Consider a classical “degree of freedom” that is linear

Chapter 6, Problem 31P

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QUESTION:

Consider a classical “degree of freedom” that is linear rather than quadratic: \(E=c|q|\) for some constant c. (An example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat the derivation of the equipartition theorem for this system, and show that the average energy is \(\bar{E}=k T\)

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QUESTION:

Consider a classical “degree of freedom” that is linear rather than quadratic: \(E=c|q|\) for some constant c. (An example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat the derivation of the equipartition theorem for this system, and show that the average energy is \(\bar{E}=k T\)

ANSWER:

Step 1 of 3

Assume that the energy is linear in \(q\), so:

\(E=c|q|\)

where \(c\) is constant. The partition function is given by:

\(Z=\sum_{E} e^{-\beta E}\)

substitute with the given energy to get:

\(Z=\sum_{q} e^{-\beta c|q|}\)

multiply and divide the RHS by \(\Delta q\), so we get:

\(Z=\frac{1}{\Delta q} \sum_{q} e^{-\beta c|q|} \Delta q\)

in limit of \(\Delta q \rightarrow 0\), the sum becomes an integral over \(q\) from \(-\infty to \infty\), that is:

\(Z=\frac{1}{\Delta q} \int_{-\infty}^{\infty} e^{-\beta c|q|} d q\)

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