Calculating the One-Particle Density of States of the ripplon Gas I'm trying to understand an example that I found in my notes but I don't understand the difference in my results compared to how my teacher did it. It's from Statistical Physics and it seems that there is always another way of doing problems which we are not taught but expected to know.
Here is the problem:
Under certain conditions, the surface waves in liquid helium are quantised and may be considered as quasi-particles with a dispersion law $ε = (|p|/α)^{3/2}$ where α is some constant and $p$ is the momentum. Such
quasi-particles are called ripplons. Their quantum statistics is irrelevant for the regime of
interest
here (they are, in fact, bosons).
Consider a two-dimensional ideal gas of N ripplons in thermal equilibrium at temperature T
confined to an area A.
Calculate the one-particle density of states $n(\epsilon)$ of the ripplon gas.
THIS WHAT I DID:
If we fit the plane waves in a square area $ A=L^2$ with periodic boundary conditions we can write:
$$(p_x,p_y) = \hbar (k_x,k_y) = \frac{h}{L}(n_x,n_y)$$
So we have one mode per every $(\frac{h}{L})^2$ of the $p$-space and thus the $p$-space density of states is $\frac{A}{h^2}$. We can use the following to calculate $n(\epsilon)$:
$$n(\epsilon)d\epsilon = n(k)dk$$
Since $p=\hbar k$, we have $k=\epsilon ^{2/3}\frac{\alpha}{\hbar}=\epsilon ^{2/3}\frac{2\pi\alpha}{h} $. Therefore:
$$n(\epsilon) = n(k) \frac{dk}{d\epsilon} = \frac{16 \pi^3 \epsilon ^{1/3} \alpha ^2 A}{3 h^6}$$
I thought this procedure was good to use but suddenly the teacher does the following:
THIS IS WHAT THE TEACHER DOES:
Same $p$-space densite of state : $\frac{A}{h^2}$. Then he states the number of microstates with energy not exceeding $\epsilon = (p/\alpha)^{3/2}$:
$$\Gamma (\epsilon) = \frac{A}{h^2}\pi p^2$$ If anyone knows also why he does that and how he gets it because I received no explanation.
Then he takes the derivative to get the density in energy-space:
$$n(\epsilon) = \frac{d\Gamma}{d\epsilon} = \frac{4 \pi \alpha^2 \epsilon ^{1/3}A}{3h^2}$$
I don't understand why he does it this way and any help would be greatly appreciated!
Many thanks!
 A: If you think back to the definition of the density-of-states function, the number of states with an energy up to $\epsilon$ is equal to the integral of the density of states function from $-\infty$ to $\epsilon$:
$$
\Gamma(\epsilon) = \int_{-\infty}^\epsilon n(\epsilon') d \epsilon'
$$
But that means that conversely, by the fundamental theorem of calculus,
$$
n(\epsilon) = \frac{d\Gamma}{d\epsilon},
$$
and so if you can find the "cumulative state function" $\Gamma(\epsilon)$, you can find the density-of-states function by taking its derivative.
As far as why your method and your teacher's method yield different results, I can't follow either method closely enough to see where you might have gone wrong.  But the units of $n$ should be those of inverse energy, your teacher's result has units of inverse energy, and yours does not;  so yours is incorrect.
This is not to say that your method is completely wrong and that your teacher's method is the only way to solve the problem.  Finding the density of modes in $k$-space and then converting the result into a density in terms of $\epsilon$ should be equivalent if you do it right.  But without more information it's impossible to say where you went wrong.
