Trigonometric function expanded in spherical harmonics

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  • #1
M_1
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Main Question or Discussion Point

Is it possible to express (cos([itex]\theta[/itex])sin([itex]\theta[/itex]))^2 in terms of spherical harmonics?
 

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  • #2
tiny-tim
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Hi M_1! :smile:
Is it possible to express (cos([itex]\theta[/itex])sin([itex]\theta[/itex]))^2 in terms of spherical harmonics?
Isn't that 1/4 sin22θ ?
 
  • #3
DrClaude
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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
M_1
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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*[itex]\sqrt{\pi /5}[/itex]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
DrClaude
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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.
 
  • #6
M_1
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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
DrClaude
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Is it important that you have no ##\phi## dependence?
 
  • #8
M_1
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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
[itex]\int[/itex]r5sin([itex]\theta[/itex])cos([itex]\theta[/itex])sin([itex]\theta[/itex])d[itex]\theta[/itex]d[itex]\varphi[/itex] where
r=r0(1+[itex]\alpha[/itex]Y20sin([itex]\varpi[/itex]t))
where [itex]\alpha[/itex] 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 [itex]\varphi[/itex]-dependence.
 
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  • #9
DrClaude
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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".
 
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  • #10
M_1
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Excellent! Problem solved. Many thanks DrClaude!
 

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