Solutions Manual for Fundamentals of Statistical and Thermal Physics by Frederick Reif

126 Pages·1965·28.121 MB·English

by  R.F. Knacke

CONI'ENTS
. . . . .
. . . . . . . . . . . . . .
Preface ••
i
. .
. . .
CRA.PI'ER l: Introduction to Statistical Methods.
1
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 ••••• 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 ••••• l07
CRAPI'ER 15: Irreversible Processes and Fluctuations ••••••• 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. The reader can refer to the
numerous treatments of these subjects. 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 attention.
R. F, Knacke
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 1,1,1 1,1,2 1,1,3 1,1,4 1,2,2 1,2,3 2,2,2
No. of
Permutatiom l 3 3 3 3 6 l
-k
Since there are a total of 20 permutations, the probability is ~~G =
1.2
(a) Probability of obtaining one ace= (probability of an ace for one of the dice) x (probabil-
5
1 1 6!
ity that the other dice do not shov an ace) x (number of permutations)= (b)(l - '6')
(5!1!)
5 5
= (5) = .4o2 •
(b) The probability of obtaining at least one ace is one minus the probability of obtaining
none, or
5 6
1 - (;-) = .667
(c) 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
1 5 1 5
less than 5 is (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
l N
N N.I
W (-2) = -----<-J-)- - (- 2)
C!)
! !
where N is even.
(b) The drunk cannot return to the lamp post in an odd number of steps.
1.5
th
(b) (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).
(c) 6
1.6
m 3 r cl r N
= m = O since these are odd moments. Using n = (p clp) (p+q)
2 2 -
2 ~2
and m = (2n - N) = 4 n - 4nN + W--
4
m =
we find m2 = N and ";f+ = 3lf - 2N.
1.7
The probability of n successes out of N trials is given by the sum
2 2
E E
w(n) = 'W
m
i=l j=l
with the restriction that the sum is taken only over terms involving w n times.
1
2 2
Then W(n) = E w. E v. If ve sum over all i ••• m, each sum contributes
. i=l i m=l m
(vi-1~ ) and
2 l
W' (n) = (w + w
1 2
N N! n N-n
W' (n) = E .....,...,,.._.__..,.. v w by the binomial theorem •
n=O n!(N-n)! 1 2
Applying the restriction that w must occur n times we have
1
N! n N-n
W(n) = n!(N-n)! "W'l w2
'We consider the relative motion of the two drunks. 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 = n • Then the probability that they meet after N steps irrespective of
1 2
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
2 2
1 1 1 N 1 N 1/2 -1/2 N
(4 4X 2) (2)
pr= x + + = (x + x )
l 2N 2N 2N ! 1/2 n -1/2 2U-n
Expansion yields P' = (2) ~ n! 2N- ! (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 n << N thus ( 1-p )N-n ~ e -Np
N!
(b) (N-n)! = N(N-1) ••• (N - n+l) ~ 1f if n << N
r
N! n ( )N-n n -Np "'n -A
(c) W( n) = n!(N-n)! P 1-p ~ n! p e = n! e
1.10
0) "'n
(a) E n! = e -t. e "' = 1
n=O
= n ->..
- " n >.. e
(b) ll = L-.," n.I
n=v
(c) n2 = n~ n~n e ->-. = e ->-. (~) 2 L. ~~ = >-. 2 + >-.
-- -
2 2
(.6n) = n - .n..!-;) = >-.
1.11
(a) The mean number of misprints per page is l.
-l
Thus W(n) = -n>-..,n e ->-. = e n!
-1
and W(O) = e = .37
3
2 -1
{b) P = 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 ) , P 1-p
n. -n •
-,.._n "-
In the lilllit ~t ~ o, W(n} ~ n! e as in problem 1.9 vhere )... = ~ , the mean number of disintegra-
tions in the interval of time.
=4-n, -4 n 0 1 2 3 4 5 6 7 8
(b) W(n) e
n.
W(n) .019 .(176 .148 .203 .203 .158 .105 .()51 .003
1.13
2 2
We divide the plate into areas of size b. Since b 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
=6,n -6
W(n) e
n.
1..1.4
We use the Gaussian approDJDation to the binomial distribution.
(21.5-2COr
W (n) = _ _;l.;;;..__ e - 2(400/4 = .013
J;;¥f
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.
rN. ___1,;; ;..__
• 01 = l. - e
n=O~ cr
vhere -n = 320C0O = 66.67,
4
The sum may be approxilllated by an integral
JN¥z° l - 2 2
.99 = -- exp [-(n-n) /2a ] dn
1 ,Ji;' a
(n-n+½;&
J _ 2
.99 = -1 e y 12 dy afier a change of' integration variable.
-1 ~
(-n- )/cr
2
The lower limit may be replaced by -eo with negligible error and the integral f'ound f'rcan the
tables of' error functions.
N--n~l
We f'ind = 2. 33
a
N = 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-N
N ! V N V o
W(N) = N!(No-N)! (-) (1 - -)
Vo Vo
0
V
Thus N=NP=No v-
0
0
V V )
N - (1 - -
oV V
(N-i)2
0 0
(b) =
Ff- i
l
(c) If V << V, :::-
0
N
~o
(d)
1.17
Since O <<; <<land since N is large, we may use a Gaussian distribution.
0
0
d]
exp [-(N-N}2/2 dN
5
1.18
N N
A
R = L. r. = tr_ ~ (S. 's are unit vectors)
i -:1. i -:i. -:1.
! £ £ ~-• ~.
R•R = .f..2 S 2 + .f..2 I = m,2 + .f..2 L. I L. cos e.j
i i i j -:i. -J i j :i.
2 2
The second te:rm is O because the directions are randcm. Hence R = Nt •
1.19
N
The total voltage is V = L. v. ; hence the mean square is
i l.
- N - 2 +rN. N
v2=L.v. L.V. V.
f
i l. i j l. J
where v. = vp and 2v . = v 2 p.
J. J.
Hence P = v2 /R = (ff/R) i [l+(l-p)/:N'p]
1.20
(a) The antennas add in phase so that the total amplitude is E = NE, and since the intensity
t
is proportional to Et2 , It= ~I-
(b) To find the mean intensity, we must calculate the mean square amplitude. Since amplitudes
may be added vectorially
2 -- --2N2 --2
Et = ~·:!<!. = ~ L. §.i + ~ L. L. §_. •Sj
-., -., i i / j 1 -
where the S. 's are unit vectors. The phases are random. so the second term is O and
-J.
1.21
a
n.
J
a
Since = O, the second term on the right is O and
ni
_.!. 1 _ .!.
(A 2)2 = N-2 (a 2)2 = N.2!. lv-~ a
n n s
_ .!.
but (An2)2 = As = Nas
6
:Equating these expressions we find N = 10 •
6
1.22
N N
(a) x=I'.s., but since s. =t , x = L t = Nt
• J.. l
i i
N N N
(b) (x-x) 2 = L, (s.-t) 2 + L, L. (s,-t)(sj-t)
J
i J.. i j ~
Since (s.-t) =t - t = O, the second term is O and (x-x) 2 =L N. cr 2 = Ncr 2 •
J..
i
1.23
N N
(a) The mean step length is t. x = I: s = V.-: Nt
i i i
N N N --.- ----....
(b) ( x-x) 2 = L, ( s . - t) 2 + I: I: ( s . - t )( s . -t ) ' ( s . -t) = t - t = 0
J
i J.. j i J.. J J..
2
To find the dispersion (si-t) , we note that the probability that the step length is between
t -
sands+ ds in the range b tot+ bis:~.
)2
1-fn.b (s. -t ds. b2
Hence ( s -t) 2 = J. J.. = -
i t-b 2b 3
2 N b2 Nb2
(x-x) = L, 3 = 3
i
1.24
(a) w(e) de = de
2JT
2
(b) w(e) de= 2JT a si~ e de sine de
2
4n a
1.25
(a) He find the probability that the proton is in e and e + de and thus the probability for the
resulting field.
From problem 1.24 w(e) de= sin/ de
Since b = L (3 cos2e-1) we have !dbl = 6µ cos e sine
a 3 de 83
and
7
(b) If the spin is anti-parallel to the field,
Then if either orientation is possible, we add the probabilities and renormalize
3,J;'
a db
2 a%'
12/µ -µ
W(b)db = ( 83~ ' + a3 ,.;-:; ,} db
a% Jµ 2-i-µa %
12Jµ2-µ 12
3Ji
a db
Jµ
2-i-µa%'
12
(b)
b
3
-2µ/a -µ/a3 2µ/a
Im
Q(k) =L oo ds e iks W( s) = =b ~ ds -eiks
~ s2+b2
11
Q may be evaluated by contour integration. For k > O, the integral. is eval.uated on the path
-kb
residue (ib) = ~ i
:Fork< O
kb
kb
-residue (-ib) = ~ Q(k) = e
8