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I have been trying to solve the following question:

The potential on the surface of a sphere is given by $\mathbf {V = V_{0} \sin^2\theta \sin2\phi,\;}$ find the potential outside the sphere

I am trying to solve it by separation of variable in spherical coordinates by using the following formula for potential outside the sphere,

$$V=\sum_{l=0}^\infty\frac{B_{lm}}{r^{l+1}} {Y_l}^m (\theta,\phi)$$

Now the potential on the surface of the sphere is given, so we can use that for r=R as,

$$\tag{1}V_{0}\sin^2\theta\sin 2\phi=\sum_{l=0}^\infty\frac{B_{lm}}{R^{l+1}} {Y_l}^m (\theta,\phi)$$

Next for the value of $B_l$ I multiply both side with ${Y^*_l}^m$ and integrate. RHS becomes $\frac{B_{lm}}{r^{l+1}}$ while LHS becomes interesting. I note that $\sin^2\theta$ $\sin 2\phi$ can be converted into $Y_2^2$ with some Constant factor. $Y_2^2$ is given as follows: $$ Y_2^2= A \sin^2\theta\ e^{im\phi}$$ So my problem is, can I some how convert this into $Y_2^2$ so that it simply gives me the left hand side of equation ${(1)}?\;$ I see that $sin2\phi$ is the imaginary part of $e^{im\phi}$ with $m=2$. Please guide me through this.

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  • $\begingroup$ After multiplying equation $(1)$ by $Y^{*m}_{l}$, then you would still need to integrate both sides over the solid angle $d\Omega$ - which would give you the $B_{l}$ coefficient. Also, I'm presuming the right hand side of equation $(1)$ is the potential resulting from a charge distribution on the surface of the sphere, and you're trying to solve for the potential outside the sphere, $$\Phi(r,\theta,\phi)=\sum^{l=\infty}_{l=0}\sum_{m=-l}^{m=l}B_{lm}r^{-(l+1)}Y_{lm}(\theta,\phi)$$. $\endgroup$ Nov 21 '19 at 12:22
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You already noticed $\sin 2\phi$ is the imaginary part of $e^{2i\phi}$. Another way to say this is $$\sin 2\phi=\frac{i}{2}\left(-e^{2i\phi}+e^{-2i\phi}\right)$$

From the table of spherical harmonics you have: $$Y_2^{+2}(\theta,\phi)=\frac{1}{4}\sqrt{\frac{15}{2\pi}}\sin^2\theta \ e^{2i\phi}$$ $$Y_2^{-2}(\theta,\phi)=\frac{1}{4}\sqrt{\frac{15}{2\pi}}\sin^2\theta \ e^{-2i\phi}$$

Putting this together you find (even without doing any integral): $$\sin^2\theta\ \sin 2 \phi= 2i\sqrt{\frac{2\pi}{15}} \left(-Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right)$$ You see, one spherical harmonic was not enough. You needed two of them.

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What if you assume that the initial potential has its imaginary component, just for the sake of the argument Vo $\sin^2\theta$ $\sin 2\phi$ = Re{$U_0Y_2^2$} where $U_0$ is some constant you get from the definition of your potential and $Y_2^2$? Then you indeed can prove that the potential outside the sphere is the same $Y_2^2$ with some constant factor. The imaginary part has no physical meaning then, you use only the real one since all equations must hold if you apply Re{} or Im{} to them. My guess you will come to something like C$\cdot$Vo $\sin^2\theta$ $\sin 2\phi / r^3$

Spherical harmonics are orthonormal. If you have a spherical harmonic on one side, you have to have the same one on the other, no more and no less.

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  • $\begingroup$ Absolute beauty! Answer indeed turned out to be $R^3$Vo $\sin^2\theta$ $\sin 2\phi / r^3$ Totally enthralled by the way math unfold and then folds itself! Thank You. $\endgroup$
    – Fasi
    Nov 23 '19 at 5:59

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