MCE nr. of microstates & proof that entropy is extensive (ideal gas in a box)

Question

I am trying to find the nr. of microstates inside a box of dimensions $L_1,L_2,L_3$ of a hypersphere in the phase space.We have a gas of N particles in 3D. While we have a 6N dimensional phase space because we assumed that the gas is ideal, we do not have potential energy in the Hamiltonin, only kin. energy:

$$H = \Sigma_1^N \frac {(\vec p_i)^2} {2m}=\Sigma_1^{3N} \frac {(p_i)^2} {2m}$$

The volume of an n dimensional sphere with radius R is:

$$V= \frac {\pi^{\frac n 2} R^n}{\Gamma(\frac n 2 +1)}$$ In our case in phase space, $R=E$ where E is the energy of the states in the hyper-surface of the hyper-sphere (in phase space) and $n=3N$. Then I get:

$$V(R=E)=\frac 1 {N!} \frac 1 {2^{3N}} \frac {(E \pi)^{\frac {3N} 2}} {(\frac {3N} 2)}$$

Then I need to find the volume of the microstate. I initially consider a single particle in the box, for which we would have:

$$E= \frac {\vec p^2} {2m}= \frac {\hbar^2 \pi^2}{2m} [(\frac {n_1}{L_1})^2 + (\frac {n_2}{L_2})^2 + (\frac {n_3}{L_3})^2] \longrightarrow \vec p= (p_x,p_y,p_z)^T= \hbar \pi(\frac {n_1}{L_1},\frac {n_2}{L_2},\frac {n_3}{L_3})$$ where $n_1,n_2,n_3$ are the quantum numbers. The the volume of the microstate of 1 particle is :$\Delta p_x \Delta p_y \Delta p_z$

$$V_{microstate}=\Delta p_x \Delta p_y \Delta p_z = (\hbar \pi)^3\frac {\Delta n_1 \Delta n_2 \Delta n_3}{L_1L_2L_3}= (\hbar \pi)^3\frac 1 V_{box} $$ For N particles then you'd have:

$$V=(\hbar \pi)^{3N}\frac 1 {V_{box}^N}$$

Then in order to find the nr. of microstates inside the volume in phase space which corresponds to that of a hypersphere with radius $R=E$:

$$\Omega(E)= \frac {V_{sphere}} {V_{microstate}}= V^N (\frac {E}{4 \pi \hbar^2})^{(\frac {3N}{2})}\frac 1 {N!}\frac 1 {(\frac {3N} 2)!}$$

But the actual formula is:

$$\Omega(E)= \frac {V_{sphere}} {V_{microstate}}= V^N ( \frac {mE}{2 \pi \hbar^2})^{(\frac {3N}{2})}\frac 1 {N!}\frac 1 {(\frac {3N} 2)!}$$

This implies that I need a $2m$ factor multiplying the energy $E$. And I think it comes from the definition of the momentum I need to include the $\sqrt{2m}$. But that makes no sense since that's not how the momentum is defined. Therefore I have 2 questions:

1. Where does $m$ the mass comes from, so why is my result different?
2. Why do we say that if we do not include $N!$ then entropy is not extensive quantity, is there a way to show this?

Answer 1

1. $R$ is not $E$, since you are thinking about the $3N$-dimensional hyper-space of momenta, and $\sum_i \mathbf{p}_i^2=2mE$, so the radius is really $\sqrt{2mE}$, which supplies the missing $\sqrt{2m}$'s.

2. If you compute the entropy $S(E, V, N)$ by $S=k_B\ln \Omega$ without the $1/N!$ factor, you will see that the entropy is not an extensive function of the three extensive variables. Namely, $S(\lambda E, \lambda V, \lambda N)\neq \lambda S(E, V,N)$.