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?

  • $\begingroup$ @Tyberius Please post answers as answers, not as comments. $\endgroup$ Apr 13, 2021 at 22:44

1 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.

  • $\begingroup$ I have no doubt your answer is the correct one, but you are implicitly assuming that I, and also any other future reader, am familiar with the formulas, the theorems and the notation that you are using. I think this answer needs a little more clarity on what you are doing and what theorems you are using. (For example: what is the $Z$ function you are mentioning?) $\endgroup$
    – Noumeno
    Apr 14, 2021 at 10:16
  • $\begingroup$ @Noumeno Please find the edit $\endgroup$ Apr 14, 2021 at 10:44

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