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Consider a classical “degree of freedom” that is linear
Chapter 6, Problem 31P(choose chapter or problem)
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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