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Consider a small system of energy $\epsilon$ in contact with a heat bath of energy $E$. $\Omega_b(E)$ represents the number of accessible microstates of the bath. Then the probability of finding our system with energy $\epsilon$ is

$$P(\epsilon) \propto \Omega_b(E - \epsilon)$$

Consider the Taylor expansion of $\ln\Omega(x)$: $$\ln\Omega(E - \epsilon)\approx \ln \Omega(E)-\frac{\partial \ln\Omega(x)}{\partial x}\bigg|_{x=E}\epsilon + \frac{\partial^2 \ln\Omega(x)}{\partial x^2}\bigg|_{x=E}\frac{\epsilon^2}{2} + ...$$

The usual derivations neglect second and higher order terms. Why is this the case? I know $\epsilon$ is small but small here means small compared to $E$. I don't see how to use that to show that the higher order terms can be neglected.

user1936752
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    If $\epsilon$ is small, then $\epsilon^2$ is even smaller. – Kyle Kanos May 18 '18 at 16:20
  • $\epsilon$ is small compared to $E$ but is not small by itself e.g. a glass of water in thermal equilibrium with a room. I would have to rewrite it to get terms of the form $\frac{\epsilon}{E}$ in the Taylor expansion but I don't see how. – user1936752 May 18 '18 at 17:21
  • Actually, essentially a dupe of Q2 in https://physics.stackexchange.com/q/372397/25301 – Kyle Kanos May 18 '18 at 17:27

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