I'm trying to prove the following relation, related to the Ising model.
$S = -\frac{\partial F}{\partial T}$, where $S$ denotes the entropy, $F = -kT \ln Z$ the free energy, and $kT = 1/\beta$. It is worth mentioning that this is my first encounter with the Ising model. I think I am overlooking something obvious, and I'm hoping someone could provide me with a hint or explanation.
If we just compute the derivative using the formula's given, we find: $$\begin{eqnarray*} \frac{\partial (-kT\ln Z)}{\partial T} &=& -k \ln Z - kT \frac{\partial \ln Z}{\partial T}\\&=&-k \ln Z - kT\cdot\left( \frac{1}{Z(\beta)} \cdot \frac{1}{kT^2}\cdot \sum_x E(x)\exp(-\beta E(x)) \right)\\&=&-k \ln Z - \frac{1}{T}\cdot\left( \sum_x E(x)\cdot p(x) \right)\\&=&-k \ln Z - k\beta \cdot \bar{E}(\beta) = -k\cdot S\end{eqnarray*}$$
As you can see, I'm off by a factor $k$ on the derivative. Sure, for $k = 1$ [which is the case the literature considered so far] the relation holds. I've proven before that $S$ satisfies $$ S = \ln Z + \beta \bar{E}(\beta) $$.
Can someone provide me with a hint as to why my derivation is wrong?
Entropy has units of J/K, so your previous derivation is wrong. Everything else here is fine.