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.

  • 1
    $\begingroup$ 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 .. $\endgroup$ Commented Jan 14, 2014 at 12:57
  • $\begingroup$ ..check the spherical harmonics wiki page: there are specific harmonics at the bottom of the page for low orders to help you out. $\endgroup$ Commented Jan 14, 2014 at 12:59
  • $\begingroup$ 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 $\phi$. $\endgroup$
    – alexjo
    Commented Jan 15, 2014 at 17:19

1 Answer 1


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 $$\small \Phi(r,\theta,\varphi)=\frac{Q}{a}\left[\frac{1}{r^2}\left(\sqrt{\frac{2\pi}{3}}Y_1^{-1}(\theta,\varphi)-\sqrt{\frac{2\pi}{3}}Y_1^{1}(\theta,\varphi)\right)+\frac{\sqrt{\pi}}{r}Y_0^0(\theta,\varphi)+\frac{1}{r^3}\left(\frac{2}{3}\sqrt{\frac{\pi}{5}}\right)Y_2^0(\theta,\varphi)\right] $$

  • 1
    $\begingroup$ 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. $\endgroup$
    – Neuneck
    Commented Jan 14, 2014 at 21:21
  • $\begingroup$ @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 $a$? $\endgroup$
    – peripatein
    Commented Jan 15, 2014 at 13:13
  • $\begingroup$ This isn't the final solution, right? I mean, isn't normalisation still necessary? $\endgroup$
    – peripatein
    Commented Jan 15, 2014 at 14:47
  • $\begingroup$ @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. $\endgroup$
    – alexjo
    Commented Jan 15, 2014 at 14:55
  • $\begingroup$ I believe relation (2) ought to be emended. $\endgroup$
    – peripatein
    Commented Jan 15, 2014 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.