# Trigonometric function expanded in spherical harmonics

## Main Question or Discussion Point

Is it possible to express (cos($\theta$)sin($\theta$))^2 in terms of spherical harmonics?

tiny-tim
Homework Helper
Hi M_1! Is it possible to express (cos($\theta$)sin($\theta$))^2 in terms of spherical harmonics?
Isn't that 1/4 sin22θ ?

DrClaude
Mentor
Yes, but it depends on what you want. For instance,
$$-\frac{32 \pi^2}{9} Y_1^{-1} Y_1^1 \left(Y_1^0\right)^2 = \cos^2 \theta \sin^2 \theta$$
But if you require a result in terms of a sum of spherical harmonics, then this requires more work.

Hi!
Thanks! Yes but then the next question is if it is possible to express 1/4 (sin(2θ)))^2 in terms of spherical harmonics.

For example we can write cos2(θ)=1/3(4*$\sqrt{\pi /5}$Y20+1),
but is such an expansion possible also for 1/4 (sin(2θ)))^2? I need an a linear expansion in only spherical harmonics (not combined with trigonometric functions).

DrClaude
Mentor
I need an a linear expansion in only spherical harmonics (not combined with trigonometric functions).
Telling us what you need this for will help us help you.

Hi DrClaude,
Thanks! But as you correctly guessed I need a sum of spherical harmonics. I would be glad to do the work but I don't know how to start and more scaringly I'm not sure if it is possible. That's why I first would like to find out if it is possible.

DrClaude
Mentor
Is it important that you have no $\phi$ dependence?

Hi DrClaude,
I'm trying do calculate the exact solution for the vibrational energy of an inviscid droplet with a predetermined surface tension, radius, and amplitude.

In order to do so I need to solve the integral
$\int$r5sin($\theta$)cos($\theta$)sin($\theta$)d$\theta$d$\varphi$ where
r=r0(1+$\alpha$Y20sin($\varpi$t))
where $\alpha$ is the relative amplitude.

I try do use the nice methodology of Baxansky and Kiryati
http://dx.doi.org/10.1016/j.patcog.2006.06.001 [Broken]
but I get stuck on this issue.

There can be no $\varphi$-dependence.

Last edited by a moderator:
DrClaude
Mentor
I've got it. The answer is
$$\frac{4}{15} \sqrt{\pi } Y_0^0(\theta ,\phi )+\frac{4}{21} \sqrt{\frac{\pi }{5}} Y_2^0(\theta ,\phi )-\frac{16}{105} \sqrt{\pi } Y_4^0(\theta ,\phi )$$
You can check it for yourself in WolframAlpha. Look at the "Alternate forms".

• 1 person
Excellent! Problem solved. Many thanks DrClaude!