Statistical Mechanics - Physics at Oregon State University

A.1. SOLUTIONS FOR CHAPTER 1. 231
(1a) A = car, B = 0, C = 0 Host chooses B Probability 1
6
(1b) A = car, B = 0, C = 0 Host chooses C Probability 1
6
(2b) A = 0, B = car, C = 0 Host chooses C Probability 1
3
(3a) A = 0, B = 0, C = car Host chooses B Probability 1
3
which are not all equally probable anymore. If we now label the door which the
host chose by B, we only have sequences 1a and 3a and we have
(1a) A = car, B = 0, C = 0 Host chooses B Probability 1
3
(3a) A = 0, B = 0, C = car Host chooses B Probability 2
3
and therefore the probability that the car is behind the other door is twice as
large as the probability of the car being behind the original door. Another
way of saying this is that when the car is behind the original door A the host
will choose door B only half the time and door C the other half. If the car is
behind door C he has to choose door B. This reduces the probability in case 1
by one-half.
Problem 6.
Define the number of atoms with spin si = s to be Ns. Therefore:
S
s=−S
S
s=−S
Ns = N
sNs = M
and the number of ways to find the set of values N−S, N−S+1, · · · , NS−1, NS is
which leads to
g(N, M) =
N−S,N−S+1,···,NS−1,NS
N!
N−S!N−S+1! · · · NS−1!NS!
N!
δ S
N−S!N−S+1! · · · NS−1!NS! s=−S Ns,N δ S
s=−S sNs,M
where we used the Kronecker delta to limit the sums to the right cases.
We now approximate the factorials, keeping only terms that depend on N in the
exponent, N! ≈ N N e −N , and get