Canonical Internal Energy

1. Apr 3, 2013

Abigale

Hi,
I regard an equation for an canonical ensemble.
I do not understand why both equations should be equal.

$\bar{E}= -\frac{ \partial \ln{(Z)}}{ \partial \beta} \overset{\text{?}} = k T ^{2 } \frac{\partial \ln{(Z)}}{\partial T}$

$Z$ is a canonical partition function.
$\beta$ = $\frac{1}{kT}$.

Thx
Abby

Last edited: Apr 3, 2013
2. Apr 3, 2013

Jorriss

Try the chain rule.

3. Apr 3, 2013

vanhees71

Do you under stand the first equation? That simply comes from the definition of the partition sum
$$Z=\mathrm{Tr} \exp(-\beta \hat{H}),$$
where $\hat{H}$ is the Hamiltnian of the system. The Canonical Statistical Operator is
$$\hat{R}=\frac{1}{Z} \exp(-\beta \hat{H}).$$
The mean energy is
$$\overline{E}=\mathrm{Tr}(\hat{R} \hat{H})=-\frac{1}{Z} \frac{\partial Z}{\partial \beta}=-\frac{\partial \ln Z}{\partial \beta}.$$
As it turns out by comparing with the first Law of thermodynamics, you have
$$\beta=\frac{1}{k T},$$
where $T$ is the temperature (contrary what you've written in your posting!).

Then of course you have for any function of the temperature
$$\frac{\partial f}{\partial \beta}=\frac{\partial f}{\partial T} \frac{\partial T}{\partial \beta}.$$
From the above formula you find $T=1/(k \beta)$ and thus
$$\frac{\partial T}{\partial \beta}=-\frac{1}{k \beta^2}=-k T^2.$$
This gives
$$\overline{E}=k T^2\frac{\partial \ln Z}{\partial T}.$$