Undergrad Orthogonality of spherical Bessel functions

[sunrah]

at what value of k should the following integral function peak when plotted against k?

<br /> 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}}<br />

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 = k_i due to orthogonality of the j_l

 

[Twigg]

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.

 

[sunrah]

If that wasn't clear let me know and I'll write up what I mean in detail.

Thanks for replying.
If I understand correctly, I should try integrating by parts. So doing this I find my function can be approximated as
<br /> 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<br />

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

are progressively much smaller than first-order integrals (if that's clear??). My opinion is that for this function I_{\ell} to peak at k=k_i 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 j_l.

Is this what you meant?

 

[Twigg]

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!