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I'm trying to calculate this:

$$\sum_{l=0}^{\infty} \int_{\Omega} d\theta' d\phi' \cos{\theta'} \sin{\theta'} P_l (\cos{\gamma})$$

where $P_{l}$ are the Legendre polynomials, $\Omega$ is the surface of a sphere of radius $R$, and

$$ \cos{\gamma} = \cos{\theta'} \cos{\theta} + \sin{\theta'}\sin{\theta}\,\cos({\phi' -\phi}) $$

I am told that only the $l=1$ term survives due to orthogonality of Legendre polynomials (of course $\cos{\theta'} = P_{1}(\cos{\theta'}) $), but I'm don't see why, since the Legendre polynomials have different arguments.

How can I show that this is true?

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{\ell = 0}^{\infty}\int_{\Omega'}\dd\theta'\,d\phi'\,\cos{\theta'} \sin{\theta'}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \\[5mm] = &\ \sum_{\ell = 0}^{\infty}\int_{\Omega'} \overbrace{\bracks{% 2\root{\pi \over 3}\,\mrm{Y}_{10}\pars{\Omega'}}}^{\ds{\cos\pars{\theta'}}}\ \overbrace{\bracks{{4\pi \over 2\ell + 1}\sum_{m = -\ell}^{\ell} \mrm{Y}_{\ell m}\pars{\Omega}\,\mrm{Y}_{\ell m}^{*}\pars{\Omega'}}} ^{\ds{\mrm{P}_{\ell}\pars{\cos\pars{\gamma}}}}\ \dd\Omega' \\[5mm] = &\ 4\pi \sum_{\ell = 0}^{\infty}\sum_{m = -\ell}^{\ell} 2\root{\pi \over 3}{\mrm{Y}_{\ell m}\pars{\Omega}\, \over 2\ell + 1}\ \underbrace{\int_{\Omega'} \mrm{Y}_{10}\pars{\Omega'}\,\mrm{Y}_{\ell m}^{*}\pars{\Omega'}\dd\Omega'} _{\ds{\delta_{\ell 1}\,\delta_{m0}}} \\[5mm] = &\ 4\pi\bracks{{1 \over 2 \times 1 + 1} \,2\root{\pi \over 3}\mrm{Y}_{10}\pars{\Omega}} = \bbx{{4\pi \over 3}\,\cos\pars{\theta}}\quad\mbox{because}\quad 2\root{\pi \over 3}\,\mrm{Y}_{10}\pars{\Omega} = \cos\pars{\theta} \end{align}

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