Legendre Polynomial Integral

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Homework Statement


As part of a larger problem, I need to compute the following integral (over [itex]-1<\theta<1[/itex]):

[tex]\int \sin \theta P_{l}(\cos \theta) d (\cos \theta)[/tex]

Homework Equations



[tex]\int P_{l}(x) P_{l'}(x) dx= \frac{2}{2l+1} \delta_{l',l}[/tex]

Also, solutions are known to the following integrals:

[tex]\int x P_{l}(x) P_{l'}(x) dx[/tex]

[tex]\int x^{2} P_{l}(x) P_{l'}(x) dx[/tex]

The Attempt at a Solution



Trig identities for converting [itex]\sin \theta[/itex] into [itex]\cos \theta[/itex] don't seem to help, here - which is why I'm confused.

I would presume to convert the integrand to a form that only has cosines. I would then use the fact that [itex]P_{0}(x)=1[/itex] with orthogonality (listed above) to get the explicit solution.

How do I get rid of that [itex]\sin \theta[/itex]?!

EDIT:
Alright, I just found another formula online that was not in my text:

[tex]\int g(x) P_{l} (x) dx = \frac{(-1)^{l}}{2^{l} l!} \int (x^{2} - 1)^{l} \frac{d^{l}}{dx^{l}} g(x) dx[/tex]

If I were to use this, I might first convert the sine using

[tex]\sin \theta = (1-\cos ^{2} \theta)^{\frac{1}{2}}[/tex]

and then make that my [itex]g(x)[/itex].

I don't think this would really help me, though, since that would leave me with an awkward expression involving the l'th derivative inside an integral. So I'm still stuck.
 
Last edited:

Answers and Replies

  • #2
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Look for the orthogonality of 'associated Legendre functions' in any mathematical physics textbook.
 
  • #3
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Alright, the associated Legendre polynomials are given by:

[tex]P_{l}^{m}(x)=(-1)^{m}(1-x^{2})^{\frac{m}{2}}\frac{d^{m}}{dx^{m}}P_{l}(x)[/tex]

With [itex]x=\cos \theta '[/itex], the integral I'm having trouble with is given by:

[tex]I=\int (1-x^{2})^{\frac{1}{2}}P_{l}(x)dx[/tex]

The associated Legendre polynomials for m equal to 0 and 1 are:

[tex]P_{l}^{0}(x)=P_{l}(x)[/tex]
[tex]P_{l}^{1}(x)=-(1-x^{2})^{\frac{1}{2}}\frac{d}{dx}P_{l}(x)[/tex]

Since [itex]P_{1}(x)=x[/itex], we could write the integral as

[tex]I=-\int P_{1}^{1}(x) P_{l}^{0}(x) dx[/tex]

but then we can't use orthogonality, given for associated Legendre polynomials by

[tex]\int P_{l}^{m}(x) P_{l'}^{m}(x) dx = \frac{2}{2l + 1} \frac{(l+m)!}{(l-m)!} \delta_{l',l}[/tex]

I'm still working on it; I've got some ideas to play around with.
 
  • #4
gabbagabbahey
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Homework Statement


As part of a larger problem, I need to compute the following integral (over [itex]-1<\theta<1[/itex]):

[tex]\int \sin \theta P_{l}(\cos \theta) d (\cos \theta)[/tex]

This integral seems a bit unusual to me....what was the larger problem that this came out of?...Are you sure you don't actually want to compute [tex]\int \sin \theta P_{l}(\cos \theta) d \theta[/tex] over [itex]-\pi<\theta<\pi[/itex] instead?
 
  • #5
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This integral seems a bit unusual to me....what was the larger problem that this came out of?...Are you sure you don't actually want to compute [tex]\int \sin \theta P_{l}(\cos \theta) d \theta[/tex] over [itex]-\pi<\theta<\pi[/itex] instead?

I need to compute (in spherical)

[tex]\int \frac{1}{|\vec{x} - \vec{x'}|} \sin \theta' \delta (\rho' - a) d^{3} x'[/tex]

Originally, I thought to write out the difference between the two vectors in spherical, and do the integral (as you may have noticed by my question in the Calculus forum). After a bit of a struggle, I realized that I could just use an expansion in spherical harmonics.

The problem has azimuthal symmetry, so the spherical harmonics can be reduced to Legendre polynomials. I am then left with (where I will write out the volume element explicitly):

[tex]\int \int \int \sin \theta' \delta (\rho' - a) \frac{\rho_{<}^{l}}{\rho_{>}^{l+1}} P_{l}^{*} (\cos \theta') P_{l} (\cos \theta) \rho' ^{2} \sin \theta' d\rho' d\theta' d\phi'[/tex]

And using [itex] \sin \theta' d\theta' = -d(\cos\theta')[/itex], I'm left trying to solve the angular integral mentioned above.
 
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  • #6
gabbagabbahey
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I need to compute (in spherical)

[tex]\int \frac{1}{|\vec{x} - \vec{x'}|} \sin \theta' \delta (\rho - a) d^{3} x'[/tex]

I assume this integral is over all of space?
 
  • #7
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I assume this integral is over all of space?

Yes, sorry. I've been lazy with the bounds of integration.

Also, I forgot the prime on the rho. It's been fixed.
 
  • #8
gabbagabbahey
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Are you trying to find the potential at a point [itex]\vec{x}[/itex] do to a spherical shell of mass (or charge) of radius [itex]a[/itex]?....If so, just orient your co-ordinate system so that [itex]\vec{x}[/itex] lies on the positive z-axis.
 
  • #9
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Are you trying to find the potential at a point [itex]\vec{x}[/itex] do to a spherical shell of mass (or charge) of radius [itex]a[/itex]?....If so, just orient your co-ordinate system so that [itex]\vec{x}[/itex] on the positive z-axis.

I'm trying to find the vector potential everywhere for a charged rotating spherical shell of radius a. It would be nice if I could just concern myself with the z-axis, though!
 
  • #10
gabbagabbahey
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Okay, and the shell is rotating about the z-axis?....Is it uniformly charged?
 
  • #11
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Yes. Actually, I'm really looking at this now and thinking that the expression I came up for the charge density might be wrong. I think all of this discussion has suggested that the above integral is "too difficult/complicated to be correct" and that I might look at the charge density first. I'll take a little time and see if I can remedy the problem myself, and if not, I'll come back here. Thanks.
 
  • #12
gabbagabbahey
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Yes, your expression for the current density looks very wrong to me....you need to first correct its magnitude, but also consider it's direction! (which varies over the surface).....If I were you, I'd divide the surface current up into a bunch of infinitesimally thin loops, find the vector potential due to one such loop and then integrate it over all the loops on the sphere's surface.
 
  • #13
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Got it! Thanks.
 

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