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Consider a classical "degree of freedom" that is linear rather than quadratic: E = c|q| for some constant c

Physics Sep 29, 2020

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 kT.

Expert Solution

The partition function is:

Z = Sum over all states r of exp[-beta E_{r}]

In the classical limit you can replace the summation by an integration. To do that you need to know how many states there are inside a volume element in a region of momentum and ordinary space. The number of states for a particle inside an n-dimensional volume d^n x and d^n p in momentum space is:

d^n p d^nx/h^n

The partition function can thus be written as:

Z = Integral over x and p of d^nx d^n p/h^n exp[-beta E(x, p)]

If the energy does not depend on position and the particle is confined in a volume V and the x-integration yields a factor of V, where V is the volume:

Z = V Integral over p d^n p/h^n exp[-beta E(p)]

In this case E(P) = c|p|. If we first integrate over all directions in momentum space we are left with an integration over the absolute value of p:

Z = V Integral over |p| from zero to infinity of dp Omega |p|^(n-1)/h^n exp[-beta c |p|]

where Omega p(^n-1) is the area of an n-dimensional sphere

Substitute x = beta c |p| in this integral to obtain

Z = beta^(-n) *const.

where const does not depend on beta.

The average energy is given by the derivative:

E = -d log[Z]/ d beta = n/beta = n kT

In one dimension this is indeed kT. In general this is twice the average energy for a nonrelativistic particle.

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