Past exam of the thermal and statistical physics course of the University of Cambridge / 2015 / Paper 1 / 1 / c / Solution

Let $E$ denote the fluctuating microscopic energy and $U=⟨E⟩$ its mean, at fixed $N,V$. In a canonical ensemble with $\beta =1/(k_{B}T)$,Since $dT/d\beta =-k_B{T}^2$, this givesHere $\Delta U$ denotes an energy fluctuation about the mean, as in the question. The relation is not restricted to ideal gases when the Hamiltonian operator has no explicit temperature dependence. Alternatively, the stated equilibrium free-energy curvature gives a Gaussian energy weight proportional to $\exp [-(\Delta U{)}^2/(2k_B{T}^2{C}_V)]$, with the same variance.

For the relative energy fluctuation of a monatomic ideal gas, $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$, henceTaking room conditions as approximately $T=300$ K and $p=10^5$ Pa, with $V=10^{-6}$ m${}^3$, givesThe mean energy is about $0.15$ J and its standard deviation about $2.49\times 10^{-11}$ J. The tiny relative fluctuation explains why macroscopic equilibrium internal energy behaves as a sharply defined thermodynamic quantity. The estimate uses a thermal reservoir, not a particle reservoir; allowing $N$ itself to fluctuate changes the relevant fluctuation problem.