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Thread: 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.
    $\displaystyle \int_{S^2}P_n(x\cdot v) P_k(x\cdot w) dx$
    where $\displaystyle v$ and $\displaystyle w$ are two vectors from the unit sphere $\displaystyle S^2$ and $\displaystyle x\cdot v$ denotes the standard inner product on $\displaystyle \mathbb{R}^3$.

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

    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.

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

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

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

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

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

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

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

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

    $\displaystyle \displaystyle \int_{-1}^1\left(\frac{x}{\sqrt{\frac{2}{3}}}\right)^2dx= \frac{1}{\frac{2}{3}}\int_{-1}^1 x^2 dx=\frac{1}{\frac{2}{3}}\left[\frac{x^3}{3}\right]_{-1}^{1}=\frac{1}{\frac{2}{3}}\left(\frac{1}{3}-\frac{-1}{3}\right)=1$
    Last edited by dwsmith; Nov 22nd 2010 at 05:50 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 $\displaystyle \frac{2}{2n+1}$), proving the orthogonality was not my question, I take this property for granted.

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

    First of all the integration is over the unit sphere $\displaystyle S^2$. 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 $\displaystyle v = w$ first?
    Still it is kind of a transformed integral:
    $\displaystyle \int_{S^2} F(\Phi_v(x))dx$ with $\displaystyle F(y) = P_n(y)P_k(y)$ and $\displaystyle \Phi_v(x) = x \cdot v$
    But the transformation function $\displaystyle \Phi$ is not bijective as it is not injective, mapping all vectors $\displaystyle x$ with the same angle between $\displaystyle x$ and $\displaystyle v$ 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 $\displaystyle w=v$, as you can see by using spherical polar coordinates $\displaystyle (\cos\theta\sin\phi, \sin\theta\sin\phi,\cos\phi)$, where the coordinate axes are chosen so that the final coordinate is the direction of $\displaystyle v$. Then the integral becomes
    $\displaystyle \displaystyle\int_0^\pi\!\!\int_0^{2\pi}\!\!\! P_n(\cos\phi)P_k(\cos\phi)\,\sin\phi\, d\theta d\phi =2\pi\!\! \int_0^\pi\!\!\! P_n(\cos\phi)P_k(\cos\phi)\,\sin\phi \, d\phi = 2\pi\!\! \int_{-1}^1\!\!\! P_n(t)P_k(t)\,dt = 0$ (on making the substitution $\displaystyle t = \cos\phi$).

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

    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 $\displaystyle \theta$ and $\displaystyle \phi$.
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