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$$r>1$$

The following is applying gauss' law by explicit integration.

$$\int_0^{\pi } \frac{\sin (\theta )}{\sqrt{r^2-2 r \cos (\theta )+1}} \, d\theta=\frac{2}{r}$$

The following is finding the potential of a charged spherical shell by explicit integration. Mathematica takes forever to solve this and I cannot figure out how it would be done.

$$\int _0^{2 \pi }\int _0^{\pi }\frac{\sin (\theta ) \cos (\theta )}{\sqrt{r^2-2 r (\cos (\psi )\cos (\theta ) +\sin (\psi )\sin (\theta ) \cos (\phi ))+1}}d\theta d\phi=\frac{4 \pi }{3}\frac{\cos(\psi)}{r^2}$$

Gappy Hilmore
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2 Answers2

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We begin with the integral $I(r,\psi)$ given by

$$I(r,\psi)=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\sin \theta \cos \theta}{\sqrt{r^2-2r\left(\cos \theta \cos \psi+\sin \theta \sin \psi \cos \phi\right)+1}}\,d\theta\,d\phi$$

Using spherical harmonics, we can expand the denominator of the integrand as

$$\frac{1}{\sqrt{r^2-2r\left(\cos \theta \cos \psi+\sin \theta \sin \psi \cos \phi\right)+1}}=4\pi\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}\frac{1}{2\ell +1}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}Y_{\ell m}(\theta,\phi)Y_{\ell m}(\psi,0)$$

The integration over $\phi$ is trivial and reveals

$$\begin{align} I(r,\psi)&=2\pi \int_{0}^{\pi}\sin \theta \cos \theta \sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \theta)P_{\ell}(\cos \psi)\,d\theta\\\\&=2\pi \sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \psi)\int_{0}^{\pi}\sin \theta \cos \theta P_{\ell}(\cos \theta)\,d\theta \tag 1 \end{align}$$

Noting that $P_1(\cos \theta)=\cos \theta$ and exploiting the orthogonality of the Legendre Polynomials, we obtain

$$\begin{align} I(r,\psi)&=2\pi\sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \psi)\left(\frac{2}{2\ell +1}\delta_{\ell,1}\right)\\\\ &=\frac{4\pi}{3}\frac{r_{<}}{r_{>}^{2}}P_{1}(\cos \psi)\\\\ &=\frac{4\pi}{3}\frac{r_{<}}{r_{>}^{2}}\cos \psi\\\\ \end{align}$$

For $r>1$, $r_{<}=1$ and $r_{>}=r$. Finally, we have

$$\bbox[5px,border:2px solid #C0A000]{I(r,\psi)=\frac{4\pi}{3} \frac{\cos \psi}{r^2}}$$

Mark Viola
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    It's always a bit disappointing when we have to resort to orthogonal polynomials. But I can't see a better way, so (+1). – Chappers Oct 01 '15 at 19:04
  • @Chappers Sad? I become happy and warm inside when I see this solution. – mickep Oct 01 '15 at 19:17
  • I'm with @Chappers. Orthogonal polynomials, while correct and worthy of praise, do not inspire me to think further about the problem. Maybe because I am a physicist and these polynomials seem more like the drill I use during my day job: messy and boring. Still, and I repeat, good for MV for applying them so skillfully. – Ron Gordon Oct 01 '15 at 19:21
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    @rongordon In my former life, I was a scientist also. And you're correct. The use herein is a bit mind-numbing. But, as I stated earlier, brute force works well when it works. Mark – Mark Viola Oct 01 '15 at 19:29
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    @mickep Well, pleased that this warmed you. ;-))... Mark – Mark Viola Oct 01 '15 at 19:40
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    @Chappers Well, I confess .... I took the easy way here. There might be much more creative approaches, but brute force works when it works. And thank you for the up vote! Much appreciative. – Mark Viola Oct 01 '15 at 21:05
  • How did you decompose that 1/sqrt into spherical Y's ? Is that an identity? – Gappy Hilmore Oct 01 '15 at 21:31
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    @grdgfgr The square root term is a Green Function (or Green's Function, depending on whom you wish to believe). It is a solution to Laplace's Equation for $r\ne 1$, $\theta \ne \psi$, $\phi \ne 0$. It can be decomposed, therefore, in terms of spherical harmonics, which one can show by using separation of variables (for example). – Mark Viola Oct 01 '15 at 21:37
  • @Dr.MV but how do you find the coefficients? – Gappy Hilmore Oct 01 '15 at 21:39
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    @grdgfgr Ah, good question. We find the coefficients by enforcing boundary conditions across $r=1$. These are $(i)$ continuity and $(ii)$ discontinuity of the partial derivative with respect to $r$. The discontinuity manifests in the $r_{<}$ and $r_{>}$ terms. – Mark Viola Oct 01 '15 at 21:47
  • @Dr.MV I'm sorry I have another question, it is difficult to follow without being familiar with spherical harmonics. I understand integrating wrt $\phi$ gets rid of all the terms with $e^{i m \phi},m\ne0$ But I do not understand where $\frac{1}{2l+1}$ went and why $4 \pi$ became $2\pi$. Is that how $Y_l ^m$ is normalized? – Gappy Hilmore Oct 02 '15 at 14:47
  • @grdgfgr Let me see if I can help. Here is an article that might answer the new questions. – Mark Viola Oct 02 '15 at 15:16
4

Substitution Method

Let $r^2-2r\cos \theta+1=u^2\implies 2r\sin\theta d\theta=2udu$ or $\sin\theta d\theta=\frac{udu}{r}$ $$\int_{0}^{\pi}\frac{\sin\theta}{\sqrt{r^2-2r\cos \theta+1}}d\theta=\int_{r-1}^{r+1}\frac{1}{u}\frac{udu}{r}$$ $$=\frac{1}{r}\int_{r-1}^{r+1}du=\frac{2}{r}$$

Harish Chandra Rajpoot
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