1
$\begingroup$

On the unit sphere, each square-integrable function can be expanded as a linear combination of spherical harmonics :

$$ f(\theta,\phi) = \Sigma_{l=0}^\infty \Sigma_{m=-l}^{+l} f_{lm} Y_{lm} (\theta,\phi)$$

With, reciprocally : $$ f_{lm} = \int_0^{2\pi} d\phi \int_0^{\pi} d\theta sin(\theta) Y_{lm}^* (\theta,\phi) f(\theta,\phi)$$

Which are the explicit values of $f_{lm}$, as expressions depending of $l$ and $m$, for the following function:

$$ f(\theta,\phi) = \frac {cos\theta}{1 - e^{i\phi} sin\theta}$$

Hint : The function seems to be a linear combination of only the $Y_{l\space{l-1}}$, with $l >0$ (so $m = l - 1$), but the $l$-dependence of the coefficients is not so easy.

See also Table of Spherical harmonics in Wikipedia

See also Digital Library of Mathematical Functions, for instance Refs 1 et 2 and all the chapter 14

See also Abramowitz and Stegun Ref 3 (and following pages)

$\endgroup$

1 Answer 1

10
$\begingroup$

Note that $$Y^*_{lm}(\theta,\phi)=C_{l}^m P_l^m(\cos\theta)e^{-im\phi}\tag{1},$$ with $\displaystyle C_{lm}=\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}$. Therefore in the double integral giving $f_{lm}$ we have in particular an integral over $\phi$ of the form $$\int_0^{2\pi}f(\theta,\phi)e^{-im\phi}d\phi=\begin{cases}2\pi \cos\theta\sin^m\theta,\qquad &m\geq0,\\0 \qquad &\mathrm{otherwise}.\end{cases}$$ which is calculated easily (e.g. by residues). But for $m\geq0$ the non-zero expression in the last formula is proportional to $P_{m+1}^{m}(\cos\theta)$: $$ C_{m+1}^m P_{m+1}^m(\cos\theta)=A_m\cdot2\pi\cos\theta\sin^m\theta,\tag{2}$$ where the coefficient $$ A_m=(-1)^m\frac{2^{-\frac{m}{2}}}{4\pi}\sqrt{\frac{1}{\pi}\frac{(2m+3)!!}{m!}}$$ can be calculated by calculating the norms $\int_0^{\pi}(\cdot)^2\sin\theta\,d\theta$ of the left and right sides of (2) and equating the corresponding expressions. This means that $$f(\theta,\phi)=\sum_{m=0}^{\infty}\frac{1}{2\pi A_m}Y_{m+1,m}(\theta,\phi)=\sum_{m=0}^{\infty}(-1)^m2^{\frac{m}{2}+1}\sqrt{\frac{\pi\cdot m!}{(2m+3)!!}}\;Y_{m+1,m}(\theta,\phi).$$ Hope this is useful.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .