0
$\begingroup$

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:

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

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.