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Trigonometric function expanded in spherical harmonics

  1. Oct 18, 2013 #1

    M_1

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    Is it possible to express (cos([itex]\theta[/itex])sin([itex]\theta[/itex]))^2 in terms of spherical harmonics?
     
  2. jcsd
  3. Oct 18, 2013 #2

    tiny-tim

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    Hi M_1! :smile:
    Isn't that 1/4 sin22θ ?
     
  4. Oct 18, 2013 #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.
     
  5. Oct 18, 2013 #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).
     
  6. Oct 18, 2013 #5

    DrClaude

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    Telling us what you need this for will help us help you.
     
  7. Oct 18, 2013 #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.
     
  8. Oct 18, 2013 #7

    DrClaude

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    Is it important that you have no ##\phi## dependence?
     
  9. Oct 18, 2013 #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.
     
    Last edited by a moderator: May 6, 2017
  10. Oct 18, 2013 #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".
     
  11. Oct 21, 2013 #10

    M_1

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    Excellent! Problem solved. Many thanks DrClaude!
     
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