Question

Apply the result of Problem 6.18 to obtain a formula for the standard deviation of the energy of a system of N identical harmonic oscillators (such as in an Einstein solid), in the high-temperature limit. Divide by the average energy to obtain a measure of the fractional fluctuation in energy. Evaluate this fraction numerically for N = 1, 10⁴, and 10²⁰. Discuss the results briefly.

Short answer

Therefore,

$$\begin{gathered} \frac{{\sigma }_E}{\overline{E}}=\frac{1}{\sqrt{N}} \\ {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N=1}={\left.1\frac{{\sigma }_E}{\overline{E}}\right|}_{N={10}^4}={\left.0.01\frac{{\sigma }_E}{\overline{E}}\right|}_{N={10}^4}=1.0\times {10}^{-10} \end{gathered}$$

Step-by-step solution

Given information

A formula for the standard deviation of the energy of a system of N identical harmonic oscillators (such as in an Einstein solid), in the high-temperature limit. Divide by the average energy to obtain a measure of the fractional fluctuation in energy.

Explanation

In terms of the heat capacity $C_V$, the standard deviation ${\sigma }_E$is given by:

${\sigma }_E=T\sqrt{C_{v}k}$

Consider a system of N harmonic oscillators operating at high temperatures, where $C_V=Nk$is the heat capacity at constant volume.

$$\begin{gathered} {\sigma }_E=T\sqrt{N{k}^2} \\ {\sigma }_E=kT\sqrt{N} \end{gathered}$$

The standard deviation divided by the average energy $\overline{E}=NkT$equals the fractional fluctuation in energy, so:

$$\begin{gathered} \frac{{\sigma }_E}{\overline{E}}=\frac{kT\sqrt{N}}{NkT} \\ \frac{{\sigma }_E}{\overline{E}}=\frac{1}{\sqrt{N}} \end{gathered}$$

The energy fluctuation for N=1

$$\begin{gathered} {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N=1}=\frac{1}{\sqrt{1}}=1 \\ {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N=1}=1 \end{gathered}$$

The energy fluctuation for N=10⁴ is:

$$\begin{gathered} {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N={10}^4}=\frac{1}{\sqrt{{10}^4}}=0.01 \\ {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N={10}^4}=0.01 \end{gathered}$$

The energy fluctuation for N=10²⁰ is:

$$\begin{gathered} {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N={10}^{20}}=\frac{1}{\sqrt{{10}^{20}}}=1.0\times {10}^{-10} \\ {\left.\frac{{\sigma }_E}{\overline{E}}\right|}_{N={10}^{20}}=1.0\times {10}^{-10} \end{gathered}$$

Because the energy fluctuation diminishes as the number of subsystems increases, we can utilise the energy E instead of the average energy $\overline{E}$for large N.