Why are the equations for mean energy in the canonical ensemble equal?

[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

 

[Jorriss]

Try the chain rule.

 

[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
$$\overline{E}=k T^2\frac{\partial \ln Z}{\partial T}.$$