# Why isn't the Boltzmann factor the inverse of what it is defined as?

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:

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

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.