Let me explain a bit more what I mean. To derive the Boltzmann factor, one usually talks about the ratio of the probability that a system is at some specified energies $E_1$ and $E_2$. This is taken as:

$$\frac{P(E_1)}{P(E_2)}= \frac{g(E_0-E_1)}{g(E_0-E_2)}$$

And then you proceed using the fact that $S=k_B \log g$ and taylor expanding this. However, given that the probability that a system is in some particular state is $P=1/g$ where $g $ is the number of states, shouldn't this imply:

$$\frac{P(E_1)}{P(E_2)}= \frac{1/g_1}{1/g_2}=\frac{g(E_0-E_2)}{g(E_0-E_1)} $$


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Note that in your first line, $g(E)$ is defined as the number of states with the environment having energy $E$, not the total number of states. Then your equation for $P$ should really be $P(E)=\frac{g(E_0-E)}{g_{\text{tot}}},$ where again $P(E)$ is the probability that the system occupies a state of energy $E$, not just any random state.


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