A conducting 2-D cavity (of area $L^2$) contains a photon gas that satisfies the dispersion relationship: $$\omega^2 =\frac{4\pi^2c^2}{L^2}(n_x^2+n_y^2)$$ and I wish to find the density of the orbitals per unit energy.A friend and I have two different ways to solve it, and different answers. But which one is correct?

1) My way is simple, just take the Debye formula for density of states in 2-D $$D(k)=\frac{L^2}{2\pi^2}k$$ and finally dividing by the energy $\frac{\hbar^2k^2}{2m}$

2) my friend's approach is as follows: The dispersion relationship defines a circule of radius $r=\frac{\omega L}{2\pi c}$. Then, the number of states can be approximated by the area of the circumference times the polarization number, which is two. $$N(\omega)=2\pi r^2=\frac{\omega^2 L^2}{2\pi c^2}$$ and by using $\omega =ck$: $$N(k)=\frac{k^2 L^2}{2\pi }$$

Then, orbit density is defined as $\frac{dN}{dE}=\frac{dN}{dk}\frac{dk}{dE}$ with $E=\frac{\hbar^2k^2}{2m}$. Thus, $$\frac{dN}{dE}=\frac{mL^2}{\pi\hbar^2}$$

I understand his logic, but I can't definitively say which view is correct. What do you think?

  • $\begingroup$ Seeing $\omega=ck$ and $E=\frac{\hbar^2k^2}{2m}$ in the same solution gives me a headache $\endgroup$
    – acarturk
    May 26, 2019 at 0:11
  • $\begingroup$ Regarding the comment above: Sorry if that came out dismissive. Tried to give a decent answer below. $\endgroup$
    – acarturk
    May 27, 2019 at 19:14

1 Answer 1


As I mentioned, there is no mass for photon gas. Instead, we should use the the relation, $$E=pc=\hbar\omega=\hbar ck$$

Your (not your friend's) solution

With the correction above, $$\frac{D(k)}{E(k)}=\frac{kL^2}{2\pi^2}\cdot\frac{1}{\hbar ck} = \frac{L^2}{h\pi c}$$

Your friend's solution

With the correction above, $$\frac{dN}{dE} = \frac{dN}{dk}\frac{dk}{dE}$$ $$\frac{dN}{dE} = \frac{kL^2}{\pi}\frac{1}{\hbar c}=\frac{2kL^2}{hc}$$

My attempt

Rewriting the dispersion relation in terms of energy, $$E=\hbar\omega=\hbar\cdot\frac{2\pi c}{L}\sqrt{{n_x}^2+{n_y}^2}=\frac{hc}{L}\sqrt{{n_x}^2+{n_y}^2}$$ $$\frac{L}{hc}E=\sqrt{{n_x}^2+{n_y}^2}$$

Now assuming $n_x,n_y\in\mathbb{Z}^+$, the $\sqrt{{n_x}^2+{n_y}^2}$ denotes a location vector on the first quadrant of $\mathbb{Z}^2$.

Interval $(E,E+dE)$ covers an area of $\frac{\pi}{2}\cdot(\frac{L}{hc})^2\cdot E\cdot dE$ on this space. Then (because there is 1 node per unit area), $$dN=\frac{\frac{\pi}{2}\cdot(\frac{L}{hc})^2\cdot E\cdot dE}{1}$$ Then $$\frac{dN}{dE}=\frac{\pi}{2}\cdot(\frac{L}{hc})^2\cdot E$$

$E=\hbar ck$, so $$\frac{dN}{dE}=\frac{\pi L^2}{2h^2c^2}\cdot\frac{hck}{2\pi}=\frac{kL^2}{4 hc}$$


My solution is almost the same as your friend's, so I think their point of view was correct.

I don't know how Debye formulae work, so I cannot specifically comment on your approach, sorry.


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