internal energy

For a system in thermal equilibrium with a heat reservoir at constant temperature $T\,$ , the instantaneous energy of the system, $E\,$ , will fluctuate around its mean value $\left\langle E \right\rangle\,$ , and we refer to

$U = \left\langle E \right\rangle\,$

as the internal energy of the system.

Energy fluctuation

The mean square deviation of the internal energy, by which we mean

$\sigma_E^2 = \left\langle( E - \left\langle E \right\rangle)^2 \right\rangle\,$ ,

can be found as follows. Note that

$(E - \left\langle E \right\rangle)^2 = E^2 - 2 E \langle E \rangle + \langle E \rangle ^2\,$ ,

$\sigma_E^2 = \langle E^2 \rangle - \langle E \rangle ^2\,$ .

Now

$\langle E \rangle\,$	$= \sum_i \frac{ E_i e^{-\frac{E_i}{k_B T}} }{ Z }\,$ ,
	$= k_B T^2 \left( \frac{\partial \ln Z}{\partial T} \right)_V\,$ ,

$\langle E^2 \rangle\,$	$= k_B T^2 \sum_i \frac{ E_i^2 e^{-\frac{E_i}{k_B T}} }{ Z }\,$ ,
	$= k_B T^2\left(\frac{\partial \langle E \rangle}{\partial T}\right)_V + k_B T^2 \frac{1}{Z^2} \left( \frac{\partial Z}{\partial T} \right)_V \sum_i E_i e^{-\frac{E_i}{k_B T}} \,$ ,
	$= k_B T^2 \left(\frac{\partial \langle E \rangle}{\partial T}\right)_V + \langle E \rangle^2\,$ ,

hence

$\sigma_E^2 = k_B T^2 \left(\frac{\partial U}{\partial T}\right)_V = C_V k_B T^2\,$ .