3
$\begingroup$

Consider a spherical harmonics expansion/series like this: $$f(\theta,\varphi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m \, Y_\ell^m(\theta,\varphi)$$ Presumably if we take two functions on the sphere $f$ and $g$ that both have such expansions, then their product $h(\theta,\varphi)=f(\theta,\varphi)\,g(\theta,\varphi)$ also has an expansion. Now, my question is: how do the coefficients $f_\ell^m$, $g_\ell^m$ and $h_\ell^m$ relate?

Since the spherical harmonics expansion is quite similar to a Fourier series, I would expect some sort of convolution theorem to apply, but I can only really find one for the other way around. That is, that spherical convolution corresponds to multiplication of the coefficients in the expansion. I am interested in multiplication of spherical functions, and what that does to the coefficients.

I did find this question, but it does not really seem to be answered. In particular, the accepted answer refers to spherical convolution (which is the wrong way around).

$\endgroup$
  • 1
    $\begingroup$ Such relation involves Clebsch-Gordan coefficients. See, e.g. here. $\endgroup$ – Start wearing purple Jun 13 '13 at 13:32
  • $\begingroup$ Thanks! That's pretty much what I was looking for (except I'd prefer it in a form that gives me $h_\ell^m$ based on the coefficients of $f$ and $g$, but I'll try to figure that out myself). You may want to consider posting it as an answer, so I can mark the question as answered. $\endgroup$ – Jasper Jun 13 '13 at 14:08
4
$\begingroup$

I think I found what you were looking for (I've been looking all day!). See equations A.33-A.38 in http://geo.mff.cuni.cz/~lh/phd/cha.pdf : $$ h_{nm} = \sum_{n_1m_1} \sum_{n_2m_2}f_{n_1m_1}g_{n_2m_2}Q^{nm}_{n_1m_1n_2m_2} $$ where $$ Q^{nm}_{n_1m_1n_2m_2} = \sqrt{\frac{(2n_1+1)(2n_2+2)}{4\pi(2n+1)}} C^{n0}_{n_10n_20} C^{nm}_{n_1m_1n_2m_2} $$ and $C^{nm}_{n_1m_1n_2m_2}$ are the Clebsch-Gordan coefficients as mentioned by user "Start wearing purple" above

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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