I am working out an average number N of bosons of spin $S = 0$ connected to a two-dimensional domain with surface A. The gas is ultrarelativistic with a single particle energy $\epsilon = cp$.

The first thing that came to my mind was: this problem takes relativity into account; we are dealing with photons! Note that this energy is the Plank's one:

$$\epsilon = cp = c \hbar k = \hbar \omega$$

So I had a clear idea in mind; it was just about doing:

$$N \approx \int_0^{\infty} \frac{g(\omega)d\omega}{e^{\beta(\epsilon - \mu)} - 1}$$

But, surprisingly, I was suggested to integrate over momentum instead:

$$N \approx \frac{2 \pi A}{h^2} \int_0^{\infty} \frac{p dp}{e^{\beta(pc - \mu)} - 1}$$

Then, the suggested $g(p)$ is:

$$g(p) = \frac{2 \pi Ap}{h^2}$$

The first thing I did was to use dimensional analysis on the given density of states: $\frac{T }{ML}$. A priori made sense, because if the density of states in function of the energy $g(\epsilon)$ has dimensions of $[\epsilon]^{-1}$ , why not $g(p)$ having dimensions of $[p]^{-1}$?

So let's assume the latest integral is OK. What I did to try to get it was calculating the density of states $g(\epsilon)$ with the idea of make it at the end function of p. So I got:

$$g(\epsilon) = \frac{32 \pi^3 L^2 \epsilon}{h^2 \lambda^2 \omega^2}$$

Which has dimensions of $[\epsilon]^{-1}$. As $p =\frac{h}{L}$ (note that here $L = \lambda$):


1) Energy:

$$\epsilon = cp = c \hbar k = \hbar \omega = \frac{p \lambda \omega}{2 \pi} = \frac{h \lambda \omega}{2 L \pi}\sqrt{n_x^2 + n_y^2}$$


$$\boldsymbol{n}=\left(n_{x},n_{y} \right)$$

$$R=\left|\boldsymbol{n}\right|= \frac{2 L \pi \epsilon}{h \lambda \omega}$$

R turns out to be dimensionless, which is a good sign.

2) The number of states $N(\epsilon)$

$$N(\epsilon) = 4\pi R^2 = \frac{16 \pi^3 L^2 \epsilon^2}{h^2 \lambda^2 \omega^2}$$


$$g(\epsilon) = \frac{dN}{d\epsilon}$$

$$g(\epsilon) = \frac{32 \pi^3 L^2 \epsilon}{h^2 \lambda^2 \omega^2}$$

$g(\epsilon)$ turns out to have dimensions of $[\epsilon]^{-1}$, which is a good sign.

Now we want to get $g(p)$ out of $g(\epsilon)$ so we use the energy:

$$\epsilon = cp$$

So the $g(p)$ I got is:

$$g(p) = \frac{32 \pi^3 L^2 cp}{h^2 \lambda^2 \omega^2}$$


As you can see they do not match.

Note that the dimensions for my $g(p)$ are $[\epsilon]^{-1}$ in contrast with the suggested $g(p)$, which has dimensions of $[p]^{-1}$. Are both expressions OK? I would say no, because dimensions do not depend on the approach used; they cannot change, as both approaches work on momentum space.

Then, how can I get the density of states in function of momentum?

  • 1
    $\begingroup$ Where did you get your function $g(\epsilon)$ from? $\endgroup$ Dec 13, 2018 at 22:46
  • 2
    $\begingroup$ You are confusing terms here. Why is $p=\frac{h}{L}$ a constant? Why don't you write the density of states $g(\varepsilon)$ with $\varepsilon$ instead of $\lambda$ and $\omega$? I would suggest you to show your entire calculation, step by step, so we can find where exactly you have a problem. $\endgroup$
    – eranreches
    Dec 14, 2018 at 12:28
  • 1
    $\begingroup$ First of all, you are mixing too many parameters. You need to use $\varepsilon$ or $p$ or $k$ or $\lambda$ or $\omega$ - not all of them at the same time. Then, imposing the boundary condition you can quantize $k$ or any other variable. This is then used to calculate the volume $\pi R^{2}$ in the phase space of this variable! $\endgroup$
    – eranreches
    Dec 14, 2018 at 14:30
  • 1
    $\begingroup$ Hint: you can use $g(\epsilon)d\epsilon = g(p)dp$. $\endgroup$ Dec 19, 2018 at 0:06
  • 1
    $\begingroup$ @Dani Thank you for your help, my doubt has been solved. $\endgroup$
    – JD_PM
    Dec 19, 2018 at 18:42


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