Fundamentals of Statistical and Thermal Physics  Reif  Solutions Manual_text
Short Description
Solution of fundamental of istatistical and thermal physics  F. reif...
Description
CONTENTS
Preface.
i
CHAPTER 1:
Introduction to Statistical Methods
CHAPTER 2:
Statistical Description of Systems of Particles
11
CHAPTER 3:
Statistical Thermodynamics
l6
CHAPTER
Macroscopic Parameters and their Measurement
18
CHAPTER 5:
Simple Applications of Macroscopic Thermodynamics
20
CHAPTER 6:
Basic Methods and Results of Statistical Mechanics
32
CHAPTER 7:
Simple Applications of Statistical Mechanics
4l
CHAPTER 8:
Equilibrium between Phases and Chemical Species
5^
CHAPTER 9:
Quantum Statistics of Ideal Gases.
6k
!):
1
CHAPTER 10: Systems of Interacting Particles
8^
CHAPTER 11: Magnetism and Low Temperatures
90
CHAPTER 12: Elementary Theory of Transport Processes
93
CHAPTER 13: Transport Theory using the Relaxation Time Approximation
99
CHAPTER l *: NearExact Formulation of Transport Theory
107
CHAPTER 15: Irreversible Processes and Fluctuations
117
1
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 time would probably 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 thank Dr. Reif for the help and encouragement he 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 greatly appreciate
your cal ling errors or omissions to my attention.
R. F. Knacke
CHAPTER 1 Introduction to Statistical Methods
1.1 There are 6*6*6 = 216 ways to roll three dice.
Throw
The throws giving a sum less than or equal to 6
1,1,1
1,1,2
1A,3
1,1,2*
1,2,2
1,2,3
2,2,2
1
3
3
3
3
6
1
No. of
Permutations
Since there are a total of 20 permutations, the probability is
^
1.2
Probability of obtaining one ace = (probability of an ace for one of the dice) x (probabil
(a)
ity that the other dice do not show an ace) X (number of permutations) = 5
=
= () (b)
.
1*02
.
The probability of obtaining at least one ace is one minus the probability of obtaining
none, or
^ 1 (c)

(§)
= *667
By the same reasoning as in (a) we have 1
2
£
(£) (f)
4
g,
57ir
=

040
1*3
The probability of a particular sequence of digits such that five are greater than less than
5 is
1 5 i 5 (— . () v 2 y '2 )
5 and.
five are
Then multiplying by the number of pe imitations gives the probability
irrespective of order 5151
(W(h' '2'
~
2 46
1.2*.
(a)
To return to the origin, the drunk, must take the same number of steps to the left as to the
right.
where (b)
Thus the probability is
N is even. The drunk cannot return to the lamp post in an odd number of steps.
1
1.5 r
«
(a)
()
(b)
(Probability of being shot on the
trial) = (probability of surviving Hl trials) X
IJ
(probability of shooting oneself on the N
th
N_1 trial) =
1 (g)
6
(c)
~
v
m
5 (g)
= nr = 0 since these are odd moments.
m
and
i? = (2n 2
ve find m
 N)
2
4
Using n
= (2n

u)
2
= (p
S
r
N (p+
NJ = nl(Nn)i
'
V
Ve consider the relative motion of the two drunks.
n „ Nn V
1
2
With each simultaneous step, they have a
Probability of l/b of decreasing their separation, l/k of increasing it, and l/2 of maintaining it by takin g steps in the same direction.
and n
,
respectively.
Let the number of times each case occurs be n^, n 5 ,
Then the probability of a particular combination in N steps is n n n 2 l 3 t B(a n n 3>  11 ( ) () i a n:ln ; () 5
V
W°3
3
2
Then the probability that they meet after N steps irrespective of
The drunks meet if n^ = n^.
the number of steps n^ which leave their separation unchanged is
n
N P = S
^
n::nfc'
12
„ n =0 3
3
n
l
^
where we have inserted a parameter x which cancels if n^n^. sum over n^ n^ n^ and choose the term in which x cancels. 1
p’ =
x +
N
i
1
+
2N
,
/
= ()
)
(
x ^
®
21T
/
+ x
hMtt
2
We then perfoim the unrestricted
By the binomial expansion,
)
2U 2N Expansion yields
2
° (^1/2 )
Since x must cancel we choose the terms where n = 2Un or n = H (
P .
Thus
(a)
In (lp)
00
InS}T
r\ (c)
W
w
N_n = p (Un) ~ Up
=
•
•
•
N
n
\
n=0
^T
»
—
2
11
15
X e=e
U
ST P
X
.n X 
K, l
(mf
«U
thus (lp)
^
if n
n
e
X
Up

n
51
n \
.
v
"A.
o
X
X — = X ni
y A
^X
2 =
2
/ \2 = (An)
n
?
Z
(B)
^
nI
n=0
4
o
 n~ = X
The mean number of misprints per page is 1 . ..
= W(n) v '
«
X y hT=ee=l
nI
n=0
a
n
n+i) “
"
Nl _n ,, _ % Un (1 p ) = niTunVi p
y S
(a)
(

() 2
1
n
— ni7
W( 0) = e”
=
i
e
—
= .37
3
ni
n:
e
XX e
2  h * X
= X
If
Nn
« e”^
2
P = 1
(*)

1
E ^r=.08 n *
n=0 1.12 (a)
Dividing the time interval t into small intervals At we have again the "binomial distribution
for n successes in N = t/At trials
In the limit At
>
0 , W(n)
’
X* — n« nJ
e
^
n: P nX( lJn )~T
W(n > = X
Nn
as in problem 1.9 where X =
S
f
the mean number of disintegra
"t
tions in the interval of time.
(b)
W(n) =
e'
n
4
1
0
W(n)
.019
.076
2
3
4
.148
.203
.203
5
.158
6
7
8
.105
.061
.003
1.13
We divide the plate into areas of size b
2
2
.
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.
Clearly n
«N
so we may use the Poisson distribution
W(n) fj
e
n
6
o
Hil
.003
3
1
1
.086
— .162
1.14
We use the Gaussian approximation to the binomial distribution. v2 (215200)*
1
W(n) =
2(400/1+)
= .013
1.15
The probability that a line is in use at any instant is l/30 .
We want
II
lines such that the
probability that N+l or more lines are occupied is less than .01 when there are 2000 trials during the hour.
IT
.oi = i 
n=0‘/^r
where

~^
2000 n = = 66.67,
2a
Z a
15 a =
= 8.02
The sum may he approximated by an integral
2
.99 =
exp [(nn) /2a
%/&?
§
p(l •
99 =
7=^ e V27T 
1
~ M/ 4(ngj/CT The lower limit may he replaced hy
2 ]
dn
a
2y2
dy
'
after a change of integration variable.
with negligible error and the integral found frcm the
°o
tables of error functions. *
n
i
=
!Z l
We find
2.33
N = 66.67 + (2.33) (8.02)
 .5
*
85
1.16 (a)
The probability for N molecules in V is given by the binomial distribution
N =
N_N
N
I
^
NI(N N)i K ' o
o
v
o
o
Thus
2
2
(Nfl )
0>)
_
2
H
jj
(c)
(d)
V
If
If V
»
V
« VQ

IT
N
1
2
2 H
2
(AMP ,

o
o
..
2 M
1
M
—9 IT
Q,
1.17 Since 0
« — « 1 and o
since M
q
is large, we may use a Gaussian distribution,
P(tf)dN =
—
1
2
2
(2?r
AM
2
exp [(NN) /2 AN

i
)
5
]
dM
1.18
N
U R = Z r. = _ i a
p2 ^ rr = tr Z
„ 2 s.
(S.’s are unit vectors)
v* §.• Z '
v Z
p2
.
+
1
i
§. Z — —
t
t
i
/
=#+rZ
A
„.p2
§. " J
X
p2
,
i
o
Z / 2
Hence R
The second term is 0 "because the directions are random.
cos e
0
=
2 Eft
.
1.19
Z
The total voltage is V =
v. 1
i
hence the mean square is
;
s =
v
N
—=
Z
v.
1
1

where v. = vp and v.
2
=
x
i
v
2
N + Z
N _
Z /
.
1
X
J
0
p.
2 2 P = V /r = (u
2
^/R)
Hence
v. v.
.
/
p ri+(lp)/lTp]
1.20 (a)
= ME, and since the intensity
The antennas add in phase so that the total amplitude is E t
is proportional to E^ (h)
2
2 ,
1^
= H !.
To find the mean intensity, we must calculate the mean square amplitude.
Since amplitudes
may he added vectorially E
2
=
2 = E
E, *E,
Z
2
S
2 + E
where the S^'s are
2
= HE
—g A
n
Since a
2
or
N a
n. i
.
i
HI t =
N _
N + Z
^
Z
=
I
/
.
i
Zan. .
/
i
0
a
n. 0
= 0, the second term on the right is 0 and i i
(A
l
1
2 .2
= H2 (a
)
1
0.2 2
= H2 103 a
)
i_
"but
S.
*S,
i/O
The phases are random so the second term is 0 and
unit vectors.
E
1.21
Z
Z
i
(A
2
2 )
=
A„
Equating these expressions we find N = 10°.
6
=
Ha c
1.22
N _
x
(a)
=Es.,
N x = E
_ but since
(xx)
=
'L,
=
E
(s
i
Since
(a)
(s
£) =
£
 t
— i
+
2
IT
=
E
(s.
£)
s
(s.t)(s, £) 1 0
j
0**0
2
u + E
+ ds in the range
a
o
= Na
.
i
ii
E (s.l)(s.l)
j/i
,
)
(s.~V)=tt
,
1
J
=
C>
we note that the probability that the step length is between
tbtO'L
+ b is
7—^2
Hence

•
4 —®— r
Ul°
'v1
)
2 ds
—\ 2
,
(XX)
=
2
E
b
3
i
\1(9)
‘
2
T
— Nb
2
3
H
d9 =
w(9) de =
2
r =
*
i
,
(a)
S
N _ II x = E s^ = E^=
t.
To find the dispersion (s^E and
/
—
i
s
N
E
i
i
(xx )
II
E
= 0, the second term is 0 and (xx)^ =
The mean step length is
(b)
nL
=
L
a
27T
2
sin © d9
sin 9 d9
We find the probability that the proton is in 9 and 9 + d9 and thus the probability for the
resulting field. sin 9 d9 W(9) d9 =
From problem 1.2!
2_ ,\ u Since b = >— (3 cos 01) we have ,
db —
6u cos 0 sin 9 = =
a3
W(b)db = 2W(0)d0 =
a3 a3
6a/
7
a/T db ia\
(b)
If the spin is anti parallel to the field,
W(b)db =
Then if either orientation is possible, we add the probabilities and renormalize
W(b)db =
1.26
0“
,W
{
.
48 '
S
Q may be evaluated by contour integration.
V( s )
=
iks
a«
dS TT
For k > 0, the integral is evaluated on the path
kb
residue (ib) =
For k < 0
8
e ^r
Q(k) = e
9
*
Thus Q(k) = e
kb ,, dk e
C/W fj i
*
ikx
IT
Q
'Leo
r Jn
dk e
Nkb
«tr
dk e
ikx
e
Hkb
r
ikx + i. dk e ar Jr, ,
Mkb
,
cos k x
This integral may he found from tables.
dx 2 2 x + N *) does not become Gaussian because the moments s n diverge.
(P(x) dx
1 
=
TTb
2
7r
1.27
From equation (1.10.6) and (1.10.8) we see that
View more...
Comments