# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 320,607

## Unresolved question about energy distribution around mean energy

Posted by peeterjoot on April 10, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

In [1] is an expansion of

\begin{aligned}P(E) \propto e^{-\beta E} g(E),\end{aligned} \hspace{\stretch{1}}(1.0.1)

around the mean energy $E^{*} = U$. The first derivative part of the expansion is simple enough

\begin{aligned}\frac{\partial }{\partial E} \left( e^{-\beta E} g(E) \right) &= \left( -\beta g(E) + g'(E) \right)e^{-\beta E} \\ &= g(E) e^{-\beta E}\left( -\beta + (\ln g(E))' \right)\end{aligned} \hspace{\stretch{1}}(1.0.1)

The peak energy $E^{*}$ will be where this derivative equals zero. That is

\begin{aligned}0 = g(E^{*}) e^{-\beta E^{*}}\left( -\beta + {\left.{{(\ln g(E))'}}\right\vert}_{{E = E^{*}}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.3)

or

\begin{aligned}{\left.{{\frac{\partial }{\partial E}\left( \ln g(E) \right)}}\right\vert}_{{E = E^{*}}} = \beta\end{aligned} \hspace{\stretch{1}}(1.0.4)

With

\begin{aligned}S = k_{\mathrm{B}} \ln g\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\frac{1}{{k_{\mathrm{B}}}} \left( \frac{\partial S}{\partial E} \right)_{E = U} &= \frac{1}{{k_{\mathrm{B}} T}} \\ &= \beta\end{aligned} \hspace{\stretch{1}}(1.0.5b)

We have

\begin{aligned}\left( \frac{\partial \ln g(E) }{\partial E} \right)_{E = U} = \beta\end{aligned} \hspace{\stretch{1}}(1.0.6)

so that

\begin{aligned}E^{*} = U.\end{aligned} \hspace{\stretch{1}}(1.0.7)

So far so good. Reading the text, the expansion of the logarithm of $P(E)$ around $E = E^{*} = U$ wasn’t clear. Let’s write that out in full. To two terms that is

\begin{aligned}\ln e^{-\beta E} g(E)= \underbrace{\ln e^{-\beta U} g(U)}_{- \beta U + \frac{1}{{k_{\mathrm{B}}}} S}+ {\left.{{\frac{\partial }{\partial E} \left( \ln e^{-\beta E} g(E) \right)}}\right\vert}_{{E = U}}+ \frac{1}{2}{\left.{{\frac{\partial^2 }{\partial E^2} \left( \ln e^{-\beta E} g(E) \right)}}\right\vert}_{{E = U}}(E - U)^2.\end{aligned} \hspace{\stretch{1}}(1.0.7)

The first order term has the derivative of the logarithm of $e^{-\beta E}g(E)$. Since the logarithm is monotonic and the derivative of $e^{-\beta E}g(E)$ has been shown to be zero at $E = U$, this must be zero. We can also see this explicitly by computation

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{-\beta e^{-\beta E} g(E) + e^{-\beta E} g'(E)}{e^{-\beta E} g(E)}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{-\beta g + g'}{g}}}\right\vert}_{{E = U}} \\ &= -\beta+{\left.{{(\ln g)'}}\right\vert}_{{E = U}} \\ &= -\beta + \frac{1}{{k_{\mathrm{B}}}} {\left.{{\frac{\partial S}{\partial E}}}\right\vert}_{{E = U}} \\ &= -\beta + \frac{1}{{k_{\mathrm{B}} T}} \\ &= -\beta + \beta \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.7)

For the second derivative we have

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{\partial }{\partial E} \left( -\beta + (\ln g)' \right)}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{\partial }{\partial E} \frac{g'}{g}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{g''}{g} - \frac{(g')^2}{g^2}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{g''}{g}}}\right\vert}_{{E = U}} - ((\ln g)')^2 \\ &= {\left.{{\frac{g''}{g}}}\right\vert}_{{E = U}} - \beta^2.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Somehow this is supposed to come out to $k_{\mathrm{B}} T^2 C_{\mathrm{V}}$? Backing up, we have

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{\partial^2 }{\partial E^2} \ln g}}\right\vert}_{{E = U}} \\ &= \frac{1}{{k_{\mathrm{B}}}}{\left.{{\frac{\partial^2 S}{\partial E^2}}}\right\vert}_{{E = U}}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

I still don’t see how to get $C_{\mathrm{V}} = {\partial U}/{\partial T}$ out of this? $C_{\mathrm{V}}$ is a derivative with respect to temperature, but here we have derivatives with respect to energy (keeping $\beta = 1/k_{\mathrm{B}} T$ fixed)?

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.