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- Thread starter M_1
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- #2

- #3

DrClaude

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$$

-\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

- #4

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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 cos

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

- #6

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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

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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]r^{5}sin([itex]\theta[/itex])cos([itex]\theta[/itex])sin([itex]\theta[/itex])d[itex]\theta[/itex]d[itex]\varphi[/itex] where

r=r_{0}(1+[itex]\alpha[/itex]Y_{20}sin([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.

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]r

r=r

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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$$

\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

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Excellent! Problem solved. Many thanks DrClaude!

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