So, I'm doing an extensive homework of electromagnetism and we are searching for the total electromagnetic angular momentum of the Thomson dipole. In the end, there is one integral we cannot solve. By '"we" I mean the entire class: nobody is getting how to solve it. Professor swears there are no expansions or approximations, so...

The integral is, in spherical coordinates

$$\int_0^{2\pi} d\phi \int_0^\pi \frac{r^2(r-q\cos\theta)\cdot \sin\theta \cdot \cos\theta}{(r^2+q^2-2rq\cdot \cos\theta)^{3/2}}d\theta.$$

The solution is supposed to be $\frac{8\pi}{3}\frac{q}{r}$ for $r > q$ (q is the distance separating the monopoles). I could find only $\frac{4\pi}{3}\frac{q}{r}$, and with a negative sign. Can anybody illuminate me on a possible solution?

  • $\begingroup$ Use the substitution $$t=\sin\theta$$ $\endgroup$ – Shreyansh Pathak Sep 23 '19 at 6:35
  • 2
    $\begingroup$ @Unique That substitution will make the integral more complicated. $\endgroup$ – G. Smith Sep 23 '19 at 17:18

This is an easy integral, not a hard one, because there are two standard transformations that reduce it to trivial form. (“Trivial” means one you can remember.) You can use these two handy transformations in future integrals.

The integration over $\phi$ is already trivial.

For the integral over $\theta$, first change the variable of integration to $u=\cos\theta$, the usual substitution for polar-angle integrals. Also, let $\alpha$ be the dimensionless ratio $q/r$ because the integral depends only on this one parameter. Then one gets

$$2\pi\int_{-1}^1 \frac{u-\alpha u^2}{(1+\alpha^2-2\alpha u)^{3/2}}du.$$

Now change the variable of integration a second time to $v=1+\alpha^2-2\alpha u$, in order to simplify the denominator. It is usually better to have a complicated numerator than a complicated denominator. The result is three trivial integrals,

$$\frac{\pi}{4\alpha^2}\left[(1-\alpha^4)\int_{(1-\alpha)^2}^{(1+\alpha)^2} v^{-3/2}dv+2\alpha^2\int_{(1-\alpha)^2}^{(1+\alpha)^2} v^{-1/2}dv-\int_{(1-\alpha)^2}^{(1+\alpha)^2} v^{1/2}dv\right].$$

A little algebra gives the expected result,



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