# Spherical Harmonics

• Mar 26th 2013, 04:17 AM
zook104
Spherical Harmonics
Let

$\displaystyle h(\theta, \phi)= 1 + sin\phi sin\theta$

By taking linear combinations of these spherical harmonics, find Anm for all n
and m such that

$\displaystyle h(\theta, \phi)=\sum^{1}_{n=0}\sum^{n}_{m=-n} A_{nm}Y^{m}_{n}(\theta, \phi)$

Where $\displaystyle Y^{m}_{n}(\theta, \phi)$ is the first few spherical harmonics.

Any and All help is appreciated :)
• Mar 26th 2013, 03:15 PM
TheEmptySet
Re: Spherical Harmonics
Expand it just like a Fourier Series.

$\displaystyle Y_{0}^{0} (\theta,\phi) =\frac{1}{2}\sqrt{\frac{1}{\pi}}$

$\displaystyle Y_{0}^{0}h(\theta,\phi)=A_{00}Y_{0}^{0} (\theta,\phi)$

Now integrate both sides over a sphere of constant radius.

$\displaystyle \int_{0}^{\pi} \int_{0}^{2\pi} Y_{0}^{0}h(\theta,\phi)h(\theta,\phi)\sin(\theta)d \theta d\phi = \int_{0}^{\pi} \int_{0}^{2\pi} A_{00}Y_{0}^{0} (\theta,\phi)\sin(\theta)d\theta d\phi$

$\displaystyle \int_{0}^{\pi} \int_{0}^{2\pi}\frac{1}{2}\sqrt{\frac{1}{\pi}}(1+\ sin \phi \sin \theta)\sin(\theta)d\theta d\phi = \int_{0}^{\pi} \int_{0}^{2\pi} A_{00}\frac{1}{2}\sqrt{\frac{1}{\pi}}\sin(\theta)d \theta d\phi$

Now just integrate and solve for $\displaystyle A_{00}$

Then rinse and repeate with the next few spherical harmonics.