Legendre polynomial.

Question

How may I express $$V(\theta,\phi) = (Q/a)(\sin(\theta)\cos(\phi) + 0.5\cos^2(\theta))$$ as a Legendre polynomial? I am confused by the presence of $\cos(\phi)$.

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 $$ \cos^2\theta=\frac{1}{3}P_0^0+\frac{2}{3}P_0^2 $$ 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 $$ \cos^2\theta=\frac{1}{3}P_0^0(\cos\theta)+\frac{2}{3}P_0^2(\cos\theta)=\sqrt{4\pi}Y_0^0(\theta,\varphi)+\frac{4}{3}\sqrt{\frac{\pi}{5}}Y_2^0(\theta,\varphi).\tag 6 $$ Finally, putting together (5) and (6) in (1) we obtain $$ f(\theta,\varphi)=2\sqrt{\frac{2\pi}{3}}Y_1^{-1}(\theta,\varphi)-2\sqrt{\frac{2\pi}{3}}Y_1^{1}(\theta,\varphi)+\sqrt{4\pi}Y_0^0(\theta,\varphi)+\frac{4}{3}\sqrt{\frac{\pi}{5}}Y_2^0(\theta,\varphi)\tag 7 $$ so that, comparing (7) and (2), the coefficients $f_\ell^m$ are $$ f_1^{-1}=2\sqrt{\frac{2\pi}{3}}\qquad f_1^{1}=-2\sqrt{\frac{2\pi}{3}}\qquad f_0^{0}=\sqrt{4\pi}\qquad f_2^{0}=\frac{4}{3}\sqrt{\frac{\pi}{5}}\tag 8 $$ So you have $$ V(\theta,\varphi)=\frac{Q}{a}\left[\sqrt{\frac{2\pi}{3}}Y_1^{-1}(\theta,\varphi)-\sqrt{\frac{2\pi}{3}}Y_1^{1}(\theta,\varphi)+\sqrt{\pi}Y_0^0(\theta,\varphi)+\frac{2}{3}\sqrt{\frac{\pi}{5}}Y_2^0(\theta,\varphi)\right] $$