Statistical Mechanics - Physics at Oregon State University

6.3. MAXIMUM ENTROPY PRINCIPLE. 133
Replacing the sum by an integral yields:
C −1 =
N
Nɛ
ω
and with x =
i xi we get
C −1 =
· · ·
This reduces the entropy to
S(U, N) = −kB
1
dx1 · · · dxN
ɛ √ π
N
Nɛ
ω
dx g(x)
−( dx g(x) e U
ɛ −Nx)2
log
e−( U
ɛ −Nx1−···−NxN ) 2
1
ɛ √ U
e−( ɛ
π −Nx)2
C 1
ɛ √ π
dx g(x) e −( U
ɛ −Nx)2
− ( U
ɛ − Nx)2
(6.81)
(6.82)
(6.83)
In order to prepare for taking the thermodynamic limit we now express the
entropy as a function of the energy per particle u and the number of particles.
The thermodynamic limit is taken in such a way that the energy per particle is
constant. This gives for the entropy per particle, s = S
N :
s(u, N) = −kB
dx g(x) e −N 2 ( u
ɛ −x)2 1
N log
C 1
ɛ √ π
dx g(x) e −N 2 ( u
ɛ −x)2
− N( u
ɛ − x)2
(6.84)
This expression can be split into two parts, by separating the two parts of the
integral in the numerator, and we have s(u, N) = s1(u, N) + s2(u, N) with
1
s1(u, N) = −kB
N log
C 1
ɛ √
(6.85)
π
and
s2(u, N) = NkB
−N dx g(x) e 2 ( u
ɛ −x)2
u ( ɛ
dx g(x) e−N 2 ( u
ɛ −x)2
The second term can be written in the form:
s2(u, N) = 1
2 kB
∂
∂N log
− x)2
dx g(x) e −N 2 ( u
ɛ −x)2
(6.86)
(6.87)
In the thermodynamic limit the exponent is sharply peaked and the integral is
proportional to g( u 1
ɛ ) N . Hence s2(u, N) is proportional to ∂
1
∂N log(N) or N , and
goes to zero in the thermodynamic limit.
Using (6.82) for C we find
1
s(u, N) = kB
N log
Nɛ N
dx g(x) e
ω
−N 2 ( u
ɛ −x)2
(6.88)