This is a really basic question, but I'm kind of confused. I have this integral $$\int_{0}^{\infty}\frac{p^{2}dp}{e^{\alpha+\beta p^{2}/2m}+1}$$ where $p:=|\mathbf{p}|=\left(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\right)^{1/2}$ is the magnitude of the momentum of a particle of mass $m$. This integral represents the total number of particles in a Fermi-Dirac gas.

Now, I want to convert this integral to something of the form $$\int n(p_{x})dp_{x}$$ where $n(p_x)$ is the number of particles with momentum between the interval $(p_x,p_x+dp_x)$. To this end I must use cylindrical coordinates but I don't know how to convert the term $dp$ in something like $dp_xdp_r$. Can you help me please?


1 Answer 1


You have spherical coordinates but you want cylindrical coordinates with $p_x$ the axial direction (commonly labelled $z$).

When you do an integral, you want to integrate over an infinitesimal volume element, $d\tau$. For your spherical coordinates, this is $d\tau = p^2\sin\theta dp d\theta d\phi$ and for cylindrical coordinates it is $d\tau = p_\perp dp_\perp d\phi dp_x$ where $p_\perp^2 = p_y^2+p_z^2$. For pictures of these volume elements and discussion, see e.g. Griffiths, Electrodynamics, Section 1.4.

Your integral is spherically symmetric, so the angular variables can be integrated out. For the spherical integral we get:

$$\int d\tau\, f(p) = \int_0^\infty p^2 f(p) dp \int_0^{\pi}\sin \theta \,d\theta \int_0^{2\pi}d\phi = 4\pi\int_0^\infty p^2 f(p) dp $$

and for the cylindrical integral, we get:

$$\int d\tau\, f(p_x,p_\perp) = \int_{-\infty}^\infty dp_x\int_0^\infty p_\perp dp_\perp \,f(p_x,p_\perp)\int_0^{2\pi}d\phi = 2\pi\int_{-\infty}^\infty dp_x\int_0^\infty p_\perp dp_\perp \,f(p_x,p_\perp) $$

So, replace $4\pi\int_0^\infty p^2 f(p)dp$ with $2\pi \int_{-\infty}^\infty dp_x \int_0^\infty p_\perp dp_\perp f(p_x,p_\perp)$.

Your integral is then:

$$ 4\pi\int_0^\infty \frac{p^2 dp}{e^{\alpha+\beta p^2/2m}+1} = 2\pi\int_{-\infty}^\infty dp_x\int_0^\infty \frac{p_\perp dp_\perp}{e^{\alpha+\beta (p_x^2+p_\perp^2)/2m}+1} = \int_{-\infty}^\infty dp_x n(p_x) $$

where $n(p_x) = \frac {2\pi m}{\beta} \ln\left[ e^{-\alpha-\beta p_x^2/2m}+1 \right]$.

  • $\begingroup$ Can you explain the mathematical part, I mean the change of variables that you use? $\endgroup$
    – Ana S. H.
    Apr 6, 2013 at 23:40
  • 1
    $\begingroup$ I added some detail to my answer. I'd be happy to add more if wanted. $\endgroup$ Apr 7, 2013 at 10:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.