# PHYS6562 M5 Daily Question

## Undistinguished particles and the Gibbs factor

(a) $\Omega = \frac{V^N}{N!}$

$S = k_B \space \log(\Omega) = k_B(N\log(V) - B\log(N) + N) = Nk_B\left(1+\log\left(\frac{V}{N}\right) \right)$

The average volume of a region for single particle is $$v = V/N$$ with the entropy of $$s = k_B\log(v)$$. Therefore, the configuration entropy is related to N times the entropy of a particle in its region with average volume.

(b) $S = Nk_B\left(\frac{5}{2}-\log(\rho\lambda^3)\right) = Nk_B\left(\frac{5}{2}+\log\left(\frac{V}{N}\lambda^3\right)\right)$ The configuration entropy for an ideal gas particle is also related to the small average volume $$v$$. $S_c = Nk_B\log(v)$

But the ideal gas particles also occupy the momentum space and each particle has the average length of $$\lambda$$ in each dimension. So

$S_p = 3Nk_B\log(\lambda) = Nk_B\log(\lambda^3)$

Combine the configuration and momentum entropy for one particle, $S = S_c+S_p = Nk_B\log(v*\lambda^3) = Nk_B\log\left(\frac{V}{N}*\lambda^3\right)$ The total entropy is related to the N times the space occupied both in configuration and momentum space.