The microcanonical ensemble
Entropy is given by s = k ln R∆Γ.
We have that ∆Γ = V N d3N p and E(p) =
1
2m
Calculating ∆
PN
i=1 |pi |
2.
Geometrically, ∆Γ = VDN where ΩD is the √
volume of the D-dimensional
sphere in momentum space, with radius R = 2mE0 and thickness ∆.
Write the volume of a spherical shell in 3N dimensions as V (R) =
C(D)RD . Then the volume of a shell of thickness ∆ is
V∆ = V (R) − V (R − ∆)
= C(D)R
D
"
∆
1− 1−
R
D #
For a gas, D ≈ 3 × 1023 so V∆ ≈ C(D)RD and ∆Γ ≈ V D C(D)RD
We need C(D). Get this by considering the integral
I(D) =
Z
∞
dx1 . . .
−∞
Z
∞
2
2
dxD e−(x1 +...+xD ) = π D/2
−∞
In polar coordinates this is
I(D) =
=
Z
∞
2
e−r r D−1 C(D)dr
0
D
1
Γ
+ 1 C(D)
2
2
where r D−1 C(D) is the surface area of the D-dimensional sphere. Then
comparing results for I(D) we have
C(D) =
and
∆Γ =
2π D/2
Γ
D
2
2π D/2
Γ
D
2
+1
+1
RD V N .
D
Using the result Γ D
2 +1 =
2 ! and Stirling’s approximation, N ! =
√
N
−N
2πN
the result follows. (Note that when taking logs of N ! the
√ N e
ln( 2πN ) term is often ignored as its contribution is small compared to the
other terms in the log.
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