Statistical Mechanics - Physics at Oregon State University

2.2. ENERGY AND ENTROPY AND TEMPERATURE. 27
and
we find
− U
kBT
− U
kBT
=
log(P (s)) = − ɛs
kBT
s
− ɛs
P (s) (2.16)
kBT
− log(Z) (2.17)
= (log(P (s)) + log(Z)) P (s) =
P (s) log(P (s)) + log(Z) (2.18)
s
where we used P (s) = 1. This leads to
∂
−
∂T
U
=
kBT
∂ log(Z)
∂T
V,N
s
+ ∂
∂T
and using the partial derivative relation for the entropy
∂S
=
∂T V,N
∂
−kB P (s) log(P (s))
∂T
s
P (s) log(P (s)) (2.19)
s
(2.20)
The result of the integration is
S(T, V, N) = S0(V, N) − kB P (s) log(P (s)) (2.21)
where we still need to find the constant of integration S0(V, N). This is easiest
if we take the limit that T goes to zero. In that case the system is in the ground
state. Suppose the degeneracy of the ground state is g0. Then P (s) = g −1
0 for
all states s with the ground state energy and P (s) = 0 for other states. The
logarithm of zero is not finite, but the product of the form x log(x) is zero for
x = 0. Hence we have
S(T = 0, V, N) = S0(V, N) − kB
sing
s
g −1
0 log(g −1
0 ) = S0(V, N) + kB log(g0)
(2.22)
From the formal definition of entropy we see that the second term is equal to
the ground state entropy, and hence we find S0(V, N) ≡ 0. As a result we have
the useful formula
S(T, V, N) = −kB P (s) log(P (s)) (2.23)
Up to now we have considered three variables describing our system: S,
T, and U. These variables are not independent in thermal equilibrium. If we
s