Orthogonality of spherical Bessel functions

In summary, the integral function I_{\ell}(k,k_{i}) should peak at k = k_{i} due to orthogonality of the spherical bessel functions, which can be approximated by expanding the inner integral using integration by parts. However, for higher order spherical bessel functions, there may be a dependence on the order, so it is important to examine these higher-order integrals to ensure the accuracy of the approximation.
  • #1
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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
 
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  • #2
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.
 
  • #3
Twigg said:
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
[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?
 
  • #4
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!
 

1. What is the concept of orthogonality in spherical Bessel functions?

The orthogonality of spherical Bessel functions refers to the property that states the inner product of two different spherical Bessel functions is equal to zero. In other words, the product of two distinct spherical Bessel functions integrated over a certain range is equal to zero.

2. How is the orthogonality of spherical Bessel functions used in mathematical calculations?

The orthogonality of spherical Bessel functions is a crucial concept in solving differential equations involving spherical coordinates. It allows for the separation of variables in a differential equation, simplifying the solution process.

3. Can the orthogonality of spherical Bessel functions be generalized to other types of functions?

Yes, the concept of orthogonality can be applied to other types of functions, such as Legendre polynomials and trigonometric functions. However, the specific properties and equations may differ.

4. What is the significance of the orthogonality of spherical Bessel functions in physics?

The orthogonality of spherical Bessel functions plays a crucial role in solving problems in physics involving spherical symmetry. It is commonly used in fields such as electromagnetism, quantum mechanics, and fluid dynamics.

5. How is the orthogonality of spherical Bessel functions proven mathematically?

The orthogonality of spherical Bessel functions is proven using integration by parts and the properties of spherical Bessel functions. The details of the proof can be found in mathematical textbooks on differential equations or special functions.

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