Fall 2007 17 Sm

A system is composed of N identical classical oscillators, each of mass m, defined on a one-dimensional lattice. The potential for the oscillators has the form

(1)
\begin{align} U(x)=\epsilon \left|x/a\right|^n,\;\; \epsilon > 0, \; n>0. \end{align}

(Thus, the oscillators are harmonic for n = 2 and anharmonic otherwise).
Find the average thermal energy at temperature T.

Hint: An integral that appears in the course of evaluating the partition function cannot be computed in terms of elementary functions. Fortunately,it amounts only to an unimportant overall coefficient.

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Answer

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Classical partition function for a single oscillator is

(2)
\begin{align} \zeta=\int_{-\infty}^{\infty}dp\exp{\left(-\beta p^2/2m \right)}\int_{-\infty}^{\infty}dx\exp{\left( -\beta\epsilon\left|x/a \right|^n \right)} \end{align}

By change of variables, we bring this to the form

(3)
\begin{align} \zeta=\Gamma(1/2)\Gamma(1/n)\frac{2a}{n}\left(\frac{2m}{\beta} \right)^{1/2}\left( \frac{1}{\beta\epsilon}\right)^{1/n} \end{align}

where

(4)
\begin{align} \Gamma(z)\equiv\int_{0}^{\infty}dt\; t^{z-1}\exp{(-t)}. \end{align}

A learned reader would recognize this as the Euler Gamma-function. However, knowing this is not necessary. The product $\Gamma(1/2)\Gamma(1/n)$ is just a numerical coefficient, which will disappear from the final result.

The average energy is given by

(5)
\begin{align} E=-N\frac{\partial}{\partial \beta}\ln{\zeta}=\left(\frac{n+2}{2n}\right)N kT. \end{align}

This result resembles the equipartition theorem in the sense that material constants do not enter.

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