# How do I derive the critical temperature for bose condensation in two dimensions?

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.

• 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 $4\pi$. – wsc Apr 18 '11 at 2:34
• @wsc I'd post that as an answer as it probably answers the question satisfactorily :) – Lagerbaer Apr 18 '11 at 3:20
• 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 $\int_0^{2\pi}d\theta\cdot\int_0^{\pi}\mathrm{sin}(\phi)d\phi=2\pi\cdot2$ – ShadSterling Apr 18 '11 at 3:43
• ... 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! – ShadSterling Apr 18 '11 at 3:54
• @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. :) – wsc Apr 18 '11 at 3:54