# Spherical harmonics

Given the following potential: $$V(\theta,\phi)=\frac{Q}{a}\left(\sin\theta \cos\phi+\frac{1}{2}\cos^2\theta\right)$$ on the surface of a sphere of radius $a$ I am trying to solve Laplace's Equation outside the sphere (where there aren't any charges). I know the general solution to Laplace's Equation outside the sphere is given by: $$\phi(r,\theta,\phi)=\sum_{l=0}B_l r^{-l-1}P_l(\cos\theta).$$ I am not quite sure how to proceed as am very new to spherical harmonics. Does the next step involve expressing the given potential as a Legendre polynomial? I'd appreciate some guidance.

• Your general solution isn't quite general enough: it should be $f(r, \theta, \varphi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m \, r^\ell \, Y_\ell^m (\theta, \varphi )$ where $Y_\ell^m (\theta, \varphi ) = Y_\ell^m (\theta, \varphi ) = N_{\ell,m} \, e^{i m \varphi } \, P_\ell^m (\cos{\theta} )$ and $P_\ell^m$ is the associated Legendre "polynomial". So you have one term $\sin\theta\cos\phi$, so you're going to need terms of the form $P_1^{\pm1} (\cos{\theta} ) e^{\pm i \varphi}$ to represent than and terms $P_2^0 (\cos{\theta} )$ and $P_0^0 (\cos{\theta} )=1$ and go from there .. – WetSavannaAnimal Jan 14 '14 at 12:57
• ..check the spherical harmonics wiki page: there are specific harmonics at the bottom of the page for low orders to help you out. – WetSavannaAnimal Jan 14 '14 at 12:59
• Observe that $\Phi(r,\theta,\phi)=\sum_{l=0}^\infty B_l r^{-l-1}P_l(\cos\theta)$ only if there is no dependence on $\phi$!! in your case you have to consider the azimutal coordinate $\phi$. – alexjo Jan 15 '14 at 17:19

Let be $$\frac{2a}{Q}V(\theta,\varphi)=f(\theta,\varphi)=2\sin\theta\cos\varphi+\cos^2\theta.\tag 1$$ The Laplace spherical harmonics form a complete set of orthonormal functions and thus form an orthonormal basis of the Hilbert space of square-integrable functions. On the unit sphere, any square-integrable function can thus be expanded as a linear combination of these: $$f(\theta,\varphi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m \, Y_\ell^m(\theta,\varphi)\tag 2$$ where $Y_\ell^m( \theta , \varphi )$ are the Laplace spherical harmonics defined as $$Y_\ell^m( \theta , \varphi ) = \sqrt{{(2\ell+1)\over 4\pi}{(\ell-m)!\over (\ell+m)!}} \, P_\ell^m ( \cos{\theta} ) \operatorname{e}^{i m \varphi } =N_{\ell}^m P_\ell^m ( \cos{\theta} ) \operatorname{e}^{i m \varphi }\tag 3$$ and where $N_{\ell}^m$ denotes the normalization constant $N_{\ell}^m \equiv \sqrt{{(2\ell+1)\over 4\pi}{(\ell-m)!\over (\ell+m)!}},$ and $P_\ell^n(\cos\theta)$ are the associated Legendre polynomials.

The Laplace spherical harmonics are orthonormal $$\int_{\theta=0}^\pi\int_{\varphi=0}^{2\pi}Y_\ell^m \, Y_{\ell'}^{m'*} \, d\Omega=\delta_{\ell\ell'}\, \delta_{mm'},$$ where $δ_{ij}$ is the Kronecker delta and $\operatorname{d}\Omega = \sin\theta \operatorname{d}\varphi\operatorname{d}\theta$.

The expansion coefficients are the analogs of Fourier coefficients, and can be obtained by multiplying the above equation by the complex conjugate of a spherical harmonic, integrating over the solid angle $Ω$, and utilizing the orthogonality relationships. This is justified rigorously by basic Hilbert space theory. For the case of orthonormalized harmonics, this gives: $$f_\ell^m=\int_{\Omega} f(\theta,\varphi)\, Y_\ell^{m*}(\theta,\varphi)\operatorname{d}\Omega = \int_0^{2\pi}\operatorname{d}\varphi\int_0^\pi \operatorname{d}\theta\,\sin\theta f(\theta,\varphi)Y_\ell^{m*} (\theta,\varphi). \tag 4$$ where $Y_\ell^{m*} (\theta, \varphi) = (-1)^m Y_\ell^{-m} (\theta, \varphi)$.

The evaluation of the expansion $f_\ell^m$ may be very long in this way...

We can use some tricks in your case for $f(\theta,\varphi)=2\sin\theta\cos\varphi+\cos^2\theta$ observing that $$\sin\theta=-P_1^1(\cos\theta)$$ and $$Y_{1}^{-1}(\theta,\varphi) - Y_{1}^{1}(\theta,\varphi)= {1\over 2}\sqrt{3\over 2\pi}\cdot e^{-i\varphi}\cdot\sin\theta-{-1\over 2}\sqrt{3\over 2\pi}\cdot e^{i\varphi}\cdot\sin\theta =\sqrt{3\over 2\pi} \sin\theta\cos\phi$$ so that $$2\sin\theta\cos\varphi=2\sqrt{\frac{2\pi}{3}}\left(Y_1^{-1}(\theta,\varphi)-Y_1^{1}(\theta,\varphi)\right)=-2P_1^1(\cos\theta)\cos\varphi\tag 5$$ and observing that $$\cos^2\theta=\frac{1}{3}P_0^0+\frac{2}{3}P_2^0$$ and using the relation $Y_\ell^0(\theta,\varphi)=\sqrt{\frac{2\ell+1}{4\pi}}P_\ell^0(\cos\theta)$ where $P_\ell^0(\cos\theta)$ are the ordinary Legendre's polynomials $P_\ell(\cos\theta)$, we obtain $$\cos^2\theta=\frac{1}{3}P_0^0(\cos\theta)+\frac{2}{3}P_2^0(\cos\theta)=2\sqrt{\pi}Y_0^0(\theta,\varphi)+\frac{4}{3}\sqrt{\frac{\pi}{5}}Y_2^0(\theta,\varphi).\tag 6$$ Finally, putting together (5) and (6) in (1) we obtain $$f(\theta,\varphi)=2\sqrt{\frac{2\pi}{3}}Y_1^{-1}(\theta,\varphi)-2\sqrt{\frac{2\pi}{3}}Y_1^{1}(\theta,\varphi)+2\sqrt{\pi}Y_0^0(\theta,\varphi)+\frac{4}{3}\sqrt{\frac{\pi}{5}}Y_2^0(\theta,\varphi)\tag 7$$ so that, comparing (7) and (2), the coefficients $f_\ell^m$ are $$f_1^{-1}=2\sqrt{\frac{2\pi}{3}}\qquad f_1^{1}=-2\sqrt{\frac{2\pi}{3}}\qquad f_0^{0}=2\sqrt{\pi}\qquad f_2^{0}=\frac{4}{3}\sqrt{\frac{\pi}{5}}\tag 8$$ So you have $$\small V(\theta,\varphi)=\frac{Q}{a}\left[\sqrt{\frac{2\pi}{3}}Y_1^{-1}(\theta,\varphi)-\sqrt{\frac{2\pi}{3}}Y_1^{1}(\theta,\varphi)+\sqrt{\pi}Y_0^0(\theta,\varphi)+\frac{2}{3}\sqrt{\frac{\pi}{5}}Y_2^0(\theta,\varphi)\right]$$ The general solution to the Laplace equation outside the sphere is $$\Phi(r,\theta,\varphi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell \frac{B_\ell^m}{r^{\ell+1}} \, Y_\ell^m(\theta,\varphi)\tag 9$$ with $B_\ell^m=\frac{Q}{a}f_\ell^m$, that is $$\small \Phi(r,\theta,\varphi)=\frac{Q}{a}\left[\frac{1}{r^2}\left(\sqrt{\frac{2\pi}{3}}Y_1^{-1}(\theta,\varphi)-\sqrt{\frac{2\pi}{3}}Y_1^{1}(\theta,\varphi)\right)+\frac{\sqrt{\pi}}{r}Y_0^0(\theta,\varphi)+\frac{1}{r^3}\left(\frac{2}{3}\sqrt{\frac{\pi}{5}}\right)Y_2^0(\theta,\varphi)\right]$$

• Nicely explained! In practice, it is very useful to understand that the $Y_l^0$ are independent of $\phi$, and terms like $\cos(n \phi)$ (or $\sin(n \phi)$) can be constructed as the linear combination of $Y_l^n \pm Y_l^{-n}$. Thus the experienced eye can see that the term with $\cos(\phi)\sin(\theta)$ most certainly involves $Y_1^1$ and $Y_1^{-1}$ and the $\cos^2(\theta)$ involves only $Y_l^0$s. – Neuneck Jan 14 '14 at 21:21
• @alexjo Alright, first, thank you so much! Second, is your final expression for V($\theta$,$\phi$) the final answer to this problem? Is that the solution for the potential outside the sphere of radius $a$? – peripatein Jan 15 '14 at 13:13
• This isn't the final solution, right? I mean, isn't normalisation still necessary? – peripatein Jan 15 '14 at 14:47
• @peripatein the expression of $V(\theta,\varphi)$ is the expansion for $V$ in terms of spherical harmonics. I added the final expansion for $\Phi(r,\theta,\varphi)$. No normalization is needed; it's still inside. – alexjo Jan 15 '14 at 14:55
• I believe relation (2) ought to be emended. – peripatein Jan 15 '14 at 16:13