c)
Cp -Cv =
1.7
1.7 The second law asserts that the reciprocal absolute temperature
[(~~)t +p] G~\·
liT is an integrating factor of ctQ, the resulting function of state being called the entropy S, so that for quasistatic changes dS = ctQ/T. This
Solution
(a) From
dU+pdv
=
[(~~)p +p] dv+ (~~)v dp
one finds
mp =
(~~)o'
mv =
(~~)p +p.
(omop
(~~)s = -(:~)v' (~~)s (:~t, (~~)p = -(:~)T' (~~)v = (:~)T' [F is the Helmholtz free energy, H is the enthalpy, and G is the Gibbs free energy. J
~_(3) avop - ov p =
I
adds a further relation to those given at the beginning of Problems 1.3 and 1.5. Returning to these, adopt the absolute temperature as the most convenient empirical temperature scale t to establish the following results. [Note that in a quasistatic adiabatic change the entropy is constant.] (a) By considering dU = TdS-pdv, dF == d(U-TS), dH == d(U+pv) and dG == d(U+pv TS) establish Maxwell's relations =
Also, differentiating, we obtain
v)
-1.
(~~)t dv + (~~)v dt
(~~)t[(:;)tdP+(:~)p dt]+(~~)vdt = (~~)t dp+ [(~~)t (:~)p +(~~)Jdt.
(~~)p - (~~)v = (~~)J~~)p'
(c) The first equations of Problems 1.3 and 1.5 yield whence
d:Q = Cpdt+lpdp = dU+pdv, Cp =
=
T(::)v'
G~
Iv
=
T(:~)v'
Ip
-T(:rt,
T(~;)v'
m"
T(~~)p'
0
=
Hence
Cv
,:i
(b)
(b) The independent variables (v, t) and (p, t) are involved, suggesting that a substitution for one in terms of the other two is required SOme where. We have
dU =
9
THE SECOND LAW
=
oQ
The laws of thermodynamics
(~~)p +p(:~)p'
This, combined with the relation proved in (b) and with C v = (oU/ot)v, yields the required result.
mp (c)
Ipmv
=
-TCp,
lump
=
T(::)p'
=
TC o ,
Ivlp = -T(Cp-Cv)·
(d) In view of these relations, and earlier ones, which of the six functions C, I, m give a set of independent quantities in terms of which the others can be expressed? Solution
The stated equation for dU clearly yields the first required result by the process indicated in Problem 1.2(b). This can be applied next to
d(U-TS)
dU-TdS-SdT
=
-pdv-SdT.
This yields the last equation. The other two results are established analogously. (b) These results follow immediately from the equation at the begin ning of Problem 1.3, except that in the case of Iv and lp a Maxwell relation is also needed. For example, TdS = CIl dT+l vdv implies
Iv T(~~)T = T(:~)v' =
1.7
Chapter 1
10
lpmo = (:~)p -TCp, lomp = T2(:~)v = TCv' It was shown in Problem 1.3(a) that
mv mn Iv lp . relations for mo and mp _TCp+TCt} = 1 Ip Iv Ivlp , -+----
V2 -VI
"1-1
= --dUo
n-l
Vaporization Melting (normal) Melting of ice
Thus if 1 < n < "I the work done in an increment exceeds the drop in the internal energy of the fluid. Hence dQ = dU+dW is positive while dT is negative. Now C
Ap AS AT Av 0; then we have
T-=-=
Hence "1-1
VI
where A}'" 2 is the heat required to take a mass m of material isother mally from the state of aggregation 1 to the stage of aggregation 2, and V 1 and v 2 are the volumes occupied by this material in the two states of aggregation. [If m is taken as the unit mass, then A is the latent heat, and the v's are specific volumes.] (b) Show that increase of pressure raises the boiling point and the freezing point of a normal liquid, but that, in the case of water, pressure lowers the freezing point.
AT =
TI-bvg is a constant.
V2
dQ = dU+pdv = Cv dT+pdv ,
v
=
1.14 (a) If a fluid is in equilibrium with its saturated vapour pressure, then the pressure in the system is independent of volume and depends only on temperature. Show from the Maxwell equation given in Prob lem 1.9(a) that under these conditions AI'" 2 dp T dT = ~ (Clausius-Clapeyron equation),
Tv n- I (pvn)(n-I)/n
whence dU = Cv dT. Also, from pvn = B?, pv
= f ,(z) .
b-=-= T Cv z
and this makes the fluid isotropic of index n. The last result required follows from T prn:TfTii
= z· Z-I f'(z)
The change taking place is defined by TdS = bCv dT, whence dT dS dz
.
It follows that another constant is
B? == Atv n- I
21
The laws of thermodynamics
1.15
n -"I
= --ICv n
state 1
state 2
liquid solid solid
vapour liqUid liquid
dp/dT V2 V2 V2
> VI > VI
< VI
positive positive negative
JOULE-THOMSON EFFECT
1.15 If gas is allowed to flow slowly through a porous plug between two containers, which are otherwise isolated from each other and from their surroundings, the enthalpy HI == U I + P IV 1 before the process is equal to H 2 , the value after the process. The temperature change is measured by the Joule-Thomson coefficientj == (aTjap)H.
and one observes that, for 1 < n < "I, C is in fact negative. (d) In the solution of Problem 1.12(b) it was noted that if z == Tv g then Uv g = fez), and also dS = z-ldf(z) = z-If'(z)dz = z-lCv dz .
\
(IJ
22
1.16
1.15
Chapter 1
The laws of thermodynamics
(b) From part (a) and a Maxwell relation we obtain
(a) Show that dB
= TdS+vdp, j=
and hence that
.
J
v
= -(Tcx C p p
-1) •
j
=-
[T(:i)
v
J
+v(~~) jC(~~) p
=
(2av _3ab -b P) v 2
j
v
v
p
PiO
=
2a 3a a 3b 2- 9b 2 = 3b2
.
=
0, i.e. at Vi
a
PiO a 27b 2 Pc - 3b 2 ' -a =
= ViO = 3b.
The
p
It follows that
j < 0
3b 2
9.
Also
31'.:.
A1io
a) 8a a = ( 3b2+ 9b2 (3b -b) = 9b
and
lio 8a 27Ab 1'.: = 9Ab . sa- = 3 .
Solution
(a) Using dU = TdS-pdv, we obtain
= dU+pdv+vdp = TdS+vdp,
3b
v
Figure l.lS.1
THERMODYNAMIC CYCLES
1.16 (a) In an incremental process a fluid gains heat energy aQ at temperature T. Establish that in a closed cycle
as required. It follows that dB
T
[T(~) -v] aT p
(f) Its maximum occurs at dpddvi corresponding value of Pi is
[The maximum of the inversion curve lies at a pressure and tempera ture which are above their critical values.]
dB
p
2a 3a Pi = bv.1 - v?-1 .
Show also that =
J
(~~) jC(~~)
(e) From the result in part (d) the curve is
PiO = 9pc'
1io
+v
= _1 Cp
2ab)c . j (p_~+ 2 3 ~
v
T'
(e) Obtain the curve on a (p, v)-diagram separating the region j > 0 from the region j < 0 for a van der Waals gas. This is the inversion curve. [Expansion leads to cooling only for states lying in the j > 0 region.] (f) If Pw is the pressure at the maximum of the inversion curve, show that for a van der Waals fluid.
[T(:i)
This is more convenient than the result in (a) if p = p(v, T) is known.
(c) From Problem 1.9(d) we have Tcx p = 1. (d) The expression is obtained by a direct calculation of
(c) Establish that j = 0 for an ideal classical gas. (d) Show that for a van der Waals gas introduced in Problem 1.11 j
[T(:~)p(~~t -v(~~)J/Cp(~~)T
=-
(b) Show that, alternatively,
fdi .:;
= [V+TG!)JdP+T(::)p dT,
so that j
23
(Clausius inequality),
where the equality holds for a quasistatic cycle. (b) In a general quasistatic cycle a working fluid receives heat at various temperatures and gives up heat at various temperatures. The efficiency 1/ of the cycle is defined as the mechanical work done divided by the sum of all the (positive) increments of heat gained. Prove, using
== (aT) ap H = _ [T(as) ap T +v] jT(as) aT p = ...!...IT(~) Cp aT p -v],
where one of the Maxwell relations of Problem 1.7(a) has been used. The required result follows.
\
0
,
(-J
Chapter 1
24
1.17
1.16
1't-12 ----y;-
11c, the Carnot efficiency),
the heat gained is equal to the work done. Hence
where Tl is the highest temperature at which heat is gained, and 12 is the lowest temperature at which it is given up. (c) Show that the maximum efficiency 11c can be attained by using as a working fluid an ideal classical gas of constant heat capacities, which is working between isothermals at temperatures Tl and 12 separated by an adiabatic expansion and an adiabatic compression. rThis is the Carnot cycle.] (d) If you have used the properties of the ideal classical gas in (c), generalise the argument so that it applies to any fluid describing a Carnot cycle quasistatically.
QI =
==
= f
jT
Hence Q2/Ql ~ p
dQ-f dQ +T
_T
12/Tl , so that 11 ==
~
1 dQ_,
-12
1't
12'
l-Q2/Ql ~ 1-12/1't.
(c) With the notation shown in Figure 1.16.1, TV'Y- 1 is constant on the adiabatics be and da [Problem l.lO(a)]. Hence
a
1 = 'F-V'Y-I T,V'Y-l T,V'Yla 2 d , lb
c
.... v
Figure 1.16.1
=
Wbe
--
W~
:::: -
pdv
La
Ve
Vd
= -I
dU = C v
~
dU:::: Cv(TI -12),
(12 -1't).
As in the solution of Problem 1.16, the amount of heat absorbed from a reservoir during the isothermal change ab is
1 'F-v'Y2e,
Qab
The heat rejected on cd is Va
~
Solution
i.e. Vb
Vb
-A 12 In- :::: A121n
1.17 The working fluid of a thermodynamic engine is an ideal classical gas of constant heat capacity Ct). It works quasistatically in a cyclic process as follows: isothermal expansion at temperature TI from volume VI to V2; cooling at constant volume from temperature 1't to 12; (iii) isothermal compression at temperature 12 from volume V2 to v I; (iv) heating at constant volume from temperature 12 to TI . Obtain expressions for the amount of heat (Qt) supplied to the gas in steps (i) and (iv) and the amount (Q2) rejected in steps (ii) and (iii). Show that for the above cycle the efficiency 11 < (1't -12)/TI . How is the efficiency affected if (ii) and (iv) are replaced by adiabatic expansion to volume V2 and by adiabatic compression to volume v 1> respectively.
f dQ-f dQ = QI_Q2
+1't
Va
Vd
(d) From the first law work done by the engine is equal to the heat gained by the fluid in a cycle: W = QI Q2. Also from part (a) we have Q'l = 1't 12 Hence W Q I -Q'l _ l_Q2 12 1-11 = Ql QI Q1 Tl
the integrals extending over the appropriate increments. QI is the total (positive) heat gained; Q2 is the total heat rejected, counted positively. Energy conservation gives the mechanical work done as W = QI - Q2 ; the efficiency is 11 (QI - Q2)/Ql' Now
o idQ
pdv =
a V
e
For a quasistatic cycle the equality holds for each increment and hence for the cycle as a whole. (b) Divide the increments of the cycle into those in which heat is gained, dQ+, and those in which heat is lost, dQ_. Then define Q2
d
Vb A1't fbdV -:::: ATlln,
e
fd~ ':;;;0.
1 dQ+,
a
pdv =
Hence 11 = 11c. Note that the work done on the adiabatics cancels out:
We have dQ/T':;;; dS, where dS is the incremental change of the entropy of the fluid. The entropy is a function of the state of the fluid and in a cyclic process its final value equals its initial value, whence
==
f f b
Q2 = -
Solution
QI
25
It follows from Joule's law [Problem 1.9(b)] that the internal energy remains constant on each isothermal, so that dQ = dU + (tW implies that
the Clausius inequality, that 11 .:;;;
The laws of thermodynamics
Qed
-=
('j
Vb
AT1ln-. Va
Vd
= A12lnVe
Vb
A12ln
Va
1.17
Chapter 1
26 since Va
Vd
and Vb =
V c'
1.19
During the cooling bd the heat rejected is
The internal energy gained by F2 is
C,,(TI -1;) .
Qbc
Q2 = C2 (To 1;).
During the heating da the heat supplied is
The overall loss of internal energy is
Qda = C v (11-1;) .
QI -Q2 = CITI -~12 -(CI +C2)To . It follows from energy conservation that this must be equal to the total amount of work done by the Carnot engine. (b) In the absence of the performance of work, energy conservation yields CI (I1-To) = C2 (To-1;) so that
To aTI +b1; (a+b I) .
The entropy lost by FI is
Hence
= Qda+Qab AI1 1n(vb/va)+C,,(TI -1;) QI -Q2 = A(I1-1;)ln(vb/va)' QI Q2 TI -72 11-1; 11 = < -TI QI T, + Cv(TI -1;) QI
A In (V2/V I) If steps (b) and (d) are changed as suggested, a Carnot cycle results, and the conclusions of Problem 1.16(c), (d) apply. I
1.18 (a) Two fluids FJ, F2 of fixed volumes and constant heat capaci ties CI , C2 are initially at temperatures TI, 12 (11 > 1;), respectively. They are adiabatically insulated from each other. A quasistatica11y acting Carnot engine E uses F 1 as heat source and F2 as heat sink, and acts between the systems until they reach a common temperature, To say. Obtain an expression for To and for the work done by the Carnot engine. (b) If a common temperature is established by allowing direct heat flow between F I and F 2, what is the final temperature and what is the change in entropy? (c) Show that for all positive C I , C2 this change is an increase. Solution
(a) Since CI = T(aS/aT)", the entropy lost by FI is _ fTo dT TI SI = -CI T = ClInT, . T,
0
The entropy gained by F2 is S2 == C21n(To/1;). If the working substance of the Carnot engine is in the same state finally as it was initially, the entropy gain of the whole system is S2 -SI = In
l(~r2 (~rlJ =
0.
Hence To where
l1a1;b,
CI
a == CI+C.l '
b
==
C2 CI +C2
The internal energy lost by FI is in the form of heat QI = -CI
f
To
T,
27
The laws of thermodynamics
dT = CIOi-To) .
f
~dT
f
To
~==-~
l
The entropy gained by
~
11
~=~~T,' 0
is
,,)
S2
C2
dT
T,
To
= C2 ln T, . 2
The gain of entropy for the whole system is S2- S 1
To)b (To)aJ = (CI +C2)ln [( 1; TI
To (CI+C2)ln11'1i"
Hence S2 -Sl
aTI +b1;
= (CI +C2)ln 11'1i'
(c) For aU positive a, b such that a+b = I, consider y = all +b12 -l1a
This quantity has a minimum when 11 = 1; and is therefore never nega tive. Hence S2 -SI ;;;.. O. 1.19 (a) Show that the Jacobian
a(p, v) _ (a p)
aCT, S)
=
(av)
aT s 'as
(a p) T -
as
T
(av) aT
s
I.
(b) A thermodynamic engine uses a cycle which is represented by closed curves on a (p, v)· and on a (T, S)-diagram. Use the fact that the work W done by the working fluid is equal to the heat Q adsorbed, expressed in the same units of energy, to obtain the result of part (a).
28
1.19
Chapter 1
Solution
J
(a) We have dp
p
)
Solution
l
)
T -
T
(b) We have for corresponding domains of integration in the (p, and (T, S)-planes a(p, _, W aCT,S) dTdS
ffdpdV
where the Jacobian arises because of the change of variables. Also
Q=
Hence, since W
njdlli = O.
among the intensive variables.
(aTa s (av) as (aas (av) aT s p
=
(d) Establish the Gibbs-Duhem equation SdT-vdp+
Git (~~)v +(~~)TG!)V
G:)v
29
The laws of thermodynamics
Gi)sdT+(~~tdS.
Hence, using two Maxwell relations of Problem 1.7(a) =
1.20
~
The equation states that, if the independent variables (which can be chosen to be all extensive variables) are multiplied by a factor a, then the entropy is multiplied by the same factor, and is therefore also extensive. A similar equation' for U can be deduced from this equation, as is necessary since U is also extensive. For suppose S(U, v, nl, ... ) = So can be solved for U; then if g is some function, the solution is, say, U
It follows that the solution of S(aU, av, anI' ...) aU
Q for any domain, the result follows. But aU
CHEMICAL THERMODYNAMICS
1.20 If two identical systems are joined together and considered as one system, the quantities whose values double are called extensive, and those whose values remain the same are called intensive. A phase is a thermodynamically homogeneous region of space, thermodynamically uniquely specified by its internal energy U, its volume v, and the num bers of molecules n I, n2, ... of the chemical species contained in it. (a) Verify that the equation for all a > 0
= g(So, v, nl' ...) . = aSo is
= g(aSo,av,an1, ... ).
= ageS, v, nt, ... ), so that we have g(aSo, av, an 1> ... ) = ag(So, v, n 1> ... )
•
This is the analogue of the given equation for S. Similar equations hold for v and the nj. (b)
TdS = T
(auM) .
v,ni dU+T
(M) a V
U,ni dv+T~ I
M
an,.
dnj
- L Iljdnj .
S(aU, av, anI, an2, ... ) = as(U, v, n l , n2, ... )
If Sa
is consistent with the extensive nature of S, U, v, and the nj. (b) Assuming the entropy to be a function of the state of a phase, and that it satisfies
G~)v.ni
T'
== S(aU, av, anI' ... ), then dSa = aSa d(aU) + aSa d(av) + L aSa d(anj) da a(aU) da a(av) da I a(anj) da
(~) U,n = f,
Hence
i
S
even if the nj are variable, show that
I
= r(U+pv -
L Illnl) . j
TdS = dU+pdv- LIl;dnj, j
Ili
== _T(;S)
nj u, v, other n's
.
(d) From part (c) we have
Ilj defined here is called the chemical potential of species i.] (c) Show by differentiating with respect to a, that G
== U-TS+pv
dU-TdS-SdT+pdv+vdp- L(lljdnj+ntdlli) i
From part (b),
Lllinj. j
dU-TdS+pdv- Llljdnj
[G is called the Gibbs free energy.]
i
lu
O.
= O.
O.
oUfpov - ToS
I
'"\ 1.21 Each of n phases I, 2, ... n consists of one and the same chemical species. A new system without inhibiting constraints is formed of these n phases by combining them so that the internal energy U, volume v, and particle number n of the resulting system are each the sum of the original quantities for the phases:
L~,
V= LVI,
1= I
i
=I
n =
L nl' 1= I
Show from the result in Problem 1.20(b) that for equilibrium in the new system
Ii
T2
1;"
PI =
pz
= ... = Pn,
III = lIz
=
=
LlltOn/; LPjOVj ~
0,
f
where the two sums involve pressures and volumes which can change owing to internal processes in the system(4). (a) A system consists of two phases, labelled I and 2, which are . .. u... u~.uJ separated mechanically at pressures PI and P2- If they are then coupled mechanically, but their compositions are fixed, show that the phase whose pressure is greater will expand. If the volumes of the phases are fixed, but the molecules of a certain chemical species can suddenly be transported from one phase to the other, show that the molecules will migrate from a region of higher to one of lower chemical potential. [The tendency to equalisation of intensive variables in equilibrium is illustrated by both Problems 1.21 and 1.22.J
n
n
n
U=
31
whose mass is fixed). Suppose that, with a generalised interpretation of 0, one has
Hence
SdT-vdp+ Ln/dlll =
The laws of thermodynamics
1.23
1.20
30
= lIn·
Solution
Solution
(a) If one phase expands, it does so at the expense of the other, -ovz. Hence so that ov I
The entropy change for phase i is
OSI
I
= r;(oUj+pIOVj-lliOnj).
-P Z OV2 = -(PI-P2)OVI ~ 0. then PI > P2; if ov I < 0, then PI < P2' -PIOVI
I
Thus if OVI > 0, In either case the result stated in the problem follows. (b) Let the phases be denoted by I and 2, so that on I -on2, since the particles lost by one phase are gained by the other. Hence
We seek to maximise LoSt subject to i
j = LOVt = L ont = LOU i t i
°.
1l1 0n l +ll zonJ, = (Ill -1l2)onl ~ 0.
Using undetermined multipliers, consider
If on l > 0, then 112 > Ill; if onl < 0, then 112 result stated in the oroblem follows.
~ [(*-a)Ui+(~-~)Vi-(~-~)njJ
f
Here Ti, Ill, PI, a, ~,~ are constants. The quantity fhas to be maximised for arbitrary and independent variations of £1;, VI, and nj. One finds
t [(~-a)OUt+(~-~)OVj-(~-~)ontJ PI = PI
III
= II:!.
= ... =
= O.
PV
gU=AT,
Cv
~,
1;, = l/a,
= ... = Pn
T"t
= ~Ti =
= ... = lIn = ~Ti
In either case the
1.23 An ideal classical gas of constant heat capacities is considered in this problem and it is recalled that for such a system
This implies
TI = T2
< Ill'
=
Cp =
[S
exp C
(l+~)A,
J
-(~-I)i ,
g =
~-l,
}
(Problem 1.9)
(Problem 1.10)
v
where i is a constant (and not V-I !).
1.22 A form of the second law suggested by Problems 1.5 and 1.16 is 0, where 0 denotes an incremental change the end points of which are equilibrium states of a closed system (i.e. a system
8U - pov - ToS ~
(4) For a discussion of this generalisation see P.T.Landsberg, Thermodynamics with Quantum Statistical Illustrations (lnterscience, New York), 1961, p.156.
I
Chapter 1
32
1.23
(a) Two such systems having identical heat capacities satisfy pv = AT, pv A'T. Given only that A is an extensive variable, prove from Problem 1.10 that Bland B2 are extensive while B3 and i are intensive. [Because A is extensive, it is written as kN, where k is Boltzmann's constant and N is the number of molecules.] (b) Show that for one such gas
S = A [(1+
~)lnT-lnp+iJ = A(~lnT+lnv+i-lnA).
[The above is a generalised Sackur-Tetrode vapour pressure equation, and i is here called the chemical constant.] (c) Show that the chemical potential of the gas satisfies the condition
:T
= lnp+
(I +~)O-lnT)-i.
[It is not possible to identify the chemical constant thermodynamically. But it is possible to do so from statistical mechanics. See Problem 3.13.] Solution
1.24
The laws of thermodynamics
(c) Since pv = gU
G
U+pv-TS
Since G
B1
AB?-l
!:iN (Problem 1.20) and A
(B~)"r BI
=
X(B~ - )"rB2
X"r (B~ B3 )
l
=
(1)
= p&"r-I)I"Y
.
THIRD LAW
1.24 Suppose the entropy remains finite and continuous as T -+ O. [This is from reference 4, p.112 equivalent to Nernst's heat theorem.] Show that then X = Cv , Cp , III, Ip, m v , mp each tends to zero in this limit. Solution (i) Recall from Problem 1.7(a) that
dF = dU-TdS-SdT = -SdT-pdv ,
rr = Bl = p"r-l
(BA2 )"r-
(:~)p = -S =
= exp
J
[S CII -('Y-1);
dH
.
'YlnT-('Y-l)lnp = C -('Y-1)I.
dG
= A/g (Problem 1.9), it follows further, on multi
S = A ( g+ I InT-lnp+i) .
g
T->O
[(aF) - -(au) ] aT aT v ' II
= TdS+vdp
= CpdT+(lp+v)dp,
-SdT+vdp
[Problem 1.7(a)] .
Hence the following quantity is indeterminate as T -+ 0:
_(~;)p = S =
H;G .
Hence
;i~ [- (~;)J = ~~ [C (~;)J
v CII
=lim II
whence Cp = (aH/aT)p' Also
It follows that
Since'Y = g+ I and plying by A/g, that
F-U .
i.e. ST= 0 = ST= o+Cv, T= o. Hence Cv -+ O. (ii) From H = U + pv, we have
= B3
S
(aF)
limaT
T->O
for both gases so that B3 B~. Hence B~/BI = B~/B2 X, and they are therefore extensive. The substitution of i for B2 through B2 AeSIA-i was therefore reasonable, and i is intensive. (b) In Problem 1.10 it was shown that l
I
Now, as T -+ 0, F -+ U for finite S, so that the right-hand side becomes indeterminate. Differentiating numerator and denominator and sub stituting the values appropriate to T = 0, we obtain
The adiabatic through a given point (Po, To) satisfies t to
ph-Oh
Nk [part (a)], this yields
Ii kT = Inp - 1+g InT-i+ I + g .
A"r Bl .
Hence, if A '/ A = X, we have
= AT, the Gibbs free energy of the system is
(1 +~)pv -TS = AT [1 + ~ - (1 + ~ )tnT+lnp-i].
=
so that
(a) We have
33
p -
so that Cp -+ O. (iii) From Problem 1.3 we have
TdS
= CvdT+lvdv = CpdT+lpdp
'
= mvdv+mpdp.
Chapter 1
34
1.24
If S remains finite and continuous
T(~!t = CO(~~)y +Iv(~:) y . , = ... 40. Consider various cases as T (x,Y)
Result
4
O. One finds
(T,v) (T,p) (v, T) (p, T) (v,p) (p, Cv 4 0 Cp 40 Iv 4 0 Ip 4 0 mv 4 0 mp 4 0
1.27
The laws of thermodynamics
(a) Show that the slope PL of the line L satisfies op) (op) , (aatp ) T = -PL' ( oT t = aT L PL; (b) Show also that for any function of state y
(:~)p
(it)p;
X/To
Oy) (oT
t
, (Oy)
(it)
= PL ap t;
G;)
=
T
(c) If Wand X are two functions of state for the
This argument incorporates independent proofs of (i) and (ii), above. 1.25 Take a strong heat theorem in the form (as/ax)y 40 as T 40, where x, yare any two independent variables (other than S). Show that the result of Problem 1.24 can then be strengthened to yield the vanishing of
35
(OW) (aw) oT x = aT
establish that
(OW) (OX) oX ap t'
I
t -
T'
PL
T
Write down the equations resulting from (W, X) = (S, p), (S, v), (v, p) and from the first and last of these deduce that
Solution
S. _ (OS) T - oT t +PLVO:p ,
We have now (as/ox)y 4 0 in the solution (iii) of Problem 1.24, so that it is clear that the quantities (Cv/T), (Cp/T), etc. will vanish.
O:p
I
I (ov)
= -;; aT t
I +KTPL,
. 1.26 Show that an ideal classical gas of constant heat capacities (defined in Problem 1.9) cannot exist indefinitely close to the absolute zero, even if the third law is taken in the weak form given to it in Problem 1.24. Show that an ideal quantum gas (defined in Problem 1.9) does not violate the third law even if it is taken in the strong form given to it in provided one takes (x. y) (v, T). Problem I
where O:p and KT have the meanings given to them in Problem 104. Interpret these results in terms of curves which might be plotted. Cd) Recover the result
Solution
Solution
Consider an ideal classical gas. The result Cp -C v = A of Problem 1.9 will be violated near T 0, since from Problem 1.24 we have Cp 40, Cv 4 O. If pv gU (ideal quantum gas), then
g oP) ( aT v = -;;Cv 40 from Problem 1.24. Hence a Maxwell relation yields (as/OV)T 4 0 as T 4 O. Note that for (x, y) == (T, IJ) the strong theorem fails already for an ideal electron gas for which (as/aT)v approaches a non-zero value as T 4 O. To show this, use Problem 3.15(d).
C-Cv p
=
T(:~},(:;)
{Problem 1.8(a)]
p
by choosing the line tOto be a line of constant volume. (a) Taking the two independent variables to be T and p, 1 dt = dT- - I dp. PL The stated results follow. (b) We have
(Oy) at
and from part (a)
(OyaT)
p
oT at
(Oy) oT
p
p
Oy)T = (Oy oP) (Oy) (at op at T = - PL op T (Oy oP) (Oy) ay ) (oT op aT PL op I
'
I
t
PHASE CHANGES
1.27 Let Tdp) be a line L in the phase space of a fluid with two independent variables which assigns a temperature TL to any pressure p. Let temperatures T be measured from this line: t T- TL(p).
t
t
.
(c) For two independent variables the following result is general:
(~~)x = (~~t -(~~)y (~~t
.
Putting Y
('OaTW)x
~
=
T, Z
~
Solution
t
(a) Integrate
('OaTW)t - (apax) ('OW) ('OW) , ('OW) (ax) aT ap t ax T aT t -PL ax T ap t
The three special results are, using Maxwell's relations, as)
( aT
~
_
_ (as)
p - T - aT
t +PL, (av) aT p p\
=
t
=
T
=-
(1.27.2) (1.27.3)
I
Ca~ t = (aa~}~,
.
Choosing W = T and X = S or V, we obtain
(~~tp
(~~)s (~~)II on L .
This ratio is l/p~ on L. Also
(aa~) p~
p
=-
(~~)
II ( : )
= -
T
pl~ ( : )
T
is finite and non-zero and the left-hand side vanishes on L, = 0 on L. Also
(ap/aV)T
1.28 Suppose the line L in the preceding problem marks a transition from a phase I to a phase 2. (a) Assuming (as/aT)t to be continuous across L, recover the Clausius Clapeyron equation of Problem 1.14 by integrating the Cp/Tequation of Problem 1.27(c). (b) Show that, if p~ is finite and non-zero, on the line L
(:) T
aT t+PL (av) aT v
Since P is the only variable on L, the last term vanishes, and
As
II
(as)
(~~t (~~)Cp (~~)p (~;)C~
II
p
aT p
The first term in (as/aT)t does not contribute. (b) As in Problem 1.27(c), we have
p (ax) aT t('OaxW)T ~ (aaT) (av) aT p .
p ap)s = (ap) (aT aT c = (aaT) (~~)p (~~)p
(as)
(d P)
The last equation yields, on multiplying by V-I, the connection between CY.p and K T . Equation (1.27.1) shows that when Cp/T is plotted against (av/aT)p one obtains a curve which close to the line L approaches a straight line of slope pL and intercept (as/aTk· The notation (k means: evaluated at t :=: 0, Le. at the line L. Equation (1.27.2) shows that the same line is an asymptote of the curve obtained by plotting C./T against -(ap/aT)II(av/aph· The last relation under (c) shows that the curve of CY.p versus KT has an asymptote of slope p~ and intercept V-I (av/ aT)L' (d) Choose W = S, x = p in the first equation of part (c), noting that
, ('oaxW) (ax) ap t
~ T
at constant pressure from a point just inside phase I to a point just inside phase 2. Then 2 A}'" (V2- V I ) ' T dt L
(1.27.1)
,
I TCv (as) aT -PL, (aaT)1I (av) ap t ' (;;)p (;;)t -p~ (~;)T . as) ( aT
-PL
37
The laws of thermodynamics
1.29
1.27
Chapter 1
36
(~~)p = Ca~)p
0 on L.
THERMAL AND MECHANICAL STABILITY
,
1.29 Two macroscopic thermodynamic fluids I and 2, free from external forces, are in thermal and mechanical contact and the total system is isolated. The four independent variables, say the volumes and entropies VI> V2, SI, S2, are constrained by the given total volumelJ the given internal energy U: 0.29.1) () IJt+V2,
= PL ' O.
spite of the divergence of Cp , CY.p , and KT on L, which follows from
part (b), the equations between them given in Problem l.27(c) remain
valid. These are generalisations of relations due to Pip pard (5) • J
U
(5) M.J.Buckingham and W.M.Fairbanks in Progress in Low Temperature Physics (Ed. c.J.Gorter).
since for each fluid U variables.
3, 89 (North·Holland, Amsterdam), 1961. and A.B.Pippard, Phil.Mag., 1,473 (1956). See also
J.Wilks, The Properties of Liquid and Solid Helium (Oxford University Press, Oxford), 1967,
p.302.
\
=
U,+U2
,
(1.29.2)
U(S, (), n) there are no other independent
38
Chapter 1
1.29
VI
(a) Taking the remammg two independent variables as establish the useful results
aV2) 81 = -1, (~ as2) f(aU2) (au J/(au2) (~ 81 = L aV2 82 av I 81 aS2 aV2) ( aS =0, as2) (aUI) /(au2 ) ( asI = - asI aS2 l
I
(1.29.3)
vI
0.29.4)
)
V2'
(1.29.5) 0·29.6)
V2·
l )
i.e.
PI
=
(au2 ) aV2 82'
0.29.7)
P2 (= Psay) (mechanical equilibrium) and (au 2 ) ( au asI = aS 2 ~ (= T say) (thermal equilibrium).
(1.29.8)
V2'
i.e. TI = (c) Show that the conditions for mechanical and thermal stability, i.e. for S to be a maximum rather than only an extremum, include
p
p
- [( -a ) + ( -a ) av I 81 aV2 8
J = [(-vK81) I+ (1) J>0 VK8 2 0
39
1+ (-av2)
aV I 81 = 0, (aU2) (av2) (au2) (as 2)
( au aV I 81 + aV2 82 aVI 81 + aS 2 ~ 81 = O. l)
0.29.13)
The first of these is already Equation 0.29.3). Using it in the second one yields Equation 0.29.4). Next, differentiate Equations 0.29.11) and 0.29.12) with respect to SI at constant v I:
avI) + (av2) asI ( as I
(av2) asI = 0 , 2 -as2) = O. ( au as I + (aU2) a;-2 82 (av2) asI + (au as 2) (as I l
0
[(:~)v+(:£)Jo= [(~)I+(~)JO>O'
0 29 9)
'
•
.
(1.29.10)
where the suffix 0 indicates evaluation at equilibrium conditions. Solution
VI +V2[V I, Sd = v, UdvI,Sd+ U2[V2(VI, SI), S2(VI, SI)] = U.
0.29.11) (1.29.12)
Since V and U are constant, and S I and v I are independent, the partial derivatives with respect to S I and v I are independently zero. Differen tiating Equations 0.29.11) and (1.29.12) with respect to VI at constant
= 0 ,
VI
. I.e.
VI
vI
v2
0.29.14)
vI
The first of these is already Equation 0.12.5). Using it in the second one yields Equation 0.29.6). (b) Rewrite Equations 0.29.4) and 0.29.6) in the forms
(:'~)81 [(~~: t2-(~~:)8J /(~~:) (:%JVI
1-
2 f(au L asI /(au aS 2) J = )
v\
V2
PI-P2 ~
v2
1-
TI~
0.29.15)
,
;
both quantities must vanish for equilibrium. (c) Necessary conditions to ensure oS < 0 include that the second derivatives of S with respect to S I and v I be negative. Differentiating Equation 0.29.13) with respect to v I, at constant S I, we obtain 2 2I f(a2U2) (aV2) (a U2 ) (as 2) J(av2) a U ) ( avi 81 + L av~ 82 aV I 81 + aV2aS2 ~ 81 aVI 81 2 2) (a2v2) f(a U2) (as2) (a 2U2 )(av2) J (as 2) au +( aV2 82 avi 81 + L as~ ~ 8 1+ as 2aV 2 av I 8\ ~ 81 2) (a2S2) + -2 =0 ( au aS-2 -aVI 81 • V2
v2
On using
(a) Write Equations 0.29.1) and (l.29.2) as
VI
)
l
l )
VI
SI yields
VI
(b) The appropriate form of the second law of thermodynamics is, from Problem 1.22, 0 U+ pov - ToS ; ; . 0, i.e. oS at constant Uand Vis to be a minimum. Show that for equilibrium between fluids 1 and 2, i.e. for S = S1+ S2 to be an extremum, the necessary conditions are
( au av I 81
The laws of thermodynamics
v2
vI
VI
SI
and
1.29
av2) (-aVI 81
=-1
(a2v2) --2 'avi 81
=0 '
and Equation 0.29.15), the results of parts (a) and (b) yield the simplification, valid at equilibrium,
22 2) (a--2 a2UI) + (a2U2) (au --2 --2 + -~( aV I 81 aV2 82 aS2 aVIS ) 81J 0 v2
=0 '
40
1.29
Chapter 1
whence 2 {(-aZs) aVI 81
(aZsz) --z aVI
81
1 == - T
[(aZUl) (aZu z)]} --2 aVI + --z avz 81
82
.
(1.29.16)
0
Next, on differentiating Equation (1.29.14) with respect to omitting now terms which will not contribute, we obtain
aZul (aZuzas z azuz avz)asz au a s aSi + as~ as! + aszavz aS I aS + aS 2 aSi 2
Sl, and
2
2
I
1.31
The laws of thermodynamics
constraints TJ = constant, Tz = constant, on intensive variables. This reduces the number of independent thermodynamic variables by one. Whereas in Problem 1.29 two independent variables lead to two stability conditions, one independent variable in this problem may be expected to lead to only one stability condition. Of the four variables, /)1' /)z, TI , Tz only one is independent, and we chose Ii J. Condition (1.29.3) becomes
O.
dll z -= -1
dV I
Using the results of parts (a) and (b) we find
aZUl azuz auzaZs ) ( aSi + as~ + asz aSiz
0
whence
{( aZs) aSi
VI
(aZS2) aSi
'VI
-
1 T
[(aZUl) aSi
VI
+
For equilibrium at constant temperature the free energy may be minimised. So, in analogy to S = SI + S2, we have
= 0 ,
(a 2U2)]} as~
V2
o'
0.29.1
Now the condition for S to be a maximum, rather than just an extremum,
is that the quantities (1.29.16) and (1.29.17) be negative, whence
identities (1.29.9) and (1.29.10) follow.
[Notes.
1. We have shown that the sum of two terms must be positive from both Equation 0.29.16) and Equation 0.29.17). Each of the four terms depends on the variables of one system only and it therefore is reasonable to infer that the following inequalities hold for each system individually:
(v~Jo = (-~~t,o (~:~)8.0>0, (~)o (~~)v.o (~~~)v,o > O.
(1.29.18)
2. For completeness the sign of the cross term a2 Slav Ias I must also be considered (see Problem 1.31). 3. The important case of stability in a two-fluid system in the presence of mass transfer is treated in Chapter 7. Note particularly Problem 7.3.] 1.30 Reconsider the derivation of the stability conditions in Problem 1.29 on the assumption that, instead of the whole system being isolated
energetically, each part system is ins contact with a heat reservoir at
temperature T.
[Hint: The appropriate form of the second law is 6F+ S6T+ p61) ;;;.. 0.]
P = FI(vl, 1',. F~(1J2' T2 ), and Equation ( I .29.1 5) is replaced by
f(aFI LaVI
a/'~) d1!2] + ( aV2 T2 dv 0 J
This problem differs from Problem 1.29 in that one condition + Uz = U on an extensive variable has been replaced by two
=
(aF'z) av.
f(aFl)
L 3vJ
T]-
T2
l
0
O.
Thus the variability of II I and I) 2 yields o
P2,O
as an equilibrium condition (free energy an extremum). For a minimum rather than an extremum, the yields
p)
a2 ( avt
Thus
(iJ
2
2
f
f2)
( aav~
. 'l
ad
TI
condition
>0. T2
ap 2
+ ( aU2
o
dv
> o.
0.30.1 )
1.31 The relation between the conditions (1.29.9), (1.29.10), and 0.30.1) is elucidated by considering one of the fluids in the preceding two problems. (a) Let a, b. h be real constants, let x, y be variables, and let I(x. y)
+ 2hxy + by2
0=
.
Show that I(x. y) has one and the same sign for all real x, y if ab > h 2 • (b) If a function f(x. y) is subjected to a Taylor expansion about its value at (x o, Yo), say, show that the condition (a) applied to the second order terms is eauivalent to a condition on the Jacobian
a f) 2 ( ax 0 a2f ( axay 2
Solution (6)
(6) J.T.Lopuszanski, Acta Phy~. Polonica, 33,953 (1968).
41
J
a(fx , fy) I I a(x. y)
f)
a2 ( axay 0 a2f) ( ay2 0
> o.
1.31
Chapter 1
42
Here I(x, y) is to be reinterpreted as f(x, y) f(x 0, Yo) and the suffix 0 denotes an evaluation of derivatives at the point (x 0' Show that the conditions of Problems 1.29 and 1.30 for thermo dynamic stability applied to a single phase are met by the appropriate form of the condition (b) above, together with the condition that one of the diagonal terms of the determinant is positive.
1.31
The laws of thermodynamics
We conclude with some remarks concerning the relationship between the results (1.31.1) to (1.31.3) of this problem and the result (1.31 A) of Problem 1.29. An interpretation of 1 > U is obtained as follows: a(-p, T) a(v, S)
a(Uv , Us) a(lI, S)
= - -T
Solution
(a) Write 1 a [(ax
I(x,
+ }zy)2 + (ab -
nc vK
v
an interpretation of a, b, and and the result follows. above is fxxfvv > (fxy}2, i.e. The condition of parts (a) and
1 == I
fxy fyy
fyx
s)
one must have written as
~
1
[1> 0,
fxx
>
01
> 0]
>
=>
C
v
yy
T v
>0
(1.31.1)
-
(1.31.2)
> 0,
ap ) ( an s
In Equation (1.29.18), on the other for a single phase, fxx>O,
( 1.31.5)
IJ
T
,
T
>0
=>
T -C
>0
=>
> 0,
v
> 0,
vKs
vK T
> 0,
>0
vKT
>0
(1.31.6) ( 1.31
where the last inequalities in (1.31.6) and (1.31.7) are clearly fulfilled. Indeed, Problem IA(c) enables one to infer also that in both cases C 2C > 0 , i.e. TICp > 0 . K v s ~v
TIC v
>0
=>
>0
but because of Ks
a2
+ T/1:::::E C
[Problem 1.8(b) 1
this requires a condition for the right-hand side of this equation (e.g. TIlICp > 0). The case of two fluids IS considered in Problems 7.3 and 7.4. GENERAL REFERENCES
= C
=>
>0.
v
p
i.e. f
TIC v -K
y~s
T
f yy
CT v KT
v 1 0,
X~IJ,
>0,
T
> 0,
oKs> 0,
Then f, fxx, and fyy have like signs. For thermodynamic stability the sign itself is prescribed. For example, with the interpretation F~U,
l(al1) ap T
The result (1.3104) does not go quite as far. In this case one would like to argue
> O.
I
(ap ) /( as) all T aT v
a(-p, T)/a(IJ, s) a(v, T) a(v, T)
The implications (1.31.2), ( 1.31.3) become
}z2)y2] .
of a if ab > h 2 • This implies that a and b have like Thus I has the signs. (b) We have, if x and yare measured from x 0 and Yo respectively, and first order terms are zero, 2 2 2 ( a f ) a2f) (a f) f(x, 0) = ( ax2 y, 0 x + 2 axay /y+ ayL x, This
43
I
vKs
>0.
(1.31
Problem 1.29, interpreted
fyy>O.
(1.31
It is now seen that these conditions are to be supplemented by 1
which implies a condition on the cross term a 2 u;avas.
> 0,
M.W.Zemanski, Heat and Thennodynamics, 5th Edn. (McGraw-Hill, New York), 1968. R.Becker, Theory o[Heat, 2nd Edn. (Springer-Verlag, Berlin), 1967. M.E.Fisher, Rep.Progr.Phys., 30,615 (1967). J.s.Rowlinson, Liquids and Liquid Mixtures, 2nd Edn. (Butterworths, 1969.
2.2
Statistical theory of information and of ensembles
45
~Suppose that an extensive variable x takes on the value XI when ~ system considered in the preceding problem is in its ith state. Recon sider the search for the maximum &ntropy distribution as carried out in
2
the preceding problem if the average value of X( =
Statistical theory of information and of ensembles
~:
;11
1 \1
P,T,LANDSBERG
(University College, Cardiff)
ENTROPV MAXIMISATION: ENSEMBLES
In these problems a statistical &ntropy is defined, and is distinguished from the thermodynamic entropy by the symbol for its initial letter, Though k can here be any constant with the dimension of entropy, it is usually taken to be Boltzmann's constant.
~.1
L i=
Pi
where Z(x) == L exp(-t3xj)' Also show that
Solution
The preceding solution is generalised by considering
I. Use the
1
f = -kL(PilnPi
where 13 is the second multiplier. We have
I
af
I
.., = PN
=N
= -k(lnp;+ I -a+{3xj)
apj
and
I Z(X)ex p (-t3x j)
Pj
Solution
We need to maximise S = -kLP;lnp; subject to LPt l. Let a be an undetermined multiplier; then consider the maximisation of
Z(X) == L exp(-t3xj) , j
i
Also
with respect to each Pi' One finds the condition
, -k(lnpj+l
a)
0,
Xo
\;
= LPjXj =
I
Z(x)~xjexp(-t3xj) J
= [alnZ(X)] a{3
Lastly
Hence lnpj = a-I for all j. Therefore all Pi are equal. Normalisation yields the required result for Pj' The maximum &ntropy is
S2 == Smax = +kLPj[{3xj+lnZ(x)] i
SI == Smax
= -k f (-kln~) = klnN. 44
(j = I, 2,
where
f = -kL(p/lnp/-apt)
apj
,
It follows that
S = SI == klnN.
af
apj+t3PiXj)
i
&ntropy S == -k LPi InPi of the probability distribution,
= P2
[alna~(.~211,x2''''
S = S2 == kt3xo+klnZ.
method of undetermined multipliers to show that for the maximum
PI
=
and
N
I, 2, .." N), where
= I, 2, .." N and an undeter
I
Pj = Z(x)ex p (-t3x j)
Xo
A system can be in anyone of N states. The probability of it
being in its ith state is PI (i
have the given value xo' Show that for j mined multiplier 13, one finds
it/IXi) is known to
= kt3xo +klnZ ,
l
~
x I,X2,
...
46
Chapter 2
2.3
v 2.3 The extensive variables x and Y take on values Xi and Yi, respec tively, when the system considered in the preceding two problems is in state i. If the average values of x and yare fixed at Xo and Yo for the system, show that for the maximum s,ntropy of the distribution function I Pj = :::exp(-tJxj -')'Yj) (j = I, 2, ... N) , where:::
2.5
47
The term grand ensembles is sometimes used if only the mean total number of particles is fixed. In petit ensembles the total number of particles itself is fixed. With this interpretation, verify that the following identifications are consistent with thermodynamics: tJ
L exp (-(3xt - ')'Yi), and tJ and')' are undetermined multipliers. i
Statistical theory of information and of ensembles
-!>
I kT '
')'-!>
Jl
-kTlnZ
-!>
F,
kTln:::
-!>
pv ,
where F is the Helmholtz free energy, provided only that the statistical s,ntropy and the thermodynamic entropy can be identified.
Show also that 0 In:::) x - -( ootJXl>
Yo
",YN,'Y
=- (
In:::) ar 0
Solution X"""YN,{J
In the case of S2,
and S
S3
==
S2 ktJxo+k')'Yo+kln:::.
is now the internal energy U, and we write
XO
k~U+klnZ
U-F T
[cfProblem 1.7(a)].
In the case of S3, Yo is the mean number of particles n, and we write
Solution
Proceeding as in the preceding problems, we find
S3 = ktJU+k')'n+klnZ
U-I.m+pv T
(cf Problem 1.20).
f = -k L(PilnPi-apj+tJPiXi+')'PiYi), j
The identifications proposed in the problem follow.
of
:;- = -k(lnpj+ l-a+(3xj+')'Yj)' UPj
VI' n = n l' 2.5 (a) A system is specified by the values of U = UI> V Suppose its entropy is then SI' Under different conditions it is specified by v = VI' 11 11 1, and a temperature 7;. Suppose its average internal energy is then U = U2(7;) and its entropy S2(7;). Choosing the appro priate ensembles, prove that if U2(7;) ~ UI then S2(7;) > SI' (b) Discuss this result qualitatively in terms of probability distribu tions(1).
Hence 1
Pj
= Zexp(-tJxj -')'Yj)
where
Z
Also Xo =
~PiXi 1
== Smax
L exp(-tJxi-')'Yi)' j
0 In:::) = - ( ~ x"Y,,'Y
Lastly S3
==
'
Yo = - (
Solution
OlnZ)
ar
x"Yj,{J'
(a) We have, using a canonical ensemble, for the second condition of the system
kLPi(tJXi+')'Yi+lnZ) = k(3xo+k')'Yo+klnZ,
S2 =
i
2.4 A collection of copies of the system which are at any given time distributed over their states in proportion to the probabilities Pi obtained in the preceding three problems is called an ensemble. Problem 2.1 describes a microcanonical ensemble, Problem 2.2 a canonical ensemble if x is the internal energy of the system, and Problem 2.3 a grand canonical ensemble if x is the internal energy and Y the number of (identical) particles in the system. Z and Z are called partition functions.
=
U T;+klnZ2 .
Now, instead of summing over states in Zz (as in Problem 2.2), one can sum over energies E j by inserting the degeneracies gj of these energy levels. Then, if the suffix '0' refers to a particular energy ,
Z2
tgjexp
(-{f;) > goexp (- :~).
Choose now Eo as the energy U1 of the first description of the system. (1) See also E.A.Guggenheim,
,Research. 2,450 (1949).
2.5
Chapter 2
48
Then go = g 1 becomes the N value of Problem 2. I for the microcanonical ensemble appropriate to the first condition, and Sl = king!. Hence S2
U2
> 12 +klngl 12
# Sl .
Statistical theory of information and of ensembles
2.6
49
(b) From Problem 1.7(a) we have
dF = d(U-TS)
= -SdT-pdv+p.dn
and from Problem 2.4
In the second condition the energy of the system can fluctuate and the probability distribution is spread over many more states than is possible when the energy is fixed. Hence S2( 12) > St.
dF
-d(kTlnZ)
Hence
-klnZ dT kTd(1nZ).
U
P
p.
kT2dT+ kT dv - kT dn ,
dlnZ PARTITION FUNCTIONS IN GENERAL
2.6 (a) If the number n of identical particles in a system, its internal energy U, and its volume v are given, the microcanonical ensemble of Problem 2. I is appropriate. If these quantities are varied one has to consider a set of ensembles with neighbouring values of n, U, 0, which become the independent variables. Show from Problem 1.20(b) that the partition function k InN satisfies alnN) ( ~ V,n
p
= kT'
(alnN)
~
V,v
=
kT'
(
P,
n
alnZ) ( av T,IoI
L kT'
an
v, T
=
(alnZ)
ap'
v,T
p.
fl (dlnZ) = U-p.n
kT' aT V,1oI kT 2 '
.
(
PV)
d kT
=
n i p kTdP.+ U-p.n)dT+ kTd/J .
(d) This follows since the last result is an exact differential. See also Problem 20.4. d,7 A system is kept at fixed chemical potential and temperature. Show that the -logarithm of its grand partition function is proportional to the volume. [Hint: The first result of Problem 2.6(c) is useful.]
'I.
Solution
From Problem 2.6(c) x
InZ satisfies
(
ax)
av
x
p fJ., T
v
= kT
Hence for systems at constant p. and T x canst. v Alternatively, if p. and T are constants, the Gibbs-Duhem relation of Problem 1.20(d) shows that p is constant. Hence
(d) Establish
e~)T' v-p.
S)
(P.fl U == d k1'- kT+I
Now write down the derivative on the right and use Problem l.20Cb) to simplify the result to
= kT'
U alnZ) ( aT P, n (c) For a grand canonical ensemble the independent variables are v, T, and p., only average values of U and n being now given. Systems specified in this way are supposed to be in equilibrium with a large heat reservoir at temperature T and a large particle reservoir at chemical potential p.. Show that the partition function satisfies
(alnZ)
(pv)
d kT
dlnN)
au
Systems specified in this way can be regarded as isolated systems. (b) For a canonical ensemble the independent variables are 0, n, and T, only an average value of U being now given. Systems specified in this way can be regarded as isolated apart from a large heat reservoir at temperature T with which they are supposed to be in eauilibrium. Show that the partition function satisfies dlnZ) ( ---a;- 1;n = kT'
and the results follow. (c) From Problems 2.4 and l.20(c) we need
T,v
T(:;)IoI,v'
In.:.
kT
0:
v.
Solution
(a) The results follow from Problems l. 20(b) and 2.4 which yield TdS
kTlnN = dU+pdv -p.dn .
2.8 The replacement of the partition functions Z and Z in Problem 2.6 by their thermodynamic equivalents, given in Problem 2.4. leads to thermodynamic results of some interest.
50
Chapter 2
2.8
(a) Infer from
Statistical theory of information and of ensembles
2.9
EN}R0PY MAXIMISATION. PROBABILITY DISTRIBUTIONS
( that
n
alnZ)
a;-
n
/2.9 A one-dimensional normal distribution of zero mean and standard deviation 0 is given by
V.r = kT
v(~~)
OlnZ) =
( oT V,!l
show that
-----ar
S = k [ InZ+ T ( OlnZ)V,!l
(a) Show that its G;ntropy is ~kln(21Te02), where e is the base of the natural logarithms. (b) Show that for given L:x 2p (X)dx == 0 2 the normalised probability distribution having the largest G;ntropy is the one-dimensional normal distri bu tion.
l.
Hence establish the relation
Solution
S = v
(a) The G;ntropy is
(:i) !l . v,
(c) Establish the thermodynamic relations as found in parts by purely thermodynamic methods.
S
and (b)
(a) Replacing InZ by
we obtain
0fJ.
v, T
as required. (b) From the equation stated OlnZ)
( aT
!l
which is the first result stated under (b). we obtain O(PvlkT)] S = T kT [ aT pv
= r+ vT
[Ir
(oP) oT
_
-rln .::.
S =
-1 v. J J .
v(~~)T
V
•
We must maximise with respect to arbitrary variations of p(x) in the integrand of f[p(x)] == -kL:p(X)lnp(X)dX-aJ:p(X)dX-t3L:x2p(X)dX
=
(a
p ) v aT
(:i)!l .
_~-t3x2 =
-k-k
l'. JJ. .
O.
It follows that
p(x) = aexp(-t3x 2 )
for the maximum. The normalisation
SdT-vdp+ndfJ. = 0
n
x 2p(x)dx
where ~, t3 are Lagrangian multipliers. Hence
!l
(c) From the Gibbs-Duhem relation of Problem I
and
1
Replacing InZ as in part (a),
v,
f
~ k In (21Te0 2 )
v, T
S pv S kT - kT2 = kT
TS v,
kT 0fJ.
In 21T02 x 2 ) 2 -202 dx
= '2k ln (21T0 2 )+ 20k 2
= ~(op)
[O(PvlkT)]
kT
-kfp(x)lnp(x)dx
= -k fp(x)
Solution
~=
-ex> <x <ex>.
(21ra2fVzexP(-2:22)
p(x)
v, T'
(b) From
one finds
51
a (2)
f '" exp(-bx
00
-00
x 2r exp(-/lx 2 )dx
For r = 0 the result is (1T/(3)Y'.
1X
2
)dx
= a (%)Y' =
= ~~-"":"':--':':'':'::'-'''--'''':''':-'-I
I. for r
=
1,2,
52
Chapter 2
2.9
Also 2 0
f
=a
Hence
= 2
2
_oox exp(-bx )dx
p(x) =
="2a(7r)Yl [33 =
-r~-_-_-,12'-'-'-'-Y2exp (-
I
Statistical tl1eory of information and of ensembles
2.11
(a) Maximise as in Problem 2.5 the integrand in
2[3
;;2) .
under the conditions stated.
f
p(x)[ -klnp(x) -a'-[3lx Ir] dx ,
f whence
-k-klnp(x)-a'-[3lxl' = O.
The distribution of largest &ntropy compatible with the specifications is Po(x)
2.10 Let By p(x)
2rllT
==
(I + +), where r
"" j~ M, == [f x I'p(x)dx S[p(x)]
is the Gamma function,
I = f-:P(X)dx M r = 2a r
11,
(r> 0), the rth moment, and
1
= aexp(-[3lx
The constants a, [3 can be identified by
pC-x), a probability distribution,
_00
53
f" 0
= 2a
too exp(-[3xr)dx = 2ar (I ++)[3-1Ir ,
2a (r+l) x r exp(-[3x')dx = -[3-('+l)/T -- = I r r r[3
It follows that _
I -lIr
a- 2r. (r. + I) M r
= -kIp Inpdx, the &ntropy of probability distribution.
r
Br
My '
r
Show that
M, 9~' Byexp {S[P(x)J k
_l.}
and that the equality sign holds when Br
Ir)
(IX uexp rM;
=
p(x)
and
r
.
p()(x)
k
The following integral will be needed: irs
==
L""x ex P (-[3x r )dX. S
Put y
x r , dx
I I = -X-(r-I)dy = _y-(r-1)/rdy. r
Then i r. s =
r
If""Po(x) (B, Ix-Ir) dx--I = In Mr In--
0
r
_bO
Mr
rMyr
r
Br
It follows that for given Mr all other distributions have a smaller &ntropy, i.e. Mr exp(So_!);;, exp{§[P(XH_!}
Br k r k r'
which is the required result. MOST PROBABLE DISTRIBUTION METHOD
If""
-;:
(Ixlr)
as required. The &ntropy of the distribution Po satisfies the relation So -.-- = -
Solution
By
M;.ex p - rM; ,
yP exp (-[3y)dy
If'{p+l) [3P+ 1
=r
(S+ I)
I = -;:[3-(st l)/T -r- .
S+ I ) (P -r--I
2.11 The states of a quantum-mechanical system are labelled by a complete set of quantum numbers. Suppose one of these determines its energy Ej and thatthe corresponding energy degeneracy is gj (j = I, 2, ... ). Consider an ensemble of N copies of this system in the sense of Problem 2.4. Let one State of such an ensemble be specified by the numbers (n 1> n2, ... ), where nj is the number of systems with energy E j • The capital S is a reminder that a State of an ensemble is considered.
54
2.11
Chapter 2 (a) Prove that the number of ways of realising this State is G
n!
=
(c) From part (b) and the definition of "'(n), we have
Nfgrtlgrt2 I 2'" nl!nry! ...
d "'(N) """ dn[nlnn-n+!ln(21Tn)]
(b) If n is large enough, check from a book on special functions that Stirling's approximation holds:
r(l +n)
nrt e-rt (21Tn)Y'
where r is the Gamma function and n is a positive integer. (c) Make the continuity assumption (3) with regard to n, and define Gauss' "'-function by
dnr(l+n).
"'(n)
From tables we find "'(3)= 1·2561,ln(3·5) 1'2528, so that the error in "'(3) """ In(3·5)isO·0033/l·2561 = 0·26%. It is less for n > 3. (d) For the most probable State of the ensemble, we have to consider InG(n 1 ,n2, ... }-exLn;-{3LE;n;,
f
j
J
n· ::1
N
J
J
= -"'(nj)+lng;-cx Hence, if (n!,
{3Ej = O.
ni, ...) is the most probable State, nt = gjexp(-cx-{3Ej )-!
J
where (3 is a Lagrangian multiplier. (e) Discuss the relation between this result and that of Problem 2.2. (a) In any State of the ensemble, let us give systems with the same energy the same letter, and systems with different energies different letters. Hence G(n I, n2, ... ) gives the number of distinguishable arrange ments of N letters, n 1 of one type, n2 of another type, etc., and is equal to N!/n I !n2 L.. if the degeneracies are neglected. As a result of the degeneracy gi' each of the G arrangements gives rise to a number of further arrangements equal to the number of ways of assigning nj systems to gj states, i.e. to gfi arrangements. The result of part (a) is thus obtained. This matter is discussed in many books. 'l!(n)
was pointed out in
.
If these are M energy levels, cx may be identified by summing over the energy levels to find
N+!M
Solution
(3) The importance of this assumption and the usefulness of hoc. Natl. Acad. Sci., U.S., 40, 149 (l954).
j
Instead of maximising G with respect to each ni' it is more convenient to maximise In G. We have af a an. = an J-Inn;! + nj Ingj - an; - {3E.jn;]
i.e.
Lg.e-iJbj+O( j
LEjn; = Eo·
w(nt) = In[gjexp(-ex-/3Ej ] ,
n'!' """ _ _.:::.1-_ _ ::L
N
;
where cx and {3 are undetermined multipliers to take account of the conditions that N and Eo are given, i.e.
2: n; = N,
Verify, by the use of tables of "'(n), that for n ~ 3 this approximation holds with an error of less than 0·26%. (d) Make an assumption of equal a priori probabilities of different ways of realising a State, and assume that the energy of the ensemble is given. Hence show, using part (c), that for the most probable State of the ensemble the probabilities nj!N are given by
./
= Inn+ 1-1 +fn
""" Inn+ln(l +(n) = In(n+!).
Prove from (b) that "'(n) """ In(n+-!).
55
Statistical theory of information and of ensembles
2.11
= exp(-cx)4.gjexp (-/3E;)
Zexp(-cx) ,
J
where the canonical partition function has been introduced. It follows that '!l _ ( ~) g; exp (-/3}'j) N- 1+2N Z When the terms in 1/N are collected together, the desired result is obtained. (e) The most probable State of the ensem ble leads to the most probable values nt. The canonical ensemble of Problem 2.2 yields the mean values of the n; as averaged over all States of the ensemble. In this average the most probable State makes a dominant contribution, but less probable States will also contribute, and hence the d·ifferent results which are, strictly speaking, obtained by the two methods.
56
Chapter 2
2.11
The present method requires the nj to be large enough for the assump tion of continuity to be justifiable. This condition is difficult to fulfil unless N ~ 00. Although M is often also infinitely large the usual result follows only if MIN ~ O. These difficulties are often overlooked. It is particularly easy to overlook them if the coarser approximation \}f(n) - Inn is used, which yields at once
Solution
(a) Let u refer to the upper level, and I to the lower level. Then nl +nu
G( )
N! n = [!(N+n)]![!(N-n)]!
ni _ exp(-{3Ej)
Z
Hence using Problem 2.11 (b) InG(n)
2.12 (4) Develop the ideas of Problem 2.11 for an ensemble of N identical systems each of which has two non-degenerate states. Take N to be even, and specify a State of the ensemble by the number n (which is even) giving the difference between the num ber of systems in the lower energy state and the number in the upper energy state. Assume that N, but not the total energy, of the ensemble is given. (a) For n 0 for all i, then K(p,pO)
3.1 (a) A system is specified by variables Xl' Xl' .." XN which have in dependent normalised probability distributions PI (x 1), Pl(X2), "', PN(XN)' Show that the entropy may be written as
== k~PilnpP~ I
S
I
PP for all 0, [Suppose the PP are the equilibrium probabilities of a possibly non isolated system, for example it could be the canonical distribution if the system is in contact with a heat reservoir. Then the knowledge that the actual distribution is (say) Pi under these conditions, represents an 'information gain' K(p, pO). Conversely, the passage from Pi to PP corresponds to an internally produced entropy K(p, pO). The quantity K(p, pO) was introduced by A. Renyi (6).] is positive unless the distributions are identical (Pi
(b) The Xi are interpreted as the three Cartesian velocity components V; of a particle in a gas with point interactions (Le. the particles are points and interact only if they are at the same point) in equilibrium at temper ature T. Assuming that the mean kinetic energy associated with each component is'; kT, show that for a state of maximum entropy
V;) =
(a) Let y == lnx - 1+ l/x. Then
dy I-x
dx
This is negative for 0 < x < I and positive for I y occurs therefore for x = I, so that y ~ O. (b) With Xi pdp?,
k~ p;oxilnXi ~ k ~PiOXi (1- xl) I
I
m)Y2 exp(mv,2) (21rkT - 2kT '
where m is the mass of a molecule. (c) Derive from (b) the probability that a molecule in this gas has a speed in the range (V, V + d V) is p( V)d V, where V can have any value from 0 to 00 and ,I
Solution
K(p,pO)
= -kitJJpi(Xi)lnPi(Xi)dXl
I
< x. = k
p(V) = 41rV
The least value of
2;. (Pi -
2
Vl) ( m)" exp (m 21rkT
2kT .
This is the Maxwell velocity distribution. (d) Discuss the range of validity of these results by inspecting the assumptions needed to obtain them.
pP) = 0,
I
Solution
Hence K is non-negative. If all x/s are unity, K = O. If one Xi =1= I this term will contribute a positive quantity to K and K > O.
(a) The independence of the probability distributions Pi(Xt) implies that the probability of finding a state XI' X2, ... , Xn of the system is
(5) P.T.Landsberg, Phys.Rev.• 96, 1420 (1954) and Section 34 of P.T.Landsberg, Thermo dynamics with Quantum Statistical Illustrations (lnterscience, New York), 1961.
N
P(Xl> X2, ... , XN)
(6) For recent discussions, see F.Schlogi, J.Phys.Soc.Jap., 26 Supplement, 215 (1969).
=
n
Pi(Xi)'
i = 1
63
I
64
Chapter 3
3.1
Hence the entropy is
r
I
Statistical mechanics of ideal systems
65
(d) The main assumption is that PI (Vd, P2 (V2), P3( V3 ) are independent probability distributions and that the point particles have only point interactions. In a dense gas the interactions cannot be approximated in this way. It has also been assumed that one particle can be treated separately from the rest. This is invalid for a system of indistinguishable particles when exchange effects must be expected.
S = -kf..JPlnPdT = -kf..JPI ...PNln(Pl",PN)dx1 ... dXN
-k
3.2
~ (fPi InPidX;).
(b) The following constraints apply to the p;'s:
L~Pi(V;)dV;
J~Pi(!mW)dV;=!kT
I,
~
(i
3.2 Obtain the following quantities for a Maxwell distribution of velocities: (a) The average of the nth power of the velocity is
1,2,3).
-~
Maximising the entropy subject to these constraints as in Problem 2.2,
consider f= Hence
-kitl
(fp;lnp;dV;+o:;jPidV;+13Jpi WdV;).
:;;
-k (Inp; +0:;
f
W+ 1) dlj
where n is real, n > I, and (b) The average speed is
exp(-131 W-0:;),
O:j
f This yields
~
2
lj exp(-13llj
2
1,
i.e.
( ~)
8) .
«V-(V»)2) = kT( m 3--; (d) The 'fluctuation' in the kinetic energy is
\6
exp(-O:j)
1,
(;m)2«V2_ is not generally con served, but it does remain constant in a quasistatic adiabatic change, since S is then constant. The generalised black body relations of part (b) do not presuppose a boson gas and the question arises if they can be illustrated also by a simple model of a fermion gas; see Problem
Inp+(s+2)lnT-(s+2)+i].
s,-)
Statistical mechanics of ideal systems
(a), (b) Use Problem 3.12 to find
3.14 Develop the implications of a zero chemical potential for the model of Problem 3.12. Ca) Show from Problem 3.11 (c) that
U
3.15
c.
where Ct and Cl are the velocities of sound for transverse and longitudinal directions. [For black body radiation based on bosons the characteristic laws U 3pv = iTS = 3vDT 4 hold. If one puts U = (4a/c)vT 4 , then
a) ( aT P. a1 ) ( av T, ('I) 1
(aTal)
1
V
I(s) I(s - I) ,
s+
I I(s)
v
v,(n)
I(s-I)'
(b) The coefficient of volume expansion
1 v
(av) Jl(s-1)' I(s)
v aT
T
(av) aT
p,
(s+ I
,
(s+ l)pv.
kT(n)
I(,)"s+l,a) I(,)"s+l,a) T( ') -Jl(n)+kT(n)Cs+2)-----r(- - ) l',)"S,a l'')',s,a
/C')', s+ I, a) 1(,)" s, a)
92
Chapter 3
For a
Fora =
I the solution of part
3.17
yields
z
no+
:z =
II(I-tjf l .
These are the partition functions for fermions and bosons. CONSTANT PRESSURE ENSEMBLES
3.18 The (petit) pressure ensemble is one for which p, T, and the num ber of particles n are given, together with the mean energy and the mean volume Vo. (a) Show from Problem 2.3 that the probability of a state (E, v) is exp l-(E + pv)/kT] pee, v) = Zp ( p, T , ,n) where Zp is a normalising constant related to the Gibbs free energy by exp(-G/kT).
(b) Discuss the difficulties associated with the partition function ,
L.
[Em(VI)+PVI] exp kT .
m, I
(c) Show that these difficulties do not arise if one takes Zp
T, n) = VoI
J' r ;;-
exp .. Em(V)+PV] kT dv
I roo ( Pv) voJo Z(v, T, n)exp, - kT dv,
where Z is the canonical partition function, and v 0 is a suitably chosen constant volume, the definition of which presents difficulty. (d) Find an expression analogous to that under (b) for the classical partition function Zp , and verify that there are no difficulties in this case. Show also that the ensemble average of the volume is -kT [
1)
alnZp(p, T, ap
n)]
alnZ(V, T, av
In Problem 2.3 identify x with the energy E and y with the volume v. A state i of the system in the ensemble is specified by the volume VI and one of the energies Em(v/) available for that volume. Thus i imolies two specifications. The result of Problem 2.3
S kt3Eo + k')Vo + kIn Z and the thermodynamic result TS = Eo+pvo-G
must be equivalent for all Eo and Vo and this leads to -(E+pv)/kT P(E,v) = exp Z ( ) p P, n (b) In general the 'eigenvolumes' VI form a continuous set so that the {-summation diverges. (c) By integration over v one avoids the divergence, but shifts the difficulty to the question of how Vo should be defined. However, the precise value of Vo will not affect those thermodynamic quantities which depend on a differentiation of InZp. The reference cited gives an intro duction to the current situation. (d) In the classical case E may be replaced by the classical Hamiltonian H expressed in terms of the f generalised coordinates and momenta. Fol lowing the procedure in Problem 3.4,
.
- -If... fexp [H(PI .kT .. qr)+pv] dpl ... dqf·
Zp(p, T, n) - h
There is no divergence and no dimensional difficulty here. The differen tiation indicated in the problem now yields the average volume. RADIATIVE EMISSION AND ABSORPTION
3.19 In a certain fermion system transitions occur from a group i of single-particle states I(ei) to a group j of states J(ej) at a rate Uij
T,n
in analogy with the result P = kT [
93
Solution
j
Zp(p, T, n)
Statistical mechanics of ideal systems
(b) and (c) are discussed by Lloyd and O'Dwyer (5) who give references to earlier work.]
00,
Zp(p, T, n)
3.19
n)] T,n
of Problem 2.6(b). [The classical ensemble is used in Problem 9.4. The difficulties under
=
L [PrSuqJ-pJSJIqrl
f(ei) J(ej)
,
where PI is the mean occupation number of a quantum state I, (I -qJ) is· the mean occupation number of a quantum state J, and Su is a transition probability per unit time. (5) P.Lloyd and J.J.O'Dwyer, Mo/ec.Phys., 6, 573 (1963). See also D.R.Cruise, J.Phys.Chem., 74,405 (1970).
94
Chapter 3
3.19
(a) Show that for free fermions in equilibrium at temperature denoted by C.. )0,
to be
I)J
[N (qrPJ II PlqJ
has the value unity. (c) Assuming the result (a) and that Su probabilities, show that
Pi/
= (Su)o for all transition
Su
SJI Su
. Vl.l.UVa
number of any
Xl
qJ PI
kT
- exp Il/k-:i)
Hence Uj!
[ ( E I -llr+IlJ-EJ ) kT 0 -1
exp
Pij - Pji,
J.
In thermal equilibrium all chemical potentials are equal, as follows from Problem 1.21. This establishes the result.
11; - Ili kT
These results are of use in the recombination theory of semiconductors where the Il/s are driving forces for transitions. They are then referred to as quasi-Fermi levels, see Chapter 16.
3.20 Stimulated and spontaneous emission rates and absorption rates of photons due to single particle transitions between states I and J are, respectively, in the notation of Problem 3.19:
u sp AUPIqJ' u abs == BJIN"PJqI ' Here J corresponds to a lower energy, and it will be assumed that Bu
(Bu)o.
= (Au)o . Bu = BJ[ may
Au
The quantum-mechanical result that be used in these equations. Note that the spontaneous rate does not depend on (a) Show from Problem 3.19 that Au Bu. (b) Show that u st -u abs u'P
= I + XI
The result follows. (b) Radiation in thermal equilibrium corresponds to black body radiation (see Problem 3.14), Hence, using also part (a), 1 exp
kT
E - Er + 111 - E J + Er - Ili exp J
= -PJ -qI
Pi/
(a) Writing Xl == exp [(EI -111)/ kT] one has essentially from Problem 3.12
= 1 (PI)O
= exp EJ -EI
(PIqJ) PJqr 0
0
where
Solution
1 Y"u = exp(hv/kT)
= (SJI)
= I,J LPISUqJ
that the ratio of the reverse rate
Since one fermion is the maximum mean state I,
Hence, since
(d) Reverse to forward rates are in the ratio
0
[They are equal in thermal equilibrium in accordance with the principle of detailed balance.]
1-'
J is zero.
It follows that
- Ili ) exp (111kT .
(PI)O
~
the net rate I
95
= 11/,
(EJ -EI) kT '
exp
where EI is the energy of the state /. Show with the assumptions of to forward rate 0" is
!!Ji.
In thermal in equilibrium Ili
SJ[ Su
where III is the chemical potential appropriate to state I. Let N" be the photon occupation number of a mode of frequency v where hv = EI-EJ . Assuming (a) valid, show that the quantity Y"u defined by
SJI Su
Statistical mechanics of ideal systems
EI -Ill) -u
exp (
Y "u
3.20
exp
IlJ -1l1+ hV )
kT
'
(c) Show that the ratio of reverse to forward rate is N" N + I exp
"
+hv kT
(d) Show that the result (c) satisfies the equation of Problem 3.19(d) only if the radiation field is that of a black body at temperature T. [All these results are important and are used in semiconductor theory, Problem 16.3, and in the theory of the laser, Problems 15.6 to 15,8.]
3.19
Chapter 3
94
(a) Show that for free fermions in equilibrium at temperature T, to be denoted by (.")0, EI-J.J.I) =exp ( ~ where J.J.I is the chemical potential appropriate to state I. Let N v be the photon occupation number of a mode of frequency v where hv = Assuming (a) valid, show that the quantity Y vIJ defined Y vIJ
I)]
PlqJ
exp (
J.J.'-J.J.') Tr .
Solution
(a) Writing Xl == exp [(EI - J.J.I)/ kT] one has essentially from Problem 3.12 l'
Since one fermion is the maximum mean occupation number of any state I, 1
SJI = (SJI) SIJ SIJ 0
--r
J is zero. Hence, since
= (PlqJ) = expEJ-EI PJql
.
kT
0
(d) Reverse to forward rates are in the ratio SJI = PJ ql = expEJ-EI+J.J.j-EJ+EI-J.J.i SIJ q, PI kT . J . j -J.J.i) Uij -_ LPISIJqJ (1J-exp----;(T I,J
Hence Uij
where
=
The result follows. Radiation in thermal equilibrium corresponds to black body radiation (see Problem 3.14). Hence, using also part (a),
I (EI - J.J.I + J.J.J - EJ) ] 1 kT 0 -I . YvIJ = exp(hv/kT)-llex p In thermal equilibrium all chemical potentials are equal, as follows from Problem 1.21. This establishes the result.
= Pij-Pji,
P'i J.J.·-J.J.i ~=exp~ Pij kT
,
[They are equal in thermal equilibrium in accordance with the principle of detailed balance.]
(PI)O
(c) In thermal equilibrium the net rate I in equilibrium J.J.i = J.J.j,
0
where EI is the energy of the state I. (d) Show with the assumptions of (c) that the ratio of the reverse rate Pji to forward rate Pij is Pij
95
Statistical mechanics of ideal systems
It follows that
has the value unity. (c) Assuming the result (a) and that SIJ = (SIJ)o for all transition probabilities, show that RJ -EI) exp ( --;:y-
3.20
. of semiconductors These results are of use in the recombination where the J.J./s are driving forces for transitions. They are then referred to as quasi-Fermi levels, see Chaoter 16. 3.20 Stimulated and spontaneous emission rates and absorption rates of photons due to single particle transitions between states I and J are, in the notation of Problem 3.19:
== BIJNvPIqJ,
u
sp
== AIJPIqJ,
u
abs
== BJIN"PJqI .
Here J corresponds to a lower energy, and it will be assumed that BIJ = (BIJ)o,
AIJ = (AIJ)o·
The quantum-mechanical result that B IJ = B JI may be used in these equations. Note that the spontaneous rate does not depend on N". (a) Show from Problem 3.19 that AIJ = B IJ. (b) Show that ust-uabs
(
N" I -exp
J.J.J-J.J.I+hV) kT .
(c) Show that the ratio of reverse to forward rate is N" N
"
+ 1exp
J.J.J -J.J.I+hv kT
(d) Show that the result (c) satisfies the equation of Problem 3.1 only if the radiation field is that of a black body at temperature T. [All these results are important and are used in semiconductor theory, Problem 16.3, and in the theory of the laser, Problems 15.6 to IS
96
Chapter 3
3.20
Solution
3.21
Statistical mechanics of ideal systems
Solution
(a) In this case the rate is zero in thermal equilibrium, i.e.
(a) Let XIJ == PJqJip[qJ. Then Au/Bu (xIJBJJiBIJ)-1
(BIJNvo+AIJ)(pJqJ)o = BIJNvo(pJqJ)o·
It follows that (
NvqJPJ) = N + AIJ . vo PJqJ 0 BIJ
It follows from Problem 3.19(a) that in equilibrium
Applying the result of Problem 3.19(b), we find BIJ = AIJ. (b) The ratio of the net stimulated emission rate to the spontaneous emission rate is BIJNvP[qJ(l-PJqJip[qJ) BIJp[qJ
( I -exp
= N v ( I exp
EJ - EJ -Ih +IlJ ) kT IlJ-IlJ+hV) kT .
(c) The forward rate is B IJ(Nv + l)PJqJ' The reverse rate is BJ./VvPJqJ. Hence the required ratio is reverse rate Nv IlJ - Il[ + hv --exp forward rate N v + 1 kT (d) For black body radiation N
Nv v
+I
( hV) = exp - kT .
The results of Problem 3.19 apply in this special case with S IJ = B IJ(Nvo + I) , SJ[ = B IJNvo .
3.21 In Problem 3.20 (Nv)o was assumed known because the result of Problem 3.19(b) was used. Assume now the existence of the three rates, with unknown (Nv)o and unknown. temperature-independent transition probabilities BIJ , A IJ, BJJ ; assume also that fermions make the transitions with EJ hv and that the system is in equilibrium at temperature T. (a) Show that BIJ BJI by supposing that (Nv)o -+ 00 as T -+ 00. (b) Assuming Wien's law (Nv)o <X v 3 exp(-hv/kT) for v -+ 00 show that <X
v3
[This is Einstein's original argument (6) which established the existence of spontaneous emission. The coefficients introduced here are called Einstein's A and B coefficients.] (6) A.Einstein, Ann.Physik, 18,121 (1917).
hv (XIJ)o = exp kT
Hence as T -+ 00, (x/J)o -+ I; so that the result follows. (b) One finds (Nv)o
AIJ/BIJ exp(hv/kT)-1
so that the result follows.
GENERAL REFERENCES
G.H.Wannier, Statistical Physics (John Wiley, New York), 1966.
97
4.1
Ideal classical gases of polyatomic molecules
99
Solution
(a) The total energy E of any system can be written as the sum of the E(t) + E(2) + €(3)' The total partition different types of energy, E function is therefore
4 Ideal classical gases of polyatomic molecules
L exp[(-E(I)f- €(2)j- €(3)k)lkT] ilk
Ztot =
,
where i, j, and k are the numbers of energy levels associated with the energies of types (I), (2), and (3). We can write Ztot in the form
C.J.WORMALD
(University ofBristol, Bristol)
Ztot
=
Li exp(-€(I)dkT) L exp(-E(2)dkT)Lk exp(-€(3)klkT) , j
so that
THE TRANSLATIONAL PARTITION FUNCTION
4.1 From Problem 2.2 we have that the total energy E of an assembly of N systems is given E
OlnZ) = -N ( of)
V,N
= NkT2 (OlnZ) - oT
(b) We calculate first the Helmholtz free energy F InZN
V,N'
L exp(-EjlkT) , j
I+
in which Ej is the energy of the jth quantum state. (a) For an assembly of systems each of which possesses energy levels of the form E(I)i+ E(2)j+ E(3)l" where i, j, k label the levels of essentially three independent spectra, show that the total partition function Ztot can be written as the product Ztot = Z(1 )Z(2)Z(3)' (b) For one mole of gaseous molecules confined within a container of fixed volume V we have from Problem 3.5 the translational partition function for one mole of an ideal classical gas in the form ZN(T, V) = (
2'1fmkT)3NI2 VN h2 i(T~ ,
where i is the internal partition function for a single molecule and No. Show that the molar Gibbs free energy and the entropy of the gas are given by the equations
N
G= -
S
-NokT[tlnT+~lnM
= Nok[iInT+~lnM
InP-3'6605]-NokTlni(T) ,
InP-I·1605]+Nok Ilni(T)+ T
Olni(T)] :>''1'
'
where P is in atmospheres and M is the molecular weight of the gas. [Note: I atmosphere = 1·01325 x 106 dyne cm- 2.] (c) Would you expect these formulae to be valid at all temperatures and pressures?
98
I
InN! + ~ Nin
2'1fmkT
1.2
-kTlnZN'
+ Nln V + Nlni(T) .
Using Stirling's approximation that InN! NlnN - N [Problem 2.11 (b)], putting M mN and V NkTIP we obtain
where the partition function Z of a single system is given by
Z =
= Z(I)Z(2)Z(3) .
~ In(~:2) + Hnk-ln(1'01325 x 106 ) +HnT+~lnM-lnP+lni(T)
Using N = No = 6·0225 X 10 23 mole-I, h k = 1·38054 X 10- 16 erg deg- I we obtain F
Finally
6·6256
X
.
10-27 erg s, and
= -NokT[~InT+iInM-lnP-2·6605]-NokTlni(T).
G=
F+ NokT, so that
G=
-NokT[~lnT+~InM
InP-3·6605]
NokTlni(T).
The entropy S is obtained using S = -(dG/dT)p,N' Noting that the term T x ~ In T yields ~ In T+~, we have
S
Nok[~ In T+ ~ InM -lnP- I ,1605] + Nok
T
Olni(T)] oT
.
(c) We note that as T approaches zero the entropy approaches -00, whereas it should approach the value zero, which corresponds to the system being in a single quantum state. However, if we have only one quantum state we cannot write a partition function for the canonical ensemble of the form ZN = ZNIN! as this formula is conditional upon the number of available quantum states being much greater than the number of molecules. For gases at low temperatures and high densities it is necessary to replace Boltzmann statistics by Bose-Einstein or Fermi-Dirac statistics, and when this is done the paradox is resolved.
100
Chapter 4
4.2
THERMODYNAMIC PROPERTIES AND THE THEORY OF FLUCTUATIONS
4.2 (a) For a single molecule the distribution function for the energy is broad, whereas for an assembly of many (N) molecules the distribution function for the total energy is sharp. This contrast arises because in the assembly the relative fluctuations in the energies of individual molecules almost cancel each other out, and only a small relative fluctuation in the energy remains. For an assembly, in which there is a Boltzmann distribution, show that the mean square deviation of the energy £ of the individual molecules, which is fluctuating about the average energy (£) of all the molecules, is given by (£- (£»2
= N«e 2 ) - «(;-)2),
where e is the energy of an individual molecule and (e) is its mean value. (b) Show that the heat capacity C v of an assembly can be written in the form
~
=mrTZ
N 2 kT 2 «e)
Show that the fluctuation in the total energy £ of an assembly from its equilibrium value V is the heat capacity Cv given in part (c). Solution
£
= Ie;,
(£)
= I
(el)
=1 (note that £ is the internal energy V), where ej is the energy of the ith molecule and (ej) is the energy of the ith molecule averaged over the whole assembly. 1= 1
so that
N
j
'
N
I I «ej-(ej»(ej-(ej»). 1=lj=1
(£-(£»)2
In a Boltzmann distribution the molecules i and j are statistically independent, and we can readily show that terms for which i =1= j average to zero. Using the Boltzmann equation to express the fraction of the molecules i and j in the assembly which are in the states PI and Pj we have «ej - (ej »( ej - (ej»)
= [dI(e i -
(€j »exp(-ej. pJ kT)
".J Pi
x
-f-
L(ej - (ej»exp(-€;.
= (€i -
(€j»(€j - (€ = «€i)- (€i»«€;)- (ej» = 0 . Retaining only terms for which i = j we have N
(£_(£»2
= L
(€ j -(ej)}2
= N<e-(e»)2
1
j
N«e 2 ) - 2(e)(e)+ (e)2)
= N«e 2)
(e)2).
Alternative expressions for the energy £ can readily be obtained from £ F + TS, which we can write in terms of Z as £
ZN -kTln N!
£
NkT
hence
(olnZ) J
r
ZN + T LklnN! + NkT --aT v ;
2(0InZ) __ Nkl~\J __ Z oT v a(T7TJ v - Nk Z
From this we have
C" = ( oT v
I ( = -
o£) Nko(Z'/Z) oOIT) v = T2 oOIT)
Differentiation using the quotient rule immediately yields Co
=
Nk [Z"Z' (Z'Z )2J . T2
The average energy of a molecule is given by (1) This holds for a system in contact with a heat reservoir at temperature T. It can be obtained from the modification of the result p 0: cxp[S(X)/kj (Problem 22.4) to this situation, when it reads p 0: exp [-F(X)/kTl, where F is the free energy.
pJkT)]
Pj PI
0£)
N
N
N
I (ej-(el» 1= 1
E-(£)
2
1(
101
We can write
(e».
(d) The fluctuation at eqUilibrium in any thermodynamic quantity x from its equilibrium value xo, at constant y, is given by the general formula (1) 02£ 02S) (X-x o)2 = kT ox2 - T ox 2 Y •
(a) We have
Ideal classical gases of polvatomic molecules
(Z)2] Z '
where Z is the partition function of a single system, Z' = oZ/o( liT), and Z" = 0 2 Z/o( I/T)2 , both taken at constant volume. (c) Use the formula obtained in (b) above to show that the heat capacity C v of an assembly is proportional to the rate of change of the average energy (£) with temperature so that (see also Problem 22.3) Co
4.2
(e)
Lniet j
4.2
Chapter 4
102
Ideal classical gases of polyatomic molecules
103
THE CLASSICAL ROTATIONAL PARTITION FUNCTION
where exp(-edkT)
ni
4.3 (a) For most polyatomic molecules at temperatures above the normal boiling point of the fluid, the rotational energy levels are so closely spaced that they approximate to a continuum, and rotational energy may be calculated accurately on the basis that the molecule is a rigid body obeying the laws of classical mechanics. Show that the rotational kinetic energy of a heteronuclear diatomic molecule about its centre of mass is given by
L exp(-edkT) i
so that
Leiexp(-edkT)
=
(e)
---"i_ _ _ __
Lexp(-edkT) i
From
4.3
Z=
L exp(-t;/kT) we have
_~( 2 ~) e - 21 Pe + sin 2 e
i
dZ dOlT)
, / k1L..tiexp(-ei kT)
=-
so that
1
k(e) =
and
[ ~J Z
aO/T) v
=Z ,
= Z = z" _ (Z')2 Z
Z
Zro!
,,_ az' _
1, 2 - aO/T) - k2 'lei exp(-edkT) ,
L elexp(-edkT)
(e 2 )
i
k2
k 2L exp(-edkT)
,
i
and the ratio
Z')2 (
=
Z
1
= h2
Joe
1T
f J-~J-~ 1T
0
r~ r~
(e)2
k2
Now
Zro!
Cv
=
Nk (Z" T2 Z
-
Z') Z
2
=
Nk(e ) (e)2) T2 122 -122
=
N kT2(e2)- (e)2).
(d) Putting x = E and y = V and noting that the average value of the total energy (E) is the internal energy U, we have (E-U)2 =kT
Using
(:~)v = ~
( a2S)
S)
E a2 aE2- T aij2 v'
a2 (
I
aij2 v
=-
P2)] dPe dPq, de dljJ .
=
~~(81T2kT)'/,(!a/b/etl
,
where a is the symmetry number of the molecule and is defined as the number of values of the rotational coordinates which all correspond to one orientation of the molecule. The quantity la/b/e is the product of the three principal moments of inertia of the molecule. Obtain expressions for the rotational energy, heat capacity, free energy, and entropy from the above partition function. Solution
1 (aT) T2 au v
we obtain (E - U)2
[
I ( exp - 21kT PJ+ sin~ e
(c) The general quantum mechanical solution for the rotational coordinates of a rigid body with three degrees of rotational freedom is complicated but, since the moments of inertia of all but the very lightest of molecules are large, and as the quantum levels are closely spaced compared with kT at temperatures above the normal boiling point of the substance, the classical partition function may be safely used. Evaluation of the classical phase integral yields
the ratio Z" Z
'
where I is the moment of inertia, and Pe and Pq, are the moments conjugate to e and 1jJ. (b) Obtain the rotational partition function for the rotation of a heteronuclear diatomic molecule by evaluating the phase integral introduced in Problem 3.4.
Z'
Z dOlT)
= [a(Z'/Z)J
k aO/T) v
Here
dZ
'
= kT 2 C v .
Using the result obtained in part (a) we find N Cv = kT 2(e 2 )- (e)2) .
=-
1 T 2C v
(a) The molecule rotates about the centre of gravity c. The total energy is the sum of the precessional energy, in which the atoms of mass mA and mB follow the circular orbits shown in Figure 4.3.1 with velocity dljJ/dt, and the energy of end-over-end rotation with velocity de/dt. The moment of inertia is 2
2
l=mArA+mBrB
mAmB + (rA+rB) 2 mA mB
4.3
Chapter 4
104
= mA(rAsinO)2+ mB(rB sinO)2 = IsinzO
IQ = mA(Aa)2+ mB(Bb)2
.
The precessional kinetic energy E", is
HmA(rAsinO)2+mB(rBSinO)21~~ VSin20(~~y
2
dO)Z +tmB (dO)2 rBdt V (dO)2 dt
The total energy €
is therefore
+
=
V(IP+ tihin 2 0) . Differentiating with respect to () and if; we obtain the angular momenta €
Po and Prp. Po
O€
M
10
P
O€
• I¢sin 2 0
f"'foo exp (= 211' h2 (211'IkT)'h 0
3.
Zrot
4.
Zrot :;
-QO
211'
€
(211'IkT)'h .
Zrot
{'II'
Jo
P~ ) 2IkTsin2 0 dPrp dO . sinO dO
=
811'2IkT h2
811'2
21 [, I )V'(kT)% = -(811' ah3 abc .
We at once obtain the molar rotational thermodynamic functions -
E rot
-
NokT
J
_ L~Jl 811'2 'h 3 aT - NokT dTL1lnah3(lalbIJ +:2"lnkT ,
2 0lnZ
whence E = !NokT and F rot
~j+
= I
h2 (211'IkT),-I'(211'IkT)Y'
(C )
_
so that
L~ exp ( - 2~!T) dPo
Using the same standard integral again we have
For the end-over-end rotation energy Ell we have simply EIl=tmA ( rAdt
105
molecules
Ideal classical gases of
so that
The precessional inertia is
E .. =
4.4
tv
~Nok.
2
811' k) = -NokTlnZ = -NokT ( ~ In T+ -! Inlalblc -In a + t In 11' + ~ Infi2
and
_ Srot
+! In T+ :Pnlalblc
E
+ NokInZ
2 811' k) Ina + tIn 11'+! Infi2
Collecting together the numerical constants we obtain ~
.srot
InT+:2"1 In Ia Ib Ie
Ina+ 134·684).
THE QUANTUM MECHANICAL ROTATIONAL PARTITION FUNCTION
Figure 4.3.1.
(b) The fourfold integration is straightforward.
[211'
Jo
1.
2.
roo
d¢
211'
(pj)
J~ exp - 2IkT dPe .
This has the form of the Gaussian error integral
f:
eXP (-dx 2 )dx
t(~)~
4.4 Polyatomic molecules in the gaseous phase are free to rotate about the centre of mass of the molecule. This rotational motion is strictly independent of the translational motion, and at moderate temperatures in which the vibrational modes of the molecule are excited only to a negligible extent, it can be assumed to be independent of vibrational motion within the molecule. A heteronuc1ear diatomic molecule has two degrees of rotational freedom, and the rotational energy is quantized such that the energy €J in the Jth rotational state is given by h2 €J J(J+l)811'2I=J(J+I)kOr,
where Or = h 2/811'2Ik and has the dimensions of temperature. Each rotational state except the first (J 0) is doubly degenerate, as the
4.4
Chapter 4
106
rotation can be either left or right handed, and the degeneracy WJ of the Jth level is in this case (2J + 1). (a) Assuming that the rotational energy levels are close enough together to be approximated to a continuum (i.e. when 8 r lT < 1) show that the quantum mechanical rotational partition function is the same as the classical rotational partition function obtained in Problem 4.3(b). From this partition function obtain expressions for the rotational energy, heat capacity, and entropy. (b) At low temperatures when 8 IT ~ 1 the rotational energy cannot be approximated to a continuum and a summation must be used. Investigate the form of the rotational energy and heat capacity in the low temperature limit. Sketch the temperature dependence of the rotational energy, and deduce the form of the rotational heat capacity curve.
Ideal classical gases of polyatomic molecules
4.4
107
The rotational entropy is given by
2 - = T E rat Srat + NoklnZ rat = Nok+ Nokln (81T h2IkT) 81T2k) = Nok ( 1 + lnIT+ lnll"2 ' which can be written in the alternative form
Srot
= Nok ( 1 + In;) .
(b) The rotational energy is given by
E
rat
= NkT2dlnZrat dT
Nk8
dZ
-Z d(81T)
Using the rotational partition function in the form
Solution
(a) The rotational energy €J and the degeneracy WJ are given by
h
€J = J(J+ 1)81T 21 ' WJ = (21+ 1). Hence
r
Zrat = ~(21+ 1)exp L00
When 81T
1000 cm- I . The remaining 2n 5 are bending modes, and generally have frequencies considerably lower. Vibrational modes are excited at high temperatures, and an error in the assignment of these modes does not usually introduce an appreciable error into the calculated thermodynamic quantities. It was shown in Problem 3. 9( e) that the heat capacity of N identical quantum_Ul-oscillators is ("'vib = NkE(2x) where and
(eT )3 + r
(iY- (ir+ . -]. 8
Note that as T -+ 00 the energy approaches the limiting value of Nk(T-je r ) rather than the classical value NkT. Terms in ie r do not occur in either the entropy or the heat capacity:
r
I
Cv = Nk [1 + 45
Using S S
1)2'
By graphical interpolation of the Einstein function E(y) tabulated below calculate the molar vibrational heat capacity and entropy at 298·15°K of 02, C1 2, and Br2 for which the fundamental vibrational frequencies are 1580, 565,and 323 cm- 1 respectively.
(e)2 16 (e )3 i + 945 i +
E/T+ NklnZ we obtain
2 8 (e )3] I e I e e I (e Nk [ 145 T T -Nk ln T - 3T 3T I
r
r)
r
r
8 - 2835 16
(i) 3+ ..]
y
r
L
Nkll-lni-9~(ir-
(eT
r)
.
The limiting value Nk of the heat capacity as T -+ 00 is approached from above. This is in accordance with the ideas of Problem 4.4. THERMODYNAMIC PROPERTIES ARISING FROM SIMPLE HARMONIC MODES OF VIBRATION
4.6 (a) In Problem 3.9 it was shown that the partition function for a [I)-oscillator of angular frequency w = 27rv is given by exp(-!hw/kT) Zvib = I exp(-hw/kT) Show that the heat capacity and entropy of a [I)-oscillator are independ ent of the amount of energy which may be possessed bv the oscillator in its lowest vibrational state. (b) The vibrational partition function for a polyatomic molecule with i vibrational modes may be written as the product of the functions of i distinguishable [I)-oscillators [see Problem Zvib =
111
Ideal classical gases of polyatomic molecules
r;) -eXp(~hwJkT) .
A non-linear polyatomic molecule has 3n - 6 vibrational modes, and a linear molecule has 3n - 5. Of the 3n - 6 modes n - I are stretching
y
Nok
1·000 00 0 0·50·979 ],704 1-0 0·921 1 '041 1·50·8320·683 2·0 0·724 0-458 2·5 0-609 0-309
Cvib
SVib
Nok
Nok
3'0 0'496 3·50·393 4·0 0-304 4'50'230 5·0 0·171
0'208 0-)39 0·093 0-066 0-041
y
Cvib
Svib
Nok
Nok
5'5 0·125 0·028 6'00-090 0'017 6- 5 0-064 0'011 7,00-0450,007 7·5 0'031 0-005
y
8·0 8' 5 9'0 9·5 10·0
Cvib Nok
0·021 0'015 0-010 0·007 O-OOS
0'003 0- 002 0'001 0·001 0·000
(c) By expanding the exponential term in the vibrational partition function and retaining only terms up to (e v /T)2, obtain simple approxi mate formulae for the heat capacity and entropy of a [I)-oscillator which are valid in the region of temperature for which (e v /T) < 1. Use these formulae to calculate the vibrational heat capacity of iodine vapour at 100°C and compare with the values obtained using the Einstein functions. (The fundamental vibrational frequency of 12 is 214·6 cm- I .) Solution
(a) Rather than rewrite the numerator of the function in the form exp (E min/ kT), we will carry the calculation E min thv. The energy of the [I)-oscillator is given by _
EVib -
NkT
(a InaTZVib)
2
v
so that = NkT2 =
d~{ln I
.. Nkl-;-I
eXP/-hv/kT) + In[eXp(-thV/kT)J} I
_] +!Nhv.
4.6
Chapter 4
112
The right hand term, which is the minimum (zero point) energy of the oscillator, disappears on subsequent differentiation. For the heat capacity we have Cvib
aE) = (aT
v
(hV) d I = Nk k dTexp(hv/kT)
For the entropy of the [1 ]-oscillator we have alnZVib ) aT + NklnZvib,
so that
_ hv 1 ~_1 Svib - NkT kT2exp(hv/kT)- 1 + NkT dT1n[exp( !hv/kT)]
+
Nhv
-Nkln[l
Nhv exp(-hv/kT)l-!r
Terms arising from the zero point energy cancel out. (b) We will denote the spectroscopic units, which are wave numbers, by w (cm- l ). Multiplication by the velocity of light e converts wave numbers into frequency v. Using the numerical values of the funda mental constants we have that hw hv hew Ov 2x = kT = kT = = I· T ' where Ov is the vibrational characteristic temperature of the [I)-oscillator and Ov/T is y in the table given. For oxygen at 298·15°K, Ov/T 1·4387 x 1580/298·15 7·62. Graphical interpolation of the Einstein functions gives Cvib/Nok O' 027 and SVib/Nok = 0'0044, so that Cvib 0·054 cal mole- 1 deg- 1 and Svib = 0·0088 cal mole- 1 deg- 1 • Similarly, for chlorine and bromine we obtain Cvib = 1·095 and 1·64 cal mole- 1 deg- l , and Svib = 0·52 and 1· 32 cal mole- 1 deg- 1 , respectively. These values agree well with experi mental values. However, at temperatures for which Ov/T ~ 1, thermo dynamic functions calculated from the harmonic oscillator model are not in good agreement with experiment. (c) Expanding Zvib = [1 - exp(-8 v/T]-1 we obtain Z-l
= 8;
[1
t(8;)+ie;y
...
J.
x
r,
In ~ 2_
2 •• _
The free energy F, energy E, and vibrational heat capacity Cvib, and the entropy Svib are now readily obtained.
F E
= -NkTlnZ
NkT[ln (8;)
NkT2 d~lnz
CVibG~) Svib =
hV) I S· =Nk ( -kT exp(hv/kT)-l Vlb
2 x3
{(8) t ; - Ii (0)2 ; + ... -t (0;) + i (8;) k(0;)
In(1- x) = -x
-Nkln[l- exp(-hv/kT)]+Nkln[exp(-thv/kT)] , which yields
113
We use the logarithmic series expansion (which is valid only when Ov/T < I) to obtain InZ- 1 -lnZ;
1
hV)2 exp(hv/kT) -Nk ( kT [exp(hv/kT) 1]2
Svib = NkT (
Ideal classical gases of polyatomic molecules
4.7
-"'1,
(8;) + -b c;r-.. J, Nk [1 - -b (0; + ... J,
NkT[l-! v
-t (8;) +f4 (8;)2
NkTe~/)+ NklnZ
r
Nk [I-In (8; )+ f4 (8;) 2- ...
For iodine vapour (12) we have 214·6 x 1-4387/373,15 From the approximate equation we have
J.
= 0·827_
Cvib = Nok[ I -b(0-827)2] = 0·943Nok = 1·873 cal mole- 1 deg- 1 and by graphical interpolation of the Einstein function we have
,
CVib 0·945Nok = 1·877 cal mole- 1 deg- 1 • At 8v /T 1 the approximate equation yields Cvib/Nok = 0'917, which differs from the value given by the Einstein function (0-921) by about a quarter of one percent. _ Similarly, for the entropy the approximate equation yields SVib/Nok 1·218 whereas the value obtained by graphical interpolation of the Einstein function is 1 . 215. CORRECTIONS TO THE RIGID ROTATOR-HARMONIC OSCILLATOR MODEL
4.7 (a) To treat the vibration of a diatomic molecule as simple harmonic motion is unrealistic because on this model dissociation of the molecule can never occur. A more realistic potential function was given by Morse, who proposed the equation u(r)
= D e {1-exp[-(3(r-re )j2},
where De is the dissociation energy of the molecule, (3 is an empirical constant, r is the separation of the nuclei of the two atoms, and re is the equilibrium separation.
~r
t 4.7
Chapter 4
114
Show that at low amplitudes of vibration this model allows simple harmonic oscillation, and show that the constant {3 is then given by 21T2 P.c 2 )% {3 = We ( -D- e where We is the frequency of vibration of the molecule, expressed in wave numbers, about its equilibrium position, and where p. is the reduced mass of the molecule. (b) When the amplitude of vibration of a diatomic molecule becomes large the simple harmonic oscillator model is inadequate for precise work and a contribution to the thermodynamic properties due to anharmonic vibration must be considered. The vibrational spectra of diatomic molecules are often represented empirically by energy levels which are described by the equation Ev
= hVe(V+!)-xehve(V+!)2+Yehve(V+!)3,
where V is the vibrational quantum number and where Xe , Ye, etc. are called anharmonicity constants. When the Morse potential is put into the Schrodinger equation, the following approximate expression for the allowed energy levels is obtained Ev
= h ve (V+ !I) _
( hv )2 ( V + 1)2 e
4D e
- exp(ixe u - !u)
Zanh-
l-eu
r
o
where 0 == 6!:
InZo =eo/T-I '
r(W;~e f - I J
81' _ Be InZ)' = e/T' where l' = We Be is the rotational constant (h/81T 2 /) calculated with the atoms at their equilibrium separation re. For the above terms in X e , 0, and l' obtain expressions for the corresponding terms which must be added to the heat capacity of the molecule. (e) The equilibrium separation of the nuclei in a chlorine molecule is 1·988x IO- B cm and the corresponding value of Be is 0·2438cm- l . Calculate the contribution to the heat capacity of chlorine gas at efT = 2 (i.e. 133· 2°C) arising from the above terms, and express these contributions as a percentage of the heat capacity obtained from the simple harmonic oscillator model. Solution
(a) At low amplitudes of vibration we may expand the exponential of the Morse potential and examine the leading term
= D e (l-exp[-{3(r-re )j2}. = l-x+!x2-i x 3 and putting X = (3(r-r e ) we obtain u(r)
= hv(V+!)-x ehv(V+!)2,
the corresponding partition function is
115
of the partition function:
2"
Given that the fundamental vibration frequency of 35Cl2 is 564·9 cm- I and the coefficient (3 of the Morse potential is 2·05 X lOB cm- I , calculate the molar dissociation energy Jj e in kilo calories and the anharmonicity constant Xe of chlorine gas. (c) Show that for an anharmonic oscillator, whose energy levels are described by the equation Ev
Ideal classical gases of polyatomic molecules
4.7
Using
e- x
u(r)
= -De[{32(r-re)2-{33(r-re)3+-&{34(r-re)4- ... ].
From the leading term for the potential energy we obtain the force F: F
= -2De{32(r- re)
.
The equations of motion for the two atoms of mass m 1 and m2 at distances r 1 and r 2 from the centre of mass are given by d2rl _
2
m l dt2 - -2D e{3 (rl+r2- re) ,
2xeu
]
2 d r2 _ _ 2 _ m2 dt 2 2De{3 (r l + r 2 r e ),
1+(eU-I)2,
where u == hv/kT == efT. [Hint: Expand the exponential expression containing Xe and retain only the leading term. Simplify the summation over this term by utilizing the first and second derivatives with respect to u of the simple harmonic oscillator function.]
which can be written
(d) It can be shown that terms which allow for the fact that the moment of inertia of a vibrating rotator is greater than that of a nonvibrating rotator (0 term), and which allow for the centrifugal stretching of the molecule (1' term), must also be added to the logarithm
wherer l +r 2 = r,and l/ml+l/m2 = l/p.. When a mass p. is constrained by a force constant of -2De{32 the
so that
2 d (rl +2 r 2 ) dt
=
-2D {32(r + r _ r )(_1_ + _1 ) e 1 2 e m 1 m2
d2 r p. dt 2 = -W e{32(r- re) ,
I 4.7
Chapter 4
116 frequency of vibration v is
=
CWe
1 (2{32De)~ 21f
v
Using v =
Ideal classical gases of polyatomic molecules
4.7
The first term is the simple harmonic oscillator (SHO) partition function equal to I/O - e- U). We write the anharmonicity term in the form
J.L
we obtain
xe u
{3 =
We
L Ve- uv + V 2 e- uV .
Now
(b) The reduced mass J.L of 3sC12 is given by
"
and
)2 (2
X
3.1416 2
10 12 erg mole-I
X
2~~
(1
e_U)3
which can be written X
17·5
X
2.9979
2
X
10
20
)
2·9979
X
1010
(1
so that
56·34 kcal mole-I.
Comparison of the empirical equation for the vibrational energy levels of a diatomic molecule with the form obtained from the Morse potential gives hv" hcw e xe = 4De 4De/N
10-
e- U )2+
)3 ,
564.9 ( 2.05 X lOS
X
~
Using these expressions the anharmonicity term becomes 2e- 2u
w
6·6256
LVe-uv = - (1 _ e-U )2
2 e- uV = e- U (1 "e-uv = "V du 2 L.. L..
= {32 (21f 2J.LC 2 )
27
U
e- U
~ Le-uV = du
2
X
-e
and
where m l = m2 = 35/No, so that J.L 17· . We note that if in the numerator of our equation for {3 we use 17· SINo, then in the denominator we use the dissociation energy per molecule De De/No, and the No cancels out. Hence we obtain for the molar dissociation energy
= 2· 3574
=
L..
ml m 2 J.L = ml+ m2 '
De
117
X
564·9
X
6·0225
X
10
2xeu e-u )(e U
-
1)2 ,
we obtain - exp
(d) The correction term which must be added to the SHO partition function is obtained by quite straightforward differentiation: ~
23
lnZcorr = !xe u
tu+
u(e U
-
1)2+
b U
e -
1+
u
Using = 7·166
X
10- 3
NkT
•
(c) The partition function for the anharmonic oscillator is
we obtain for the four terms
L exp[-u(V+-!)+UXe(V+-! V=o
Zanh
= NkO(-!x e +!)+Nk
..
=
L
v
exp{(!xeu-tU)[-uV+XeuV(V+ 1)J} 0
..
= exp(!xeu-tU)
L e-UV{l+exp[xe(V+ I)]}. v 0
Expanding the exponential term in xe we obtain Zanh = const[
£ ev=o
uv
+
£ xeuV(V+ l)e- UV] v=o
2dlnZcorr dT
.
e 8IT(20/T- 1 + 1 x 2xeO .. ---:-....,..,.,.,-~
+NkbOe8IT(eOIT-l)2+NkT2 x
and finally Ccorr
(dE (ff"'" corr
) V
Nk x 4xe(O/T)2e O/T+ 1 +Nkb(O/T)2 e O/T e (e'l/T - 1)3
2
+ Nk
x 16')' O/T
J
4.7
Chapter 4
118
(e) The constants 'Y and 0 for chlorine are
Be 0·2438
'Y = - = = 4·316 X 10-4 We 564·9
o=
J
r(WeXe)n 6 Be L T -1 We
7·979
10- 3
X
Using Nok = 1 -987 cal mole- 1 deg- 1 , (J /T :::: 2 -0, e OIT 7 -389, we obtain the following values for the contributions to the heat capacity arising from the terms in X e , 0, and 'Y. CXe
::::
0·0190 cal mole- t deg- 1 1
Co = 0·0151 cal mole- deg-
= O· 0071
C"(
1
cal mole- 1 deg- 1
L '/
I
4.8
Ideal classical gases of polyatomic molecules
(d) The nitric oxide molecule has two doubly-degenerate energy levels which are separated by the unusually small gap of e/k 174°K. The oxygen molecule has two levels which are separated by an energy gap of e/k = 11 300o K, and at high temperatures the lower level is triply degenerate and the upper level is doubly degenerate. Estimate the temperature at which the electronic heat capacity has its maximum and calculate the corresponding molar electronic heat capacity for these molecules. [Slide rule accuracy is adequate for parts (c) and (d) of this problem.] Solution
(a) The partition function for a system with two energy levels is
,
Taking the zero of energy eo = 0 we have
_
Z = woe 1+ we- elkT ) ,
t
where w = w d Wo and e = e 1 - eo. The energy and heat capacity follow directly £
CONTRIBUTIONS TO THE THERMODYNAMIC PROPERTIES ARISING FROM LOW LYING ELECTRONIC ENERGY LEVELS
= NkT21
,~.Jnwo+ d~ln(l + we- e/kT )] = wNe------"r-::;
and 4.8 (a) For certain molecules, notably oxygen and nitric oxide, there is an additional contribution to the thermodynamic properties, which arises from the presence of two electronic energy levels separated by an energy which is small compared with kT, so that thermal excitation is sufficient to populate appreciably the upper level. Construct the partition function for this two-level system and show that the electronic contribution to the heat capacity is given by
Cel
wNke
IkT(
C=
C At low temperatures e/ kT C
(~kT+2)
v
wNe +w
= wNkee/kT(---,e,.-;;/k"T_)2
ee/kT+ w
~
1, and
~ WNk( le~k~r ~ 0 . ~
1, and
~ wNk(~kT)2e€/kT --+
O.
We locate the maximum in the heat capacity by putting the temperature derivative of In C equal to zero:
where w is the ratio wdwo of the degeneracies, e is the energy of the upper level, and the energy of the lower level is taken as zero. (b) Investigate the high- and the low-temperature limits of the electronic heat capacity, and show that lnCe ) has a maximum value when e
(_a£) aT
(b) At high temperatures e/ kT
e/kT )2 eelkT + w '
kT = lnw+ In e/kT- 2
= woe-eolkT+ wle-el/kT .
Z
,
The heat capacity of a simple harmonic oscillator with OfT 2·0 (from Problem 4.6) is 1·438 cal mole- 1 deg- 1 , so the above terms contribute 1 ·32%, 1 -05%, and 0·49% respectively, a total of 2· 86%.
119
~ _ ee/kT (e/kT)2 wNk -
(ee/kT + W)2
In(~) = ~+ 2ln (~)- 2In(ee/kT + w) wNk kT kT
.
d (C) e 2 dT ln wNk = - kT2 - T
(c) Obtain an expression for the electronic heat capacity of a molecule with three equally-spaced singly-degenerate energy levels. By taking numerical values of e/kT between 1·0 and 4·0 compare the magnitude of the maximum in the electronic heat capacity of this molecule with that of a molecule in which the lower energy level is singly degenerate and the upper level has threefold degeneracy.
+
2(e/kT2)ee/kT ~€/kT ..L • •
multiplying by T we have
+ 2) (e€/kT+ w) = 2~e€/kT (~ kT kT I
I(
'
0;
,
I
4.8
Chapter 4
120 and
2(e/kT)e €jkT e €jkT + w = --"--'----'-- e/kT+ 2 2{e/kT) w 1+ = (e/kT+ 2) ,
so that we
~
4.9
4.9 The fundamental vibration-rotation spectrum of isotopically pure H 35 Cl is shown in the lower part of Figure 4.9.1. The relative intensity I rel and frequency w of each line, expressed in wave numbers, is indicated below the spectrum. This type of spectrum arises in the following way. When the vibrational quantum number V increases by unity (~V = + I) as a result of the absorption of radiation, the rotational energy levels may be affected in two ways. In the transition to a higher vibrational level, the increase in bond length of the molecule is accompanied by an increase in the moment of inertia. If the molecule absorbs only a small amount of energy, then it may fall into a lower rotational energy level (AI = -1) and give rise to the P branch of the spectrum. If a large amount of energy is absorbed, there may be an increase in the rotational energy, sufficient to raise the molecule into a higher rotational level (AI + I). This gives rise to the R branch of the spectrum. The upper part of Figure 4.9.1 shows the first eight rotational energy levels of the molecule and the double headed arrows indicate the transitions which give rise to the line in the spectrum below the arrow. For ~ V = + I the rotational transitions which we have just discussed are indicated by upward pointing arrows, whereas for ~ V I the downward pointing arrows will pertain. The selection rule for vibration-rotation spectra is AI 0, ± 1. Now the vibration-rotation energy of a diatomic molecule on the rigid rotator-harmonic oscillator model is given by
taking logs we obtain as the condition for the maximum
(e/kT+ 2) Inw+ In e/kT- f
.
(c) The energies of the three level system are 0, e, and partition function is Z = I + e-ejkT + e-2ejkT .
and the
The energy and heat capacity follow directly _ e-ejkT + E = Noe I + e-ejkT + e 2ejkT' , from which we obtain _ ( e )2 e-€jkT(1 + e- 2€jkT + 4e-€jkT) C = Nok kT (I + e-€jkT + e-2ejkT)2 Working to three significant figures we calculate for the two-level system with w = 3 and for the three-level system with w = I the following values: €/kT
1·0
1'5
2'0
2'5
3'0
3'5
4'0
0'249 0·424
0·540 0·602
0·821
0·991
1·02
0'933
0·790
0'634
0·578
0'486
0·390
0·303
C/Nok (tWO-level) (three-level)
121
CALCULATION OF THE THERMODYNAMIC PROPERTIES OF HCI FROM SPECTROSCOPIC DATA
-€jkT _ 2(e/kT) _ _ (e/kT- 2) - e/kT+ 2 I - (e/kT+ 2)
e kT
Ideal classical gases of polyatomic molecules
Ev, R
(V+t)hew+J(J+ l)heB,
where w is the fundamental vibrational frequency expressed in wave numbers and B is the rotational constant h/8rr 2 Ie for a particular vibrational state. The constant e is the velocity of light so that B is also in wave numbers. The above equation represents just one energy level, so that the total vibrational-rotational energy of the molecule is given by the sum over all V of such terms. We now use the equation for EV,R together with the Bohr frequency condition
The maximum for the two-level system is 1·023 at e/kT = 2·8 and the maximum for the three-level system is 0·637 at e/kT = 1·9. (d) We can locate the maximum in the electronic heat capacity using the equation e kT = Inw+ln(e/kT+2) e/kT- 2 .
.
E'y
R
E'~, R
= hew ,
where the single prime and double primes indicate upper and lower vibrational states, and w is the frequency expressed in wave numbers, in conjunction with the selection rule (~V = + I and AI = -I) for the P branch, to obtain a general expression for the difference of energy between any two levels. The set of frequencies P(J) in the P branch is then given by p(J) = (B' - Bn)J + (B' B").f2,
For nitric oxide w 1, and for oxygen w 2/'l...j By trial-and-error substitution we find that the electronic heat capacity of nitric oxide has its maximum value at 72° K and the corresponding maximum heat capacity is O' 875 cal mole- 1 deg- 1 • The maximum for oxygen is at 4900° K and it has the value 0·616 cal mole- 1 deg- 1 •
w
where J may have any integral values I, 2, 3 etc. other than zero.
'tl
I
122
4.9
Chapter 4
,I
(
4.9 by
8
R(J)
7 6
4
3 2 01
I
I
,
_
I
I
I
I I I
I
I
I
I
I
I
1
,
,
i
'I
8
'
,
~
- 61
5 -4
'I
_0 1
J
'"..... .::,
'"
M 0
00 N
M
N
N
N
'"t-
N
N
N
v v 00 N
00
t-
'"
on
V
M
N
M 0
'"
'"
r-
V'l
6
v
0"1
6
.::,
on tt-
00
00 0\
.::,
N", 0"1
'--' "---' '-"' \...J ' - - ' \...J ' - ' ' - -
r-
t"'VI r- 0\
on
~
M '"
6'::'
.::, .::,
OMM
0
'"
00
v v
N
0 00
;...::, 6
1: ...,
t-
t-
..... on t-
P branch
.....
oo
00 '" N l:sl
-
'" '" N
N
:;1l
0
'" N '" N'"
N
M
Solution
(a) Using the equations for R(J) and P(J) we readily obtain the combination terms
v_N \0 v N
66'::'
00 v 0 "'_ M "'00 NM M
V
R branch
on '" t
B").! + (B' + B").f2 ,
where J may now have any integral value including zero. Analyse the spectrum of H35CI in the following way. Calculate the combination terms R(J) - P(J) and R(J - I) - P(J + I), and by considering them as a function ofU+ I determine B' and B" respectively, hence obtain the fundamental vibrational frequency w. (b) The rotational constant in a vibrational state may be related to the equilibrium value Be by the equation B v Be - (V + t)a where a is a vibration-rotation constant. Calculate Be and hence obtain the equilib rium moment of inertia for H35CL [BI h!8rr 2 = 5· 0553 X 1011 atomic mass units A2 Hz.] (c) Given that the degeneracy of the Jth rotational energy level is 2J + I (cf Problem 4.4), derive an expression, based on the Boltzmann distribution law, for the population density of the rotational energy levels. Show that the most densely populated level is that for which kT )V2 J max = ( 2Bh -t·
R(J) - P(J)
'E
= w+ 2B' + (3B'
(d) The area beneath an observed peak is proportional to the intensity of the line, and this is in turn proportional to the population density of the energy level. Absolute intensities are difficult to measure, but relative intensities are readily available from the observed spectrum. Show that a plot of In(IreJ/U+ 1) againstJ(J+ I) has a slope of-hB/kT. Analyse the R branch of the spectrum and determine the temperature of the gas. (e) Calculate the entropy, Helmholtz free energy, and heat capacity of H 35 CI at 3000 K and 1 atm pressure. [The molecular weight of H 35 CI is 35 ·9877.J
-3
-2
' - - - - ' \...-.....J " - - ' " - - ' '--'''--------I
123
In the same way the set of frequencies R(J) in the R branch is given
J
5
Ideal classical gases of polyatomic molecules
= 2B' + (3B' + B").! + (B' + B").! + (B' -
B").f2 - (B'
= 2B'(J+ 1), and
..
R(J-I) = w+J(B'+B")+.f2(B'-B"), p(J+ 1) = w+J(B'
3B")+J2(B'-B")-2B";
on subtraction we obtain
Figure 4.9.1. The vibration-rotation spectrum of H35 0 showing the energy levels and transitions which give rise to the P and R branches.
R(J-I)-p(J+ 1)
L
=
2B"(U+ 1).
B").f2
4.9
Chapter 4
124
We can obtain the combination terms from the spectrum in the following manner: 0
1 3 2865 p(J) 2865 2844 p(J+ l) R(J) 2906 2926 R(J-I) 2906 R(J)-p(J) 61 R(J-I)-p(J+ I) 62 J
21+ 1
2 5 2844 2821 2944 2926 100 105
3 7 2821 2798 2962 2944 141 146
4 9 2798 2775 2980 2962 182 187
5
6
11
13
2775 2751 2998 2980 223 229
2751 2728 3014 2998 263 270
4.9
"
J(J+ I)B .
7 8 17 15 2728 2703 2703 3030 3014 3030 302 311
The total number of molecules occupying all the rotational levels is simply the rotational partition function, which can be approximated by the integral kT (21+ l)exp -J(J+ I) hB] kT dJ hB'
Jor'"
h B = 8rr 2Ic
n = (21+
substitution
,("dO
n
d (nJ) dJ
so that ]2+J+
a
(i- 2hB kT)
gives
X
J
=
0
kT)'h - ~ . ( 2hB
(d) For convenience we select the line J(lmax) of greatest intensity the R branch, and we scale all other intensities relative to it IJ
hB/kT(21+ 1) exp[-J(J+ l)hB/kT] rel
I = [max = hB/kT(21m + I) exp[-Jm(Jm + l)hB/kT] so that hB InJ;.el In(21+ I)-ln(21m + 1)+ [Jm(Jm + l)-J(J+ 1)) ,
Imax in
I
where the integer J m is a constant.
Analysing the lines of the R branch we have:
10- 40 g cm- 2 .
(c) The relative population of the rotational energy levels is given by the Boltzmann distribution law n
-(21 + 1)(21+ 1) hB kT exp [ -J(J+ 1) hB] kT + 2exp [ -J(J+ 1) hB kT
o
16·8628 Ie = 10.4888 = 1·60770 atomic mass units A2
nJ
and of
= 16·8628 atomic mass units A2 cm-
1-60770 x 10- 16 6-0225 X 1023 = 2-6695
hB] .
[
The value of J for which nJ/n is a maximum is given by
we calculate the equilibrium moment of inertia to be
-
hB
1) kT exp -J(J + 1) kT
Jmax
5·0553 x 10 11
=,.,. "(""'"' '
back
[
The fraction of molecules in the Jth rotational state is therefore
-10·0558 == -Be+(l+!)a; and
125
so that
We now plot R(J) - P(J) to 21+ I and obtain from the slope B' = 10·0558 cm- I , and from a graph of R(J - 1)- P(J+ 1) to 21+1 we obtain B" = 10·3448 cm- I . Using these constants and the value of P(J) 2865 cm- I at J = 1, we obtain w = 2885·7 cm- I . (b) For V = 0 we have Bvo 10·3448 cm- I , and for V 1 we have I BVI = 10·0558 cm- . We obtain the vibration-rotation constant a by eliminating Be between the equations 10·3448 = Be - (O+!)a, addition gives a = O· 289, Be = 10·4888 cm- I . Using
Ideal classical gases of polyatomic molecules
J(J+ 1)
0 1 0·76 In 100Irel /2J + I 4"33
21+ I I rel
= gJ exp(Er/kT) .
For Hel the statistical weight gJ is simply the degeneracy of the level is given by 21 + I, and the energy J(J+ l)h 2 Er rr2I
2 3 0'95 3-94
6 5 1'00 2'99
12 7 0·93 2'58
20 9 0·83 2·22
30
42
11
13
0'64 1'76
0'41 1 ·15
56 15 0·22 0'40
Plotting In 100fret/21 + I against J(J + 1) for values of J > J(l max) we obtain a straight line of slope 0·05 ± 0-001, so that Bhc 10·488 x 6·6256 x 10-27 x 2·9979 x = 302 ± 15°K. k x 0·05 1·38054 x 10- 16 x 0-05
L
126
4.9
Chapter 4
The intensity of the spectral lines is not a simple exponential function as the Boltzmann distribution law alone suggests, as there is a term which involves the dependence of the transition probability upon J which we have not considered. However when J > J(I max) this term is small compared with the Boltzmann term (3). (e) Formulae for the contributions to thermodynamic properties of a gas which arise from translational, rotational, and vibrational motion of the molecule have been obtained in Problems 4.1, 4.4 and 4.6. We shall consider first the vibrational contributions. From the fundamental 2885·7 cm- 1 we obtain the ratio vibrational frequency w Ov/T 1·4387 x 2885· 7/300 13· 839. The value of exp(Ov/T) which occurs in the formulae for the vibrational thermodynamic functions is then approximately 1 x 10 6 , so that the magnitude of these thermo dynamic functions is exceedingly small, and they can be completely neglected. For the translational and rotational contributions we have for the entropy: Strans = 1'98717[~ln300
= 36·71
cal
mole- 1
Inl +!ln35'9877-1'1605] deg- 1
,
Srot = 1·98717[1+1n2·66948 x 1O-4o+1n 300+1n(8'lT 2 k/h 2 )] =
1·98717[1-91·1215+5'70378+88
:;;: 7·9288 cal mole- 1 deg- 1
,
_ -1 Strans + Srot - 44 • 6391 cal mole- 1deg .
= 10118·8 cal mole-I, F rot =
-1,98717 x 300[ln300+ln2'66948 x 1O-4°+ln(8'lT 2 k/h 2 )] -1782'5 cal mole-I,
= -11901 cal mole-I.
Finally for the heat capacity Cp we have:
Cv+R
= iNoK:.rans+!Nokrot+Nok
6·956 cal
mole- 1
127
Ideal classical gases of polyatomic molecules
THERMODYNAMIC PROPERTIES OF ETHANE
4.10 At 500 0 K and a pressure of I atm the calorimetric heat capacity Cp of ethane is 18·66 cal mole- 1 deg- 1 and the calorimetric entropy is 62·79 cal mole- 1 deg- 1 • The molecular weight of ethane is 30'047, and the principal moments of inertia of the molecule are 42'23, 42·23 and 10·81 x 10-40 g cm 2 • The following vibrational frequencies (cm- 1 ) have been assigned from the spectrum: C-H stretching C-C stretching C-C bending CH 3 group deformation
2955 993
2954
2996 d
2963 d
1375
S2I d 1472d
1190 d
1375
1460 d
Frequencies with the subscript d are doubly degenerate. (a) The only assignment which has not been made is that of the twisting mode about the C -C bond. Calculate the contribution to the heat capacity and to the entropy which arises from this mode. Investigate the following three models (denoted b, c, and d below) which purport to describe the nature of the motion in this mode by choosing (where possible) the parameters of the model to fit the heat capacity, and using these parameters to calculate the entropy: (b) The free rotational model in which the methyl groups rotate freely about the C C bond. The reduced moment of inertia lr of the rotating group is given by 1m Ir = Igy+Im , where Ig is the moment of inertia of the rotating group about the axis of rotation and 1m is the moment of inertia of the rest of the molecule about the same axis. The kinetic energy € of the rotating group is given by 1
-1'98717 x 300[!ln300+!ln35'9877-1nl-2'6605]
Ftrans
Cp
4.10
g
This value is in excellent agreement with the calorimetric value of 44·5 cal mole- 1 deg- 1 obtained at 298· 10 K for a mixture of H 3s CI and H 37CI in their natural abundancies of 75·4% and 24·6% respectively. Similarly for the Helmholtz free energy we have:
Frot
(
= 3·5 x 1·98717
deg- 1 •
The calorimetric value of the heat capacity is 6· 96 cal mole- 1 deg- 1 • (3) G.Herzberg, Molecular Spectra and Molecular Structure, 2nd Edn. VoU, Spectra of Diatomic Molecules (Van Nostrand, Princeton), 1950, p.l25.
€
= 21r P~
where Po is the momentum conjugate to rotation about this axis at an angle O. [Note: Include a symmetry number in the derived thermo dynamic formulae.] (c) The harmonic oscillator model in which the two methyl groups undergo torsional oscillation. (d) The restricted rotator model in which at low temperatures only torsional oscillations take place, but as the temperature is increased the amplitude of these oscillations increases until a certain energy barrier is overcome, and thereafter rotation of the hindered group becomes possible. The potential barrier V is given by V
! Vo(l -
cosnO) ,
where n = 3 for ethane and Vo is the maximum energy of the barrier.
4.10
Chapter 4
128
Pitzer and Gwinn (4) tabulated thennodynamic functions for the hindered rotator model, and a section of their tables is reproduced below. In these tables Z is the numerical value of the classical partition function for the group with reduced moment of inertia I r , and the tabulated functions have the dimensions of cal mole- 1 deg- 1 • RT 0·25 3-355 3'180 3·008 2·838 2·678
2-0 2·5 3·0 3·5 4'0
0'30 3'004 2·836 2-667 2'500 2·343
0'35 2'709 2·548 2-380 2'218 2'069
0-40 2'458 2-303 2'138 1'978 1-834
0-25 1'632 1-840 1·996 2-106 2'168
0-30 1·606 1-801 1-952 2'054 2·110
4.10
c.nblNok = 1-181, w(cm8:yIT
0-40
Total for doubly degenerate vibrations:
1'541 1·717 1·846 1·934 1'980
so that
CviblNok
= 3· 246, SINok
Cvib = 1,9872(1·181+3-246) Svib
=
0-430_
1190 1460 1472 2963 2996 3-42 4-20 4'24 8'51 8-63 0-410 0-275 0-013 0'270 0-010 0-148 0-079 0-074 0-002 0·002
0-35
= 38·75 cal mole- 1 deg- 1 , Ctrans
C~b
1-9872 (0·430+ 1·302)
= 1· 302 ,
8-796 cal mole- 1 deg- 1 3-45 cal mole-I deg-
1
,
_
= ~Nok = 2-980 cal mole- 1 deg- 1 _
For rotational motion we have [cf Problem 4.3(c)]: 3 I = Nokb-lnT+!lnI"IbIc
Ina+ 134-68] = 1·9872 x 2·3026[i Ig500+! Ig(42·23 x 42-23 x 10-81 x 10- 120 ) -lg6] + 1-9872 x 134-68 17-83 cal
= tNok
mole- 1
deg- 1 ,
= 2·980 cal mole- 1 deg- 1 _
For vibrational motion [cf. Problem 4.6(a), (b)] we obtain the entropy and heat capacity of each vibrational mode by graphical interpolation of the hannonic oscillator functions tabulated in Problem 4.6_ Now = I -4387w/500 so that we have w(cm- 1 ) 8:yIT C~tllVok
SVib/lVok
993 2-86 0-525 0-232
2955 1375 1375 2954 8·50 8'50 3-95 3'95 0·013 0-013 0-315 0-315 0-002 0-002 0-097 0-097
(4) K.S.Pitzer and W.O.Gwinn,J. Chern. Phys., 10,428 (1942).
= (Cp -Nok)expt-Cvcalc = 16-67-(2'980+2-980+8-796)
= 1·91 cal mole- 1 deg- I Svib
,
= 62-79-(38'75+ 17-83+3'45)
= 2·76 cal mole- 1 deg- 1 _
= Nok[~ In T -lnP+ ~ InM - 1-164] = 1·9872 x 2-3026[~ Ig500 -Ig I + ~ Ig30-047 - 1-164]
Crot
821-5 2·36 0-645 0-346
)
SINok
For motion about the C-C bond we therefore have,
Strans
=
1
1'574 1-756 1'900 1-995 2·048
(a) We calculate first the contributions to the heat capacity and the entropy which arise from motion within the ethane molecule about the c-c bond. This is done by subtracting the statistical thermodynamic quantity from the calorimetric quantity_ For translational motion we have [cf.Problem 4_I(b)]:
-
129
Total for singly degenerate vibrations:
S~b/lVok
Solution
Srot
Ideal classical gases of polyatomic molecules
C~b/lVok
Heat capacity Z-l
Entropy Z-l
Vo
c
We can now investigate the proposed models. (b) The free rotational model. We must first obtain expressions for the thennodynamic functions of a rotor with one degree of freedom_ We follow the methods of Problem 4.3. The kinetic energy of a part of a molecule which is rotating about a single axis with respect to the rest of the molecule is given by 1 e = 2l/~ . The classical partition function is therefore Z
= -I 1211' ah
0
J-
-_
(p2)
21T exp _ _8_ dO dP8 == -(21TI kT)'h. r 2Ir kT ah
3
(81T !r kT)'Iz = __
The relevant thennodynamic fonnulae follow at once: -_ 2 alnZ _ E - NokT aT - !NokT , Cv = !Nok,
F = -NokTlnZ _ E
= -NokT ( !lnT+!InI;.
(
S= T+NoklnZ=Nok !lnT+!Inlr
81T2 k)
Ina+!lnJiZ ' 2
81T k ) Ina+!lnJiZ+! .
130
Chapter 4
4.10
The smallest moment of inertia listed above for the ethane molecule is 10·81 x 10-40 g cm 2 , and this clearly corresponds to joint rotation of the two methyl groups along the axis formed by the C-C bond. The reduced moment for one methyl group is consequently (5 ·405 x 10-40 )2 Ir = 2 x 5.405 x 10 40 2·702 X 10-40 g cm 2 In the course of the complete rotation of one methyl group with respect to the other we see that there are three identical configurations; hence (J 3. From the above equations we obtain Crot 0·99 cal mole-I deg- I , Srot = 2-46 cal mole-I deg- 1
0·7 ------:----lnO·49659 = 1·377 . Syib = 2·736 cal mole- 1 deg- I .
(d) The restricted rotator model. We calculate first the numerical value of the classical rotational partition function obtained in part (b). 3, we obtain Putting Ir = 2·702 X 10-40 g cm 2 , T = 500 o K, and (J I/Z = 0·292. Plotting and cross plotting the tabulated thermodynamic functions we find that a heat capacity of I ·91 corresponds to a value of V/RT = 2-80, and that 2-66 cal mole- 1 deg- I is the corresponding entropy. On comparison of the calorimetric entropy, 2·76 cal mole-I deg- I, with that obtained for the three models: (b) 2'46, (c) 2'74, Cd) 2·66 cal mole-I deg- I , it would appear that the harmonic oscillator model is marginally better than the restricted rotator model. However when the above comparison is made over a wide temperature range it becomes evident that the restricted rotator model is superior. In Problems 4.9 and 4.10 we have calculated some thermodynamic properties of ideal classical gases of polyatomic molecules from statistical
4.10
Ideal classical gases of polyatomic molecules
131
mechanical formulae based on simple molecular models. We have used in these formulae constants derived from experimental spectroscopic data, and have made a comparison of the calculated values with those obtained from direct calorimetric measurements. Agreement between the values obtained via these two routes validates the molecular models upon which the statistical mechanical formulae are based.
~I
•
For this model there is no parameter which we can adjust to improve agreement with the experimental values. (c) The harmonic oscillator model. Using the Einstein functions (Problem 4.6) we find that CyiJNok = 1·91/1·987 = 0·96 corresponds to () /T = O' 70, and this in turn is equivalent to assigning a frequency of 243·3 cm- 1 to the torsional oscillation mode. We obtain an accurate value of the vibrational entropy using the formula obtained in Problem 4.6(a) rather than by interpolation of the tabulated Einstein functions, so that for ()y/T = 0·70 we have, Svib ()y/T
Nok = ~~_{IJ 17'\_ 1 -In[1 exp(-()y/T)]
Hence
t.
~
(
5.1
Ideal relativistic classical and quantum gases
133
Solution
(a) The grand partition function is, from Problems 2.4 and 3.12,
5 P.T.LANDSBERG
(University College, Cardiff)
5.1 A system of non-interacting identical particles of rest mass mo is in a cubic box of side 1. In a single-particle quantum state (j I, h, h, a) the energy is e(jl, h, h) and the momentum components are Pr = (h/L)ir (r = I, 2, 3), where the j's are integers which cannot all be zero. The chemical potential J.I. is written as ')'kT, T being the temperature of the system. The spin label a assumes g values if each single-particle state has spin degeneracy g. (a) Prove that
pV = 'l7kTiLL L In[l +'I7tCiI,i2,ja)]
hhh where '17 = + 1 for fermions and '17 = -} for bosons, and
(. . .)
t h,/z,/3
[e U kT ,i2,h)] .
exp ')'
1
(b) Using a continuous spectrum approximation, assuming the energy to depend only on the magnitude of the momentum, and treating the magnitude of the momentum, pee), as a function of the energy e, show that the pressure is 41TgfP "" [p(e)pde p = 3h 3 P= 0 exp(e/kT-,),)+'17
j
where the sum extends over all many-particle states of the system. If the system is in a typical state i, it contains a total number of particles and a total energy given respectively by Cil = i2 i3 a excluded) g
= E;
= kTg (
(1
=
L L = 1 i1
::::::-00
L
L
it ,h,i,
(1
""
""
'L
L
h
::::::-00
h
:::::-QCI
n;CiI,i2,i3, a)
e(jl,h,i3)njCiI,i2,j3, a).
Hence
L nnnn [t(jl,i2,h)r M1 ,!2,h,a)
Z
;
i l h i3
(1
A many-particle state i can for indistinguishable particles be specified by the set of occupation numbers n;(l,a,a,a l
n;(a,
a,
I,a l ), nj(a,l,a,a l ),
(S.1.I) These njCil>h,j3, a) can have values a or I for fermions and a, 1,2, ... ,00 for bosons. A summation over i is then equivalent to summing over all admissible values of the numbers (S, 1.1). Hence ),
lor""
Z
I or"
= n(l, 0, 0,LI) = 0
L
n(O, 1,0, I)
....
... nnnn [t(jI.i2,ia)]n(j"h,i"a) i h h
O.
(1
.
l
The product goes over all quantum numbers. Carrying out the summa tions first
(c) Show that in the non-relativistic case P
= Lexp{(pNj-Ej)/kT}
Z
Ideal relativistic classical and quantum gases
Z
21TmgkT)';' h I(,)" " ±)
= nnnn[I +'I7tUI,h,h)]'1 (1
h i2 ;,
n
[I +'17fCil>/z,ia)j'1l.
j"j2J,
Hence
in agreement with Problem 3.12(f). (d) Given that in the relativistic case e2 = p 2C 2 +e5, where eo = moc2 show that 41Tg r~ (E 2 + 2Eeo) 'hdE 3 3 p = 3h c exp[(E+eo)/kT-,),] +'17 whereE == e-e o. (e) Show from n = v(op/oJ.l.h (Problem 2.8) that the mean number of particles in the above case is (n) _ 41Tvg r""(E+eo)(E2 +2Eeo)'hde - h 3c 3 exp[(E+eo)/kT-,),] +'17
Jo
Jo
132
kTlnZ
pv
= 'l7kTg L
LLln[ I +'17fCi I ,iz,ia)] .
i l h ia
(b) We can put
L~~ ... II
h
IJ
-+
f
oo
41TL3f""
~ ... dildi2dj3 =Jl3
... p 2 dp.
0
Applying this to Equation (S.1.2), with V = L 3 , we obtain
p
=
41T'I7kTg roo 1.3
Jo In[1 +'I7t(p)]p
1
dp
(S .1.2)
Chapter 5
134
5.1
t
5.2
Ideal relativistic classical and quantum gases
135
form suitable for substituting the L/s. In this way one finds (n >and pv. where t is regarded to be a function of p through its dependence on the The entropy is found, by using S v(op/oT)v,/.L (see Problem 2.8) from Also the expression for pv, t(p) = exp['Y €(p)/kTj.
jr'
TS = ~{r"(E2+2E€0),J,EeXP(-a+E/kT)dE Integrating partially one finds the stated relation. (c) We have pee) = (2mo€)¥. so that the pressure is kT [exp(E/kT-a)+llF 41rg roo (2mo)'/'€%d€
(E2 + 2EEo)'hexp (-a+ E/kT) dE} -akT '
p = 3h 3 exp(€/kT-'Y)+ll o [exp(E/kT-a)+llJ2 whence the result follows. = (€2 -€6)'A" so that (d) We have cp(€) B[ 4L4 + 13EoL3 + 10€ -akT(3L 3 +9€ 2 +6E6L 2 c 3 [p(€)]3 [(€ - moc 2 )( € + moc )]'h It is convenient to make a table of coefficients as follows: = E'h(E + 2€0)'/' = (E 2 + 2E€o)'h . This gives the result. L4 L3 L2 L1 (e) The stated result is found after a partial integration. TS 4 13€ IO€6 9akTEo -6akTEa S.2 Let B E'dE Lr== o'~o,~~ ,1;,r (r=0,1,2,3,4) l1(n> 0 311 9€ 6€ B where a (11- m oc 2)/kT, and B 41rvg/3h 3C 3. _pv (a) Establish -1 - 4E o -4E~ 0 B 3B[L3+3€oL2+2€6Ld, (n) _ (n)€o 0 - 3Eo - 9E6 -6€ 6 pv = B[L4+4€oL3+4€aL2] ' B
Jo
i
Jo
f
.
/r/L'7"
OO
_.\1._
Ld , B[4L4 +(I3€ 0 -3akT)L3 +(1 O€o -9akT)€oL2 -6akT€6 (11 ~€o)(n)+ TS-pv 3 9€o B L 2] ' 2 €a V = 3B[L4+3€oL3+
TS
6E6
0
where V is the internal energy excluding the energy due to the rest mass The last line gives V/B. This shows that the relativistic theory leads to a of the particles. renormaIisation of the chemical potential from 11 to 11' :::: 11- moc 2 • (b) Verify that Alternatively one can keep the 11 of non-relativistic theory and renormal V-TS+pv = (11-€o)(n) ,
ise the internal energy from V to 0' = V +€ by the energy due to and discuss this result.
the rest mass of the particles. (c) Verify that this system is not an ideal quantum gas according to the (c) Neither the relation pv gV nor the relation pv gO' is satisfied definition in Problem 1.9(e). with a constant g. Cd) A system is said to be ultrabaric if its energy (including its rest (d) A calculation of V+Eo(n)-pv yields the following sum of positive mass) per unit volume is exceeded by its pressure. Show that this does terms: not apply here. B(2L 4 +8E OL 3+ 1 IEaL2+6E6Lt). [If interactions are taken into account then a system may become ultrabaric. (1) 1 5.3 Using equations of the preceding problem, establish the following' results: Solution (a) In the non-relativistic limit, U == moc 2/kT;p I, (a) and (b) If one multiplies the integrand in the expression for (n) in 2 Problem S.l(e) by (E +2Eo)'l'!(E2 + 2Eo)¥. one has the denominator in a kT )1':1 Lr "'" (kT)r ( 2moc2 f'(r+,)I(a,r-,,±). (1) S.A.Bludman and M.Ruderman,Phys.Rev., 170,1176 (1968).
5.3
Chapter 5
136
In the extreme relativistic limit, u
« I,
Lr "'" (kTyr(r)l(a, r-l, ±) .
Discuss the meaning of these approximations. (b) Use these approximations to establish the results given in the fol lowing table. u~1
(n)
u
2rrm okT)% vg ( h 2 l(a,!,±)
kT)3 8rrvg ( hc l(a,2,±)
2rrm okT )'Iz ,vg ( 1.2 kTI(a,,, ±)
kT)3 24rrvgkT ( hc l(a, 3, ±)
~u
pv
u
u«l
1U
(c) Discuss the approximation e'" 1, and show that it leads to
« 1 when made in conjunction with
(
5.4
Ideal relativistic classical and quantum gases
(c) The approximation implies 'non-degeneracy' as used in connection with gases in statistical mechanics. It means that moc 2 -p. ~ kT or u - p./ kT ~ I. Since u ~ I is given, the restriction on p. is that it can be negative, but if it is positive it must satisfy p./kT « u. 5.4 A four-vector (cb, a) in an inertial frame I and the corresponding one (cb', a') in an inertial frame I' are related by
(5.4.1)
a' ;:;:: f3(a+v' b),
V'bl=f3(v'b+~:a)
, (5.4.2)
v x b'
V
x b,
where f3 = (l -v 2/c 2 rv>, v is the velocity of the origin of I in I', and c is the velocity of light. (a) Prove the reciprocal relations
~
a = f3(a ' -v • b') ,
kT )V> Lr ;:;:: (kT)r ( 2moc2 r(r+!) e '" . [The results of Problem 3.12 for an ideal quantum gas have here been recovered with stand s = 2 respectively. The extreme relativistic thermodynamic properties of a Fermi gas are seen to be rather similar to the thermodynamic properties of black-body radiation. An important difference is, however, that a = 0 in the latter case. The theory can be taken further by obtaining expressions for other quantities. (2)]
137
(5.4.3)
2
v'b
I v ') . f3v'b-c2a (
(5.4.4)
(b) An inertial frame 10 can be defined by specifying that the vector b shall be zero, Le. b o O. If the origin of 10 has velocity w in I' and the corresponding value of f3 is denoted by f3w, prove that
a' ;:;:: f3w ao , b'
Solution
a'
w,
(5.4.5) (5.4.6)
The integral is L = kT r -
(
)
r
f
~
0 (x 2
xr dx +2xu)V,[exp(x-a)+17]
The square root in the denominator is (2xurV> or X-I in the two approxi mations. The approximation u == moc2/kT ~ I implies relatively low tempera tures, or relatively heavy particles, or both. The thermal velocities of these particles will therefore be small enough for a non-relativistic treat ment to be valid. In the limit mo -+ 0, however, any temperature above absolute zero is capable of imparting high thermal velocities to the par ticles, and such a situation may be called 'extreme relativistic'. (b) The results stated are simple algebraic consequences of part (a) and the results of Problem 5.2. (2) P.T.Landsberg and J.Dunning-Davies, in Proceedings of the International Symposium on Statistical Mechanics and Thermodynamics (Ed. J.Meixner) (North-Holland, Amsterdam), 1965.
a' = ; : +w • b ' .
(5.4.7)
If I~ be a frame with velocity w+ dw in I so that (a, b) in I corresponds to + b o+ db o) in I~ then, assuming an equation of the type (5.4.7) to apply to I and I~, establish
(a o dao,
-aod(L)
b ' • dw
(5.4.8)
and hence that I da o I da = -+w· db f3w
(5.4.9)
[A physical application of these results is given in the following problem.] Solution
(a) From Equation (5.4.1) substitute for V' b in Equation (5.4.2). Similarly from Equation (5.4.2) substitute for a in Equation (5.4.1).
138
5.4
Chapter 5
(b) Equations (5.4.5) and (5.4.6) follow immediately from Equations (5.4.1) and (5.4.2). Equation (5.4.7) is then a useful consequence of these results. (c) Use the relation
-d(JJ~) = ~~
W'
(5.4.10)
dw .
f
[These considerations, basic to relativistic thermodynamics, have been recently a subject of controversy. (3) Note that the transformation of dQ is independent of the value of A.] Solution
(a) From Equation (5.4.7) it follows that
Now note that
,
E'+'Ap'v'
a' b' • dw = c2 W • dw
JJwao
=7
( 1) w • dw = -ao d JJw '
which is Equation (5.4.8). Now differentiate Equation (5.4.7) to find da' =
~o+aod(L) +b' • dw+w' db' .
The second and third terms cancel as a consequence of Equation (5.4.8), yielding Equation (5.4.9). 5.5 The quantity a = E+ APV and linear momentum b = P of a system transform like a four-vector (cb, a) of Problem 5.4. The suffix '0' refers in this problem to the inertial frame in which the centre of mass of the system is initially at rest, JJ := (1 -w 2/c 2 and w is the initial velocity of the centre of mass in an inertial frame I. For a fluid in a container A = I, and A = 0 for a solid, a box, or a fluid with the energy and momentum (but not the rest mass) due to the stresses in the container included with the system. Assume v = vo/JJ, p = Po for the transforma tion of volume and pressure. (a) Establish the result E-w' P = Eo JJ (b) Establish the relation
r'h,
,
dE-w'dP+Apodv+
(1 - A)Po
JJ
I
~
dvo=l3(dEo+Podvo).
(c) Justify physically the following expression for the compressional work done on the system:
dWc = -Po
-(l-A)vOd(~ )]
( I-A) .
= -Po Adv+-JJ-dvo
(d) Justify the expression for the translational work dWtr W' dP if the non-relativistic definitions of force and work are adopted. (e) From the first law in the form dQ = dE-dW = dE-dWc -dWtr show that the heat transforms as dQ = dQo/JJ.
139
Ideal relativistic classical and quantum gases
5.5
EO+JJ~ovo+w'
P'.
Hence the transformation of P and v makes the terms involving A cancel out. On changing the notation to the present problem, this yields E Eo/JJw +w • P. (b) We use Equation (5.4.9) and then evaluate da' -da o/(3 = W' db': 1 AVo W' dP = dE+APodv+Avdpo-j3dEo dVo-Tdpo
= dE ,
1 , APO JJ dEo+APodv-Tdvo I
P
= dE-j3dEo -{fdv o +A Po dv+
(1 -A)pO
t3
dv o
as required. (c) An increment of volume transforms as dv
= idvo+vod(i)
so that there are two causes of compressional work in frame I for A = 1. One is compression of the system; the other is due to Lorentz contrac tion of the volume when it suffers acceleration. As the system is 'bare' and the walls are not included, the only way of producing an accelera tion is by regarding the pressures on the system as doing the work. If A 0 the system behaves like an accelerated box and compressional work arises only if dv o =1= O. This work is -Podvo in 10 , but -Podvo/JJ in I. So we must combine -Podv for A = I with -Podvo/JJ for A = O. dWtr = f· ds
dP dt . ds = w . dP .
(e) The right-hand side of part (b) is I
I,
I
j3(dEo+ Podv o) = j3(dEo -dWo) = j3d Qo . The left-hand side is dE-dWtr-dWc = dQ. (3) P.T.Landsberg and K.A.lohns, J.Phys.Soc.Jap., 26 Supplement, 310-312 (1969) andAnnals of Physics. 56, 299 (1970). P.T.Landsberg, Essays in PhYSics, 2, 93 (1970) and papers in A Critical Review of Thermodynamics, Eds. E.B.Stuart, B.Gal-Or, and A.I.Brainard (Mono Press, Baltimore), 1970.
t 6 Non-electrolyte liquids and solutions A.J.B.CRUICKSHANK (University ofBristol. Bristol)
6.0 The condensed fluid is generally discussed in terms of so-called configurational thermodynamic properties, this designation implying the parts of the corresponding total properties which may be attributed to the inter-relations between the molecules. Unfortunately, the conven tional definitions (I) of the configurational properties give rise to difficulties which are inimical to the purposes of this book. For indistinguishable molecules possessing neither rotational nor vibrational degrees of freedom, the canonical partition function takes the form stated for proof in Problem 3.S(c), namely, _ (2rrmkT)3N/2QN ZN h2 N! '
where QN is the configurational integral defined by QN=
f..Jexp(-~;)drl ... drN; v
here the potential energy M of Problem 3.S(c) is identified as the configurational intrinsic energy U*, and the r i are three-dimensional position vectors. ZN is dimensionless, and it may be used to study solutions either as it stands, or by factorizing it in either of two ways: 2rrmkT )3N/2U N QN _ (a) ZN = ( h2 N! uN = ZtrsZ * ;
=
ZN
(b)
(
2rrmkT)3N/2Q , ; ; == h2
ZmolZconf'
The first factors represent the translational or molecular part, and the second factors the configurational part. The Helmholtz free energy F (= -kTlnZN, cf Problem 2.4) can accordingly be exhibited as a sum in either of two ways:
+ F* ,
(a) F
=
(b) F
= Fmol + Fconf
~rs
Ftrs == ,
-kTlnZtrs ,
Fmol == -kTlnZmol ,
F* == -kTlnZ* ; Fconf == -kTlnZconf'
6.0
Non-electrolyte liquids and solutions
141
If necessary we may define, in addition, rotational and vibrational Helmholtz free energies. The corresponding partition functions may be assumed independent of molecular environment in the fluid state, at least in the absence of hydrogen bonding etc., so they include no volume dependent terms, and do not contribute to the change in the free energy on mixing. The procedure usual in solution theory is based on (b), the justification being that (2rrmkT/h 2 )3N/2 is independent of volume, and so cannot change in an isothermal mixing process; its dl~merit is that the factors of ZN are not dimensionless, but have the dimensions of [urN and [vjN respectively. The corresponding 'free energies' are, however, individually extensive, cf solution to Problem 3.4(c). The consequent logical difficulties can be resolved by using molar quantities and referring volumes to unit molar volume, but since the procedure based on (a) avoids these difficulties altogether, it will be adopted here. The two procedures necessarily give equivalent results for the thermodynamic functions of mixing, as does using ZN itself, see solution to Problem 6.11. To describe liquid-vapour equilibria, i.e. the possibility of two phases of different densities at the same p, T, an equation of state t(p, u, T) = 0 must have three real roots in u over some region of P and T (see Chapter 7). It is consequently convenient to use T, u rather than T, p as independent variables, and we therefore separate p (rather than u) into configurational and translational parts. Thus we define
_ (aF* ----a;- )
p* = -
T'
_ (aF au
Ptrs = -
trs )
T'
respectively, as the configurational or internal (2) pressure and the translational (3) or kinetic pressure. The latter is the quantity described by the kinetic theory of gases, i.e. Ptrs == NkT/u. F* may be evaluated for the ideal gas by putting U* = 0 (the necessary and sufficient condition to define the ideal gas) in the configurational integral, giving QN = v N , whence Z* = I. Thus the configurational Helmholtz free energy for the ideal gas, F.d = 0, whence ptd == -( aF.dl aU)T = 0, and the total pressure for the ideal gas is Pid ==Ptrs+pTd = NkT/u, see Problem 6.I(a). It follows that all the configurational thermodynamic potentials for the ideal gas are zero; since 'configurational' is always taken to mean 'pertaining to or resulting from the forces between molecules', this conclusion has at least the virtue of semantic consistency. For the general case of the non-ideal fluid, F* may be evaluated (i) by proceeding in the same general way as above, but using a molecular model to specify U* in the configuratio'nal integral; or (ii) for a specific
(I) R.H.Fowler and E.A.Guggenheim, Statistical Thermodynamics (Cambridge University Press,
(2) J.O.Hirschfelder, C.F.Curtiss, and R.B.Bird, Molecular Theory of Gases and Liquids (Wiley, New York), 1954, Chapter 4, Section 2, p.255.
Cambridge), 1952, pp.700-701. J .S.Rowlinson, Liquids and Liquid Mixtures, 2nd Edn. (Butterworths, London), 1969, pp.250-252.
(3) J.O.Hirschfelder, C.F.Curtiss, and R.B.Bird, Molecular Theory of Gases and Liquids (Wiley, New York), 1954, Chapter 2, Section 6, p.116.
140
II
6.0
Chapter 6
142
(
f:
P = NkT,
+
pr v ;
Ut
pt dv .
F; + TSt .
F*
The configurational properties X; as defined above are identical to the 'residual' properties (4) obtained by comparing the volume derivative of the total property to that for the ideal gas over the range 00 to v, id ) ] Vr(ax~) X*==I -av T -(aX dv, ~ av T as may be seen on subtracting (aXtrs/aV)T from both terms of the integrand, since (aXid/aV)T is zero. For example, in the case of G F+pv,
p
( aG)T = (aF) & T + P + v (aav ) T and
p
v (aav
f
(a
pd v (apt) Gt = '"vrIV (apav~ ) T V a;J. ) ] dv "" v Tv dv, implying Gt = + pt v. In this argument, however, it is assumed that (aG trs! av h v(aptrs /al)T, i.e. that Gtrs == Firs + PtrsV, which is equivalent to defining G* == F* + p*v at the outset.
f
CELL THEORIES OF THE LIQUID STATE
"
= NkT-3NkTln[ 1-11(v:
f'J,
(6.1.1)
where 11 == (-./211'/6) '/', and Va is the volume of the system at maximum Na 3/..J2. density (close packing) Va (c) Derive the equation of state, P = p(v, T,N) corresponding to the configurational Helmholtz free energy expression in part (b) and prove that the isothermal bulk modulus B T -v(ap/aV)T is I (Va) B T = P [ I +}11 -;;-
pv ] NkT .
'Il
Solution
(a) Assume each cell to be of average size v/N. Consider first a particular arrangement, X, of the molecules among the cells, and evaluate the contribution Q~) of this arrangement to QN' The N independently ranging vectors r l , ... , rN are weighted by exp(-U*/kT), and this is zero whenever rj is outside the cell to which molecule i has been assigned. Thus Q~) breaks up into separate integrals each over a cell of volume v/N. Then
Q~)
= (
dr l )
(LINdr2)..·(LINdrN)
=(Nf
There are N! such arrangements avoiding multiple occupancy, so
6.1 Simple cell theory. The cell theories all assume that each molecule is, during most of the time, confined to a cell whose boundaries are determined by the potential due to the neighbouring molecules. For a 00 whenever a molecule overlaps hard-sphere fluid, this implies that U* its cell boundary, and U* = 0 otherwise. Suppose the cell boundary to be defined by planes which perpendicularly bisect the lines joining the centre of the molecule to the centres of its neighbours when all are at rest at the centres of their cells, so that the cells completely fill the volume without overlap. (a) If the volume of each molecule is negligibly small, and the mole cules are identical, prove that the configurational Helmholtz free energy, P
N!
and compare the model to the ideal gas. Remember that there are ways of arranging N particles among N cells. (b) Assuming that the polyhedral cell may be approximated by the sphere of the same volume, Le. of radius b, ~1I'b3 == v/N, and that the molecule, diameter a, cannot approach within ~a of the boundary, prove that
Since all equations of state necessarily approach ideality as v -+ 00, it follows that F{(oo) = Fj~(oo) O. The other configurational thermo dynamic. potentials are then defined as, for example,
Gt
143
is, for this model,
equation of state, Hp, v, T) = 0 which yields a total pressure p~(v, T), by integrating the corresponding internal pressure, pt == p~ - NkT/v, with respect to v, as
F{(v) - F;(oo) =
Non·electrolyte liquids and solutions
6.1
QN
= -kTln vN ,
(4) l.S.Rowlinson, Liquids and Liquid Mixtures, 2nd Edn. (Butterworth., London), 1969, pp.57-58.
QN =
~ Q~) = N!( ~
f'
whence QN N! vN =
using Stirling's approximation, QN v
-N
-W = e
,
which leads directly to the required result. In the case of the ideal gas, the total volume is accessible to every molecule, i.e. exp(-U* /kT) is unity for all values of each rj' Then QN VN, and Z* QN/V N = 1, so P = U* = O. The difference in P between this cell model and the ideal gas corresponds to S* = -Nk in
11
6.1
Chapter 6
144
the former case. This is the so-called communal entropy (5), which would be gained by the assembly in the cell model if the molecules were allowed freely to interchange between cells, i.e. if multiple occupancy were allowed. (b) Since U* = 00 whenever a molecule overlaps its cell boundary, its centre cannot approach within! 0 of the boundary. The average volume accessible to the centre of the molecule-the so-called free volume per molecule-vf , is Vf
3
Now j1Tb == vIN. and Vf
0
3
=
so
= ~[V IIl_ (j1T)
The
1 N -1'IVa IV
1'1 (
= -(v'
gives
of
QN =N! whence
=
r
N{ ~ [I
1'1(
v: )
V
v: yilT
3N
QN = NW-N [1- (Va) IhJ3N v 1'1 N
a
V
(Va) J-\ . NkT{ (I) )1/,[ I 1'1 (I)/; )lhJ-l + I }, P = P* + Ptrs = -v- 1'1 /; P*
and
V
= NkT -1'1 (Va) V V
13
III [
1-1'1 -
V
1\ f ~I
6.2 Hirschfelder's cell theory for hard spheres assumes that the cell is bounded by the polyhedron whose apices are the centres of the neighbouring molecules when the latter are at rest at the centres of their own cells, mutually distant a, in close-packed array; it is further assumed that this cell may be approximated by the sphere of radius a, and that the centre of the molecule may not approach within 0 of the cell boundary (6). (a) Taking a3 ..j2(v/N), 0 3 ..j2(v a /N), as for close packing, prove that the configurational entropy is
-Nk+ 3Nkln [1
Compare this to the configurational entropy according to Problem 6.I(b). [Hint: it is necessary to take account of the fact that this way of defining the cells leads to a multiple overlap, so that each element of volume is counted more than once. J (b) Express the isobaric thermal expansivity and the isothermal (Xp
(5) J.O.Hirschfelder, C.F.Curtiss, and R.B.Bird, Molecular Theory of Gases and Liquids (Wiley, New York), 1954, pp.213-216.
I
== -;;-
(av) aT
Ky == -
p ,
~(:;)T
(cf Problem 1.4) as functions of (T, v), (p, v), respectively, and compare these to the corresponding properties derived from the equation of state obtained in Problem 6.l(c). Solution (a) Dr
= (j1T)(a - 0)3, Le. the volume over which the centre of the molecule can move. With
(v )Ih '
V)lh
a~ll_1'I(V: )
II -1'IC:) lhft} , IVa)
' 0 == V2 N ..j32 x ! 1T(V I/, - V:')3 .
a == V2 (N As in Problem 6.1,
= -P
145
and the given result follows. It is clear that for this model By approaches (from above) the ideal gas value in the limit of large v. As (v decreases towards unity, BT values lie in the range found experimentally for solids rather than in that for liquids.
'
To obtain the isothermal bulk modulus, differentiate both sides with respect to V at constant T:
and solutions
(V: ) I/,J
P{I+
I
PV = NkTll-1'I(V:) V'TI P+V(~)y = -NkT[I-1'I(v:)
-V(~)T
S*
(V )IIlJ-l 1-1'1 (-al'av*) -_-3NkT-3v1'1 (Va)lhl T
Non-electrolyte
'
and the given result follows. (c) From the result of part (b) we have
hence
Hence
j1T(b-!0)3.
==
6.2
Ntl r
QN N!vf,
=
NW-N(N;r)N
(6) R.J.Buehler, R.H.Wentorf, LO.Hirschfelder, and C.F.Curtiss,J.Chem.Phys.. 19,61 (1951).
146
6.2
Chapter 6
so
P
= NkT[ I-In(v
V:,)3+
\I,
Inv -In(v32 x J1T)]
.
To obtain the given result, we have to remove the term -NkTln(v32 x J1T) .
In close-packed array, the polyhedral cells defined by the twelve neighbouring centres clearly have a fourfold overlap. The actual figure here is 5 ,92; the difference arises from approximating the cells by spheres, increasing the overlap. Thus the normalized configurational free energy is
3In [I _ (
N kT{ I -
F*
v; ) },
and the required result follows directly from S* From Equation (6.1.1)
3In
S* = -Nk{ 1-
-(aF*/aT)v'
r 11 (v; )\/']} . I
The only difference between these equations is in the value of the coefficient -q, which increases from 0·905 to 1·0 as the amount of overlap between neighbouring cells is increased from zero. (b) It is simplest to consider the general equation of state for both models as p -_ NkT[ - - 1-11 v V or NkT
(Va )II>J-I
V
'I1V "1)';'
a
'I
= -
p
6.3
V
whence I
(Xp
(av )
== -;;- aT
Nk pv
p
QM = MM(e- a)M .
Use this expression to obtain QN for a K-tunnel model, where N == MK. For the derivation of QM see Problem 9.3. (b) For given 1)IN, only one of band e can vary independently, but the ratio ble can take any positive value consistent with b, e > a; thus QN must be maximized with respect to ble. Show that this maximization gives
v%
In Q
Similarly,
KT
NkT
-Vp2
i[I-11(v~)
-i11(v; f'J
=
Solution
increases.
(a) There are clearly N!/(M!)K ways of arranging N molecules into K sets of M; thus, by an argument exactly parallel to that of Problem 6.2(a) we have
-j11 (va)' hJ-I I)
![I_11(v;)'h][I_i11(v;) (Xp
rLI -v~ () Vl~ '('J
Va
rjl-i11(V; ) %]-1
It is clear that at given T and v, p increases and
3Nln
where, as in preceding problems, == Na3/v2. (c) Comment on the general features of the equations of state of Problems 6.1 to 6.3.
p '
and substituting for pv from the equation of state, (Xp
147
6.3 The tunnel theories for the hard sphere fluid replace the regularly packed polyhedral cells of the preceding problems by hexagonal prisms or tunnels, stacked parallel in two-dimensional array. The longitudinal motions of the spheres in neighbouring tunnels are taken to be mutually independent, the transverse motion of each sphere being limited only by the geometry of the tunnel. Barker's tunnel theory (7) is analogous to the cell theory of Problem 6.2: the tunnel cross-section is defined by the hexagon whose apices are the centres of neighbouring tunnels; it is assumed that the cross-section may be approximated by the circle of radius b, b being the centre-to-centre distance in the two-dimensional array, and that U* = 00 whenever the centre of a molecule approaches within a of its tunnel boundary. (a) Formulate the configurational integral for transverse motion in terms of a free area af per molecule. For motion parallel to the axis of the tunnel (longitudinal motion), the problem is that of M hard spheres, diameter a, arranged with their centres on a straight line (the tunnel axis) so that they lie always within a distance eM, where e, the mean distance per molecule, is greater than a. If the positions of two spheres, i, j, are defined by Xi' Xi' then U* = 00 whenever IXi - Xj I < u. The configura tional integral, QM, for one tunnel is
Differentiating with respect to V we have
(Va) 1/3 Nk(aT) 1- 2 -p av 3 11 -
Non-electrolyte liquids and solutions
'IlJI
QN(transverse) whence QN
and KT decrease as 11 (7)
= [afN(e-
I.A.Barker, Australian J.Chem.. 13, 187 (1960).
N!
af (M!)K (6.3.1 )
r
i
6.3
Chapter 6
148
j
(b) The area of the hexagon of side b is h/3 3 b 2 , so the volume per tunnel is h/3 3b 2 cM, and the volume of N/M tunnels is h/3 3 b 2cN; there is a threefold overlap, so v (6.3.2) N = h/3b 2 c == exb 2c .
6.4
and so, for the total pressure, PV
= [7rN(b -
The three equations of state of Problems 6.1-6.3 differ only in the value of 11, which is determined by the degree of overlap of neighbouring cells or tunnels. The index of (va Iv) is! because the restriction on molecular movement is expressed by a radial quantity, i.e. in dimensions oflength.
Substituting for c according to Equation (6.3.2) into Equation (6.3.3) we obtain N QN
a)2(N~b2
= [7rN(b
It is convenient to maximize lnQN rather than QN:
~_,,'_v b - a IV\Nexb 2
dlnQN) ( db v This is zero when
_a)-l
6.4 Smoothed-potential theory. Suppose that the molecules of Prob lem 6.2 interact according to a spherically symmetrical potential tjJ(r),
2v
tjJ(r) -+ 00 as r -+ 0 , '; ~,
j!
(7rN)N(b*-a)3N
Now
= L7rNb*3(1-:* YJN
so
a ( b* =
V2 {/3
(
Va )'h
V
{/~
and b*3
w = Wo,
v 1 Na.
whence
Q = Nln7r+Nlnv'3 2 + 3Nln Inv~
w
rI-V!: (v ) 'Il] '
and the required result follows on elimination of the term Nln(27r/v'3), which is due to incomplete normalization, cf.solution to Problem 6.2(a). Note that this model includes the communal entropy; this is because the summation of QM allows a molecule to be anywhere within its tunnel. (c) Differentiating In (QN/V N ) with respect to vat constant T, we find P*
NkT (v )'1> [ 1-11: (v) =-v-11:
t LtjJ(r;j) . i, i
The configurational intrinsic energy w(s) of a molecule in its cell obviously is negative when the molecule is at the cell centre, increasing toward +00 as the radial displacement from the cell centre approaches the nearest-neighbour distance a. The smoothed potential, or square well, theory approximates w(s) by
a- -N
Va) 'I, '1.j2
V
u*
_(VO)'/'
(~r'(±
b*
< 0 for a < r < 00
and that the configurational intrinsic energy is the sum of the interaction potentials of all pairs of molecules without extra contributions from groups of more than two, i.e.
or Ncxb alv = a; cxb v/N. But v/N exb c, so the extremum of InQN is at b = c. The second derivative confirms that this is a maximum. Then QN
tjJ(r)
'~
2
3
3
tjJ(r) -+ 0 as r -+ 00 ,
~
3 Nexb a v
b-a=b
(v: f'Tl
13 I ()' NkTll-11 1): J-I
pv
(6.3.3)
a)2(c-
-{It
=
149
This is clearly a member of the class
Given af = 7r(b - a)2, we have QN
Non-electrolyte liquids and solutions
1\
=
00,
< s < (a- a) (a - a) < s , 0
;
where Wo = Wo (a) is the configurational intrinsic energy of a molecule at rest at its cell centre, and a is defined by tjJ(a) == O. (a) Let z be the coordination number of the cell lattice, i.e. the number of neighbours of a particular molecule with which it is in contact when a = a. Then z has the maximum value 12 for face-centred cubic and hexagonal close-packed lattices. If tjJ(r)
4e*
l( 7y2 (7 rl
where -e* is the minimum value of the pair potential corresponding to
\
150
6.4
Chapter 6
6.4
Non-electrolyte liquids and solutions
151
r = r* = !<j2a, prove that if only nearest neighbour interactions are counted, either
when the numerical term h/3211", due to overlap, is dropped. Differen tiating with respect to v at constant T, and adding Ptrs gives, cf.Problem 6.3(c), U* = 2zN€* (v ~ (V4 V2) P NkT[ V 1+ 2zN€ 4 v~ - 2 vU3 , or U* +00, whence the required result follows. according to whether or not any molecule has Sj > (a - a); Vu has the (c) Putting P = 0 in this equation of state gives same meaning as in the preceding problems. (b) Derive the explicit form of QN = QN(v, T) for this theory, and [1_(V~ = 4z€ _2(V~) thence prove that the equation of state is
r
[(V; _(V;)
)';'J-l
l
pv
V NkTll -( ;
Y13Tl
(vu/v) )'/,J-I + 4zN€*2 [- (V)4 ; -(I); )2J .
Settingy =
p
Sj«a
a),
r(~r2 -(~rJ
whence
[(V;
!zrp(a), with
Wo
= 4€* [( v;
r (V;) 2J '
t:,
r-( rJ .
(b) Since the cell is bounded by an infinite potential barrier at a - a, the free volume is, as in Problem 6.2(a), Vf = ~1I"(a - a)3, with a V2(v/N) '13, a == V2(v u /N) 'h. Then 4211"( 'il '13)3 3N V -Vu
JI
.
(NWo) '
QN = N!(Vf) N exp - kT
'Ii
v
a
IJ
Sj
In Q v~ = -N+ 3Nln [
+
I I I
+
= 2z€* v;
and
+
i=I,2, ...,N,
pair-interaction distances are rij = a. Then
Wo
kT
4z€
2y2)(l- Y
c
In the first case, corresponding
U* has the same value as when every molecule is at its cell centre, Le. all
rp(a) = 4€*
4J .
+
Solution (a) It follows from the postulates of the model that U* has only two
= Nwo or U* = 00.
r
transforms this into y2(l
(c) Prove that equation of state of part (b) predicts 'condensation', i.e. the possibility of co-existence of stable high and low density phases at the same pressure and temperature. Note that it is a sufficient, though not a necessary, condition for two-phase equilibrium that the equation of state has two real positive roots in IJ for p = 0 over some range of T.
values open to it, U* to
[(V;
+
Figure 6.4.1. Comparison of van der Waals and Dieterici equationfl of state for two substances having similar vapour pressure (indicated by the broken lines): a, van der Waals; b, Dieterici; c, ideal gas. The vapour pressure has been located by the equal area rule, see Problem 7.5(b), and Figure 11.14.1.
There is clearly a range of T, 0 < T < 1", for which there are two real roots, 0 < y < 1/v'2, while for T> 1" there are no real roots. Note that the van der Waals equation, see Problem l.ll(a), also shows this behaviour; the Dieterici equation of state,
RT exp (a) P v-b -RTv '
(V) Ill] ----;:r2zN€* [(V)4 has no real roots in v for P 0, but nevertheless predicts condensation. 1-; ; - (v; ) The difference is illustrated by Figure 6.4.1.
I
152
6.5
Chapter 6
6.5 Hole theory. Instead of accounting for variation of total volume by variation of the cell radius, we consider an assembly of fixed-volume cells which may be either occupied or vacant. With N molecules distributed among (N + No) cells in close-packing, the total volume cf Problem 6.1, (N+ NO)a3 N+ No . _ Na 3 v ..)2 = Va ----y;r- , wIth Va = ..)2 where, as in the preceding problems, a is the parameter of the cell lattice. It is assumed that only nearest-neighbour attractions are significant, and
that the configurational energy per cell has the value corresponding to the molecule being at the cell centre [i.e. !(a) per neighbour] every where within a free volume Vf, and the value +00 outside Vf. (a) Show that for this theory the general form of the configurational integral is Nzx(a)] exp QN vdi, 2kT '
~ [tUI
/\ f
6.5
(b) In the expression for QN, put Vr (i, X) = Vr, and take the average coordination number for every configuration to be that for all configura z zN/(N+ No). Then tions, Zx
I\1
l
Vr
QN = -In--Nv Nv (NV Vr zNva(a) Inr;r - - N ) In (NV) - - N +Nln--N, V Va Va Va Va V 2kTv
r
V V (V = NkT .va .-In-Va Va
P
* = NkTlln_V_ Va V-v a
= NkTf_ln Va
_
and the
result follows.
(I
V
- 1] .
z(a) (Va) 2kT V
Va) + z(a) (Va)2 _ NkTJ. V 2kT V V
"'" NkT(1 _ Va 2v
)-1 + Nz(a)V -2-:
Recall that (a) is necessarily negative for a pair potential like that of Problem 6.4. The corresponding form of the van der Waals equation is
Solution
f-
NVf In (V-Va) - - + In-Va V
Adding Ptrs gives the required result. Expand the logarithmic term in the equation of state, retaining the second power in (va/v), since this occurs in the second main term; the result reduces to 1+ Va)+ Nz(a)Va pv 2v 2 V
where z is the coordination number of the cell lattice. (c) Establish the approximate form of this equation of state appro priate to V IVa ~ 1, and compare it to the van der Waals equation, see Problem 1.11 (a), and to the equation of state of Problem 6.4(b). (d) Show that the equation of state of part (b) predicts a two-phase region; the argument of Problem 6.4(c) is not appropriate because of the logarithmic term; use instead the condition (ap/av h = 0, when two real roots ensure phase separation.
m:
.
Differentiating with respect to V at constant T, we find
2kT V
(a) The assumption of uniform potential over each Vr (i, X) for the ith molecule in the configuration X allows the configurational energy = zx (a)N/2kT to be taken outside the integral. As in the solution to Problem 6.1 (a), each integration dr j yields Vf(i), within the configura tion X. Then (X) _ Nz x(a)] flN . QN - exp 2kT i = 1 Vf (I, X) ,
J~ I
r
ZN2 (a) eXPL - 2(N + No)kT
The number of arrangements of N molecules among N+ No cells is clearly (N+ No)!/(No!), so, since N+ N_ = _ (Nv/v a )! N QN - (Nv/v a N)!vr exp Using Stirling's approximation in the logarithmic form and simplifying, we obtain
QN kTlnr;r V
NkT[ z(a)(Va)2] p == - -In ( 1 - Va) - +- V
N
QN
where zx is the average number of occupied cells neighbouring an occupied cell in the configuration X (a member of the set appropriate to N,N+No)· Assuming that z x may be replaced by its average over all configura that Vr (i, X) may be treated likewise, and that the average, Vr, so obtained is independent of overall density, i.e. is a function only of the lattice parameter a, confirm that the equation of state is
Va
153
Non·electrolyte liquids and solutions
pv
'\
b
RT ( 1--;
)-1 -~.
Thus the statistical theory produces a result formally the same as that obtained by empirical modification of the ideal gas law. If we compare this equation of state with that obtained in Problem 6.4(b), we find that the salient points are that the first term in the former suggests a second class of equations of state similar to those of the cell theories, but having the index of Va /v equal to unity rather than
Chapter 6
154
6.5
In fact, one may construct models for which this index is 1, so that defining Vr by a radius gives (va/v)v\ by a cross-section gives (va/v)';', and by a volume gives (Va/v)l. In the low density limit the cohesive energy term from the equation of state of Problem 6.4(b) takes the form -a/v 2 instead of -a/vas in the hole theory. Thus a family of equations of state, having the same sort of 'theoretical' justification as the van der Waals equation, may be set up; cf Problems 9.7,9.17, and 11.l2 to ILlS. (d) Differentiating the equation of state of part (b) with respect to volume, we obtain ap = NkT[ZIj>(a)v~ _ Va ] av Va kT v 3 v 2 -v a v .
II
1.
This is zero when 2
V
+
What are the corresponding extensive quantities? (b) Problem I.S(a) states for proof
cp -Cv where I
t;
KT == -
!(:;)
T •
C;
C:
Note that while is defined by (au* /aT)v, c; identity given in Problem 1.1 (a), a p ) (al)) (aT) ( av T aT p ap v = - I
(aH* /aT)p' The
is useful in avoiding the difficulties due to p being treated as the dependent variable.
EQUATION OF STATE TREATMENT OF LIQUIDS
Solution
(a) It is simplest first to establish F* , and then to use G* == F* + p*v, U* = F* + TS*, etc., for the others; but all four may be established independently. From the equation of state,
6.6 The van der Waals liquid (i). The van der Waals equation of state is usually written for one mole, as (P+;2)CV-b) = RT
p*(vdW)
with R == Nok, No being the Avogadro number, and v the molar volume, v == Nov/N. It is the simplest explicit equation of state capable of describing qualitatively the observed phase behaviour of fluids and their mixtures (8). (a) Use the methods suggested in the Introduction to this chapter to prove that for one mole of fluid obeying this equation of state,
(v Ib -ifI) -v V( _ I -=I)dV-a v-b v v a
RT
and F*(vdW) = -RT
2
f
co
since F*CV
= (0)
=
~ =
O. Hence
F*(vdW)
.
V RTln=--b v-
a
v
v' 2a
_V
Since S* == -( a F* / a T)v, we have V +RT_- U*(vdW) S*(vdW) = -Rln=--' v-b v-b b
H*
=
F*
Similarly = -=V + RTln=- v-b
a
v
II
a
= (F*+TS*)vdW=-=v
v a b a G*(vdW) = RTln=-b -=+ v v RT=--b-= v v b v G* = -V + RT=--b v - + RTln =--b v a b a H*(vdW) = -=+RT=--b-= v vv . R.L.Scott and P.H.van Konynenberg, Disc.Faraday Soc., 49, 87 (1970).
2a
(8)
KT '
Thus (Cp - C v ) is specified completely by the equation of state and derivatives. Use a simple adaptation of the method of part (a) to prove that for one mole of the van der Waals fluid 2aR(v-b)2 (C; - C~) = -v~3R-T---"-2a-CV =- bP
This has two real roots for T < -zlj>(a)/4k, otherwise two imaginary roots; the two real roots are coincident for T = -zlj>(a)/4k. The locus of the real roots is the boundary of the region within which a single phase is unstable, and -zlj>(a)/4k defines the 'critical temperature' above which phase instability does not occur.
U*
Tva;
~(:~
ap
II
vazlj>(a) v~zlj>(a) _ kT V + kT - 0 .
a
155
Non-electrolyte liquids and solutions
6.6
156
6.6
Chapter 6
1\
-p+
T(:i)
(~~t v(~~)T' =
v '
so
we have
I V[-P{+TC:j)JdV.
U*
I
For the van der Waals fluid, p
-p*+T (-a
*)
aT
v
*
Gt For the van der Waals fluid
G*(vdW)
=-R
The expression for identities,
v
a
v
f f
D[ V I RT = (v-b)2dv
(aH/av)T
,~
thence, using the identity
1b V+(V-b)2 I bJ dv
= - v
is obtained from the partial differential
(~~)T = [,J-TGi)J (~)T;
..
2a v b -=- + R Tln=--b v - + R T=--b v-
v
(~~t (~:t(~)T; (~:)T = (~:)s + (~~)p G~t which give
2a
-
_ J.. _ap* v Ov dv .
2a RT -v-
and U*(vdW) =
157
Non-electrolyte liquids and solutions
From the solution to Problem 1.24,
Alternatively, for U*, since Problem 1.9(a) gives
(~~)T
6.6
I
,
The corresponding extensive properties may be discussed in terms of G* , since it includes all three types of term: 2a/v 2an/v; the correspond ing extensive term must be linear in n, i.e. v b nb In=-- = RT--- = RT- v -b v- b v nb Thus 2an 2 v nb G*(vdW) = + nRTln-+ nRT- v v-n b v-nb
(b) The quantity f:(~~) do used in the argument of the introduc (~t(:it(~~)v 1, p) tion gives precisely the volume-dependent part of X. If, as with aH) (ap) (a -v +T ( av (C C there is no other part, the definition av aT
I
T -
and so
H{ =
T
f:[v(a:J)T +
v '
T(W)J dV.
f [(ax) -(aXav au =
id )
T
] dlJ
T
reduces to (Cp
-
C v );
(Cp
-
Cv )
Now
=
(Cp
vTa:~
Cp)t - (Cp
~ = -T
-
Cv)id .
(a/))2( aavp ) aT p
T'
but this involves differentiation at constant pressure, as with Cp itself. This is avoided bv substituting for (al)/aT)p according to
TC:;)v = 2a
(!~)p
(!i))( ~~)T : yielding 2a b -v + RT=--b C -T(!i):; (a:v) v b
= v2 - R T-(v---b-)-::"2 ,
H*(vdW)
v ),
t
ve:;)T = ~-RT[(v~b)2-~ J, ve:;)T+ RT[(V~b)2 -~J+R~V~b -~) and
p -
X*
For the van der Waals fluid,
2a
T
p -
'I
=
T .
158
)\
Chapter 6
For the van der Waals fluid, R
(:~)v
( oP)
v-b
aV
6.6
Cv
RT
T
2aJ-I
R2T [RT rv-b)2-,J3
= rv-b)2
For the ideal gas, yielding
.2a
+ -3 V
Y(r) = I
=R
[1-(#)r]'I2.
,
(b) Apply this expression for G*/RT to two van der Waals liquids, 0, I, at the same reduced temperature r, and thence prove
[2a rv -b)2]-1 1---=v"'"3R=-T=
Gj(T)
= flG~ T
,
where
(:~)v
(oP)
R
ov
v
(Cp -Cv )
=
h == Vel ~ I VcO
'T'
RT
fl
_1£!
=
T
'T'
leO
(c) Show that the analogous relations for the configurational enthalpy and the molar volume (at p = 0) are
R
and
(C;
159
where = _
yielding Cp
Non·electrolyte liquids and solutions
6.7
C:)(vdW)
LI -
R
Hi(T)
2a(v- b)2]-1
v 3 RT -R
[o(a1v)] - ---aT
* C (vdW) v
v
Solution
(a) When p = 0, v/(v - b) a/RTv. This equation is clearly quadratic in v, having two real positive roots for T less than some determinate value: the smaller represents the stable phase, for which (ap/av h > O. Substituting into the expression for G* given in Problem 6.6(a), for v/(v b)-since a/RTv is the simpler function-we obtain
0.
[Note: The methods explored in this problem are applicable to any explicit equation of state Hp, v, T)
G*(v T) /,
6.7 The van der Waals liquid (ii). The 'liquid' phase of the van der Waals fluid at a temperature below the boiling point may conveniently be specified by the condition p = 0, taking the smaller of the two positive roots in v. In this problem, since we shall deal in molar quantities throughout, we shall omit the bar signifying this. (a) Use the equation given in Problem 6.6(a) to express G*, the molar configurational Gibbs free energy for the van der Waals liquid at p = 0 as a function of T, the liquid molar volume VI> and the parameter a. Thence use the reduced variables
r
T
8a 27Rb'
Vc
is the smaller root of the quadratic:
[I (I _4RTb) a
On transforming into reduced variables, this becomes
a
VI
3b
[( I
)l J I - 8a/27Rb h
#T
or 9 16r
Ve
cf> = - [ I Vc
for the
or
a v/RT
3b,
(I
32 r)'I2]
'ft,
2 l-(l-#r)'iz
2
= Y(r)
Considering the other terms in G*(VI> T) we have
to show that for the van der Waals liquid G* RT=
v/
a ) = RT(-RTv/ -2a - + a -I +In- RTIJ/
v = _a / 2RT
cf> = -
where the critical critical molar volume van der Waals fluid. cf.Problem 1.11 (b), are
Te
where
v
Te ' temperature, Te, and
(f).
hIV/o(~).
_ [ 2a rv-b)2][ _ 2a rv-b)2JI - R v 3 RT I v 3 RT It is obvious that
flHti
2
a In-v/RT
InY(r)+ln2-1,
t
In2-1n[I-(l-~r)'I:t] == In2
InY(r).
160
Chapter 6
6.7
)
Then, substitution into the expression for G*(vj, T) gives the required result. (b) At temperature T, for substance I
GT
RT
2 yeT) -In Y(T)+ In2
r
f l G6
Y(T)-lnY(T)+ln2-1
, 'J
2a RTb H* = --+-v v-b
0, this reduces to H* RT
= 4e*
RT'
and the required result follows. From Problem 6.6(a), f0r the van der Waals liquid 2a RTv = --+ -RT v v-b
III
IJIRT
a
2
1)/RT
yeT)
f
\ I
*
2RT yeT) - RT ,
U*Cl\)
I
I~
and
m; ( IIT) = 2RT yeT)
f1
RT = H'(T) .
Similarly, since
= 1\ yeT)
we have VII(T) = v CI
vlO(~) = hIVIO(~)
VeO
9 Y(T), 16T
hi
=!:'..s.! VeO
(aaol)3
vlI(T).
! i,2.(rij) , j
cf Problem 6.4.] (b) Show that, at p = 0, the result of part (a) implies, for one mole, N=No, F,(T,No ) =
fIF6(~' No);
Gj(T, No)
f1G'tJ(
f,
No) ;
just as when the comparison is made in terms of a specific equation of state, cf Problem 6. 7(b), , . (c) Show that the corresponding relation for the Ft, defined as the sum of the translational and configurational parts of the molar Helmholtz free energy, is
I!TYCT)'
VCII!TY(T)
Tel _ eT
Teo - e6
[Hint: consider a particular configuration of the N molecules of species fl> ... , fN, in volume Vi. and take the configurational intrinsic energy in this configuration to be
, 2RT' , R*(T) = - - - R T o yeT)
1
I
i,
whence HdT)
h~Q~)(~, N),
where
.
>n,
m
then the critical temperature Te correlates directly(ll) with e*lk, and the critical volume v c with a 3 . (a) By considering the general form of the configurational integral, QN, show that if the form of (r) for two substances, I and 0, is as 0, above, then at p
Now, from part (a),
and
r(~ r-(;rJ,
Q}P(T,N) =
a ---1
161
6.8 The principle of corresponding states (i). The relations proven in Problem 6.7(b), (c) can be derived also by considering the configura tional integral itself(iO). Problems 6.4(c) and 6.S(d) exemplify the proposition that if (r) takes the form
2
RT
and at p
Non-electrolyte liquids and solutions
irrespective of its explicit form. The second provides a powerful method of testing whether two liquids do in fact follow the same reduced equation of state, at least in the region p - 0(9).
I.
For the reduced temperature to be the same, T = T/Tel T'ITeo, so we = Tlfl' must consider the reference substance, 0, at temperature Then
6.B
(T)
=fIFJ(~)-RT1nhl-!RT1nfl ~RTln~l
(9) A.J.B.Cruickshank and C.P.Hicks,
,
Disc.Faraday Soc., 49,106 (1970).
(10) K.S.Pitzer, J. Chem.Phys., 7, 583 (1939).
These two relations, together with that proven in part (b), are valid for any pair of liquids which follow the same reduced equation of state,
(11) J.S.Rowlinson,
pp.265-266.
Liquids and Liquid Mixtures, 2nd Edn. (Butterworths, London), 1960,
6.8
Chapter 6
162
where MI and p = 0,
( , i~
are the two molecular weights; thence prove that at
6.8
Ft(T,N)
=
r-
(a) Any pair of molecules of species 0 mutually distant r contribute $o(r) = 4€6 [(
to
:0 (:0r]
E'trs =
t'l
Similarly, a pair of molecules of species I mutually distant contribute
to Vr Then $I(radao) = II$o(r). Thus if the volumes occupied by No molecules of species 0 and species 1, respectively, are so related that in geometrically similar configurations the separations of all pairs of species 1 are v I / a 0 times the corresponding separations of all pairs of species 0, then for that configuration (A),
F' = -RTln.[ (
"l,1
this condition clearly implies that every dimension of the container of I is VI/VO times the corresponding dimension of the container of O. This hIVo, or Vo vI/h l . Thus, if the assembly of in turn implies VI species I is defined by T, I), N, and that of species 0 by Till' VIh I, N, then Vti(vlh l , A,N) Vi(v, A,N) kTlit kT and this will be true of every configuration, as long as we may assume a I : 1 correspondence between configurations in the two assemblies. Since QN has the dimensions of v N , it follows that in this case
(I) _ QN (v, T,N) $(r) and
2rrmkT)% v ] h2 N -NkT,
2rrmkT)'/' v ] h2 No
~(27rmkT)'!,V] L h2 ~
{ RT = -RT In
N
hi
rl = T/'Fcl
Q)..P(T,N) =
(O)(~!. ) QN hI' II ' N . = r2
= Till
hrQ~)(f,
so N).
I F1(T) = -RT { 1n$+ 1 + In [(2rrmkT)'hvcI]} h2 No '
and for substance 0 at the same reduced temperature, i.e. at Till' ,(T) Fo it =
RT{ [(2rrmokT)'hvco]} 7: In$+ I+ln 1h No ' 2
whence fiFo,(T) h
= -RT { In$+ I + In
Q}J)
hl>( Q~)
vN
---;;n-
Q~)
= (vlhd N
[(27rmlkT)'l:VCI]} h2 No + !RTIn mIll mo
+RTln~
(';
VcO
and I 3 mIll FI(T) = lIFor(T) h - zRTln mo - RTlnhi .
Now, from part (b), Ft(T)
IIFti(f) ,
and simple addition gives the required result. Differentiating, (aF' I av h, gives p Irs N kT/v, so G,
(b) It follows directly that
+ 1n$+ I} .
Then, for substance I at T, p = 0,
q
VnA) = II uti (A) ;
= 0, $
-NkTln [(
4€i[(:0)m (:or]
$1('::)
(f, N).
when Stirling's approximation is used for N!. Then, for one mole, denoting per mole by F',
v~.
At P
Q}J) IlkT QJS) =
-kTIn-;:v- = --y:ln(vlh)N
Changing N to No, with NokT = R T, the required result follows. (c) From the translational partition function,
Generalize these relations to the case p =1= O. Solution
163
whence
III!J(~).
(T)
Non-electrolyte liquids and solutions
V ] = -RT1n [( 2rrmkT)% 2 h
No
and the same argument gives , GI(T)
IIGo,(T):3. fi -"RTln mIll mo -RTInh l
•
6.8
Chapter 6
164
~ I
f We can now either differentiate the given expression with respect to T, = -st and obtain U t == Ft + TSt , or obtain S' == -caF'/aT)v
6.9
Non-electrolyte liquids and solutions
165
The rigorous derivation for C T is as follows:
(aFt laT)v
cl
as 2rrmkT)'12 v ] S' = Rln [ ( h 2 No +RT(r'iz x ~TY')+R,
Fl +lJIP = Fl +¢VCIP = Ft +h 1
Ct = Ft o
+ voP
Pot + ¢vcoP
kl
0
k
0
'
l
whence
whence TS' = RTln [( 2rr:2kT
v
J+ ~RT
= II f ~t0 + ¢VCOp/I kI
t
II C 0
=
'Pot
JI
0
+ h I ¢v coP
.
Now U'(T) = ~RT;
Ft == IIFJ -RTlnh l -
as H'(T) we have 3
J.RT
.:i RT Uo'( T) 21;7: h '
6.9 The principle of corresponding states (ii). The necessary condition for the relations proven in Problem 6.8 to be useful is that there exists (over the relevant range of T and v) an approximation to QA?) or Fti, which can be expressed by a Taylor series expansion about a datum v. Because QW) determines f7;, but not vice versa, we examine first the expansions for the configurational thermodynamic potentials. For simplicity, we shaH consider only the case P O. (a) Expand Hti(T/f) about T == () to obtain
and similarly for H'(T). Adding Ur(T) and Hr(T), respectively, gives the required result. (d) When p =1= 0, v is no longer a function only of T, but is either itself an independent variable, or a function of T, p, according to ~(p, v, T) = O. Then, to ensure I and 0 are at the same reduced volume, compare F!(T,v) and FJCT/II,v/h 1 ). Similarly compare CiCT,p) and CJ(T/II' p/k l ), where kl == Pcdpco. The procedure is exemplified by extending the van der Waals case, cf Problem 6. 7(b), as F*(T, v) = RT(-
v~T+ Inv ~b)
whence Fr(T, v) T
f6 ( h Thus t
Fl (T, v) Ci(T,p)
8:r + In¢~1) RT( 9 7: - 8¢r + In ¢ ¢ t ) .
' hiv) =
(T v) liFo h' hI t
RTlnh
l
~RTln/l
I\UJ(f,:J.
o
co
ml 2RTInmo
p,
tn TO'
T) rn co (-ey-n(arHti) = L--L I n=orn!r n (r-n)! aTr p,T 00
JI.* ( o
Ht ( -IT) 3
~-
where t == 1- T/el; and thence prove the alternative expansions: i)
=/Ic;r(f ' :J-RTlnhl-~RTln/l-!RTln::
_ t( IIT ' k\12-)
Ht) L (-e)n (an -aT" (IT) = non!
11.* -
8:r + In¢ ¢ t)
= RTC
(T, p) - IIHo (T,lJ)
RT(-
1
o and addition of hi ¢v coP to both sides converts this to the required result for ct.
~RT
U'(T)+ p'v == U'(T)+ RT
Tln/l-!RTln:
0
Tn (anHti) L= 01-n-, --n. n. aT p, T = 0 co
n
(b) Use the Gibbs-Helmholtz relation, C = H+ T(ac/aT)p, to show that for n > 2, anc) ( - aTn p,T=O
(
I )n-I (n ()
--
n-I(-eY-I(arH) 2)' L . . 'r= I (r l)! aT' p,T=O'
and thence prove
f
C* (I) = R*«()+I(ac.w/kT)
QxNQ(I-x)N
Nl!N2!
I!
Ix and, since UM UM
the mean, and remembering that there are N! arrangements,
F~d)
whence
= RT[xlnx+(l
orU ») v
Y
Yw =
since fi. but U6'(O) this approximation
v'
since R Tin E 11 obtain
x), and the mean number of z(i - x), so the total number is
)Nz+ !x 2Nz+ !Nz - xNz
Y = x(l of temperature, the differential equation of part (b) satisfied by
and the required results follow immediately.
+ TC:o(O)lnEj+ F~d)
0 for this process.
,
Neglecting the second term we
1
of I, I interactions is ! zxN I , and x)N2 • The total number is then
x)N
xf,- (I - x)fi 1
= !zNoEo in the regular solution model, so to
FM = US(O)f E
U
Y = Y = x(l
X)2 fi ,
UM = x(1 - x)Now for one mole of solution. For the Helmholtz free energy, isochoric process, it is clear from Problem 6.9(c) that
+!x 2 Nz
is
x 2 fl + 2x(l - x )f12 + (l
1 - x)w/!ZEo ,
- (dY)
(x
x)ji1·
I - x) - f2(1- x)x + 2fl2X(l - x)j
Y T dT .
x 2 )Nz+ !x 2Nz+ !(l- x)2Nz
t(f),
X)f2 U
+ 2x(l- X)fI2+ (I
x)ln(i-x)1,
OYW) -T (aT dYW) T ( dT
(c) In the case of random mixing ex I, 2 interactions per molecule of 1 is Nl z( 1 x) = x( I - x)Nz. The number the number of 2,2 interactions is !z(1
and
u~(l) -xfl ut(~) - (1
fx
-FrS/d) ,
Yw, so that Yw
(x
= fx
then
d
-T ( aT
f~ U6(0),
U~,
Iffx is taken as
Yw/kT)N! NI!N2!
and the required result for FM follows at once.
But
U6{X)
U6'(OHfx -xfl-(l
we obtain QxNQ(1-x)N
181
Non-electrolyte liquids and solutions
,
f \
6.13 The regular solution (ii). In the model of Problem 6.12, instead of random mixing, the quasi-chemical equilibrium: (1, I) + (2, 2)
~
2(1, 2) ,
for which the energy increase per molecular unit of reaction is clearly = 2w/z. The equilibrium 'concentrations' of the three types of interaction are assumed to accord with a quasi-chemical equilibrium constant K, defined by
flu
flF
==
-NokTinK ,
where fll" is the Helmholtz free energy change per molar unit of
6.13
Chapter 6
182
Non-electrolyte liquids and solutions
6.13
reaction, or, equivalently, by
holds at T = 00. Is this last assertion necessarily valid? Suggest an alternative way of evaluating FM which avoids the logical difficulties associated with the reference temperature T = 00.
-t::.U) (-t::.U) K == nexp ( NokT = nexp kT ' where t::.U, the intrinsic energy change per molar unit of reaction, is given by the usual relation
Solution
(a) From Problem 6.12(a), the numbers of interactions are
t::.U = t::.F-T ( dt::.F) dT ' and n is the change in the degeneracy factor such that
2,2:
!z(N2 - Y); zY,
4Y2 (Nl - Y)(N2 - y)
(
= nexp -
2W) zkT .
For a I, I interaction, there is clearly only one distinguishable arrange ment, and likewise for a 2, 2 interaction; but two I, 2 interactions may be achieved in four ways. This is confirmed by the fact that for w/z kT = 0 we must have y2 = (N l - Y)(N2 - Y) to conform to random mixing, Y = N l N 2/(N l + N 2 ) = x(1- x)N. On putting exp(-2w/zkT) = 1/7'/2, the equilibrium relation becomes I y2 (N l - y)(N2 - y) = 7'/2 giving (N l - Y)(N2 - Y) = 7'/2 y2 .
[C 12 F [C l l ][C22 ]
(a) Express the concentrations of the different types of pair inter actions in terms of z Y [see Problem 6.12(a)], the number of I, 2 interactions, and show that the equilibrium value of Yobeys (Nl - Y)(N2-Y)-7'/2y2
1Z(Nl- Y);
and the concentrations are given by dividing each number by v. Then
K is equal to the ratio of the product of the concentrations of the reaction-product species (each raised to the power equal to its numerical coefficient in the reaction equation) to the corresponding product of the concentrations of the reactants. Thus, denoting the equilibrium concen tration Cji , we have for the above reaction
=
I, I:
1,2:
kTlnn == Tt::.S.
K
183
=0
where 7'/ == exp(w/zkT). If UM be written 2 UM == ((3+ l)x(1- x)Nw ,
Putting Y
= 2x(1 -
x)N/((3+ 1), so that,
- = N rLX - 2x(1X)] (3+ I ' - r 2X(1-X)] N2 - Y = N LI - X (3+ I '
Nl - Y
so that UM
-+
x(1-x)Nw as (3
(3
-+
I, prove that
= [I +4(7'/2_I)x(1-x)]Y'
.
(b) Since (3 js a fU!lction of T, Y is also a function of T, and the relation Y = Y - T(dY/dTlcannot be solved for Y directly. Verify the alternative (1 5) form Y = d(Y/T)/d(1/T) and show that substituting for Y and for d(1 / T) in terms of (3, and integrating between (3 = I and (3 = (3 gives for FM _
(id)
1
_
FM -FM +"J.zNokT(1
4x 2(1 - X)2
_
y2
=
N2
((3+ 1)2
,
the equilibrium relation becomes N2
r (3 + I - 2x (3 - I + 2x] x) ln L(1-x)((3+1)+xln x((3+I) ,
((3+ 1)2[((3+ 1)x-2x(1-x)][((3+ 1)(1-x)-2x(1-x)] 47'/ 2x 2(1 - x)2N2
,I
when Y((3 = 1) i.§.. put equal to zero. (c) Show that Y((3 = I) = 0 is the necessary condition that
-
-
((3+ 1)2 which reduces to
dY
(32-4(7'/2_1)x(1-x)-1
Y = Y-T
dT
and the required result follows.
(15) E.A.Guggenheim, Mixtures (Clarendon Press, Oxford), 1952, pp.38, 39.
v
\
=
0
184
Chapter 6
6.13
6.14
(b) The required form follows exactly the standard alternative form of the Gibbs-Helmholtz relation: H = [3( G/T)I 30IT)]p' Thus -
dY T dT =
y=
dY
jd(1/T)
Td(1/T),~
1 dY Y+Td(1/T)
T(dY/dT) at T
(32
4(17 2 -1)x(l-x)- I
d{3 dT
= 0
dY
132_ (1- 2X)2 17 = 4x(l x)
T dT
2
2w zkT = In[j32
When (3 (1 - 2x)2 d(3 .
df
zNkx(1
2w
X){
(l-X Xl)
Integrating between 13 I and (3 13 we obtain = zNkT Y = ~[(1- x)ln(j3+ 1- 2x)+ xln(j3- 1+ 2x)-ln«(3+ 1)]1 + Y«(3 = 1) .
r
=
=
Y - Y( 13
l)x(l
x)+ I .
=
I)
X)N(
13
I, then T ==
00,
1
I
x
{3+ I
I)
x 2w 2x + {3- 1 + 2x - 13+ I exp zkT '
exp(2w/zkT)
==
Y({3 = 1)-
I, and
(3
=
_
1)- Y«(3
1).
Thus.llt T =_00, Y({3 = 1) = 0 is the only assumption which satisfies Y Y - T(dYldT). In fact, to assume that the general relation [3(AG/T)/3(l/T)]p = AH holds at T 00 implies categorically that AG(T = 00) = O. This is to assert that AG/T remains differentiable with respect to I/T at liT == O. There is no simple way to establish the validity of this assertion. It can be avoided, in principle, by_choosing a reference temperature T = e, other than T 00. Then Y must be evaluated at T = e through the ratio QN/QXNQ(1-X)N' Various approxi mations have been used on this problem, and it has been proven(16) that the result of part (b) is obtained by assuming that all pairwise interactions are mutually independent. Higher-order approximations give slightly different results; the algebra is too tedious for inclusion here; the methods are described in the reference cited. The result of part (b) is important because it suggests a temperature dependence of FM - F~d) which is different from that of UM , as is usually observed when experimental results are converted to give FM (isochoric) and UM (iso choric). Note the qualitative similarity to the inclusion of the term TC~o(O) In E f in the corresponding-states formula [cf Problem 6. 12(a) ].
4j3dj3 } [13+ l][(3L (1- 2X)2 J .
Separating the integrand on the right-hand side into partial fractions, we find 4(3 [(3+ l][j3+(l-2x)][j3-(l-2x)] I x I ) x(1 x) 13+ 1 2x + - 13+ 1 whence Y ZNk( 1 x d T = 2w (3+ I - 2x + 13- 1 + 2x - 13+ I d(3.
=~L(l-x)ln
4(17 L 2
dY) ( T dT /3=
Substituting for Yand for d( liT),
zNkT
=
2x)2]-ln4x(1-x) ,
2(3
(32
) d{3
2x(l - x) (d17 ) = 4x(1 - x) (_~) 2w 13 dT (3 zkT2 exp zkT '
-
whence 2w zk d(l/T)
I
x
1)+~ {3+1-2x+(3-1+2x {3+1 dT'
2x(1 (1
I- x
ZNkT2(
=
Y-Y({3
Hence
and, from part (a),
Y
From the result of part (b)
(32
'
giving
00.
185
From the result of part (a)
y == 2x(1 x)N
13+ I
=
dY TdT
d(Y/T)
=
Non·electrolyte liquids and solutions
6.14 Solutions of chain mo1ecules (i). Problem 6.10 shows that the ideal solution formulae may be derived from the simple cell model. For the case Val = tJ a2 Vax the same result is obtained from the models of Problems 6.2 and 6.3. There is thus some justification for using the simple cell model to derive the thermodynamic functions of mixing for
(3 + I - 2x (3 - I + 2x ] = I-x +xln x -In((3+I) +Y(j3= 1).
The given result follows at once when Y«(3 I) is put equal to zero. 1 corresponds to 17 2 I, 2w/zkT = 0, i.e. T 00, in the (c) (3 general case w =1= O. The simplest way to proceed is to evaluate
(16) E.A.Guggenheirn, Mixtures (Clarendon Press, Oxford), 1952, pp.42-47. I
1
186
Chapter 6
6.14
I
for the first segment, there are
solutions of non-interacting chain molecules. The essential assumption is that the molecules of both species are chains whose links, or segments, can each, interchangeably, occupy a single cell. The simplest case is a solution of 'monomer' (species 1) and 'r-mer' (species 2), the molecules of the latter each occupying a chain of r cells. The total number of cells is then N NI + rN2 • As in Problem 6.1 (b) and Problem 6.2, QN
z z - I (N - rn)
=
=2:A
(_Z_)(I 0, T 1 (7.1.5) ->0. Co ' vKT Solution
(a) The Taylor expansion for U/3 is
=
LlS +
(au) au
sLlv+
(a iJ)
/Lll)2.
and similarly for Ua., but with oS, ov, instead of LlS, Ll,). The net change in U (total) is o~)Ua. + o~l!a -
(~¥)v[(n - O~)OS+ o~LlS] + (~~)sr(n - OnOl) + o~Ll/)]
(a U) iJ.r 0 ,
as v (au)
== 0;
Clearly the first two terms in OU are zero; using Equations C7.1.6) 1.8) as appropriate on the last three tenns gives
2
aV 2
onSv+o~Llv
whence
1.I)
(a U) ( a U)
195
Phase stability, co-existence, and criticality
S
+
(~~) vG~)
J
and the change-of-constraint formula
(~~)v [(~)
S
+
(~~) v(~~)J
gives the required result.
7.1
Chapter 7
196
7.2
Phase stability, co·existence, and "riti""lit"
197
An alternative procedure is apparent on writing condition (7.I .1) as
02U) U2S == ( OS2 v'
>0,
(7.2.3) or
Continue via the Jacobian, as
U2S
I Uvs
Usvl U2v
=O(T,-P)/O(S,IJ)=_(OT) /(Ov)
o(S, p) o(S, p) oS p op s
or equivalently,
oeUv, Us) o(-p, T)/O(I!, S) (Op) /(OS) O ( I I , S ) - O ( v , T) O(V,T) = - a;; T oT v' Note that the phase stability conditions (7.1.4) or (7.1.5) may be derived by considering the case where the total intrinsic er,ergy and the total volume are kept constant(2). The results (7.1.4) or (7.1.5), whose equivalence reflects the identity KslKT = CvlCp , indicate that in a one-component system diffusional stability is secured by the conditions for thermal stability (TICv > 0) and for mechanical stability (III) KT > 0). It will be seen (Problem 7.7) that this is not so for a two-component system, an additional condition of diffusional stability being required.
Solution
(a) Proceeding as in the solution to Problem 7.1 (a), write the Taylor series for and Fi3 about FJO), including terms up to (04FloV 4 )T' Then
ljF
(n
o~)Fa. + lj~p;, - n/?JO) ,
7.2 Higher order stability conditions. For simplicity we consider uniform systems in which one intensive variable is, additionally, held constant. The appropriate forms of the equilibrium condition are, cf.preamble to Problem 1.22,
( ~~) t;
> 0, T, v
F == U- TS, or
(OO~) > 0, S, p
H
U + PI) .
It is necessary also to validate the first-order stability conditions thus
by comparison with conditions (7.1.4) and (7.1.5). Prove that the first-order stability condition for a one-component, one-phase system constrained to constant volume and uniform and constant temperature is
02F) ( -ov 1 T, n
- (OP) (1)
I
T,n
liKT
and that the condition for stability when IIIJKT
02p) =0, ( ov 2 T, n
03 p ) ( ov 3
>0
(7.2.1)
I (02F) -lj~)O/)+O~d/J]+2! av 2 T[(n
I(a F) I(a F) + 4! 'av 3
+
a/J3
4
(7.2.2)
T, n
(~~)T
F)
(aaiJF) a F)
( aiJ
." .
4
0,
1
2
no~
2!(dV) n-lj~
T
I 3 n2lj~ ( 3!(Llv) (n-lj~)2 I
T
I n3lj~ [ 4! (dv) (n _lj~)3 1
3
3
I
T
4
(2) l.Prigogine and R.Defay, Chemical Thermodynamics (trs.D.H.Everett), (Longmans, London), 1954, pp.209-212.
-lj~)(O/J)4+ O~(d/! )4]+
T
Using the relations (7, 1.7), together with the corresponding higher-order relations for increasing powers of dv gives the coefficients of the derivatives of FJO), leaving off the redundant indices and suffixes, as
2
0 is clearly either (7.2.1) or, if (a 2F/aV 2)T = 0, since (LlV)3 c~n take either sign, (7.2.2). In the general case, the condition for stability is that the first non-vanishing derivative of p must be (i) of odd order, (ii) negative. (b) The proof is exactly analogous to that of part (a). (c) The various sets of conditions are related as follows. That either condition (7.1.4) or condition (7.1.5) ensure all four of I/KT' I/Ks, T/C p , T/C v positive is obvious from the relations cited. Note that condition (7.1.1) alone does not ensure all four parameters positive; nor does I/Ks, I/KT > 0, etc. The relation between condition (7.2.1) and conditions (7.1.1), (7.1.2), (7.1.3) is obvious on writing I T/CvvKT a 2F) (7.2.5) ( av 2 T = vKT = T/C v > 0 .
whence
7.3
n
Sv F 2v U2S = U2S U2V - UvS UvS = U2S U I . I USv U2V (d) The conditions (7.1.4), or (7.1.5), equivalently imply that all four of the reciprocals of Cp , Cv , KT , Ks are positive; (ap/aT)v, n being positive and non-singular implies that (Xp has the same sign as KT , and has the same zeros and/or infinite discontinuities. Then, in virtue of the relations Cp = Cv + T(X~v/ KT and KT = Ks + T(X~v/Cp, a stable phase has Cp > Cv > 0, T/C v > T/C p > 0; KT > Ks > 0, l/vKs > l/vKT > O. This suggests that in the absence of an infinite discontinuity in T/C v , stability is likely to fail first by T/Cp , I /v KT going through zero simultaneously. If Cp , KT go through zero, however, then Cv , Ks approach zero faster. (e) In the van der Waals equation, using molar quantities,
ap ) (aT
v =
R v- b
If v = 3b, T
0
for all v
(aavp ) T
>b.
= -
RT 2a (v - b)2 + -;?
8a/27 b, we have for the uniform-density state (3)
ap ) (-ao T >0,
I -K
V
T
Cv > 0 for all T, v > 0, i.e. T/C v > O. Further, Cp - Cv < 0 at the point specified, since I /v KT < 0, so Cp may have either sign, and Ks has the sign opposite to that of Cp . Along this isotherm, (ap/aV)T has two zeros which, with C v > 0 everywhere, necessarily define stability limits. Thus the stability condition which fails, at the so-called spinodal locus, is T/C p > 0, l/vK T > 0, and it fails by passing through zero; if Cp is positive anywhere in the unstable region, the part of that region within which Cp > 0 is bounded by zeros of Cp , the locus coinciding with zeros of Ks. It is the fact that phase stability normally breaks down by T/Cp , 1/ vK T passing through zero simultaneously, which justifies using conditions (7.2.1) or (7.2.3)-usually the former-rather than the complete conditions (7.1.4) or (7.1.5) when searching for critical points or for spinodal loci (see following problems). 00
7.3 Co-existence of two phases (i). A one-component system, con strained to constant total entropy, volume, and mass, is supposed to comprise two phases, (X and (3, in thermal, mechanical, and diffusional contact. The phases are both internally uniform, but they may differ in (3) That this state has a higher free energy than the two-phase state may be deduced from the results of Problem 6.6(a). For an analogous problem in statistical mechanics see Problems 11.14 and 11.15.
7.3
Chapter 7
200
T, p, and IJ.. The perturbation to be considered is the transfer of oS, Oli,
7.3
201
Phase stability, co-existence, and criticality
where
on (extensive quantities) from phase (3 to phase ex.
(aaSU)v,n'
(a2F) av T.n'
(a G)2 2
This is equivalent to a generalization of Problem 1.29. The general condition for equilibrium, as in the preceding problems, is (0 U)s, v > o. (a) Taking S"-, Ii a., and na. as independent variables, express Ua. and by Taylor series expansions about U!O) and 0°), and thence show conditions for equilibrium, Le. for U Ua. + UR to be an extremum, are (4) Ta. = 1/3, Pa. Pp, 1J.a. 1J.{3; (7.3.1)
(d) Use the same general method as in part (b) to prove that the conditions for stability under constant and uniform temperature,
and that those for stability, i.e. for U to be a minimum rather than merely an extremum, are
are secured by
U;s
ifsv
ifsn
U:;s
U;v
U:;n
u:.s
ifnv
U;n
I U;s U:;s
ifsv U;v
I >0 ;
U;s=
eu.) 2 ast 0;
2
v, n
+
Va.
I > 0;
IJ.
(7.3.3)
U)
+ (a~ (3
I)
(
~~) s,
=
pl"TP
, Fvn Uw F2v Fnv F2n
I
v. n
,
etc.
p
(
~~)
T, v
(
~~
G ) T, p
(:~)
I=
>0.
'
p
(~~)T
=
n
0, that
= V ;
and that the molar entropy and volume may be defined also by
s = (~~t. p' (7.3.5)
U2sF2vG2n = 0,
(7.3.6)
av) ( an T, p
v
,
with
U= Solution
(a) Since in OU are:
oU =
That the co-existence condition p. = p~ may be deduced equivalently by maximizing the canonical partition function for non-uniform density at T < 7;, is seen from Problem 11.14. A related analysis using the grand partition function establishes also that,... ,..~.
(4)
(~~)s,
v
(7.3.4)
The results of Problem 7.I(c) are useful. (c) To prove that condition (7.3.2) also is secured by condition (7.3 .5) is both formidable and tedious, but the method is similar to that of part (b). Prove the following general proposition, which is a pre requisite for proving that condition (7.3.2) is secured by condition (7.3.5),
U2S Us"
UvS U2v Uvn UnS Uno U2n
etc.
T,p'
> O.
V{3
also, it follows from the Gibbs equation, SdT- II dp + ndlJ.
> 0;
TIT -C' > 0; v -;;va.
an
2n
lla.KTa. '
(7.3.2)
G2n =
2
Fv~ I > 0, F~+n > 0 , F+
F;v
IFnv
(b) Prove that the conditions (7.3.4) and (7.3.3) are secured by I ' K T,,-
F2v =
2
Note that it follows from Problem 1.20(b) that
U;s where
2
U2S =
oSp
-OS,,-,
ua.) v,n [(-aasa.
oVp
=
(~~
p
-ov"-, onp
=
-on,,-, the first-order terms
(au) ]oS a. + [(aua.) ]00 " ~ - (au.) ~ asp v,n _ S,n allp S.n aV,,-
+
f,(aua.) v - (~) lona. + .... anp s.
Lana. s,
Clearly, conditions (7.3.1) are the necessary and ensure 0 U 0 to the first order.
conditions to
202
7.3
Chapter 7
When these conditions are met, I
ou = "2
[(aZu,,-) as~
+ as~
(32v CX2SCXlv", ..... 20
(oS,,
v,
(oS,.01J,.) I
+2
I·
+"2l(
n
aU,. an~
(a~ anffi 1
s,
II
+
0
)]
s,
II
2
na.)
(
+
I~~( a2 ua.
~
ana.as,.
CX2v
+ (32S(32va1'20
+ «(32S(32V
_
2 CXS v _ (3sva I'Sv
2 CX2v (3sv) a I'2v
2
+ CXSv
((32V _ (3SV) CX
2v
CX
Sv
2
(CX2V _ CX s v )
+ (3sv a
I'2v
a
I'Sv
.
(7.3.7)
aS2"
CX2S,
"
~
asp a,) p
(3sv ,
we can write the determinant (7.3.3)
= (CX2S+
+ (32V)
(cxs v + (3sv)2 •
out and collecting like terms we obtain cx~v + (32S(32V - (3~v + CX 2S(32V
cxs v(3sv
+ CX2v(32S -
cxs v (3sv •
According to the solution to Problem 7.1 (c), the first four terms become
T T --,:-:----- + -=--=- v a.KrfL C v"
vpKrpCvp
CX2V)2 ( CXSv - (3SvR I'2v
(32 v a' 2v
( 1 + x)A + ( I +
~ )B + xC 2 ,
Thus D has the form D
Arraying the coefficients in the expression (7.3.7) so that the three 'even' coefficients, i.e. those of (OS,,-)2, (8V,,-)2, and (on,.)2, are diagonal, gives the determinant on the left hand side of inequality (7.3 .2). Since 0 U is to be positive for all possible combinations of oS,., 81)", on", it must be zero in particular for any two of them being zero, so that the three diagonal elements must be positive; when anyone is zero Equation (7.3.7) reduces to a simple quadratic, and the argument of Problem 7.l(b) thus adds that the 2 x 2 minors of the determinant shall also be positive. Since the determinant comprises the totality of the coefficients in Equation (7.3.7), it follows that the conditions (7.3.2), (7.3.3), and (7.3.4) are both necessary and sufficient. (b) Using the notation
a2 u a2 u,
CX1SCX 2v -
2 (32 v CXS v )::;;""20
v
+ terms in (OS,.)3 , etc.
D =
( CX 2SCX 2v
2 (3sv CXSv", .....So
)
+ (~) anpasp v ](On,.oSfL)
D
203
2
+( a~:~JJ (olJ,.on fL ) )
Phase stability, cO'existence, and criticality
The last two terms rearrange to the quadratic form
I' aZU,.) ,.) + I av ,. an ,. S
+
Z
7.3
The last four are partially separable as
(aZup)
v, n
11
A
K
T
C
,etc.,
_ v(JKsf3 K
x -
r,. v" Va. S" Evidently D > 0 is secured by x > 0, [(1 + x)A + (1 +1/x)B] > O. According as x ;:s; I, however, the latter condition may be met with either A or B < 0, and this is not pre-empted by condition (7.3.4), T/Cv,,- + T/C vp > O. Thus the conditions (7.3.3) and (7.3.4) are secured by the condition (7.3.5), though not by (T/v a. Kr"C v,. + T/v pKrpCvp ), > 0; but conditions (7.3.3) and (7.3.4) do not require (T/C v ,.+ as necessary. The further inequalities from condition cond!tion (7.3.2) also are secured by condition (7.3.5). Note that condition (7.3.2) includes terms in I)"
a2 U) - - I ( 1+---4- Scxn T) (avan S - nKs Cp ' 2 a ( u)
S. v
-;(~+ TS2).
n
Kr
Cv
A two-way perturbation always leads to a condition like (7.3.3), implying through its cross-terms that some property has the same sign for cx and (3, whereas the diagonal terms of condition (7.3.2) being positive implies only that the sum for cx and (3 is positive. This is because a determinant of sums is not equal to the sum of the corresponding determinants for cx and (3 separately. (c) The form in which the problem is stated suggests working from right to left in Equations (7.3.6). The method suggested by the solution to Problem 7.2(c) is to use the change-of-constraint formula. To prove (a2G/an2>r,p 0, start with the definition G == U- TS+pv, whence given the Gibbs equation [Problem 1.20(d)] for one Mtnl">Af'lpn1 SdT-v dp+ ndJ.l. 0, we have dG
-SdT+vdp+J.l.dn
204
7.3
Chapter 7
whence
( OG) on T.p =IA,
(02G) on 2 T,p
=
(01A) on T,p;
but dIA -SdT+ 15 (dlA)T, pO. 2 To relate (02G/on )T, p to second derivatives of F, write
( 02G) on 2 T, p
(01A) on T, p
(01A) (01A) (ov) -on T, v+ & T, n on T, p ,
7.3
Phase stability, co-existence, and criticality
Forming the determinant in F and multiplying through by (0 2 U/oS 2 )v, n gives the required result. The result (7.3.6) has the following implication in regard to expression (7.3.7). If the latter is separated into two quadratic forms, one for each phase, then the corresponding pair of 3 x 3 determinants are each zero by (7.3.6). This means that for each phase there is a particular combination of oS, (1), on (such that oS/on S, ol)/on = Ii) for which the quadratic form has two coincident real roots. Thus, provided that the principal minors of the determinant for that phase are positive, Le.
(02F) (01A) ( 02F) or12 T, v + ovon T op T, n ' (01A) l(oP) 2 T,V+ (02F) ( 02F) =(jn ooon T OV T,n & T,n' 02F) = ( On 2 T,v
(02F)2j(02F)· ol)on T 00 2 T,n'
and the required result follows on multiplying through by (02F/OI)2)T, no Alternatively, use the method of Jacobians,cfsolution to Problem 7.1(c),
F2n Fvn I
f~vl
F2v
o(IA,-ph O(IA,-p)/o(n,l) (01A) /(OV) o(n,t)T = o(n, -p) o(n, -p) = - on T, p op T, n 02G) (02F)
= ( on 2 T,p Ot)2 T,n'
The third step necessitates expressing the three second derivatives of = F(ll, n) at constant temperature in terms of the second derivatives of U = U(S, P, n). The above procedure gives
U2S Usvl Uvs U2v
I
> 0,
U2S
v-
(02U) /(02U) oSon v OS2 v, n '
and from Problem 7.1(c) we obtain
Similarly
( 02F) Ot) 2 T, n
(02U)
S, n -
(02U)2/(02U) oSOV n OS2 v, n .
2 (01A) (01A) (01A) (OS) 0 F) ( Of) on T = OV T = OV S + oS v, n a;; T, n 02U) (02U) (op) = ( ovon s + oSon v oT v,n =
02U) (02U)(Op) (OS) ( ovon s + oSon v oS v,n oT v,n 02U) (02U) (02U) /(02U) ( ovon s - oSon v ovoS n OS2 v,n'
> 0,
the quadratic form for that contribution to 0 U is everywhere positive or zero. These are the conditions (7.1.1)-(7.1.3) secured by conditions 1.4) or (7.1.5). The same applies to the other phase; but unless the two phases are identical, there is no combination of oS, ov, on for which OU overall is zero, since non-identity implies oS/on = Sa. S/3, etc. Thus condition (7.1.4) or (7.1. 5) for each phase secures overall stability. (d) Using the notation
*
2
Fa.) = a , 2n ( 0on! T, v we can write the determinant
02p, ) ( 01J/30~/3 T
(a2n+{32n)(a2v+{32V) (a vn +{3vnf
= a2na2v+{32n{32v
F
(02U) ( 02F) on 2 T, v = on 2 S,
205
{3v. n, etc., a~n-{3~n
+a2v{32n+ a2n{32v- 2a vn {3vn . The first four terms are zero by Equation (7.3.6). That equation enables the last three terms to be put into quadratic form as:
(avn{32n - {3vn a 2n)2 a 2n {3 2» This is positive if, and only if (02Fa./on;,)T, v and (02f~/on3h, v have the 2 {32n a vn ",....2n
+ {32
_ 2
(01A)
(011) T, n on T, p
vn{3 2n
{3
a vn vn
=
same sign. Now
( 02F) on 2 T, v = cfpart (c), so
(1)
1")
02 ( on 2 T,
(01A) (on) (oP) - op T, n on T, p on T, n' v-
v =
vKT
v
n2KT
Thus the determinant is positive if, and only if, KTa. and KT/3 have the same sign, negative volume being excluded, and the other condition
02F) + (02P,) ~ >0 (_a. on~ T, v on! T, v
206
Chapter 7
'.
7.3
requires that sign to be positive. Compare this result with that of Problem 1.30; again, a two-way perturbation requires for stability that the two phases or fluids be individually stable, whereas a one-way perturbation requires only that they be collectively stable. 7.4 Co-existence of two phases (U). Reconsider the system of Problem 7.3, but use molar rather than extensive properties as the variables. The perturbation may then be described in terms similar to those of Problem 7.1, i.e. as o~ mole passing from phase (X to phase {3, with consequent changes in the molar entropies and volumes of the phases imposed by the constraints to constant total entropy and volume. (a) Show that (n«-o~)oS(t+(np+o~)oS(3-(S£O))o~ = 0, (7.4.1) (n(t
O~)ov",+(np+o~)ov(3-(v~O)-vJO»o~
OU
n",(U(t -
U~O»+
np(Up -
~O»+
oHU(3 -
0,
7.4
where the constraints on the partial derivatives refer to the molar quantities for phase (X; and similarly for . Substituting in 0 U we obtain
-o~ [u,(0) + (~) '" as",
r
-
oU = (n ... - o~)U'" + (n{3 + o~)Up- n... mO) - n(3~O) . Note that if n(3 -l" 0, and ~O) -l" U£O), this reduces to the corresponding relation of Problem 7.I(a). The Taylor expansion for
-(0) ) (au... ) u.. _-u. . + (au... as... F OS ... + av,,-sOv..
Then, by virtue of Equations (7.4.1), (7.4.2),
OU
[(n", -
o~)oS", - S~O)o~{e~:) F - (~~)
+[(n", - onov", -v~O)o~] [(~~:)s -
(a 2v.. ) +2 as! F(OS",) + 1
as... av",
~SbO) + pp
(UJO)- T",S~O)+ p",v~O»]
°
(:~) (~~)p (:;) v
T
-I
to obtain [.1 U- TS+ pv. Thus the third condition is [.1", = [.1p. (c) The second-order terms in 0 U are
2-
...) (0- )2+ ... os.. ov ... + ~(a2U 2 av! s 1)... ,
(~)s J
+second-order terms. (7.4.4) Clearly, for OU = (to first order) T", = ~ and p", ~pp. _Now G == U- TS+ pv so, on dividing by n, we obtain [.1 = U- TS+ pv; Alternatively, take (au/an)s v = [.1, change variable twice to give (au/an)r,p = U, and use .
-
2 2 (a u... )
(au.)
(all)
U",) _ (0£0) i +v~O) ( au",) av", s - v&O} (aUp) avp "8] + second-order terms.
a 2u'" (n",-0~)L2(aS!)/oS",)
U.. is
(a~",) Of) J a!) '" s '"
+ o~ rU!O) - uJO) + S~O) (a S&O) as", F as(3
+ o~[ (uJO)
Rearrangement gives the required result, and similarly for 00 = 0. Since U... = U... (S... , v.. ),
'"
+[(np+o~)oS(3+SbO)o~] ~ ii+[(n(3+0nOI)(3+V&0)o~] ~ S
Solution
°
+
oS if
+o~ ~O)+ (~~)OS{3 + (~~)sOVIlJ + second-order terms = [(n", - o~)oS'" -SiO)o~] (~~:)ii + [(n", - onov", -~O)O~](~~:)s
(7.4.3)
_ (a) !nitial state: S = n",S!;.O) + n(3S~O); perturbed state: SIt = S~O)+ oS", Sp = S!O) + oS(3, nIt ot etc., and this specification is complete, since oS"" O!)(t' o~ are independent variables. Then oS = (n ... - o~)(S~O)+ oS(t)+ (np + o~)(SbO) + oSp)- n... S~O)- npSbO) .
[(~) as(3 i OSp + (~) aV(3 O!)(3Jl
J
i
oU -- n... [(au«) as... OS ",+ (au... av", ) sO!) ... +np
(7.4.2)
where the indices and suffixes have the same meanings as in Problem 7.1, with the incidental difference that neither oSp nor OS(t necessarily tends to zero as S~ tends to zero ( and simila~ for the 01) t;.l 01) (31 (b) Use Taylor series expansions for U(t' U(3 about U~o>, UJO) respectively to confirm that the conditions for equilibrium are (7.3,1). (c) Thence show that thermodynamic stability is secured by the condition that each phase be stable by the criteria (7.1.4) or (7.1.5), (d) Give a physical interpretation of condition (7.3.1) in terms of the tangent planes to the surfaces U'" = U(t(S(t' v",), Up = U(3(S(3' vp). What is the maximum number of phases which can, in general, co-exist in a one-component system?
207
Phase stability, co-existence, and criticality
+
2
J
a 2U'" ) 1 ( a 2-) U'" 2 + (as",av« oS",ov"'+2 av", 8(0/) ... ) 2~
( 2-)
~ lau(3 o~) [2lau, (a:sl)F(OS(3) + (aspavJOSpOV(3 + 2 avp 2
/Ol)p)
J
2
(7.4.5)
The simplest condition ensuring that this sum is positive is that both square brackets be positive, i.e. that each phase fulfils (7.104). Since 5S(3' 5v (3, are dependent variables, it follows that while this condition is sufficient, it may not be necessary. Substituting for 5S(3 and 5v (3 according to (704.1) and (704.2) does not reduce the second-order terms to a single bracket, except if the two phases are identical. Thus the conditions which may be formulated in terms of sums
1
[(o;f£) + e;~~)J ' etc.,
are not sufficient. (d) The stability conditions (7.104), (7.1.5) ensure that for a stable
phase the U U(S, v) suiface is concave upwards. For two co-existing
phases, the equili~!ium c0I!.ditions Ta..:::' 1{J, Pa. ;= P(3 ensure that the
tangent planes to Ua., U(3 at Sa., Va. and S(3' v(3 respectively are parallel; it
is evident from the third term of expression (70404) that 11", = 11(3 ensures
additionally that the tangent planes are identical. Thus two phases can
co-exist in a one-component system over a range of temperature, but
fixing the co-existence temperature fixes the co-existence pressure.
There is in general one, and only one, way of putting a common tangent
plane to three uniformly curved surfaces. Thus three phases can co exist at a determinate temperature and pressure. It might seem that four phases could co-exist fortuitously, but since the tangency must be geometrically exact, this occurrence has zero probability. 7.5 Criticality, continuous equation of state. The region of liquid vapour co-existence of real fluids is invariably bounded in temperature, and the same is true of the van der Waals and similar continuous equations of state, ~(p, ii, T) ;= O. Designate this bounding temperature 'Fe. Then for all T > 'Fe the fluid is stable, by the criteria of Problem 7.l(b), ( c), over the accessible range of volume; and for T < Te, the continuity of ~(p,fJ, T) = 0 implies a region of instability bounded by the so-called spinodal locus, U2S _
IUvs
_Us.
I
= 0,
U 2V
cf.Problem 7.2(e). The locus joining the molar entropies and volumes of co-existing phases is called the binodal locus, and defined by the co existence conditions (7.3.1). (a) Consider two co-existing phases, ex and ~. Use the three co existence conditions in terms of the F = F(T,v) surface to show that if the tie-line connecting Fa.(T,va.) to F(3(T,ff(3) becomes vanishingly short as T -:;. Te, then at T = Te
OF) ( oU T
< 0 , (02P) ov 2 T
=
(03P) (04F) 01)3 T = 0 , 01)4 T
>0,
Phase stability, co-existence, and criticality
7.5
7.4
Chapter 7
208
I
I
i
209
and that the binodallocus is tangential to the spinodal locus. (b) Consider a temperature !:J.T below 'Fe. The co-existing phases are defined by !:J.p, 1:J.1),., !:J.v(3; and since the equation of state is continuous, !:J.p is expressible as an explicit function of !:J.v, !:J.T along this isotherm from & a. to !:J.1)(3' Use a Taylor series expansion for !:J.p in terms of !:J.v and !:J.T, together with the co-existence condition 11a. = 11(3 to show that p\ 2 {(02p) 1 (03 2 [(!:J.va.) 2 -(!:J.v(3)] ovoT c!:J.T+4' ov3)c[(!:J.Va.) + (!:J.V(3) 2 ] } = O. Derivatives with respect to T of order higher than first may be neglected. (c) Use Taylor series expansions for !:J.p;= !:J.p(!:J.v a.' !:J.T) and !:J.p !:J.p(!:J.v(3,!:J.T), together with the result of part (b) to show that p 02p) 1(03 ) JYz !:J.v IX = -!:J.v(3 [ -6 ( oToiJ c!:J.T oU 3 T, C i.e. that the locus of the mid-points of the tie-lines connecting the molar volumes of co-existing phases passes through the critical point, the binodallocus near the critical point being quadratic in volume, and that
~~ = (~~) v
.'1
c'
i.e. that the vapour-pressure curve is co-linear with the isochore passing through the critical point. (d) In real fluids, the binodal locus near the critical point is described better by !:J.va. :::::: -1:J.1)(3 :::::: X(!:J.T)'J, than by the result of part (c); what does this imply about Hp, D, T) = 0 in the critical region? Solution
(a) The co-existence conditions are:
Ta.
1{J,
Pa.=P(3,
F)=F(3-p(v(3-va.),
~ that at each T,
and P(3 have a common tangent of slope -po If F F(T, v) is continuous betwe~ v_and v(3 there is necessarily an intervening region of instability (F = F(iJh convex upwards) bounded by points of inflexion (02P/ov 2h = 0, and these are necessarily both between iJ a. and U(3, i.e. the spinodal locus is necessarily inside the binodal = F( T, v) being everywhere locus. If the binodal locus is to close, analYtic implies that the points Va. and v{3 and the two points of inflexion (a 2 F/oV 2 )T = 0 must coincide at T = Te, i.e. the spinodal locus coincides with the binodallocus at T = Te, with
02F) ( -ou 2 T,
-
c
(op) -ov T,
-0
c -
•
7.5
Chapter 7
210
J
If this point is to be stable, then the given conditions on the higher derivatives of F must hold. The remaining condition, p > 0, follows
from the requirement that all continuous equations of state necessarily converge to pu = RT as fj -+ 00. Thus a dilute phase (vapour) can never sustain negative pressure, and so liquid-vapour co-existence can only occur at p > 0; and this applies also at the liquid-vapour critical point. The critical point is the only point on the spinodal locus at which Cd 2 F/ai5 2 h and (a 3 F/afj3)T are simultaneously zero. (b) Since
p (aau ) T,
C
(a2p) au2 T, =
0,
C
7.5 or, since 11/J"
f
udp =
i
I
t
2 } + 41 [ (I1v,,) 2 +(l1v{3)]
;1
1
Ll.va
&
/J
T, C
d(l1v)
r.
0,
3p
(aarau2p ) l1TJ + 811 f(03au c
v..
p)
Vp
3
J
v.
e(l1v)4 V~
,
and the required result follows. (c) The Taylor expansions are as in part (b) except that & is replaced by !1t)", 11/J{3' respectively. Subtracting the second series from the first eliminates I1p as
(a 3p
0,
03 p ) ( OV3 T, c (I1v" + &(3)(111) " _11/J(3)3 = 0, and 111)" =
-&{3 .
02p ) c(&,,-!1t){3)I1T+ 1 oiJ3 ) e[(l1v,,)3 0= ( aTov 3
T,
C
(!1t) ,,)2 ,
and the required result follows. To obtain I1p/I1T, add the two Taylor series expansions for I1p, and put &" = -l1u(3; since (110,,) 3 -( I1v {3 )3, the second and third terms on the right hand side are each zero, and the required result follows. (d) The results of part (c) demonstrate that the physical consequence of the assumption that is a continuous differentiable function of T and v is that the binodal locus has a rounded top which is quadratic in v. The order of the curve depends on the order of the first non-vanishing derivative of F with respect to u at constant T. If the fourth derivative is zero, then stability requires that the fifth be likewise, with the sixth being positive. If the first non-vanishing derivative is of the order 2n (since it must be even), an analysis similar to the above must show that !1t)" is proportional to (11 T) 1/(2n - 2). In other words, no continuous 0 can give I1v" proportional to (11T)'j, or equation of state ~(p, iJ, T) to any similar power of I1T. Note that if the equation of state is not continuous from liquid-like to gas-like states, the spinodal locus ceases to have any simple meaning; but the critical isotherm is necessarily continuous, even if non-analytic, and, at least
Substituting into the argument of the integral and integrating gives for the first two terms I
(I1V{3) 2 ](110,,-11/J{3) 2
02 p \ I (03 p) (oTou) c I1T = - 3! ov 3
a -) I1T+-1 (a- ) , (11/J)2+ .., a(I1P») = ((a( I1v) T. c aTau T, c 2 aiJ 3 T, c
2L(I1/J)2
3p) (aou) T}(&") 2
Substituting back into the equation for (a2pjaTor5)cI1T, we obtain
and [a(l1p )/a(l1v )JT. c is obtained by differentiating the Taylor series as
2p
0,
whence I
The physical meaning of this condition-known as Maxwell's equal area rule-is obvious from the diagram of Problem 6.4. Expressed in terms of & as independent variable it becomes
f
r) (I1v ,,)2+ I1v ,,11/J{3 + (11/) (3)2] .
p [(11/),,)2 - (111) (3)2] ( 03 ov 3 ) T, e { - 3!I [(l1v ,,)2 + I1v ,,&{3 + (11/J (3)2]
I'
&dp = O.
[a(I1p)J '0(11)
1 (03 p ) 3! av3
Substituting for (o2p/orafJ)cI1T in the result of part (b), we obtain
P~
Ll.V.
211
*" I1v{3'
a2p) (oTou CI1T
the expansion for I1p is
2 3p ( a p\ ap) I (a ) I1p = ( aT ii, CI1T+ arav) C I1v 11T+ 3 au 3 T, c(I1v )3+ .... The condition IJ." = IJ.{3 has to be expressed in terms of I1v" and I1v {3'
given I1p" I1P{3' and I1T. Now (aIJ.japh = iJ, cfProblem 7.3(d), so IJ." = IJ.{3 implies Ll.v.
pa
Phase stability, co-existence, and criticality
p
(aoiJ)
(I1V{3)3],
t
= (02p) C
OU 2
c
= 0
•
Chapter 7
212
7.6
7.6 Stability at a critical point. The preceding problem examines the critical point in terms of derivatives of F = iJ), i.e. in terms of the volume derivatives of pressure. To elucidate the relevance of the condition of thermal stability, T/C v > 0, requires the [1 = [1(8, surface. (a) Express the slope of the projected tie-lines (on the 8, v plane) by alternative equations derived, respectively, from the co-existence 1{J, Pa. = Pp, by using Taylor series expansions for AT conditions Ttl. and Ap in terms of AS a. , A,}tI.; ASp, Avp. Thence show that the binodal locus at T = 'Fe is characterized by -(Op/aU)T = 0, irrespective of whether or not T/C. = 0; i.e. that the binodallocus on the [J = [1(8, surface coincides, at the critical point, with the spinodal locus as specified by the condition of mechanical stability alone. (tJ} Us~t!!.e equation of the tangent at T = 'Fe to the spinodal locus on the U = U(S, v) surface, D
s
I r:: Uvs
~Sv I
U2iJ
0,
c
= 0 .
Solution
(a) The expansions for AT and Ap are [leaving off the bar, since all extensive variables are to be read as molar (intensive)]: 2 (a2U) (a U) AT = A as • = aS2 v, cAS + asa,) c A!J + ... , 2 2 -Ap = A ( -a u) = (a - - AS+ (a -2 • Av + .... av s asov c ou s, c
(OU)
U)
U)
Putting Ttl. = 1{J, Ptl. = PP' and subtracting the series for {3 from that for ex, and neglecting terms of order higher than the first, we obtain 2 a U) (a2U) ( aS2 v, c(AStI. - ASp) + aSa,} c (Av... Aup) = 0, 2 (a2U) a U) ( asav e (AS... ASp)+ ov2 s)Av tI. - Avp) = O. At the critical point, the limiting forms of these equations equivalently define the tangent to the binodallocus: 2 (a2U) (a U) OS2 v,e dS+ asav e dv = 0, oSov c dS+ ov 2 s,c dlJ O.
(02U)
(02U)
02u
asov =
(aT) av s
-
Phase stability, co·existence, and criticality
213
The conditions of stability above the critical point require that (a 2 p/avanc < 0; dividing through we obtain
(~): dS- (~): (~~)
v, c dv
+
I
l(~) a~:TJ(~:~)T' c
c th)
= O. (7.6.3)
Comparing Equation (7.6.3) with (7.6.1) and (7.6.2) evidently justifies the assertion of the problem. It is sometimes argued (5) that the identity of (7.6.1), (7.6.2), (7.6.3) requires that the ratio (ap/av h c/(TICv)c = 0, so that if Co is infinite at the critical point, the infinity is of a lower order than that in KT • Certainly, for (T/Cv)c finite, even if vanishingly small, the first and second isothermal derivatives of p with respect to v must be zero, and criticality is determined by the conditions of mechanical stability, i.e. in the one-phase region near 'Fe,
TTl >-C ~O, >-K p v T
~O
cf.solution to Problem 7.2(d). It is doubtful, however, whether the foregoing analysis, being based on a continuous analytic equation of state [for which (T/Co)c is finite, cfProblem 7.2(e)] is applicable to real fluids in which, apparently, (T/C,)c = 0, because the critical isotherm in such systems is non-anaiytic at the critical point.
to show that at the critical point
~ (~~~) r,
7.6
p (as) (aT) T (a ) ov T as v = - c. aT
v '
BINARY SYSTEMS
7.7 Diffusional stability of the single phase. In a binary system the local composition is completely specified by a single intensive variable, the mole fraction of component 1,
n - n+m
x=-
N N+M'
where n is the number of moles and N the number of molecules of component 1, and m the number of moles and M the number of molecules of component 2, within an arbitrarily small element of volume. We therefore expect stability with respect to diffusion of either or both components in a binary system at uniform temperature and pressure to be ensured by a single condition. Suppose a two-component, one-phase system to be maintained at constant and uniform temperature and pressure. The appropriate form of the general equilibrium condition of Problem 1.22 is 8Gr,
p
>0.
Consider two regions, ex and {3, each internally uniform with respect to the chemical potentials III and 1l2' ex containing n ... and mtl. moles of the (5) J.S.Rowlinson, Liquids and Liquid Mixtures. 2nd Edn. (Butterworths, London), 1969, p.83.
7.7
7.6
Chapter 7
214
111 = 111(T, p, x);
- .G(O)+ G(O) -- n..111.. + m ..112.. + n{3111{3 + m{3I1~{3 . . {3 G(O) The perturbation to be considered is the passing of on mole of component 1 from a to 13 and, simultaneously, the passing of om mole of component
=
/I
,...1{3'
"
(7.7.1)
/I
,...2..
,... 2{3 •
(b) Use the definition G nl11 + m112, together with the Gibbs-Duhem equation, cf Problem 1.20Cd), SdT+lJdp+
Li nidl1i
11 1
( '0ax )
_
T, p
T, p, m
0111 ) ( om n(0111)
an
am
(7.7.2)
0,
(7.7.3)
0111) an
I x Ol1l
m =
0111 ) ( om n 02111) ( on 2
n + max; -
x 0111 n+ m
ax
an
=0.
(7.7.4)
(0 112) om
n
om~
n
02111 --= omon 02112
x oil' = - n+ m
-a:x:
ndl11 + mdl12 = O.
Put
an
dill = ( 0111) m dn + (0111) om
2 ( x ) 2 0 11 2x Oil 2 n+ m OX22 + (n + m)2 ox
x(1-x)02
2x-I oI12
n
dm ,
dl12 = ( -0112) dn+ (OI1Z) an m am n dm;
x(1-X)02 111 2x-lol11 --+ (n + m)2 oX (n + m)2 OX2
112 - - - (n+ m)2 --+ omon ox 2 (n+ m)2 ox
(7.7.8)
The last two terms include only the independent variables, so to ensure that the first-order terms sum to zero, these terms must be independently zero. The condition (7.7.1) follows. Thus, in addition to being uniform with respect to temperature and pressure, the system has to be uniform with respect to each of the chemical potentials. Note that condition (7.7.1) must apply also in the case where a and 13 are co-existing phases. (b) For a binary system at constant temperature and pressure the Gibbs-Duhem equation reduces to
n + m ax ;
(I-X\202111 2(1-X)0111 n+ m) ox 2 (n+ m)2 ox 2 = (1- x) 2 0 112 _ 2(1- x) 0112 02112) ( on 2 m n+m ox 2 (n+m)2 ox 2 02111) . (X)2 0 111 2x 0111 ( om 2 n· \n+m ox 2 + (n+m)2 ox (
(7.7.7)
+(1111l 111Jon - (1121l 112",)om.
m
0211 )
>0
oG = (n", - on)ol1l", + (m", + om)0I12", + (nil + on)ol1lll + (mil - om)oI1Z{3
1 x 0112
m =
(7.7.6)
•
(a) Write down the exptession for G = G", + Gil (subsequently to the perturbation) corresponding to the given expression for G (0). By inspection
T,p,n
(0 112)
>0
Solution
(c) Formulate the first and second derivatives of x with respect to n and m, at constant temperature and pressure, and thence prove
(
T, p
and express it also in extensive variables.
T.p,m
+m(Ol1l) T,p,m
aX
02e
+m(0112) = 0, on 1', p, m
_ (0 112) T,p,n on
(0 112)
(f) Show that condition (7.7.6) is exactly equivalent to
= 0,
to prove the results n(Ol1l) on
(7.7.5)
112 = I1zeT. p, X) .
(d) Expand 0111 .., 01111l' 0112 .. , 0l121l as Taylor series in powers of on, om, and use the results of part (b) to show that the condition (7.1.1) being met ensures that the first-order terms in oG = oG(on, om) sum to zero. (e) Use the results of part (c) to convert the second-order terms in oG = oG(on, om) into derivatives with respect to X; derive a result from Equation (7.7.2) which reduces the second derivatives of 11; with respect to X to first derivatives; and thence show that for stability (terms second-order in on, om, to sum to a result greater than zero)
2 from 13 to a, without restriction on the ratio on/om. (a) Express oG in terms of on, om, and thence show th~ t a necessary condition for equilibrium (terms first order in on, om to surrJ to zero) is "
215
Thence establish that 111 and 112 are completely specified by
two components respectively, and 13 containing n{3 and m{3 rJ'loles, so that the extensive Gibbs free energies are, initially,
""1..
Phase stability, co-existence, and criticality
then
l
111 ( 0an )
m
an J [('0am111) n + m (aamI12)
+ m (0112) m dn + n
I dm
= 0 .
Chapter 7
216
7.7
7.7
2
ax
dS- (T) (a (:£) C C aT p
2n ~]
+
(-CT)
v c
J
an
(-CT) (a-aTp) v
= (1
m
oM! x)a:;,:-,
(~)
aM! -X-, ax
+m) (-aMI) am II
5G
=
(n",
lX
m
1 8n + (aM2 0;
[(~) IG~) T, c
= 0
II,
through
J
dv
O. (7.6.2)
(aT)2 alp as vavaT .
v
- ( as
- [ n rL (an)
aT
0, changing the differentiating variable from S to
8G = (aM2/i) m +m rL
(as)
p aD) (aT) [a (a ) ( a;;8 = as II av av
the terms which are actually first order are
aMIIX\
T
p
dS+ - ) alJ
(~) (:~)
and since (ap/a!) h T, we obtain
·anj
[e:~I!) m 8n -
c
(av) as
-(~~)II(~)T' e~) v= - (~DJa~(~~) Jv - (~~)v (~~)T'
M ] + (nil + 8n) I,(a !(3\ m 8n - (aMI~\ am} n 5m + ...
Hm" - 8m)
'
D
J
r(
dv = 0
(ap/a/) h, c must be zero irrespective of (T/Cv)c' (b) From Problem 7.
5n+ (aMI 0,
or
(OJ.l.I) om n
< 0,
or
(0on J.I. z) m
( 2-) 0 G
T, p
>0.
>0
_ OJ.l.2 ox
>0
for the system.
T, p •
oZG x ox2
= 0,
2
ax '
7.8 Criticality, continuous equation of state. The region of two-phase co-existence in a binary system is frequently bounded in temperature, either by an upper critical solution temperature (UCST), as with liquid vapour co-existence, or by a lower critical solution temperature (LCST), as when a one-phase binary system separates into two liquid phases with increasing temperature. In either case, if G = G(p, T, x) is continuous, there is necessarily a region of instability bounded by a spinodal locus, (ozG/ox2h~ p = 0, analogous to the locus (02F/oij2) 0 of Problem 7.5; the corresponding binodallocus is defined by the co-existence conditions, cf conditions (7.7.
(7.7.13)
0 (~ uX 1: p
I Gnm
for each region, Le.
It follows that
azo x)ox 1
Gmn
OJ.l.l) , ( on m ( OJ.l.2) om n
(OJ.l.~
(OJ.l.l) T, p
T,p
< O.
so that, in respect of two regions, [cf solution to Problem 7.3(c)] stability is ensured by either G 2n > 0 or Gzm > 0, for each region. Note that for a homogeneous region, adding material at the same T, p, and x is a reversible process. Note that proceeding by analogy to Problem 7.2(a), Le. using o G(T, p, x), and the perturbation Dna., -Dma., leads directly to
- J.l.l ,
by Equation (7.7.2) ,
J.l.l - J.l.2 2
(0 l}
ax "
or omon
A third procedure is to truncate the expansions for J.l.l a., J.l.l/3' etc. at the first derivative. This gives, for the second order terms, (7.7.10) rather than the full list (7.7.12). Separating the terms (7.7.10) into two inequalities, one for each region, gives simple quadratic forms for both. The corresponding determinants are zero by Equation (7.7.4), and, by an argument similar to that of the solution to Problem 7.3(c), the stability condition is
and hence
x (OJ.l.l)
G2n
2
we have
(00) ox
02G
> 0,
H(n ... + ma.)(Dxa.) + (nB + ma )(DXB)] ox2
cannot be negative, the stability condition is
G
(02G) om2 n
or
0 ;
2 +(I_X)a J.1.z=_(OJ.l.I "OJ.l.2\. ax 2 ax ox)
= - - (OJ.l.l -OJ.l.2) - [(1- x)Dn+ xDm]2
> 0,
conditions correspond to the fact, cf Equation (7.3.6),
These that
02J.1.2 + (1 - x) oX 2
Thus expression (7.7.12) becomes
DG
221
Phase stability, co-existence, and criticality
O. n
T...
i'.
= 1/3,
P...
P/3' J.l.la.
= J.l.l/3'
J.l.za.
J.l.z/3·
(7.8.1 )
Write Equation (7.8.4) as
(a) Consider two co-existing phases, O! and (3. Use the conditions (7.8.1), together with the relations derived in the solution to Problem 7.7(0 to show that the tie-line connecting Ga.(T, p, xa.) to Gp(T, p, xp) has the equation
Ga.
aG p _ aGa.
= 0,
(03G) ox 3 T,
P
4 (0 G) 3x 4 T,
= 0,
P
and the equation to the tie-line is Ga.- Gf3
and that the binodallocus is tangential to the spinodal locus. (c) Use a Taylor series expansion for G about the critical point, at constant pressure (omitting powers of l1T higher than the first), with the results of part (a) to show that (7.8.2)
ox4 (04~
3
(l1X)2
T,Ac
i.e. that l1x is quadratic in (T-
G)
3 = -6 ( 32xoT
AC
l1T,
(aG/ax)e
(7.8.3)
Solution
(a) The solution to Problem 7.7(0 gives [leaving off the bar, since all extensive variables are to be read as molar (intensive)],
= (PI-P2)e'
4
33G ) I ( aG\
( ax 2aT c l1T+ 2! ax 3 aij e l1T(l1x tt + l1x(3)
3G tt
I (aax4G) c[(l1xa.) + l1xa.l1xf3 + (l1x{3) ) = o.
P2tt+3x IX
4
2
+31
whence (1- xa.)P1a.
= (1- Xtt )P2tt+ (l -
3G tt
xa.) 3x
a.
=
'
2
2
oG .. G.. + (l - x .. ) ox a.
Ga.-Gp
l1x a.- l1x {3)+ (a G) e l1T(l1x tt -l1x{3) ax c( axaT ( aG) 3
Thus, from the co-existence condition PItt
oG tt
Ga.+(l-XaJ-~-=Gf3+(l
uXa.
a G ) c l1T[(l1xa.) 2 -(l1xf3) 2 ] + I ( ax20T
PI{3'
oG@
uXf3
X(3)~
(7.8.4)
+
Similarly _ oGa. P2 .. - G.. -Xa.-~uXa.
= G{3
3G
x ~ f3 aXf3
(7.8.8)
The forms obtained using Equation (7.8.5) are more symmetrical than those obtained from Equation (7.8.7). The relation may be set up as follows
3Ga. x .. )ox ' a.
Pltr.=X"Pla.+(l-Xa.)P2tt+(l PI tt
(7.8.7)
X(3)G;)a.,f3·
(c) The co-existence conditions should be used in terms of G and its derivatives; Equations (7.8.6) and (7.8.5) or (7.8.7) are suitable forms. On using Taylor series expansions to fourth order for (aG/ax), and recalling [cfProblem 7.S(c») that l1xa. =1= l1xp, and that (a 2 G/ax 2 )e, (3 3 G/ax 3 )c = 0, we obtain from Equation (7.8.6), on dividing through by (l1x a. -l1x (3 )
T.J near the critical point.
PIa.
(x ..
Thus the tie-line is the common tangent at the points x tt' x{3 to the curve G = G(x), for constant temperature and pressure. Since, for stability, (a 2G/ox)2 is positive at X tt , xf3' and the continuity of G = G(x) requires a range of x for which (3 2G/ax2) is negative, this range must be between Xtt and xf3' and be smaller than xf3 - x tt · (b) The argument is exactly parallel to that of Problem 7.S(a), except for the last part of the latter; there is no restriction on the sign of
> 0,
l1xa. = -l1x p ,
(7.8.6)
oXf3 - oXa. '
and that the binodal locus spans a larger range of x, (xp - xa.) than does the spinodal locus. (b) Thence show that, if the tie-line becomes vanishingly short as T -,>- 'Fe, then at T 'Fe , p
_ _ 3G@_aGa._ aG p 3G .. Gf3 - ~ ~ X{3 ~ + Xtt ~ vX{3 lIX a. uXf3 vX a.
On comparing this with Equation (7.8.5) we obtain
Gp-Ga. _ (OG) . xp-xa.- ox a.,p'
2 0 2 T, ( oxG)
223
Phase stability, co-existence, and criticality
7.8
7.8
Chapter 7
222
I (
4 a G\ 3 ax3aij c l171(l1x tt ) -
3
(l1x{3) )
(a3xG)4 c[(l1Xa.)4_( l1X f3 t)+ ..·,
+ 4!1
(7.8.5)
t
4
1.8
Chapter 1
224
and
2
aG", (aG) -x'" ax", = - ax eX",
( a G)
1.9
Phase stability, co-existence, and criticality
(a) Prove that
3
( a 2G ) axaT c fj"Txa. ax aT c fj"Txa.fj.xa.
a
(a
4
F215 Pox
IFxo
4
G ) c fj"Tx a.(fj.xa.)2- 311 ax4 G) c X a.(fj.xa.)3+ ... , - 2'1 ( ax 3aT 2
G)
+xll ~ aXil -_ + (aG) ax c XIl + (a axaT c fj"TXIl+ ...
where
4
o.
(fj.x1l)4- 4Xa.(fj.xa.)3+ 4xll(fj.x1l)3]
(7.8.9)
Substituting for (a 3 G/ax 2 oT)c from Equation (7.8.8) gives finally 3 3 a4G ) 2 (ax 3 aT c fj"T[ (fj.x a.) + (fj.xll) ] 4
+
G)
a ( ax4
-
c [(fj"x a.)2 -
(fj.x1l)2 j[fj"x a. - fj.xll j2
0
3
3
c
2
-(fj"xa.)(fj"xll)+ (fj"xll)
(aax4G) c[fj.x '" + fj.xIl j[ fj.x
2]
T. 0'
etc.,
G) T, P. c
(- ) 3
=0
.
Solution
(a) The change of constraint formula gives, in molar quantities,
4
+
,
(- )2
a (-ax3
or
04G) , 2 (ax aT fj"T[fj"xa. + fj.xll][(fj"xa.)
T. x
and thence deduce the stability conditions for a phase at constant temperature and volume. (b) Proceed by analogy to Problem 7.6 to show that for the alternative equations for the (ii, x) projection of the tangent to the binodallocus at the gitical point to be equivalent requires that G2X be zero whether or not F 20 is zero. (c) Differentiate the determinant of part (a) to obtain the equation for the tangent to the spinodal locus at the critical point, and use the relations obtained in the solution to part (b) to reduce it, for (ap/au)c =1= 0, to F xv Fxv F xo - 3F2xo F + 3Fx2v F - F 3v 1 = O. 2v 2v 2v Prove that the result of part (c) is simply
(fj.xIl)L 2xa.fj.xa. + 2xllfj.xlll
4
c [(fj.x",)4_
== ( a-IJ 2 2
G ) c fj"T[(fj.xa.)L (fj.xIl)L 3xa.(fj.x"Y+ 3xll(fj.x1l)2] + 31I ( axa3 aT
1(aax4G)
a2 2
3
+ 4'
,
P2x
p) _F 2x == (aaxp) F 20
and, on addition, Equation (7.8.5) becomes 1( aG ) 21 ax2aT c fj"T[(fj.xa.
225
IX -
fj"x (3 po,
(~~)
and relation (7.8.2) follows unambiguously. Substitution from Equation (7.8.2) in Equation (7.8.8) gives the result (7,8.3). The analogy to Problem 7.5(b) and (c) is obvious. Note that, extending Equation (7.8.8) we have 4 I (a G) I ( aSG ) a3G ) ( ax 2aT c fj"T+ 31 ax4 c (fj"X)2+ 3 ox4aT c fj"T(fj.x)2
I (a G) (fj.x)4 c
T. p
+ (~~) T. v (~~) T. x T. v '
and from F == G - PI},
( aF) ax T, Hence
O.
7.9 Stability at a critical point. To elucidate the relevance of the condition of mechanical stability at a critical point in a binary system ,x) surface. In terms of this surface, the evidently requires the P = phase ccrexistence conditions include, in addition to conditions (7.7.1), the condition p", = Pll'
(~~)
(ap) +v( -aG) ax T, p ax
6
+ 51 ax 6
T. v
2 (aaxG) T. p 2
af}) ( ax T,
"~Ii
I \
p
v
=
(aG) ax T.
v-
p (a )
v ax T.
(aG) oX
v
2 (a2F) F) (av) = ax 2 T. v + OVOX T ax
(a
=
(a2G) axap T
=
la: (~~) J G;)
= T,
ra(aF) Op ax T. T. x
T, P •
T, p ;
J
v T, x
(a: I(~:~) 2
T. x
::) T
T. x
226 and
7.9
Chapter 7
2 2 2 (3 F) (3 F) (3 F) 2 2 2 OX T,v 3v T,x- 3v3x T
2 2 T,p (32F) ( 33xG) 3v 2 T,x
IT
3(Fx, Fv) _ o(x, v) -
Phase stability, co-existence, and criticality
we have
2 3 F) ( -0 2 X
Alternatively, use the method of lacobians:
F2x Fxv Fxv F2v
7.9
v)
1L2), -p ]/ 3(x, v) o(x, -p) o(x, -p) 2 3 G) (02F) = ( 3x 2 T,p ov 2 T,x'
= _ r3( ILl .
1L2 )]
oX
T, p
/(31))
op T, x
-
dx ( -0 2G) 2 ox T, p, c
I)
->0 vKT
T, p
T, p
>0 ,
V
I K
T
(ILIa. _ 1L2a;) -_ (OFa;) ox a.T,v
T >0 . > 0, -C v
(7.9.2)
(OFfl) 3vfl T, x = -Pfl '
(~) 3x
flT,"
-_ (1L1fl _ 1L2fl)'
(~) oVfl T, x
~) T, v (oXfl
1L2fl,
p' The Taylor series expansion for -t::.p,
t::.v, is
02F) dx+ (02F) -2 dv = 0 (3v 3x T, c (1) T, x, C
t::.(1L1 -1L2)
=
o.
(7.9.4)
dx
(~~)
dv = 0,
r(3P)2 I(a- p) ]d x (ap) -ox T, D, av T, x, ax T, v, ..
T, v, c
T. x. c
c
C
(7.9.5) dlJ C
0 -
(7.9.6)
ap
ap
(7.9.7)
and equivalence of Equations (7.9.3), (7.9.5) requires (02G/ax 2 ) = 0; if (op/ovlr x, c = 0, then (op/ox)T, v,c = 0, and Equation (7.9.6) requires (02G/OX 2 )T, p, c = O. The situation differs from that in a one-component system (continuous equation of state) in that
02G) ( ax 2 T,
p, c
(ap) T, x, (1)
c
= 0
(-3D) oX or, explicitly,
(OD)
T, v, c
dx+ -
av
T, x,
dv = 0 C
'
(F3x F2v + Fx2vF2x - 2F2xv Fxv)dx + (F3v F2X+ F 2x vF2V - 2Fx2vFxD)dv
O.
(7.9.8)
-t::.p = t::. ( -3F) = (02F) -t::.x+ (02F) -2 t::.v+ ..·. 3v T, x OV oX T, c ov T, x, C Subtracting the series for (3 from that for l), where x = plpo. The second of these equations becomes
qL
Sj
1S1
(10) For an experimental test of this equation see F.e.Tompkins and D.M.Young, Trans.Faraday Soc., 47,88 (1951).
r
242 (since Sj _ I =
8.7
Chapter 8 XSj _ 2,
f
etc.) and hence
alb""
g
-
a.. b I exp
(ql kTqL) . + cx
:E: x)'
x)
noc
+
(1 + Aq)n 0
;
= (l+AAqA +ABqB)n O ;
:E:
(c) multilayer adsorption: :E:
c noc
(l+Aql+A2qtq2+A3qtq2q3+ ... )no;
where qi is the partition function for a particle adsorbed in the ith layer. Solution (a)
Experimental data giving n as a function of x at constant T can thus be plotted as x/n(1- x) against x and, from the slope and intercept of such a plot, the quantities no and c can be calculated. If the area per site is known, no may in turn be used to evaluate the surface area of the solid adsorbent. (12)
n
A~[noln(l+Aq)l = nO OA
oln:E:
I
Aq + , Aq
as before. For either species considered separately, In:E:) AA- ( aOAA AB, 1,
nA
8.8 The results of the preceding problem may also be obtained by statistical methods. For example, if the surface is assumed to contain no independent sites, then, for the case of monolayer adsorption, each site may contain 0 or 1 particles. Let <Xo and <XI be the respective probabili ties of zero and unit occupancy of a site. Then, if A is the absolute activity of the adsorbed species, and q the ordinary partition function for a single particle, we have <Xo l. <Xl Aq
leading to
V, A
AAqA(l-(J),
(JA
where
e ()
+ (JB .
A
Hence,
e (c) With the assumption
e, the fraction of filled sites, and using <Xo+ <Xt = ()
.
(b) mixed monolayer adsorption (species A and B):
For x > I/O +yc), () > 1, corresponding to multilayer adsorption, and (J -+ 00 as x -+ 1 (i.e. p -+ pO). This isotherm equation, known as the Brunauer, Emmett, and Teller (BET) equation (II>, may be rearranged in the linear form x
T, V,A
Re-derive the results of Problem 8.7 by this method, given the following expressions for the grand partition function: (a) monolayer adsorption (single species):
cx
n - = () no
On identifying <Xt with it follows that
Oln:E:)
A~ (
n
The summations for n can now be effected, yielding eventually
nO -
243
number of filled sites, from the identity
_gl _
C -
Surfaces
cxis o ,
Si
where
8.8
ql
'Aq 1+'Aq
solution to Problem 8.7) that
= cq, q3 = ... = q,
q2
the grand partition function becomes
Since A is proportional to p, and q is independent of it, this is the Langmuir result. A more general method consists of expressing :E:, the grand partition function, in terms of A and q for a given model, and determining n, the
Z
(l+Acq+A2cq2+A3cq3+ ...)no = [
Ifwe use
e=
(11) S.Brunauer, P.H.Emmett, and E.TeUer,J.A mer. ChemSoc.• 60, 309 (1938).
n no
= ~(Oln:E:)
no
OA
p/po
= x
and assume further that
(12) S.J.Gregg and K.S.W.Sing, loc.cit.
Aq ~
l+(C-I)AqJn o I-Aq
Chapter 8
244
we obtain
o=
8.8
ex +ex-x) ,
as before. (l3)
8.9
Surfaces
Hence 11 kT
-=
and
(alnQ)
-an
0
Solution
Let the fractional surface occupation be 0 = nino; then, of the z sites surrounding any adsorbed particle, an average number znlno will be occupied by other particles, yielding altogether zn2/2no interacting pairs. (The 2 in the denominator accounts for each pair having been counted twice.) Thus, the average total interaction energy will be wzn2/2no, where w is the interaction energy for one pair. The canonical ensemble partition function is then no! ( wzn2) Q = n!(no- n)!ql'l exp - 2nokT . It has been assumed that the distribution of filled sites is random, in evaluating both the pre-exponential factor and the number of inter acting pairs (Bragg-Williams approximation). Use of Stirling's approximation yields
InQ
wzn 2 - nlnn - (no- n)ln(no- n)+ nlnq - 2nokT
(13) For a more detailed statistical treatment see T.L.HUl, J.ChemPhys., 14,263 (1946). See also E.A.Guggenheim, Applications of Statistical Mechanics (Oxford University Press, Oxford), 1966.
A.
If qA.
0
-= exp kT
wzO
In(l_ O)q + kT
T
1'10,
11
8.9 A more realistic model of monolayer adsorption would permit a particle in a surface site to interact with particles in neighbouring sites. Introduction of an appropriate interaction potential might be expected to reproduce in the model such co-operative phenomena as surface phase transitions (cf Problem 8.5). An exact treatment is impeded by the difficulty of calculating the configurational part of the partition function, and averaging the inter action energy, whilst allowing for the re-arrangement of the adsorbed particles under their mutual influence. The Bragg-Williams approximate treatment simply assumes that the arrangement of particles is random, as it would be in the absence of interaction. This somewhat severe approximation, which is the same as that underlying the van der Waals treatment of gas imperfection Problem 1.11) is not quantitatively satisfactory; nevertheless it succeeds in predicting the occurrence of surface phase transitions below a certain critical temperature, as does van der Waals' equation. Write down the canonical partition function for such a system, calculate the chemical potential of the adsorbed species, and hence derive the adsorption isotherm equation and find the two-dimensional critical temperature.
245
wzO
(1- O)qex p kT
plpo (see solution 8.8) then
p pO
0
nC1Tnllir
wzO
form for w
p(O)p( I
0) =
O.
rp(t >F ,
where p(O) denotes the pressure at which the fractional coverage is 0, we deduce that the critical point lies at Oe = t. From the condition it follows that
dP) ( dO 0 T.
== 0 Oc
zw
c
Thus a phase transition, giving a vertical discontinuity in the isotherm, is to be expected for an attractive interaction (negative w).
The imperfect classical gas
9.3
247
9.2 Assume that for a class of substances the pair potential is of the form IP d.
O O. (c) In the case I{) = 00 for r < d and zero otherwise, we obtain
D=
= /v2N - t{3q N
This yields the pressure
exp[-J3pu - /3Ip(u)] du
1{)(lrj-rjl)
251
= 4k
The configuration integral may diverge for configurations in which oppositely charged particles coalesce. If qi -qj then the integrand contains the divergent factor rt!q2. The (two-dimensional) integral over r/> say, exists only if J3q 2 < 2, or q2 T> 2k = 2To •
For T < 210 the gas collapses into neutral pairs. Three isochores at densities P3 > P2 > P 1 are shown in Figure 9.5.1 where also the pressure of N neutral noninteracting pairs is indicated. p
P3
Solution
In the configuration integral
Q
=
f..·fdr1 ...
~
drwexp(J3 L qiqjlnrjj) i 0 . (2)
1f: .0
The two necessary and sufficient conditions for a maximum in B2 (T) are thus
< 0 for some distances, but
-00,
as it must be since both these limiting cases correspond to a hard sphere
r>d
.p
for c
={ j1TR 3 i1Td
is always positive if .p(r) is non-negative. (c) Since it follows from part (b) that a maximum is only possible when the potential is negative somewhere we have for T --J>. 0 that
259
Note that
ip(r)exp(-{3ip)dr,
r> d
B;(O)
The imperfect classical gas
9.13
\O\r)
(c)
for r";;; d
< r";;; Xd for Xd < r.
ford
~ {-U(2:-r1d)
for
r";;;
d
< r";;; 2d
for 2d < r.
ford
r>d
Solution
9.11 Calculate the second virial coefficient for two gases, one with the intermolecular potential
'Pl(r) =
{
rz)
Integration over all momenta and over r), r4, ...rN yields the
Per I, r2) .£l.4~!.L2f finding particle I in a volume element dr 1 at r I and paftTCIe2 in drJ at rA
dr) ... drNexp(- U/kT)
nZ(r12) .
P(rl, rz)
nz(r) is called the pair distribution function and the function g(r)
Wi! obtain
vZnz(r) - I
is often called the correlation function. Show that g(r) vanishes for an ideal gas. Explain why g(r) should vanish when r -+ 00 in a homogeneous system (one phase). (b) Consider a system consisting of a gas and a liquid phase, both of macroscopic extent. Give a probabilistic argument showing that one expects the following form for the pair distribution function in the two phase region: n2(r; v) V-I[XlvlnZ(r; VI)+ XgVgnl(r; Vg)] . Here VI and i)g are the specific volumes of the coexisting liquid and gas phases (see Figure 9.14.1). Xl and Xg are the corresponding mole frac tions so that v = xlvl+xgV g • The pair correlation function nz(r) de pends clearly upon the state variables (ii, T), and the notation nz(r; iI) is used above to Indicate explicitly the value of ii. p
drz ... drNexp(- U/kT)
(v ==
QN dr3 ... drN exp(-U/kT)
The probability of finding one particle (not necessarily particle I) in dr l and another in dr2 is simply !!.(N-I)P(tlt2)dr,dr21 since there areN choices for the particle in dr I, and then (N I) possibilities for the second particle. Hence n2(rlr2)
... drNexp(- U/kT) .
For the ideal gas this reduces to n2(r l ,rZ) since U
=
1)
= 0 in this case.
Taking the thermodynamic limit we obtain 1 n:z(r) ; g(r) =v 2nz -1 = 0 .
Qne expects two particles in a gas or liquid to become completely inde endent as the distance between them Increases. Hence the JOInt proBa bilIty 0 In Ing one par IC e In drl and another particle in dr2 approac es :tEe product of the probabilIties of each event, namely (
Figure 9.14.1.
~ dr I) ( ~ dr2) = vI2 dr
I
dr 2
•
(b) The probability of finding a particle in drl is still O/v)dr l , since the positions of the two phases are random when there is no outside field of force. Now Xg and XI are the a priori probabilities that the first particle belongs to the gas and liquid phase. If the first particle belongs to the gas phase then the other particle at a microscopic distance r will also be in the gas phase (of macroscopic size). Hence the conditional
9.14
Chapter 9
264
probability of finding this second particle in drz, given that the first particle is at rb is Vg112(r; vg)dr2' The same argument applies to the liquid phase. We obtain thus for the joint probability distribution func tion in the two-phase system the following expression: _ Xg l)gn2(r; Vg )+ Xl Vl nZ(r; VI) nz(r;v) = _ v 9.15 Prove that
If
kT P = ij-6
n{)
I
9.16
The imoerfect classical gas
265
In the thermodynamic limit the pair distribution function is a func r. The virial theorem of sta tis tical mechanics, tion of r 12 p
kTp
-i
f
drrr.p/(r)n 2 (r) ,
also called the thermal equation of state, follows. 9.16 (a) Prove Clausius' virial theorem in classical mechanics (for bounded motion) N
Ekin = -! E r n F n ,
(r)nz(r)dr
n" I
by differentiating the configuration integral with respect to the volume. Here n2(r) is the pair distribution function of Problem 9.14 and r.p(r) is the intermolecular pair potential. To facilitate the differentiation procedure introduce new vari ables ii = riv-1I3 in the configuration integral. The above result is often called the virial theorem of statistical mechanics.] Solution
Introduce the new variables ii = rJL, with L 3 tion integral: QN
f
= L3N
di l ... diNexp [- i
v, in the configura
~kr.p(Lfik)/kT]
where Fn is the force acting upon particle n. The bar denotes an aver age over time. (b) Apply Clausius' theorem to an imperfect gas in equilibrium, assum ing that the total force can be derived from the wall potential and the intermolecular potential r.p(rik) acting between each pair i, k of particles. Assuming that the time averages involved may be evaluated as phase averages in the canonical ensemble, show that the result of Problem 9.15 follows. [For related considerations see Problems 3.7 and 3.8.] Solution (a) Use
Note that the integration limits are now independent of v. Hence
1 aQN
aQN =---=--L kT QN av 3VQN aL
__ I
f
~
rnFn
I
A
[N
3N _
- QN dr l ... drN v L
L 3N+ I ar.p(Lfik )] [_ '\' ..p(Lfik)] 3kTv aL exp /(~k kT
In the second term use a..p(Lfik ) ~ ar.p(Lfik ) aL = rik a(Lrjk) ,
r
when T
E i
r 2 ) -- k Tp--6 Z 12 v r l2
a
exp
E i i~, ... , i'tvl i I> i 2, ..., iN) = N'
and a completeness or closure relation. We assumed here that the index If it is a discrete one, the Dirac 0 functions must be replaced by Kronecker ones. (b) If 1'11) is a properly symmetrised wave function of the N-particle system, express 1'11) in terms of the set (10.11.1 ). Also give an expression for an operator n operating on functions in the Hilbert space spanned by the set (10.11.1).
i is a continuous parameter.
with the boundary condition (corresponding to the vanishing of the wave function at the boundary of the volume) R(ro) = R(O)(ro) = 0, where ro is large. This boundary condition leads to
= mr ,
kr 0 + 71(O)(k, I) = mr .
L €po(i I -i~l) ... o(iN-i~N) , (10.11.3)
.p
For large values of r, U(r) will vanish, and the asymptotic solution for R is R = sin [kr+71(k, I)] ,
kro +71(k, I)
(10.10.4)
Solution
f...
If 11k and I1k(O) are the changes in k when n changes by unity, we have clearly 1 1 p(k, I) = 11k - I1k(O) ,
where
and from Equation (10.10.4) we have
(and if necessary summing over its spin variables).
SECOND QUANTISATION FORMALlSM(2)
)
= N!~€Pl{Jil(P1}.pi2(P2) ... l{JiN(PN).
fdil ... diNlil, ... ,iNHiI, ...,iNI'I1),
di indicates integrating over the coordinates of the ith particle
... fd''I'"
d'IN d"'I'" d'" . )(.'I"",'N~" . Inl"'I"",'N ")(.,'1"",IN ., I . 'Nill,··,IN
10), Ii>, li l ,i 2 ),···, li l ,i2 ,···,iN ),···,
a+(i)lil, ... ,iN ) =(N+1)Y>li,i l , ...,iN )·
(10.12.2)
Express the Iii' ... , iN) in terms of the a+(i) and the vacuum state. (10.11.1)
(b) If [A, BL = AB - BA is the commutator, and [A, B]. = AB+BA the anticommutator, of the two operators A and B, prove that
This is a properly symmetrised set. The bra set corresponding to the ket set (10.11.1) is
[a+(i), a+(j)L = 0
for bosons,
[a+(i), a+(j)]. = 0
for fermions.
(10.12.3)
and 1 (il>i 2, ... ,iN i = N!~€pl{J~(p1)I{J~(P2) ... I{J:;/PN).
(10.12.1)
where 10) is the vacuum state. We now introduce creation (or construc tion) operators which will produce an eigenvector corresponding to N + 1 particles from one corresponding to N particles, as follows:
10.11 For many purposes it is convenient to use a formalism which is independent of the number of particles in the system. Let I{Jn be now a complete orthonormal set (c.o.&) of single-particle functions (including the spin dependence where necessary). For a system of N identical systems one can then use as a c.o.s. of basis functions the set ••• ,iN
=
10.12 (a) Consider systems with an arbitrary number of particles, that is, consider the Hilbert space which is the (direct) product space of the Hilbert spaces corresponding to 0, 1, ..., N, ... particles. The c.o.s. span ning this Hilbert space will be the set
and hence Equation (10.10.1) follows.
lil>i 2
J
n f ~,,=
=~'
1
1'11)
(b)
(ro+ ~~)l1k = ~, d71(O») - - I1k(O) ( r o + dk
279
The imperfect quantum gas
(10.11.2)
(c) We now introduce an operator a(i) by the equation (2) For a general reference, see l.de Boer, Progress in Low Temperature Physics, Vol III, (North-HOiland, Amsterdam), 1965, p.2l5.
(il, ...,iNla(i) = (N+1)Y>(i,il> ... ,iN I.
,i,
(10.12.4)
280
Chapter 10
10.12
10.14
Prove that [a(i), a(j)L
(10.12.5)
[aU), a(j)l.
0
n/ = -
forfermions.
(d) Find an expression for a(Oli l , ... , in>. (e) Prove that
a+(j)L = aU for bosons,
and [a(t), a+(j) 1. a(i - j) for fermions.
Solution
(10.12.6)
,n =
i
i,;
n ij =
V(r/;),
(10.13.2)
2 2 "fi k + L. 2m
k
where
"nP)+l " n(~) L. I '! L. I, '
fi2 2m vj2+U(r) ,
find an expression for n in terms of the at and ak'
(f) If n is an operator of the form
n
281
In that case, one usually writes a; and ak rather than a +(k) and a(k). If n/ and nil are of the form
= 0 for bosons,
and
The imperfect quantum gas
U(q) =
(l 0.12.7)
where the n}l) and n~2) are, respectively, single-particle and two·particle operators which differ only in the particles on which they operate, express n in terms of the a+(i) and a(i).
+ 1" + 1 '\ + + akak+-; L.U(q)ak a k-q+2v f.t V(q)akak,ak'+qak_q, k,q k,k ,q
f
d 3 rexp[-i(q· r)]U(r),
V(q)
f
d 3 rexp[-i(q· r)]V(r).
10.14 Give an expression for
a +(i)a(i) Ii I> ..., iN> , and discuss the physical meaning of the operator
Solution (a)
IiI, ..., iN>
nO)
1
N! a+Uda+(i2) ... a+(iN) 10)
a+(i)a(i)!il, .. ·,iN )
where the upper (lower) sign again refers to the case of bosons (ferm· ions). From this equation it follows that the aU) are annihilation operators. (e) This follows from Equations (10.12.1), (10.12.4), and the equation given under (f) n = fdidi'{a(i - i d li2' ... , iNH ... +(± l)N -I aO - iN)Ii 1> ••• , iN. I >},
+! di dj di' dj' (iii n(2)! i'j')a+ei)a +(j)a(j')aU') .
(10.14.1)
Solution
(b) This follows from Equations (l 0.12.1) and (10.11.1). (c) This follows from Equations (10.12.4) and (10.11.1 ). (d) By considering the matrix element (i'J, ... , i~_1 Ia(t)Ii'1> ... , iN> and bearing in mind that the resulting expression is valid for any
a +(i)a(i) .
i.
t
.I'
11.1
Phase transitions
283
and show that the equation of state follows from the equations
= f(y)
(3pvo
11 Phase transitions
Vo
where
where n
(~~ ) v,
(10.2.7')
2m)'/'V q = -21T ( h2
(11.1.8)
0
,jEln(1- e" -(3E)dE .
(11.1.9)
(11.1.10)
From Equation (10.2.7') it then follows that 21Tm) '/' - e nil (11.1.11) N = ( (3h 2 V L ----;;, n=1 n and Equations (11.1.6) and (11.1.7) follow immediately. To obtain the virial expansion, we must express y as a power series in Vii from Equation (11.1.11) and substitute this into Equation (11.1.10). This can be done either by tedious sorting out, or more elegantly as follows. Ifwe write x = volvl' we have
11.1 For a perfect boson gas with single-particle quantum states j of energy Ej we have (see Problem 3.l2a) (11.1.1 )
/
tj = exp(ex-{3Ej ).
f-
Expanding the logarithm, and integrating, we have 21Tm)'/' - en" q = ( (3h 2 V n~1 n'/,
EINSTEIN CONDENSATION OF A PERFECT BOSON GAS (I)
where
yn
n'/, '
Combining Equations (11.1.1), (11.1.2), (11.1.3), and (10.2.6) and going over to an integral from the sum in Equations( 11.1.1), we find
= Nlv = llvl, with VI = vlN the specific volume.
q = -~ln(1-tj),
L n=I
Solution
(10.2.6)
T '
(11.1.7)
and where Vo is given by Equation (10.2.16). Obtain a virial expansion in the form of a series expansion of (3pv I in a power series of Vii.
and
=;
-
f(y) =
11.0 Let us remind the reader that for a grand canonical ensemble we have (Problem 10.2) q = In:=: , (10.2.5)
n
= yf'(y),
VI
D.ter HAAR (University of Oxford, Oxford)
q = {3pv ,
(11.1.6)
,
(11.1.2)
If we consider the case of spin-zero point particles in a force-free volume, the number of energy levels d.Z lying between E and E+dE is given by the expression (compare Problem 3.l3c)
x = yf'(y) ,
(11.1.12)
and we want to find the expansion dZ = 21T ( 2h":. )'/'vEY'dE.
(11.1.3)
f(y)
From Equation (11.1.2) it is clear that we must always have
< {3Ej
(11.1.4)
so that tj is always less than unity. Put y = e"
(11.1.5)
ex
f
--=-=La xn yf'(y) x n'
(11.1.13)
Let Yo be the value of y satisfying Equation (11.1.12). From the theory of functions of a complex variable it follows that 1 "r f(y)dln[yf,(y) -xl 21Ti
=
[(Yo) ,
(11.1.14)
where the contour lies in the complex y-plane and is going around the origin and around Yo.
(I) For a general discussion see, e.g. D.ter Haar, Elements of Thermostatistics (Holt, Rinehart, and Winston, New York), 1966.
282
j
284 If we write
= yf'(y)
[I Yf~y)J
we can expand the logarithm and we find I
I
= 27Ti jf(Y) d
{Inyf'(y) - J;~ 0 n +I 1
l J x yf' (y)
From Equations (11.1.1), 01.1.2), and 00.2.7') we have
OLLIS)
n+
=_1_ ~
xn+'llf'(y)rndy 27Tin~on+lj .. n+1 ,
1
N = ~ -ex-p-(--a-+ (3Ej )
I}
(3pv I
= N [1 -
11.2 It has been shown(2) that when y can be written in the form
;/2 + .. J.
ijN;N; . " "
(11.12.9)
11.12
Phase transitions
295
The partition function will thus be
- ~" ~ n [n. J {-
QN - NIL... .
N;!'
w(Ni ) exp
'!(34 1 " rJ>ijN;N; } ,
01.l2.1O)
q
i
which is of the form (11.12.2) and where the sum is over all configura tions satisfying condition (11.12.8). The function cp{N;} in Equation (l1.l2.2) is then given by cp{N;}
=
Ii
[N;ln(Ll- N;o) - NilnN; + N; ]-!(3 L rJ>ijN;N;, 01.12.11) ij
where we have used Equation (11.12.1) and Stirling's formula for the factorial. The maximum term in Equation (11.12.10) with the Ni satisfying condition (11.12.8) is found in the usual way by varying the Ni and using a Lagrangian multiplier 'Y, to be determined from the condition (11.12.8), to take the subsidiary condition into account. The result is Ll-No No In .,.' Ll-'No (3"i;.rJ>ijN;='Y. 01.l2.12)
,
"
If the density is homogeneous we put
NLl N·=, v'
and if we determine 'Y from the equation v -No No N In~- v-No-(3rJ>o-;; = 'Y,
01.l2.13)
01.l2.14)
with rJ>o given by Equation (11.12.4), all Equations (11.12.12) are satisfied. This means that relation 01.12.13) is a possible solution of Equation (11.12.12), and that the uniform-density states are thermo dynamic states as they make QN stationary. It remains to be deter mined whether these states are stable, metastable, or unstable. Substi tuting Equation (11.12.13) into Equation (11.12.11) we obtain the result (11.12.3). If CPs, given by Equation (11.12.3), is an absolute maximum, we have QN = expcps and the free energy F is given by -(3F = CPs, so that one finds p by taking the derivative with respect to v from which the relation (11.12.7) follows. We can write this equation in the form
(3(p-!rJ>o~:)(V-NO) =N,
01.12.15)
which is the usual van der Waals form. We see clearly the excluded volume term and the term due to attractive forces (rJ>o is negative!): -~rJ>o is the work that a particle has to do against the attraction of the other particles to reach the boundary.
o
296
11.13
Chapter 11
I-n) fen) = nln ( -n- -!tl4>on 2 ,
which is related to 4>s of Equation 01.12.3), when we put write n = N/v. (a) Prove that, if A is a real symmetrical matrix of the form
= ail)ij-b ij ,
bij
b ii
> 0,
01.13.1) l) =
I and
(11.13.3)
Ai> Lbij j
for all i. Prove also that A is not positive definite, if for all i. (b) Use the results of (a) to show that, when f"(n) is negative, the solution leading to condition 01.12.3) corresponds to a (relative) maxi mum of 4>, but, if f"(n) is positive, this is not the case. (c) Show that the condition on tl that f"(n) is always negative is
-tl4>o <
Solution
If we write n = N/v and use units in which 0 = 1 so that 0 < n < I, we have 4>s = vf(n) with f given by Equation 01.13.1). Equation 01.12.14) for'Y then has the form
f'(n)
= 'Y,
01.13.5)
I-n n
= In----64> n I -n . 0 n.
(11.13.6)
In the following we also need f" (n) which is given by the equation ( ' (n)
= nO -1 nF tl4>o·
01.13.7)
To prove this let A be an eigenvalue and {Xi} the corresponding eigenvector of A so that we have Axi
= ~Ajjxl I
= d{X; "-'
< XI Lb j; < xlal
.
(11.13.9)
Hence each eigenvalue is positive, and A is positive definite. On the other hand, if one considers the vector {I, I, ..., I} == {yd we if
< 0 if aj
we construct the matrix of the second functional derivatives 2 a24> . 0,
VII> 0,
VI+Vn = v.
01.15.1)
For any value of VI we have a solution of condition (11.15.1) correspond ing to the same 'Y 'Yo. The number of particles satisfies the equation N = nIvI + nnVn ,
01.15.2)
so that for all values of N/v in the range ,
o
I
_
nJ
11
01.15.3)
there is a two-density state. The densities nI and nIl are determined, as we have seen, by the conditions
flU
III
< N/v < nIl
f'(nd
= f'(nIl)
(11.15.4)
and )-n1f(nd = f(nn)-nIlf'(nn).
From Equation (11.14.8) it follows that, neglecting the transition region, we have for the free energy
Figure 11.14.2.
This is an equation for 'Y. There is thus only one solution 'Y = 'Yo such that (11.14.11) f(nd-f(nn) = 'YO(nI -nil), or
f
ilII
f'(n)dn
= 'Yo(nll -nd,
(11.14.12)
111
a! LI(g).
Instead of the Hamiltonian (12.1.1) we now have H
-(M' B eff )
However, for temperatures below Tc given by the equation
02.3.1)
,
= gIlBLSf,
M
Te (12.3.2)
2 B +- (S) Ll(g) .
Beff
gllB
(12.3.3)
g
From Equations (12.2.5) and 02.2.6) we get a system of N spin-! particles (M)
= NgIlB(S)
Te (M) Td (M)3
02.3.4)
! Ngll B tanh(! BgllBB eff) •
Hence from Equations (I equation for (M) (M)
or, using the fact that T
02.3.5)
~
.4) and (12.3.3) we get the following !NgIl B tanh [!{3gIlB(B +q(M»] ,
T
2LI(g)
0,
= Mo(Tc ~T)
(M)
Mo
(12.3.7)
(M) I 02.3.8) = tanh Mo = -1 Ngll B in the magnetisation at T O. We note first of all = 0 is always a solution. Also, we see that as the curve
~
.i\
lh
[1 -2exp (- i) J' gllBB
(M)
2kT
l
Ll(g)
y = tanh --L- (M) 2kT -Mo
(12.3.13)
(12.3.14)
+
or ) ,
diagram, while the curve
(12.3.12)
while experimentally it is found that Mo - (M) behaves as T'12. (iv) At temperatures above Tc there is no spontaneous magnetisation . When T ~ Tc , we can expand the tanh in Equation (12.3.6) and we have
y
has unit slope in a y versus
(12.3.11 )
As T....;. 0, (M) ....;. Mo. Expanding the tanh for large values of its argument, we find
(12.3.6)
(i) To find the Curie temperature we must solve Equation (12.3.6) for the case where B = 0, that is, we must solve the
Mo - T3 MJ + ...
Tc and (M) (M)
where
q
g
=
or (M)
(12.3.
= LI(g)/2k ,
there will also be a non-vanishing solution to Equation 02.3.8). One can show by evaluating the free energy (compare Problem 12.4) that this solution is the equilibrium solution, so that the system shows spontaneous magnetisation below the Curie temperature Te. Near, but below, Te, (M) will be small so that we can expand the tanh in Equation (l2.3.8):
f
and where
(1
g
where the magnetic moment M of the whole system is
where that
0 will be the only one as
(M)
(12.3.15)
whence we get for the susceptibility per spin X X
]
the so-called Curie-Weiss law.
(12.3.16)
308
12.5
12.4
Chapter 12
12.4 An Ising ferromagnet contains N spin--l particles. Let be the number of spins with z-components (-!) and let the up- and down-spins be distributed randomly over the lattice. Let
R
N.-N_
(12.4.1)
-kN[-l(1 +R)ln10 +R)+!(l E -izINR 2 ,
In!(1-R)],
!zN_p_
(12.4.3)
Q._
F== E
(12.4.7)
E == -VCQ•• +Q-- -Q.-) ,
R
where Q++ (Q__ ) is the number of nearest neighbour pairs where both spins are in the positive (negative) z direction while Q._ is the number of pairs with one spin in the positive and the other spin in the negative
1+R)+!(l-R)ln!( I-R»). 02.4.13)
(12.4.14)
tanh
zI R. 2kT
(12.4.15)
Bearing in mind (i) that for the Ising model I:.I(g) zI, and (ij) that R from expression (12.4.1) is directly proportional to (/11). we see that we have rederived the molecular field equation (12.3.8). 12.5 Use the results of Problem 12.4 to find a parametric expression for the specific heat of an,. Ising ferromagnet in the molecular field approximation. Find the jump in the specific heat at Te , and the behaviour of the specific heat as T -+ O.
(12.4.8)
( 12.4.9)
TS
or
whence introducing R from Equation (12.4.1), and using the fact that N.+N_ == N, we get expression 02.4.2). To find the free energy, we must evaluate the internal energy. We have
where the summation is only over nearest neighbour pairs and where we have introduced new variables f.1r which are + 1 (-I) if the spin on the f site is in the positive (negative) z direction. From Equation (12.4.8) we can write the energy in the form
(12.4.12)
0 we then get
I +R zJ
In I -R
we have
g ,
N.N_
From the condition 'OF/'OR
Using the Stirling approximation (see Problem 2.11) in the form
S == k(NlnN-N.lnN+-N_
!zN+p-+t.zN_p.
== -!zJNR2 +NkT[!(1 +R)
(12.4.5) (12.4.6)
=
Combining Equations (12.4.9), (12.4.12), and (12.4.1), we obtain expression (12.4.3). The free energy is thus given by the equation
(12.4.4)
InN! == NlnN - N ,
(12.4.11 )
and
and the entropy S is given by the expression
-HIr,g f.1ff.1
02.4.10)
N '
(12.4.2)
If the distribution of + and spins is assumed to be completely random, the probability W for N. + spins and N_ - spins is
II
N
2 ----.:':
where p. is the probability that a spin is in the positive z direction. Similarly we have
Solution
S == 'dn W == k(lnN! -InN+! -lnN_!} .
1
2Z
Q++ =
where z is the coordination number, that is, the number of nearest neighbours per spin. Minimise the free energy with respect to R to find the equilibrium value of R.
N! W == N.W_! '
309
z direction. On the assumption that the + and spins are randomly distributed, we find for Q++ (z is the coordination number, that is, the number of nearest neighbours per spin):
be an order parameter. Show that the entropy and internal energy are given by S
Cooperative phenomena
~!
Solution
From Equation (12.4.13) we find for the specific heat 1
Cv
dR
-2 zINR dT '
Cv :
02.5.1)
while R satisfies Equation (12.4.15) which for our present purpose we shall write in the form RTe R tanh (12.5.2) T ,
12.5
Chapter 12
310
whence RTc /,(
dR
- T2f I cosh
2R
TC)
T'
(12.5.3)
so that the specific heat is determined by the parametric equations Nkr 2 (12.5.4) v 2 C cosh r-rcothr' rcothr
=
(12.5.5)
T '
where r is related to R by the equation r R Tc/T. To find the behaviour near we note that as T -+ Tc, r -+ 0, so that we can expand the various hyperbolic functions occurring in Equation (12.5.4); the final result is Cv -+ ~Nk
as T
-+
(12.5.6)
Tc .
As Co 0 for T> Tc in the molecular field approximation [E = 0; see Equation (12.4.3)] the jump in C v is ~Nk at Tc . When T ~ ,we see from Equation (12.5.5) that r "'" Tc/T and we thus obtain from Equation (12.5.4) Co "'"
Tc)2 exp (2Tc) 4Nk ( T -T
.
II
M·1
{Sf1
•
-gIlBL{Sf2
'
f).
[B+2 LI(f) -g) )S"g, + 2 LICf) -g2)SI2]} gllB II .' gllB 12
(12.6.2) (12.6.3)
gllsLSf' fi
I
and
>,}
Bm
B ql<M»
B~ii
B -q2(M»-ql(M2> '
q2<M 2
(12.6.4)
with (M j
)
( 12.6.5)
-!Nglls(Sj>,
the extra factor -! deriving from the fact that the N spins are divided evenly over the two sub lattices, and qI
_ "I(f) -gl) __ "I(fz-gz) } 2 L. N 2 2 L. N( 2, II (gIlB) 12 gllB) "Ie f 2-gd "I(f) -g2) - 2 L. N(g )2 = - 2 L. N(g )2 I, IlB 12 IlB
(12.6.6)
where we have included an extra minus sign to take into account that the dominant interaction leads to antiferromagnetism. As the two sublattices are now decoupled in Equation (12.6.2) the results of Problem 12.2 can be used for each of the sublattices, and we get > = iNgllstanh[!j3gIlB(B-q)<M»-q2(M2»],} _ .1) (12.6.7) - 'lNgIl B tanh[2j3gIlB(B-Q 2 <M»-q)(M2>)]· At high temperatures we can write tanhx "'" x, or "'" !j3N(gIlB)2(B-Q)<M»-Q2<M2~r;r "'" !j3M(gIlB)2(B -Q2(M 1 >-Q) (M2»
(12.6.8)
,f
so that we get for the total magnetisation
Solution
= -gllB L
-(MI' Bm) -(M 2 • B~if) ,
where
q2
Taking into account that there are two sublattices, we write the Hamiltonian in the form
311
where the f), gl (f2' g2) are lattice sites on the first (second) sublattice. In the molecular field approximation we replace S'I and S'2 by their averages (S» and (S2>, which will now be different, and we rewrite the Hamiltonian (12.6.1) in the form
(12.5.7)
12.6 In a ferromagnetic substance I(f - g) is positive when f and g are nearest neighbours. If this quantity is negative, we are dealing with an antiferromagnetic. To consider an antiferromagnetic we assume that the crystal can be divided into two sub lattices such that all the nearest neighbours of 3. spin on one sublattice are on the other sublattice. We shall now have two averages which will be different for the two sublattices, as a negative value of I(f - g) implies a preference for an antiparallel alignment of nearest neighbours. Use the molecular field approximation to find the susceptibility of an antiferromagnetic at high temperatures and to find the Neel temperature, that is, the temperature below which both sublattices show spontaneous magnetisation. Consider again the spin--! case.
II
Cooperative phenomena
12.6
<M>
=
"'"
2B9 T+9) ,
(12.6.9)
with 2LI(g) Q = QI+Q2 = -N{gIlB)2 ,
(12.6.10)
which is the same as expression (12.3.7) apart from the sign, and f2
[B+2LI(fZ-g)Sgl+ 2 gllB II gllB
ICf2-g2)SIJ}, (12.6.1)
9
QN(gIlB )2
(12.6.11)
Chapter 12
312
12.6
To find the Neel temperature, TN, we put B = 0 and expand the tanh, as just below TN the sublattice magnetisations will be small [compare the solution of Problem 12.3 (in]. We then have . (MI> ~ -T(ql(M I >+q2(M2»+term
q
of third order in (Mi>,
e
I
12.7
Cooperative phenomena
313
parallel to (M 2 > (see Figure 12.7.1). The total magnetisation (M I >+(M 2> will be parallel to B, and its magnitude, divided by B, will give us Xl From Figure 12.7.1 we then see that Xl
.
(12.7.3)
q2
(12.6.12)
(M2) ~ - qT(q2(M2 >+ql(M I »+·...
(M 1 )
For an antiferromagnetic we always have a vanishing total magnetisation when there is no external field so that we can put in Equations (12.6.12) (M I >::; - (M2 >and we then see that the sublattice magnetisations vanish at TN given by the equation (12.6.13)
TN =
We note that if we had assumed that there were only nearest neighbour interactions q I would be zero and q2 = q so that TN and e would be the same. We also note that in the unlikely case where q I > q2 we would have no transition; this is not surprising, as q I > q 2 would mean that there would be a strong tendency for spins in the same sublattice to align antiparallel. In that case, the model used would clearly be a very poor one. 12.7 Find for an antiferromagnetic of the type considered in the preceding problem the perpendicular and parallel susceptibility Xl and XII (for the cases where the applied field is perpendicular or parallel to the sublattice magnetisations) at temperatures below the Neel temperature. Solution
Let
(M 2 >
Figure 12.7.1.
Consider now We then expect all vectors to be parallel (or antiparalIel) and, moreover, for small B we expect (M 1 > and -(M 2> to be nearly equal to m so that we can write (12.7.5) (M 1 > = m+oml , (M 2> = m+om2' We use again Equation (12.6.7) and write for the argument of the first tanh !{3gI1B(B -q I (M 1 >-q2(M2»
(M1>o =
(M 2 )o
m,
(12.7.4)
Bllm.
=
(12.7.1)
! (3gl1Bm(q2 -q I) +!(3gI1B(B -ql om. -q2 om 2),
(12.7.6)
where the index 0 indicates the absence of an external field. Consider first the case B1m. (12.7.2)
and similarly for the argument of the second tanh. Expanding in terms of the small quantity which contains B, omh and om2, we obtain from Equation 02.6.7)
The effective fields are again given by Equations (12.6.4). We are interested in the susceptibility, that is, in the value of the magnetisation when there is a small external field present, and as B1 m, we expect that (M 1 > and (M 2 ) will be no longer strictly antiparallel. Let € be the angle between (M t ) and -(M 2 >. We expect € so that from Equation (12.6.4) it follows that B -q2(M 2>must be parallel to (M 1 >and B -q2(M 1 >
om l om2
=
i{3N(gI1B)2(B -qlomt -q2om2)cosh- 2[!{3gI1Bm(q2 i{3N(gI1B)2(B -ql om2
-qdJ , }
ql om I) cosh- 2[! {3gl1Bm(q2 -qdl .
(12.7.7)
Adding these equations, we find the total magnetisation (MI>+
om l +om2 ,
12.7
Chapter 12
314
and dividing by B we find Xl!. The result is
2A q(T+ A) ,
XII
where
ecosh- 2 rtiJgJ.LBm(q2 -ql)] ,
A
(12.7.8) (12.7.9)
q2
XII
Xl) ,
(12.7.1
where we have used Equation (12.6.13). As T -+ 0, A vanishes exponentially so that XII -+ O. Finally, we note from Equation (12.6.9) that as T approaches T,.. from above, the susceptibility also approaches 1/q 2' 12.8 Consider the Ising model. We can write the Hamiltonian in the form H = -!gJ.LBLJ.Lr f
L lJ.LfJ.Lg ,
= -!gJ.LBBLJ.Lr-!zLIJ.Lfii, r
f
where J.L is the average value of J.L and z is the coordination number, that is, the number of nearest neighbours per spin. This approximation essentially reduces the problem to that of a single spin problem, or, as we saw in Problem 12.4 to assuming that the + I and -I values are randomly distributed over the lattice. The next approximation takes into account that because of the interaction there will be a preference for ++ pairs or -- pairs, rather than + pairs. Let Q be the total number of pairs in the lattice, so that Q =
HZ+l =
(
!gJ.LBB J.Lo+
tzN.
( 12.8.3)
Let Q++, Q+_, Q__ be, respectively, the number of ++, + ,and -- pairs. Prove the following relations for Q++, Q+_, and Q--:
2Q+++Q+_ : zN+, } 2Q__ + Q+_ - zN_ ,
(12.8.4)
z) -!gJ.L
L; J.Lj j l
where 0 indicates the central spin and (i) Prove that in this approximation
(12.8.1)
(12.8.2)
( 12.8.6)
A given state of the system is now characterised by the long-range parameter R and the short-range parameter a. To proceed further we consider instead of a one-spin system the system of z + 1 spins where we take into account explicitly the interactions between a central spin and its z neighbours, but take the influence of the other spins in the lattice into account through a mean field B'. That is, we consider the Hamiltonian. z
BB'J J.Lj-HJ.L0 1 j
i
z
L; 1J.Lf,
(12.8.7)
= I, ..., z its nearest neighbours.
· ~
(12.8.10)
316
Chapter 12
12.8
If we write
Cooperative phenomena
12.8 We note that when T
K
~{3gPsB,
K'
!(3gPs(B+B') , J=~{3I,
>
,R
Z
!J.oL±'1
!J.J±.l
"·!J.z
~±'1 exp (KPO+ K'~Pj+JLPOPj)
02.8.12)
Z =
exp(Kpo+K'Pl +JpopdD exp[(K'+Jpo)Pj] (12.8.13) I
!J.O=±.1 !J.l " " 1.... 1....
/ , z exp(Kpo+Kpl+JpOP1)[2cosh(K+Jpo)]
= exp(K - K' -J)Z~~l
1 l
,
exp(-K + K' -J)~-) 1, Z-- -- exp (-K-K' + J)Z(-) z- 1 , Z_ +
where
j
(12.8.15)
exp(Kpo)[2cosh(K + K' +JpoW
= Z++Z_,
1.02.8.14)
Let us denote the four terms corresponding, respectively, to the Po, Pj values: + I, + I; + I, -1; 1, + I; -I, -1 by Z++, Z_ +, Z__. We have Z++ = exp(K + K' +J)Z~+ll ,
L
02.8.22)
!J.o ±.l
2
!J.o ±.1!J.I =±l
(12.8.21)
To find the transition temperature, one carries out first the Pl summation in Equation (12.8.14) to find
z
=
I-x 1 +x .
o
L
=
= 0 so that then
(12.8.11)
we have
311
( 12.8.23)
with Z±
= e±'KZ~±') .
(12.8.24)
From Equation (12.8.12) it follows, on the one hand, that
( ) _ olnZ _ Z+-Z_
(12.8.25) Po Z On the other hand, we have Z+tanh(K'+J)+Z_tanh(K'-J)
" _ olnZ 02.8.26)
L,.. (Pj) - oK' z Z I
Z~~)l = l2cosh(K' ± J) jZ-
(12.8.16)
1 .
From the consistency condition
From the general theory of partition functions (compare Chapters 4 and 6) it follows that Z++:Z+_:Z_+:Z__
(Q++):~(Q+_):!(Q_+):(Q __ ).
02.8.17)
Equation (12.8.8) follows immediately. In this derivation we have used a symmetry argument to argue that Z+_ must equal Z_+. One could say that this is the argument which determines K' in terms of K and J, and we shall see presently, that indeed, a consistency requirement leads to K' such that Z+_ Z_+ (see part (iii)l. From Equations (12.8.3), (12.8.4), and (12.4.1) we find easily Q++ Q--
= izN(l +0+2R), = izN(1 +0 2R),
1
02.8.18)
Q+- = !zN(l -0) , and hence from Equation (12.8.8) we find (1 +0+2R)(1 +0-2R)
(1-0)2
whence
=
l-x 2
1 L(Pj) ,
z
(12.8.27)
j
we then find Z+ I +tanh(K' -J) Z_ = 1 tanh(K' +J) ,
or, using Equations (12.8.24) and 02.8.16), cosh(K' +J) 2(K' - K) - exp cosh(K' -J) z 1
(12.8.28)
(12.8.29)
from which the equality of and Z_+ also follows. To find the transition temperature we put B = 0 in Equation (12.8.29) so that now (12.8.30) K' = ~ {3gP s B' , while the equation for determining K' becomes
x
-2
,
1 +x 2 -2x[I-R 2 +R 2x 2 j'h o
=
(12819) ..
2K' cosh (K' +J) cosh(K' -J) = exp z -I
(12.8.20)
We note that K' 0 is always a solution. The transition temperature is that temperature for which a non-vanishing solution for K' becomes
(12.8.31 )
Chapter 12
318
12.8
Expanding both sides of Equation (12.8.31) for small values of K' up to K'3 we obtain, after taking the cosh(l + K) 2K' Incosh (J - K') = Z (12.8.32) whence In (I + 2K' tanhl + 2K '2 tanh 2 1 -1 K'3tanhl + ...) ( 12.8.33)
13
Green junction methods(1)
or
D.ter HAAR (University of Oxford, Oxford)
, 2 '3 2 2K' 2K tanhJ-JK' tanhl(l +2tanh l)+··· = z-I . (12.8.34)
As K'
-+
MATHEMATICAL PRELIMINARIES
O. it follows that tanhlc
=z I '
13.1 Show that, if [A,B1_ ==AB-BA :::: K, where A and Bare operators and K is a c-number, and if A is a c-number, e X(A+B) e Me AB e-Y>x2K (l3.1.1)
(I
or I +xc - z
whence
(12.8.36)
I '
Let .p(A) == eXAe AB • We then have
2 Xc::::
At temperatures just below K'2 satisfies the equation
K'2
Solution
a~~A):::: AeMe
(12.8.37)
z
, we find from Equation (12.8.34) that
We further have ~
[eM,BI_=
3(tanhl - tanhlc ) tanhl + 2 tanh 3 1
(12.8.38)
eMBeAB = (A+B).p(A)+[eM,B1_eAB. 03.1.2) An
L n on
becomes and the definition (12.8.9) of molecular field approximation Xc
= exp(-/3cJ)
X
it follows that in the
exp(-2/z) ,
( 12.8.39)
which is the same as expression (12.8.37) in the limit as z 00. To find a in that limit we note that it follows from Equations (12.8.9) and (12.8.11) that X exp(-21) ~ 1-21+ ..· (I in the limit as / -+ 0 (and also 1 -+ 0). We then get from Equation (12.8.20) after some expansions a ~ R2 . {I 2.8.41)
I ..
l\1KAn-l=AKe M . (13.1.3)
Hence or
a.p = [A+B+AK1.p, aA
(13.1.4)
.p = eX(A+B)+'I\X 2 KC ,
(J 3.1
where C is an arbitrary constant operator. By letting A -+ 0, we find that C is the unit operator, which concludes the proof. If [A, B1- D, where D is an operator, expression (13.1.1) becomes more complicated. We note that we can only commute e A and e B , if A and B themselves commute. 13.2 Show that if
is the unit step function, 1,
t>O;
e(t)
0,
t
< 0,
(13.2.1 )
we have
We note that we can write Equation (12.4.9) in the form E = -VQa,
L n 0 =
(iv) From Equation (12.3.10), the fact that for the Ising model L/(g) g
An
~
[An,B1_=
( 12.8.42)
which is now the same as Equation (12.4.3), if (12.8.41) holds. In fact, one can prove generally that all results in the present approximation become those of the molecular field approximation in the limit as z -+ 00.
8(t-t')
== :t[8(t-t')1
-(}(t'-t)
= o(t-t').
(13.2.2)
(1) For a general discussion see: V.L.Bonch-Bruevich and S.V.Tyablikov, Green Function Methods in Statistical Mechanics (North-Holland, Amsterdam), 1962, and D.ter Haar in Fluctuation, Reio.xation and Resonance in Magnetic Systems (Ed.D.ter Haar) (Oliver and Boyd, Edinburgh), 1961. ~HI
320
Chapter 13
13.3 Make it plausible that
+
== lim -+0 X
y
E
13.3
f!II(~) + i1l"D(X), X
- IE
(13.3.1)
Solution
To prove the equality (13.3.1) we consider the integral < .. 0
J
f(x)
- - . dx,
-00
X
03.3.2)
IE
= lim lim €
-+ 0 8 -+ 0
(i
-8
+
f+OO) +8
-00
f(x)
--. dx + lim lim X -IE
€ ..,. 0
8 -+ 0
i
+8
-&
f(x)
- - . dx.
X -IE
ACt)
=
321
The first term gives the principal value of
I
r
iCH - JiNo )tJ i(H - p.IVo h P A exph P
)t] ,
(13.4.3)
a
are functions of t
(13.3.3)
exp
with JJ. = O:/{3. The generalisation from the usual Heisenberg operators is necessary whenever Hand Nop do 110t commute, as should be clear from the solution of the present problem and the result obtained in Problem 13.1. Prove that «AU); B(t'»), as well as the correlation functions FBA
where we assume that has no singularities on the real axis. We write this in the form I
Green function methods
according to our convenience, and the A(t) and B(t') are time-dependent (Heisenberg) operators, defined by the equation
where x and E are real and (!jJ indicates that in an x-integration the principal value of the integral must be taken.
I = lim
13.6
= (B(t')A(t»
and
F~B
= (A(t)B(t'»
(13.4.4)
[' only.
Solution
This follows by writing out explicitly the grand canonical averages, using the facts that (i) inside a trace operators can be cyclically permuted, and (ij) the various factors commute.
f~f(X~dX 13.5 Prove that the «A motion:
and the second term gives
. .f
hm hm
+1
+ «[A, H - JiNop]-(t);
ih«A(t); B(t'»)
r,
I
-rn=o\r I 1-
K '
L'"
dr'
L~rdre-"r
S:
(14.5.2)
dr' e- Kr ' (14.5.3)
14.6
Chapter 14
336
14.6 Estimate for the Debye theory the average number no of negative particles in the Debye sphere.
14.10
The plasma
337
Use Equation 04.9.1) to derive the following plasma equation of state: (14.9.3)
P = nokT [1 - 18!zo ] '
Solution
no ;::::; trrnorb
I
3V41T
(kT) 'h e 2 n 'Il
with no the quantity defined in Problem 14.6.
,
which, as follows from Equation (14. 2.6), is large compared to unity for all cases where the Debye theory is applicable. 14.7 Estimate for the Debye theot:"Y the ratio of the energy found in the preceding problem to the fluctuations in energy in the Debye sphere. Solution
From Problems 14.4 and 14.5 it follows that the energy attributable to each particle is of the order of e 2 #< which is of the same order as Eo· The total electrostatic energy in th~ Debye sphere will thus be of the order noe 2K. and the fluctuations of the order vnne2K., so that the required ratio is of order which, as we saw in Problem 14.6, is a large number.
vnn
14.8 Find for a hot dilute plasma an expression for the Debye length in a mixture of ionised gases.
Solution
For a plasma the evaluation of <W)tav is simplified by the fact that all forces are Coulomb forces. For W we have W
kr j
I
I'
where the summation is over all ki:J1 d S of ions while ej and nj are their respective charges and densities. 14.9 In Problem 9.16 it is shown that one can use the virial theorem to write the equation of state in the form pv = NkT +~<W)tav , (14.9.1) where W is the virial deriving from in termolecular forces, that is,
W=
L (F
j •
rj) ,
(14.9.2)
Li (rj . 'ViU) ,
-
i
(14.9.4)
where the superscript C indicates that where we are dealing with Coulomb forces and where U is the total electrostatic potential energy. As U is a homogeneous function of degree -1, we thus have from Equation (14.9.4) W= U.
(14.9.5)
To get the time average of U, we write it in the form -
t
1
U-
2 .. I
I
ejej
r.. II
I '\'
2
'
i,..ei"'; ,
(14.9.6)
I
where e; is the charge of the ith particle, rij = Irj-rjl, and "'; is the potential at ri due to all other ions, so that we can write,
-e;J
, ["'i(r) ",,(ri) = Ir-r.1
Solution
Using arguments similar to those 'Which led to expression (14.1. 2) we find 2 411"L e·n· (14.8.1 ) ~ K. 2
L(ri· Ff)
I
r
',.
'
04.9.7)
which is the total potential acting at r; less the potential due to the ith ion itself. For "'i(r) we can take the Debye potential (14.1.1) and taking the average [compare expression (14.4.1)] we get <W)tav
tav indicates an average both over the time and over all particles in the system.
P
Pc o+pe,
(14.10.1)
where the first term is the perfect gas term (we assume that the only interactions are Coulomb interactions) while the second term derives
338
Chapter 14
14.10
14.12
from the electrostatic interactions. Prove that
The plasma
Solution
If we use expression (14.10.2) and write
(14.10.2)
ue
where ve is the total electrostatic potential energy, by considering the configurational partition function, introducing new variables r; where L3 = the volume occupied by the system, and finally using the fact that ve is a homogeneous function. Solution
We have the following relations:
,•
with the solution ve
~I
d3rl ... d 3 rN exp(-{3ve) ,
and
(iJiJvS
_ ~~
12
I
ve(rJ = ve(r;L)
a
Q 3L2 aL L
3N
f' d3r
l ...
I
,
(14.10.6)
r
J' (14.10.7)
(3ve(r·) I L I
where the integration is now over a unit volume. The only L-dependence behind the alai sign lies now in the factor L3N and in the L in the index of the exponential, and there is nowhere an implicit dependence. Hence we get, after taking the derivative with respect to L and changing back to the old coordinates, {3p
QIf3 Jd r l
...
(14.11
(14.11. 7)
3 1 I ne d rN;-[N+ J {3ve]exp(-(3u ),
I
I
14.12 Considering the charges in a plasma to be quantities which can be changed adiabatically (scale transformation), and considering the total electrostatic potential energy ve as a function of the entropy S, the volume, and the magnitude of the charges e, find a general expression for Uin the form Ne 2 ve f(x) (14.12.1) where
po = NkT+~(ve).
(14.10.9)
prove that for a hot dilute plasma
= adiabatic invariant.
I)
CiJ~t'lJ
(14.1
04.11.1)
= IN, and x is a dimensionless adiabatic invariant.
From Equation (14.10.4) it follows that 2ve e
iJve) __ I ve ( iJv S.e 1 v
(14.12.2)
Applying a general procedure, explained for instance in Problem 1.20, we have
14.11 From Equation (14.10.2) and the thermodynamic equation of state Problem 1.9a) ave) (iJpe) ( Tv T = T iJT IJ
VI
Solution
or
vT 3
(iJpe) iJT IJ'
which proves the expression (14.11.2).
I
d 3rNexp
T
(14.
1 = -Lve(r;)
I
)
and Equations (14.10.2) and (14.1 1.5) we get se = 4ve jaT 3 v,
(14.10.4)
,
j
rtlL, and using the fact that
_ I _ I _
(14.11.5)
Using the Maxwell relation (see Problem I
(14.10.3)
alnQ {3p=-- . av
(3p
aT 4 1).
uev
e
i
we have
=
1_,1
ve =
= U, r;
(14.11.3)
(14.11.4)
where the integration is over the volume v of the system,
Putting v
ve v '
we can write Equation (14.11.1) in the form
;1"
f
Q
339
I
2ve ve -de--dv+TdSe e 3v
dve or
oev
I;'
=
adiabatic invariant.
Relation (14.12.1) now follows immediately.
(14.11.2)
\!
(14.12.3) (14.1
14.13
Chapter 14
340
14.13 Use the result of Problem 14.11 and Equation (14.9.8) to find first an expression for x and then the form of f(x) in the Oebye approximation. [Use dimensional analysis.]
15
Solution
From Equation (14.11. 2), and the fact that the only natural constants which can occur in x are e and k, we find by dimensional analysis {3e2 ---v,
x
Negative temperatures and population inversion
(14.13.1)
VI
It is of some interest to write this adiabatic invariant in a form by using Equations (14.1. 2) and (14.
pi
J
x :::.::
2
')'J
K VI
2rD
(14.13.2)
The adiabatic invariance of v/rb is plausible, as one would expect that a simultaneous slow change in the volume and in the Oebye cloud would preserve the degree of order and hence the entropy. From Equations (14.13.1) and (14.9.8) it now follows at once that in the Debye approximation f(x)
=
U.M.TITULAER (Rijksuniversiteit, Utrecht)
different
04.13.3)
15.0 The concept of negative temperature is introduced by means of a statistical model of a spin system. The thermodynamic peculiarities are discussed and the possible use of negative temperature systems and systems with non-thermal level occupation as amplifiers of radiation is indicated. Ramsey's classical paper on the subject(l) may serve as a general reference, especially for subjects treated in the first few problems in this chapter. 15.1 Consider a system of n identical weakly interacting spins. Each of the spins occupies one out of 2s + I equally spaced non-degenerate energy levels with energy mW (m = -s, -s+ I, ..., +s). The average energy of each spin is uo. Find the maximum entropy distribution and the relation between Uo and the temperature parameter fl Calculate the partition function of the system and the specific heat as functions of {3 for positive as well as negative values of {3. [Use the relation {3 = 1/kT to define temperature for negative {3.] Find the simplified expressions for the case s = !. Solution
As for an ideal gas the partition function is simply the product of n identical factors, one for each spin. A straightforward application of the procedure used in Problem 2.2 gives for the probability Pm of occupying level m of any single spin -I e-/lmW Pm = Z({3)
with Z({3)
-/lmW _
m
e
-
sinh(s+-!){3W . hi RW gn ~~
(3 will be determined from the relation Uo
=-
alnZ({3)
:\(./
=
W[(s+!)coth(s+!){3W
(1) N.F.Ramsey, Phys. Rev., 103,20 (1956).
341
coth-!{3W].
342
Chapter 15
15.1
15.3
As a result of the concave character of the function cothx the sign of U o is negative when ~ is positive and conversely. For the specific heat we obtain by differentiating once more auo c = -nk~23i3 =
For s
Negative temperatures and population inversion
Solution
(a) Let p(E)dE denote the number of states with energy between E and E +dE. The average value of the energy is given (in a canonical ensemble) by
nk~2W2L!cosech2!~W-(s+!)2cosech2(s+!)~W].
Uo
! we obtain the simplified expressions Z(~) = 2cosht~W ,
The entropy for a system with s
S
!nk~2W2sech2!~W.
t is given by
= n(k~uo+klnZ) nk[~Wtanh!~W+ In(2cosh!~W)] .
For the system with s = ! the expressions are simpler and their general features (signs of Uo and ~, zero of C at ~ = 0) are more easily found. On the other hand any system which can be described by means of a set of occupation probabilities P-'h and P+'h can also be described by means of some inverse temperature k~; the additional information that lnpm is proportional to the energy of level m, which follows from entropy !. maximisation, does not imply any additional constraint in the case s This was our reason for not taking s = t from the beginning. 15.2 (a) Show from general formulae for the canonical ensemble that negative temperature distributions are possible if and only if the density of possible states of the system decreases fast enough as a function of energy; specify 'fast enough' quantitatively. (b) Show that this condition is violated if the energy of the system includes the kinetic energy of a particle or the energy of a mode of the radiation field. [The results of Problems 3.13(c) and 3.14(c) may be used.] For a spin system this implies that negative temperatures are meaningful only if the coupling of the spins to the kinetic degrees of freedom in the 'lattice' and to the radiation field can be neglected. In practical terms this means that the time in which the spins reach equilibrium among themselves should be much shorter than the time in which equilibrium with the lattice or the radiation field is established. J (c) Consider a system for which the density of possible states increases that only temperatures exponentially: peE) 0: e o£, a > 0; show T < l/ka are permissible. [An exponentially rising density of states was postulated (2) in a statistical model of strong interactions at high energies. The possible consequences of the resulting maximal temperature were also discussed. J (2) R.Hagedorn, Nuava Omenta Suppl. III, 147 (1965).
= fe-IlEEP(E) dE
I
f
e- IlE p(E) dE .
Unless peE) decreases at least exponentially with E this expression diverges for any negative value of~. Even the total probability that the system occupies any state with energy lower than any given value Eo would be vanishingly small. Such distributions are physically unaccept able. (b) For the kinetic energy of a particle we found in Problem 3.1 that the number of states with a velocity between Vand V +d V is proportional to V2d V. This corresponds to EY'dE, which is not exponentially decreasing. For the quantum mechanical case a similar result was derived in Problem 3 .13( c). For a single mode of the radiation field peE) is a constant; for the photon field as a whole peE) 0: E2 [see Problem 3.14( c) J. Finally, if a system consists of two parts such that
uo(~) = - Wtanht~W ,
C
343
E
E(1)+E(2), then peE) = IE dE'p(1)(E')p(2)(E-E').
In this formula
p(l)(E) and p(2)(E) are the density of states for the subsystems. 1t is obvious that peE) will only decrease exponentially if both p(l) and p (2)
decrease at least exponentially. (c) If peE) 0: e aE a comparison with (a) shows that values of ~ such tha t ~ .;;; a are unacceptable. Temperatures higher than 10 = 11 ka are therefore impossible in this model. 15.3 Suppose that the level spacing constant W in a system of n spins t is proportional to an external magnetic field: W = /lH. Calculate the amount of heat and work which has to be supplied to the system if we want to keep it at a constant temperature Tl while the field is changed from HI to H 2 • Find also the amount of work needed to increase the field adiabatically and the relation between temperature and field during such a process. What is the net effect of a Carnot cycle operating between two negative temperatures? [The heat absorbed is given by the relation dQ = TdS, for both positive and negative temperatures. J Solution
From the formulae derived in Problem 15.1 we find for the increase in U and S during isothermal magnetisation ~
(ll.U)TI
= n/l[H2 tanhx 2 -
(LlSh l
=
k~l(ll.U)TI +nk[lnZ(x2)-lnZ(xdJ ,
=
!/l~IHI,2 .
,
'i
I
\
Xl.2
HI tanhx d
,
15.3
Chapter 15
344
The amount of heat supplied to the system equals TI (LlS)Yl' The amount of work exerted on the system is given by (M)Yl
=
-nkTIllnZ(x2)-lnZ(x l
)]·
The relation between T and H along an adiabatic line follows from the observation that S depends only on the combination x = !{3J,J.H. Thus x must be constant along an adiabatic and Hand T are proportional to one another. The amount of work absorbed during an adiabatic increase of His (LlA)SI
=
nJ,J.(H2 - Hdtanhx
=
T2 -TI
nJ,J.HI-------y;-tanhx .
We now consider the Carnot cycle
(H2, T I )
adiabatic
(
~ H3
=
T2 ) TI H 2, T2
X2
= !{3IJ,J.H2
j iw'''om,,'
adiabatic
(
H4
=
~
TIHI' T2
)
XI = !(3IJ,J.HI
TI-~
(LlA)tot = -r,-(LlQ)Yl = -r--(LlQ)Y2 . I
345
15.4 (a) Two systems of n l and n 2 spins! respectively, are initially at different temperatures and are then allowed to exchange energy. Determine the final equilibrium temperature and the direction of heat flow. [The level spacing may be taken the same in both systems.] (b) Combine this result with that of the preceding problem and see whether Kelvin's and Clausius' formulation of the second law can be maintained if negative absolute temperatures are admitted. If not, propose a modification. It is reasonable to call a temperature Tl hotter than T2 if the energy content of the system at TI is higher than at T2. With this convention, negative temperatures are hotter than positive ones and -0 indicates the hottest conceivable temperature. Like +0 it cannot be reached in a finite number of steps. (a) The total energy content of the system is equal to n lu({3d+ n 2u({32)' At equilibrium each spin is at the same temperature; this final tempera ture corresponds to u({3d
By adding the amounts of work done on the system in each of the four stages one finally obtains ~-TI
Negative temperatures and population inversion
Solution
1i,o,"'=" (HI> T I )...
15.4
2
Both for positive and for negative values of TI and ~ there are two possible results: (1) an amount of heat LlQ is absorbed from a bath with temperature I;.; a fraction (1 - ~/Td is converted into work, the rest is discarded into a bath with I ~ I < I Til; (2) work is exerted on the system while heat is transported from a bath with low I TI to one with higher ITI. The meaning of these results and their relation to various formulations of the second law will be discussed in the next problem. Notice that we do not consider Carnot cycles between positive and negative values of T. In our model such processes would be possible if the adiabatics would cross at H = 0, T = O. However in this region our model is no longer thermodynamically correct. In real spin systems the interaction between the spins would no longer be negligible and some sort of magnetic ordering would occur in the immediate neighbourhood of H = 0, T = O. A similar phenomenon was observed in the case of the classical ideal gas (Problem 1.26).
=
[n IU({3I) + n2 u({32)]
(n l +n2)
__ ~ (_ nltanh!{3lw+n2tanh!{32W) wargtanh +n . n
(3f -
l
2
Since u({3) is a monotonically decreasing function, equilibrium can be reached only if energy is transported from lower to higher values of {3. If we adopt the definition of hotter and colder proposed in the problem, then the natural direction of energy flow is from hotter to colder temperatures. If we recall that dQ = TdS and notice that (dU/dS)w < 0 for T < 0 we see that the same is true for the natural direction of heat flow. The sign conventions used for dQ and the ordering of temperatures according to 'hotness' are reasonable but not unavoidable (3). (b) With our definition of 'hotter' the possible results of Carnot cycle between negative temperatures as calculated in Problem 15.3 can be viewed differently: (1) when heat is converted into work there is a heat flow from a cool to a hot reservoir; (2) when work is converted into heat some heat has to flow from a hot to a cool reservoir. By combining this with the possibility of natural heat flow from a hot to a cool reservoir we see that Kelvin's principle can be violated. It must be changed into the modified statament: It is impossible to devise an engine which, working in a cycle, would produce no effect other than: (1) the extraction of heat from a positive temperature reservoir and the production of an equal amount of mechanical work, (3) The consequences of alternative choices are discussed in Section 13 of P.T.Landsberg, Thermodynamics with Quantum Statistical Illustrations (lnterscience, New York), 1961.
15.4
Chapter 15
346
15.7
Negative temperatures and population inversion
347
(2) the rejection of heat into a negative temperature reservoir while an equal amount of work is done on the engine. Clausius' formulation of the second law remains unaltered.
The model discussed in this problem was introduced with reference to thermodynamical aspects of laser action (4). The relation with laser amplification will be discussed in the next problems.
DYNAMIC POLARIZATION
15.6 A beam of N photons is directed at a sample containing n two level systems at temperature T. The frequency I-' of the photons is resonant with the level splitting (hI-' = W); the probability that a single photon is absorbed by a system in its lower state and excites it to its upper state is equal to A. Using Problem 3.20, calculate the attenuation or amplification of the photon beam. Neglect spontaneous emission by the system and effects of the size of the sample. [The attenuation or ulI;phft...:ation of resonant radiation is a sensitive indicator of the temperature of a spin system. The presence of negative spin temperatures in a sample and the relaxation to positive temperatures J were first demonstrated in this way by Purcell and Pound
The most direct way of bringing a spin system from a positive to a negative temperature is by means of a sudden reversal of the magnetic field. The spins cannot follow the reversal and are left with a polarization opposite to the field. Since this method is essentially not quasistatic and not completely reversible it is not suitable for obtaining very hot negative temperatures. The alternative method of dynamic polarization, sketched in this problem, is analogous to the techniques of optical pumping. 15.5 Consider an impurity in a magnetic crystal with three relevant energy levels. The transitions between the levels may occur via different mechanisms. Those inducing transi (3) • tions between levels 1 and 3 and between levels 2 and 3 are repre sented by heat baths with tempera tures 11 and ~. The transition between levels I and 2 can occur by means of energy exchange with the spin system. Find the temperature --+--(2) of the spin system at which there is no longer any net heat flow between the reservoirs, and between the (I) I f reservoirs and the soin system. Solution
The transition between levels 1 and 3 is in equilibrium with a heat bath at temperature 11 if the occupation probabilities of the two levels satisfy the relation P3 PI exp(-.81 WI)' In the same manner we find P2
=
P3 exp( +.82 W2 )
= PI exp(-.81 WI +.82 W2 )
•
Equilibrium is possible only if the temperature of the spin system satisfies the relation .81 WI -.82 temperature will be negative if .81 is small enough This . compared to .82 ; in terms of the temperatures TI and ~ we must have WI
11 > 12W2 .
Solution
According to Problem 3.20 the probability of absorption by a system in the ground state is equal to the probability of stimulated emission by the same system in the excited state. The net increase (or decrease) in the number of photons is therefore equal to I:J..N == NnA(-p_% +p+%).
In this formula P-lh and P+% are the probabilities that a two-level system is in its upper or lower state. A comparison with the result of Problem 15.1 gives I:J..N == -nA t anh 1.8W . If the temperature of the two-level systems is negative the beam of photons is amplified by it. This is the principle of maser or laser amplification. Neglect of spontaneous emission is justified if the number of photons in the relevant modes of the radiation field is large compared to unity. For radio and microwave frequencies this is usually the case; in the optical region it is often necessary to include effects of spontaneous emission. 15.7 In Problem 15.6 we saw that the energy in a two-level system at negative temperature can be extracted by means of stimulated emission. In this way it may be used for the build-up of a coherent electromagnetic oscillation. The system discussed in Problem 15.5 could be operated in such a way that heat from the reservoir at T, is partially converted into coherent oscillation. Show that the efficiency of this energy conversion cannot exceed that of a Carnot process operating between TI and ~. ( 4) P.Aigrain in Quantum Optics and Electronics, Les Houches 1964, Eds. C.DeWitt, A.Blandin, and C.Cohen-Tannoudji (Gordon and Breach, New York), 1965, p.S27. (5) E.M.Purcell and R.V.Pound, Phys. Rev., 8t, 279 (1961).
Chapter 15
348
15.7
15.8
Negative temperatures and population inversion
349
For a stationary state we must have
Solution
From the total energy furnished by the heat bath at TI a fraction WdW I has to go into the reservoir at~. Stimulated emission can only exceed absorption if at least the equilibrium temperature of the spin system is negative. According to Problem 15.5 this implies that W2 /W I > ~/Tt· Consequently the conversion efficiency 17 has to obey the relation (6) W3 72 17';;;;- but not the number of photons N. Next we consider the equations for the three-level systems. Suppose the temperature Tl is so large that the upward and downward transition rates between levels I and 3 caused by this pumping system are equal: A 13 A 31 = P. Suppose further that the reservoir at 72 and the systems causing transitions between levels I and 2 (other than the radiation field) are so cold that the upward transition rates can be neglected. Give equations for the change in occupation probability of the three level., for given values of P, A 32, A Zb and B21 and a given number N of photons in the resonant cavity mode. By putting the rates of change equal to zero the degree of inversion in the stationary state can be derived. Simplify this expression by assuming that A 32 is much larger than the other transition rates. (c) By comparing the results of parts (a) and (b) the number of photons in the cavity N can be found. For which values of the pumping rate P is laser action possible? Express the laser output in terms of P-Per • Solution
(a) By adding a term for the losses to the result derived in Problem 15.6 we find for the change in the number of photons dN dt
Llner
N
- Tc + nB21 (P2 -PI)N .
(6) H.E.D.Scovil and E.O.Schulz-DuBois, Phys. Rev. Letters, 2, 262 (1959).
Putting the left hand sides equal to zero and using PI + P2 + P3 = 1 we obtain for the stationary value of the inversion P(A 32 - A zl ) - A21 A 32 (P2 - PI},t = (P+ A 32)(2NB21 + A 21 )+ P(2NB21 + A32 + A2d . If we assume that A 32 is much larger than all other transition rates, this expression becomes P-A 21 (pz-pdst 2NB21 +A 21 +P' (c) A stationary regime is only possible if this inversion rate is equal to the critical inversion rate calculated in part (a): I P-A 21 nB21 Te 2NB21 +A z1 +P or nTc 1 N T(P-A 2d 2B21 (P+A 2 tl· The critical pumping rate is equal to (put N
0):
1 + l/nB21Tc -1/nB21 Te The power output of the laser equals N /Tc • It can be rewritten in the form N/Tc Hn-Llncr)(P-Pcr )' p. - A cr -
211
Notice that t (n - Lln er ) is exactly the number of systems in level I when the critical inversion is reached. The quantity nplP is the number of transitions to level 3 induced by the pumping heat bath. A fixed number of these excited systems is evidently needed to overcome losses due to spontaneous transitions; the remaining ones give rise to photons leaking out of the cavity. Modifications and improvements of this laser model can be found in any textbook on the subject (7). (7) e.g. A. Yariv, Quantum Electronics (John Wiley, New York), 1967, Chapter 15.
16.1
16
Recombination rate theory in semiconductors J.S.BLAKEMORE (Florida Atlantic University, Boca Raton, Florida)
Recombination rate theory in semiconductors
enjoy the same translational wavefunction; we regard these as distin guishable quantum states.) Fermi-Dirac statistics are appropriate for describing the occupancy of 'one-electron' states in accordance with the Pauli principle. An electron which occupies a 'one-electron' state acts dynamically as an independent particle, not interacting with the other electrons in 'one-electron' states, although the properties (e.g. energies) of the 'one-electron' states are affected by the average behaviour of all of the electrons in the system. As discussed in Chapter 3, the time-averaged probability that a state of energy E be occupied by an electron is of the form 1
ftE) 16.1 (a) For non-interacting or 'one-electron' states at energy E, write down an expression for the probability that any such state is occupied. You may employ the knowledge of Fermi-Dirac statistics gained from Problem 3.12. In the present chapter, we shall want to express occupancy probabilities in terms of an electrochemical potential or Fermi energy EF for equilibrium situations. (b) Given that a band of permitted electron states extends upwards )'h, express in from energy Ec with a density of states geE) = A(E integral form a condition for EF at temperature T when no electrons occupy the band at equilibrium. How can this be simplified for the case of no quite small? (c) The kind of band we have considered is usually referred to as the conduction band of a semiconductor, containing a few electrons in the lowest states, and many empty states. When thermal equilibrium is perturbed, the conduction electron density n may well be different from no. Use a quasi-Fermi level Fn as a normalizing parameter for the electron conduction density under such circumstances. At an energy well below Ec in a semiconductor we expect to find a band of allowed electron states which is almost completely full, extending downwards from energy Ev' Describe the total number of 'holes' in this valence band in terms of the Fermi energy EF for equilibrium (Po for holes) and of a hole quasi-Fermi level Fp for a non-equilibrium situation (p holes). Show the sense of departure of Fn and Fp from EF when the free electron and hole populations are enlarged. (d) Show that at eqUilibrium the product nopo depends on tempera ture but not on no or Po provided that the Fermi level lies within the intrinsic energy gap, i.e. between the top of the valence band and the bottom of the conduction band. Why is the quantity nj = (nopo)'h called the intrinsic pair density? Relate the energy difference Fn - Fp for a non-equilibrium situation to the product np (mass action law). Solution
(a) According to the Pauli principle, no two electrons in a system can be in the same quantum state. (Two electrons of opposing spin can 350
351
(16.1.1)
which has the necessary maximum occupancy of one electron per state for states of low energy. Conformity with Boltzmann statistics for the high-energy limit of small occupancy enables us to identify the parameter ~ with kT. The parameter a in Equation (16.1.1) performs a normaliza tion function, for if geE) dE distinguishable one-electron states are found in an energy range dE at energy E then in equilibrium at temperature T we can always find one (and only one) value for a such that
-
g(E)f(E)dE
N
(16.1.2)
is equal to the total number of electrons in the system. The normalization concept can in practice be expressed much more conveniently in terms of a parameter with the dimensions of energy. This is the electrochemical potential, or Fermi energy EF
= -kTln(a)
.
(16.1.3)
This can replace a in Equation (16.1.1) in writing the equilibrium proba bility of electron occupancy as feE)
=
I +exp[(E-EF)/kT]
(16.1.4)
We may note from this that the occupation probability is 50% for a state at energy EF itself. States well above EF are sparsely occupied with electrons. States well below EF are almost all filled; they contain few 'holes' among the ranks of filled states. (b) Given that a band of states starts from minimum energy Ec and that geE) = A(E-Ec)Y' for higher energies, we know that at thermal equilibrium the distribution of the total electron supply no over states of various energies must conform with _ no - I
A(E-Ec)Y' , l+exp[(E-EF)/kT]dE.
(16.1.5)
352
Chapter 16
16.1
This inserts the form of Equation (l6.1.4) into Equation (16.1.2). Equa tion (16.1.5) serves as a definition of the resulting Fermi energy for any combination of no and temperature. Given further that no is very small, we can see that Equation (16.1.5) must be satisfied with much lower in energy than Ec. In this asymp totic case, no
A(kT)'hexp[(E F -Ec)/kT]
r"
16.1
The same figure shows a valence band of states lying below the equi librium Fermi energy. Such a band will be almost filled with electrons at equilibrium, i.e. will have a relatively small density Po of 'free holes'.
exp(y)dY ... Ec :u~ fn.
= A (kT)'12 (:br)Yz exp [(E F -Ec)/kT]
Nc exp [(E F
-
geE)
1+exp[(E-Fn)/kT]dE
(16.1.7)
as an adaptation of Equation (16.1.5). Equation (16.1.7) places no apparent restriction on the value of n. The quantity Fn serves as a quasi Fermi level for non-equilibrium situations, and becomes coincident with EF for equilibrium itself. Provided that the total non-equilibrium population is rather small (so that Fn is lower in energy than Ec), the procedure of Equation (16.1.6) can be applied to Equation (16.1.7) in writing n
= Nc exp [(Fn -
Ec)/ kT]
5 t'F
= §g FP
(16.1.6)
Ec)/ kT] .
It may be noted that no of Equation (16.1.6) is the same as though the band were replaced by Nc states, all at energy Ec. The conditions of Equation (16.1.6) are met provided that no 4;. Nc ; this is a temperature dependent inequality which requires that EF be several kT lower than Ec· (c) We now consider the possibility of a departure from equilibrium, to provide for n =1= no electrons in this conduction band. It is apparent that a normalizing parameter Fn with the dimensions of energy can be used to define the relationship of n and the crystal temperature by
n
353
Recombination rate theory in semiconductors
(n 4;. Nc ) •
(16.1.8)
It is worth observing that the electrons distributed over band states for a non-equilibrium situation may have a velocity distribution which is incom patible with a thermal one for any real temperature. However, Equa tions (16.1.7) and (16.1.8) are concerned only with characterizing the total conduction electron density for a given crystal temperature T and a given density of states. In practice, we hope that excess electrons thrown into a conduction band will thermalize their speeds within a time of 10- 11 seconds or less, while in many semiconductors the time taken for excess conduction electrons to return to states in lower bands is very much longer. When equilibrium is disturbed in a semiconductor, it is usually in the sense of making n > no. From Equations (16.1.7) or (16.1.8), this makes Fn higher in energy than the equilibrium E F , as shown in Figure 16.1.1.
ELl Ev
Density of states g(E), and density of occupied states (shaded) Figure 16.1.1. An idealized yalence band (almost full) and conduction band (almost empty) for a semiconductor with an intrinsic gap extending from energy Ev to energy Eo. 'The equilibrium Fermi energy EF is compatible with the densities no, Po for free electrons and holes at temperature T. For any non-equilibrium densities n, p, which represent an enlargement of the free carrier densities, the quasi-Fermi leyels separate from EF in the sense Fn > EF > Fp.
From the preceding arguments, it will be clear that Po, E F , and Tare related by a condition
f
geE)
Ev
Po =
--«>
1+exp[(E F
-
(16.1.9)
E)/kT]dE ,
since the factor of g(E)dE is clearly the probability of a state not being occupied by an electron. When the Fermi energy is considerably higher than Ey (so that even the uppermost states in the band have rather few free holes), then we can expect to apply the procedure used in connection with Equation (16.1.6) for electrons, and write Po = Nyexp[(Ey -EF )/kT]
(Po
4;.
Ny).
(16.1.10)
In this limiting situation, the free hole density is the same as though the valence band were replaced by Ny states all at energy Ev' Evidently, if the valence band contains P =1= Po free holes for a non equilibrium situation, the density p can be described in terms of a hole quasi-Fermi level}t~ by
f
Ev
p
-00
g{E)
I +exp[(Fp -E)/kT]dE
p = Ny exp [(Ey -}t~)/ kT]
(arbitrary p) (small p)
1
(16.1.11)
Chapter 16
354
16.1
16.2
as adaptations of Equations (16.1. 9) and (16.1.1 0). The quantity Fp will coincide with EF at equilibrium, but will be lower than EF (as shown in Figure 16.1.1) when P > Po. (d) For equilibrium conditions, we can use Equations (16.1.6) and (16.1.10) to write nopo = NcNyexp[(Ey -Ec)/kTj (no ~Nc, Po ~Ny) (16.1.12) which depends only on the width of the valence band/conduction band intrinsic gap, on the density of states configurations near the extrema of these bands, and on the temperature. The product nopo does not depend explicitly on the value of no or of Po if the cited inequalities are met, and these equalities require that the Fermi energy should be appreciably higher than Ey yet appreciably lower than An intrinsic semiconductor is one for which the densities of free electrons and free holes are equal, which of course is the case if excita tion of electrons from the valence band into the conduction band is the dominant reason for the existence of free holes and free electrons. An intrinsic semiconductor is for practical purposes independent of the existence of any extrinsic features (such as the presence of foreign impurities or lattice defects which may in less perfect materials provide localized states for the provision of either free holes or free electrons). Thus in an intrinsic semiconductor, the densities of free holes and free electrons are each the quantity nj
(nopo),/,
(NcNv )% exp
-Ec )/2kTj.
NcNvexp[(Fn -Ec)/kTjexp[(Ev -}~)lkTj
= nrexp[(}~ -}~)lkTj
at
..
f\
I
( II
(16.1.1
by employing Equations (16.1.8), (16.1.11), and (16.1.13). Instead of nr on the right side of Equation (16.1.14) we could equally well write nopo, remembering that nand p must be written in the integral forms of Equations (16.1.7) and (16.1.11) if the free carrier densities are large enough to bring a quasi-Fermi level close to the edge of a band. At any rate, when p~ and Fp sti11lie within the intrinsic gap, -}~
kTln (np/nopo) .
(16.1.15)
16.2 Consider a semiconducting material for which the natural pro cesses of energy transformation at a certain temperature are sufficient to maintain a population of no free electrons per unit volume in a conduc tion band which can accommodate many more electrons. The density no results from a balance at thermal equilibrium of two opposing
355
processes: (i) the generation (at rate = g) of conduction electrons by excitation from filled electron states at lower energies, and (ii) the recombination (at rate r) of free electrons by their de-excitation to any empty states at lower energies. Now suppose that thermal equilibrium is perturbed, and the conduction electron density changed to n no + ne' We refer to ne as the excess electron density. Whenever ne is non-zero, then g =1= r, and usually both g and r are then modified from their equilibrium values. Such a pertur bation may result from externally induced excess generation at a rate ge, or by a flow of excess electrons from elsewhere. The net effect of the various influences can be expressed in an electron continuity equation an -=
(16.1.13)
For a semiconductor (which mayor may not be intrinsic) containing excess electrons and holes as a result of some departure from equilibrium, the product np can be written as np
II I
Recombination rate theory in semiconductors
I
=ge +g-r+-V-I e n
where In denotes the current density due to electron flow. In discussions of recombination in a semiconductor, it is convenient to define the electron lifetime (I) T as the excess electron density per unit rate of (r In terms of T, the electron continuity equation is ane ne I at ge --+-V-I Ten' (a) Comment on the conditions under which this becomes an ordinary and linear differential equation. What is then the general form of solu tion for any period of time-invariant excess generation? (b) Describe the build-up of ne when ge starts abruptly at time t = TI and continues at a constant rate for a long time thereafter. What happens when ge ceases equally abruptly at time t = 12? (c) From these results, show how ne varies when ge changes abruptly T3 . Sketch the time dependence for ge2 larger from gel to ge2 at time t than or smaller than gel' (d) When in practice the excess generation rate ge is an arbitrary func tion of time, express ne as an integral with respect to prior generation. Use this to show how ne responds to a pulse of generation which is an isolated half sine wave as a function of time. (e) In the same way, show how ne declines when ge is an exponentially decreasing function of time. Solution
..,i
(a) The continuity equation for electrons becomes an ordinary differ ential equation if the excess electron density ne is a function of time but not of location. This can be true if we consider a region well within a large, homogeneous single crystal, a region far from any surfaces, p-·n junctions, or contacts. It must also be presumed that the mechanism (I) The various meanings of 'excess carrier lifetime' are explored in detail with respect to the various processes of generation and recombination in the books by Ryvkin and by Blakemore cited in the bibliography at the end of the section.
Chapter 16
356
16.2
responsible for the excess generation rate ge is spatially uniform; thus if ge results from the absorption of photons of suitable energy directed from without, the photons must be ones which are absorbed rather weakly by the solid. Since we assume spatial uniformity and remoteness from surfaces and contacts, the electron current will be solenoidal (non-divergent) and the continuity equation reduces to dne ne (16.2.1) dt = ge--:;: This equation is also linear if the lifetime T is a constant, i.e. if the difference between the natural rates of recombination and generation is directly proportional to the departure from an equilibrium free electron density. (In practice it is found that T has some dependence on ne for any dominant recombination mechanism, though for certain recombina tion regimes T behaves as a constant over a fairly wide range of ne') When conditions of spatial uniformity and constant T are imposed, Equation (16.2.1) can be integrated immediately for any period of time in which ge is maintained at a constant value. The general solution is of the form (16.2.2) ne = geT+Cexp(-t/T) where the quantity C is determined by the initial conditions. (b) If. there is no excess generation prior to time t = T" and the constant rate ge thereafter, then C in Equation (16.2.2) must have such a value that ne = 0 at t = T,. This condition is satisfied if
357
Recombination rate theory in semiconductors
16.2
acts of generation prior to T3 , and the latter to the consequences of the more recent generation. Thus ne
= ge,T[exp(13/T)-exp(TdT)]exp(-t/T) +ge2T{ l-exp[(T3 -t)/T] (16.2.5)
(t> T3 ) .
This can be separated into the sum of a constant component and a transient component in the form ne
= ge2 T -
[(g e2 - ge,)Texp(13/T) +ge,Texp(TdT)] exp(-t/T) (t> T3)' (16.2.6)
The consequences of this equation are illustrated by the curves of Figure 16.2.1 for situations in which ge increases, decreases, or remains unchanged at the time 13. 2'0 ~!....
ge,T
1'5
1'0
0'5
C = -geTexp(TdT)
so that ne
= geT{l-exp[(T, -t)/T]}
(t> T,) .
(16.2.3)
This has reached the value n e (T2)
= geT{l-exp[(T, -12)/T]}
at the time 12 when excess generation abruptly ends. From Equation (16.2.2) it is clear that ne will subsequently decay in accordance with ne
= Cexp(-t/T) = n e(T2)exp[(12 -t)r] = T[exp (12/T) -exp(TdT)] exp(-t/T)
ge
(t
>
T2)' (16.2.4)
(c) In this problem, we assume that the lifetime is independent of n e , so that the continuity equation (16.2.1) is linear. Thus if ge operates at a value gel during the period T, .;;;; t';;;; T3 , and at a different value ge2 for t> T3 , then ne at times later than T3 can be described as a simple sum of contributions in the forms of Equations (16.2.4) and (16.2.3). The former of these will describe that component of ne which derives from
0 0
4 3 2 Time scale (t - T 1 ) in units of T
5
6
Figure 16.2.1. The transient behaviour of the excess pair density when excess generation starts at a time Tl and changes to a different rate at time T3 · The curves here follow Equations (16.2.5) and (16.2.6) after time T3 for situations in which ge2/gel = 2, I, and 0·3.
(d) We must now consider the solution of Equation (16.2.1) wheng e is an arbitrary function of time. This solution can be envisaged by considering ge to be a sequence of instantaneous generation events, each of which can be described by a delta function. Thus suppose that
l:ge dt = NfJ(t - to) . For this generation alone, Equation (16.2.1) has the solution ne(t)
= Nexp[(to-t)/T].
Now since the lifetime t is assumed to be a constant in this problem, Equation (16.2.1) is linear, and the solution for ne(t) for many delta function acts of generation at different times is simply the sum of the
358
16.2
Chapter 16
values for
lle (t)
16.2
the integral representation for ne(t) is
=
o·s
I~ge(to)exp[(to-t)/r]dto
(16.2.7)
~
Gr
which is a very simple example of a Green's function form of solution. We can use Equation (16.2.7) to determine how ne responds to sinusoidally varied creation. Suppose that we have a single half sine wave of generation: ge(to) = Gsin(wt o )
(0
0·6
0'4
< to < 1r/w) ,
\ g,
\
and that there is no excess generation at other times. Then, from Equation (16.2.7), we have ne(t) = Gexp(-t/r)
I
sin(wto)exp(to/r)dto
(0
\
\
< t < 1r/w)
0
4
5
6
As a corollary of the preceding discussion, it can readily be established that generation in the form of a continuous sine wave results in an excess carrier density comprising the sum of a constant term and a sinusoidal term. The sinusoidal component lags behind the phase of the generating sine wave by the angle 8 = tan-1(wr). (e) We should now like to know how ne decays when ge is itself an exponentially decreasing function of time. Suppose that ge has been maintained at the constant value G for all negative time (so that ne = Gr at time t = 0), and that ge = Gexp(-t/T) for all positive t. Then from Equation (16.2.7) we have
ne(t) = G exp(-t/r) fwtexp(x/wr)SinXdx 0
< t < 1r/w) , (16.2.9)
nell)
while the generation is in progress, and
(Gr)exp(-t/r) +
f:
Gexp(-to/T)exp[(to -t)/r]dt o (t
> 0) .
Provided that T and r are not exactly the same, the integration yields
in
G ne(t) = -exp(-t/r) exp (x/wr) sinxdx w 0 Gwr 2 . ~ ~ [1 +exp(1r/wr)]exp(-t/r)
ne(t) = GrTeXp(-t/Ti=;exp(-t/r) (t
> 1r/w)
(16.2.10)
while for the special case of T
for the subsequent monotonic decay. The form of the response while the generation is still in progress can be seen more clearly by defining a phase angle 8 = tan-l (wr), and rewriting Equation (16.2.9) as ne(t) = Grcos8[sin(wt-8)+exp(-t/r)sin8]
3
Figure 16.2.2. The response of the excess carrier pair density to a half sine wave of generation, in accordance with Equations (16.2.9) through (16.2.11). The dashed curve illustrates g. for a situation of WT = I, and the solid curve is the result for n,.
and Equation (16.2.8) is of this form when we change to the dimension less variable x = wt o. Thus
rSi~~t) -COs(wt)+exp(-t/r)]
\ 2
0
Time t in units of r
f
-----:;:--..-
\
0·2
(16.2.8) as a description of ne while the generation process is still going on. In order to describe ne after the half sine wave of generation has ceased, we must write Equation (16.2.8) with 1r/ w as the upper limit of integration. Now through integration by parts we can establish that exp(ax) . exp(ax)sinxdx = 1 +a2 (asmx cosx) ,
w
359
Figure 16.2.2 illustrates the rise and fall of ne in accordance with Equations (16.2.9) through (16.2.11) when wr = 1, to make the phase angle 8 = tan-I (wr) = !1r.
derived from each generative act separately. Thus when
ge is any arbitrary (not necessarily continuous) function of time, then
ne(t)
Recombination rate theory in semiconductors
(0
ne(t)
(16.2.11)
\
r the result is
G(t+r)exp(-t/r)
(t> 0) .
(16.2.13)
Figure 16.2.3 displays the decay of ne according to Equation (16.2.13) for the particular case of T = r, compared with the conventional expo nential decay for T = 0 and the greatly slowed decay when T = 3r. Only if T is smaller than, or equal to, r is it possible to determine r from the decay rate at large values of t.
< t < 1r/w).
:~
(t> 0), (16.2.12)
,
360
Chapter 16
16.2
16.3
Recombination rate theory in semiconductors
361
stimulated recombination? (See also Problems 3.20, 15.6, and 15.7 for related considerations.)
0'6
r.:.
0'4
~
'"
the number of electrons capable of excitation, and the thermal environment for such excitation. However, it will not depend on the occupancy of states within the band, provided that this occupancy is small for all band states. Thus g = A(Nd -Na -n) (16.6.3) where the quantity A should depend on the temperature, on the minimum excitation energy Ed, and on the density of conduction states, but not on Nd , N a , or n. Since g r at equilibrium, Equations (16.6.2) and (16.6.3) can be equated for this condition to yield no(no+Na ) A (16.6.4) Nd - Na -no = iio{3. ' which is of the form of Equation (16.6.1). Evidently we should identify A/iio{31 with the mass action constant n 1, so that the rate of generation is g = iiO{3l n lCNd -Na -n)
(16.6.5)
whether or not the conditions are of eqUilibrium. It may be noted that Equation (16.6.1) would be recreated from mass action considerations even if generation and recombination processes involving two or more electrons were considered; but then Equations (16.6.2) and (16.6.3) would involve higher powers of n. (b) The difference between the rates of natural recombination and generation can be used in the definition of an excess electron lifetime T in the continuity equation ne dne dt = ge +(g-r) = ge -T"
66
(1 . .6)
for excess electrons in a situation of spatial uniformity. Here ge is once again an excess generation rate caused by an external influence. Provided that rand g are given by Equations (16.6.2) and (16.6.5), then ne ne T r-g = iio{31[(nO+n e )(no+n e +Na)-n1(Nd -Na -no-nell There is a major cancellation among the terms in the denominator of the right side, since noCno + Na ) is equal to n 1(Nd - Na - no). Thus I T = vo{31(Na + 2n O+nl +ne) (16.6.7) in the continuity equation (16.6.6).
16.6
Chapter 16
372
ne(Na +2no+nl +ne) = _~ep' va"'l as a response to spatially uniform generation. Now suppose that excess electrons are created at the rate ge (per unit volume and time) as a result of an incident stream of 10 photons (per unit area and time), each with an energy of at least Ed' Excitation can take place only from neutral donors, thus we may expect a quasi uniform (7) excess generation rate aphNdn/o,
where aph is the photo-ionization cross section of a neutral donor. If not all incident photons have the same energy, then aph is a suitable a\erage with respect to the excitation probabilities to the various upper statesEu = hv-E d • SinceNdn = Nd -Na -n, then
ne(Na +2n o +nl +ne) _ !. ~ 0 Nd-Na-no-ne va""p.
Recombination rate theory in semiconductors
373
the condition
This continuity equation requires in steady state that
ge
16.7
(16.6.8)
controls the relationship between 10 and ne' This relationship is shown graphically in the log-log plot of Figure 16.6.1. Part (a) of the curve shows the linear dependence of ne on light intensity for weak illumina tion. The curve changes character at point (b), which corresponds with
(d) (e)
,,: bJl
.£
10gIo
Figure 16.6.1. Variation ofne with incident photon flux 10 as required by Equation (16.6.8).
(7) In a more rigorous version of this problem, we should have to concern ourselves with the attenuation of the photon flux with increasing distance through the crystal, to provide a depth dependent ge'
ne "'" Na +2n o +n l !. "'" o
}
2va(3,(Na + 2n O+nl)2
----:-::-':-'--'---"--:--::---=-----'-''------: aph(Nd -2Na - 3n o-nd
(16.6.9)
For a semiconductor crystal with relatively weak compensation, in which Na ~ N d , and for a temperature low enough to make the mass action constant n I small compared with Nd , it is possible for ne to be substantially larger than Na + 2no + n I yet substantially smaller than N d . Such a combination of requirements is necessary in order that the region (c) in the curve of Figure 16.6.1 be prominent. Within that range (when it exists), the left side of Equation (16.6.8) is approximately n: /Nd ; thus under these conditions ne varies essentially as 16 The upper limit of the range occurs at the point (d) on the curve, when 2
•
10
"'"
ne
~
NdV(31~ aph
}
(16.6.10)
Nd -Na -no
and as section (e) of the curve indicates, a further increase of photon flux cannot make no + ne any larger than the available electron supply of Nd - Na . The behaviour of the system when no+ ne becomes comparable with Nd - Na would be more complicated if stimulated recombination played a significant role. 16.7 Consider steady state conditions in a semiconductor crystal which has free carrier densities no and Po at equilibrium. We shall assume again that the semiconducting medium is spatially homogeneous, and shall assume further that the free carrier populations are non-degenerate (so that nopo = n:). Electron-hole recombination in this material is assumed to be dominated by the activity of N r monovalent recombina tion centres. Each centre has just two states of charge: 'empty', or 'filled' with a single electron. We assume that the centre can be filled in only one way, so that the filling probability is a Fermi-Dirac occupancy factor. The energy of the localized state is such that at equilibrium the ratio of filled centres to empty centres is PI/PO' Along with the definition of the quantity p" we can define the companion quantity n I = nt/PI' so that the ratio of filled to empty centres is also no/n I' It will be obvious that the free carrier densities are just n I and P I when the Fermi energy coincides with the energy of the localized state. Suppose that an empty center has a cross-section an for the capture of a free electron of speed v n . Then we say that the centre has a capture coefficient vna n for a free electron of this speed. Averaged over the
374
Chapter 16
16.7
thermal distribution of electron speeds, the capture coefficient of an empty centre may usefully be written Cn == I). The effective rigid-sphere radius r is determined by these two energy terms. The only dimensionless combination of r, C, and kT is a function of C/(kTrn I). Hence
17.5 (a) Use the results of Problems 17.3 and 17.4(b) to show that the coefficient of viscosity, at temperature T, of a gas of atoms which behave as rigid elastic spheres of radius r and mass m is approximately
(3mkT)Yl 241Tr2 (b) How should the coefficient of viscosity depend on the pressure of the gas? (c) The variation of the coefficient of viscosity of 4He gas with tem perature is given in Table 17.5.1. Explain qualitatively this temperature variation. Assuming that the force between two helium atoms is predomi nantly repulsive, and varies inversely as the nth power of their separation, use a dimensional argument to determine n from the data.
rex:
and r} ex:
Table 17.5.1. Coefficient of viscosity of 4He gas as a function of temperature.
15'0 29'5
75·5 81·8
171 139
291 197
457 268
665 339
1090 470
2
t,
~
p
6r
Tv,.+2/(n -I)
0·66 ± 0·02
whence n
r}
ex:
-+-= 2 n-I
Solution
From Problem 17.4(b),
Tv,.
I )I/(n-I) (T
From the data, I
TCK) 1/(gcm- 1 S-I) x 10 6
387
13·7±1·6.
[The value n 13 is used for the repulsive part of the familiar Lennard Jones potential. The dimensional argument was first given by Lord Rayleigh (3) .]
')0.,2 ~-=D
6r
'
where D is the coefficient of self-diffusion of the gas. From Problem 17.3(c) and (e), 0:
= 61Tn(2r)2
D
Hence
P
r}
~ 241Tnr2
(3kT)v,.
m
(3kT)Yl
m
'I
(3mkT)'h
(b) r} is independent of density, and hence of pressure, at a given temperature. (c) For a gas of rigid spheres r} ex: • From a plot of 19r} versus 19T for 4He gas, we have 19r}
~
slgT+ constant,
where s ;: O· 66 ± 0·02
> O· 5 .
4He atoms interact through a repulsive force which varies rapidly with their separation. As the temperature is raised the atoms have greater mean kinetic energy and therefore approach more closely at collisions.
'v
17.6 (a) Deuterium and helium, both of molecular weight 4·0, have coefficients of viscosity at O°C of 1·2 X 10-4 and 1·9 x 10-4 g cm- I S-I respectively. Estimate the ratio of their second virial coefficients Ba at this temperature, assuming that their molecular interactions may be treated as those of rigid spheres. [Note from Problem 9.9 that B2 is proportional to the molecular volume.] (b) An ideal gas of given molecular weight at a given temperature and pressure may be considered as the limit of a series of hypothetical imperfect gases with the same molecular weight, temperature and pressure. What is the theoretical value of the coefficient of viscosity of the ideal gas, considered as this limit? (c) The coefficient of viscosity of a gas may be determined from its rate of flow through a capillary tube of diameter d under a pressure gradient. Outline qualitatively the result of a series of such determina tions on the hypothetical series of gases. Cd) In what range of (mean) pressure would 4He behave like an ideal gas in this experiment, if d = 10-4 cm? (3)
Lord Rayleigh. Proc.Roy.Soc., 66, 68 (1900).
388
Chapter 17
(a) From Problem 17.5(a)
r
r(D 2 )
J
1· 2
2
X
10- 4
Ibe second virial coefficient B2 is proportional to the molecular volume for a rigid-sphere interaction (cf Problem 9.9). Hence the estimated ratio is
B(He) _ (1.2)' h ...__ 0 50. 1.9
I[
ff(V)dV = n ,
2
B 2 (D 2 )
-
(b) In the ideal-gas limit, B2 (and higher fore r ~ O. Hence the theoretical coefficient of [cfProblem 17.5(a)J 1/
~ fVf(V)dV = v= 0, is given IfV 21'( v) dv
;;
y-+O
i.e. 1/ ~ 00 as r ~ O. (c) 1/ ~ 00 because A. ~ 00 as r ~ 0 [cfProblem 17.3(a)]. In a tube of diameter d, the coefficient of viscosity is a valid concept only when A. ~ d. In this case, 1/ a:: l/r2. For A. ~ d the flow rate is determined by the collisions of molecules with the tube walls, and is independent of r. Thus the apparent viscosity (as judged from the flow rate) would reach a finite limiting value as r ~ O. (d) 4He would behave like an ideal gas in the pressure range for which intermolecular collisions were unimportant, i.e. for which A. ~ d. From Problem 17.3(a),
of state pRT
p::::: --::::: nkT M
where the symbols have been defined in previous required pressure range is kT
p ~ 41l'r 2d ~
61/ (kT)Yz
d
3m
= V- 2 = V2 .
In a collision between two atoms, the distribution of e is independent of their relative velocity, and the mean value of () is ¢. Calculate the average value (il) of Iill for the collisions of an atom with initial velocity u = 0, to first order in ¢. [ denotes an average over collisions.] (d) Hence show that a group of atoms initially having zero velocity will begin to 'diffuse' outwards in velocity space, with a 'diffusion coeffi cient' Dv equal to A 2 V2¢2/ T , where liT is the mean collision frequency for an atom with zero velocity, and A is a dimensionless constant. [Use the results of Problem 17.1.] Solution
(a) In the centre-of-mass reference frame the atoms have initial veloci ties ±WI = ±t(V'1 -vd. From conservation of total momentum, the final velocities also are equal and opposite. From conservation of total kinetic energy the magnitudes of the initial and final velocities are equal. Hence the initial velocities are equal and opposite, and so are the final velocities, and all four lie on a sphere in velocity space of radius WI' centred at the centre-of-mass velocity.
1 A. ~ 41l'r 2n
From the ideal gas
'
since
Hence the
~
where
2W2
W2
W2
=
! Iv'l -VII
WI
and =
WI W 2 cos
Resolving ~ parallel and perpendicular to -Wl(l
With the given figures for 4He this gives the pressure range ~
5 x lOs dyn cm- 2
WI,
is the relative velocity after the collision, i.e.
WI' W 2
(3mkT)'h, 1/ ~ 241l'r2
p
~
389
17.7 (a) Two atoms of equal mass, with initial velocities v 1 and respectively, collide elastically. What velocities are possible after the collision? (b) If the angle of scattering in the centre-of-mass reference frame is e, what is the change .6. in the velocity of one of the atoms? (c) A spatially uniform monatomic gas of number density n has an isotropic velocity distribution f(v) with the following properties:
Solution
r(He).
Tr:mc:nnrt in gases
17.7
17.6
tatm .
ill
WI
e.
we 0 btain
cose) ,
390
Chapter 17
17.7
(c) To first order in 0, A1 = w10,
All
17.8
Transport in gases
17.8 (a) For the situation of Problem 17.7(c), calculate the average value of ~ for the collisions of an atom with low velocity u, to second order in q'>. (b) Hence show that a group of atoms with low velocity u will acquire through collisions an average acceleration
O.
Therefore
AI=w,O,
and after averaging over 0
IAI = w1q'> •
2
(Ii> = _ 8 q'>2 -7
The probability per unit time that an atom of velocity zero is struck by one of velocity v is proportional to v f(v). Thus the mean value of IA I is
f
fAVf(V)dV (A)
(averaged over 8) .
vf(v)dv
Since
WI
391
U
where B is a dimensionless constant. (c) Assuming that the expressions obtained for Dv and (iJ> are correct for arbitrary u, show that in general there is a flux of atoms (per unit 'area' of velocity space) given by
(A 2V2!! + B2Vf)
=v 1 f
(A) = -q'>
2
v2f (V)dV = I
J72
fVf(V)dV
Now Ii = (v 2
l'" = .!:::. A'
,
per unit volume of real space per unit time. Hence show that the velocity distribution function f(v) in the equilibrium state of the gas is the Maxwellian distribution of Problem 3.1. (e) From your results identify fo and estimate t* in the following common approximation for the effect of collisions on f(v):
of
where A' is a dimensionless constant. Hence
f-f o
ot=-~
= !A'q'> V . (d) Because the gas is isotropic, .1 is random in direction. Hence an atom with zero velocity begins to perform a random walk in velocity space with steps of average length (.1>, occurring with mean frequency liT. In consequence its velocity departs from zero, and the step length may change as the random walk progresses. The initial diffusion coeffi cient is, from Problem 17.1,
D :::::; = -i=
(5) It is discussed further in F.Reif,loco cit., Ch.13.
-i(8 2 )(u -(v x » .
392
Chapter 17
17.8
From here the result follows by a dimensional argument, or from the following more detailed analysis:
17.8
Transport in gases
The net drift calculated in part (b) causes a flux through velocity space given by
B21jJ2
f(v) (v) = --vf(v) ,
fV xIv - u If(v)dv
r
(v)=-"----=----
For small u,
in the same units. Hence the result for the total net flux follows. (d) In equilibrium there is zero net flux for all v. From the result of part (c), af/avllv. This is consistent with the assumed isotropy of f(v), i.e. f(v) is a function of Iv I only. Thus
f,v -u If(v)dv
x
av 2 Vx = v -u-+O(u ) = v -u-+O(u 2). ax v
Iv-ul
A2J72
Hence fIV-Ulf(V)dV = n[V+O(u 2 )]
(17.8.1)
vx/v = O.
fVx
f[Vxv -u v! + O(U 2)] f(v)dv = n [-u( v~) +O(u 2]
=
Also
(~)
C~;)
!(v:) = 1v.
Therefore (v ) x
Hence (~) _
2). = -1(8 4
B 2v 2 In f = - 2A 2J72 + constant
or
Since f(v) is isotropic,
(V~2)
f(v)
f(v)
(u)
=
I
r(u) = -;:-+O(u
where r = r(O). Hence (Ii) =
( mv2)
constant x exp - 2kT .
ff(V)dV
(~)/r(u)
,
where I/r(u) is the mean collision frequency for an atom with velocity u. I/r(u) is proportional to the integrated molecular flux given by Equation (17.8.1). Hence _
=
The constant can be determined, from the normalisation condition
where B2 is a dimensionless constant. (b)
B2V2)
constant x exp ( - 2A 2J72
But v 2 = J72 by definition. Therefore A 2/B2 = j. Using the theorem of the equipartition of energy to relate J72 to the temperature T,
•
=-B 21jJ2 U+O(U 2 ) ,
=
It is easily shown that, with this f(v), 3A 2J72 v 2 -- B2
nuv
= ------::. + O(u 2 ) 3nv
~U+O(U2) 3
af +B 2vf = O. av
i.e. ,
since, by symmetry, Iv -u If(v) dv
393
2
) ,
B21jJ2 ---u+O(u 2 ). r
(c) The diffusive flux of atoms through velocity space is given by the analogue of Fick's law (in three dimensions) as -Dvaf/ov atoms per unit 'area' of velocity space per unit volume of real space per unit time, where Dv has been calculated in part (a).
as
=
n ,
( )
'/'
n 2::T
.
Thus the equilibrium distribution is Maxwellian. [The result is correct despite our approximations.] (e) fo(v) is the equilibrium form towards which f(v) relaxes through collisions, i.e. the Maxwellian distribution from part (d). (For a moving gas, modification of (d) is required.) The relaxation time t* is the time necessary for local departures from fo(v) to be communicated, by a diffusive process, to the whole velocity distribution, which has linear dimensions of order V in velocity space. From Problem 17.1 (b) and (c), an initially localised irregularity is distri buted after time t, over a region of mean square width of order 2Dvt.
Chapter 17
394
17.8
Hence
2Dut* : : : ; V2
i.e.
t*::::::;
r
r ¢2
~-
(f) For rigid spheres, ¢ ::::::; I radian. Hence the condition ¢ satisfied. To order of magnitude, however,
-
2
From Problem l7.11(d) D
t*kT
18
Transport in metals (J)
m J.M.HONIG
whence Jl _ e r jj - kT3A 2rJ>2 t *
This gives t*
r
IThe previous estimate was t* :::::; r/A 2rJ>2.J GENERAL REFERENCES
A comprehensive account of elementary (mean free path) kinetic theory is given in: J. Jeans, Kinetic Theory of Gases (Cambridge University Press, Cambridge), 1940. The following textbooks deal with the Boltzmann equation: E. A. Desloge, Statistical Physics, Part III (Holt, Rinehart, and Winston, New York),1966. F. Reif, Fundamentals of Statistical and Thermal Physics (McGraw-Hill, New York),1965,Ch.12-14. K. Huang, Statistical Mechanics (John Wiley, New York), 1963, Ch.3-6. The classical treatise on the subject is perhaps S. Chapman and T. G. Cowling, Mathematical Theory of Non-uniform Gases (Cambridge University Press, Cambridge), 1939. For recent advances (c.g. the treatment of dense gases) consult: L Prigogine (Ed.), Pmc. International Symposium on Transport Processes in Statistical Mechanics (lnterscience, New York), 1958. W. E. Brittin (Ed.), Lectures in Theoretical PhYSics, VoLlX C (Kinetic Theory) (Gordon and Breach, New York), 1967.
(Purdue University, Lafayette, Indiana)
INTRODUCTORY COMMENTS
18.0 In Chapters 18 and 19 we consider problems which deal with the response of mobile electrons in solids to externally applied electric fields, magnetic fields, and temperature gradients. In the standard approximation utilized here the electrons are considered to be independent particles subject to Fermi-Dirac statistics. In zero order approximation the solid is viewed as a 'box' or container, within which the electrons move as a 'gas'; this is the so-called Sommerfeld modeL The effect of the solid is introduced more realistically in first order approximation by regarding the periodic potential of the lattice as a perturbation on the nearly free electrons. Alternatively, one may proceed from the opposite assumption: the electrons are considered rather tightly bound to the atomic cores in the solid but able to move through the lattice by virtue of some overlap among orbitals associated with adjacent atoms. In either case the following conclusions are reached: there is an alternation between closely spaced energy levels (energy bands) and forbidden energy ranges (energy gaps), corresponding to ranges where the Schr6dinger wave equation does or does not admit of solutions. The demarcation line between allowed and forbidden levels is termed a band edge. The if; functions can always be represented as free electron wave functions modulated by a function which has the lattice periodici ty. Of cardinal importance is the specification of the energy 8, of the electrons in the solid and of the dependence of 8, on independent variables and on parameters. As in the case of free electrons the energy depends on the wave number vector k. In what follows we shall always treat a very special case, namely bands of standard form for which h2 k 2 8, =: 8,c + 2m = 8,c + EO (18.0.1) where 8,c is the lower band edge and the second term is a kinetic energy formally identical to the expression obtained for free particles, in which m is the mass of the particle, h Planck's constant, and h == h/27r. (\) In Chapters 18 and 19 Boltzmann's constant is denoted by modulus of the wave vector k = Ik I.
401
Ii;
to avoid confusion with the
r
402
Chapter 18
18.0
However, in the present context, m is not the free electron mass but, rather, an effective mass which depends on the band structure of the solid. By this method it is possible to dispense with the explicit consideration of interactions of charge carriers with the lattice. The dynamics of the particle is introduced as follows: it may be shown from very general considerations that the velocity of an electron in the crystal is specified by Vk &(k) (18.0.2) v h where Vk is the gradient operator with respect to the independent variable k. The time derivative Ii of the wave number vector is related to the externally applied force F through the equation Ii = F/h. The interaction of an electron to an externally applied field is handled in the present context by setting up and solving a Boltzmann transport equation, in which the electric, magnetic, and temperature fields appear explicitly as parameters. In a perfect periodic lattice the electron encounters no resistance to its motion; however, impurities, lattice vibrations, and other types of imperfections introduce scattering mechanisms which must also be handled through the Boltzmann equation. A standard procedure here consists in introducing a relaxation time T related to the mean free path I by T 1/\ v I. This approach can be shown to apply under certain rather restrictive conditions, and the results are equivalent to the linear theory of irreversible thermodynamics. Adjustment of T along the lines shown in later problems simulates different scattering mechanisms. As will be shown in the last problem of this section, a set of band states that is very nearly completely occupied may equally well be described in terms of the remaining unoccupied band states which may be associated with fictitious particles, termed holes. These may be regarded as charge carriers that have a positive charge and a kineti," energy, and Fermi level which is referred to the upper band edge. Thus, for holes, h2 k 2 (18.0.3) & 2m :;: &o-e
18.1
Transport in metals
403
(a) The electrochemical potential t for electrons is defined by
t :;: f.l n
(
-e!{:ls
(18.1.1)
where f.l n is the chemical potential, -e the charge on the electron, and !{:Is the electrostatic potential. Express the chemical potential in terms of the activity of the electrons in the system, which is related to the concentration of electrons, C n , by an = 'Yn C n , where 'Yn is the activity coefficient. Show that for a uniform material at constant temperature the gradient of the electrochemical potential per unit electronic charge coincides with the electrostatic field. (b) Let I" be the length of the sample specimen in the direction X = x, y, z; let v:;: Vet/e), U:;: V!{:Is; let fA and CA be the current density and heat flux along the direction X; and let T be the temperature. Referring to Figure 18.3.1 (page 414), note the orientation of the rectangular parallelepiped relative to the Cartesian axes; let L be the distance between the voltage probes at C and D, and let Hz denote the magnetic field, aligned along the z direction. The following definitions will now be introduced for uniform isotropic materials under the open- . circuit conditions fy = fz 0 and for the isothermal conditions VyT = VzT = 0: 1. The electrical resistivity Vx(t/e) . p= J wlthVxT=O. (18.1.2) x
Show that this reduces to the common formulation of p in the terms of the sample resistance Rs. 2. The Hall coefficient = Vy(t/ e ) With . _ R - J 11 VxT - o. (18.1.3) x
z
Show how this quantity is related to the potential difference across y, and to the current along x. Demonstrate why it is desirable to make the samples as thin as possible along the magnetic field direction. Interpret the results. 3. The Seebeck coefficient _ VxG/e). _ IX= VxT wlthfx=O. (18.1.4)
where &v is the upper band edge. The reader should carefully distinguish in subsequent sections between the total energy & of a charge carrier and its 'kinetic energy' h2k2/2m which is designated as e. Finally, a specimen is said to be n type or p type according as electrons or holes are responsible for conduction .
State whether IX may be re-expressed in terms of electrostatic potential differences; discuss the significance of Equation (18.104). 4. The Nernst coefficient Vy(t/e ) . _ N H V T Wlthfx = o. (18.1.5)
18.1 This problem is designed to acquaint the reader (a) with the concept of electrochemical potential which is extensively used in Chapters 18 and 19, and (b) with methods commonly used to describe on a macroscopic thermodynamic scale the response of charge carriers in conductors to externally applied forces.
%
x
State whether N may be re-expressed in terms of electrostatic potential ;I
18.1
Chapter 18
404
differences; discuss the significance of Equation (18.1.5). 5: The thermal conductivity is defined by
-c
" == V'
x
Twi th Ix
0.
(18.1.6)
Interpret this relation. Solution
(a) From standard thermodynamics we find that I1n = 11;; + RTlna n
(18.1
where an is the activity of the electrons in the system under study, and 11~ is the standard chemical potential of the electrons when an = 1. Taking the gradient of the potential (18.1.1) at constant temperature we find RT (18.1.8) V'(~/e) = -eV'(lna n ) V''Ps where V' 'Ps is the gradient of the electrosta tic potential. Since T is constant and the material is uniform, V'lna n == 0; then U V V'(~/e) -V''Ps -1=E (18.1.9) where U is the electrostatic potential difference, L the distance over which this quantity is measured, and E is the electrostatic field. (b) 1. Let Ix be the current along the x direction and utilize the Ixllvlz and defining symbols introduced earlier. Then, with Ix Iylz == Ax we may rewrite the definition (18.1.2) as _ Vxlylz _ Uxlyl z _ UxAx I PI I L I L (18.1. x
On introducing Ohm's law,
x
x
Ux /Ix = R s ' we find
P=
z
Transport in metals
405
to the definition (18.1.3) based on a phenomenological approach to irreversible thermodynamics, the application of a current along the positive x direction, and of a magnetic field along the positive z direction, leads to the establishment of a gradient in electrochemical potential (or of a difference in electrostatic potential in the case of uniform materials) along the y axis. The magnitude of this gradient (or difference in potential) is given by the Hall coefficient under the assumed boundary conditions. 3. Even for uniform materials we may no longer set V'x(~/e) = -Uxllx because in the presence of a temperature gradient it is no longer appropriate to drop the first term on the right in Equation (18.1.8). According to the definition (18.104) based on the phenomenological approach to irreversible thermodynamics, the establishment, within a sample, of a temperature gradient, subject to the conditions detailed earlier, leads to the existence of a gradient of electrochemical potential Fermi level) within the sample, whose magnitude is specified by (1'. In view of Equation (18.1.8) this results in the concomitant establishment of a gradient in activity (or concentration) and of an electric field for electrons within the sample. 4. For reasons discussed above, it is not permissible to replace V'y(~/e) with -Ully here. According to Equation (18.1.5) based on the phenomenological approach to irreversible thermodynamics, the establishment of a temper ature gradient along x and of a magnetic field along z leads to the presence of a gradient in electrochemical potential along the y axis. For reasons given in Part 3, this produces both an electric field and a non uniform distribution of activity (or charge density) along y. The magnitude of this effect in isotropic materials is specified by N. 5. According to irreversible thermodynamics the thermal conductivity measures the heat flux in a material in response to an imposed tempera ture gradient. Note that the heat flow is in a direction opposite to the temperature gradient.
(18.1.11)
L
which is the required formulation connecting p and Rs· 2. In terms of the symbols introduced earlier, we have from definition (18.1.3) and Figure 18.1.1 Vylylz Uyl z R I R I R (18.1.12) x
18.2
x
18.2 In Problem 18.1 we considered the phenomenological description of the response of charge carriers to applied forces. Here we attack the problem from the microscopic viewpoint. It is assumed that the model described in the introduction holds and that the Boltzmann transport equation also alluded to there has been solved (2) for the distribution function
z
where lz is the thickness of the sample. Note that according to Equation (18.1.12), for a given material with fixed R, the potential difference is proportional to l/lz; hence, the thinner the sample, the larger 1Uy I, and the easier it is to measure Uy • Difficulties arise, however, when lz becomes comparable to the mean free path of the electrons. According
A
to
a ) to - V' '" (ae
(18.2.1 )
(2) T.e.Harman and J.M.Honig, Thermoelectric and Thermomagnetic Effects and Applications (McGraw-Hili, New York), 1967, pp.178-183.
18.2
Chapter 18
406 with
1 +exp [(€ - 1l)lkT I
(18.2.2)
and q, = T[F+(ZeTlmc)F x H+(eTlmc)2H(F· H)] 1+ (eTHlmc)2 -
(18.2.3)
and F
ZeVr(sle)- (€ - Il)Vr T
T
(18.2.4)
The quantity A represents the probability that an electron with wave number vector k is actually encountered in the crystaL As is seen from Equation (18.2.1), this quantity is given in terms of the equilibrium Fermi-Dirac distribution function (18.2.2) and further involves a term which represents in first order the departure from equilibrium. Here v is the carrier velocity, € the energy, and Il the Fermi energy relative to the lower band edge if one deals with electrons (Z = I) or relative to the upper band edge if one deals with holes (Z = + 1); K is Boltzmann's constant, and T is the temperature. The correction term in Equation (18.2.1) also involves a function q, given by Equation (18.2.3), in which the electric charge 1e I, carrier mass m, velocity of light c, applied magnetic field H, and relaxation time T appear explicitly. q, also depends on a quantity F which in turn involves the spatial gradient of the electro chemical potential per unit charge V r (S Ie) and of the temperature, Vr T. It is through the quantity F that the externally applied forces are introduced into the problem. As stated in the introduction, the resistive effect of the medium enters through the relaxation time T. A detailed theoretical analysis shows that where this concept is applicable the relaxation time is specified by T = To€r- v,, where To is a collection of constants and r is a scattering parameter which has the values 0, 1, or 2 according as scattering through acoustic vibrational modes, optical vibrational modes, or ionized impurities predominates. In proceeding with the problems cited in this section the reader should note how the function (18.2.1) is used to construct an expression for the current density and heat flux in (a), and how the definitions of Problem 18.1 are used to formulate the transport coefficients in terms of a certain set of integrals, in (b). These integrals are then evaluated under the special set of conditions referred to in (c) and (d). On the basis of the above: (a) Derive phenomenological equations which specify current and heat flux in a crystal subjected to magnetic fields and to gradients in electro chemical potential and in temperature. Utilize the distribution function specified above.
18.2
Transport in metals
407
From (a) identify in terms of appropriate transport integrals the following transport coefficients: 1. the resistivity for H = 0; 2. the resistivity for H =1= 0; 3. the Hall coefficient; 4. the Seebeck coefficient for H = 0; S. the Seebeck coefficient for H =1= 0; 6. the Nernst coefficient; 7. the thermal conductivity for H 8. the charge carrier density. (c) Specialize part (b) as follows: introduce the mobility defined by u eTlm. Further, specify the relaxation time through the assumed relation T To€r- y, referred to in the introduction. Finally, take the limit Hz ~ 0 and apply the limiting case of either classical (-11 ;p 1) or highly degenerate (11 ;p I) statistics in Equation (18.2.2). Tabulate the resulting transport coefficients in terms of atomic parameters. Note that for highly degenerate statistics and for r = 0 the results turn out to be identical with those based on the Sommerfeld model. (d) Repeat (c) for the limit Hz ~ 00. Solution
(a) The rate of transport of charge past unit cross-section is given by
"r Zevdk 41TzejvkA d k .
J
=
3
3
(18.2.5)
On the right hand side the summation over discrete k is replaced by the integration
4~3 fd k; 3
the numerical factor arises from the counting
of states with discrete pseudomomentum hk (3). Now write out the dot product in Equation (18.2.2) and switch from Cartesian to spherical coordinates in k space; the latter step is permissible if the material under study is isotropic. Thus, we replace v x , v y , V z with v,: == V, Ve, Vq,; we then introduce expression 08.2.2) in Equation (18.2.1) and use kr == k, Ok, Tc this holds for all except negligibly small portions of the range of t values between 0 and S) one can extend the range of integration of the integral over T to -00 and +00 without changing the value of the whole expression. Thus
23
ily(t)
Time dependence of fluctuations
t 1, this becomes
l/Jy(T)dT.
472
, 11
(1) It can be shown by general arguments or-as we shall see later (Problem 24.1)-in special cases by detailed calculation, that the macroscopic-decay correlation-function assumption made here is equivalent to assuming that the fluctuations can be regarded as due to the action of a fluctuating force with correlation time 7 c which is negligible on the time scale of the macroscopic decay.
Chapter 23
474
23.5
= 0 is easily found to be
The solution of this with (J(t)
23.2
2DS,
and comparing with the previous expression for x~ gives
where {3 = "Y/21 and w' = (41c-"Y 2 )Y'/21.
It follows that the correlation function is given by
Wo(r)
D
kT. J.1 This of course, the Einstein relation, which can be ways, see for example Problem 17.12.
= k;exP(-f3l r l) LcosW',r,+ ~,sinw'lrIJ.
in other
23.4 Assuming the relation between correlation function and macro scopic decay for the current in a circuit of inductance L and resistance R, determine the mean square fluctuation in the charge passing round the circuit in time S.
23.3 Use the results of Problems 23.1 and 23.2(a) to determine the mean square x displacement in time S of a particle moving at temperature Fx T in a medium such that its mobility is J.1 Combine this with the solution per, t)
475
Comparing this with the standard normal distribution exp(-x 2 /2Llx2) gives 2Dt for the mean square displacement. Thus
f3. ,t ] , + w,smw
exp(-{3t)
Time dependence of fluctuations
Solution
= n(4nDtr'hexp (-r 2 /4Dt)
The macroscopic
of the diffusion equation for n particles concentrated at the origin at t = 0 to relate mobility J.1 and diffusion constant D.
for the current 1, L
d1 dt
+ R1
0,
gives Solution
1
Since the x component of the displacement in time S is related to the x component of velocity by Xs
=
f:
we have (f:1YxU)dt
Combining this with
vx(t)dt,
r
= 2S
t""
x~
00
.
tLf2
tkTwe get kT WI(r) = yexpr-(RIL)lrl].
r
It follows, using the result of Problem 23.1, that
Q2 == (foS 1(t)dt
w(r)dr,
where w(r) is the correlation function for V x ' Since Fx = (l/J.1)vx we see that 1/J.1 plays the part of the damping constant K, so that the result of Problem 23.2(a) gives here kT w(r) = -exp(-irl/J.1m). m The integral of this from 0 to xi becomes
= 1oexp[-(R/L)t]
2S
t""
2kTS R
23.S Our previous discussion of the fluctuations in charge passing round a circuit of resistance R was based on the relation between correlation function and macroscopic decay function. We now consider a kinetic theory calculation which verifies the result of Problem 23.4 for a very particular model of the circuit and the resistive element in it. Suppose that the electrons in the circuit can be treated classically and that the resistance arises entirely from the presence of a symmetric potential barrier of height Vo. It will be supposed that the velocity distribution of particles in planes on either side of the barrier can be treated as equilibrium distributions for the purpose of calculating the rates at which particles move into the barrier region. Collisions in the barrier region will be supposed negligible. Show that the number of particles capable of surmounting a barrier of height V which pass into the barrier from one side per unit time is equal to V/kT) where C is independent of V. Determine, in terms of
is J.1kT and the above expression for
= 2J.1kTS.
Turning now to the given solution of the diffusion equation and y2+ Z2 we note that integrating over the y and z substituting r2 coordinates gives that the probability distribution at time t of the x coordinate of particles at the origin at t 0 is proportional to exp(-x 2 /4Dt) .
t
J
Chapter 23
476
23.5
23.6
Vo and C, the net current across the barrier when an electrostatic pot~ntial
Time dependence of fluctuations
and this will also be the mean square fluctuation in the number. If the numbers are denoted by n 1 and n 2 , and the charge passed by Q, then
20E is applied across it and hence determine the resistance of
the barrier. Determine also the fluctuations in net current across the barrier integrated over the time S and so verify that the previously derived relation is satisfied.
i:l.Q2
=
Q = enl-en2, 2 e i:l.ni+ e2i:l.ni = 2e 2 Cexp(- Vo/kT)S .
Comparison with the expression for 1/R gives
Solution
The number of particles per unit volume with momenta in the range dpx, dpy, dpz at Px, Py, pz is
i:l.Q2
A exp [-(p;+ P;'+ p;)/2mkT]dpx dpy dpz .
The number per unit time with z component of momentum in this range which pass into the barrier (supposed of area a) will be the number having z component of momentum in the required range which lie in a volume (pz/m)a. Define p~ by p? /2m = V. The number of electrons with momentum greater than p~ which pass into the barrier per unit time will be
m
=
y(t)
=
v
i:l.n r2 =
.
i:l.nri:l.n s
=
M
=0
r =F s ,
where A is the mean number of pulses per second. Deduce that y(t)
Ceexp[-( Vo eoE)/ kT] - Ceexp [-( Vo+ eoE)/ kT] .
= Al~~ f(t')dt'
and
For small oE this becomes
l/Jy(r)
oE 2Ce 2exp(- Vo/kT) kT
The resistance R of the barrier is therefore given by I
11r
and
If an electrostatic potential 20E is applied across the symmetric barrier, the heights of the barrier viewed from the two sides will be Vo eoE and Vo+ eo£. The mean net current across will therefore be
=
AL~ f(t)f(t+r)dt.
Solution
The first result is easily obtained:
2
R=
L nrf(t - t r ) .
Since the quantities nr are independent and each has a Poisson distribu tion we have
Baf~ exp(-x/kT)dx
= BakTexp(- V/kT) = Cexp(- V/kT)
S,
23.6 A fluctuating quantity y(t) is made up from a series of identical pulses of form f(t) randomly distributed in time. If we suppose the time scale divided up by a series of closely and equally spaced instants t" such that tr+ 1 - tr = 0, and that nr denotes the number of pulses 'centred on' times between tr and tr+ b we may write [if f(t) is 'centred on' t = 0]:
Bexp(-p;/2mkT)dpz .
Bf~ exp(-p;/2mkT)PZ adpz
2kT
=R
This agrees with the previous result.
The number with pz in dpz at pz (z will be supposed normal to the plane of the barrier) will therefore be
p',
477
Ce kT exp (- Vo/kT) .
y(t)
Because of the independence of the classical particles we can suppose that the number of particles crossing the barrier from either side will have a Poisson distribution. For each direction the mean number crossing in time S will be
=
~nrf(t -
tr)
= M ~ f(t -
tr )
= AL~ f(t')dt'
,
where we have let 0 tend to zero to turn the sum into an integral. The second is derived as follows. We have i:l.y(t)
CSexp(- VolkT)
t I
Ii
= y(t)-y(t) =
Lnrf(t-tr )- Lnrf(t-tr )
=
Li:l.nrf(t-tr )
23.8
23.6
Chapter 23
478
r~Llnrf(t
~ Llnsf(t+ T- t s) ]
t r )] [
LLlnrLlnJ(t
tr)f(t+T-ts)
"s
Solution
We have, from a standard Fourier transform result,
= ALOf(t-tr)f(t+T t r )
i+= _~ I Ys (w)1 dw.
+=
AL~ f(t)f(t+T)dt,
=
2
In the limit of large S-.!he left-hand side, which is the integral of y2 over a YS'(-w) so that time S, becomes Sy2. Since YsU) is real Ys(w) YS (w)1 2 = I YS {-w)1 2. Thus the right-hand side can be expressed as an integral from zero to infinity and we have
where we have again let 0 tend to zero to make the last step.
-
23.7 A current I(t) consists of a random series of pulses occurring at a mean rate A per second, each consisting of a constant current 10 lasting for a time t". Find the correlation function.
y2
= lim -2f"" s .. oo S
From the previous result we have
= {~6Uo-ITI)
ITI < to otherwise.
.
So""Gy(w)dw .
y(t)
=
r
L
f(t
r
the summation being over all r such that tr lies in the interval of length S used in defining y s (t). This gives
f+= Ys(w)exp(iwt)dw . _00
Ys(w) =
=
= lim ~.,----
f+=
I
~.J2rr _~ f(t- tr)exp(-iwt)dt
f
1 _= f(t- tr)exp[-iw{t- tr)]d(t- tr)exp(-iwtr ) ~v'21T
s... ~S
= F(w) Lexp(-iwtr ) ,
where the bar denotes an ensemble average. Show that y2
2 dw
YsU) = Lf(t- t r ) ,
Then Gy (w) is defined by
Gy
1
where the tr occur at random, there being on average A values of tr per unit time. If S is very long compared with the duration of a pulse we may write (neglecting only very small end effects)
23.8 The power spectrum Gy(w) of a real fluctuating quantity y(t), for which y has been made zero by suitable choice of origin, is defined as follows. We first define a truncated quantity y s (t) which is equal to y(t) in a range of t of length S and zero elsewhere. The Fourier transform of ys(t) is denoted by Ys(w), so that 1 = .J2rr
Ys{w)
y2 =
Suppose now that
since the integrand has the value IJ throughout the region for which two pulses, displaced with respect to one another by an amount T, overlap, and is zero elsewhere. The length of the overlap region will be to - IT I if IT I < to and it will be zero otherwise.
YsCt)
I
0
Clearly, taking an ensemble average of the right-hand side will not invalidate this equation, so we get
Solution
\fII(T)
479
Evaluate Gy (w) for the case in which y(t) is formed from a series of identical pulses occurring at random times, a pulse being assumed for will be zero and simplicity to have zero time integral, so that readjustment of the zero is not necessary. Show from this example that it is necessary to take both the limit of large S and the ensemble average in order that the definition of G(w) should be ';'Ilti"f",..tnru
and so . " . , . '=
Time dependence of fluctuations
= tOO Gy(w)dw
r
where F( w) is the Fourier transform of a pulse 'centred on' the origin,
.
:1
480
23.8
Chapter 23
Le. F(w)
23.10
Time dependence of fluctuations
For sufficiently long S the output for the truncated input will differ from the truncated output ys(t) only by a negligible end effect. We can, therefore, write
1
I _~ f(t)exp(-iwt)dt . v2rr
Ys(w) = A(w)Xs(w)
This gives
I Ys(w)1 2
I F(w) 121
~exp(-iwtrf
and the output power spectrum Go(w) is then given by 2 -~~~~~:--
If S is very long compared with l/w, the summation on the right-hand side will be over complex numbers of modulus unity with random phases, there being on average AS terms in the sum. The square of the modulus of this sum will be the square of the distance from the origin to the final point of a two-dimensional random walk consisting of on average AS steps of unit length. No matter how large S, these distances squared will fluctuate wildly. It is only if we average over the directions of the steps by averaging over the possible values of the tr (i.e. taking an ensemble average) that we get a definite value, namely the mean square displace ment in a random walk of A.') unit steps, which is iust AS. Thus for sufficiently large S I Y (w)1 2 = ASIF(w)1 2 S
and so
2-----=
Gy(w) = 51 Ys (w)1 2 = 2AIF(w)1 2
•
It is clear that it has been necessary both to take S large and to take an ensemble average. Thus a definition of Gy (w) which did not include both the limit of large S and an ensemble average would not lead to a sensible result in the particular example considered, and so would not be a satisfactory definition.
23.9 A linear system has a response function A(w), i.e. an input exp(iwt) gives an outputy(t) = A(w)exp(iwt). Determine the power spectrum Go{w) of the output, if a fluctuating input with power spectrum G1(w) is applied. Hence express the mean square value of the output in terms of the -input power spectrum.
xU)
Go(w)
2
lim SIYs (w)1 2 = IA(w)1 2 lim SIXs(w)
s~~
s~~
If the Fourier transform of the truncated input (defined as in Problem 23.8) is denoted by Xs(w), we have 2
lim S IXs{w) 12 .
s-+""
The Fourier transform of the output produced by this truncated input will be A(w)Xs(w) .
= IA(w)1 2 G1(W).
The importance of the power spectrum concept stems in large measure from the simplicity of this relation between output and input power spectra. The relation may of course be regarded as obvious since one may think of G( w)5 w as being the mean square value of the narrow band fluctuations obtained by eliminating all Fourier components of the fluctuations except those with frequency in the range 5w at w. (From the point of view adopted here, this follows from the expression which has been derived for the mean square value in terms of an integral over the power spectrum; one may, alternatively, start the discussion from this less formal definition of the power spectrum.) Since a narrow band fluctuation is sinusoidal with slowly varying amplitude and phase, IA( w) I will give the ratio of the output amplitude to the input amplitude for the frequency range considered and the result follows at once. The mean square value of the output is given by
L~Go(w)dw = I
IA(w)1 2 G1(w)dw.
23.10 Calculate the response function A(w) where the system is a capacitance and resistance in series, the input being the voltage across them and the output the current through them. Calculate also the response function for a damped suspended system, the input being the applied couple and the output the deflection. Solution
In the case of the capacitance C and resistance R in series, the voltage V = VOe iwt and the current 1 = loe iwt are related by
VO Solution
G1(w)
481
Z(w)/o ,
where Z(w), the impedance, is given by elementary a.c. theory, 1
Z(w) =
R+-.
10
A(w)Vo
IWC
But, by the definition of A so that I A(w) = Z(w) = R
I
+ l/iwC
.
482
23.10
Chapter 23
For the damped suspended system the deflection applied couple P by an equation of the form 18+K8+ce
If we put P(t) by
e iwt, then
e is
23.13
related to the
It follows that
P(t).
so that
+ iKW
The inverse of this relation then gives
.J2i -2- Gx (w)
23.11 Calculate A( w) for the case in which the input xU) produces an output y(t) xU + T) x(t). Hence obtain the mean square value of the output if the power spectrum of the input is GA w), and so obtain the correlation function for x, 1/1 x (T), in terms of the power spectrum of x, Gx (w). This gives a somewhat unorthodox _derivation of the__Wi~l1~r Khintchirie--relation between correlation function. andpQwer spectrum. -snow·· that the relation obtained can be inverted to give the power spectrum in terms of the correlation function. Note: No physical system could have the response assumed here being assumed positive), since it implies a response to, say, an input [) function which occurs before the input is applied. This is, however, unimportant for the purely mathematical discussion involved in the question.
or
Gx(w)
G(w) = 2 1r
0
23.13 Evaluate the power spectrum of the fluctuations associated with a random series of identical pulses by applying the Wiener-Khintchine relation (Problem 23.11) to the expression for the correlation function obtained in Problem 23.6. Check that the result agrees with that obtained by direct application of the definition of the power spectrum in Problem 23.8: notice that in this alternative derivation we do not make the assumption (introduced in Problem 23.8 for simplicity) that the time integral of a pulse is zero.
cosWT)GAw)dw.
= 2x L
foo 1/I(T)dT
for values of w up to a value less than (but of the order of) liTe· This expression is independent of w.
But the left-hand side is X(t+T)2+ X(t)2-
0
Because 1/I(T) is negligible for T > Tc , we can restrict the range of integration over T to that between zero and Te' If w ~ I lTc, WT will be much less than unity for all T in the above range, so that cos WT may be replaced by unity in the integrand. Thus to a good approximation
L~leiWr - 112 GAw)dw (1
0
2f'"1/I(T)coswTdT. 1r
e iwr - 1
I
= -2 foo 1/Ix(T)coswTdT.
G(w) =
Expressing the mean square value of the output in terms of its power spectrum, IA(W) 12 GAw), gives
= 2
.
1/Ix(T)e- 'Wr dT
We have the Wiener-Khintchine relation derived in Problem 23.11
so that
=
_00
Solution
eiw(t+r)_eiwt = (eiWr-l)eiwt,
[x(t+ T) - x(t)]2
1-
I .J2i
23.12 Show that if the correlation function has dropped practically to zero for T > Tc , then the power spectrum will be constant up to frequencies of the order 1ITc . Tc is called the correlation time.
An input x(t) = e iwt will produce an output
=
=
1r
Solution
A (w)
Gx(w)coswTdw.
f-
c Iw 2 + iKw '
c
1
To invert this, starting from the more familiar exponential form of the Fourier transform relations, we may proceed as follows. We define Gx(w) for negative W by Gx(-w) Gx(w) and rewrite the above relation as 1 _'" -2-Gx(w)eiWr t::r=. 1/Ix(T) = .J2i ~ v21r dw.
e iwt
A(w) =
483
00
1/Ix(T) =
e will be given, after the decay of transients,
e
Time dependence of fluctuations
21/1x(T)
2 t~Gx(W)dW-21/1x(T).
'I
23.13
Chapter 23
484 We have from Problem 23.6
Af-~f(t)f(t+r)dt.
=
The Wiener-Khintchine relation gives
2 i~
Gy(w) = -
~
1
l/Jy(r)coswrdr = -
~
0
Ar-r
= -; J-~ J-~
f +~ l/Jy(r)eiWTdr -~
A(w) = VFiF(w)
which is the required result. 23.15 Consider a system for which the response to an input oCt) is a 'delta function delayed by a time r, oct - r). Determine the response functionA(w) for the system and show that ifit is written in the form A(w)
f(t)f(t+ r)eiWTdtdr
= A'(w)-iA"(w)
with A' (w) and A" (w) real, then A' (w) and A" (w) satisfy the relations
r-f _l ~rL~f(t)eiwtd~ lf-~f(t)e;wtdtJ i -·---xdx
A = -;J-~
=
485
Time dependence of fluctuations
But the ratio of the Fourier component of the output at frequency w to that of the input at the same frequency must be A( w), and it follows that
Solution
l/Jy(r)
23.15
,
_~ f(tdf(t2)eiw(trtl)dtldt2
= 2AI F(w)
12 ,
where F(w) denotes the Fourier transform of f(t), and we used in the last step the fact that f(t) is real. This agrees with the result obtained previously.
1 i-A"(W')dW'
A (w)
= -P
A"(w)
=
~
~
,
w-w
_~
,
pi-A'(W')dW' _~ w'-w '
if, and only if, the delay time r is positive. The symbol P indicates that the Cauchy principal value of the integral which follows is to be taken. smax [Note: _~ has the value ~, 0, or -~ respectively when a is positive, zero, or negative.]
Solution
23.14 The remaining problems of this section lead up to an important result concerning the response functions of physical systems. This result will be made use of in the following section. The response of a linear system may be specified by giving the output f(t) produced by a delta function input Set). By taking the Fourier transforms of the output and the input, when the input is Set), show that the response functionA(w) for the system is the Fourier transform of f(t) multiplied by .J2ir. The Fourier transform F( w) of f(t) is taken to be given by F( w) = - -1-
VFi
i- . -~
f(t) e- 1wt dt .
In this case f(t) A(w)
= =
o(t- r) and, we have from Problem 23.14 PC y2~F(w)
e- iWT A'(w)
i-.
VFi
-~
1 S(t)e-1wtdt =---
vr:rr
=-1
f
VFi
+00
-~
f(t)e-iwtdt.
,
0(t-r)e-1wtdt
= coswr- isinwr .
i
P
-A"(w')dw' , w-w
_~
f f
=P
-
{
A"(w)
= sinwr .
-sinw'rdw'
_~
--~,----
w-w
-sin[(w'- w)r+ wr]d(w'- w)
,
w-w
~
= coswr
The Fourier transform of the output f(t) will be F(w)
1-
VFi~
= coswr ,
It follows that
The Fourier transform of the input o(t) is 1 ---
-- 1 v2~-
Thus
=P
Solution
=
f-' _~
smrx ----dx+ sinwrP
x
~coswr
r
>0
0
r
=0
-~coswr
r
./Ix(r)
-l X
Z
pz -_-exp(-PIlri) _ pz PI pz PI
with pz - Kim, PI - elK, and Xl = kTle.
502
Ir
--xz K
o·
Now doing a further average over x 0 gives the required result. It should be noted that we get the correct result by the formal procedure of multiplying the given macroscopic equation by x and averaging. It will be useful to make use of this formal procedure in the slightly more complicated cases to be considered later.
504
25.3
Chapter 25
25.3 X(t) and YU) are two variables associated with a system both of
which behave classically. The statistical character of the fluctuations will be unchanged by time reversal [i.e. X'(t) XC-f) and Y'(t) y(-t) will have the same statistical properties as X(t) and Y(t)]. Deduce that AX(t)[AY(t+r)-AY(t)] = AY(t)[AX(t+T)-AX(t)],
so that dividing by r and taking the limit of small AX(t)A}Tct)
T
gives
L\Y(t)AX(t).
Here, and in the examples which follow, it is to be understood that T becomes small on one time-scale but remains large on another (see Problem 25.2). Solution --------
Adding AX(t)A YU) to both sides of the first form of the result shows that what we have to prove is AX(t)AY(t+r) = AY(t)AX(t+r).
But the invariance of the statistical properties with respect to time means that such average values are independent of time f, so that we can write in particular AX(t)AY(t+r)
AX(t-r)AY(t).
The invariance under time reversal means that we can replace r by -r on the right-hand side, so that this result becomes AX(t)AY(t+ r) = AY(t)AX(t + r)
as required. 25.4 Suppose that the rates of change of the variables X(t) and YU) are related to their deviations from their mean values by AX(t) = aAX + bAY, A1Tct)
cAX+dAY.
25.5
Onsager relations
This is a special case of the type of relation derived by On sager. It is special because the rates of change of AX and A Yare expressed, in the equations under consideration, in terms of AX and AY themselves. More general equations, to be considered later, relate AX and AY to the departures from the equilibrium values of other quantities. If it happens that AXA Y = a and AX 2 A Yz then the above rela tion reduces to the typical On sager relation form be. It must, however, be emphasised that the physical content of the relation between the coefficients is in no way diminished by its not taking this special form. The same is true in the case of the more general equations mentioned above: the fact that. it is ahyays possible to choose the quantities in terms of which AX and AY are expressed so that the Onsager relation takes its typical form is to some extent incidental.
25.5 A passive electrical network has two pairs of terminals, the d.c. voltage and current at one pair being Ii'; , II and at the other V;, I z . These are related by the equations II ali'; + b V; , Iz =
qi
ClkT,
and qlqz = O.
(b) Express the equations giving II, I z in terms of Ii';, V; as equations relating qI, q2 to q I, qz. Use the time reversal result qlqz
AXAY.
Solution
aAXAY+bAY z = cAX2 +dAXAY.
+ d V; .
q~ = ~kT,
to obtain the required relation.
Using the formal procedure discussed in Problem 25.2 we multiply the first equation by AY(t) and average, the second by AX(t) and average. The left-hand sides are then equal by the time reversal result established in the previous problem, and equating the right-hand sides gives
C VI
The network is assumed to be of such a nature that the determinant of the coefficients is non-zero: this means that I, and 12 cannot both be zero unless Ii'; and V; both vanish. Obtain the On sager relation for the coefficients in these equations by the following steps. (a) Suppose capacitances CI and Cz are connected across the two pairs of terminals. Denote the charges on these capacitances by q I and q2' Show that if the whole system is at temperature T,
Obtain an equation which is satisfied by the coefficients a, b, c, d, the only other quantities involved in the equation being AX z , A y2 and Solution
505
q2ql
(a) The generalised forces associated with q I and qz are voltages, v I and V1 say, introduced in series with the capacitors (the voltages being so connected that the work done by them is VI Dq I in one case and V1Dq1 in the other). The voltages across the terminals will then be 11 I - q dC] and 1I1-q2/~' and when the currents are zero (as they must be in equilib rium) these voltages must be zero. Thus ql
=
CIVI,
qz
=
C2 V z .
506
Chapter 25
25.5
It follows that (ql and q2 are, of course, zero for zero applied voltages, so that t:..ql = ql' t:..q2 == q2)
qlq2 (b)
q1
kT
a (-a I) I
kT
q2 (aav 2) T
= kTC2 ,
kT
aq l ) (-a 1)2 T
= O.
)
= kTCI
T
= -q2
II
-ql,
12
V.
ql CI
V2 =
'
,
,
q2 C 2
The equations relating II and 12 to VI and V; can therefore be rewritten . qI =
a b C q 1 - c,. q 2
c tt2
,
I
d
= - C1ql- C q2' 2
(c) As in Problem 25.4 multiply the first of these equations by q2, the second by q I, and average b
a __ 1= -
Clqlq2-
c2qi,
d_ - C QIQ2'
25.6
Onsager relations
gradient d T! dx in a metal: the metal is assumed isotropic so that it is not necessary to work with vectors and tensors. The coefficients are those in the equations dV dT I = -a dx + a dx ' dV dT J=bdx+c dx '
where a is the electrical conductivity. To obtain the required relation consider, as in Problem 21. 7, a piece of the same metal connected by a wire of the metal to a very large block, also of the same metal, which acts as a heat bath (temperature T) and electron reservoir (chemical potential /l). Suppose that the wire has cross section s and length I and that the capacity of the first piece of metal in the presence of the large block is C. Write down, in terms of the coefficients defined above, equations relating the rate of change of the number of electrons on the piece of metal, and of the energy of the metal to the excess number of electrons t:..n and excess temperature t:..T for the piece of metal. Multiply the first equation by t:..E (the excess energy of the piece of metal), the second by t:..n, and average. Apply the time inversion relationship (Problem 25.3) and some of the mean values determined in Problem 20.5 to obtain the required relationship. Solution
We have Lln =
-~
dV t:..v dx = -[-
t:..E = -Js,
e
=
dT dx
t:..n e IC
2
The left-hand sides are equal, and equating the right-hand sides gives, on substituting from b c,. kT C2
c
= - CI C1 kT ,
that is
b
=
c.
In this case, then, we get the standard Onsager reciprocal relation. 25.6 This problem illustrates the use of the Onsager procedure to obtain a relationship between transport coefficients for a material. The coefficients considered will be those relating current density I and thermal current density J to voltage gradient dV!dx and temperature
501
t:..T
The equations given for I and J can, therefore, be rewritten as
sa
sa
t:..n = +-t:..n--t:..T IC el ' . esb sc t:..E = - TC t:..n - T t:.. T .
Multiplying the first and second equation by t:..E and t:..n respectively, averaging, and using the time inversion relationship t:..ELln = t:..nLlE gives
a--
- Ct:..Et:..n +
-eat:..Et:..T =
-
+ ct:..nt:..T
We saw in Problem 20.5 that t:..Et:..T = and t:..Tt:..n = O. We also saw 2 2 in Problem 21.7 that t:..n = (C/e )kT. This leaves only t:..Et:..n to be put
Chapter 25
508
25.6
25.8
Onsager relations
509
in appropriate form, and we had in Problem 20.5 that
all
all
-MDon = kT (aE) T = kT (aE) an T (an) T· But
= T(as) + 11 = _T(a ll ) + 11 = ( aE) an T an T aT n
-T2~(~) , aT T n
and at constant temperature 2 dll = e d V = e dn C
so that
(~:)T = ~
Thus
-MDon =
.
2
-kT 3 - a (11) C aT T n e 2
3 a (11) a b 2 = -kT +akT -2- +-kT
i.e.
aT T n e e
-(::)n
'
Fn = - (
This relation is not of the standard reciprocal form, but it contains all the physical information derivable from Onsager type considerations. 25.7 Deduce from Problem 22.5 that if we write the rates of change of the coordinates Xr as linear combinations of the associated forces Fs ' defined by Fs = -as/axs , then the array of coefficients is symmetric, i.e. if we write Xr = Lf3rs Fs s then f3rs = f3sr . This is the orthodox form of the Onsager relations. Multiplying the equation for XI by Xm and averaging gives Lf3ls X m Fs s
=
Lf3ls ko ms s
Similarly XIX m
=
kf3ml .
So the time reversal relation gives 131m
=
f3ml .
~~)
E
=
Do (~) .
Re-express the equations relating E and Ii to Don and DoT, with coefficients as given in Problem 25.6, as equations relating E and Ii to FE and Fn. Show that the Onsager relation, when applied to these equations, yields the same relation between transport coefficients as was obtained before. Solution
DoS in the expressions for FE gives a s ) Don = - aEa(as) a(as) aE M- an aE Don (aaE2s)nDoE- (aEan
Substituting the expression for 2
FE = -
2
= -Do (::) = -Do(~) =
~~ .
Similarly
Solution
=
2
Show that the forces associated in the Onsager theory with E and n, FE and F n , are given by DoT FE = T2 ,
aT2 a (11) -e-aT T n +aT = b.
XmX I
2
rI(a s) nDoE 2+2\aEan ( a s) DoEDon+ (aan2s) E Don2]. DoS = !~aE2
The relation between the coefficients is therefore
e
25.8 The aim of this problem is to derive, by an application of the On sager relations in their conventional form, the relation between the electro-thermal transport coefficients of a metal already derived in a less orthodox way in Problem 25.6. As in the earlier problem consider a metal body connected by a metal wire to a much larger metal body, all the metals being the same. If the energy and number of electrons on the smaller metal body increase from their equilibrium values by DoE and Don respectively, the increase in entropy of the whole system will be, to lowest non-vanishing order, a quadratic function of DoE and Don,
=
kf3lm .
F
n
= _~(aS)DoE_~(aS)Don = _Do(as) = aE an
an an
an
Do '\T) (Il~
It is necessary to express both FE and Fn in terms of Don and DoT. FE is
already given in this form and Fn is easily expressed similarly if we recall (Problem 21.7) that, for the system considered, (all/anh = e 2/C, where C is the capacity. Thus
a (11) a (11) T nDoT+ an T
Fn = aT
a (11\ e Don T) nDoT+ Tc . 2
T
Don = aT
510
25.8
Chapter 25
Solving for l:1T and l:1n in terms of FE and Fn gives
l:1T
T2FE'
l:1n
TC ~
(71 -
T
2
a (11.) aT T
71
26 Stochastic methods: master equation and Fokker-Planck equation
]
FE
In the notation of Problem 25.6 we had sa sa .::In ICl:1n - eZl:1T ,
. 1lE
-
esb
1.0PPENHEIM (Massachusetts Institute of Technology. Cambridge, Massachusetts) K.E.SHULER (University of California, San Diego, California) G.H.WEISS (National Institute of Health, Bethesda, Maryland)
sc
Tc l:1n - T l:1T .
In terms of FE and Fn these become
l:1fl =
s T
1 e2aFn
J
(Jl)
s lr:~ a I e 2 aT\T
+
s I
71
lT3 (11.) e aTa \r
T2 a+ea 71
b-
]
Pc FE .
In this case (Problem 25.7) the Onsager relation is just the equality of the off-diagonal coefficients. This gives at once b
a (11.) = Ta + e aT T
26.0 In these problems (1) the following notation is used: Wen, tim, s) is the conditional probability that a system is in state n at time t, given that it was in state m at time s < t. It is normalised so that L W(n, tim, s) = I, where the sum is over all possible states n. 71
Pen, t) is the probability that the system is in state n at time t. For all cases of present interest the condition LP(n, t) = I is valid. 71
P2 (m, s; n, t) is the joint probability that the system is in state m at
T2
71
a,
which is the result obtained before (Problem 25.6).
time s, and that it is in state n at time t. , 1 A(n,m;t) = hm A W(n,t+l:1lm,t)-onml
(26.0.1)
A"'O.u.
is the transition probability per unit time (i.e. the transition rate) for a transition from state m to state n at time t. All of the processes to be considered will be such that A(n, m; t) exists. The probabilities Pen, t) and P2 (m, s; n, t) are related by
pen, t) LP2 (m,s; n, t).
(26.0.2)
m
The conditional probability wen, tim, s) is related to P and P2 by
P2 (m, s; n, t)
Wen, tim, s)P(m, s) ,
(26.0.3)
The reader is referred to Problem 2.14 for an introduction to the idea of master equations.
26.1
Show that the master equation has the form ap(n, t) A(n, m, t)p(m, t) .
L m
(26.1.1)
(I) A general reference to this section is I.Oppenheim, K.E.Shuler, and G.H.Weiss, "Stochastic
theory of multistage relaxation processes", in Advances in Molecular Relaxation Processes, Vol. 1, 1967-8, pp.13-68.
511
512
26.1
Chapter 26
In a radioactivity decay process the probability of emission of a single particle in the time interval (t, t+dt) is Adt, where A is a constant and the probability of emission of two or more particles in the same interval is zero. The state of the system is described by the number of particles n that have been emitted between time 0 and t. (a) Calculate the conditional probability Wen, t+dtlm, t) for this process. (b) Calculate the transition rate A(n, m, t) for this process. (c) Show that pen, t) satisfies the master equation open, t)
at
= A[P(n -1, t) - pen, t)],
n
1,2,3, ...
Stochastic methods
513
(b) By combining expressions (26.0.1) with (26.1.7) one obtains
1
A(n+l;n;t)=~~
A(n,n; t)
A
-A
A(n,m; t) = 0
(26.1.8) form =1= n, n-I.
(c) Substituting the transition rates (26.1.8) into the general master equation (26.1.1) one obtains the specific master equation (26.1.2) (d) Define a generating function
(26.1.2)
ap(o, t)
at
26.2
G(Z, t) =
L pen, t)zn .
(26.1.9)
n=O
= -AP(O, t) .
(d) Calculate the probability pen, t) from these last equations assuming that P(O, 0) = 1.
If the nth term of the master equation (26.1.2) is multiplied by zn and the sum taken, it is found that G(z, t) obeys the equation
aG at =
Solution
By definition
A(Z - l)G
(26.l.1
which has the solution
L m
p(n,t+~) =
(26.1.3)
G(Z,t) = G(z,O)exp[-(1-Z)At].
L onmP(m, t) , m
(26.1.4)
where onm 0 for n =1= m and onm = I for n One then obtains
m, and divide by ~.
For the initial conditions P(O, 0) = I and pen, 0) 0 for n = 1, 2, 3, one finds G(z, 0) = I and G(z, t) = exp[-( I-Z)Atj. Expansion of this function in a power series in Z and comparison with Equation (26.1.9) yields Atn (26.1.12) pen, t) = exp(-At)-,
t+ ~ jm, t)P(m, t) .
Subtract from this the identity pen, t)
pen,
t+~)-p(n, t) ~
Taking the limit one finds
=
L ±[W(n, t+~lm,t)-onmjP(m, t).
(26.1.5)
m
~
= 0 and using the definition of Equation (26.0.1) oP\n, t) =
L A(n, m; t)p(m, t)
(26.1.6)
m
as asserted. (ii) Since the state can only change by I during the time interval (t, t+ dt) one has
(26.1.11)
n.
for the probability that n particles have been emitted at time t. The distribution function of Equation (26.1.12) is known as the Poisson distribution. 26.2 Consider a master equation characterised by a time independent matrix A = (Anm) in which the elements Anm are the rates of transition from states m to states n. Define the vector of state probabilities P(t) by P(O, t)
P(l, t)
P(t) =
I P(2, t)
(26.2.1)
W(n+l,t+dt!n,t) = Adt Wen, t+dt[m, t)
0
Wen, t+dtln, t) = I
(26.1.7)
for m =1= n,n-I W(n+ 1, t+dtin, t)
I-Adt.
Let the matrix A have non-degenerate eigenvalues Aj (j = 0, I, 2, ... ),
~
26.2
Chapter 26
514
right eigenvectors Rj and left eigenvectors Lj , defined by ARj = AiR"
26.2
Stochastic methods
Substitution of this solution into the master equation (26.2.2)
L;A = A,L, .
L [a·(t)-A.a.(t)JR, = I I ,
;=0
The master equation can be written in terms of pet) and A as
P = AP.
(26.2.3)
Show that pet) has the formal solution P(t) =
"" .L (L P(O»Rj exp(A; t) ,=0 i
(26.2.4)
Define a conservative system as one in which A(n, m) n =1= m, and for which A(n, n)
-
L
I
A(m, n),
~
(26.2.5)
00
L pen, t)
,.=0
I
(26.2.6)
= I
(26.2.7)
for all t [the A(n, m) may depend on time]. Most systems of physical interest are conservative, and all systems to be discussed here are conser vative. (iii) Show that there exists a zero eigenvalue. (iv) If Ao = 0 and Al =fo 0, show that P(oo)
= Roj i~O (Ro}j ,
P(O) =
L a,(O)R,
(26.2.13)
in which it is assumed that P(O) is a known vector. To find the a/CO) we show that the inner product L,. R, = 0 when n =fo j. From Equation (26.2.2) it follows that L,. AR/ = \L,. R/ (26.2.14) L,. AR/ A,. L,. R/ or, subtracting, o = (A,. -X;)(L,.R,). (26.2.15) From the hypothesis of non-degenerate eigenvalues it follows that (L,. Rj ) 0 when n =fo j. In all cases of interest the vectors L,. and R,. can be scaled so that L,. R,. = I, and we can assume that the eigen vectors are orthonormal. If we now multiply Equation (26.2.13) by L,. and use the relation of orthonormality we find that a/CO) = L,P(O)
(26.2.8) P(t) =
Solution (i) Assume a solution for P(t) of the form
L a/(t)R, .
(26.2.11)
Thus, the a/CO) can be calculated from the relation
(26.2.16)
,=L (L/p(O»R, exp(A/t) . 0
(ij) Write the master equation out in full as
p(O,t) = AooP(O, t)+A 01 P(l,t)+A 02 P(2,t)+···
PO, t)
= A lOP(O, t) +A llPO, t) +A 12P(2, t) +...
p(2, t)
= A 2oP(0, t)+ A 21 P(l, t)+A 22P(2, t)+···
Show that, when Equation (26.2.9) is satisfied, the eigenvalues Ai are real and negative, and Ao = O. [See also Problem 2.14 for additional explanations of detailed balance.}
; =0
O.
which leads to the desired formal solution (26.2.4)
where (RO)i is the ith component of Ro. Notice that the equilibrium distribution P(oo) is thus independent of the initial condition P(O). (v) The condition of detailed balance at equilibrium can be expressed as A(m, n)P(n, 00) A(n, m)P(m, 00) . (26.2.9)
P(t) =
AP yields
f =0
where the prime denotes omission of the m = n term. Show that if
it follows that
P=
Since the R/ are independent when the Ai are non-degenerate, one may set each term separately to zero and solve the resulting equations a,(t) A/ai(t) to obtain a/(t) a/(O)exp(A/t). (26.2.12)
0 for
m
L P(n,O) = ,.=0
515
(26.2.17)
If these equations are added, taking account of the condition (26.2.5), it
is found that (26.2.10)
..
L Pen, t) = O.
,.=0
(26.2.18)
26.3
26.2
Chapter 26
516
L pen, t) = constant = L P(n,O) =
n~O
One can now verify from Equation (26.2.9) that Snm = Smm Le. S is a symmetric matrix. It is possible to conclude at this point that the AI are negative for j;;;' I if they were not, P(t) as given by expression (26.2.22), could not represent a vector of probabilities for large t. However, the fact that the Aj must be negative can also be proved by showing that S corresponds to a negative semi-definite quadratic form. This can readily be done by expressing the matrix elements A(n, m) in terms of non-negative elements B(n, m) by
n~O
L A(m,n)
= a (I, I, I, ...) .
(26.2.20) (26.2.5),
O,whichisequivalentto
m=O
can be written I)
LoA = 0;
hence the eigenvalue 11.0 corresponding to Lo is O. The existence of the non-degenerate eigenvalue 11.0 = 0 insures that the master equation (26.2.3) with the general solution (26.2.4) has a non-zero equilibrium solution P(oo). (iv) Recall the expansion of the solution (26.2.4) which expresses P(t) in terms of eigenfunctions and eigenvalues. It must be the case that Re Aj ~ 0 for all j, as otherwise P(t) would have unbounded components for sufficiently long times. One can therefore separate out the j = 0 term to find that P(t)
L [LjP(O)]Rjexp(Ajt),
[LoP(O)1Ro+ j
=1
where all terms in the second sum tend to zero as t can be written
Lo = a(l , I, I, ...) , where a is a constant, so that
-jo
00.
S(y) =
(26.2.25)
= I ,
l)-eO(OO)(l-e-O(O»
(26.3.12)
(26.3.
The coefficient of z" in Equation (26.3.12) is Pen, r) which can be written Pen, r) [1 - A(r)]A"(r) = (l -e-O(r»e- IIO (r) , (26.3.14 ) where O(r)
This equation can be solved by the method of characteristics, which requires the solution of the system of ordinary differential equations
1
where
(26.3.4)
0
-+(l-z)(ze-O("")-l)- = e-O(oo)(z
r z
= -logA(r) .
(26.3.15)
Thus, Pen, r) can be written in the Boltzmann form for all values of the time. The result in Equation (26.3.14) can be interpreted by saying that a vibrational temperature T(r) can be defined at every instant of time by
(26.3.6) T(r)
=
hv kO(r) .
(26.3.16)
It is easily verified from Equations (26.3.13)-(26.3.16) that
lim T(r) = T(oo)
(26.3.7)
r~oo
with the solution
as required. (eO(CO)-z)G(z,r) = KI ,
(26.3.8)
where K 1 is a constant of integration. The first and second equations of (26.3.6) can be solved in a similar way and lead to eO(CO) -z exp[-T(l-e-O(oo»] K2 (26.3.9) where K2 is also a constant of integration. The general solution to Equation (26.3.5) by the method of characteristics can be represented by Kl = 1/I(K2 ) where 1/I(x) is any arbitrary differentiable function.
26.4 Derive the second-order partial differential equation to which the set of differential difference equations (26.3.1) reduce in the limit as 0(00) ~ O. This differential equation is called the Fokker-Planck equa tion for this particular system which corresponds to the classical con tinuum limit of the harmonic oscillator. Solution
The singlet probability Pen, t) is also a function of the parameter 0 so that we rewrite it as Pen, 0, t) which, in turn, we can write as some function of nO, 0, t as P(nO, 0, t). We define a function p(x, 0, t) of the
T
26.4
Chapter 26
520
26.5
(26.4.1 )
Op(x, 0, t) = P(nO, 0, t)
p.(t) =
when x = nO. Equation (26.3.1) can now be rewritten as ===
521
Give necessary and sufficient conditions for the first moment
continuous variable x which has the property that
ap(::., 0, t)
Stochastic methods
f~xp(x, t)dx
(26.5.5)
to have the simple relaxation form given in Equation (26.5.3). Solution
j.{xe-8 p (x - 0,0, t) +(x + O)p(x + e, e, t)
(i) It follows from the master equation (26.5.1) that p.(t) satisfies the differential equation (26.4.2)
-[x+(x+e)e-ejp(x, e,
[J,(t)
=
""
L
P(m, t)
m"O
Expansion of the right hand side of Equation (26.4.2) in a power series in e yields
L nA(n, m) .
(26.5.6)
n=O
From this equation it is clear that if
a2p(x, t) apex, t) 2 e [x "'.. 2 +(x+ 1) "'.. +p(x, t) I +o(e ) ,
nA(n,m)
a-bm
(26.5.7)
n
where p(x, t) p(x, 0, t). We define a new time variable r = et, and take the limit as e -+ 0 to obtain the Fokker-Planck equation apex, r) ---
apex, r)
ax
p(x, t)+(x+ I)
aZ
for all values of m, where a and b are constants, then [J,(t)
azp(x, r) +x
ax2
a
x
lim (et)
r
and
lim (ne)
e->o
(26.4.4)
p(t)lt=o
e->o
t .....
n->""
(26.5.8)
which gives rise to a simple exponential relaxation of the first moment. The identification b = A, alb = p.(oo) then leads to the desired form. In addition to being sufficient, the condition (26.5.7) is necessary. Con sider the initial condition per, 0) = 1, P(j, 0) = 0 for j =1= r. From Equation (26.5.6) it follows that
(26.4.3)
= axz[xp(x,r)]+a)(x-1)p(x,r)] ,
where
a-bp.(t)
n~onA(n,r).
(26.5.9)
But if pet) is to be the solution to Equation (26.5.8) it must satisfy 26.S (i) Let the probability pen, t) satisfy a master equation Pen, t) =
L A(n, m)p(m, t) , m=0
[J,(t)
(26.5. I)
t=0
n
L0nP(n, t)
(26.5.2)
to have an exponential relaxation of the form p(t)
= p(oo)+[p(O)-p(oo)]e- At
(26.5.3)
valid for all values of t. (ii) Let the probability density p(x, t) be the solution to a FokkerPlanck equation apex, t)
at
=
a -ax[b
2 a (x)p(x,t)]+!ax t
t
2[b z(x)p(x,
(26.5.4)
(26.5.10)
at t O. Equating (26.5.9) and (26.5.10) we find that the condition (26.5.7) is necessary as well as sufficient. The preceding calculations can be generalized to derive necessary and sufficient conditions for the vector of the first k moments, p(t) Pt(t), P2(t), ..., Pk (t) to be the solution to [J,(t) A - Bp.(t), where A and Bare constant matrices. Multiplication of the Fokker-Planck equation (26.5.4) by x and integration over all x, followed by a partial integration of the right hand side of the resulting equation, leads to
where the transition ratesA(n, m) are independent of time. Give neces sary and sufficient conditions for the first moment pet) =
=a-br
pet) =
'I
1
"f
[ a b 2(X) apex, t) _""b1(x)p(x, t)dx+; xp(x, t)--ax-+xbz(x) ax -b z(X)P(X,t)-2xb t (X)p(x,t)11".
(26.5.11)
J
\
522
Chapter 26
26.6
26.5
In order that the moment relaxation satisfies ii(t) = a-bIlU)
f~bl(X)P(X,t)dx.
ris
(26.5.12)
T/(T)
a
T/(T) = D(T)dT+T/(T+dT) .
(26.5.13)
(26.6.3)
(26.6.4)
That is, if the molecule is not dissociated at time T, it either dissociates in (T, T+dr) or it is still undissociated at T+dT. Passing to the limit dT = 0 one finds
(26.5.14)
bx
= P(O,T)+P(l,T)+"·+P(N,T).
The probability that dissociation occurs in the time interval (T, T+dT) will be denoted by D(T)dT and can be calculated from the identity
From Problem 26.S(i) it is clear that the form bl(x)
523
The probability that the molecule is undissociated at dimensionless time
we must impose the condition that p(x, t) goes to zero sufficiently rapidly as x --,\> ± 00 so that the bracketed terms go to zero at too. Equation (26.5.11) then reduces to iiU)
Stochastic methods
D(T) = _ dT/(T)
with b = A and alb = Il(oo) is the necessary and sufficient condition for exponential relaxation of the first moment.
(26.6.5)
dT The dimensionless mean time TD to dissociation is(2)
26.6 Assume that an ensemble of diatomic molecules can be modelled by a system of one-dimensional harmonic oscillators whose time depen dent distribution function over the semi-infinite energy level system n = 0, 1, "', N, ... is determined by the master equation of Problem 26.3. Assume that an oscillator dissociates irreversibly whenever its vibrational energy reaches (N + 1)hv. Calculate the mean time to dissociation for an initial delta function distribution of oscillators, pen, 0) = on" where r is an integer satisfying 0 ",.;; r ",.;; N.
TD
JorooTD(T)dT =
-
S" d~dT 0
T
d
=
JorooT/(T)dT
= PO+Pl +P2 +"'+PN,
(26.6.6)
where we have set Pi =
Loop(j, T)dr .
(26.6.7)
Solution
The PI can be calculated by integrating both sides of Equation (26.6.2) over r from 0 to 00. One first notes that
Since for the harmonic oscillator employed here transitions can take place only between nearest neighbour levels, an = ± I, one need calculate only the mean dissociation time from level O. To see this, let (1',. N+ 1 > be the expected time for the energy to reach the value (N + 1)0 st~rting from rO. Then
= (To,,}+
0 one finds the following set of differential
P;(t)
= kMF;_l(t)-(kM+'Y)P:;U)
P" (t)
= 'YP:;(t)
n~1
(26.7.2) (26.7.3)
so that
from which Po is found to be Po = SN+ 1 .
(26.6.14)
P,,(t) = 'Y J:P:;(t)dt.
Using Equations (26.6.11), (26.6.12), and (26.6.14), we can now rewrite Equation (26.6.6) as n N N N+l ejll "L . , . (26.6.15) TD -_ " L., Pn "-nB( L., e SN+l-Sn ) " L., e-nli
" =0 " = 0 " =0 j = ,,+ 1 J
Carrying out the indicated summation yields I N+l e i 6 -1 TD 1- -II e i=1 J
for the mean dimensionless time to dissociation.
L -.
(26.7.4)
Equation (26.7.2) can be solved through the introduction of the generat ing function
L p:;(t)zn .
G(z, t) = "
The function G(z, t) satisfies aG -
at
(26.6.16)
(26.7.5)
I
= [kM(z-l)-'Y]G
(26.7.6)
with the solution G(z, t) .e" '.'.. ' ','
= G(z, O)exp{-[-y+kM(l- z)]t}.
(26.7.7)
~ 526
Chapter 26
26.7
one finds G(z, 0)
I for n = 1 = { 0 for n *- 1
*
(26.7.9)
raG(z, t) (taG(z, t) ]
az +'YJo az dt Z=I' (26.7.10)
= n~ln[Pn(t)+Pn(t)j = L
The second equality on the right hand side of Equation (26.7.10) follows immediately from the definition of the generating function, Equation (26.7.5). Evaluation of Equation (26.7.10) with G(z, t) given by Equa tion (26.7.9) yields III (t)
= e-'Y t (1 + kMt) +'Y
I
t
o
e-'Y t(1 + kMt)dt
=
kM
1 + -(1 - e-'Y t )
'Y
(26.7.11)
for the mean chain length of the polymer at time t. The higher moments Ilr(t)
z)' G(z, t) +'Y J: (z :x)' G(z, t)dtl
= nt Inr[p;(t) + Pn (t)j = [(z aa
= I
(26.7.12) can also readily be obtained so that it is possible to determine the disper sion of the mean polymer length. The probabilities P;(t) and Pn(t) can be obtained by expanding the generating function G(z, t) in Equation (26.7.9). This yields P;(t)
=
(kMt)n-1 (n -I)! exp[-('Y+kM)tj
(26.7.13)
and from Equation (26.7.4) Pn(t)
=
527
and
!(l+~).
The possible positions of the particle are limited by the condition
= z so that
The mean chain length III (t) is given by 00
!(l-~)
(26.7.8)
G(z, t) = zexp{-[-y+kM(l-z)jt}.
Ill(t)
Stochastic methods
probabilities of moving a distance e to the right or left are, respectively,
For the assumed initial condition P;(O)
26.8
(kM)n- I (t 'Y (n -l)! Jo rn-Iexp[-('Y+kM)rjdr.
(26.7.14)
26.8 Consider a one-dimensional random walk where a particle can move the distance e either to the right or to the left. The duration of each step is I::.t. The probability of moving in either direction depends upon the position of the particle; if the particle is at the point k, the
- N ~ k ~ N, where k and N are integers. (i) By appropriate scaling of the variables and in the limit N
~
00,
derive a partial differential equation, the Fokker-Planck equation, in continuous time and space for p(x, r) dx, the probability of finding the particle between x and x + dx at time r. (ii) The one-dimensional Ornstein-Uhlenbeck equation can be written in the form a2p ap(x,r) 1 a (26.8.1) ar = -Yax[F(X)p(X,r)j+D (x,r) 2
ax
where f is the 'friction coefficient', F(x) is an outside force acting along the x-axis, and D is the diffusion coefficient. Compare your Fokker Planck equation with the Ornstein-Uhlenbeck equation and from the explicit expression for F(x) discuss the physics of the random walk of part (i). Solution
(i) Let peke, sl::.t) be the probability that the particle is at point ke at time sl::.t. Since the probabilities for moving right or left add up to unity, the probability that the particle remains at point k during the time interval I::.t is zero, and the difference equation for the random walk therefore is
P(ke,sl::.t)=!
Ne P[(k-l)e, (s-l)l::.tj {rLl - (k-l)e] l)e] } +Lrl + (k+ Ne P[(k+l)e,(s-l)l::.tj.
(26.8.2)
Subtracting P[ke, (s -l)l::.tj from both sides and dividing by I::.t we obtain, after some algebra, P[ke, sl::.tj-P[ke, (s-l)l::.tj I::.t = ~{P[k -l)e, (s -l)l::.tj- 2P[ke, (s -l)l::.tj + P[(k+ l)e, (S-l)l::.t j } 21::.t e2 _l_{(k + 1)eP[(k+ l)e, (s -1 )I::.tj- (k - 1)c:P[(k - l)e, (s - l)l::.t j} + I::.tN 2e .
(26.8.3) In the limit as I::.t
~
0,
e
~
0,
N
~
00 ,
(26.8.4)
T
528
Chapter 26
26.9
26.8
and subject to the validity of the limiting processes IltN -+ 'Y, silt -+ T,
k€
-+
(26.8.5)
X ,
PI(Xt> t l )
~[P(
'Y ox XX, T
).]+IDOZP(X,T) ~ oxz
(26.8.6)
-'Yfx
-ax
- 2al
a Z = <XZ(t»
[X (p/a Z)yj2 }, (26.9.4) 2a Z(l-p2/a 4 )
{
= <XZ(O» = constant
P
= <X(t)X(s»
= p(lt-sl) = p(T).
(26.8.7)
peT) = exp(-~!TI), peT)
== O.
(26.9.8)
Solution
(26.9.1)
P(t3 -ttl
<X(t3)X(tI»
or more.explicitly as
The term wrdx, represents the probability that x, ,X(tr ) < xr+dx, given the information that X(td = XI, X(tl) = Xl, ... , X(tr_ d = X,_I' A Markov process is defined by the requirement that
p(t3- t d = fff-:XIX3P3(Xl,tl; X2,t l ; X3,t 3 )dx l dx l dx 3 tl
h
t,
1)
(26.9.2)
< t z < t3 .
a Zp(t3 -td = P(t3 -tz)p(t z -tl)
or
= PI(Xt> tl)WZ(Xl, tllxI' t 1 )Wl(X3, t3lxz, t l ) x ",w2(xr,trlx,-bt,-I)
for all r
(26.9.9)
If use is made of Equations (26.9.3) and (26.9.4) the integrals can be evaluated explicitly as
for all r> 2. The combination of Equations (26.9.1) and (26.9.2) for a Markov process yield the general relation p,(X 1 , t I ; ... ; X" t,)
(26.9.7)
(J> 0
or
We shall derive and solve a functional equation for the correlation function peT). One can compute p(t 3 -tl) as
x Pr-I(XI,tt; ... ; Xr-t,tr-t)·
= Wz(x"t,lx,
(26.9.6)
The brackets indicate an ensemble average. Correlation functions are of great importance in stochastic theory and statistical mechanics. In particular, transport coefficients can be written in terms of time integrals of correlation functions, see Chapters 13, 23, and 24. Show that for a Gaussian Markov process the correlation function peT) of Equation (26.9.6) has the form (~ is a constant) of Problem 23.3, i.e.
wr(xr,trlx,-l,tr-I; ",XI,tI}
I; ,,,XI,!l)
(26.9.5)
and the correlation function
26.9 For a general stochastic process in continuous state space the state probabilities are defined to be Pr(xt> II; Xz, tl;'" Xr , t r ) where Pr dx I dx 2 .. , dx r is the probability that the random function X(t) satisfies Xl ,XUI) < XI +dx1,xz ,X(tl) < Xz +dx l , ..., Xr ,X(tr) < Xr +dx r . Conditional probabilities wr(x r , tr i x r _ 1 ,tr_ I; ... X I, t 1) can be defined in terms of the Pr by
W,(x"I,!x,_h t ,
,
where the variance
so that the random walk (i.e. the Brownian motion) is that of an elasti cally bound particle.
=
(Xi)
a[27T(l-pl/a4)]% exp
Wl(X, tly,s)
where x and T are the continuous space and time variables. [Note that the passage to the limits as given by Equation (26.8.4) is the crucial step in transforming a discrete difference equation to the corresponding par tial differential equation.] A comparison of Equation (26.8.6) with the general Ornstein Uhlenbeck equation shows that F(x)
1 p = ay'TiTex
I
the above equation reduces to the Fokker-Planck equation OP(X,T) OT
529
A stationary Gaussian Markov process is defined by requiring that
I
€z
Ilt -+ D,
Pr(Xt>t I ; ... , xr,t,}
Stochastic methods
aZp(T+T') = p(T)p(T'),
T, T'
> O.
It follows from Equation (26.9.11) that either peT) peT) must have the form peT) = a 2 exp(-{JT),
(26.9.3)
> 2.
..41.
j
(26.9.
(26.9.11)
== 0 for all T or that (26.9.12)
530
Chapter 26
26.9
where (J is a constant. Since for a stationary process per) = p(-r), Equation (26.9.12) is more properly written per) = a 2 exp(-(Jlrl).
1
(26.9.13)
27
The constant (J is chosen positive in all cases of physical interest so that the correlation does not go to infinity as r -l>- 00. The correlation func tion for the important class of stationary Gaussian Markov processes is thus a simple exponentially decreasing function of time.
Ergodic theory, H-theorems, recurrence problems (I) D.ter HAAR (University of Oxford, Oxford)
27.1 In the absence of external forces, the number of atoms of a monatomic gas whose representative points lie in a volume element dudvdw d 3c of velocity space isf(u,v, w)dudvdw f(c)d 3cl, where u, V, ware the Cartesian components of the velocity c of an atom. Let A(B) be the number of atoms which leave (enter) this volume element per unit time due to collisions. Assuming (i) that the atomic collisions are equivalent to collisions between elastic spheres, and (in that the expression f(c)d 3c is correct for any point in configuration space, obtain expressions for A and B and show that
af
-d 3c
B-A
at
.
(27.1.1 )
The assumption (ii) is the assumption of molecular chaos or the so-called Stosszahlansa tz. Solution
Consider the collision between two identical atoms. Let Cl> C2, C'l> c~ be the velocities of the two atoms before and after the collision, respectively, and let w be the centre-of-mass velocity,
w
HCI +C2) .
(27.1.2)
The velocities c~ and c~ are not completely determined by CI and C2, since we only have four equations, cr +c~ = C'!2 + C~2 ") C! +C2 = C'I +c~
(conservation of energy),
(27.1.3)
(conservation of momentum)
(27.1.4)
for six components. (I) For general references see: R.Jancel, The Foundations of Classical and Quantum Statistical Mechanics (Pergamon Press, Oxford), 1969; D.ter Haar, Elements of Statistical Mechanics (Holt, Rinehart, and Winston, New York), 1954, especially Appendix I; D.ter Haar, Rev.Mod. Phys. , 27, 289 (1955); I.E.Farquhar, Ergodic Theory in Statistical Mechanics (Interscience, New York), 1964.
£,1
531
532
Chapter 27
27.1
l
Let w be the unit vector in the direction of the line of centres, that is,
the vector connecting the centre of atom 1 with the centre of atom 2
:c,ee Figures 27.1.1 and 27.1.2); we have then
c, -C't w ICt -c'd . (27.1.5)
27.1
Ergodic theory, H-theorems, recurrence
533
There is a one-to-one correspondence between the original and the inverse collision, since c" cz, and w completely determine C'I, c;, and w'. ~ e rel I
,..JII' U 2 ",'"
""
,.'"
,."
,."
-",,"w
u;
'" Figure 27.1.3.
u, Figure 27.1. L
Figure 27.1.2.
For B we now get
In the centre-of-mass system the description is much simpler. If u"
Uz, u;, u~ are the velocities in the centre-of-mass system, we have Uj
Ut+UZ ui+u~
= Cj -W,
,
Ui
Cj-W
U'I+U~
= 0, U;2+U;Z, U,
w
,
B (27.1.6) 1.8)
-u,
1.9)
Iul-u'll
ff(C,,)d3C~f(C~)d3c~ fal'2'
-+
tZd2w' ,
1.12)
where the prime on the integration sign indicates that the integration over c'" c~, and w' is such that one of the final velocities falls into the previously fixed volume element d3cI' Let J be the Jacobian of the transformation of C'l' c;, w' to c" cz, w, au; av; aw~ aw; aw;· aUl au,····· au, aUl au,
(27.1
,
Figure 27.1.4.
Simple considerations now show (see Figure 27.1.3) that A
= f(Cdd3CJf(cz)d3C2fal2-+
l'z,d zw,
(27.1.10) J =
where we consider in Equation (27.1.1) a value CI of the velocity, and where a 12'" t' z' is given by alZ-+
1'2'
= DZCrelCOSO { =0
(cosO> 0) , (cosO
< 0) .
ac C'l> c~, w') acc cz,
---'-...::..:..-=..;- w) t,
aU't awz
, (27.1.13) aW 2
(27.1.11) au', aW2
Here D is the diameter of the elastic sphere, Crel C1 C2, and 0 is the angle between wand crel . [Note that in writing down Equation (27.1.10) we have tacitly used the Stosszahlansatz.] To find B we must consider inverse collisions, which are collisions for which the velocities after the collision are c, and C2 and before the collision C'l and c~. The line of centres will now be -w (see Figure 27.1.4 which gives the inverse collision in the centre-of-mass system).
aw; . aW 2
where u, v, ware the Cartesian components of c and where WI and W2 are two quantities which determine the unit vector w (we could, for instance, use the polar angles {} and 'P). In the simple case considered here, where we have two identical atoms in collision, we easily find J = 1,
&,1
T 534
Chapter 27
27.1
a result which is generally true and is a consequence of Liouville's theorem (see Problem As a I' 2' -+ 12 and a 12 -+ l' 2' depend only on C reI and (), which are the same for the original and the inverse collision, quite clearly we have al'2'-+12 == al2-+I'2', and changing in expression (27.1.12) from C'l , c~, Wi to C\, c 2 , W, we obtain B
f
d3clf f(C'I)f(c~)d3C2 ad 2w ,
where.r.
=- fCc;),!;
at =- fCc;).
(27.1.15)
27.2 Show that with the assumptions of Problem 27.1 the Maxwell distribution of Problem 3.1 is left unchanged by collisions. that this result is necessary for the Maxwell distribution to be an equilibrium Solution
If f is the Maxwell distribution, we have (" temperature) from Problem 3.1.
( 2: )'h exp(-~~mc2) ,
f =n ~
where T is the (27.2.1)
(27.2.2)
Hence from Equation (27.1.15)
of
at ==
0,
535
Solution
From Equations (27.3.1) and (27.1.15) we have dB = fOf -(lnf+ 1)d3 c
at
or d 3cd 3c I d 2 w .
dB = f(!'f: dt
(27.3.3)
If we interchange c and C I and bear in mind that this leaves crel and thus a unchanged, we can also write
: = fC!'f; -1i'1)(1nfl + l)ad 3cd 3c 1 d2w.
(27.3.4)
If we now transform from c, CI, W to c', C'I' w' and use the fact that the jacobian of this transformation is equal to 1, we can write instead of Equations (27.3.3) and (27.3.4)
:
= f Ufl -!'fD Onf'+ l)ad 3cd 3c\d 2 w,
(27.3.5)
:
=
fUfl -fU;)(lnf; + l)ad 3cd 3c l d 2 w.
(27.3.6)
or
Taking the average of Equations (27.3.3) to (27.3.6) we have dB d3cd3cl d 2w . )In dt
(27.3.7)
As for p and q non-negative,
and by virtue of Equation (27.1.3)
fti2 == fU; .
Ergodic theory, H·theorems, recurrence problems
C27.1.14)
where we have dropped the index of al'2' -+ 12' Combining Equations (27.1.1), (27.1.10), and (27.1.11), we finally have
f2 -fU;)a d3c :zd 2w ,
27.4
(27.2.3)
p{> 0 (p'* q), q = 0 (p = q),
(p-q)ln
and as a is a positive quantity (compare Problem 27 .1), we see that dB dt < 0 . (27.3.8)
which proves that the Maxwell distribution is left unchanged by collisions. 27.3 If H == fflnfd3c,
prove that for a gas of elastic spheres dB -~O dt ...... and discuss this result. [H is Boltzmann's H function.]
(27.3.1)
One can prove that H is a bounded function and the result (27.3.8) thus means that H will decrease until the distribution function satisfies Equation (27.2.3). Equation (27.2.3) is thus not only a sufficient. but also a necessary condition for f to be an equilibrium distribution. 27.4 The equilibrium distribution function of a gas of non-interacting particles is Maxwellian. Show that the entropy of such a system is
(27.3.2)
S
-kH+K,
where K is some additive constant.
C27.4.1)
Chapter 27
536
27.4
Solution
For a gas of non-interacting particles, the distribution function f is the Maxwell function given by Equation (27.2.1), whence Inf = Inn+~In{3-~{3mc2+const.
(27.4.2)
If v is the volume per unit mass,
27.6
Ergodic theory, H-theorems, recurrence problems
537
27.6 Defining the equilibrium distribution as the one which makes W(Nk ) of Problem 27.5 a maximum for given values of N and of the total energy E, find the equilibrium distribution and find also a connection between the corresponding probability and Boltzmann's H function for the case of a perfect classical gas. [Hint: see Problem 2.11.] Solution
v
(27.4.3)
nm
we find from Equations (27.4.2) and (27.3.1) H - = -Inv In{3 n
From
(27.5.1) we have In W
= In N! + L [N; In Zj -lnN;!] + const.
(27.6.1 )
i
(27.4.4)
If the N; are sufficiently large that we can use the Stirling formula Inx!
where the bar indicates an average,
= xlnx-x,
(27.6.2)
We have G
fGfd3c.
(27.4.5)
N· In W = ~ N j In NZ. + const I
Evaluating mc 2 we find from Equation (27.4.4) H
-n = -lnv+Vn{3+const
or
H
n = -lnv -i In T+ const.
where we have used the fact that the total number of particles is fixed, (27.4.6)
N=L (27.4.7)
(27.4.8)
where the constant D involves the chemical constant. Thus, apart from a possible addition constant, Sv -kH.
E = 2.N;€i ,
oN
=a=
a=
oE
t oN; (In ~i + 1) ,
"oN L /,
(27.6.7)
L€joN; .
(27.6.8)
i
;
toN;(-In:ij +a-{3ei) = 0, = CN!
where C is a normalisation constant.
n i
(27.6.6)
Taking (27.6.6)+(a+ 1)(27.6.7)-{3(27.6.8) we get
Solution
W(Nk )
(27.6.5)
where €i is the energy of an atom in the cell Zj. Looking for a maximum of In W under the conditions (27.6.4) and (27.6.5), we use the method of Lagrangian multipliers. For a variation oN; in the N; we have •
oln W = a = 27.5 Divide velocity space into non-overlapping cells of size Zb each cell being a volume element d3 c. Let the representative points of the N atoms in a gas be distributed over the cells in such a way that there are Nk points in the kth cell. Let W(Nk ) be the probability for a given Nk distribution, defined as the fraction of all possible arrangements for which this Nk distribution is realised. Find an expression for W(Nk ) assuming that the a priori probability for a representative point to into the kth cell will be proportional to its size. rHin t: see Problem 2.11.]
(27.6.4)
The other condition to be satisfied is that of a fixed total energy,
For the entropy Sv per unit volume of a perfect classical gas, we have (see Problem 1.23 with A = nk and g 1) Sv = nk(} In T+ Inv)+ D ,
(27.6.3)
I
(27.6.9)
whence (27.5.1 )
N;
NZ j exp(a-{3ei) ,
which is the Maxwell distribution.
(27.6.10)
538
Chapter 27
27.6
For the case of a perfect gas we have 3
d c,
(27.6.11 )
I 3 N = nf(c)d c ,
(27.6.12)
!mc 2
(27.6.13)
IV; Ej
=
.
From Equations (27.6.3), (27.6.10), (27.6.11), (27.6.12), we then get In W = - fi1nfd3c+const,
I
I
27.8
Ergodic theory, H·theorems, recurrence problems
function will consist of the four numbers fl, fz, f3, f4 which are the numbers of P molecules per unit area in the four possible directions. If we assume that the number IV;jt:.t of P molecules which during a time interval t:.t change direction from ito j per unit area is given by the Stosszahlansatz expression Nilt:.t = fiSiln , (27.7.1) where Sjl is the area of a parallelogram of length ct:.t on that edge ofone of the Q molecules which is in the -i,j quadrant (see Figure 27.7.1), prove that the fi will approach equilibrium values. As there is no preferential direction, we expect that the equilibrium distribution is given by the equation
klnW.
ffq = j!fq
(27.6.14)
27.7 Consider the following simplified model of a gas, due to the Ehrenfests. In the plane of the paper there may be a large number of point particles, N per unit area, which are called P molecules. They do not interact with each other, but they collide elastically with another set of entities which are called Q molecules; these are squares of edge length a, distributed at random over the plane, and fixed in position in such a way that their diagonals are exactly parallel to the x and y axes. Their average surface density is n, and it is assumed that their mean distance apart is large compared to a. Suppose that at a certain moment all the P molecules have velocities which are of the same absolute magnitude, c, and limited in direction to (1) the positive x axis, (2) the positive y axis, (3) the negative x axis, and the negative y axis (see Figure 27.7.1). As a result the distribution y
t
f;q
=
f: q = ~N.
(27.7.2)
From Equation (27.7.1) it follows that we have (Ail = S/in/ t:.t = A) dfl
dt
-N12-N14+Nzl+N41 = A[fz +f4
2fd
dfz = AU1 +f3 -2fz]
dt
df3
dt
A[fz+f4- 2f 3 ]
df4
= AUI + f3 - 2f4]
dt
with the solutions
fi(t) =
fl'q +
-fr]e- 2At •
3 t2 I
27.8 Consider the Lorentz model of a metal in which the electrons, which are supposed to be non-interacting, are scattered in such a way that the number per unit volume Nw.w dtd 2 wd 2 w' which during a time interval dt change their directions from within an element of solid angle d2 w to within an element of solid angle d 2 w' is given by the equation
~4
d 2 d2 ' 2 2 ,_ ...-!:::!.~ Nw,w,dtd wd w - Af({},i()) 41f 41f dt,
......--
-
... x Figure 27.7.1.
539
Solution
and hence, using also Equation (27.4.1),
-kH= Sv
1
(27.8, I)
where f({}, i())sin {}d{}di()/41f f(w)d zw/41f] is the number of electrons per unit volume with velocities in a direction within the solid angle d 2 w. In this model all electrons are moving with the same speed. Find the equilibrium distribution of the electrons and show that the scattering mechanism will produce an exponential approach to this equilibrium distribution.
Chapter 27
540
27.8
Solution
As there is no preferential direction, we expect that the equilibrium distribution is an isotropic one:
27.11
Ergodic theory, H-theorems, recurrence problems
The difference oN arises from the flow of the points in phase space and, by analogy with ordinary flow of a gas or liquid, we find
q
j
j"q(w)
N,
(27.8.2)
where N is the number of electrons per unit volume. From Equation (27.8.1) we find df(w)
A'ff(WI)d2W'-f'(
L
dt
41T
w
N-[N
~=A(N-f)
J
o(w)]e- At
'
aD'
at+~
This equation shows that the reaction in flow and is a form of Liouville's theorem.
r
r
space which
o.
(27.9.1)
aH . aH 7 2) a' qj a ,i 1,2, ... , sN , (2 .9. qi Pi where to simplify matters we have numbered the p's and q's continuously. If dD / dt is the convected rate of change in D, that is, the rate of change when we follow a representative point along its orbit in r space and aD/at the local rate of change, we have
N
"laD.
aD'J
(27.9.3)
Let N be the number of representative points in dn at time t so that = Ddn. At t+ ot we have N+oN = (D+
dt
~~ ot) dn .
(27.9.4)
(27.9.7)
O.
27.10 Let v be the 2sN-dimensional 'velocity' vector in r space with components qb ...,qsN,p" ... ,psN,and let a be the 'projection' ofa cell on in r space on a plane at right angles to v. Find an expression for the mean time T spent by a representative point inside on. Solution T
is clearly given by the equation I V
+ f Lapl i + aqiqj .
(27.9.6)
dD
T=
. Pi
aD
(27.9.5)
whence
space is an incompressible
If a 80int in r space is determined by the values .of the 2s/l! variables q~'. pi (k = 1, ... , s; j = 1, ... , N) where the q~) and p~) are the generalised coordinates and momenta of the jth particle, the motion in r space is governed by the Hamiltonian equations of motion
dD
,,(aD. aD.) aqi qi + aplj ,
at = f
The time
Solution
dt =
aD
(27.8.4)
fIdp;dqi) of
(aD. aD.) apli+aqjqi
dnot.
we find from Equations (27.9.4) and (27.9.3) that
i
dt
-+-=0. aqi api .
(27.8.3)
represent an ensemble of such systems, prove that dD
Pi
a ) (aD. aD.)] f,,[-D (aaqj + api + aqjqi+ apli
aqj api
27.9 A classical system contains N particles, each of s degrees of freedom so that its development can be described by a (representative) point in the 2sN-dimensional phase space (r space) with coordinates PI, ... , PsN, q" ... , qsN' If D(PI, ... , qsN, t)dn is the number of represen
tative points in a volume element dn
oN = -
As it follows from Equation (27.9.2) that
whence j(w)
541
'
(27.10.1)
where v [qi +'''+PiN]'h and [the mean segment of the orbit inside on. From the definition of a it follows that on 1= (27.10.2)
a
and hence on av
T=
(27.10.3)
27.11 Consider a system with a fixed energy E. We shall assume that the energy surface E(p, q) = constant is an invariant indecomposable region n of r space, which means that (n for any point P also the total orbit through P lies in nand (li) it cannot be divided into two parts n' and nil which are separately invariant. We shall not prove, but only state, the plausible fact that indecomposability of n is equivalent to the transitivity of the motion, which means that the orbit from any point P in n will come arbitrarily close to any other poin t P' in n. Not only will any point come arbitrarily close to any other point in n, but if we look at a cell on in r space, the point P will traverse this cell over and over
Chapter 27
542
27.11
again, provided the total volume of n is finite as we shall assume to be the case (Poincare's recurrence fheorem). Give an expression for the mean recurrence time T and estimate this quantity for the case of a gas of 10 18 atoms of mass 1 g in a volume of 1 cm 3 with an average velocity of 104 cm S-I where the cell on has dimensions 10- 7 cm for its position coordinates and 102 g cm S-I for its momentum coordinates.
27.13
Ergodic theory. H·theorems, recurrence problems
Solution
In Equation (27.12.1) we write
f
with
n
t
T=
n
~ (l0-7)3N(l02)3N,
and
n Hence
~
T ~ (3
E )3N/2
l3N ( m X
10 18 )(1'5
~
V
~
104 (3N)"",
(3N)3N/2(l04)3N.
x 10 18) S ~ 10(10
19
)
(27.12.4)
(27.12.5)
.
n/on
L f(tn)Ot n
j
Ni
L= 0 r=L0 f(Pj)otjr.
(27.12.6)
where r numbers the recurrences of passage of P through on; and i numbers the cells, and where N; is the number of recurrences in the time interval (0, t). We have thus n/6n Ot. (27.12.7) f* = lim lim f(~) tiN:.
To estimate T for the case considered, we note that
o
= LOtn
f(tn)Ot n
Choosing for the Ot n the transit times of P through the cells on; and using Poincare's theorem that every cell will be traversed several times, we have
(27.11.1)
av
= lim L
t f(t)dt
o
Solution
The volume swept out per unit time by the orbits passing through on is clearly 01) and by Liouville's theorem this will be the volume swept out by those orbits at any time. Moreover, if n is indecomposable, all orbits will pass through on and they will fill n completely. This means, there fore, that the mean recurrence time will be
543
t .. ""
We clearly have lim Ot;
years.
.L
on .. 0 I = 0
I
mean life time of P in onj, which according
t .. ""
on/ov. We also have lim tiN; = mean t .. "" recurrence time, which according to Equation (27.11.1) is equal to n/ov. to Equation (27.10.3) is equal to
27.12 Using the assumptions of the previous problems, we can now attack a simplified proof of the ergodic theorem, that is, the equality of ensemble and time averages (2). Consider a phase function f(P) which is a function of the representative point P in r space. Its time average f* which is the average which is measured physically-is given by the equation 1 f* = lim fIP(r)] dr, (27.12.1)
t ft
t .. ""
0
while it is usually only possible to evaluate the phase average f given by
f
~L f(p)dn.
(27.12.2)
The ergodic theorem now states that
f* =f.
(27.12.3)
By dividing n into cells on; and the integral in Equation (27.12.1) into time intervals corresponding to the passage of P through the different cells prove the theorem (27.12.3).
Hence we have
f*
on; lim "L. f(Pj)n
6n .. 0
i
See KWergeland,Acta Chem Scand., 12,1117,1958.
f
n
f(p)dn
= f.
27.13 To study the approach to equilibrium and the recurrence problem, let us again consider the Lorentz model of Problem 27.8. We have already seen that the system in that model has an equilibrium distribution (27.8.2). We now divide phase space into 2m + 1 cells of equal volume, which we number from -m to +m. Each cell corresponds to an element of solid angle ow with 411' ow = 2m+ 1 (27.13.1) The distribution function is now a function of a discrete argument, -m, ..., +m), and its equilibrium value is
fv(v
I'eq (2)
= n1
Jv
-
_N __
2m+ 1 .
(27.13.2)
544
Chapter 27
27.13
and assuming Ifv - f:;q 1/fv ~ 1,
The 2m + I fv satisfy the condition +m
2: fv(t) V=-m
== N.
(27.13.3)
To measure the departure from equilibrium we introduce a function A given by the equation
if;(fv)
2m+ I)m [ (f, -rq)2] =(~ (2m+ l)Yl exp -(2m+ 1)2: v 2N . (27.14.4)
2:
(fv_t;q)2 .
Assuming that Ifv - t;q I ~ f:;q, find a relation between If and A, where we define If as If =
A(p) =
(27.13.4)
v=-m
2: fv 1nfv .
v
We then find for the function A(p)
+m
A
545
Ergodic theory, H·theorems, recurrence problems
27.16
f...J
if;(fv)exp(-ipA)
JI
df-vdtv ,
(27.14.5)
where the integral is 2m-fold since the fv satisfy condition (27.13.3). The integration is straightforward, but tedious, and the result is Np
(27.13.5)
A(p)
= (1-2i 2m+ 1
)-m
(27.14.6)
Solution
If
= 2:fv Infv == v
A
Substituting Equation (27.14.6) into Equation (27.14.2) and evaluating the integral by contour integration, we find finally 2m+l)m Am-l [(2m+l)A] w(A) = ( (m_l)!ex p 2N ' (27.14.7)
Ifeq+ 2N '
where Ifeq ==
2:v fv
eq
----m-
Inf:q .
which shoWS that w(A) decreases steeply with increasing A, provided N is large.
27.14 Use the result from probability theory that if
8,2, 8,3 are respectively the electric field components along the x, y, z directions. The rate of entropy production per unit volume 0 is given by 3·
T
L Rstasat -2
s, t
(28.4.5)
.
(28.4.7)
with Rand Q given by Equations (28.4.2) and (28.4.5). Alternatively, Equation (28.4.7) can be derived by equating to zero o(Oj -20 e )/oas for I < s < T, with OJ -20e given by Equation (28.4.6). In the steady state, the rate of entropy production is given by Equation (28.4.4), where as is obtained from Equation (28.4.7). Thus, in the steady state,
,
p'~IMpqJpJq = pJ=IMpqctlasr~S»)(JlatrJt»)
Qs
t= 1
T
o=
555
L= Qsas '
(28.4.6)
To =
P
I
Now, it readily follows from Equations (28.4.2) and (28.4.3) that R st is a symmetric positive definite matrix, since Mpq has these properties. The expression (28.4.6) is therefore of the same form as the expression for OJ - 20 e in terms of the original fluxes Jp , if we identify a with J, R with M, and Q with X. Thus, by employing the same proof as used in the last
L
Jp 8,p ,
(28.5.1)
I
where T is the absolute temperature, and in the steady state 8, and J are connected by the general relationship 3
&p =
L rpqJq , q = I
(28.5.2)
r ·it
28.6
28.5
Chapter 28
556
Ta = 3·40
&~m r = ( :
(
.
Obtain expressions for aj and a e in terms of J1> J 2 , J3 • Hence, if an approximate solution of Equation (28.5 .2) is given by Jp = O!rp ' where
r
~(~)
and O! is an arbitrary constant, use the method of the last problem to estimate a value for Ta in the steady state. Compare this with the true value and comment. Solution
aj
=
I
T
1 a e = -T
L3
I
p,q= 1 3
L p=l
rpqJpJq = -T(3.!f+3J?+3.!1+ 2J1J 2 + 2JZJ 3 + 2J3 J 1 )
I
tnpJp ::;: -T(il + 2J2 + 3J3 )
(28.5.4)
and, as expected, the estimated value (28.5.3) is less than the true value. The numerical error in the approximate steady state current is quite large since the latter equals
}
3
557
On making use of Equation (28.5.3), we then find
where rpq (l .,;;; p, q .,;;; 3) is the resistance tensor. In a particular case, with a given set of units, the electric field vector tn is
and the resistance matrix takes the form 3 I
Variational principles and minimum entropy production
•
Using the given trial solution, we let J I = O!, J2 20!, J 3 = 30!, and obtainT(aj 2a e )::;: 640!2_280!. To find the minimum value ofaj-2a e , we equate o(aj-2a e )/oO! to zero and obtain 12801-28=0, whence O! = ;,;. With this value of O!,
T(2a e -aj) = Ta = 3·06.
(28.5.3)
To obtain the true value of Ta we must first solve Equation (28.5.2) for Jp , and this gives
J~{:)
0'22) (-0'1) 0·44 while the true solution is 0·4. 0·66
0·9
Nevertheless the error in Ta is much less, being about 10%, and this agrees with the discussion at the end of the solution to Problem 28.4. ELECTRON FLOW PROBLEMS
28.6 We now wish to consider the formulation of variational principles concerning entropy production, based on a description of the system at an atomic level, rather than at the macroscopic level employed hitherto. To be quite definite we shall confine ourselves to particles, such as electrons, which obey Fermi-Dirac statistics, but similar results apply to particles obeying both Maxwell and Bose-Einstein statistics. In a metal each electron is characterised by a three-dimensional wave vector k and the number of electrons of given spin with wave number k is denoted by an occupation number f(k). In thermal equilibrium at temperature T, f(k) is equal to the Fermi-Dirac distribution function [derived in the solutions to Problem 3.12(a)], 1 (28.6.1) Jl) + 1 j'O(k) = exp "T
(E -
where E(k) is the energy of an electron with wave number k, Jl is a constant-the Fermi potential, and" is Boltzmann's constant. In the steady state, however, where a net flow of electrons occurs, f(k) differs from the equilibrium value j'O(k). Now it is known that the entropy S of an assembly of electrons is given by S::;: -"jU{k)ln.t\k)+[I-f(k)]ln[l-f(k)]}dk
(28.6.2)
where the integration is taken over the whole of k space (3). For f(k) close to the Fermi-Dirac distribution we may then let f(k) = fO(k) - ,,-lj'O(l - j'O)r{>(k) ,
(28.6.3)
where r{>(k), which measures the deviation of ffromj'O, is small. (3) P.T.Landsberg, Thermodynamics with Quantum Statistical Applications (lnterscience, New York), 1961, p.233.
r
Chapter 28
558
Making use of Equations (28.6.1), (28.6.2), and (28.6.3) show that the rate of entropy production (J (= as/at) is given by
f
af If(E-p)-dk af
(k) -rt>(k')]rt>(k) dk dk'
(28.7.9)
since it may be assumed that the second term in Equation (28.6.7) gives no contribution to as/at) coli, On exchanging k and k' in Equati0J1 (28.7.9) we obtain
~~)COll =
If
L(k, k')[rt>(k')-rt>(k)]rt>(k')dkdk'
561
The analogue of the positive definite character of Mpq and its symmetry are the relations (28.7.3) now satisfied by L(k, k'), so that for all O(k) and rt>(k), fIL(k, k')(O(k)-O(k')-rt>(k)+rt>(k')Fdkdk'
~ 0.]
Consider a general function O(k) and a function rt>(k) satisfying Equation (28.7.4). Then since L(k, k') ~ 0 for all k, k',
ff
tfL(k, k')[rt>(k)-rt>(k')j2dkdk' .
28.8 It was shown in Equation (28.7.5) that -u conv calculated for electron flow is identical with the form for U e as given by Equation (28.5.1) for the case of macroscopic variables. This suggests that just as for macroscopic variables it was possible to formulate variational principles for the relationship between forces and fluxes, based on a stationary value of the entropy production, so similarly it may be possible to obtain the Boltzmann flow equation (28.7.4) by a similar device with rt>(k) as the analogue of Jp • Now for the steady state, it follows from Equation (28.7.4) that -u conv = Ucoll and therefore by analogy with the macro scopic approach we suggest the formulation of variational principles by using -U conv instead of Ue and U coli instead of Uj. Prove that the Boltzmann Equation (28.7.4) is given by minimising U coil + 2u cony with respect to varying rt>(k), or by maximising Ueoll subject to it equalling -u conv , or by minimising ucon/u2onv. [Hint: Follow the same approach as in Problem 28.2 using the continuous variable k instead of the discrete variable p, and replacing summations by integrations.
~0
L(k, k')[O(k) -O(k') -rt>(k) + rt>(k')j2 dk dk'
~
and thus
f
f
L(k, k')[O(k) -O(k')j2dkdk' - 2 L(k, k')[O(k) -O(k')][rt>(k) -rt>(k')] dkdk'
+fL(k, k')[rt>(k)-rt>(k'Wdkdk'
~ O.
(28.8.1)
Now
f
L(k, k')[O(k)-O(k')][rt>(k) -rt>(k')]dkdk' =
(28.7.10)
L(k', k). Finally by adding Equations (28.7.9) and since L(k, k') (28.7.10) we find
~St) coil =
••
Variational principles and minimum entropy production
Solution
From Equations (28.6.7) and (28.7.2) we have Ueoll =
28.8
(28.7.7)
ev(t- to) dk ,
making use of Equation (28.6.3). The integral in expression (28.7.7) is just the total charge transported per unit area per second and is thus equal to the current density J. We obtain the required result
as) at cony
"
f
L(k, k')O(k)[rt>(k) -rt>(k')] dk dk' -
J
L(k, k')O(k')[rt>(k) -rt>(k')]dkdk'
(28.8.2) and exchanging k and k' in the second term, bearing in mind that this leaves L unchanged, we obtain the right hand side of Equation (28.8.2) as 2fO(k){fL(k, k')[rt>(k) -rt>(k')]dk' }dk .
On substituting from Equation (28.7.4) this becomes
ajO f aE O(k)dk .
2e8.. v
We substitute this expression in the second term of the ihequality (28.8.1) and obtain
f
L(k, k')[O(k)-O(k')j2dkdk' -4e&. Iv
~ =
-I I
~:O(k)dk
L(k, k')[rt>(k)-rt>(k')]2dkdk'
L(k, k')[rt>(k)-rt>(k')j2dkdk'-4e&'
f ~rt>(k)dk v
-n . ... . I
562
Chapter 28
28.8
on applying the above transformation to the case e =
