# Finding potential using spherical harmonics

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.

• 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, $$\Phi(r,\theta,\phi)=\sum^{l=\infty}_{l=0}\sum_{m=-l}^{m=l}B_{lm}r^{-(l+1)}Y_{lm}(\theta,\phi)$$. Nov 21 '19 at 12:22

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.

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.

• 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.
– Fasi
Nov 23 '19 at 5:59