# Massless 3D Density of States

1. Apr 5, 2014

### jammydav93

1. The problem statement, all variables and given/known data
Calculate the single particle density of states for massless particles with dispersion E=h_bar ck for a 3D cube of volume V

2. Relevant equations
E=pc, p=E/c,
dp=dE/c, d^3p = 4pi*p^2 dp
k=sqrt(k_x^2+k_y^2+k_z^2)
k_j = 2pi/L l_j (j=x,y,z)

3. The attempt at a solution
I have tried calculating the density of states in the exact same way as I do for a massive particle but using different energy relations.

Sum(all K)
= sum(all kx,ky,kz)
= int(dl_x dl_y dl_z)
= int ((2pi/Lh_bar)^3 d^3p)
= int ((2pi/Lh_bar)^3 4pi*p^2 dp)
= int ((2pi/Lh_bar)^3 4pi*E^2/c^3 dE)

D(E) = (2pi/Lh_bar)^3 4pi*E^2/c^3

The powers are correct for E and C however I seem to have a dependce on the volume (1/L^3 = 1/V) which I should not be getting - does anyone know why I am getting this?

Thanks,
James

2. Apr 5, 2014

### TSny

From $k_j = \frac{2\pi}{L}l_j$, what do you get for $dl_j$ in terms of $dk_j$?

3. Apr 6, 2014

### jammydav93

I get:

dl_j = L/2pi dk_j = L/(2pi*h_bar) dp_j

Sorry that must have been a typo, I still seem to have an L^3 term which won't go away though.

I now get:
= int(dl_x dl_y dl_z)
= int ((L/2pi*h_bar)^3 d^3p)
= int ((L/2pi*h_bar)^3 4pi*p^2 dp)
= int ((L/2pi*h_bar)^3 4pi*E^2/c^3 dE)

D(E) = (L/2pi*h_bar)^3 4pi*E^2/c^3

4. Apr 6, 2014

### TSny

OK, I think that's corect. But the density of states is often defined on a "per unit volume" basis. If that's what you need, how would you fix your result so that D(E)dE represents the number of states per unit volume with energy between E and E+dE?