Why can we use Bose-Einstein statistics in this expression for number density

Question

In a system with $N$ particles in some volume $V$ in contact with a reservoir of temperature $T$, we find that $$\bar{n_i}=\frac{g_i}{e^{\frac{{\epsilon}_i -\mu}{kT}} \pm 1}$$

depending on whether the particles are fermions or bosons. In the case of low $T$, we get $\bar{n_i} \approx g_{i} e^{\frac{\mu -{\epsilon}_{i}}{kT}}$ .

In some cosmology textbooks I've looked at, they state a formula for the number density of some species as $d_i=g_i e^{\frac{\mu}{kT}}\int{e^{\frac{{-\epsilon _i}}{kT}}d^3 p}$, which is essentially integrating over the approximation above. If I am not mistaken, this refers to the number of particles of species i (not energy level) per unit volume. There's two things I am confused about: where is any reference to volume made in the equation? The BE/FD distributions should give the total number of particles in a certain energy state over the total volume, correct?

The second thing im confused about: we have $\sum \bar{n_i}=N$ where we sum over all distinct energy states. If we assumed the energy spectrum was approximately continuous, shouldn't we have something like $N=\int{\bar{n_i} dE}$, not $N=\int{\bar{n_i} d^3 p}$? I dont get why the equation integrates over momentum instead of energy--its not like $\bar{n_i}$ is the average number of particles in some region of phase space, so why integrate over momentum?

Overall it seems like the equation for $d_i$ is treating the BE/FD distributions as the number density over phase space (i.e. the expected number of particles to be in a certain volume of phase space), while the equations dont seem to describe this. Am I wrong?

Answer 1

This is usually discussed toward the beginning of any solid state textbook. The authors of those cosmology books probably assume that you have a comprehensive education in physics up to that point...

First of all, I see $\mu$ in the formulas, and I assume that we are using the grand canonical ensemble. Your system is in contact with a heat bath and a particle reservoir; it cannot have a fixed N. Instead, $\mu$ controls the equilibrium number of particles in your system.

Let's start with a very simple toy system, where a boson can only ever have energy $\epsilon$. We agree that the equilibrium number of boson is $$ \bar{N}(\epsilon) = \frac{1}{\exp\left[(\epsilon-\mu)/kT\right]-1}. $$

What if we have a bunch of energy levels $\lbrace \epsilon_{i}\rbrace$? Easy. The total number of particles is just $\sum_{i} \bar{N}(\epsilon_{i})$.

But here's the first caveat. If there is degeneracy, for example if $\epsilon_{1} = \epsilon_{2} = \dots = \epsilon_{k}$, you count the same energy k times in the sum. Now you see why the sum/integral is over momentum, not energy. For particles in a box, the energy levels are given by $\epsilon(\vec{p}) = p^2/2m$, and each different $\vec{p}$ counts as a distinct level.

And finally let's see how a number becomes a density. Consider a cubic box of side $L$. The quantized momentum will be of the form $$ \vec{p} = \frac{2\pi}{L} (k_x, k_y, k_z) $$ where the $k_{i}$'s are integers. You sum over all allowed $\vec{p}$ for the total number inside the box. But summation is difficult, if only we can do an integral instead...

Notice that, in each direction, the allowed momentum changes by increments of $\Delta p_i = 2\pi/L$. Assuming $L$ is large, we do the approximation $$ \sum_{\vec{p}} = \left(\frac{L}{2\pi}\right)^3 \sum_{\vec{p}} \Delta p_x \Delta p_y \Delta p_z \approx V \int \frac{d^3p}{(2\pi)^3} $$

And that is where you get the volume of space out of seemingly nowhere.