TITLE: Derivation of the Boltzmann factor in statistical mechanics
QUESTION [12 upvotes]: I have seen similar derivation of the Boltzmann factor many times before, (http://micro.stanford.edu/~caiwei/me334/Chap8_Canonical_Ensemble_v04.pdf , just for example), which I think is incomplete.
The argument is as follows:
Consider the system consisting of our small object in contact with a big reservoir. Let the total energy be $U$. Then when our object has energy $E$, the reservoir has energy $U-E$. Let the number of accessible state of the reservoir as function of its energy $x$ be $\Omega(x)$. Then the probability of finding the object with energy $E$ is
$$p(E) \propto \Omega(U-E)$$
Consider the Taylor expansion of $\ln\Omega(x)$:
$$\ln\Omega(U-E)\approx \ln \Omega(U)-\frac{\partial \ln\Omega(x)}{\partial x}\bigg|_{x=U}E$$
Define
$$\frac{1}{k T}=\frac{\partial \ln\Omega(x)}{\partial x}\bigg|_{x=U}$$
Then exponentiating both sides, we have
$$\Omega(U-E)\approx \Omega(U)\exp(-E/kT)$$
So
$$p(E)\propto \exp(-E/kT)$$
This must be incomplete because the above can be done to any functions to prove that they are exponential.
For example, we can show that for any function $f(x)$,
$$f(x) \approx A\exp[B(x-x_0)]$$
around some $x_0$ by the above "proof":
$$\ln f(x)\approx \ln f(x_0)+\frac{d \ln f(x)}{dx}\bigg|_{x_0}(x-x_0)$$
$$f(x) \approx f(x_0)\exp\left[\frac{d \ln f(x)}{dx}\bigg|_{x_0}(x-x_0)\right]$$
Besides, it it also possible to prove that $f(x) \approx g(x)$ around $x_0$ for any function $g$ you like. For example,
$$\sin(f(x))\approx \sin(f(x_0))+\cos(f(x_0))f'(x_0)(x-x_0)$$
$$f(x) \approx \sin^{-1}\left[\sin(f(x_0))+\cos(f(x_0))f'(x_0)(x-x_0)\right]$$
The flaw is obviously due to dropping the the higher order terms in the Taylor series. So there must be something very special about $\ln$ in particular, that we can drop the higher order terms in the series expansion of $\ln\Omega$, and then exponentiate the result. I suppose it's because the higher order terms all vanish in the thermodynamic limit?
Can anyone tell me what I am missing. Your help is much appreciated. Kind regards!

REPLY [2 votes]: I was seeking an answer to the same question and found this explanation in the Thermal Physics lecture slides at University Central Arkansas:

All higher order partial derivatives are zero by assumption of a large heat bath
  $$\frac{\partial^2 \sigma}{\partial U^2} = \frac{\partial}{\partial U} \frac{\partial \sigma}{\partial U} = \frac{\partial}{\partial U} \left(\frac{1}{\tau}\right) = 0$$

Where, following Kittel and Kroemer, I am writing the entropy as $\sigma(U) = \log \left[\Omega(U)\right]$ and the fundamental temperature as $1/\tau = \partial \sigma(U, N, V,...)/\partial U$ 
This is more satisfying to me than another, more common, explanation I have seen (e.g., here) that the state energy, $\varepsilon$, is much smaller than the total system energy, $U_0$, so higher-order terms can be neglected. I don't see how that argument can be made, since in the Taylor expansion of the entropy
$$\sigma(U_0-\varepsilon) = \sigma(U_0) + \frac{{\rm d} \sigma}{{\rm d} U} (-\varepsilon) + \frac{1}{2!} \frac{{\rm d}^2 \sigma}{{\rm d} U^2} \, (-\varepsilon)^2 + \cdots$$
there is no way to write the series in powers of the small parameter $\frac{\varepsilon}{U_0}$. 
Simply claiming that the reservoir is so large that its temperature is a constant seems cleaner to me.