CONI'ENTS
Preface ••
.
... . . . . . . . . . .. ... .
i
.. .. .
CRA.PI'ER l:
Introduction to Statistical Methods.
CRA.PI'ER 2:
Statistical Description of Systems of Particles ••
ll
CRA.PI'ER 3:
Statistical Thermodynamics
16
CRA.PI'ER 4:
Macroscopic Parameters and their Measurement
18
CRA.PI'ER 5:
Simple Applications of Macroscopic Thermodynamics.
20
CRA.PI'ER 6:
:Ba.sic Methods and Results of Statistical Mechanics.
32
CRA.PI'ER 7:
Silllple Applications of Statistical Mechanics • • •
41
CRA.PI'ER 8:
Equilibrium between Phases and Chemical Species ••
54
CHAPrER 9:
Quantum Statistics of Ideal Gases • • • • • •
64
CHAPrER l0: Systems of Interacting Particles
1
84
•••••
CRAPI'ER 11: Magnetism and Low Temperatures ••
90
CHAPI'ER l2: Elementary Theory of Transport Processes.
93
CRAPI'ER 13: Transport Theory using the Rel.axation-Time Approximation
99
®.Pl'ER l4: Near-Exact Formulation of Transport Theory
•••••
CRAPI'ER 15: Irreversible Processes and Fluctuations • • • • • • •
l07
117
PREFACE
This manual contains solutions to the problems in Fundamentals of Statistical and Thermal
Physics, by F. Reif.
The problems have been solved using only the ideas explicitly presented
in this text and in the way a student encountering this material for the first tillle would
probab]_y approach them.
Certain topics which have implications far beyond those called for in
the statement of the problems are not developed further here.
numerous treatments of these subjects.
The reader can refer to the
Except when new symbols are defined, the notation
conforms to that of the text.
It is a pleasure to thar,k Dr. Reif for the help and encouragement be freely gave as this
work was progressing, but he has not read all of this material and is in no way responsible for
its shortcomings.
Sincere thanks are also due to Miss Beverly West for patiently typing the
entire manuscript.
I would great]_y appreciate your calling errors or omissions to
my
R. F, Knacke
attention.
CHAP.PER 1
Introduction to Statistical Methods
1.1
There are 6-6-6 = 216 ways to roll three dice.
The throws giving a sum less than or equal to 6
are
Throw
No. of
Permutatiom
1,1,1
1,1,2
1,1,3
1,1,4
1,2,2
1,2,3
2,2,2
l
3
3
3
3
6
l
Since there are a total of 20 permutations, the probability is
~~G
=
-k
1.2
(a)
Probability of obtaining one ace= (probability of an ace for one of the dice) x (probabil1
1 5
ity that the other dice do not shov an ace) x (number of permutations)= (b)(l - '6')
5 5
=
(5)
(b)
=
.4o2 •
The probability of obtaining at least one ace is one minus the probability of obtaining
none, or
(c)
6!
(5!1!)
56
1 - (;-) = .667
By the same reasoning as in (a) ve have
l 2 2 4 6!
(b) ('6') 5!2! = .04o
1.3
The probability of a particular sequence of digits such that five are greater than 5 and five are
less than 5 is (
1 5 1 5
2) (2).
Then multip:cying by the number of permutations gives the probability
irrespective of order.
1.4
(a)
To return to the origin, the drunk must take the same number of steps to the left as to the
right.
Thus the probability is
N
W (-2)
where
(b)
N.I
l N
= --------- (-2)
C!) ! <J) !
N is even.
The drunk cannot return to the lamp post in an odd number of steps.
1.5
th
(Probability of being shot on the N trial)= (probability of surviving N-1 trials) x
th
5 N-l l
(probability of shooting onesel.f on the N trial)= (b)
(b).
(b)
(c)
6
1.6
m= m3
= O since these are odd moments.
r
Using n
2
m = (2n - N)
and
4
m
we find m2 = N and
2
=
cl
r
= (p clp) (p+q)
2 - 4nN
-
4n
N
~2
+ W--
=
";f+ = 3lf -
2N.
1.7
The probability of n successes out of N trials is given by the sum
2
w(n)
=
2
E E
'W
m
i=l j=l
with the restriction that the sum is taken only over terms involving w1 n times.
2
Then W(n) = E w.
. i=l i
2
E v. If ve sum over all i ••• m, each sum contributes
m=l m
(vi-1~2 ) and
W' (n) = (w + w
1
2
W' (n) =
N
E
l
N!
n
.....,...,,.._.__..,..
v
n=O n!(N-n)!
1
w
N-n
2
by the binomial theorem •
Applying the restriction that w must occur n times we have
1
N!
n
W(n) = n!(N-n)! "W'l w2
'We consider the relative motion of the two drunks.
N-n
With each sillr..i.ltaneous step, they have a
probability of 1/4 of decreasing their separation, 1/4 of increasing it,and 1/ 2 of maintaining
it O'J taking steps in the same direction.
Let the number of times each case occurs be n , n ,
1
2
and n , respectively.
steps is
3
2
The drunks meet if n
1
= n •
2
Then the probability that they meet after N steps irrespective of
the number of steps
vhere we have inserted a parameter x which cancels if ~=n •
We then perfonn the unrestricted
2
sum over n n n and choose the term in which x cancels. By the binomial expansion,
1 2 3
1
1
1 N
1 2N 1/2
-1/ 2 2N
(4
pr=
x +
4X
2)
+
2N !
~ n! 2N-
(2)
P' =
+ x
(x
2N 2N
l
Expansion yields
(2)
=
)
1/ 2 n
-1/ 2 2U-n
) (x
)
! (x
Since x must cancel we choose the terms where n = 2N-n or n = N
Thus
1.9
(a)
1n (1-p)N-n ~ -p (N-n) ~ -Np
(b)
N!
(N-n)! = N(N-1) • • • (N - n+l) ~ 1f
(c)
N!
n (
)N-n
n -Np "'n -A
W(n) = n!(N-n)! P 1-p
~ n! p e
= n! e
n
<<
thus (1-p )N-n
N
if n << N
r
1.10
"'n!n
0)
(a)
E
n=O
(b)
(c)
- =
ll
=
=
e -t. e"' = 1
n ->..
"L., n >.. eI
-"
n.
n=v
n2 = n~ n~n
-2 -2
(.6n) = n
e ->-. =
e ->-. (~) 2 L.
~~
...!;)
- n- = >-.
1.11
(a)
The mean number of misprints per page is l.
>-.n ->-. e -l
= -,
e
= n!
n.
Thus
W(n)
and
W(O) =
e
-1
=
3
.37
=
>-.2 + >-.
~
e -Np
{b)
P
2
=1
-
-1
2: ~ = .08
n:::() n.
1.12
(a)
Dividing the time interval t into small intervals ~t ve have again the binomial distribution
for n successes in N = t/~t trials
N!
n (
)N-n
W( n ) = n., (N -n ) •, P
1-p
In the lilllit ~t ~
o,
-,.._n "W(n} ~ n! e as in problem 1.9 vhere )... = ~ , the mean number of disintegra-
tions in the interval of time.
(b) W(n)
4n -4
n
W(n)
=-,
n. e
0
1
2
. 019
.(176
.148
3
.203
4
.203
5
.158
6
7
8
.105
.()51
.003
1.13
2
We divide the plate into areas of size b.
Since b
2
is much less than the area of the plate,
the probability of an atom hitting a particular element is much less than one.
Clearzy n << N
so we may use the Poisson distribution
W(n)
6n
=,
n.
e
-6
1..1.4
We use the Gaussian approDJDation to the binomial distribution.
W(n)
=_
_;l.;;;..__
(21.5-2COr
2(400/4
e-
J;;¥f
= .013
J..1.5
The probability that a line is in use at e:n:y instant is 1/30.
We vant N lines such that the
probability that N+l or more lines are occupied is less than .01 vhen there are 2000 trials
during
the hour.
•01 = l. -
1
r.N ___,;;;..__
e
n=O~ cr
vhere
-n = 20CO
30 = 66.67,
4
The sum may be approxilllated by an integral
.99
J
N¥z°
J1
=
(n-n+½;& 1
.99 =
-
-1
(-n- 2)/cr
The lower limit may be replaced by
-eo
~
l
- 2
,Ji;'
2
exp [-(n-n) /2a ] dn
--
a
_ 2
e y 12 dy
afier a change of' integration variable.
with negligible error and the integral f'ound f'rcan the
tables of' error functions.
- l
N-n~
We f'ind
a
N
=
2. 33
= 66.67 + (2.33)(8.02) - .5 ::: 85
1.16
{a)
The probability f'or N molecules in Vis given by the binomial distribution
N !
W(N)
Vo
(c)
FfV
If
<< V,
0
o
V
ov- 0
0
(N-i)2
N-N
Vo
N=NP=N
(b)
V
(-) (1 - -)
= N!(No-N)!
0
Thus
N
V
V
NoV
-
=
0
V )
(1 - -V
0
i
l
:::-
N
~o
(d)
1.17
Since O <<; <<land since N0 is large, we may use a Gaussian distribution.
0
exp
5
[-(N-N}2/2
d]
dN
1.18
N
R
R•R =
N
= L.i
-:1.
!
S
.f..2
i
A
= tr_i -:i.
~-
r.
2
-:1.
£I £ ~-• ~. = m,2 + .f..2 L. I L. cos
2
+ .f..
i
(S. 's are unit vectors)
i
j
i
-J
-:i.
The second te:rm is O because the directions are randcm.
e.j
j
:i.
2
2
Hence R = Nt •
1.19
N
The total volt age i s V
= L.
v. ; hence the mean square i s
i
l.
-
N -
v2=L.v .
i
2
where v . = vp and v .
J.
J.
2
N
+r.
i
l.
N
f
L.V. V.
j
J
l.
= v 2 p.
P
Hence
1.20
= v2
/R =
( f f/R ) i [l+(l-p) /:N'p]
The antennas add in phase so that the total amplitude is E = NE, and since the intensity
t
is proportional to Et2, It= ~I(a)
(b)
To find the mean intensity, we must calculate the mean square amplitude.
may be
added vectorially
Et
2
--2N2
--
= -.,
~·:!<!. = ~ L.
-.,
i
where the -J.
S . 's are unit vectors.
§.i
--2
+~
L.i
L. §_1.
/ j
a
n.
J
ani
= O, the second term on the right is O and
_.!.
(An2)2
_
but
•Sj
-
The phases are random. so the second term is O and
1.21
Since
Since amplitudes
_
1
= N-2
.!.
(a 2)2
n
-~
= N.!.2 lv
as
.!.
(An2)2
=
As = Nas
:Equating these expressions we find N = 106 •
6
1.22
N
N
x=I'.s.,
•
J..
(a)
s. =t, x =L t =
but since
l
i
(b)
(x-x)
N
2
L,
=
(s.-t)
i
Since
=t - t = O,
(s.-t)
J..
Nt
i
2
N
Jj
i
J..
N
L. (s,-t)(sj-t)
+ L,
~
the second term is O and (x-x)
2
=L.N cr2 = Ncr2 •
i
1.23
N
(a)
x = I:i si
The mean step length is t.
( x-x)
(b)
2
N
=
L, ( s . - t)
i
2
+
N
I:
J..
j
I: ( s . - t )( s . - t ) '
Ji
t -
2
( s . -t) =
J..
J
J..
t -t
= 0
we note that the probability that the step length is between
,
b tot+ bis:~.
-t) 2
(s
Hence
i
N - - . - ----....
To find the dispersion (si-t)
sands+ ds in the range
N
= V.-: Nt
=
1
J.
t-b
i
(x-x)
-fn.b (s. -t )2 ds.
2
N
= L,
i
J..
3
2b
b2
b2
= -
Nb2
3
=
3
1.24
w(e) de
(a)
w(e)
(b)
=
de
2JT
de= 2JT a
2
si~
4n a
e
de
sine de
2
1.25
(a)
He find the probability that the proton is in
e
and
resulting field.
w(e) de= sin/ de
From problem 1.24
Since b = L
a3
(3 cos 2 e-1)
we
have !dbl = 6µ cos e sine
de
83
and
7
e
+ de and thus the probability for the
(b)
If the spin is anti-parallel to the field,
Then if either orientation is possible, we add the probabilities and renormalize
a3
,J;'
12/µ 2 -µ
W(b)db =
(
db
a%'
83~
12Jµ2-µ
a
12
3
a%
Ji
a3 ,.;-:;
+
'
12
Jµ
,} db
2 -i-µa %
db
Jµ2-i-µa%'
(b)
Q(k)
L
=
2µ/a 3
-µ/a3
-2µ/a
oo
ds e
iks
W( s)
=
=Im
~ ds
~
b
11
Q may be evaluated by contour integration.
b
eiks
-
s2+b2
For k > O, the integral. is eval.uated on the path
-kb
residue (ib) = ~ i
:Fork< O
kb
-residue (-ib) = ~
8
Q(k) = e kb
Thus Q(k) = e
<?(x)
-lklb
L"°
= 1
2rr
= -1
'IT
1"°
0
L
0
1
dk e - ikx QN(k) =2rr
1
dke -ikx e Nkb. +2rr
J"°
dke -ikx e -Nkb
0
dk e -Nkb
k
COS
X
This integral may be found from tables.
C?{x)
dx =
!
7T'
X
Nb dx
_2.. 2
2
+ Irb
c,>(x) does not become Gaussian because the moments sn diverge.
1.27
From equation
(1.10.6)
and
(? (x)
(1.10.8)
= 12rr
we see that
L"°
(1)
dk e - ikx Q (k) •• -~(k)
1
where
We expand Q.
J.
(k).,
-
J.
thus fran
1L
~
(1),
J.
00
v-(x) = 2rr
1
2
exp [i s.k - - (6s. )
Q. ( k ) =
dk
J.
N_
2
2
k ]
lN
exp(i{L. s.-x)k 2 L.i (.6.s.)
i J.
J.
This integral -may be evaluated by the methods of section l.ll.
@(x) =
N
where
µ =
L.
i
&'
2
2
k]
We find
2
exp [-(x-µ) 2/20]
s.
J.
1.28
The probability that the total displacement lies in the range! to!+
@(r) d\: =
J~- J
W(~1 ) W{2.2 ) ••• W(~) d\1
(d3r)
where the integral is evaluated with the restriction
N
r <
-
L. s. < r +
i=l
-i
-
9
dr
-
dE
after N steps is
We remove the restriction on the limits by introducing a delta f'unction
In three dimensions
(? (r)
or
= -l 3
(2rr)
Leo
.
d
3k e -i.k•r
- - QN(k)
-
1.29
For displacements of uniform length but of random direction W(~) = 5 (s-~), in spherical
lii7rs
coordinates..W(s) is a properly normalized probability density since
27T'
J
7T'
J
c:o
s 2 sin e ds de dq> 5 (s-t) = 47rf
2
1
0
0
!i,,r s
0
c:o
5 (4,ii--t)
=1
O
We find the Fourier transform Q(!).
Q(~
=JJJ a
3~ e~·~ w(~
_~17T' sin4rre de
-
t:Cll
=J2rr1Jc:o
0
e
i.k-t cos e
s
2
sin
e
ds de dcp 5 (s-t) ei.ks cos
0 0
47r
s
2
=
0
From
we have, for N = 3
(?(r) - _ l _
- (2rr)3
k2 sine dk de dcp e-i.kr cos e
102zr11c:o
0 0
The angular integrations are trivial and
To evaluate the integral, we use the identities
3
sin k-t Sin kr -_ -l s i n kr sin
. k·D cl-COS 2 k-f~
:'v
~)
k2
k2
2
lO
e
=
=
1
4k2
1
[cos k{r-l)-cos k{r+t][1-cos 2kt]
2 [3 cos k(r-t)-3 cos k(r+t) + cos k(r+-3-t)- cos k(r-3-t)]
8k
The integral of the first two terms in the last identity is
3
E
lco
cos k(r-t)-cos k(r+t) dk = 3
4
k2
o
1=
sin2 !(r+t)-sin2 !(r-t) dk
k2
o
On making the substitutions x = !<r+t) and y = ~(r-t), it follows that
lco Ir+tj
Ji:
3
2
0
2
sin x
x2
J
00
dx _
3
4
0
2
sin y dy _ 3rr ( Ir+tl _ Ir-tl)
Ir-tl
2
y2
-
Io
The third and fourth tenns may be evaluated in the same way.
; Jco
0
cos k{r+¼)-cos k(r-3-t) dk =
~
0
-g,..
~
(lr-3tl-lr+3-tl)
Ccmbining these results, we have
1
O<r<t
8rr .t,3
G>{r) =
1 3 (3-l-r)
16ir-f;r
t<r<3-t
r
0
>3-l
CHAPrER 2
Statistical Description
p =
~
2!
Systems~ Particles
op = ~ o E
Hatched areas are accessible to the particle.
L
X
2.2
Since p/ + P/ = 2mE, the radius in
p space is .,;;;;;;- •
11
(a)
The probability oi' displacement x is the probability that cp assumes the required value in
the expression x = A cos (wt+ cp).
-w-(cp)dcp = ~
P(x)dx = 2v(cp)
.and
dtp =
~
7I'
vhere the i'actor oi' 2 is introduced because tvo values oi' <p give the same x.
Since
ve have
av,
(b) We take the ratio oi' the occupied volume,
in phase space vb.en the oscillator is in the
p
range x to x + dx and E to E + 6E, to the total accessible
volume, v.
ov/2
E - P2 + m ,ix2
Since
- 2m
ve have
Ae = 7r
~
,./mw2
P(x)dx
Hence
=
(J.)
2
r-
= ~E ,
and v =
2:° 6E
•
6v = 2 6:p 6x
2rr 6E
V
(a)
For arr:, x, p ~ = 6E
m
P(x)dx
= ~ dx = ~
7r
.:here ve have used
(1).
p
dx
!
/
"V2mE -
From (1) and x = A cos (wt +
P(x)dx =
Thus
7r
2 2 2'
mW X
cp), it i'ollows that E =
½w2mA.2 •
dx
7I'
'22
VA~-x~
2.4
(a)
The number of wey-s that N spins can be arranged such that ~ are parallel. and n
parallel. to the fie1d is
and
J.2
2
are anti-
n.. = -1. (N - -E ) , n
ve have
Since
Thus
n(E) =
N
E
N!
<2 - 2µH)!
.l.
2
µH
2
= -1. {N + -E)
2
2µH
6E
(~
+ 2~)!
2
µH
Using Stirling's approximation, 1n n! = n 1n n-n, ve have
(b)
1n n(E)
= - .±(N _ ~)
2
µH
1n .±(1. _ l ) _ .±(N + ~) 1n .±(1. + l ) + 1n 6E
2
NµII
2
µII
2
NµH
2µII
N!
-
We expand 1n f(~) = 1n ~ ! (N-~) ! about the maximum, n1
( c)
where
From (1.)
ftni_) is eval.uated by noticing that the integral. of f(nJ.) over all n
1
must equal. the total. number
of permutations of spin directions,~-
n
1
is found :frcm the maximum condition
d 1n f(n )
1
1.
1.
..,
----
and
N-~-
~
Substituting these expressions into (2), we have
1.3
= N/2
1
and
Since
O(E)
E )
it f'ollws that
n..J. = -2 (N - -µH '
2
E
2
N <2µH) J
exp [ -
6E
2µH
2.5
{a)
If
cl. f
is an exact differential,
Since second deri-va.tives are equal,
{b)
If
71. f
lB -a.
is e:x:a.ct,
f = f(B) - f(A) = 0
if
A= B
2.6
(a)
o2f'
2
J,2 2
J,1 ay + 1 {x -2)
(b)
(c)
dx
=
2
r
eydX
l
2 2
J,2
10 J,2 2
J,2
7
3, J,1 {x -l)dx + 1 2dx =3 , 1 (x -x)dx + 1 x dx = 3
exact
2,2
J,1,1
(d)
{a)
71.f=
[
] (2,2)
x+Z
for all :paths
=l
(1,1)
X
The particle in a state vith energy E does vork, 71.w, vhen the length of the box is changed
to L
X
+ dLX •
This vork is done at the expense of the energy, i.e.,
it follows that Fx = -
(b)
o
dX oy I
not ex.act
~~
•
"d. W =
-dE.
Since
2rr2~2
X
The energy levels are given by E = - m
Fx
:P = L L
2
p = 4ms.2 ( ~ )
mV
y
l
z
= -
'E"'"'L
y
z
vhere
L 2
V
= LX
L
y LZ
X
I f Lx
= ~ = Lz,
by symmetry
2
2 2
nx = ny = nz,
and using the expression for Ewe have
14
cl. W =
F dL ,
X
X
2.8
~
h
~
~
2
J
l
~
+
pdv
=
pdv
p dv
C
2.9
From the definition of the Young's modulus, the force is
dF-YA
- L
F2
1
W= -
F
dL
L
L
2
2
F - dF = (F
- F )
YA
2AY
l
2
l
2.10
2.11
p=a
v-5/ 3, a may be found by evaluating this expression at a point on the curve.
6 d:ynes/m2 and 103cm3, a= 32.
10
For the adiabatic process,
8
AE = ~ - EA = 32V- 5/3 <N = -36 =-36oo joules
J,
1
Since energy is a state :function, this is the energy cba.Dge for all paths.
8
(a)
W = 321 <N = 22,400 joules
Q
(b)
= AE
= -36oo
+ 22, 4CO = ll,8CO joules
31
P = 32 - 7 (V-1)
8
W=
Q
(c)
+W
fi
f
[ 32 - 3 (V-1)]
=
ll,550 joules
= ll,550 - 3600 = 7 , 95C
W =
Q =
l
joules
8
(1) dV = 700 joules
700 - 36:::,0 = -29(X) joules
15
In units of
CHAPl'.ER 3
Statistical Thermodynamics
(a)
In equilibrium., the densities on both sides of the box are the same.
750
volume, the mean number of Ne molecules is
(b)
P
=
3
(4)
Thus, in the larger
and the mean number of' He molecules is
lCXX)
Cf)
100
=
io-
l8
5
3.2
From problem 2.4 (b)
l
) ln -(l
1 - -E) - -(N
l
l 1 +E-)
1n O(E) = - -(N
- -E
+ -E ) ln -(
2
µH
2
NµH
2
µH
2
:;·; µH
{a)
where we have neglected the small term
~
=
~
ln ~ .
ln O(E)
=
..1....
ln
2µH
l - ~(l 2
Consequently
E = -NµH tanb
(b)
T
(c)
<
0
½<1 - ~)
if
E
>
l
)
NµR
~
0
M =µ(n -n ) = µ(2n -N) where~ is the number of' spins alligned parallel to H.
1 2
1
Since
~ =
l
>i'
2 (N - ~)
frcm problem 2.4
(a),
M = µ(N -
M = Nµ tanb
~
µH
N)
.I:!!
kT
3.3
(a)
for system A, we have f'rom 2.4 (c)
ln
(1)
O(E) =
where we have neglected the term
In equilibrium,
E
Hence
µ~
(b)
Since energy is conserved,
Substituting from (a) we find
E + E' = bNµH + b'N'µH
E = µ~(bNµH + b'N'µ 'H)
µ~
(c)
Q=
E-
2
bNµH = NN'H ( b'µ'µ - bµ'
µ~ + µ'2N'
2
µ)
+ µ'~'
from (2)
ln
oE
2µH
E
1
= µ'~'
(2)
75.
(d)
P(E)
a:
where E0 is the total. energy, E0 = bNµII + b 'N'µ 'H.
n(E) n' (E -E)
From (l)
0
P(E)
a:
f
exp [_
]
[ 2µ H2N
This expression may be rewritten in terms of the most probable energy
E,
equation
(2).
2
P(E)dE = C exp
t(E~!1 ]
dE
where
The constant C is determined by the normalization requirement
co
!-co
C
2cr2
J;;'cr
2
exp [- (::~) ] dE = 1
2
exp [- (E-E) ] dE where
P(E)d.E = - 1-
Thus
-
'
cr
is given above.
(e)
(3)
(:f)
Fram equation
E-
(3),
µ N(bµHN + b'µ'N'H) -b' Nµ~
µ~ +
-
µ'~'
-
l\*EI
hence
µ'
N'
>>
N
N'
>>
N
_µ_.a;;;.~-:F-N-
=
3.4
The entropy change of the heat reservoir is 68 1 = - ~' and for the entropy change of the entire
system, AS + AS' = AS -
Q
T'
Hence
~
o.,
by the second law.
Q
-v
/\es~
'->,;)
3.5
(a)
In section 2.5 it is shown tba.t n(E)a: yN'X(E), where Vis the volume and X (E) is independent
of volume.
Then for two non-interacting species with total energy E0
n(E)
=
N +N
C nl(E) n2(Eo-E) = CV l 2 Xi(E) x2(E)
17
2 atmospheres.
CRAPI'ER
4
Macroscopic Parameters and their Measurement
4.1
373
(a)
~
=me
1
273
dT
T
= 4180 l.n
Since
~s
DB
res
= -
D.stot =
(b)
fil
=
-~
~
= - ~
T .
T
~
= 1310 jou1es;°K
273
(373-273) = -1120 jou1es;°K
+ D.sres = 190 jou1es/K
The heat given off by either reservoir is
Q = mc(50) = 2.09
X
105 joules.
Since the
entropy cha.llge of the water is the same as in (a) ve have
(c)
There vill be no entropy cha.llge if' the system is brought to its final temperature by inter-
action vith a succession of heat reservoirs differing ini'initesima~ in temperature.
4.2
(a)
The final. temperature is found by relating the heats, Q, exchanged in the system.
required for melting the ice is mi.f, vbere mi is the mass of the ice.
nents of the system,
Q
The heat
Since for the other compo-
= mct.T, where m is mass and c is the specific beat, we have
mi.i + Q(melted ice) + Q(original water) + Q(cal.orimeter) = O
(30)(333) + (30)(4.18)(Tf-273)+(200)(4.18)(Tf-293)+(750)(.418)(Tf-293) = 0
Tf = 283°K
(b)
The change in entropy when the ice melts is
m/
T•
T
DB = me
1
Ti
Hence
f dT
T
For the other components
= me l.n
Tf
T
1
~
- m/
g§3_
g§3_
A,
DB - 273 + mwc l.n 293 + mic l.n 273 + mccc l.n 293 = 1 • 6 jouJ.es, r-..
18
(c)
The vork required is the 8l!IOUD.t of heat it vould. take to raise the water to this temperature.
W =
lllr/\v
(Tf-Ti) + mccc (Tf-Ti)
= [(750)(.418)
(230)(4.18)](293-283)
+
l::,S
=
RI'
V
dV
Since pV = RI'
1 cl 1·Tf
f
i
Q
c dT +
=
T
joules
by the first lav
EQ=cd.T+pcIV
= c dT +
= 12,750
T
i
T
Tf
Vf
+ R ln Ti
Vi
= c ln -
r Vf RdV
Jv
V
1
independent of process
4.4
The spins becane rand~ oriented in the limit of high temperatures.
Since each spin bas 2
available states, for N spins ve have
=k
S(T -+ m)
S(T)-s(o) =JT
But since
ve have
ln
o
S(T
-+m)
r
= JJ
n
= k ln
Q!tl.
dT
-
dT
T
~ =
(1)
Nk 1n 2
T
T
1
2T
c 1 (T
ffe'1
1
where we have set s(o) = 0 by the third law.
cl
1)
(2)
= c (l-ln 2)
1
From (1) and (2)
Nk 1n 2
- 1n 2
=1
= 2 • 27
Nk 1n 2 = c (1-ln 2)
1
Nk
4.5
The entropy at temperature Tis found by evaluating S(T) - s(o) =
J
T£.CT1
T- d.T.
Labeling the
0
undiluted and diluted systems by u and d, ve have
T
su(T1)
1
- s (o) =
u
sd(Tl) - sd(o)
l
ffe'l
= Su(o) by the third lav.
Since there are oncy-
case as in the other, we must have Sd(T ) =
1
dT
= c (1-ln
T
1
T
1
(2 - - 1) Tl
2)
T
r
2
T dT
= J1
c2 TT= c2
i'r1
sio)
c
2
io as maey
.7 8tJ(T1 ). Hence
19
magnetic atoms in the diluted
CHAPl'ER
5
Simple Applications of Macroscopic Thermodynamics
5.1
(a)
pV7 = constant for an adiabatic quasistatic process.
substituting pV = v Rr ve have
V
l-7
or Tf = Ti ( ...!.
vi )
(b)
The entropy change of an ideal gas in a process taking it from T1 ,vi to Tf,Vf vas found in
problem 4.3
t:,S
= O for an adiabatic quasistatic process.
and since 1n 1 = O, it follows that
V l-7
Ti' = Ti
5.2
J
(a) W =
(b)
p a;,
= area
enclosed by curve
joules
Since the energy of an ideal gas is only a function of the temperature and since pV
AE
( c)
= 314
...!..1 )
(V
Q
=
3
PcVc
~ 6T = 2 R (~
=
¾(6-2)
x
1a9
=
Rr,
PAVA
-R-)
-
ergs= 6co joules
= AE + W vhere AE is calculated in (b) since the energy change is independent of path.
Again the vork is the area under the curve.
Q =
600 + [ 4CO +
1
~
]
=
1157 joules
5.3
(a)
(b)
(c)
W
J
=
p a!-/
= area
under curve = 1300 joules
To find the heat, Q, ve first find the energy
change in going from A to
course, is independent of the path.
t:.E =
c.p
=
i R(Tc-TA) = i (PcVc - PAVA) = 1500 joules
20
c. This, of
Then since
Q = .6E + W
1500 + 1300 = 2800 joules
Q =
(d)
From problem 4.3
=
~
R 1n l2
X
lO~/R + R 1n
6
X
10 /R
3
X
1
;a3
=
2.84 R = 23.6 joules/°K
5.4
(a}
The total system, i.e., water and gases, does not exchange energy with the environment, so
its energy remains the same af'ter the expansion.
change is given by the heat it absorbs,
6E
Since the water cannot do vork, its energy
= Q = G,_,fT,
where
~
is the heat capacity.
The
energy of the two ideal gases is, of course, dependent o~ on temperature, .6E = C6T, where C is
the heat capacity.
6E(vater) + 6E(He) + 6E(A) = O
'1l(Tf-Ti) + CHe(Tf-Ti) + CA(Tf-Ti) = O
Thus
(b)
Tf = Ti,
no temperature change.
Let -l be the distance from the left and v the number of moles.
Renee, in equilibrium -l = 3(80-f..), -l = 60 cm.
(c)
Since there is no temperature change, the entropy change is vR 1n
vr
V..
J.
~ = ~A + ~He =
R
3
1n
20
6o
50 + R 1n 30
=
p
3.24 joules; K.
5.5
(a)
The temperature decreases.
The gas does work at the expense of its internal energy.
(b)
The entropy of the gas increases since the process is irreversible.
(c)
The system is isolated, Q = O, and from the first law
6E = -W
=-
T (Vf-Vo)
The internal energy of the gas is o~ a i'unction of temperature, .6E = vcv(Tf-T 0 )
Thus
In equilibrium p =
vey(Tf-To) = -
T
T (vf-vo)
and since pVf = vRrf , ve have
21
(1)
Substitution in (1) yields
The equation of motion is
(1)
mx=pA-mg-pA
0
Since the process is approximately adiabatic
pV7 =constant= (p 0 +
AV 7
0
mx = (p + !!!:8.) - o
A
(Ax)7
and
r)
V
7
0
(2)
mg - p A
-
o
The displacements from the equilibrium position,:, are small so ve can introduce the
V
V
dinate cbange x = Ao + 'I} and expand about : •
!_ = -""""'1~ - = (~)
xr
r
V
7
(1 -
V0
(:+'I})
ill +
V0
We keep onzy the first and second terms and substitute into
••
coor-
... )
(2).
2
m11 = -(p
o
+ 5)
A
UV
'I}
0
This is the equation for harmonic motion and the frequency may be determined by inspection.
½
Hence
V
= !_
21T
[ (p
o
+ !!!:8.)
A
A2z]
Vm
0
5.7
p(z+dz)A
The volume element of atmosphere shown must be in
equilibrium under the forces (pressure) x (area) and
z
gravity.
Then if n is the number of particles per unit
volume, m the mass per particle, and g the acceleration
of gravity,
p(z+dz)A-p(z)A = -n(Adz)mg
~
dz
dz = -
~ dz, where
NA
22
p(z)A
(1)
Since p = nkT,
(b)
vRT
Substit uting V =
-
pV 7 = constant.
p
we obtain
'
Thus
3.E_ _ ...L dT
p
(c)
- r -1
(2)
T
From (1) and (2)
dT _
dz -
.Ll:::zl
rR
(3)
µg
dT
For N , µ = 28 and letting , = 1.4, dz= -9.4 degrees/kilometer.
2
(d)
Integrating (l) with constant T we have
= p0
p
{e)
From (3)
T
=
T0
-
(; ;l)
µgz
1
P~
Thus
[-µgz/RT]
exp
f
=
p'
Po
z __-__.µc..g...,_d_z_'..---..,....R(T
0
o
-
Ii:ll
)
r ~
R
5.8
I A{P+6P
AP
Consider the element shown.
to move a distance
s and
and density, and p = P-P
0
I
6x is its initial width.
the right a distance
g-ti:i.s.
The wave disturbance causes the left face
Let P and p be tb.e equilibr i um pressure
0
is the departure from the equilibrium pressure .
tts
=
-
vJ.
6.V
P:P0
or
23
= -
J.
Ex
At:.5
P
vhere we have neglected the pressure difference D.p across the slab.
from the equation of motion.
m
025
= PA - (P+l:i.P) A = -b.PA
ot2
oP
t::,.p "' dX L:ix
Since
and
m
=
p A/::,x
cP cp _ P o2g
di= di=
ot2
we have
Then
but since the order of differentiation is immaterial,
i =l - ~
i
~
2
ot KsP ox2
5.9
(a)
For an adiabatic process pV7 = const.
Hence
oV
dp
=
V
,P
and
Ks
(b)
U =
{p KS)
Substituting
P =
vRl'
V
oV
v1 ®
= -
~
=
and
(
1
1) 1
- ~
V
P -
,
we find
1
u =
(c)
(d)
u
1
c: T2 ,
u
(zI?-)2
independent of pressure
= 354 meters/ sec.
5.10
Substituting the given values into the equation
cp - cV =
2
vr2a
-K-
we :find
Cy=
24.6 joules mole
-1
deg
-1
1
1 = l.14
5.ll
When the solid, of dimensions x:yz, expands an increment iN, ve have
24
The displacement is found
v+av = (x+dx)(y+ay)(z+dz)
To first order
V +av = V
+ yzdx + xzdy + :xydz
Therefore
For the adiabatic, quasistatic process
= (~ )
dS
dT +
p
(~!)T dp = 0
cc~\
_/ov_\p' ve have
Using ( ~cri)\ p -- ~T and the Maxwell relation dl') T = \"2ff)
l (''v_\
vhere a = V
%:r)
p
o.T
D.T = -pc
Thus
C
p
where
D.1'
~
= ...E_
p
pV
C
5.13
From the first law
cp =
T(~)P
W/2=•a,;/2~~\ =·C~\C:/2
Substituting the Maxwell relation
(~)T
= {i)p
~~~)T
and the definition
P
= -T (~)
rrT =
a= - ¼(t)p we have
2
a vr - vr :
5.14
(a)
(b)
From
(1) we
may read off the Maxwell relation
(l)L
(2)
= aT2 (L-L)
0
(3)
(~)T
Since
F
= -
(~)T
(c)
(1)
TdS = dE - FdL
=
=1
-2aT
T
To
r
bT
T'
25
(L-L
0
)
dT' = b(T-T
0
)
S(L,T) - S(L0 ,T) =I'r,L (~)T dL =I'r,L -2aT(L'-L0 )dL' = -aT(L-L0 )2
0
Hence
S (L,T)
=
0
S (L0 ,T0 ) + b(T-T 0 ) - aT(L-L0 )2
In the quasistatic adiabatic process DB = 0
(d)
S(T 0 ,L)
+ b(Tf-T)
- aTi(L~-L
)
0
0
~
0
T
Hence
=
2
= S(T0 ,L)
+ b(Ti-T)
-aTf (Lf -L0 )2
0
O
b-a(L -L )2
T
i o
i b-a(L -L ) 2
f 0
f
Thus
if L? Live bave Tf > Ti, i.e. the temperature increases.
(e)
Fram the first law
Substituting (2) and (3) ve find
~~~T = -2aT (L-L
0)
Hence
CL(L,T)
=
C(L0 ,T) + I'r,L
~~~)T
bT - aT(L-L0 )
d[, =
0
(f) S(L,T )-S(L ,T )
o
o
o
= fL (~)
JL
T
dL' = fL
JL
0
S(L,T) - S(L,T )
o
0
T
=
1
T
C dT
1
~
=
T
--2a To{L'-Lo)
1T
S(L,T) = S(T
aT
1
(L-L )
o
T1
dT'
0
= b(T-T)
-a(L-L0 )
0
'!'bus
= -aTo(L-Lo)2
2
bT' -
T
0
dL'
0
,L)
0
2 (T-T)
0
+ b{T-T) - eT(L-L )
0
2
0
5.15
cf.
(a)
(b)
Q =
TdS = dE - 2cr
From (1)
Hence
dS =
o~\
( di)T
dx +
cos\
~/x: d.T
=
t
dE
(1)
dx
2ot
T -T
! co~\
T ax)T
dx
dx +
Equating coefficients, we have
26
!
co~\
T af)x
<fr - 2ot ax
T
Fran
(1),
ve ca.n read off the Maxwell relation
Since cr = cr
0
-ctr,
If' the :film is stretched at constant temperature
= 2-to- 0
E(x) - E{o)
(c)
W (0
--t
x)
= -
By the first law
Hence
J
Jx
F dx = -
dS =
X
2ot dx' = -2otx
0
dE
T
+
zfY
T
dv
(1)
(i)T dv + (l)v dT = ¥(i)T dv + ~ ( ~),, dT + zf;{' dv
Equating c oefficients, we have
From (1), we mey- read off the Maxwell relation
Thus
If' one mole is produced
5.17
By equation (5.8.12)
substituting p = n kT (1 + B2 (T) n] we find
~
\oV)T
- p
-
+ n~ dl32 dT
27
p = n~ dl32
dT
>
0
5.18
(a)
:f'rc:m
Since
(5.8.12)
ve have
(b)
From the first law, dE = Td.S - pd.V.
0 = T
.
( c)
Hence
cos~ -
p
av)E
and
(~\
oV)E
=
~T
For a van der Waals gas
2
vRI'
v a
v2
P=~-
Thus
5.19
(a)
From
(p + 't)<v-b)
(1)
= RT
we have
(
J
2
Solving, we find a =
(b)
CV
D
2
T
9
rlT>~'
~
8" RI'cvc,
c
(v -b) 3
=
+ 2a3 =
O
V
C
2RI'
b =
-RTc
(v -b)2
-
T
C
6a
--,;:=O
V
C
C
VC
3 ·
Rr
Substituting a and bin (1) yie1ds pc= 3. _£
ff
(p'
(c)
i
:,
2)
VC •
(v' - ½)
=
~ T'
To find the inversion curve, ve must have
(l)H
= ~p
(~
<I) p
28
-
1)
= 0
or
From the van der Waal.s equation
dp = RdT + (-
v-b
R:r
_ 28:'\ dv
,.3)
(v-b)2
v-b
V
T
thus
On
:
(v;b)2 = b
i:>Jiminatillg v and puttillg the equation 1n tems of the dimensionless variables of problem
5.19, it follows that
T'
6-3/4
(9, 3)
p'
5.21
(1)
(a)
We have
a -- V
-1
(ov
""°"D -ddT..s -- a' -ddT-a
o-8
,
p
µ =
(~)H :
= µ' :
Substituting these expressions into ( l) ve find
(2)
29
(b)
1
T~
Fran (2)
T
=
T
0
T
= TO
exp
r ,s a'diS'
J 13•'
[
0
µ'C
r
~
l +
V
Jr,s
,s
a'd,S
µ, C ,
J
l + :____,J2_
0
V
5.22
The system may be represented by the diagram
w
(a)
By the second
law
41
42
.6S=T - T~o
For max:IJmJm heat, the equality holds and since
'½
= q ..Jfl, it follows that
1
41
Ti
w=Ti-To
ql
(b)
V
= 11.9
5.23
We have the system
l
(1)
(a)
(b)
"3'J the second law,
Thus
It follows that
(c)
The maximum amount of vork vill be obtained vben
Fram (1)
30
Tf = JT T ' •
1 2
To freeze an additional mass m of water at T0 , heat
resulting in an entropy change 00
mL
must be removed from the ice-water mixture
The heat rejected to the body of heat capacity C
-o
{ Tf dT
Tf
increases its temperature to Tf With an entropy change .6.82 =C IT
T = C 1n T. By the
mL
Tf
V
O
0
second law 00 + t::.S 2 = - T + C 1n T ~ O. For min:ilnum temperature increase and thus minimum
1
0
0
1
= - ~.
heat rejection the equality holds and it follm.s that
The heat rejected is Q = C(Tf-T 0 )
5.25
The probability that the system occupies a state With energy Eis proportional to the number of
states at that energy,
n(E).
By the general definition of entropy, S = k ln
probability
P
ex:
n(E) = e 8/k
n(E),
we have for the
(1)
To raise the weight a height x, an amount of energy, mgx, in the form of heat, Q, must be given
off by the water.
The temperature change is found from
Q = C'AT = -mgx
t:il: =
vhere C is the heat capacity of 'Water.
10-
4
°K.
-mgx/c
For a mole of liquid and x
= l cm,
t:iJ: is on the order of
Hence we may approximate the water as a heat reservoir and find for the entropy vhen
the weight is at height x
S=S
vhere S is the entropy at height x
0
Here we have absorbed the te:rni e
usual nomalization condition.
o
_2:.._=S
T
0
o
= O. Thus (1) becomes
sjk
P(x)
=
Ae
-mgxjkl' 0
into the constant A wbich is found to be
kTjmg
by the
Then the probability that the veight is raised to a height Lor
more is
co
(? =
1
co
1
P(x)dx =
L
L
Substituting L = l cm and the given values yields
C?
= e-1018
31
kT
-2 e
mg
-mgxjkr
O
-mgLjkr
dx = e
0
In the procesS:S a
~
-+
b and c
-+
d no heat is absorbed, so by the f'irst law 'W = -6E, and since
= mT, ·where C is the heat capacity, we have
wa"'b = -(¾-Ea) =
We~= -(Ed-Ee)=
-vc;,
-vc;,
(Tb-Ta)
(Td-Tc)
Hovever, in an adiabatic expansion TV?'-l = canst.
T
Hence
wa,b =
wc,d =
-vc;,
Tb
-vc;,
Tc
(1~
G: -1)
V
=
-vc;, Tb (1- <v:)
=
vc;, Tc (1-
(:~)
7-l
7-l
)
)
The volume is constant in :process ~c so no vo~ is :performed.
Thus
CB'.API'ER 6
Basic Methods and Results of Statistical Mechanics
6.1
(a)
The :probability that the system is in a state with energy E is proportional to the Boltzman
factor e-E/kr.
Hence the ratio of the :probability oi' being 1n the i'i.rst excited state to the
probability oi' being in the ground state is
e
e
(b)
-3~/oo
=
--Hw/~
By the dei'ini tion of mean value
1 +
3e -iiwjkr
1 + e -15.wjkr
6.2
The mean energy per particle 1s
µ He-µR/kr
e ==
Hence i'o~ N particles,
- µHeµ.H/kl'
e-µifi'ki:r +eµH)\!r
E = -NµH tanh
32
I
6.3
(a)
The :probability that a spin is parallel to the field is
eµRjkr
l
p = e -µHflci' + eµR/kT = 1 + e - 2
vhich yields
For P = .75
(b)
µR/k.T
(1)
T = k J.n[fyl-P]
and.
H = 30,CXX) gauss,
Fran (1), w"'ith µ
= 1.41
x
T
lo-23
= 3.66°K
ergs/gauss,
T
x 10-3 °K
= 5.57
6.4
The power absorbed is :proportional to the difference in the number of nuclei 1n the two levels.
This is,
·w'here N is the total number of nuclei.
Since µR << kr, ve may expand the exponentials and keep
onzy the first two tenns.
Hence,
n
(1 +
- n
+
-
~ -
1 + ~)
.., N ___
kT"'-_ ___,;;_ =
1+~+1-.@
kT
kT
1
Thus the absorbed power 1s proportional. to T- •
:p(z+dz)A
_ _ _ _...__ _ _ z + dz
The volume element of at2!1osphere shown must be 1n
...,_--+-------~ z
equilibrium under the forces o:f the pressure and
gra:nty.
mn(z)Adzg
Then i:f m is the mass per :particl.e, A
the area, and g the eccel.etation of granty, we
have
p(z+dz) A - p(z)A =~dz A= - mn(z) A dzg
Substituting the equetion of state, p =~leads to
dn(z) - - 5 n(z)
dz
Thus
-
~
n(z) = n(O)e-mgzjkr
33
p(z)A
6.6
E
(a)
As T approaches O the system tends to the low energy state, while in the limit of high temperatures all states became equal.ly probable.
(b)
The specific heat is,
C..,
oE)
= (~
The energy goes :from the low to the high temperature
, i.e., the slope of
E.
V
T
~:::1
tl+e - J
_N\.l ~~:
(c)
+_
and
To find the temperature at which
evaluate T vhere
[
E changes
limit, we
J
N e:l+e:2
E-Ne:
1 +2 -2 - e :l
which yields
34
The heat capacity is
and
f1v
(a)
as
-+ 0
T --.
o,
The mean energy
T -+
cc
•
per mole is
E
-e;....,
Oe -OJ,,.m
/A.I. + 2e:e ,~
l + 2e -e:jkT
= N
A
2 NA e:e -e:/kT
l + 2e-e:jkT
(l)
where NA is Avogadro's number.
(b) We find the molar entropy from the partition function z. Clearly, Z =
-e: .j.kT
t =~e 1
, the sum being over states of a single nucleus.
~
NA
where
i
Thus
s
and
(c)
As T
-+
=
k (1n z + ~E)
(2)
O, the nucleii go to the ground stat~, and by the general definition of entropy
S=Nklnn=Nklnl=O
A
A
as T
-+
=,
all states become equally probable, n
-+
3 and
It is easily verified that expression (2) approaches these limits.
(d)
CV
From (1) we have
T
6.8
The polarization, ~ is defined a.s
N ~ where
0
N is the number of negative ions per unit volume,
e is the electronic charge, and
i
is the vector
from the negative to the positive ion.
Then the
35
0
£,
a
0
>e
0
aean x component of Eis
p
vhere x .nd
e are
X
= !Pl
-
cose
shown in the diagram.
=
Ne
It!
lt_xl = Ne X
-
The positive ion can have energy + e;a or - e~a ,
and since there are two lattice sites for each energy ve have
ela ~
_~ ~
(~) e 2
+
(- ~) e
2
tanh
F = N e - - - - - - - - - - - - - - - -_ ~
2
x
et'.9. ~
_~ ~
e 2
+ e
2
e!a
2ld'
6.9
(a)
The electric field is found from Gauss's Theorem
Ia~ •£
da =
lm-q
A
vhere q is the enclosed charge and£ is the vector normal to the surface, a.
By symmetry,
E 2rrr L
Ea=
E(r'
- !.I
The electric potential is given by
U(r)
=
=
lm-q
=_gs,~
rL -
ir ~--~
=-
f!l
ln ~o
0
ve can evaluate q.
Since U(r) = -Vat r = R,
0
Thus
E(r) =
r ln rR
V
0
(b)
A
! ,
U(r) =
ln !,_
V
ro
- ln R
ro
The energy of an electron at position! is qU(!) = -eU(r) vhere e is the electronic
charge.
Since the probability of being at! is proportional to the Boltzmen factor, ve have
for the density
= p(O)
(0
!...
r
-ev~ ln(R/r )
o
6.10
2
mw2r 2
The centrifugal force mw r yields the potential energy - - 2- . Since the probability that
a molecule is at r is proportional to the Boltzmann factor, the density is
2 2
p(r) = p(O) exp [mw r /2kT)
(1)
(a)
(b)
and r
Substitutingµ= NAm, the molecular veight, into (1) and evaluating this expression at r
2
ve have
1
6.ll
The mean separation, x, is found from the probability that the separation is between x and x+dx,
P(x)dx
_ U(x)
e
kT dx
cc
[<!)
12
6
- 2(!)
]
X
Where
u(x) = u0
It is easily verified that the
minimum of the potential is at x = a.
X
Since departures from this
position are small, ve may expand U(x) about a to obtain
36 U0
2 252 U0
U(x) ""' -U0 + -2- (x-a) (x-a) 3
3
a
a
vhere ve have kept only the first three terms. It follows that
36 U (x-a)
P(x)dx = C exp [ - a~o
We have absorbed the irrelevent constant term,
Uc/k'I',
e-
2
252 U0 (x-a) 3]
+ a~
d.x
into the normalization constant
C.
This constant is eval.uated from the usual. requirement
00
1
1
00
P(x)dx = C
exp
[
-
O
co
The predominent factor in the integrand is
36 U0 (x-a}2]
. [ 252 U0 (x-e.)3]
exp
dx = l
2
3
a kT
a.k.T
exp t36 U0
expanded in a Taylor's series as in Appendix A.6.
extended to
-co
(x-a) 2/a~]
so the second factor may be
Furthermore, the region of integration can be
since the exponential is negligably small except near a.
(1)
Thus
vhere ve neglect all but the first two terms.
The first integral is evaluated in Appendix A.4
vhil.e the second is O since the integrand is an odd function.
6
C=a
Then
By
definition,
x=
J~
x P(x) dx,
cu;1
_?,._
iT
kT
and to the same approxillla.tion as in (1) we have
exp
36 U0
2] (
- (x-a)
l.
[ - -a 2-kT
252
u0 (x-a)3)
+ ------a 3k.T
dx
The first term in the integrand is a Gaussian times x and the integral. is just a.
may be eval.uated by making the substitution~= x-a
37
The second
-
X
=a+
K
0
u~½
7rk1'
Noting that the second integral is O since the integrand is an odd function, and evaluating the
first by Appendix
A.4, we find
Thus
a= -
1
dx
dT
X
.15 ak
= au
0
.15k
+ (.15)ald' ~ ~
» kl'
since U
0
0
6.12
(a)
Let the dimensions of tl:e box be x, y, z.
Then the volume and area are
V = xyz
A =
(1)
xy + 2yz + 2xz
EJirninating x,. we have
V =
2yz
y + 2z
A -
yz
av av
.
From the conditions for an extremum, ~=oz = O, it follO'ws that
(A-2yz) (y+2z) - 2yz (y+2z) - y (A-2yz) = 0
(2)
(A-2yz) (y+2z) - 2yz (y+2z) -2z (A-2yz) = 0
Subtracting these equations yields y = 2z.
Then substitution in (2) gives
1
A2
and from (1) we find x = ./3
(b)
l
-
.
l
A2
A2
y - -
z - -
E ' - 2J3'
From equation (1), we have
dV = yz dx + xz dy + xy dz
(3)
g(x,y,z) = xy + 2yz + 2xz - A= 0
(4)
The equation of constraint is
and
Adg = A(y+2z) dx + A(x+2z) dy + A(2x + 2y) dz= 0
(5)
Adding (3) and (4) and noting that the differentials are nO'W independent, we have
yz + A(y+2z) = 0
xz + A(x+2z) = 0
xy + ;\.(2x+2y) = 0
We multiply the first equation by x, the second by y, and the third by z, and add.
3xyz +
A(2xy
+ 4xz + 4yz) = 0
Substituting (4) yields
A=-~
2A
(6)
and substituting
;I..
into equations (6) gives
l -
~ (y +
2z) = 0
l -
i (+
2z) = 0
X
(2x+ 2y) = 0
1 - :
The first two equations show that x = y and the third yields x = ~first and second equations gives
Alfa
= .,,/3
x
Alfa
,
Y =
13
,
z =
Then substitution in the
Alfa
.fj
2
6.)3
S
l
But since L, P (l)
r r
= L. P
s
s
s
1
(2 )
+
S
+
2
= l, we
s
2
=
-k L. P (l)ln P (l) r
r
k
r
L. P (2)1n p ( 2)
s
s
s
may write
-k L. L. p (l)p (2)ln p (1) _ k L. L. p (2)p (l)
=
rs
r s
r
s r
= -k L. L. p (l)p (2) 1i.n_ p (1)+
rs
r
l:-
s
r s
-k L, L, p
r s
1n p
rs
1n
r
r
1n
p (2)
s
p (2~
J
s
-k L. L. p (l)p (2)ln p (l)p (2)
=
=
s
r
s
r
since
rs
P (l)
r
.
s
p ( 2) = p
s
rs
6.14
From the definition, we have
S + S = -k L.P (l)ln P (l)_k L. p (l)ln p (l)
1
SubstitutL".lg P (l)
r
=LPrs
and
s
2
2
P ( )
=L P
s
r
s1 + s2
r
r
r
r, r, P
-k
=
-k L, L,
s
s
(1)
in (1) we have
rs
=
s
rs
rs
p
r s
rs
r, r, P
1n P (1)_k
r
1n p
r
rs
1n? (2)
rs
s
(1)? (2)
(2)
s
For the total system, the entropy is
s
= -k
L, L, p
1n p
rs
r s rs
(3)
Then from (2) and (3)
s-(s +s )
1
2
=
-k
= •k
r, r,
rs
p
rs
LL Prs
r s
1n p
ln
rs
p
r
r, r, P
+ k
rs
(l)p (2)
s
39
rs
1n P (1)p (2)
r
s
(4)
(b) Since -1.n x
~
-x +
1 or l.n x
l.n
Fran
s-(s1+s 2)
(4),
But since
L. L.
r s
P
·r
=
p
r
k Z: I:
~
x-1,
we have
(l)p (2)
s
p
rs
- 1
p (l)p (2)
P5
~
1n r
rs
k r, r,
rs
(p (l)p (2)
p
r
rs
rs
p
s
rs
(l)p ( 2 ) = 1, it follows that
s
s-(sl+s2)
5 k
r, Lo Pr(l)Ps (2)_1
= 0
r s
6.15
(a)
S = -k L. Pr l.n Pr and So
r
Since
S-S
o
=
-k Lo Pr (O)l.n Pr (O) ' we have
r
\_p
=kL.
lnP +P lnP (O)_p lnP (O)+P (O)lnP (OJ
rLr
r
r
r
r
r
r
rJ
vhere we have added and subtracted the term k Lo P 1n P (o).
r
r
r
two tems in (1) is zero.
P (0 )
r
Since
=e
(l)
We show that the sum of the last
-,3E
r;z
1n P (O) = -f3E - 1n Z
r
r
Then using Lo P = 1, ve have
r
r
-!:
r
p
r
L Pr Er
ln p (O) = f3
r
~E
+ L. P ln Z =
r
r
r
+ 1n Z
Lop (O)ln p (O)= -13 Lo P (0)3 - Lo P (C)ln z = -p3 -
Similarzy
r
r
r
r
r
r
r
r
1n Z
Addi:cg, we obtain
_z p
r
1n
r
F (0)
=
+
:z ? (0)1.n p
r
r
r
(0)
= 0
Thus (l) becom:::s
p (0)
s-s
P (0)
(b)
Since! 1n
= k L,
0
P (0)
-f-- ~ T r
-
[
-P 1n
r
p
r
r
r
0
~
r
1n p
•
J.,eq_uation (2) yields
s-s o ~kl'. Pr
consequentzy S
+ p
~O)
r
P
r
s.
40
r
(O)] = k Lo
r
p
r
1n _r__
Pr
(2)
= 0
CHAPI'ER 7
Simple Applications of Statistical Mechanics
7.1
We label the positions and momenta such that r .. and p.j refer to the j
(a)
-iJ
i.
There are N. molecules of species i.
-1
th
molecule of type
Then the classical partition function for the mixture
J.
of ideal gases is
1
ZI
exp[- ~l (:2.1/+. • ,+E_lNl2) • • •
The integrations over~ yield the volume, V, while the E_ integrals are identical.
Since there
The term in brackets is independent of volume, consequentzy
ln Z' = (N + ••• +Nk) 1n V + 1n (constant)
1
lo
1
p = §" dV 1n Z' = (Nl+ ••• +~) fN
and
pV
(b)
For the i
th
=
(N1+ ••• +Nk)kT = (v 1 + ••• +vk) RT
(1)
(1)
gas, piV = viRT, hence by
p = I: p.
i
l.
7.2
(a)
-€
=
By the equipartition theorem the average value of the kinetic energy of a particle is
23 kT.
(b)
The average potential energy is
iL
mgz e -i3mgz dz
u
f
Le -pmgz
dz
0
Then
On
carrying out the differentiation, we find
u=kT+
mgr.
l - emgLjkr
41
7.3
{a}
Before the partition is removed, we have on the left, pV = vRT,
is
After removal the pressure
_2vRT_~
Pr -(1+1i)v - 1+b
(b)
r
r~:v i
r(l;~lV
The initial and final entropies of the system for different gases are
Si
a
vR
Sf
a
vR
N:v + i ln T+o~
r (l;:lV
+
+ vR
~ 1n T+o~
+
ln T+aJ
+ vR
+
~ 1n T+o~
Here one adds the entropies of the gases in the left and right compartments for Si while sf is
the entropy of two different gases in volume (l+b)V.
Then
(c)
!::,S
= S -s .
l.
f
= vR
t
21n V(l+b)
__.__.._ - 1n - V - 1n bV
NAV
NAV
J
NAV
= vR
1n
(l+b)2
b
In the case of identical gases, S.l. is again the sum of the entropies of the left and right
compartments.
Sf is the entropy of 2v moles in a volume (l+b)V.
sl.. = vR 11n ..,;!,__ + .11n T+cr
L
NAv
2
7 + vR L!in bNVAv + .11n
T + crl
2
~
~
= vR 1n
Thus
(14~)
2
7.4
(a)
The system is isolated so its total energy is constant, and since the energy of an ideal gas
depends only on temperature, we have
or
The total volume is found from the equation of state
Vl
RI' l
V 2RI'2
V=--+-pl
P2
Thus the final pressure is
=
42
(c)
7.5
We take the zero of potential energy so that if a segment is oriented parallel to the vertical
it contributes energy Wa and if antiparallel it contributes -Wa to the total energy of the rubber
band (Thus if the rubber band were ~ extended, the total energy would be -Nila).
segments are non-interacting,
Since the total energy is additive,
where U is t he energy of interaction, the eq_uipartition theorem still applies and
7.7
I:f the gas is ideal, its mean energy per particle is
2
PX
2
Pv
e = 2iii + 2m = :k!r
and the mean energy per mole becc:mes E = NA:k!r
Thus
43
Since the
For classical. harmonic motion the mean energy is
2
-E-L+-mw
1
2 2
x
2
- 2m
By
equipartition, each of these terms contributes½ kT.
2
Hence
o:T
P"'X
7.9
(a)
The mean elongation is found by equating the gravitational and restoring forces
cix.
(b)
= Mg,
thus
By the equipartition theorem
½a (x-x)2
=
!2
kT
(x-x)2
or
kT
a
(c)
-Jkr a'
or
M -
g
2
7.10
(a)
Then the mean energy of N particles is
Let the restoring force be -ax.
1 72
1
E = N <2 mx + 2
-E
By equipartition
=N
Thus
(b)
1 kr + -1 kT )
(-
2
-ax3,
found from
=
J~
J~
oE
=?!r=Nk
the mean energy per particle is
E:=
The kinetic energy term contributes
= N kT
2
C
If the restoring force is
2 )
ax
J_ 72
1
2 mx + 4
½kT by
exp [- ~ ]
4
ax
equipartition.
~
dX
exp [- ~ ] dX
44
The mean of the potential energy is
1 4
where we have made the substitution, y = 13 / x.
d
Thus
4
= -
Hence
O (¥
€=
Then the total energy is
¼in 13 + 1n Jco...., e -czy /4
4
1
1
dy) = 4p = 4 kT
1721"""'1i'.
+
ax = l kT+ 1 kT= 3 kT
2 mx
4
4
2
4
E = NE and
7.11
Since the :probability of excitation is proportional to e -
kT where n is an integer, it is
clear that the two parallel components are negligibly excited at 300° where i:iw11 >> 300 k.
The perpendicular component is excited since hw << 300 k; and since 300° is in the classical
1
region, the equipartition theorem holds and we have for the mean energy per mole
1 ~
E = NA (2 mx
1
+ 2 mw l
22
x )
= NA kT
Thus
(a)
K !::,;a.
0
Consider a cube of side a.
The force necessary to decrease the length of a side by .6a is
and therefore the pressure is t::,.p =
Hence
K
K
0
6a/a2 •
The change in volume is !:::,.V = -a2.6a.
1
=
a3 ( -
= -
a 26a )
a
K .6a/a 2
= Ko
(1)
0
(b)
The Einstein temperature is eE = hwjk.
Since
I
w =
~Kd1m where m is mass, we have from (1)
(2)
The length a is found by considering that the volume per atom is µ/ pi-IA whereµ is the atomic
weight, p the density, and NA is Avogadro's number.
Thus
Substitution of m
a=(_H._)
pNA
1/3
=(
63 •5
(8.9)(6X1023 )
= µ/NA and the given values yields
GE"'"' l50°K
45
1/3
)
=2.3Xl0-8cm
7.13
'We have
Mz
= Ng µ
0
0
J
B/71).
~ (71) for
By equation 7.8.14,
B_21(71)
=2
(coth 71 -
12
2
coth
~2) = 2 sinh
cash n 71
2
cosh ({/2) + sinh (//2)
2 sinh 71/2) cosh (71 2)
= 2
2 ( /2)
sinh
sinh
(71J2)
cosh
J =½becomes
cash
sinh
t~f~1
cosh
sinh
(71/2)
Thus
-
The energy of the magnetic moment in the field His
E = -1:_•1!
where
e
-µH cos
=
e
is the angle between the magnetic moment and the direction of the field or z-axis. Then
the probability that the magnetic moment lies in the range
e
toe+ de is proportional to the
Boltzmann factor and the solid a.ri..gle 2rr sine de.
Thus
P(e)de
«
epµHcose
sine de
w
J
J
1'1z = N0 µ Z =
and
SµH cos
e
n eSµH cos
e
'IT
·'o
e
sin
e de(µ
cos e)
0
sin
e de
0
No
or
H
o
~
e
1n
= No ~
1n
~A
H ~,-,
We
r -;re ;3µH
,., o
e
sin
case .
2 sinh
~µH _
~ H
pµ
F.B
de
No
=
. . -... "µH
SJ.Il.H ...,
e
i:,µ:!
- e
-;3µH
[ µHpcosn
•
. ,..," µH - sinh pµH ]
(1)
have
46
1ere by (7.8.14)
I f Tl
<<
1
B
J
and J >> 1
ll (J
(Tt) = J-
in such a way that
Bi Tt)
=
J
f
l
+ -)
2 coth
JTt >>
coth
l l~
l
(J + -)Tt
- -2
2
coth -2 Tl
1, B (Tt) becomes
3
JTt -
½(~~
= coth
1
JTt
JT)
-where we have used
coth x = e
e
X
-X
X
+ e
-X
- e
=
(l
+ X + •••~ +
-
(1 + X + •••
?l1 -- XX ++
•••~
•••
:::::
-1X
for small x
whereµ= gµ J is by (7.8.2) the classical magnetic moment.
Then
0
becomes
(a)
(1)
:i-,1z = N0 µ ~oth pµH - P~
The energy of a magnetic moment in a field is
E = -1:. • ~-
For spin
,
2 atoms
(b)
n(z )/n(z ) > 1
1
2
(c)
since
n2
> :a:1 .
Ii' µH << kl' we have
+ (l - µH 2.....R + !2 µ ~ ? 2,::2)
...,
and
+ (l - µHl~ +
1
2
µ
2
2 2
Hl (3)
7.17
To find the fraction,;, of molecules with x component of velocity between_-:;; and
integrate the distribution between these J.illlits, i.e.,
47
v,
we must
s = .±
n
;
J
P,}
_-:;,
X
Making the change of variable, y =
s = -1
X
-(mv 2/2kT)
e
x
dv
X
m
~
Jfi'
.l
=f-Jti!!. (-~-/
2rrkT
g(v )dv
vx,
we have,
(2/)
2
e- y
dy =
Ji;; --12
1
,/~
2
L
2
e -(y / 2 ) = 2 erf
42
0
Then from tables s = .84-3.
In problem 5.9 we found that the velocity of sound is
1
u = czm)2
µ
where , = C/Cv and µ is the atomic weight.
Since µ = N~,we have
1
l
u = czm) 2
=
(Z:kT) 2
NAm
m
.l
The most probable speed is ~ = (2kT/m) 2 •
For helium, 1
= 1.66
so that u
Thus
= .91 ~ and
the fraction of molecules with speeds less than u is
,.,
s =
1 1°91v
n
O
F(v)dv = 47!"
1
O
mv2
- 2kT
-::i
1
2
-
·91(2kT/m)
,,,!,.
(;kT)
2
v
2
e
dv
.l
Making the change of variable y = (m/ld') 2 v, we have
s =4trr~
1.91./2 2 - 2/2
y ey
dy
0
(2rr)2
This integral :rn.ay be evaluated by noticing that integration by parts of
1
a e-y2;~~ dy = ye-y2
0
/21 a + 1·a y 2 e-y2/2 dy
0
l
a _ 2/2
e y
dy yields
O
0
Since the integral on the right is the required one, we have,
The integraJ. on the right may be evaluated from tables of error functions.
48
Thus we finds =
.r,.
7.19
(a)
V
2
(b)
(c)
V
X
X
by symmetry
(d)
kl'
= - by eq_uipartition
(e)
= 0
m
3
(v2 V)
= (v
X
X
+ 2
V
y VX +
2
-V )
V
Z
X
By appendix
Q
A.4 we find
2m
Since the energy is €
=
+ bv ) 2 = 2
V
y
X
X
vy
= 0
+ 2b -V
X
-V + b 2V 2
y
y
3/2
V
2
V
X
y
mv
ve
2 = (k'I')
m
2
dv
2kT
= (7rkl')
l
2
, a.".ld since v =
(l/v)
(1/v)
{b)
vX 3
0
V
n:Tv;
=
= (l+b2) kT
2
m
Jco r{tl dv =J"" 4n C21rm"ld')
TT-7vY = !n
(a)
(V
= 0
(f)
7.20
(vX 3 v)
y
~
=
1.27
=
7T
1
2
2
2 mv , we have v = 2€/ m and dv
~
d€/ ...; 2m€.
=
On
substituting these
relations into the speed distribution F(v) dv, it follows that
3
€
- l
- 2
2
F(€)de = 27m (nkT)
€ e kl' de
7.21
rhe most probable energy is given by the condition ~(e) = 0
1
~
1 e
....
Thus
E
....
The most probable speed is
V
E
l
kl'
€2
-kl'
or from problem 7.20b
€
e - kl' = 0
1
=2kl'
1..
= (2kl'/m) 2
1
-2
2mv
Hence
=kl'
7.22
V
(a)
(b)
V
=
V
O
(1 + ~
) =
C
V
since v
0
X
Therms freq_uency shift is given by
(.6v) rms
=
2
'V
But
Since v
=
2
X
= 0 and v
X
= V
~
2
v
-
-21
-vJ
l
2
2
0
~+ ~ x + ~
= kT/ m by equipartition, we have
= 0
Thus
(c)
(~v)
rms
=
~
l
kT
2
v O + -2v 0
2 2
2
Vo
-v~
O
me
C
The intensity distribution is proportional to the distribution of the x component of
velocity.
I(v )dv = I 0 exp [
where I 0 is the intensity at v=v
0
2
2
- me (v-v 0 )
2kT Vo
2
•
7.23
(a)
The number of molecules which leave the source slit per second is
18 molecules/sec
x 10
= 1.1
where A is the area of the slit.
(b)
Approx:il!lating the slit as a point source, we have by
(7.u.7), the number of molecules
with speed in the range between v and v + dv which emerge into solid angle
A~ (v) d3z = A[f(v)v cos e][v
2
cos
e~
dv
an
is
an]
1 for molecules arriving at the detector slit; hence the total number which reach the
detector is
13Jnv2
e- - 2 -dv =
where we have used the results of appendix A.4 and ps
(1)
= ~.
Letting L
2
between source and detector, the solid angle nd is approx:il!lately A/L •
= 1m.
be the disiance
Substitution of the
11 molecules/sec arriving at the detector slit.
given values into (1) then gives 1.7 x 10
(c)
In the steady state, the n1.nnber of molecules entering the detection chamber per second is
equal to the number being emitted.
or
From
Pd
(1)
we find
A
= Ps
= 2. 4
2
TIL
x 10
-8 mm of Hg.
The rate of change of the number of particles in the container is given by
dN
l dt = - cpA = - 4 nvA
50
Since the pressure is proportional to the number of particles, we have
dp
dt
-=
or
vA
- 4Y
t*~
p(t) = p(O) exp
where p(O) is the original pressure.
p
Hence the time required for the pressure to decrease to
p(t)/p(O) = 1/e is
t -
4v
VA
7.25
As in problem 7.24, we have
or
where p(O) is the original pressure.
The mean speed, v, is for nitrogen at 300°K,
v =
J~
7f
k:T' =
m
.!n5 x 105 cm/sec
10-6
_ 4(4n/3)(1c?)
Thus
t =
(.475
X
105 )(1) 1n 10-l =
4007 sec
1
(a)
Since the rate of effusion is~= p/(27r m kT) 2 ,and the concentration is proportional to the
pressure, it follows that
(b)
For the uranium separation, we have
1
238 + 6 19 )2
c'235I C'238 = c235I c238 (235 + o 19
=
The rate of change of the number of particles inside the container is
dN
dt = -
thus defining i...
l
-
NA
4 nvA = - 4V
j8-rr m
k:T'
= -
;>,.:r,r
"-Jm
Since pressure is proportional to the number of particles, we find after
integrating
For Helium gas p/p 0 = 1/2 at t = 1 hour.
(1)
Su:t,stituting we have
;>,.=~ln2
51
(2)
where ~e is the mass of the helium molecule.
From (l) and (2) it follows that the ratio of the
Ne and He concentrations in terms of the initial concentrations is
~e/~e = (~/~e)
0
= 1,
we have at t
=1
1,/m;;J ~
[-1n 2 (J~./,e' -~ ~
hour,
(1- j~/~e'
(1-J~/,e,)
I
7fe-' 11He
(a)
exp [-,. ~/.;~e' - 1
(~/~e) 0 exp
=
Since (~/~e)
0
= 2
)
= 2
The rate of change of the number of particles in the left side of the box is
dNl(t)
dt
= (no. entering/sec)-(no.escaping/sec)
Since the rate of effusion is
f nvA,
dNl (t)
dt
=
we have
=: I:
Av r
;J
Ii.il,72° LN2<t)-N1<tj
where N is the total number of particles.
-
'1
r-2N1<t2J
From the equation of state,
and
P1 (0) + P2 (o)
N = ----2
V
(1)
kT
Thus
_Av
e
Integrating,ve find
(b)
Tbe initial entropy is
Si=
Nl(O)k
(1n 2N:(O)
In final equilibrium, N1 = N2 = ~
Sf = Nk
(ln
~+~
52
1n
T
+ v0 )
V
t
7.29
vz
(a)
Inside the container
{b)
The velocity distribution,
=Oby symmetry.
<1>(:!), of the molecules which have effused into the vacuum is
<I>(v) d 3v
-
-
-
= f(v)
V
z
d 3v
2
where f(v)
.... is the Maxwell distribution.
Thus
V
=
z
J~dvxf~ dvy f
f dvxf"" dvy f
00
O
00
-co
dv
<I> (v)
z
V
z
00
dvz
<l>
(v)
f
0
-co
f
=
00
dvz e
mv
z
- 2kT
0
mv
00
dvz e
-
V
z
V
z
2
2
z
2kT
0
The second equality follows because the integrations over v
and v are identical in the numerax
y
The integrals over vz are tabulated in appendix A.4.
tor and denominator and therefore cancel.
3
~ (21;)2
VZ
½(~) aJ~ 1;;'
a
7.30
(a)
In t:i.!lle dt, molecules with velocity ccmponent v
z
which are in a cylinder of height v
·with base at the aperture can escape from the container.
z
dt
Molecules of higher speed can escape
from a larger cylinder than can molecules of lower speed, and a proportionalJ.y larger share of
the fast molecules effuse.
Thus the mean kinetic energy of particles in the beam is greater
than in the container.
Since <I>(_!) = :r(;:)
EO
V COS
!2
2
mv
d! 9
(~)
d3v
l
(b)
J
=
J
2
= 2mv
e and d3v
= V2dV
-
3
.
S1!1
oolrr/2J,ro1T(v2
€0
<l>
(2::)
e de dqi, we have
mv2
i
=-------------------------f f-rr/J,-rr
J""-.? 00
O
1
(v2 dv sin
o
O
2
dv sine de d~)( 2 mv) e
e de
- 2kT
v cos
e
dcp)
mv2
~ e 2kT dv
-l m
2
0
= ---------mv2
= 2k.T
e - 2kT dv
where the integrals over v are evaluated by appendix A.4.
Since the mean kinetic energy in the
. -
the enclosure is e;i =
3
2 kT, it follows that
e:
=
0
4e;i.
3
7.31
When a molecule is scattered, it commutes momentum l:.p
l:.p /
z
at.
z
= 2mv
z
to the disk.
The force is
To find the total force, we must multiply l:.p /at by the number of molecules with
z
velocity in the range (v,
dv} which strike the disk in time dt and then integrate over all
,v
velocities.
Clearly only molecules in the cone of half-angle e = sin-l R/L will strike the
0
disk.
F Jd3~ [A~ (!) at]
=
~!z
2
= J:°"J:ei2rr(v dv sine de dcp)(A f(v) v cos
= (2mA.)(2rr)(- 1 cos 3e) le 0
I
3
0
e d_::)(2m
;/os 8 )
J°"v4 f{v)dv
0
The integral can be expressed in terms of the mean square speed inside the container, i.e.,
! °"4
v f(v)dv
=
2
~
0
Thus
F
1 nm 2
=3
v
A ( - 1 cos 3e)
3
But by eq_ui:i;,artition we have
8
= 31
0
O
nm 2
v A [ 1 - cos 3 (s-i.,..,
.,_....-1 _RL)]
2 =nkT=p
-
1
3 nmv
F = pA [1 - cos 3 (sin-l
Hence
CHAPI'ER
f)]
8
Equilibrium between Phases and Chemical Species
8.1
By equation
(8.3.8), we have the proportionality
(?{v,T) dV dT
o:
exp [ -G0 (V,T)jk.r 0 ] dV dT
where V and Tare the volume and temperature of the portion of substance of mass M, and T is
0
the temperature of the surrounding medium.
minimum at V =
V1
and T =
Expanding the Gibbs Free Energy G (V,T) about the
0
T yieldf
(1)
54
T and v
(8.4.7)
where the derivatives are evaluated at T =
(1)
the first derivatives in
By
are O.
=
v.
Since we are expanding about a mini.mum
.
cG
(dT 0)
,
(2)
V
Since this is Oat T
Thus
and at T
( c2GD
"T2
= T0 = T,
C:
V
From (2) we also find the mixed derivative
cG
(~)
ar v
By the equation preceding
cG
0
= (T-T
o
K
"s
T
= (T-T ) (¾-)
O
)(els~ ?,v2Jr
C
(ffe)
---( ~p)
,
=
oV T
1
0
(V,T ) + -CV
2T 0
0
at
T
= To ,
V
= .....V
Substitution of these results into
VK
0
V = r~/p 0
T
oV T
=
G (V,T) = G
and since
at T
av
(8.4.13)
( "-vo)
By definition,
T
cC
= (1 - ~)(~)
= 0
T
T
(T-T
)2
0
(1)
1
..... 2
+-:::-- (V-V)
2VK
and CV = McV where ~ is the specific heat per gram, it follows that
(?(v,T)dV dT • exp
t:;/
(T-Tf -
2M: k'.1'
0
0
where we have absorbed the constant term G
0
(v,
(V - Mjp 0
)j
dV dT
T) into the normalization factor.
0
Performing tl::e
usual normalization, we find
=~~;,,:;/exp t
Mc
(?(V,T) dV dT
8.2
(a)
yields
_V_ (T-T
2.kT 2
o
)2
0
For the solid, l.n p = 23.03 - 3754/T and for the liquid l.n p
triple point, the pressures of the solid and liquid are equal.
55
= 19.49 - 3(£3/T. At the
hence
(b)
Since 1n p = -
t
+ constant vhere
t
is the latent heat, we have
t suu~lima t·ion
= 3754 R = 31,220 joules/mole
tvaporization= 3c;63 R = 25,48() joules/mole
(c)
tfusion = tsubl:ima.tion - tvaporization
= 31,220 - 25,48'.) = 5740 joules/mole.
The latent heat of sublimation is equal to the sum of the latent heats of melting and vaporization, i.e.,
tS
=
t m + tV.
t
Since
= T6V (dp/dT), the Clausius-Clapeyron equation,where vis
the volume per mole, it follovs that
(:)s
a
To(vg~s- vsol) ~o (vliq- vsol)(:)m + To(vgas- vliq)(:)J
or
8.4
I
Since
=
~ and
l::,.S -+
0
as
T
-+
0
by the third law, it follovs that :
-+
0
as
T
-+
O.
8.5
(a)
Q/1 moles/sec evaporate and are svept out by the pump.
pm"/
Q
E
(b)
=
Rr
Then by the equation of state
QRI'r
or
pm= LY
r
We find the temperature at the pressure pm from the Clausius-Clapeyron e~uation
d"D
L
dT = T(vgas - vliquid)
:::::s
T
L
vgas
where we have neglected vliquid since the volume per mole of the gas is much greater than that
of the liquid.
Then using v
gas
= Rr/p we find
1
c!... - !...)
T
R T
o
m
-1
Hence
T
m
TR
o
= T O\
{ - T
pm )
1n ;;-
-o
56
8.6
The intensity, I, is given by
Differentiating with respect to T and dividing by I,we have
(1)
Since pis the vapor pressure, the Clausius-Clapeyron equation yields dp/d.T.
dp
L
dT =
L
T(vgas - vliquid) ""T vgas
Substituting into (1) and using v
gas
=
Rr/p gives
1 dI
I
L
1
dT = RI'2 -
2T
Differentiating the molar latent heat yields
(1)
(s -s ) is the molar entropy difference between the phases.
2 1
[os2
osl
(~) T = T tdP) T - (dP )~
7
~A
Thus
But (os/c)p)T = -(ov,loT)P, a Maxwell relation, and since the coefficient of expansion is
a = -1/v (cv/oT) p ,it follows that
(2)
Differentiating at constant p gives
~~\ -
(~P • (s 2 -s 1 ) + T
T(os/'oT)
p
Thus
is the specific heat at constant pressure.
(~
p
=
¥+ ( P2
C
(~)J
-
c"O
)
-1
Substituting (2) and (3) into (1) yields
1) fT +} +(cP2 - cpl )
a:-f.. = -T(a2v 2 - a1.v
d.T
Since
57
(3)
8.8
The bar moves because the higher pressure underneath it lowers the melting point.
The vater is
then forced up along the sides to the top of the bar and refreezes under ~he lo.rer pressure.
To melt a mass
dm
of ice, heat -\'..am must be conduc,ced through the bar..
it per unit time is
J:!
=
-K:
The heat passing through
(be) m
a
vhere mis the temperature drop across the bar and x: is the thermal conductivity.
the position in the vertical direction, we have
p.(bc
dm =
l.
Letting x be
dx), and equating the rate at which
heat is conducted to the rate it is absorbed in the melting process ve have
(be)~= t pi (be~)
-x:
(1)
mis given by the Clausius-Clapeyron equation
d"D
ll.p
t
dt ""m = -T-,(-v1,-1a_t_e_r___v_i_c_e,...)
vhere v is the volume per gram or 1/p.
Substitution in
(1)
Then since ll.p = 2
mg/be,
yields
dx
dt
From the Clausius-Clapeyron relation, dp/d.T = L/TAv we find
Ja: JD.;
=
dp
We know L(cp) and ti.V(cp), and we can find dp/dcp from the curve of vapor pressure vs. emf.
Thus
J
T
d.T', =
-
273.16 T
Jcp ~vf9co'j
.
L
o
I
d:p
(-)
d~
dcp'
where 273.16°K is the temperature of the reference junction of the thermocouple.
Then
8.10
We divide the substance into small volume elements.
Tle total entropy is just the sum of the
entropies of each of these, or
S = Z: S. (E.,N.)
il.
l.l.
where Ei and Ni denote the energy and number of particles in the i th volume.
58
If' we consider
an interchange of energy between the i
th
and j th volumes,we have since the entropy is stationary
els.l.
clS = 0 =
but since dE. = -dEj, we have
l.
cs.l.
2;E.
l.
2s .
~
o~i
dE. +
l.
;sj
= cE.
0i
J
T.l.
or
J
dE.
J
= T.J
Similarly, if we interchange particles
dS = 0
cs.
cs.
].
m.plies
~
oN.
dNi
or
J
8.ll
(a)
The chemical potential can be found frcr:n the free energy byµ=~:.
He consider a small
0 1{
volume element at height z over which the potential energy may be considered co?J.stant.
It is
clear that the addition of a constant term to- the energy does not change the entropy of the
substance in the small volume element.
fu..'"Plicitly, the partition f'unction will be the free
particle f'unction times the factor e-~-rnez.
Thus
8 "~
1n ~free e-;,mg~ + ~ (Ei'ree +
mg~
" k
[1n
2
free + "freJ
Hence the free energy is, letting N be the number of particles in £::.V
F = E - TS
=
_ 6N
cF -_ .2. -"'..c
and
µ -
'"'i'
2
¾1llcr
+
Nmgz -
~ ~ ~; + ¾Lri T + c
+ mgz - kT
where we have put t::.V/N = ld'/p .
(b)
Si.riceµ is independent of z,
~
kT d-o
or
dz =mg+pd; =O
a:o
. ---=
p
Integration yields
:p(z) = p(O) e
8.]2
The relative concentration remains the same.
nco
~
The dissociation is described by
o
2
- -- = K (T)
nee~
n
2 2
Generally, when the number of particles is conserved, equilibrium is inde:pendent of v,
59
J
8.13
The intensity of the I atoms in the beam is proportional to their concentration
7:
in the oven.
Since dissociation proceeds according to
We have
nI-
2
= K (T)
n_
. n
l.2
K (T)
= ~
which gives
by the e~uation of state.
(1)
= ~(T)
The total pressure is p =Pr+ Pr which yields on substitution
2
into
(1)
or
Thus, if the total pressure is doubled, the ratio of final to initial intensities is
-1
-1
+Jl
+Vl
+ 8Kpp
+ 4Kp p
Sir..ce KP is small we may expand the square roots to obtain
-1 + 1
~f
+
49
~-,,--"C""""-~-
-1 + 1 + 2Kp-n
c[J_
J.
2
=
8.14
For chemical equilibrium we have
where n is the number of molecules per unit volume.
b
••• "O
m
-m.
= (kI')
Since ni = :pi/ kT it follows that
bl+b2+ ••• +b
m K (T)
n
= K:,:i (T)
8.15
7'h
-./
tn••e "".:.i_
· .,. s..,~ 1-,
c.,r -',,
a~ = dE +p dV.
Letting d(:pV) = V dp + p dV we have
-d. G,
= dE
=
+ d(pv)-v dp
d(E + pV) - Vdp
= d.H - V dp
where the last follows from th2 definition of enthalpy, H = E + pV.
-a. Q
Then if pis constant
= d.H
i.e., in a constant pressure process, the heat absorbed is equal to the change in enthalpy.
a chemical reaction,Q = 6E =
L b.h. since enthalpy is an extensive quantity.
i
J. J.
60
In
8.16
B'J the e~uation preceding
(8.10.25)
Lb.
i
1
1n ~-
=
I
1
Lb.
i
1
(1)
1n .ni
Substitution of ni = pijki' into (1) yields
L
b.
i
1
=L
1n P.
i
1
(b. 1n ~. , + b. ln kT)
, 1
1
1
b
p m = K (T)
m
p
or
where
1n ~(T) =
_
1n ~-'+bi 1n kT)
1
d 1n K (T) = L (b. dTd 1n ~-· + bTi)
dT
p
i
J. .
1
Thus
By
Li (b.1
(8.10.30) ~
b.
Then multiplying Ti
€i
1n ~it = kT where e:i is the mean energy per molecule.
piVi
by 1 = ILkT we have
b.
d
1n K (T) = L J.2
dT
p
i kT
1
where h.' is the enthalpy per molecule.
1
~·i
L
i
b.h.
:t.
kT
1
1
2
Multiplying and dividing by Avogadro's number gives
b.h .
d
dT
p.V.
1
+ ; ) =
Ci.1
1n Kp(T) =
L ..1:...1:.
i
RI'2
.6H
= -
Rl'2
where hi is the enthalpy per mole.
8.17
We let~ be the fraction of H 0 molecules which are dissociated.
2
Since the pressure is pro-
portional to the number of molecules, we have
PH
2
2p0
2
=----
(1)
The total pressure is the sum of t~e partial pressures.
(2)
Substituting (1) and (2) into
we find
~(T)
"Dt3
=-----2
(1-g )
(3)
(2+;)
61
8.18
In problem 8.16 it was shmm that
Lili = Rr
Using ~(T) from problem
2 9:_ 1n K (T)
dT
p
8.17 would give M for the dissociation of 2 moles of ~O since the
reaction is 2H 0 -)2H2 + o •
2
2
Hence for the dissoc~ation of l mole
2
M
where ~(T) is given by e~uation
~ ~
=
1n 1),(T)
(3) in 8.17. Since
K (T)
P
P_53
(1-s/ (2+s)- 2
2
9:_ 1n Ps
2 dT
2
is found by plotting 1n
of this curve at
s vs.
1700°. We find ~T
we have
p<3
= ___
s _ _ _-
Lill= RT
Hence
s << 1,
3
=} R1'2 9:_ 1n s
2
dT
temperature using the given values and measuring the slope
1n
s = .oo64.
1700°K
Thus at
M = ¾(1.99)(1700) 2 (.cc.64) = 56,000 cal/mole
8.19
N
(a)
The partition function is Z = ~ wheres is the partition function for a single molecule.
For a li~uid we assume a constant interaction potential-~ and that the molecule is free to move
in a volume N •
V
0
Thus
s = .l:...}' d\:i
h3
0
d 3r e -(f3i /2m)+j3~
-
-
where the integration is crver N,e,v and all
O
S = N,e,v0
thus defining
.E•
/TJ h\
J
d 3_E e-f3i/2m = N,e_ ef3~ v 0 !;'
0
s' .
(1)
Hence
(b)
For an ideal gas, the partition function is
The Helmholz free energy is
Fg, -kT J.n Zg
a
-kT
~
tg(~Ng_ J.n Ng)
= -kT ~g 1n Vg
62
s
I
-
Ng 1n Ng + N~
where we have used Sterling's approximation ln N ! ""'N
g
ln
g
N - N
g
g
clF
Then
(c)
µ
=~
=
g
oNg
(ln V
-kT
For the liquid, we have by (l)
g
~' - 1n Ng)
J
Ft = -kT (1n [N-f., el3TJ v 0 ~ N,e_ - 1n N,e_,
(N,t 1n v0 ~ 1 + N,t
= -kT
clF,t
Therefore
µp
1.;
(d)
=~
oN,t = -kT
(ln
V
o
~1
0
+ N,e_)
t,T}
+ t,T} + l)
When the liquid is in equilibrium with the vapor, the chemical potentials are equal or
V
__£
Thus
Ng
= V
o
ef:>TJ+l
but since pjkr = Ngl'vg we have
(e)
E....
= e-t,T}
kT
ev
(2)
0
We calculate the molar entropy difference from the Clausius-Clapeyron relation.
where in the second equality we have assumed that the molar volume of the liquid is much less
than that of the gas.
Since pv
g
= RI' we
have
Then using (2) in this relation we find
&,
= R (l + ~)
The molar heat of evaporation is
L = T6s = RI' (l +
fu')
""~=NAT}
(f)
From
(3)
ir
=
l +
(3)
if
~ •
Taking the logaritbm of (2), we find
kT
V
1.+.!J_=ln-=ln~
kT
pV
V
0
where v
g
is the volume per molecule of the vapor.
63
0
TJ>>kT
Hence
(4)
when v g and v 0 a re evaluated at Tb·.
In (4) -we can take v to be the volume per mole instead of the volume per molecule since we
(b)
have the ratio vg/v 0 •
To est:iJnate the order of magnitude of L/RI'b we let v ::::: 22.4 liters/mole
g
3
4
3
4
-8
23
3
and v ::::: - mN ::::: - 'TT (. 5 x 10 ) ( 6 x 10 ) ::::: 10- liters /mole.
o
3 o A
3
L
20
-:::::ln--:::::10
~
10- 3
Thus
we find L/r~ from eq. (4).
(h)
V
V
0
water
.018 liters/mole
nitrogen
. 035
benzene
.C87
g
31 liters/mole
6.4
29
L/ T0
1/T0
calculated
experimental
.83 cal/ gm°K
1.45 cal/ grn°K
.37
.64
.16
. 27
These results are remarkabzy good in view of the very simple theory considered here.
Quantum Statistics of Ideal Gases
9.1
configuration
0
€
3€
xx
xx
xx
X
X
M]
BE
FD
1
1
--
1
1
1
2
X
X
no. of states
1
1
,
(a)
(b)
---
(c)
~ =
e
-€13
+ e
-3€/,
+ e
-4:i,
1
X
2
.J..
1
X
2
1
1
9.2
(a)
For an FD gas, we have
1n z = aN + Z:: 1n (1 + e
-a~E
r)
r
Hence
s = k [ln z + pE]
r
[
= k
+ ~ 1n (1 + e
64
and nr
-a-pE
=
;1
r) + P~
1
but since
N = I: iir
and E = I:
r
r
1
r
= -----
0:+/3€r
e
€,
we :find
r
[r.r nr
S = k
Fram n
nr
(ct+ /3€ )+
r
Lo 1n (1 + e -ct-/3~r~
r
J
(1)
nr
(2)
it follows that
+l
(1-n)
r -
a+ /3€ r = 1n
-a-/3€
e
r
n
1n
r
(3)
= 1-l'!
r
Substituting (2) and (3) in (1) we find
s =k
=
(b)
Gc
r.
:n.:.
nr ln(l-llr) - 1n :n~
r + r 1n (1 + 1-n
r
-k
z::
r
+ c1-n ) 1n (1-~ ~
r.
. r:J
InL> 1n :nr
(4)
For the BE case
1n Z
= aN
- Lo 1n (1-e
r
-a-/3€
r)
and
The calculation of S Froceeds in exactly the same way as in (a).
r.
s = -k r
(c)
~
:nr
1n
iir - (1+n)
r
1n
(1+n)
(5)
r
Fram (4) and (5) we see that in the classical limit where iir
s ~ -k z::
r
:nr
1n
We quote the result.
<< 1
:nr
for both FD and :BE statistics.
9.3
The partition function in
sN/NI
s=
where
Lo
it it it
X y Z
. The sum is over all FOSitive
2
exp [ - ~11
2m
(it
X
2+
it
y
2 +it 2]
)
Z
(1)
xyz when we use the standing wave solution, equation (9.9.22).
it it it
The density of state is now greater than in the traveling wave case, and by (9.9.25)
If' we assume that successive terms ins differ little from each other, we can aFFrox:il!la.te (1)
with integrals.
/31'12
cc
z:
Thus
e
-
2
=--K
2m X
io:,
e
""
K ::()
/31'12
2m
--K
·2
L
X
(~ dK)
7f
0
X
L
1
(271m)2
2mi
13
=
X
X
(A.3.6)
by
3
2
(2rr m kT)
s= 3
V
Hence
h
vhich is identical to (9.10.7).
(a)
The chemical potential can be found from the He.1Jnholtz free energy, F
N
F = -kT 1n
fi
-kT [N 1n
:::e
=
-kT 1n
z.
s - N 1n N + N]
3
vhere ve have used Sterling's approximation, 1n NI
-:sF
L
~' =
~
K
K
X
A
Hence
(c)
•
m
(1)
is the area.
y
Thus
:::e
e e: 0 }Jcc exp
.
s'
µ'
~
13
exp [- i3i'i\K 2 +tt 2 )~e: ]
2m
X
y
0
= e
2
-
~
1n N'
i3e:
= -kT
-
o. 3
eq_uilibrium µ = µ' and by (1) and (2)
N -
2
i3E:
e
o 1;3
N'
(2rr
or
N'
m kT)
. 2
n
Letting n' = N'/A and since pjkT = N/v, ve find
9.5
1
By equation (9.9.14) if L = L
X
=
L = L = v3 we have
y
Z
66
1
2
13h
2
2
A
K +K ) dK dK
2m
X
y
2rr)2 X y
2
fi'
1n e
-
IL--(
-:S .
s
3 vhere ve put A =
o
= -kT
~
2
i3e:
l -
(a)
(2rr
3
h
kT) 2 •
The partition function for a single molecule of a 2 dimensional gas vith binding energy -e:
is
vhere
1n N-N ands =
µ = ~ = -kT 1n .s..
oN
N
Then
(b)
:::e N
V
(2)
0
€
(b)
p
r
r
(v)
(1)
221-~
cc = 2rr 1S.
(_!:.)
= -
oV
~ 3 (n 2 + n 2 + n 2)
3
m
y
X
Z
(2)
(c)
The mean pressure is -P
= I:r nr P-r where iir is the mean number of parti cles in state r.
Thus
(d)
by
P=g_
3V
L.n
r r
€
=_g_3
r
The expression for pis valid for a system in which the energy of a c;_uantum state is given
(1).
For photons,
€
r =
Hence
(e)
VE
and
The mean force on the wall is found by multipzying the change o::' momentum per collision,
2mv, by the number of particles which collide with the wall per unit time, 1 nvA.
0
Hence the
pressure is
1(-)1P = A 2mv (ti nvA)
Letting -2
v
~
=
1
-2
3 nmv
2
v we have
(3)
1
?
since N(~ mv-) = E.
Equation (3) is the exact result as shmm in section 7 .14.
9. 6
(a)
µ
s
- ?
4 - 22
.,.,ro'ol
•m •
9 ,5! ,
.,.,- -a
~-?~ ~i·~
• . s to ~_,_._n.d .,u.,_ •
By !
,'=
'~ , and,_-~--=-
l\·cP) 2 -- .,.,_ 2 _ :;_2 .
The dispersion is (~
0
µ
/
4
,;-o-
=2
t..,
gv
--2
(c.p)
Hence
=
Since
E
r
€
r
';3V
'
--2
~2
0
= - ~ca. 1n Z and (c.E) =
,. ,
n
?-?
4~
(E- - E:-) = _.2 (c.E)
4
2
9V
(b)
r
2
ot3 2
1n
z,
it follows that
67
~
by
(2 ) in 9.5
c2
4
4
(6p)2 = -2- l n Z =
- 9V 2
9V orl
dE
cf3
but by (2) in 9.5, -E = 3 V -P•
2
(6p)2
Thus
(c)
Substituting
p
=
2
cp
'3"l
cl~
2k.T2
bP
3V
oT
=
-- V
~ kT , W? find
--?
(6p)- /p
2
2kT2 N
-2,
= (- kT) ~ (IT
'3"l
T V
V
kT)
2
= 3N
*9.7
The probability that a state of energy€ is occupied is proportional to the Boltzmann factor
e
e
-€/kT 0
-€/kl' O
•
Since the vibrational levels have a spacing large compared to kT, we see that
0
is small and we may neglect this contribution to the specific heat.
On the other hand,
the spacing between rotational levels is small so that these states are appreciably excited.
For a dumbbell molecule the rotational energy should be
where A is the moment of inertia and w the a.ngul.a.r frequency.
We have neglected rotation about
the axis of the molecule because the moment of inertia is small.
may use the equipartion theorem yielding €r = kT
0
•
At room temperature, we
Adding the energy of the translational
motion gives for the total energy
€=et+ Er=
i
kTO + kTO =
i
kTO
Thus the molar specific heat is
The reaction proceeds according to
21ID =
and the Jaw of mass action becomes
= K
n =
(1)
From the results of section 9.12, we can immediately write dmm the quantities ;.
notation of that section,
68
In the
[2rr(2m) kT]2
~2 - h3
~D
=
2
s
y_
h3
t3€'
l
V
R2
e
HD - h3
e
2
2fi
3
t3e~
2
[2rr(2M) kTJ2 e
V
~ kr - !2 t,nwH2
1 .
- -2 f,·i\-- w_o
~kr
2
e
2
2t?
l
[ 2rr ( m-tM )s:r ]2 e
f,€'
HD
~kT
112
1
e
- 2 t3n"1m
where we have assumed that molecules are predaminantzy in the lO'W'est vibrational state and have
treated the rotational degrees of freedom classicaJJ.y.
The electronic ground state energies€' are approximatezy equal for the three species since
this quantity is altered onzy slightzy by the addition of a neutral particle to the neucleus
(to form D).
Thus
(2)
Similarzy, the moments of inertia are of the form
A = !_ µ * R 2 = !_ ~m2 R 2
2
o
2 m -lill
o
1 2
where R
0
,
the equilibrium distance between nuclei, depends primarizy on electromagnetic forces.
It is then approximatezy the same for the three species, and we have
~
2
~
2
~
-2
=
m M mM
<2) (2) (m+M)
-
2
=
~4
(mmM+M)
2
(3)
FinaJJ.y, the frequency for two masses coupled by a spring of constant
Again,
K
K
is
must be the same for the three molecules, a.~d
w
HD
=
(4)
Substituting the ~'s into (l) and using the results (2),
Since
7I
2
=
7) , we hz.ve
2
(3),
and
(4)
yields
%D
or
-=
N
~ l(m·+M)2l
L mM j
exp
9.9
In a thermalJ.s" isolated ~uasistatic process, there is no entropy change.
Hence by the first lav,
TdS=O=dE+pdV
Substituting
E =Vu and p
=,½u where
o
Thus
-l
V
8v.!t
rN'
3 V'
=
J
du
u
=
1Tf
T.:L
u(T) is the energy density, we have
-
-
= u dV + V du +
dT'
4 '.fr since -u
or
T
ex:
lu oY
3
li.
4
T.
Ti
f = 2-
9.10
(a)
By the fundamental thermodynal!lic relation
TdS=dE+pdV
Then substituting
E=
Vu and ii = 31u,- we have
-
-
TdS=uoY+V~+
4 -
1-
3
-
urN=ruoY+V~
du
But since u is onzy a function of T, du = dT dT, a.."ld. it follows that
4
dS = -3
(1)
dV + (c8 )
(2)
oV T
2
sfev 6T)
=
2;T V
dT
and (2) yields
(~s)
Since (c
(1)
.L
oV T
Equati.Ilg coefficients in
(b)
~ dV . y d.u dm
.,.
T
T dT -
ds = (~s )
We also have
4u
= -3 -T
(3)
and
(d2S/'cT QV), it follows from (3) that
70
Tf
Integrating, we find - .,;. 1n 8 = 4 1n 3
Ti
or
1n T4
Thus
=
1n -u + C where C is the constant of integration.
Finalzy -u
~
4
T.
The energy of the black body radiation in the dielectric is
where c' is the velocity of light in the material.
vacuum by c 1
c/n, n
0
0
being the index of refraction.
2
E =V-
Thus
Here
=
7r
(kr)4 n 3
15 (ct1)3
CJ
c' is related to the velocity of light in
=
4cr n0
0
3
4
VT
C
4/6o c 2i'i3•
16 CJ n0 3 vr3
is the Stefan Boltzmann constant, cr = n2k
=-----C
Hence
(1)
Substituting v, the volume per mole, for Vin
gives the heat capacity per mole,~'·
classical lattice heat capacity per mole is~= 3R.
c'
V
-=
CV
(1)
The
The ratio of these two quantities is
16crn 3
o
vr3
3Rc
For an order of magnitude calculation, we can let v = 10 cm3/mole and n 0 = 1.5.
At 300°K, we
9.12
(a)
The total rate of electromagnetic radiation or :pover is
P
= @(4n
=
2
r )
= 4rrr2 oT h.·
6
l.L
22 ergs/sec
(4n)(l02)(5.7',<10- 5)(10 ) · = 7.2xio
7.2 x 1a15watts
(b)
Since the total energy/sec crossing the area of a sphere of radius r' = llm must be the
same as that crossing the sphere of radius r = 10 cm, we have
or
71
= 5.7
(c)
X 10llergs/sec c:m2
= 5.7
X 104 watts / c:m2
The power spectrum bas a maximum at
tiwmax
- kr
he
---=3
-krA.
max
Thus
(a)
The power given off by the sun is
P =
4
cr T 0
2
(4n R)
Half the earth's surface intercepts radiation fram the sun at a:ny time.
Renee the ~ower that
the earth absorbs is
The power that the earth emits is
In equilibrium, the power incident upon the earth is equal to the power it radiates, or P.= P
e
1
Thus
T = T
(b)
From Eq. (1)
(1)
/R
o "-J 2L
T = (5500)
9.14
We consider the liquid in equilibrium with its vapor (at vapor pressure p).
flux of moJ.ecules incident on the liquid is
of particles escaping from the liquid.
By (7.12.13), the
p/ ./27r m kr' , and this must also be the flux
j(
Since J{ cannot depend on the conditions outside the
liquid, this is the escaping flux at all pressures.
J{
Thus
=
p
.J27r m kr'
The mass of a water molecule ism= 18/6.02•1023 = 2.99-10-23.
P
4
= 23.8 mm Hg = 3-17•10
)(
=
Thus for water at 25 0 C where
2
dynes/cm we have
4
3.17•10
(27rX2.99 X 10-2 3x1.38>0..0-l~ 298)½
= 1 _14
X
1022 molec~es
sec•cm
At the vapor pressure p(T), the flux of molecules of the vapor incident on the 'Wire is by
cp.). = p/ ,J2rr m kr'. When the metal is in equilibrium 'With its vapor, the emitted flux
(7.ll.13),
must equal the incident flux,
cp
e
Then when the temperature is T, the number of particles
= cp .•
i
emitted from the 'Wire per unit length in time t is
N = p(T) 2rrr t
-J2rr m kr
1
But N = 6M/m where liM is the mass lost per unit length and m is the mass per molecule.
6M = p(T) 2rrr t
Thus
,./2:rr m kr •
m
or
~
p(T) = ~
where ve have usedµ= mNA and R = NAk.
vX
=
o by
symmetry.
J
J
d
By definition,
=
d
3y f(!)
v}
(1)
3
y f(:~:)
At absolute zero, the particles fill all the lowest states so that the distribution f(y) is just
a constant a.nd cancels in
(1).
= µ.
By sylllllletry,
and (1) becomes
9.17
(a)
At T = O all states are filled up to the energy µ.
Hence
the mean number of pa..-ticles per state is just 1 and we have
f
E=
µ e:p (e:) d€
0
1
By (9.9.19)
V
p(€)d€
=~
4?T
1
(2m) 2 2
3 €
de;
ii
where the factor 2 is introduced since electrons have two spin states.
3
3
3
5
E = v2 (2m)2
2n'
ii3
Jµ €2
0
73
d€
= v(2m.)2 µ2
5rr2 ii3
2
VX
=
12
3V
(b)
Since
2
"tl.2
µ =
2
2m (3rr
i) 3
by
(9.15.10), it follow-s that
E
(c)
Substituting the expression forµ in
(1)
2
io (3rr2)3 !
=
i
we have
2
2
E=
(1)
Nµ
(i)
3
(2)
N
As particles are added, each new particle can go only into an unfilled level of bigher energy
than the one before.
Hence the energy is not proportional to N as it vould be if particles were
added to the same level.
(a)
Differentiating E in 9.17 (c) gives
2
2 /
~E)
~
p = - ( av T 5
(b)
~2
N
m (;,)
cu
5/ 3
(1)
Again from 9.17 (c) we find
"O
(c)
3
For Cu metal,
N/v
the electron, we find
=
8.4
p::::::
X
=
g!
3
22
10
2 2/3 "tJ..2 N 5/ 3
(-)
m V
-
= (3rr)
5
V
.
3
electrons/cm •
1r? atom.
Substituting into (1) where m is the mass of
This enormous pressure shows that to confine a degenerate
electron gas, a strong containing volume such as the lattice of a solid is required even at
o°K.
9.19
From the general relation TdS =
a power of the temperature,C
a:
cl. Q
Ta
6S
= CdT, we have that if the heat capacity is proportional to
J
=
C;
a:
J
Ta-1 dT
a:
Ta
Thus the entropy is proportional to the same power of the temperature as is the heat capacity.
(a)
The heat capacity and entropy of conduction electrons in a metal is proportional to T.
Renee the ratio of the final and initial entropies is
sf
400
2CQ
-=-=2
Si
(b)
The energy of the radiation field inside an enclosure is proportional to T4 , and the heat
capacity has a T3 dependence.
Thus
9.20
(a)
At T = O, the total number of states within a sphere of radius
tt
F
must equal the total
number of particles N.
Thus
2 _V_
(2rr)3
(}± 7r
3
3)
tt
N
=
F
or
2
><2
µ =
2
~ (3rr
i) 3
(1)
The number of particles per cm3, N/v.,is NAp/µ -where NA is Avogadro's number, p = .95 gm/cm3,
andµ= 23.
(b)
From (1) ve find the Fermi temperature TF = µ/k = 36,0C0°K.
To cool 100
cm
3 or 4.13 moles of Na from 1°K to .3°K, an amount of heat
Q must be removed.
-where~ is given by (9.16.23),
Thus
Hence
Q
.0022/.8
=
.0027
2
R)(~ ~')
= (4.14)(;
cm
1i·
3
T dT
= -.0022
joules
3 of He 3 are evaporated. Note that a large amount of substance is
cooled by a very small amount of helium.
9.21
(a)
the magnetic moment.
2
/kT -whereµm is
m
In the actual case, most electrons cannot change their spin orientation
If electrons obeyed
MB statistics, the susceptibility -would be
x.._
MB
= Nµ
-when an external field is applied because the parallel states are already occupied.
electrons near the top of the distribution can turn over and thus contribute to
then, that the correct expression for Xis
\re with
x.
Only
We expect
N replaced by the number of electrons in
this region, Neff.
-where TF is the Fermi temperature.
2
µm
Thus
(b)
X
= Neff kI'
=
(1)
4 or 105, (1) yields a X per mole on the order of 10-6.
Since TF is on the order of 10
electrons obeyed MB statistics, X -would be changed by the factor
75
If
\m
-- =
X
This is on the order of lCO at room temperatures.
KINETIC AND MAGNETIC
EN:::RGY
9.22
µ
When the field His turned on, the energy levels
te
of the parallel spins shift by -!lmH vhile the
levels of the antiparallel spins shift by -tµmH·
Density of
States
The electrons fill up these states to the Fenni
energy µ as shovn in the figure.
The magnetic
moment is detennined by the number of electrons
vhich spill over from antiparallel states to
parallel states to minimize the energy.
at the Fenni energyµ.
This is (µmH)p(µ
0
)
where p(µ
0
)
is the density of states
Consequently the magnetization is
0
!:!H
V
where n =
N
V
9.23
If we choose the zero of energy to be that of the electrons in the metal, the partition function
of an electron in the gas outside the metal is by (9.10.7)
s = 2v
h3
(27r
m
k.Ti e-vjkT
where the factor 2 occurs because electrons have two spin states.
The chemical potential µ
:!Ound from the Helmholtz free energy.
F = -kl' 1n Z =
~
=
-k.T [N 1n
~ - N
~
= -k.T 1n 1 = -kl' 1n 2V3
~
N
~
1n N + N]
(2nm kT)~
e-vjkT
The chemical potential inside the metal is justµ= V - ~, and in equilibrium the chemical
0
potentials inside and outside are equal, i.e.,µ=µ.
0
0
is
Thus
Hence the mean number of electrons per unit volume outside the metal is
n
= TI
=
V
g_
h3
l
(271" m 'kr)2 e _q,/kl'
(1)
Consider a situation in which the metal is in equilibrium with an electron gas.
Then the emitted
flux is related to the incident flux by
(1)
q,e = (1-r) q,i
The electron gas outside the metal may be treated as a classical ideal gas from which it follows
that the incident flux is
3
1
•i - 4
-2
1 2
nv - 4 ~3
where we haved used (1) of problem 9.23.
(2mn l<T)
_q,
e
~
/l<T
-l""' ]l
-
8kT 2
The emitted flux is then given by (1).
But since the
emitted flux cannot depend on the conditions outside the metal, this is also the flux where
equilibrium does not prevail.
Thus the current density is
=
4"rr m e~l-r) (kT)2 e-q,jkT
h
where e is the electronic charge.
9.25
There are (27r)- 3 d 3_!t states per unit volume in the range (1t:d1t) and with a particular direction
of spin orientation.
Since the mean number of electrons with one value of
K
is ~/3(€-µ) +
~
If we take the z direction to be the direction of emission, the number of electrons with wave
vector in the range (~:d~) and of either spin direction which strike unit area of the su....-f'ace
of the metal per unit time is
(1)
Then the total current density emitted is given by summing (1) over all values of
allow the electrons to escape.
K
which will
That is, the part of the kinetic energy resulting from motion in
the z direction must be greater than V, or ~
0
2
K
Z
2
/2m
77
> V0 1 where we use the notation of problem
9.23.
In addition, we assume that a fraction r of -electrons of all energies are reflected.
Thus the electron current density is
J
= e~ = 2e(l-r)
e
(
2rr
)3
1= 1= 1=
",./2mv )11
dit
....,
(m z/m) dit z
dit
X
Y
....,
where e is the electronic charge.
2
~11
2
2
2
1 + exp [ -;:;::-(K +K +K )
=i
X
y
Z
The integral over
J
2
=
xe ;x -+a
K
z
= _ ½1n
is of the form
2
[l+e -x -+a]
e-x -+a+l
Hence
Since V
0
I~
>µand, in general,
(µ-V
0
)1
>> 1, we can use 1n (l+x)::,,, x to expand the logarithm
and obtain
J = 2e (1-r)
(2rr)3
F.a.ch of these integrals is
(a)
Since the integrand in (9.17.9) is even, one can write
= 2
where we have integrated by parts.
Since the first term on the right vanishes
I
2
(b)
=
4r = X dx
'J o, e X+1
This expression can be evaluated by series expansion.
~
Since e-x <lone can write
-x
= ~ = e-xx
ex+l
[l - e -x+e -2x-e -3x •. •]
l+e-x
=
~
n=O
Hence (1) becomes
(1)
(-l)n e-(n+l)x x dx
I
2
(-itf
~
= 4
n=O
co e
-(n+J.)x x
dx =
~
4
o
Jco e
( -J.):
n=O {n+l)
Ti.le integral is, by appendix A.4, eg_ua.J. to unity.
""'Y y
ey
o
Thus
(c)
C'
C
,,,··
,
\
Fig. l
One can use the contour
Fig. 2
C of
J
C
2
z
-.......
(1)
Fig.
to write
coI:
dz= 27T:i.
l
sin
J
or
C
Here the integrand ~ 0 for z ~ co.
'-J.'n
~
n=l n
z
71"
Fig. 3
z
dz
2
sin
=
2
coI:
= 2i
(-it
(3)
n2
n=l
71"
-2i S
(4)
z
71"
The sum in (2) is unchanged if n ~ -n.
Hence one could eg_ually
veil write the symmetric expression
J
dz
= -2i S
C' z sin 71" z
along the contour C' of Fig. (J.).
(5)
2 .
Adding (3) and (4) one gets then
dz
J
C+c' z
2
s
= -4i
(6)
sin nz
But here the contour can be completed along the infinite semi-circles as shown in Fig.(~ since
the integrand vanishes as
lzl~co.
Except for z = O, there is no other singularity in the
enclosed area so that the contour can be shrunk down around an infinitesimal circle C surrouno
ding z =
0 as
(3).
shown in Fig.
For z ~
l
2
z sin
71"
z
=
2
O,
l
l ~ 3
z [m - 0
l
=3
1
[l + b
z ••• ]
22
z •••J
71'
7fZ
Hence
f
C
2 ~z
z sin m
l
=
l 2 2
3
[l -
m
l
=7
7fZ
= <-2n:i. Hl)
0
79
-r;rr
71"
+
5z
+
z ••• ]
-l
Thus it follows by (5) that
i-rr2
-T
or
S =
= -4iS
_ ~ ( -l)n _ 7r2
n~l
n2 - 12
9.27
(a)
We expand eikx in the integral
J(k)
(1)
=!~
Interchanging the sum and integral yields
(2)
(b)
Putting x=z in
(1)
to denote a complex variable, we evaluate the integral on the closed
contour sho'Wll in the figure.
Here C' is the path where z is increased by 27ri and C" is a small
circle around the singularity at i-rr.
Then the integral on the complete path is zero since there
are no singularities in the region bounded by the contour.
C'
I
C"
C
But the integral is Oat x =±cc and cancels along the imaginary axis.
=J
e
ikz
where both integrals are evaluated from left to right.
dz
(3)
Along C' vhere z = x + 27ri, we have
since ex+27ri = ex
eikx dx
80
Hence
Thus f'rom (3) we have
The last integral is just J(k).
(l-e -2rrk)
J
J
ikz
ikz
e
dz
=
e
dz
C (ez+l)(e-z+l)
C" (ez+l)(e-z+l)
c",
To evaluate the integral around the small circle
(4)
we put z = 7Ti + t; and expand the integrand
in tem.s oft;
J
e ikz
ikl;
-7Tkf (-t;)(
e
t;)
dz
= e
I'
d~
C"
= e -7Tk
C"
(9+c2m) (-ik~
= 2-rrke -7Tk
where we have used ex= .l+x+ ••• and the residue theorem.
Expanding
(6),
(5)
Then combining (4) and (5) we find
7rk
=
(c)
[l+iks+- •• ) a" i'
-1; 2
~
= e-1Tkf
(6)
sinh 7r k
we find
2
7rk
(1rk)"'
J ( k ) = sinh 7r k = 1 +
---r
1(-gk)
3 0
4
- •••
Then com!)aring coefficients with the series in (2) we can read off the Im's.
(a)
According to equation
(6.9.4),
We find
r2
= 1r2/3.
the probability of stater of the system is
pr
c::
e
-pE -aN
r
r
Then the mean. number of particles in the states is
L.
n
s =
r
n s exp [-p(nl€1 + n2€2 + ••• )- a(n 1 +n2 + ••• )]
L.
r
exp [- ~ (nl€1 + n2e2 + ••• )- a(n +n + ••• ))
1 2
where the sum is over all n , n , ••• withcut restriction.
1 2
Letting
l
we have
=
L. exp [-i,(n1 €1 + n 2€2+••• )- a(n1 +n2+ ••• ))
r
n=
s
1
13
'a~
"'s
1nZ
where the partial denotes taking the derivative where €s occurs explicitly.
(1) in the usual way, we obtain
81
(1)
Evaluating the sum
(2)
{ 1-e -{~.l~
(- -<n"il•2i\
..
)} 1-e
})
Thus
1n
l
= -
L.
1n (1-e
-a-Be
· r)
r
and
(b)
Using the same argument vhich leads to ( 9 . 2 .10 ) v ith the function
---=-2
(lms)
'i3'
c/s
where again the differentiation is performed where es occurs explicitly.
a:-lj3e
"- ) 2 -(w.u
s
s
----e
a+i3e
(e
=
ve obtain
c,n
1
= -
£.
~
= -ns-,..,
s -1)2
Thus
1
( 1 + -_)
n
s
iis (1+ns )
The correction term in (9.6.18) is absent sho,n.ng that if on1~ the mean number of particles is
fixed, the dispersion is greater than if t he total number is known exactly.
(c)
For FD stati stics, the sums in (2) become
l
bl'
=
~
+ e -(a:+i3e1J
= L.
b. (l + e
-a-;3e
r)
r
~hus
and
:J.
(lm)2
=
s
1 ens
p oe
s
=
--~
" 1-'1
- -1 ~
p oes
l
a-lj3e
s
e
= a-lj3e
(e
s+1)2
= -n s (1-:n)
s
82
1
= a:+5e
. s
+ l
e
V
= -ns 2 cln -l
s
We can remove the restrictions on the sum over states by inserting the Kronecker
where the sum is over n
r
= 0,1, •••
for each r.
Letting o:
-
= o:+io:'
o :f'unction
and interchanging the sum and
integral yields
(l)
where
This sum is easily evaluated as in section 9.6 and
-a-f,€
lnt (!!)
(1-e -
=~ 1n
r)
(2)
r
To approximate
(1)
we eA-pand the logarithm of the integrand about o:' =
0 where
it contributes
appreciably to the integral.
1n
e<PT l
(g)
=
~+
1n
l
(g)
= (a+ia')N +
1nZ(o:) + Bl(ia') +
1
2 o:'
- - B
2
½B2(ia
2
1
+
where
i(N+J\)o:'
e
Hence
As in section 6.8, we optimize the a:pprox:i.mation by setting
!
o 1n
N +
or
o:N
and (3) becomes
Substitution into
e--
(1)
=e
~
yields
z
[ (g)
= e
o:ri ,z (
~ a
)
aN
f
7f
(o:) = 0
,2
l:g
"":2 20:
l, (o:) e
1 ]3 1 2
e
-
-
2
0:
2
aa'
4T
or
1n z "" aN +
bl (a)
= o:N -
I
-a-f3€.,..
1n (1-e
-)
r
where we have used (2).
(4)
The vo.lv.e o: is determined by (1.:.)
83
(5)
2
)
+
which gives
CHAPI'ER 10
Systems of Interacting Particles
10.1
Since cr(w) dw
2
cc K
dtt
and for spin waves ro
= AK 2
it follows that
,
Then by (10.1.20)
Substituting the dimensionless variable x = ~1'.m we obtain
The integral is just a constant not involving~.
Hence
10.2
(a)
By
(10.1.18)
3V
2 3
where
1
2iT
2
for ro <
ill
for
But since
,
~
C
ill
>~
we find
In tenns of the dimensionless variables x
1n
z
= ~:,m
= YNn - _2li
~
y3
and y
Jy
= ~~,
1n (1-e -x)
this gives
x2 dx
0
This expression can be put in terms of the :function D(y) by integration by parts.
84
Thus
= 'Y
=
~ - 3N
~
i1wD
3N
-
1n
(1-e-y)
(1-e
1n
-eJT )
+
ND(y)
+ ND ( e Irr)
D' ~
(1)
where keD = ~wD.
(b)
The mean energy is
Thus
E
=
= -N71 + ~:r D(y) = -N71 + 3N kT D(eJT)
(c)
(2)
The entropy may be found from the relation S = k[ln Z + l3E].
Then from (1) and (2), we
have
s
= Nk
[-3
1n
= Nk
[-3
1n
(1-e-y) + 4D(y)]
-eJT
(1-e
) + 4D(eJT)]
For y >> 1, the upper l.illlit in the function D(71) may be replaced by
D(y)
= 1_
y3
!co
o
co.
~dx
ex-1
h
The integral is shown to be 7T '/15 in tables.
D(y)
Hence
=
4
7T 3
y
>>
l
5Y
For y << 1, we can expand ex::= l+x in the denominator
D(y)
=;
2
x a.x
~7)
=
1
y
<< l
From the results of' problem J.0.2, we find for y >> 1, or T << eD when we use y
1n
z
=
......_4T3
N
@¾
+ -"'"-3
.<>.:.J.
5e
D
= ~~ = eJT
'srr4 NkT4
E = -N~ + - 5- ~
D
c!-/
4,n-4
S = - 5 Nk eD
For y
«
1 (T >> eD) we find, using ey :::::: 1 + y
1n Z =
~
E=
er111
- 3N 1n
+N
-IiTj + 3N 1d'
s = Nk
[-3 1n
er/1' + 4]
10.4
The pressure is given in terms of the partition function by the relation -p =
equation
(1)
of problem 10.2
1n Z
Thus
-
_ N
P-
£!1
ov
=~ -
3N 1n (1-e
-eJT )
+ ND
10oV
~
1n Z where by
(eJT)
-eJT
-eJT
+ 3N kT e
1-e
where
oD(GJT)
dD(eJT)
In terms of,= -(v/eD;(deJdT), this becomes
p = N .s.!l
+ 3,NkT D(e frr)
oV
V
D" ~
,
4
37T
p = N 2V + -5-
2!1.
4
V e 3
?' N kT
D
For T »
eD, D( eJT) ~ l
p _ N ,Q:!
-
oV
Comparing the expression for the energy
problem 10.4, ve obtain
-
E,
on
=
+ 3?' N kT
V
equation (2) in problem 10.2, vith equation (1) in
{E+Nn)
p -- N oV + ?' -------~V
86
Thus
a=
10.6
(a)
Fran electrostatics,
(b)
The induced dipole moment in terms of the polarizability a is
(c)
The energy of a dipole
12
in a field Eis
u = -g ·~
2
a:
p E
2
a:
R-6
By equation (10.3.8) and (10.3.16), the log of the partition function is
l.n Z
where n = Njv.
l.n ~!
+ N l.n V +
½NnI
(~)
The first two terms are the ideal gas partition functions while the last is the
contribution of the interaction.
1n
~~ ~
Since the interaction is represented by an additive term in
z, it must also be represented by an additive term in the entropy, S = [1n
k
Z -
~ ~~
1n
z]
Thus the entropy per mole is
S(T,n) = S (T,n) + S'(T,n)
0
where
S'(T,n) = k
[½ n
I(~) - t3
~(½ n I(B_V]
and
To show that A(T) is negative, we must show that the derivative
We
have
(~
but
c/c~
Also
is positive.
;:,. Jo:,
=
c~
o
-~ue -~u -e -~u+l = 1 _ ( ~ > 0
~u
e
since (1-+f3u) is just the first two terms in the expansion of e~u.
and A(T) is negative.
[I(~)/S]
Thus the integral is positive
10.8
From the arguments of section 10.6 ve can write down the partition function for a two dimensional
gas
z
vhere
=
1n
('h2/3
arm/ z
4
N.
zu
U
=
N 1n A +
1~
2 A I(/3)
I(/3) = J'\e -/3u_1)27rR dR
0
Denoting the film pressure by Pr, ve have
Pr =
1 cl
cA ln Z
i
Pr
N
1 if
-I
kT -- A- 2 A2
or
l
I.
2
= - -
10.9
One can write dovn the mean energy from (10.7.3) and (10.7.5)
(a)
E = -Ngµ 0
Here N is the total number of particles.
H. sj
1
Z
= -NS
(1)
lcDc B (x)
S
The consistency condition is
(2)
We solve this equation in the given temperature limits using Fig.(10 .7.1)
T
<<
T
C
Here the slope, kT/ 2n JS, of the straight line is small and the intersection with B (x)
s
is at large x vhere B (x)
s
== 1.
Thus (2) becomes
or
and
2n JS
X=-w
E = -NS kT (2nJS)(l) = -2n NS2 J
(3)
T == TC
In this region kT/2n JS is large and the intersection is at small
x.
We cannot, however,
approx:il!late Bs(x) by the linear term ½(s+l)x {eq.10.7.12) since this vould not cross the
88
line (kT/2n JS)x.
Hence ve must evaluate the second term in the expansion of B (x).
s
small x,
coth x
Substitution in Bs (x) =
B6
½8s+½)coth
~
3
x
x1 + 3 - 45
X
X
½coth ½j
(s+½)x -
(x) ~ ½{<s<½) [cs4)x + }<s<½)x -fycs<½)
Bs(x)
Then (2) becomes
=
½(s+l)x
2
or
1- E ~ ~~
3
½
(S+l)x - (s +s+½) (sit;)
2~JS x
= 52
•
<< 1
yields
2
""½
For
+
·
}
x3
2
- (s +s+½) (sit~) x3
~<½ [½ - 2n JS~+lJ
(4)
In the energy (2) we substitute Bs (x) to first order and use (4)
E
= -NS kT x Bs{x) = - ts(s+l) kT x2
= _ 5NS(S+l)kT
s2+s+½
L-
3kT
[
]
T ""TC
2n JS(S+l~
(5)
T >> Tc
From (4) we see that x=O at kT
=kTc =} nJS(S+l),
i.e., the energy becomes Oat this point,
and larger T gives imaginary x in this approx:i.:mation.
(b)
T >> TC
E=O
Hence
From equations
(3), (5), and (6) we have
~o
C =
~
2fr
=
T
2
15N k T
~- !
nJ(S2+s+½)
3
0
( c)
~
kT
<< TC
T ""T C
T>T
C
89
C
(6)
CHAPI'ER 11
Magnetism and Low Temperatures
11.1
The mean magnetic moment is
2'.(-oE_/oH)
-
lolnZ
oH
r
=~
M = ----"""t,E=-----
2'.e
r
where Z
=Le
r
-t,E
r.
r
11.2
The change in entropy per unit volume is given by
f;;B
=f
H
O
dH
(~)
oH T
o
But from the fundamental thermodynamic relation
dE = TdS - MdH, we have the Max1,ell relation
oFr
(~)
(--2)
dT H
=
oH T
where Sand M are the entropy and magnetic moment per unit volume.
0
X =
A/(T-e) we obtain
oM
Cy/)
C
H0
Thus
f;;S =
J
0
Since M = XH and
0
AH
H -(T-e) 2
=
AH
(T-e)
2 dH = -
H 2
A
0
2
(T-a)?
11.3
The change in the molar volume when the field is increased from H=O to H=H
6v =
But
from dE = TdS -
p dV
f
H
o o
"
(g)
cH T
0
is
dH
,P
- MdH, we see that
where v and Mare the molar volume and magnetic moment.
we find
AB
a H
=------2
0
[T-e (l-faP) )
0
90
Since
M = XH and x = A/[T-e 0 (l;ap) J
£:;.v
and
H
o
={
_A_e_0_a_Ha._H_ _ = _ __A_e"""0_a_ _..,,.
0
2
[T-00 (1~)] 2
[T-00 (1-+a p)J 2
o
H2
u.4
(a)
From the relation
d.E' = TdS
pdV
-
+ HdM where
cos)
=
~ V,T
In the normal state,
M = VXH""
0 since X"" O.
E' = E -
v'Ff-/&r we obtain
c~M)
aT V,H
Thus (oS/oH)V,T
""o.
In the superconducting state,
4-rrM
B=H+-=0
V
or
Thus Mis independent of temperature and (cS/oR)V,T = 0
(b)
C=
Since the heat capacity is given by
T(osjoT) we bave fram
S-S
n
s
(11.4.18)
V
dH
=r.-::HLl'1f
d.T
(1)
where His the critical field.
( c)
When T
= TC ,
the critical field is
C
o.
Thus
VT
dH 2
= ~ (-)
Ll'1f
d.T
-c
s n
11.5
From (1) in problem
11.4
and H=H
2
C
[1-(T/T ) ) we obtain
C
C -C
s
n
=-
VH
2
3VH
2
_c_ T +
c
27rT 2
27rT Ii.
C
T3
C
But since C ==aT + bT3 , we have
n
(1)
To satisfy the .condition Cs /T ~ 0 as T ~ 0 we must set the coefficient of T in (1) equal to 0 1
i.e., a= VHc
2
/2rr
Tc 2.
91
Thus
ll.6
(a)
The entropy is related to the heat capacity, dS = Cd.T/T.
Then since at T=Tc and T::0 the
entropy of the normal and superconducting states is the same, we have
s(Tc)-s(o)
=
f
f +
Tc C d.T
Tc C dT
T
0
=
0
(1)
Fram which it follows that
(b)
Since
dE =
Cd.T, the energy at T=TC and T::0 for the normal and superconducting states is
f
f
Tc
7T 2
En(Tc) - En(0) =
Es(Tc) - Es(0) =
But En (T)
= E (T) and by
C
S
C
(1)
7T dT
0
T
ctr
+
+
,_,n
u..i..
3
C
=
4
dT =
0
a= 3,/1' C2•
7T 2
Es (0) - En (0) = - ~
'+
Thus
ll-7
From the relation dS
=
CdT/1' we have for the entropy in the normal and superconducting states
T
f
=JTc
Sn (Tc) - Sn (0) =
c ,dT = )'Tc
0
S6 (Tc) - S (0)
s
O
ctr 3
efdT = _ c
3
Since the entropy is the same for both states at T=0 and T='rc' it follows that a=3r/l'c2 •
Hence at T=T
C
92
CHAPl'ER 12
Elementary Theory of Transport Processes
12.1
6
(b)
6
(c)
(a)
,E,
(b)
,E,
( C) {,
(a)
The equation of motion for an ion in a field
(a)
6
12.2
mx
Hence
1=
=
t2
is
e!
=
e
e
X=m
and
By ( 12. l. 10) ,
C:
t2e -t/r d.;
t
2
2
= 2T2
0
e!
and
(b)
X =-T
2
m
The time to travel a distance xis
t = ~ ; =£
T
Thus the fraction of cases in which the ion travels a distance less than xis
s=
f
,/2'
T
0
e
-t/T dt
-
T
= 1-e
-V2
12.4
The differential cross section is given by
cr(n) = -2rrs ds
d!2
where d!2 is the ele::nent of solid angle ands is
the "inpact parameter" as shown.
From the sketch,
Since q> =
90-e/2
s = (a + a ) sin q>
2
1
and dn = 2rr sine' de', it follows that
-(a +a ) sin 81/2
ds
an
1
=
2
47!" sin
93
e'
=
.757
a(e)
Hence
= -2rrs
(a1 +a2 )
ds
an=
2 sin
e 1 /2 cos e 1 /2
2 sine'
12.5
By arguem.nts similar to those of section
12,1, one has that the probability that a molecule
travels a distance x without collision is
where t i s the mean free path.
p = e -x/~
Letting P
= .9
and
t = (ff
ncr)-l yields
.9 = x ./2 no = x ./2' pajkI'
-4
300°K gives p = 1.14 X lJ llllll Hg.
-ln
Letting x =20 cm and T =
12.6
The coefficient of viscosity is
(1)
where we have used
t
= (
~ n cr 0 ) -l. To determine the cross section,
we
not e that the volume
per molecule is A/NAp where A= 39.9 gms, the atomic weight, and p = 1.65 gm cm- 3, the density
in the solid state.
T::us the radius of' the molecule is
r
3
=
A
1/ 3
(47r N P)
A
and
From
00 =
(1)
4rrr2
=
,,0- 4 gm cm-l sec-1 •
we find 11 -- 1.1 x ..,_
5.6 x 10-15 cm2
Th
· en ·h
.., e rat·io of the ca1 cult
a e d va 1 ue
~ 11 t o
0.1.
the observed value, 11 = 2.27 X 10 - 4 gm cm -1 sec -1.
,is
12.7
l
The coefficient of viscosity is proportional to T2 since 11 = 'IIN/ 3 ..f2 a
0
Thus if' the temperature increas s, the terminal velocity decreases.
The velocity is independent
of pressure.
12.8
The force on the inner cylinder is the pressure p times the area A= 2rr RL,
.9- = pA = 2rrRL
''"'U
11 ~
1
and v = (8kr/71Ill) 2 •
We make the approxillla.tian that the air at the outer cylinder is moving with it, i.e., at
velocity ro(R+o) ~ cuR, and the air at the inner cylinder is stationary.
Thus
and the torque is
..7R
G =
The viscosity is given by Tl = mv/3 ,.[2 a •
(b)
Since air is ma.inzy composed of nitrogen, we
0
let m = 28/6.02
X
1023 , the mass of a nitrogen molecule.
Putting a0
~
10
-15
2
0
cm and T = 300 K,
we find
T) = 2
and
x 10-4
gm
cm-l sec-l
G = 8 dyne cm
12.9
(a)
We let
wherex,y, and z are to be determined.
Letting L = length, T = time, and M = mass, the
dimensional equation becomes
But since C = FRs, it has dimensions MT- 2Ls+l
Thus
which gives
On
x+(s+l)z
= 21
x+2z
= 01
y + z = 0,
solving this system of equations, we find x = 4/1-s; hence the cross section a
0
is proportional
to v4/(1-s).
(b)
The viscosity is given by T) =
riv/3../2
a and here a
0
0
~ v4/l-s.
Since velocity depends on
the square root of the temperature, we have
1
-2)
T) ~ T
1
T4;tr.-s) = 'I!°C..+7hV2(s-1)
12.10
(a)
Consider a disk of radius rand width
at the point x.
'
t::;x.
There is a pressure difference
2
.C:.p across the disk giving force -rrr t:.p and a
.
ou(x,r)
viscous retarding force 27rT t:.xT)
or
•
\.
__,.,
\
I
Hence
~x-1~1-
the condition that the disk moves without
acceleration is
95
But since t::.p
~
- cpfx)
ox
t::.:x., where the minus sign is introduced because p(x) is a decreasing
function of x, we have
2 op(x) _ 2JT
ox -
71T
6U(r,x)
or
r7J
Assuming that the fluid in contact with the walls is at rest, it follows ihat
f
u(r)
~J
ox
::,_~/...,.\
1
du'
0
For a liquid,
~~
r,
r
r'dr'
a
The mass flowing per second across a ring of radius r' and width
dr 1 is pu(r')2rrr'dr 1 where pis the density.
Then the total mass flow is
a
f
M=
pu(r')2rrr'dr'
0
= li2 £.....
TJL
(p -p)
1
2
Ja
2
2
(a -r' )r'dr'
0
4
= ~ ~ ~ (pl-p2)
(b)
For an ideal gas, the density is p(x) = (µ/RT)p(x) whereµ is the molecular weight.
Thus
the mass flow is
.
Since Mis independent of x, we obtain
L
•f
ax
M
= -
•
M =
p2
µa
TJ
0
w},..ich yields
-s m 1
4
1f
7r
16
p
~ a
TJRT
4
L
(
p dp
1
2
2)
P1 - P2
12.ll
Let T(z,t) be the temperature at time t and position z.
thickness dz and area A.
By
We consider a slab of substance of
conservation of energy, (the heat absorbed per unit time by the slab
is equal to (the heat entering the slab per unit time through the surface at z) minus (the heat
leaving per unit time through the surface at z+dz).
me oT/ot =
pA
Since the heat absorbed per unit time is
dzc oTjclt, where c is the specific heat per unit mass and
p
is the density, we have
pA dz
c~
[Qz -Qz+dJ
= A
~
• A [-•
- ~•
~
- •
~(~)0]
and
u.12
The wire gives off heat fR per unit time and length.
In order that the temperature be indepen-
dent of time, a cylindrical shell between rand r -kir must conduct all the incident fR heat.
Thus
T
-1
o dT = r2R
2rrK
T
or
6.T = T--'l'
o
f
b dr
a
r
2
= I R 1n
27rit
~
a
12.13
(a)
The heat in.flux per cm height of the dewar across the space between the walls is
Q = -2rrr
where r is the radiu.s, r
We let
~
al'
(1)
K-
dr
10 cm.
oT
°ar
_ 298 - 273 _ 4 oK/
- 10. 6 - 10 - 2
cm
t:1r
~
t:.r
by (12.4.9)
The specific heat per molecule is c = (3/2)k while cr
7~
is v- = (8kT/ 7l!ll)-I
•
0
= 47rR
2
= h7r(lO
-8 ) 2 en.
2
Choosing an intermediate te~perature T = 286 0 K, we find v
The mean velocity
=
l.2 x lO 5cm/ sec.
Substitution of these values into (l) yields
Q
(b)_
= l.2 watts/ cm.
The thermal conductivity remains approximately constant until the pressure is such that
the mean free path approaches the dimensions of the container, L = .6 cm.
At lower pressures,
the conductivity is roug~ proportional to the number of molecules per unit volume n.
the heat influx is reduced by a factor of l.O when
n =
1
l0./2 L cr0
ham p = nl!:T, we find p ~ 3 x 10-3 mm Hg.
97
~
14
10 molecules/cm3
Thus
(d)
(e)
cr cr
~
.frr 1'I} Jµ NJ'kT
gives cr
3
.
= 7T<l2gives
~
= l. 8
l l x io-l5 cm2 ,
l =
10
X
2
cm •
•
-8
cm, d 2 = 3.0 x lO
-8
cm.
l2.l5
The gases are uniformly mixed when the mean square of the displacement is on the order of the
square of the dimensions of the container.
2z
=
We have, after N collisions,
r.
i
since the second term is zero by statistical independence.
Then
2
=
Z
where N = t/-:.
22 2
22
v -r N=
v rt
3
3
Since -r = ,f,fo = 1/,/2 n cr
v, one
has on neglecting the distinction between
2
and v,
z2
2
Substitution cf z
h
2
= -~-J v kT)t = ~
5
3412\crp
= 10 · en, cr"" lC
-l~
kT
CTp
J~
7r
kT
v2
t
m
2
"cm, and the given temperature and pressure yields
sec.
12.16
A
molecule undergoes a momentum change of 2µv z in a collision ,.;i th the cube where µ
the reduced mass.
= rnM/m+.'-1,
Then the force is given by the momentum change per collision t:imes the
number of molecules which strike the cube per unit time, summed over all velocities.
,z
~
......:Ttot = 2µn (27r)
1/2
2100 2
L
vz
0
We may approximate the exponent since
v >> V.
Thus
e
r3m
- -(v
-V )2
2
z
dv
z
- ~ (v
e
2
z
-v)2
z
m8
and
(1-1-f)mv v)
1/2
-.5tot=2µn(2rr)
2f°"vz2 e - f3m2
2
V
L
z
(1+/3mV"zV)dvz
0
The first integral is just the force on a stationary irall as in section 7.14.
force is then
The resistive
f3mv 2
--
z
2
e
2
=µnvVL
by Appendix A.4 and 7.u.13.
Since we expect M >> m,
M
~ =
dt
V = V O exp
-mn
µ.
""'m, and the equation of motion becomes
v VL2
t~
v l~
n
The velocity is reduced to half its original. value in time
C
=
M 1n 2
- 2
mn v L
CHAPl'ER 13
Transport Theory using the Relaxation-Time Approximation
13.1
By (13.4.7)
't" V
2
z
The value of the integral :is unchanged if v 2 or v 2 is substituted for v 2 •
X
y
Z
v
2
=v
2
X
+v
2
y
+ v Z 2 , we obtain
2
ael = - -e3
J
a3 v
-
Hence since
~e -rv2
The quantities-rand ogfc,e are functions o~ of v.
yields
99
Thus the substitution d 3v = v 2 dv sine de d~
The f'u.nction g becomes the Maxwell velocity distribution
c~t'£!)
In terms of ; =
(2kl'/m)1/2 ,.
__ -3/2 ne-(v;v)
0€ -
(1)
e
l
kr)
(-
this is
~
Substitution into
2
3/2 - mv
2kI'
2
....5
2rr
V
of problem 13 .1 with s = v
;v yields
2
cr = ~ {-r)
m
where
{-r)cr =
8
yJ;
f
cr
2
co
ds e -s
"'
s 4 -r(vs)
0
]3.3
Since
€
(u}+ u/ +Uz2 ),
= ½mrf = ½n
we have
o€
og _$
~ - 0€
Thus
=
T)
-mf
3u $._ u 2 u -r = -m2
- oUX__ .z X
d
f
_$mu
dUX - 0€
X
3u $ u 2 u 2 -r
0€ Z X
d
rn· spherical coordinates, the components of velocity became
u
u sine cos
=
X
r
2
Therefore
T) = -m
JC')
2
= -m 7T
2rr
cos 2 cp dcp
f
uZ
=
e
u cos
= rrdu sin e de dcp
d3E
and
cp
sin3 e cos 2 eae
1T
0
f"" dU ~o
O€
0
u6
J1r [cos e sine - cos 4e sin e]de [(X) dU fc
2
T
u6 -r
O
2
= -m
But
since
41r
15
f""
f 7rdcp f 2rrsin e de
0
dU $
0€
0
u6 -r
Im-, it follows
=
0
(l)
that
2
TJ
= - ~5 J1rdcpf ~in
0
0
100
edef
0
dU $
O€
u6-.
=
Substitu+.ing the expression
-J
i
= -2rr
-(u;v)2
ne
2
2 into equation (1) of 13.3 yields
from problem 13.2 where;= (2kI'/m)1/
nkT (,)
T] =
-where
(' )
T]
15..f;
!
co
ds e
-s
T]
2
s =
u;v
0
13.5
Ptttting '
~
-)-1
( "le. n cr v
0
into TJ = nkI' ,
and using
v=
1 2
(8kT/7!lll) / gives
This is larger by a factor of 1.18 than the mean free path calculation (12.3~18) and smaller
than th-2 result of the rigorous calculation (14.8.33) by a factor of l.20.
13.6
We assume the local equilibrium distribution
/3 3/2
1
2
g = n(~)
exp [- 2 pmv]
(1)
where the tenperature parameter /3 and the local density n are functions of z.
Then in the
notation of section 13.3,
df(o)
df(o)
~ = - dt =
0
on
~
-~ ~
- /3
on
0
~
0
dz(t)
og ~ dz(to)
o
0n
= - dn 6Z dt
- ~ z
dt
0
cg
0
where the last follows since dz ( t
0
)
/dt
O
= v z.
(13.3.5) becomes
M
df(o)
=-w-
This expression is independent of t'.
J"" -t'/-r
e
dt'
0
(2)
or
Using (l) this expression becomes
181
Thus
f
ll
Ln:
= g-g v z...
~
2
on
3 ot3 mv
oz + 2p oz - 2
(3)
c1~
It remains to determine on/clz for which we use the condition
vz
or
J
d
3
2-
g vz -
J
3
d ::. v z
2
-rg
=
J
l:n
3
-
d v f v
11
on + 1.. 06 Ln az
2p az
mv
2
z
= 0
c)~
~
2
=
0
The first integral is zero since the integrand is an odd Iunction of v.
J
d3y
where we have assumed
T
z
V/
[¾ ~ +(¾ ½- m;
g
independent of v.
integrand is replaced by vx
2
or vy 2
2
)~]
= Q
By symmetr'J this equation also holds if v
Then since v
2
=
vX
2
2
z
in the
2
+ vy 2 + vz , we have
or
on
substituting (l) arid d3y = v 2 dv sine da d9, and canceling co.Cl!Ilon factors.
The integrals
are tabulated in appendix A.4
l
or
n
Substitution of the equation of state n
05
oz
= ~
= p,!'kI' = pp
C:o
-:=-
or
The pressure is independent of z.
l
on
dZ
c:.z
(4)
yields
= 0
Then ( 3) c...'1d (r) give
c1s
f = g-gv z -r oz
[~P -m;~
g
+ gv z -r
T
dT
dz
l,2. _ l:_ mv-27
l? 2 :riJ
13.7
Since there are no external forces and f is independent of time, the Boltzmann equation becomes
of
of
f-f(o)
V • ~ = VZ a'z = -r
J.D2
where f(o) = g is the local Maxwell distribution of the previous problem.
where f(l)<< g we have
V
~
z clz
f = g-v ,.
Thus
z
=
~ on
V
+
dZ
z on
V
~~
= - f(l)
dZ
z Op
Letting f = g+f(l)
..
rlEn$ ~oz + ~clg ~oiJ7
This is identical to equation (2) of 13.6.
The ·rest of this calculation proceeds as in that
problem.
13.8
The heat flux in the z-direction is given by
=!
2
mf
d 3v f v v
z
j
~.2._!~
gv-r -clT
f=g+-zT
oz
where
2
2
2 kT
The integral over g is O since the integrand is an odd function of v.
Qz =
m,- 1 clT
4 T dZ
1
~ 3;~ f
1.:1 (27r) J
1
3
2
d Y v vz
2
We can exploit the synmietry of the integral by replacing v 2 by v 2/3.
z
-
2
Consequentzy
z
~ r..~-t3mVJ21
-
e
2
Then the onzy angular
2
de-oendence occurs in d3v = v dv sine de dcp whic.h yields 47TV dv when the integration over angles
-
is performed.
Thus we obtain
'Tr !llT
Qz =
3T
6 J I.L5 Jco
o dvv e
clT [ ~ 3/ ~
~ L(2rr)
1
2
~
-;3m
1
J
Jcoo dvv8e - ~
The int2grc.ls are tabulated in appendix A.4 •
~
"',z
.
=
IT m,3 T
oT
oz
3
C
(m3) / ~ ~-~
E
J t lo
2rr
_2_.nk2r,,1
2
~
2
( pm)
7r
7 2
/
-~m 105
...,
32
1r
-:,,
oT
oz
,-
13.9
Th~ mean free path argument has
K
since
t
=
v,- and v
1 2
= (8kT/7TII1) / •
= -3l ncv
t
2
4 nk T
= - - - -r
7r
m
Thus the constant relaxation time approximation yields a
103
result greater than the mean free path argument by ~
2
7T
=
1.96.
13.10
From problems 13.4 and 13.8 we find
~ = (5/2)(nksr/m)T
T)
where c =
(3/ 2)k,
n kl'
=
2~
=
2 m
T
the specific heat per molecule.
5C
3 iii
The simple mean free path argument gives
C
m
T)
1.3
Experiment shows that the ratio (K/ TJ)(c/ m)-l lies between
and
2.5
(see discussion after
(12.4.13)) so that the constant relaxation t:L..-ne appro::d..mation yields better agreemer.:t with
experiment than does the mean free path argument.
13.11
In the absence of a temperatur.e gradient, the eqv.ilibrium distribution of a Fermi-Dirac gas is
(1)
In the given situation, f is independent of time but ,rill depend on z because of the temperature
gradient oTfe z along the z direction.
·r his temperature gradient alone would cause the electrons
to drift, but since no current can flow, there is a redistribution so as to set up a field !(z)
to counter the effect of the temperature gradient and reduce t he mean drift velocity v
z
to zero.
Thus the Boltzmann equation becomes
V
of + e C ( z)
z
dz
m
cf
ovz
f-f
=
0
T
We ,rill assume that the temperature gre.dient is sufficie..'ltzy small so that f ca.n be written
f = g+f(l), /l) << g whe!'e /l) !'epresents a s:::iall departure from local equi.lioriu."D. •
.,.(1)
J.
Thus
(2 )
T
It is convenient to write this in terms of the dinensionless parameter
X
= ~(€-µ ) = ~( 1
2
~
Thus
~=
ov z
clg
dX
(€-µ)
ox
2 -µ )
i
~= ~
5mv og
~= .
z dX
104
z
(2)
IllV
i
Then from ( 2) we find
It remains to find
vz
C from the condition that
= !}'d3v f·v
n
-
The integral over g is O since the integrand is odd.
z
= 0 where f is given by (4).
Letting T = TF = const. for electrons near
the Fe!"!li level, we have
or
(5)
The second integral can be expressed in terms of n, the total number of electrons per unit
~
1
cg
Since ox= ~v cv
volume.
z
z
00
J
00
oo
C = -l }' dv v £5.._
d
l
1
·...,,
dvz v Z2 ~
dX pm -co Z Z dV z = -/3m [v Zg] -co - -/3m
3
2 -.;;-::-:=-clg
1
dvv
z ox
pm
J
3 v g = -nd
pm
Proceeding to the evaluation of the first integral in
only of v
2
so that we can replace v
00
-co
g dv Z
Since g=O at ±co, it follows that
where we have integrated by parts.
J
J'
2
z
by v
2
(5),
(6)
we note that x and ogfc,x are functions
The substitution d3v = v dv sine de d~ yields
/3.
2
4n v dv when the integration over angles is performed.
J
Thus
2
3
d !_ v 2 x
~
=
~7T
l"°
2
dv v x
fx
2
Using x = 13(½ mv -µ) this becomes
~
3
~
c1-//2foo dx x(x+/3µ)3/2
pm
-co
dX
(7)
From the properties of the Fermi distribution, we know that og/ox ~ 0 except where€~µ or
x
~
O.
Ti e integrand contributes appreciabzy to the integral only L'l this region so we can
expand the factor x(x+/3µ) 3/
may be replaced by
-co
2
about x = 0 to approximate the integral.
with.negligible error.
2
x(x+pµ)3/2
and,
~ (t3µ)3 / 2 (x+ l2 ~)
/3µ
e
X
from (1)
105
Also, the lower limit -13µ
Hence (7) becomes
2 5' / 2
0-,,=.!!.(-)
3 ~:m
!co
dxx(x+pµ) l / 2
r2
2
= 'IT / 3.
= -23/2
- - µ 3/2ml/2
5Tf2
X
These integrals are the
and
rx
-a,
r 1 and r 2 defined
Combining the results of
[!co
t3fi3
xe X d x +3(eX+l)2
2/3µ
-co
by e~uation (9.16.9).
(5), (6),
(8),
and
la,
x2 eX dxj
-c::o(eX+l)2
It is shown there that
r1
(8)
= O
we obtain
and (4) becomes
f = g-v
z
T
~
(9)
F dX
Having found the distribution f, ve can nov calculate the heat flux Q.
z
Qz
=n
<vz(½
2
where
2
(vzv )
mv2 )) =½nm
1
(v v ) = -n
z
J
d 3v v v
z
2
f
The integral over g is zero since the integrand is odd,
Hence
Q
= - m-rF
2
Z
~
d 3v v '2..2 x~ t3 dZ ~
dX
1 9.§.
Z
3
1
( 2µm ) l /
3
n fi3t3
2
f
7
d 3v v 2 ,l~
- Z
d~
(10)
The calculation of these integrals is similar to the steps leading to (8),
2
.
replaced by v /3 and the integration over angles performed to yield 41r,
(10) becomes
4nJ=
-
3
dv
V 6X
0
7 2
d = -2rr (-)
2
/
~
OX
3/3m
The second integral in (10) is
J
d3V V 2V2 ~
,..
Z
dX
=
47Tf a, dv V6 og
3
dX
0
1c6
1=
..0:,
X
(x+pµ) 5/2
Again v 2 can be
z
The first integral in
~
X
dx
Here we have neglected the squared term in the expansion since it is much less than 1.
integral is I
0
+ (5/2 ~µ)
r1
The
= 1.
(12)
we
find
-rr2 nk2.r
Thus
K =---T
3
F
m
13.12
Combining the results of the previous problem and (13.4.12), we find
2
K/cr
el
k 2
= 1L_
(-) T
3 e
At 273 oK this becomes KIcrel = 7.42 X 10 -u erg cm -1 deg -1 or in more standard units
KI crel = 6 •68 x 10
-6
watt obm deg
6
10
watt ohm deg -1
X
•
0
Experimental values at 273 Kare
Ag
Au
Ql.
Pb
Pt
6.31
6.42
6.09
6.74
6.85
element
K/crel
-1
w
Sn
6.88
8.30
Zn
6.31
CHAPrER 14
Near-Exact Formulation of Transport Theory
l4.l
The mean rate of momentum loss per unit volume due to collisions is, by (14.4.22)
(6E)
=
JJJ
d3! d3! 1 an' ff1 vcr (v,e')(6E)
(1)
! !1 n
where we assume central forces.
We first perform the integration over angles by changing to
center of mass coordinates.
Thus
(6p) = m (v'-v)
-
- -
107
becomes, since v
=c +m
~ V and v' = c +
-
~ V'
m_
(A;e_) = µ(V' -V)
mml
whereµ= - - , m being the mass of the ions and~, the mass of the heavy molecules. Since
m-tml
the mean values vx and vy are zero, it is clear that (Apx) = (Apy) = O. Hence we are left with
(Ap ) = µ(V ' - V ) and
z
z
z
(2)
dn' cr(V,e')(Ap )
z
V' is given in terms of V and the scattering angles.
z
Using the coordinate system illustrated
z
in Fig. 14.8.1, we see that
Vz' = V'cos(y' ,!) =
since V' = V.
V
cos (y' ,!)
Inspection of Fig. 14.8.1 yields
V cos
(Y',~)
=
v[cos(y,z) cos e'+sin e'cos ~' sin(y,z)]
=
V cos e'+ V sin 0 1 cos cp' sin (y,E._)
z
µ(V z '-V z ) = µV z (cos e'-l)+µV sin e'cos cp' sin (Y,!)
and
Substituting into (2) we obtain, since
J~os
cp'dcp' = O,
0
For hard spheres, a is constant and
27rµVzcr J'\cos e'-l)sin e'de' = -4n µVzcr = -µVzcrt
0
where a. is the total cross section.
Equation (1) then s:L~plifies t o
t.,
-(Ap z) =µat
ff
Vzv ffl d3! d3~1
(3)
! !1
where
Here u is the mean z component of velocity of the ions.
we let
a=
II$
2
To avoid carrying repititious constants
(4)
and
l'J8
f can be simplified by making the approximation that v
z
>> u, then
(5)
We again express! and !i in relative coordinates
m
1
V = C + --V
m+ml -
v =c-~V
-1
m+m -
1
Thus
On completing the s~uare this yields
(oml -o;lm)
(a-+al) [ .£ + (a+a ) (m+m )
1
(a-+al)
i2
1
001
+ - - v2
a+al
where
K =
+
-
C
(~ - a1m)
(6)
V
(a+al) (m+ml)
Also
= K
a1
+
z
a+a Vz
1
(7)
We wish to put the integral (5) in terms of y and the variable~ defined by
shown (14. 2 .6) that d3y1 d 3y = d3£ d3v.
(6).
It has been
In terms of K we have
d3£ d\ = Jd 3~ d3y
It is easy to check from (6) that J=l.
3/ 2
- (0.pz) "µat nnl
c~:)
ff
Thus (~ becoraes
_1_ v2
e-(~)
i2
e al->a
~+2a u Kz+2u ;:;l
vJ
The first two terms in the brackets give integrations over odd functions and hence do not contribute to the integral.
- (6pz ) -
~
n3 at
Changing to spherical coordinates gives
(001)5/2
nn1
(a+ai)
~
L7T
Jex, e -(a-+<-v...)x2~ J,
O
[J°' e
0
.:he integrals are found in appendix A. 4 and we obtain
109
ool v2 5
a-+a1
v dv
f
n
2
27T cos e sin
0
mml
whereµ -- m-+ml.
The momentum balance equation is then
ne e + (6.p ) = 0
z
which gives
Since j
= neu = oel C: , we have
14.2
(a)
We will assume the distribution function
(1)
f = f(o)(l+qi)
where
(2)
vz
It is shmm in problem 13.6 that the condition
= 0
yields the relation
(3)
and that the pressure is independent of z.
(b)
The Boltznann eq_uatior: (ll".7.12) becomes
~Co)
DI
of CO )
=~ +
cf(o)
~
.p
~=""-0
:Because the situation is tir.ie indepencier:t and there are no external fo:::-ces, we are left with
(~f3 ~;) f(o) v
or
on using
(2),(3),
2
(1 -
~;v-j ff
=
d3yl. dn 1 f(o)fl. (o)Vo
6~
(4)
and (J.4.7.10).
It remains to determine the :f'unction ~- The left side of (4) leads us to expect that a good
approx:i..n:.a.tion will result if we choose
110
where A and a are constants.
¢ must satisfy the restrictions (14 .7.14) to be a good approximation.
J
These become
2
d 3:::: f( O) Vz (l-av ) = 0
r
J
d 3v lf (o) V
-
V
z
(1-av2 ) = 0
S::..rnilarly only the v
The first and third integrals are zero because the integrands are odd .
component of v in the second relation will give a non -zero integral.
deten:iine a.
z
He use this condition to
That is,
2
3/2 / - ~~v
0
(:)
on
USi..."lg V
z
v e
dv
= V COS
ki'
-=-=a
m
mB
5
a=
and
0 = A
V
z
2
(1- m,5 v )
(5)
5
We will use the procedure of section 14.8 to evaluate the constant A.
Multiplying (4) o;:; i and
(6 )
'I'he integral on the left is ee.si.l~y ev~luated.
r
vv
= -n
d 3V .;.""" ( ::)
-
26m
1
- (·-)5n(Qm)
5
f3m
,.,
1·le have
[ 2 - -2Bn
_ 2 V 2 + (Br.:)
~- V
-::z
:;)
z
;
2
pm
+ (-;:-)
)
2
2
V
J 17T
2rrd~
cos 2 9 sin 9d9
0
0
r
co
mB
n (2rr)
3/2 8
v
e
dv
'-' 0
The int2gral o:f the second term in (7) was evaluated previously.
to be 7n/ 5pm.
(7)
z
The third term is easily shown
~hus the integral becomes
n
pm
(1-2 + -7) - -2 -n
5
- 5 f3m
We split the right side of (6) into an integration over the angle
lll
(8)
n' and over~ and ~1 •
We
define
J =
fu,an'
V cr 6[vz (1-
rJ
v
2
Then using (8), equation (6) becomes
(9)
f""
2To evaluate J, we express 61v (1-av )1
1...Z
j
1
a=
pm/5).
in center of mass coordinates (we have reintroduced
By (14.7.11),
Using v = c +
½y_ \v1
2
= 5:.. -
½y_,
we have
'2
L
?
v (1-o;v ) + v 12 (1-av )=(cz + 2'z)(l-ac--o:5:..•y 1
2
2c
-
z
o:2
4
)+(c 2 -
V
b
2 ,)(1-ac
2
2
V
-+a.:::"Y.. - a 4
)
c c2 - ~ c
z
2 z
v2 - o;V z-C•V-
2o;
Then c = c' and ly_l = l-y_ 1 I gives
-a (Vz '
2
6jv (1-av ~ =
L. z
.:J
(lC)
c•V' - V c•V)
z- -
CaIT'Jing out the product in ( 10) •,;e find
J =
-af
2rr
0
f
· J
0
7r
dco'
sin e'dB'aV
2
1c
[x
where the coordinate system is shown in Fig. 14.8.1.
first integral is (14.8.7).
We notice that, except for constants, the
Consequently, by (14.8.20), these terms yield
22 o;
T)
J
The second integral involves or..ly a change of the index x toy but
is otherwise the same as the first.
where cr
2
(V 'V '-V V )+c (V 'V '-V V )+c (V ' -v ~
z x
z x
y z y
z y
z z
z
cr
+ c V)
(ll)
J-rr sin e'de'a Ve z (v z ' 2 -vz2 )
(1 ?)
TJ
VV (c V
z
xx
yy
is given by (14.8.21).
It remains to find
-a
f
2-rr
0
dcp'
0
fy (14.8.18)
A A)2 -v 2
Vz ,2_v z 2 = Vz 2cos 2 e'+2V V sin e'cos e'cos~' (A
.S.'~A) + V2 sin2 e'cos 2~'(t·~
2
2
112
f
and
2
2
2
2rrdcp'(V 12 -v ) = 2rr V (cos 0 1-1) +
z
0
To eliminate
s,
z
-rrv2
z
sin2 0 1 (;·~) 2
(13)
- -
we write
or
and (13) becomes
The last term in J is then
(14)
and
(ll) and (14) give
J =
?
3
(c V V . + c V V + C V 2 aav
xxz
yyz
z z
TJ
-- J2 a a V
TJ
(V (c•V)
z - -
- C
-
C
t)
(15)
z 3
t)
z 3
We now evaluate
The evaluation of thi s integral is straightforward though tedious.
=n 2(mS)
-27r
2
v z(l-av ) =
C
z
-ac
z
C
2
a C vV)
- ac z (c•
-- - 4 z
?
3
+ l V
2
(16)
z
This expressicr.. is multiplied by J in the i~tegrc.nd .
-
a
2
V
z
JJ' -
-
TJ
v- 2 + a~ 2 C 2_2
a
V + l2
z 3
3 z
2
Since f(o)f (o) has the si.!llple fozr.:
1
(16)
2
- Ct2 V z
c •V
C
2
- Ea V2..v z
C.!l
e-~.. e:; flL~Cti8:: i::
This yields explicitly
~tz 2y Z 2-ac2c Z2..v Z2_ ~
2v2y 2_.g
i+ Z
Z
c.
-c
:i..'1 this theory .
C
z
.,,(o)_~ (o) is clee.:r-ly
-'-1
c e.nd V; hence we need keep only even terms in v (l-crv 2 )J.
I=(f'a.3c d3v f(o)fl(o)(J2 a a V)
In the center of mass system
2y 2y
(c
X
4 7a
zv +
O
C
2+c
X
Z
2
2 ?~
z v z --Y-
y
2y 2y
y
2y
2+c
Z
Z
z4)
(17)
all these integr~ls are the standard ones encou..>1.tered
We quote the result of integrating over c.
ll3
In (18) the terms appear in the
order that they appear in (17)
I = [
i a] [
n
2
~~
2
(~/] [
l
(-)
5/2]
X
~
(18)
J
d3v
~v2
~O
-
- e
TJ
2
[vv 2 _ 2~
5a vv 2 _ a v3v 2 _ £ v 2v3- y_ + 5a v3 + £.... v5 + a v3v 2]
4
2
3
6~
12
b
z
z
z
z
z
The integral becomes on subs~ituting a= ~/5,
J
=
!
00
5
V crTJe
~t
~
TJ
dV 2rr
0
f
2
~
[ _z -
3
V 2v3 - v
b
bO z
2
~
7!" (cos2e
__2
2
2- - bO vcos e
+
- b1
m8
60
v5]
II$v2)
+ bC5"""
sin
e de
O
~
=
vv
~r
d3v e -
(-
~v2
!CD
~
3) II$
v7crri e
ay
8
0
We let s =
=
½-/mi
-n-
= -
cr
and
~ (-8 m/l)(.J~y
102471"
where
V
TJ
½f
2
"a ~)
ds .-s
5
7
-
_l_
(II$)3
is given by (14.8.26).
Substitution into (18) yields
I __ 32
-
n
25
32 n cr
_
t
~?
j_
cr
25 7!"1/2(II$)3/2
and by (9)
A = _
2
(!!!:IT)
1/2
o~
~
TJ
TJ
dZ
~
25
(!!!:IT) l / \ (lv2) o~
~
z
)
oz
ncrTJ
To find the coefficient of thermal conductivity, we eV2luate the heat flux
Q
z
= n (v z (12 mv2 ))
=
½n
J
=
k2
Af d3v f(o) v 2v2 (1 - II$5 v2)
=
½m
A
d3y :f(o) (1+1>)vzv2= ½-n,
-
z
G~)2-~((~~~
ll4
J
d3y f(o)~ vzv2
Q
Thus
=_
nksr2 A =
z
m
25
(..1L)
32 cr
l/2
~
dZ
!!$5
TJ
= _
25 ~ ~7r kl' dT
32 -cr
ln
dZ
TJ
Therefore the coefficient of thermal conductivity is
(c)
:By (14.8.32)
Renee
l4.3
(a)
(b) :By (12.4.2), the simple mean free path argument gives
C
K
- = - =
TJ
m
l.50 -Rµ
Hence _the Boltzmann equation gives a value for K/TJ greater than the mean free path calculation
by a factor of
2.5
(c)
~ X lQ-7erggm1aii"]
TJ
Ne
A
Xe
l.55 .780 .237
J.. 55 .785 .250
Calculated
Experimental
14.4
3 2
Let t i ng f = n (:) / exp [ -
½Bmv2 ] ,
we have
Performing the integration over angles yields
= 47!Il
ci)
3/2
l
= 2
V
dv e
and by appendix A. 4 this is
ll5
- ~
2
[
ln n (~
3/2
-
~
2 ]
N[ V
3]
- l nN- +3
- l n T -3
- l n2rrk
-m+ - r
-v
2
2
LJComparison with (9.10.10) shows that except for a constant, this is -S/k where Sis the entropy
per unit volume of an ideal gas.
(a)
Differentiating the function
H =
dH =
dt
yields
J
3
d _!. f 1n f
J :!.. dt
d3
elf (ln f +l )
elfjdt can be found from the :Boltzmann equation.
(1)
We shall assume that there are no external
forces so that the distribution :function should depend only on the velocity .and time.
Then the
:Boltzmann equation becomes
elf
at
Substitution of
(2)
into
(1)
dH =
dt
=ff d3..:.J.v an' Vcr (f1 'f'
(2)
- f f)
1
yields
JJJ
a3-v d3-1
v an 'Vcr(f1 'f' -f1 f) (ln
(3)
f+1)
This integral is invarient under the interchange of:!.. and ~l because a is invariant under this
interchange.
When the resulting expression is added to (3) we have
(4)
Similarly, this integral is invarient if the initial and final velocities are interchanged because
cr is invarient under the inverse collision, i.e.,
Thus
~
dt
=
!fff
d3-v
2
1
d3-1
v '
an'
V'cr (f1 f-f 1 'f') (ln f 1 'f'+2)
But V=V' and by (1 4 .2. 8) , d3:!._ I d3_::1I = d3:!._ d33
1
dH =
dt
!fff
2
3
3
d v d v
-
-- 1
an'
Vo (f f-f 'f' )(ln f 'f' + 2)
1
1
1
(5)
Adding (4) and (5) gives
~~
=-
½JJJ
3
d v d~
an'
Vo
116
(r1 1 r 1 -f1 f)(ln r1 'f'-ln
f 1 f)
(6)
(b)
Letting f
1
'f'
=y
and
= x, we see that since (ln y - 1n x)(y-x);::: O, the integrand
f f
1
in (6) is greater than zero wiless f
1
1
f
Hence dH/dt < 0 where the equality holds when
= f f.
1
1
the integrand identica.lzy vanishes.
14.6
The equilibrium condition is
1n f = Af B mv
X
X
+ B mv
y
y
+ B mv
Z
Z
+
c-12 mv2
(1)
This can be written in the form
1n f = -A'(v-v)
-
2
--0
+ 1n C'
where the constants A', C' and the constant vector v
f = C' e
Hence
v
--0
-A'
1
(v-v
contain the constants in (1).
- --o
is clearly the mean velocity which can be set equal to zero without loss of generality.
determine A', we find the mean energy of a molecule.
J
J
d3!
€
d
½rrrlf
4n
f
2
""v dv
½
rrN
2
To
2
e-A 'v
0
3
! f
3 m
=
Letting
€
= (3/2):Id', we find A' =
3/2
gives C' = n(~kT)
m/2kT.
4 A'
The constant CI is easily found from n =
J
d3! ·f which
showing that f is the Maxwell distribution.
CHAPTER 15
Irreversible Processes and Fluctuations
15.1
Let the friction force depend on a, v, and
T)
in the manner
Then the dimensional equation where L = length, T = time, and M = mass becomes
Thus x+y-z = 1, z=l, -y-z = -2.
Solution of this system of equations yieldx x=y=z=l, and
f
cc
117
T)aV
15.2
Since NAk = R vhere NA is Avogadro's number and R is the gas constant, we have from (15.6.11),
t
or
Substituting R =
mole
RI'
NA=
-1
t
2
37111a(x)
8.32 X 107 ergs mole -1 0 K-1 and the given values yields NA= 7.0 X 1023 molecules
•
15.3
(a)
The energy of an ion in the field at point z is - e Cz.
~)
Hence
(b)
e
nfzT
f3e! (z+a.z)
e
1 + f3e
:::::
f3ei' z
f.
dz
(1)
The flux is given by
J
Then from
D
=-Don~ D[n(z+dz) - n(z)]
oz
dz
(1)
J
J
(c)
( d)
J+J
D
µ
D
= -nD f3e !
= nv = nµ
-nDi, e.l+nµ! = 0
µ =
e
I:!:.
Thus
kT
D
dv
d.t =
Substitution of v(t) = u(t)e-rt into
c
(1)
1
(1)
-rv + iii F' (t)
gives
au
-l r t .,..,,
("'")
t
"
dt= -!D.e
Hence
1ft . ,
e 1 " F'(t')dt'
u = u(o) + iii
0
Since u(o) = v(o) it follows that
v = voe -rt +
~
e -rt
J
t e rt' F' ( t' ) dt'
(2)
0
15.5
By dividing the time interval t into N successive intervals such that t = NT, we can replace the
integral in (2) of 15.4 by the sum of integrals over the intervals of length T betveen ~ and
118
Then t' = kT + s where k is an integer and dt'
Hence
v-v e
-rt
=·
e
-,NT N;l l
L,
k=O m
o
=
J,. e'
,y{k,-+s) F 1
-
= ds.
(k,-+s) ds
o
N-l
L, e-,,- ( N-k )
¾JT e's F'(kT+s)
k=O
ds
0
In the integral, s <Tor e's< e'T~ l since,- is chosen so that T << ,-1 .
v-v e
-')'T
=
0
=
where C\_ =
lf,.
m
It follows that
N-l
L, y
k=O
k
=y
F'(k,-+s) ds.
0
15.7
We integrate the Langevin equation
dv
dt
=
-')'V
over a time,- which is small compared to,
-1
+
l
m
F' (t)
but large compared to the correlation time,-* of
the random force
v(,-) - v(o)
=
-')'V(o),- + iiil
J,.
F'(t')dt'
(1)
O
The time interval t is divided into N smaller intervals such that t = NT.
subdivision t'
= kT +sand dt'
= ds.
Then in the k
th
Hence (1) becomes
vk = (l-n) vk-l +
¾f ,.F'[(k-l)T+s]
ds
0
We can find the velocity at t = NT by an iterative procedure.
Defining Gk as in problem
have
vl
= (1-,,-) V0 +G0
v2
= (l-')'T) vl + Gl
(l-,,-) [(l-')'T)v0 + G]
+ Gl
0
2
= (1-')'T) V0 + (l-,T) G0 + Gl
Continuing this way we see that
VN = (1-rr ) N Vo+ (l-')'T )N-1Go+ (1-,,- )N-2 Gl +••• + GN-1
N
Si.nee,,- << l, e-,,-
-~,- N
N
= (e'
) ~ (1-rr) it follows that
ll9
15.6 we
or
since
rr <<
1.
This result agrees with 15.5.
15.7
Gk = G =
J(F'
(k+s) ) ds = 0
0
__ N~l _
Hence
y -
t...
k=O
_ N;l
Yk -
-rr(N-k)
e
t...
G=O
k=O
v-ve-rt =Y=O
and
0
v = v 0 e-rt
Then as t
~o.,,
v ~ O.
15.8
From 15. 5 we have
-
N-1 -
i2=I:
N-1 N-1
k=O
It
'WaS
Thus
r
yk2+r
y
k / j
k
y
j
shown in 15.7 that yk = O.
N-1
-y2 = N-1
L e-27T ( N-k ) Gk 2 -_ G2 e -2rt ~~ e 21 Tk
k=O
k=O
Where we have used NT= t.
The sum is just the geometric series and may be easily evaluated.
y2
=
e -2rt ~-e ~""' G2tl-e - rtJ
2rr
1 -e 2n
2
where e 7T-l""' 27T by series expansion since
2
-v+
{v-v e-'v)
o
Letting t
~o.,
)'T
<< 1. To find G2 ve note that
2
= Y
2
= G
yields, by equipartition,
V
2
21
a2
2
kT
2
G
=m- =
2)'T
120
1-e - 2 rt
2rr
From problem 15.5 we ~ve
(l)
and
where we have changed the integration variable from s tot' to conform to the notation of the
2
text. G is independent of k allowing the choice of' k==O in (1).
2
G =
\JTdt'fT{F'(t') F'(t")) 0 dt"
m
o
o
The integrand is the correlation :f'unction of F 'which depends only on the time difference s = t "-t' •
Making this change of' variable, we bave
- J'r
2
G = \
m
-r-t'
dt'f
(F'(t')F'(t'+s)) 0 ds = \
o
-t'
m
vhere the last equality follows because (F'(t')F'(t'+s)) 0
T-t'
dtj K(s)ds
o
-t'
J
T
= (F'(o)F'(s)) 0 = K(s).
Since K depends
only on s, the integration overt' can be performed easily by interchanging the order of integration and reading off the limits from the figure.
2
G
~ ~2 [1_:ds L:K(s) dt' + 1:~, f-;(,)
dt]
T
t'
Since T >> T* and K( s) ~ 0 for s >> -r*, s can be neglected compared
to
T
and the limits replaced by ±
co.
Thus
121
By
2
the results of 15.8, G = 2kI'
n/m. It follows that
Jet) K(s)
= 2zDiT
7
ds
-<C
15.ll
(a)
We integrate the Iangevin equation
a..v
1
m
F ' ( t)
dt = --yv +
over the interval~-
J
v(-r)
V
=
dv
JT
-7
dt' + ~JTF'(t')dt'
V
O
0
0
Since Tis small, ve can replace v by the initial velocity v
and obtain
l:::.v
= v-v0 =
in the first integral on the right
0
--yv T + -l!TF'(t ')dt
o
m
I
0
= -rv0 -r + G
and
l:::.v
(1:::.v)2- = (7 v 0 -r)
2
""' G =
=
-1
2
- 21 v
V
0
(l)
since G =
T
0
G+
T
o.
a2
2k.Trr/m
(2)
where ve have used the result of 15.8 and kept only terms to first order in -r.
(b)
Fram (1), keeping terms to lowest order in
and noting that odd moments of Gare
T
(!:::.v) 3 = -3,
(!:::.v)4 =
2
Since G
(c)
«
-r,
~
V
2
G
Hence (!:::.v) 3 and (!:::.v)
2
4 are proportional to
r2-r2v0 /m.
15.12
From eq. (2) of 15.4
(v(o)v(t)) = (v2(o)) e--,,t +~iol
=
2
(v (o)) e-
rt
-rt
.-rti\rt• F'(t')~
+ ~ (v(o))
.
But
ve find
~
G should be proportional to T.
From (2), (.6.v) 3 = - 6kr
T
0
o,
ft
rt' F'(t'))
dt'
0
since F'(t ') is random, the second term on the right is O.
122
(e
By equipartition
-r2 •
-rt
~
{v(o)v(t)) = -m e
Thus
and since (v(o)v(-t)) = (v(o)v(t)), we have for all t
(v(o)v(t)) =
~
m
e-,ltl
15.13
Squaring equation (2) of 15.4 yields
.,,...
v 2 (t) = v 2e-2 iv+ 2v
0
·
7-2rt
0
It
.,,+ I
-2,t
e 1 v F'(t')dt' + ~
o
·
m
J 1
t
t
dt'
o
ey
(t '+t •)
F'(t')F(t")dt' 1
o
(1)
since (F'(t')) = O.
The correlation function in the integral is only a function of the time
differences= t"-t'.
Ma.king
this change of variable yields
tdt'f\,(t'+t") (F'(t')F'(t"))dt" =ftdtft-t~s e1 ( 2t'+s)K(s)
f 0
0
0
-t I
where K(s) = (F'(t')F '(t'+s)) = (F 1 {o)F 1 (s)).
that of problem 15.10.
The evaluation of this integral is similar to
Thus interchanging the order of integration and reading the limits off
the diagram of problem 15.10 with t substituted for
J\tf
Q
=J
t-t~s e'( 2t '+s)K(s)
-t I
o
J
( 2,,-t
ds K(s)e
-t
1)
_ --
2I
+
f t
ds I t dt 'ey(2t '+s\(s)
-S
\s
O
r t-:t 'e 1 ( 2t '+s\(s)
J0
ys
o
<< 1, we can put e 1s ""l in the above and find
ds K(s) (e
2rt
21
o
f -tt
-l) =
Substitution in (1) yields
v2 (t) v/
=
+J
.
( 2rt -2ys) f t
( 2,(t-s) )
ds eysK(s)e
-e
I
ds e)' 5 K(s)e
-l
2 i'
2 )'
-t
0
J
Since K(s) is only appreciable when
yields
O
-t
=
T
e-
2
(1-e -2,t)
rt + -"----,._2
2)'m
la, ds K(s)
-0::,
-2--
as t -'>a,, v (a,) = kl' /m by equipartition and
-2-~
l
v (co) = - = m
Hence
,
=
~
J~
2rm2
ds K(s)
123
!co
...,,
( 2)'t 1)
ds K(s ) ~ e
dsK(s)
as in (15.8.8)
21
-
f a,
...,, ds
K(s)
15.14
(a) We will denote the Fourier coefficient of F'(t) by F'(m) and that of v (t) by v (m).
is
oo
F'(t) =
J
•
t
eJ..lO
F'(m) am
and
v(t) =
~
f
co
_..,e
1·wt
v(w)
That
(1)
dw
Substitution in the Langevin equation gives
iwv(m) =
-tV(m)
!m F'(m)
1
m(7+iID)
()
or
+
V ID
F'()
ID
Denoting the spectral densities of F' and v by JF 1 (m) and Jv(m), we have from (15.15.u)
=
or
(b)
Thus
2
m2 (r 2 -fo))
e
v{m) 12
7T' I
(2)
JF,(co) = m2(,l-fo)2) Jv(m)
By
(15.10.9)
K(s)
=
(V(o)V(s))
JvCco) =
~
J~
= kT
m
e-rls l
(3)
K( s) e -iIDs ds
[r
• !, ~ e (-w-+, )s ds +
f
e (-W>-7)sds]
3v(a,) • : , [c/:;,,,j - (-7=:io,~ ":, [ x.,2J
(c)
(4)
Substitution of (4) into (2) gives
l
JF(w) = TI mkT y
or
JF,C+)(co) = 2JF(m) =
~ mkT
r
This result depends on co<< (T·;(- )- 1 , where T;: is the correlation ti.me of the fluctuation force,
since in deriving (3), one used a time scale such that -r >>
124
,.*.
Consequentl;y',
IDT ::::: 1
>> w-r*.