# About the width of the distribution of the kinetic energy in a gas

In this lecture about statistical mechanics, page $$10$$, the author said that the kinetic energy $$E$$ of a gas can be viewed as a random variable (because it is a sum of squared velocities, which themselves are random variables), and that its probability distribution is very sharp around its mean $$U=\langle E \rangle$$.

He said about the above distribution,

The distribution of the system's total energy $$E$$ is very sharply peaked around its mean $$U=\langle E\rangle$$: the width of this peak is $$∼ΔE_{rms}/U$$,

where $$\Delta E _{rms}=\langle (E-\textrm {U})^{2} \rangle ^{1/2}=\sigma$$.

What justifies the author's claim that the width of the distribution of $$E$$ can be approximated by $$ΔE_{rms}/U$$?

• That would be for a volume of gas in contact with a heat reservoir of a certain temperature. One can calculate the probability of energy fluctuations of the subsystem from changes in entropy. – Pieter Dec 5 '19 at 21:13
• It should be emphasized that what is plotted here is the probability of finding a particular mean particle energy. The distribution of the individual particle energies is wide. – dmckee --- ex-moderator kitten Dec 5 '19 at 21:34
• that is not even a function, it is multivalued at some parts – Wolphram jonny Dec 6 '19 at 4:26

The claim follow from central limit theorem. If you consider your energy per particle $$E/N$$ to be defined as $$\frac{E}{N} = \frac{1}{N} \sum_i \frac{mv_i^2}{2}$$ then using the central limit theorem (https://en.wikipedia.org/wiki/Central_limit_theorem) you will find that the width of the peak of the distribution of $$P(E/N )$$ distribution is then $$\approx 1/\sqrt{N}$$.
Note the supplemental factor $$1/N$$ used in my formula. This would be the correct way of expressing that the distribution of energy is very peaked.
Now in your lectures, then instead compared the root-mean-square of the distribution of $$E$$ to its average $$U$$, in this case the width of your distribution should be $$\Delta E _\text{rms}$$. As the ratio of $$\Delta E _\text{rms} / U \simeq 1/\sqrt{N}$$ is very small when you effectively draw your distribution, you will get a very peaked distribution. So the claim in your lecture notes should be that the aspect ratio of your distribution is indeed $$\Delta E _\text{rms} / U$$, not that the width of the distribution is this ratio.