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The following derivation of the Boltzmann factor is obviously wrong, or incomplete:

$$p(E) \propto \Omega(U-E)$$

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": $$\ln f(x)\approx \ln f(x_0)+\frac{d \ln f(x)}{dx}\bigg|_{x_0}(x-x_0)$$ $$f(x) \approx f(x_0)\exp\left[\frac{d \ln f(x)}{dx}\bigg|_{x_0}(x-x_0)\right]$$ The pitfall is obviously due to exponentiating a Taylor series with higher order terms dropped.

So what should be added to complete the proof?

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You've answered your own question:

The pitfall is obviously due to exponentiating a Taylor series with higher order terms dropped.

To bring this into sharper focus, witness that the method, as you say, does indeed have some validity for any function analytic in the neighbourhood of the point in question. Unless we are at a stationary point, there is a nonzero linear term, so exponential functions and linear functions look exactly the same to first order and this last statement is what your proof is actually proving. Witness that $\exp(-\beta\,E) = 1 - \beta\,E + \cdots$. Your proof is simply a fancy, and not very clear, restatement of Taylor's theorem for function away from a stationary point (where the first order term is nought).

So, if you choose your interval of interest to be narrow enough, the method is perfectly valid.

Out of interest, is this method cited as a "proof" in a thermodynamics text?

It fails when you want to extrapolate to wider intervals, so that the higher order terms are important.

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Since the reservoir is big, T is considered constant. $\frac{1}{\mathrm{k}T}=\frac{\partial\ln\Omega(U)}{\partial U}$ is constant, thus any higher derivatives vanish.

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  • $\begingroup$ Why is it special for $\ln\Omega$ so that we can drop higher order terms in the series, but not for, say $\sqrt{\Omega}$? $\endgroup$ – velut luna Feb 8 '15 at 10:09
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    $\begingroup$ Just because $\frac{\partial\ln\Omega(U)}{\partial U}$ is $\frac{1}{\mathrm{k}T}$. $T$ is constant and thus $1/\mathrm{k}T$ is constant. $\endgroup$ – Leonard Michlmayr Feb 8 '15 at 10:36

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