## Relativisitic density of states

Posted by peeterjoot on April 2, 2013

**Setup**

For photons and high velocity particles our non-relativisitic density of states is insufficient. Let’s redo these calculations for particles for which the energy is given by

We want to convert a sum over momentum values to an energy integral

Now we want to use

where are the roots of . With

Writing for the roots we have

Note that

we have

**3D case**

We can now evaluate the density of states, and do the 3D case first. We have

Observe that in the switch to spherical coordinates in momentum space, our integration is now over a “radius” of momentum space, requiring just integration over the positive values. This will kill off one of our delta functions, leaving just

or

In particular, for very high energy particles where , our 3D density of states is

This is also the desired result for photons or other massless particles.

**2D case**

For 2D we have

Note again that we are dealing with a “radius” over this shell of momentum space volume. This is a strictly positive value. That and the corresponding integration range is important in this case since including the negative range of would kill the entire density function because of the pair of delta functions. That wasn’t the case in 3D, where it would have resulted in an off by two error instead. Continuing the evaluation we have

or

For an extreme relativisitic gas where (or photons where ), we have

**1D case**

**Question**: For the 1D case, we don’t have to make a switch to spherical or cylindrical coordinates, so it looks like the second delta function has to be included, and the integration range over both positive and negative values of ?

Assuming that’s the case, we have

and for or

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