In class we derived the 3D case, but there's a step I don't understand:

$$ N = g \cdot {V \over (2 \pi \hbar)^3} \cdot \int\limits_{0}^{\infty}{1 \over{e^{\left( E_p \over{K_B T}\right)}-1}} d^3 p = g \cdot {V \over (2 \pi \hbar)^3} \cdot 4 \pi \cdot \int\limits_{0}^{\infty}{p^2 \over{e^{\left( E_p \over{K_B T}\right)}-1}} dp $$

... I feel like if I knew why that step made sense, I could figure out how to do the equivalent thing for the 2D case, but I'm stuck on that.

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    $\begingroup$ you are just going from cartesian to spherical coordinates, $(p_x,p_y,p_z)\rightarrow(p,\theta,\phi)$, which has a jacobian of $p^2 \mbox{cos}\theta dp d\theta d\phi$. The integrand only depends on the radial coordinate, so the solid angle part integrates to $4\pi$. $\endgroup$ – wsc Apr 18 '11 at 2:34
  • $\begingroup$ @wsc I'd post that as an answer as it probably answers the question satisfactorily :) $\endgroup$ – Lagerbaer Apr 18 '11 at 3:20
  • $\begingroup$ Thanks @wsc, I thought it had something to do with that. Since I'm having trouble, I'm going to walk through it 'out loud' here... If $E_p$ is only a function of $p_{\rho}$, it stays unchanged, and the $p^2$ in the numerator could be called $p_{\rho}^2$. The limits for $\rho$ remain $0$ and $\infty$, but the limits for $\theta$ and $\phi$ are reduced to their meaningful range. Using the notation more familiar to me, $\theta$ and $\phi$ factor out to $\int_0^{2\pi}d\theta\cdot\int_0^{\pi}\mathrm{sin}(\phi)d\phi=2\pi\cdot2$ $\endgroup$ – ShadSterling Apr 18 '11 at 3:43
  • $\begingroup$ ... applying that to the 2D case, converting to polar coordinates, the Jacobian is $r$, $\theta$ factors out into $\int_0^{2\pi}d\theta=2\pi$, the $r$ really means $p_r$, which is just $p$, and the whole thing goes like this: $$ N = g \cdot {A \over (2 \pi \hbar)^2} \cdot \int\limits_{0}^{\infty}{1 \over{e^{\left( E_p \over{K_B T}\right)}-1}} d^2 p = g \cdot {A \over (2 \pi \hbar)^2} \cdot 2 \pi \cdot \int\limits_{0}^{\infty}{p \over{e^{\left( E_p \over{K_B T}\right)}-1}} dp $$ That makes sense to me, now I can go trip over the next step... Thanks again @wsc! $\endgroup$ – ShadSterling Apr 18 '11 at 3:54
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    $\begingroup$ @Polyergic, that's exactly right. @Lagerbaer, I try to avoid submitting 2-sentence responses as answers, but I suppose there isn't much more to say here. :) $\endgroup$ – wsc Apr 18 '11 at 3:54

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