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I Orthogonality of spherical Bessel functions

  1. Jun 13, 2016 #1
    at what value of k should the following integral function peak when plotted against k?

    [itex]
    I_{\ell}(k,k_{i}) \propto k_{i}\int^{\infty}_{0}yj_{\ell}(k_{i}y)dy\int^{y}_{0}\frac{y-x}{x}j_{\ell}(kx)\frac{dx}{k^{2}}
    [/itex]

    This doesn't look like any orthogonality relationship that I know, it's a 2D integral for starters, but I'm told it should peak at k = ki due to orthogonality of the jl
     
  2. jcsd
  3. Jun 14, 2016 #2

    Twigg

    User Avatar
    Gold Member

    This integral is 3D if your integrand only depends on the radial coordinate, and not angular coordinates. Note it starts at 0 and goes outwards. What you need is a way to make this whole thing a single 3D integral. If you expand by parts, you should be able to isolate the integrand of the dx integral as a function of y inside the dy integral, and that should result in the orthogonality integral. If that wasn't clear let me know and I'll write up what I mean in detail.
     
  4. Jun 14, 2016 #3
    Thanks for replying.
    If I understand correctly, I should try integrating by parts. So doing this I find my function can be approximated as
    [itex]
    I_{\ell}(k,k_{i}) \approx \frac{k_{i}}{k^{2}}\int^{\infty}_{0} y^{2}(\ln{y} - 1)j_{\ell}(k_{i}y)j_{\ell}(ky)dy
    [/itex]

    to do this I'v used [itex]j_{\ell}(0) = 0 [/itex] for [itex]\ell > 0[/itex] (I'm not interested in monopols) and that the value of higher-order integrals, e.g.
    [itex]
    \int^{\infty}_{0}dz (\int^{z}_{0}dy\int^{y}_{0}j_{\ell}(x)dx)
    [/itex]

    are progressively much smaller than first-order integrals (if that's clear??). My opinion is that for this function [itex]I_{\ell}[/itex] to peak at k=ki then the approximation shown featuring the two un-integrated sph. Bessel functions must dominate. Also just plugging some values into python seems show that there are orders of magnitude difference between higher-order integrals of jl.

    Is this what you meant?
     
  5. Jun 14, 2016 #4

    Twigg

    User Avatar
    Gold Member

    That is what I had in mind, but to be honest I hadn't examined the remainders when I first replied. It might not be the best method, it's just what jumped out at me. After working it through on paper, I got the same approximation as you when I expanded the inner integral (the dx integral) to first order using integration by parts. When I did the second-order correction, I found that the next term in the part of the inner (dx) integral that is a multiple of the ##\ell##-th spherical bessel depends on ##\ell##, due to the recursion formulas. So when you run your Python calculations, you might want to examine those higher-order integrals for some high-order spherical bessels, like ##j_{10}(x)## and ##j_{100}(x)##, just to be on the safe side.

    Here's what I got for the second-order integration by parts expansion:

    ##\int_{0}^{y} \frac{y-x}{x} j_{\ell}(kx)dx = [ y (\ln y - 1) - \ell y (\ln y - \frac{3}{2}) ]j_{\ell}(kx) - ky^{2} (\ln y - \frac{3}{2}) j_{\ell + 1}(kx) + ...##

    By orthogonality, you can ignore the ##\ell + 1## term and higher in ##I_{\ell}##, but the coefficient on the first term will probably be an infinite series in ##\ell##. You might even be able to extrapolate the series.

    To get the above result, I used the following recursion relation after the first integration by parts:

    ##j'_{\ell}(z) = \frac{\ell}{z} j_{\ell}(z) + j_{\ell + 1} (z)##

    This is where I'm getting my series for the coefficient on the ##\ell##-th spherical bessel.

    Hope this helps!
     
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