Trigonometric function expanded in spherical harmonics

1. Oct 18, 2013

M_1

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

2. Oct 18, 2013

tiny-tim

Hi M_1!
Isn't that 1/4 sin22θ ?

3. Oct 18, 2013

Staff: 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.

4. Oct 18, 2013

M_1

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).

5. Oct 18, 2013

Staff: Mentor

Telling us what you need this for will help us help you.

6. Oct 18, 2013

M_1

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.

7. Oct 18, 2013

Staff: Mentor

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

8. Oct 18, 2013

M_1

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: May 6, 2017
9. Oct 18, 2013

Staff: 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".

10. Oct 21, 2013

M_1

Excellent! Problem solved. Many thanks DrClaude!