I am trying to calculate the entropy for a classical two dimensional free gas of 𝑁 particles in a box of side 𝐿 and total energy 𝐸. Using two methods, I obtain different scaling factors inside the logarithm of the number of microstates:
Microcanonical Method (Direct Counting):
Here, I compute the phase space volume directly with the energy constraint 𝐸. The number of microstates is given by:
$$ \Omega=\frac{1}{N!}\left(\frac{1}{h^2}L^2\int_{0}^{\infty}d^2p\right)^N=\frac{1}{N!}(L^2E)^N .$$
Here I kept $\frac{2\pi m}{h^2}=1$. To get the entropy, we use Boltzman Law followed by Stirling approximation and get $$S=K\ln\left(\frac{1}{N!} (L^2 E)^N\right)=Nk\ln\left(L^2E\right)-Nk\ln N +Nk\tag{1}$$
Canonical Partition Function Method:
Here, I calculate the partition function at a fixed temperature 𝑇 as $$ Z=\frac{1}{N!}\left(\frac{2\pi m kT}{h^2}\right)^N$$ where T is related to the energy 𝐸 via $𝑇=\frac{𝐸}{𝑁k}$. Substituting this into the partition function yields$$Z=\frac{1}{N!}\left[\frac{L^2 E}{N}\right]^N$$
In short taking negative KT times natural log and then differentiating with respect to temperature with a minus sign we get entropy $$S =Nk\ln\left(\frac{L^2E}{N}\right)-Nk\ln N +Nk\tag{2}$$
From (1) and (2), we can see there is a difference in the solutions. Latter contains $\frac{1}{N}$, and the former doesn't contain. Can you help me understand why there is a difference and which one is more precise?
