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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. 1