Why are Energy Fluctuations approximately equal to $k_BT$?

Question

I have read, from multiple sources, that the energy fluctuations of a single particle of a perfect gas are approximately equal to $k_BT$, where $k_B$ is the Boltzmann's constant and $T$ is of course the temperature of the perfect gas.

Why? How can we prove that this statement is true?

And also: is there a more general version of this statement? Does it only apply to perfect gases and single particles?

Answer 1

The partition function $Z$ is defined as $$Z(T,\mathbf{x})=\sum_{\{\mu\}}e^{-\beta \mathcal{H}(\mu)}$$ where sum is over microstate ($\mu$).

For an ideal gas, $$Z(T,V,N)=\int \frac{1}{N!} \prod_{i=1}^N\frac{d^3q_id^3p_i}{h^3}\exp\left[-\beta \sum_{i=1}^N\frac{p_i^2}{2m}\right]$$ $$Z(T,V,N)=\frac{1}{N!}\left(\frac{V}{\lambda(T)^3}\right)^N$$ $$\Rightarrow \langle H\rangle =-\frac{d\ln Z}{d\beta}=\frac{3}{2}\frac{N}{\beta}$$ $$\langle H^2\rangle_c=-\frac{\partial \langle H\rangle }{d\beta}=\frac{3}{2}\frac{N}{\beta^2}$$ $$\sqrt{\langle H^2\rangle_c}\propto \sqrt{N}k_BT$$ As proposed.

---

Further In general, $$\langle H^2\rangle_c=-\frac{\partial \langle H\rangle }{d\beta}=k_BT^2\left.\frac{\partial \langle H\rangle }{\partial T}\right|_\mathbf{x}$$ $$\Rightarrow \langle H^2\rangle_c=k_BT^2C_\mathbf{x}$$ The above is true in general for canonical ensemble.