Calculating energy U from $\partial U/\partial q$

Question

Imagine $N$ oscillators with only two possible energies, $\epsilon_0$ and $ \epsilon_1$, with $\epsilon_1 > \epsilon_0$. Taking $\epsilon_0 = 0$ for now

I showed $\Omega(q\epsilon_1) = \frac{N!}{(N-q)!q!}$ and then

$$\frac{\partial S}{\partial q} = k \log(N/q - 1) $$

How can I use the above equation to show that

$$U = N\epsilon_1\frac{e^{-\epsilon_1/(kT)}}{1+e^{-\epsilon_1/(kT)}} $$

I tried moving the $\partial q$ over to the right, and then have $dS = dU/T$, but i wasn't getting anything meaningful.

Answer 1

You should use $U=q\epsilon_1$.

With the total number of particles $N$ being constant, we have: $$\frac{\partial S}{\partial q}=\epsilon_1 \frac{\partial S}{\partial E}=\frac{1}{T}\epsilon_1\tag{1}$$ As you said: $$\frac{\partial S}{\partial q}=k_B\ln(N/q - 1)=k_B\ln(N\epsilon_1/q\epsilon_1 - 1)=k_B\ln(N\epsilon_1/U - 1)\tag{2}$$ $$\to \,\,\,\,\,U(T)=N\epsilon_1\frac{e^{-\epsilon_1/(kT)}}{1+e^{-\epsilon_1/(kT)}}$$