Statistical Mechanics - Physics at Oregon State University

A.1. SOLUTIONS FOR CHAPTER 1. 233
x = e −1+α
S
s=−S
se +βs
The derivatives with respect to x of the first equation is :
which gives
Also we then get
−1+α dα
0 = e
dx
dTmax
dx
S
s=−S
e +βs + e −1+α
0 = dα
+ xdβ
dx dx
S
s=−S
+βs dβ
se
dx
= −dαN
− βN − xdβ N = −βN
dx dx
which shows that this term is maximal for β = 0 and this gives immediately
x = 0. When β is small the equations for α and β give
1 = e −1+α
x = e −1+α
S
(1 + βs) = e −1+α (2S + 1)
s=−S
S
s=−S
s(1 + βs) =
1
2S + 1 β
which shows that α is constant and β ∝ x, and hence
Tmax = log(2S + 1)N − cx 2 N
S
s=−S
with c = 2S+1
S
s=−S s2 , which is again a Gaussian distribution for the multiplicity
function:
Problem 7.
g(N, x) ≈ g(N, 0)e −cNx2
Twelve physics graduate students go to the bookstore to buy textbooks. Eight
students buy a copy of Jackson, six students buy a copy of Liboff, and two
students buy no books at all. What is the probability that a student who
bought a copy of Jackson also bought a copy of Liboff? What is the probability
that a student who bought a copy of Liboff also bought a copy of Jackson?
We have two students who buy no book, two students who only buy Liboff, four
students who only buy Jackson, and four students who buy both books. Therefore
s 2