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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 $U=\langle E \rangle$.

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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 $ΔE_{rms}/U$?

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    $\begingroup$ 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. $\endgroup$ – Pieter Dec 5 '19 at 21:13
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    $\begingroup$ 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. $\endgroup$ – dmckee --- ex-moderator kitten Dec 5 '19 at 21:34
  • $\begingroup$ that is not even a function, it is multivalued at some parts $\endgroup$ – Wolphram jonny Dec 6 '19 at 4:26
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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.

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