Change of variables in the Density of States function

[AndersF]

TL;DR

Reexpress the density of states for a Fermion gas in terms of momentum in terms of the energy.

I have a problem where I am given the density of states for a Fermion gas in terms of momentum: $D(p)dp$. I need to express it in terms of the energy of the energy levels, $D(\varepsilon)d\varepsilon$, knowing that the gas is relativistic and thus $\varepsilon=cp$.

Replacing $p$ by $\varepsilon/c$ and $dp$ by $d\varepsilon/c$, I would get $D(\varepsilon/c)d\varepsilon/c$, but I'm looking for $D(\varepsilon)d\varepsilon$ instead.

I'm missing something? I know this is basic stuff, but I am stuck and this issue has me clueless...

 

[vanhees71]

You have
$$\tilde{D}(\epsilon) \mathrm{d} \epsilon=D(p) \mathrm{d} p$$
and
$$\mathrm{d} \epsilon=c \mathrm{d} p$$
So you get
$$\tilde{D}(\epsilon) c \mathrm{d} p = D(p) \mathrm{d} p=D(\epsilon/c) \mathrm{d} p,$$
from which finally
$$\tilde{D}(\epsilon)=\frac{1}{c} D(\epsilon/c),$$
which means that $D$ transforms as a density, as it must be.

You have to be careful with the notation. That's why I distinguished $D$ (function of $p$) and $\tilde{D}$ (function of $\epsilon$).

 

[AndersF]

Oh ok, I see, thank you very much!