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Math Help - How to use the orthogonality of the legendre polynomials in this integral expression?

  1. #1
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    How to use the orthogonality of the legendre polynomials in this integral expression?

    Hello,

    I'm having trouble integrating Legendre Polynomials in the following expression.
    \int_{S^2}P_n(x\cdot v) P_k(x\cdot w) dx
    where v and w are two vectors from the unit sphere S^2 and x\cdot v denotes the standard inner product on \mathbb{R}^3.

    I'd like this integral to be zero if n\neq k and I'm guessing orthogonality of the legendre polynomials P_n(x) should help, that is
    \int_{-1}^1 P_n(x) P_k(x) dx = \delta_{nk}.

    I appreciate any tips, thank you!
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  2. #2
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    The Legendre Polynomials form an orthonormal set of vectors when the Gram Schmidt orthogonalization is applied.

    \displaystyle \delta= \begin{cases}<br />
 \frac{2}{2n+1}, \ \mbox{if} \ i=j\\<br />
 0, \ \mbox{if} \ i\neq j<br />
\end{cases}

    We can show this using 2 Legendre Polynomials as a quick example  P_0=1 \ \mbox{and} \ P_1=x

    \displaystyle \int_{-1}^1 1*x dx=\int_{-1}^1 x dx=\left[\frac{x^2}{2}\right]_{-1}^{1}=\frac{1}{2}-\frac{1}{2}=0

    \displaystyle \int_{-1}^1 x*x dx=\int_{-1}^1 x^2 dx=\left[\frac{x^3}{3}\right]_{-1}^{1}=\frac{1}{3}-\frac{-1}{3}=\frac{2}{3}

    By taking the Gram-Schmidt of the Legendre Polynomials, the kronecker delta will be defined as \displaystyle \delta= \begin{cases}<br />
 1, \ \mbox{if} \ i=j\\<br />
 0, \ \mbox{if} \ i\neq j<br />
\end{cases}

    \displaystyle ||1||^2=\int_{-1}^1 1dx=\int_{-1}^1 1 dx=\left[x\right]_{-1}^{1}=1-(-1)=2

    \displaystyle ||x||^2=\int_{-1}^1 x*x dx=\int_{-1}^1 x^2 dx=\left[\frac{x^3}{3}\right]_{-1}^{1}=\frac{1}{3}-\frac{-1}{3}=\frac{2}{3}

    Therefore, the orthnormal basis after GS method is { \displaystyle \frac{1}{\sqrt{2}},\frac{x}{\sqrt{\frac{2}{3}}},..  ..}

    \displaystyle \int_{-1}^1\left(\frac{x}{\sqrt{\frac{2}{3}}}\right)^2dx=  \frac{1}{\frac{2}{3}}\int_{-1}^1 x^2 dx=\frac{1}{\frac{2}{3}}\left[\frac{x^3}{3}\right]_{-1}^{1}=\frac{1}{\frac{2}{3}}\left(\frac{1}{3}-\frac{-1}{3}\right)=1
    Last edited by dwsmith; November 22nd 2010 at 06:50 PM.
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  3. #3
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    Ok, thanks. Although I made a mistake in the orthogonality formulation (leaving out the \frac{2}{2n+1}), proving the orthogonality was not my question, I take this property for granted.

    The question is how to apply this property to the term
    \int_{S^2} P_n(x\cdot v) P_k(x\cdot w) dx.

    First of all the integration is over the unit sphere S^2. Secondly, the argument in the legendre polynomials isn't the same, but one is a 'rotated version' of the other.
    It may be a good idea to try to solve this integral for v = w first?
    Still it is kind of a transformed integral:
    \int_{S^2} F(\Phi_v(x))dx with F(y) = P_n(y)P_k(y) and \Phi_v(x) = x \cdot v
    But the transformation function \Phi is not bijective as it is not injective, mapping all vectors x with the same angle between x and v to the same value.

    I hope that I made my problem clearer, thanks again!
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  4. #4
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    The result is true in the case w=v, as you can see by using spherical polar coordinates (\cos\theta\sin\phi, \sin\theta\sin\phi,\cos\phi), where the coordinate axes are chosen so that the final coordinate is the direction of  v. Then the integral becomes
    \displaystyle\int_0^\pi\!\!\int_0^{2\pi}\!\!\! P_n(\cos\phi)P_k(\cos\phi)\,\sin\phi\, d\theta d\phi =2\pi\!\! \int_0^\pi\!\!\! P_n(\cos\phi)P_k(\cos\phi)\,\sin\phi \, d\phi = 2\pi\!\! \int_{-1}^1\!\!\! P_n(t)P_k(t)\,dt = 0 (on making the substitution t = \cos\phi).

    Unfortunately, that does not seem to give any insight into why (or whether) the result holds for general  v and w.

    Caution: I have used the standard British notation for spherical polars. I believe that in the US it is more usual to interchange the roles of \theta and \phi.
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