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Bessel functions and the dirac delta

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the scalar product of diracs delta function ##\delta(\bar{x})## and the bessel function ##J_0## in polar coordinates. I need to do this since I want the orthogonal projection of some function onto the Bessel function and this is a key step towards that solution. I only want hints and guiding, not for someone else to give me the solution :)


    2. Relevant equations

    We have the scalar product
    ##
    <g,h>= \int_0^R g h r dr
    ##

    3. The attempt at a solution

    So if I just plug in my Dirac and Bessel in the scalar product above the whole integral will be zero since Diracs delta function will pick out the value at ##r = 0## and the weight function ##w(r) = r## is zero there. This error will come from the fact that we are on the boundary of the interval right?

    Anyway, I've tried this trick below but i'm not sure that its fully correct to do it like this. Can anyone confirm and deny?

    ##
    <\delta(\bar{x}), J_0(\frac{\alpha_{0,k}}{R}r)> = \int_0^R\delta(\bar{x})J_0(\frac{\alpha_{0,k}}{R}r)rdr = \frac{1}{2 \pi}\int_0^{2\pi}d\varphi \int_0^R\delta(\bar{x})J_0(\frac{\alpha_{0,k}}{R}r)rdr = \{dA=r d\varphi dr\} = \frac{1}{2\pi} \int_{A}\delta(\bar{x})J_0(\frac{\alpha_{0,k}}{R}r)dA =\frac{1}{2\pi}J_0(o) = \frac{1}{2 \pi}
    ##
     
  2. jcsd
  3. Feb 11, 2016 #2

    Orodruin

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    The delta function is not just given by ##\delta(r)## in polar coordinates. You may want to read up on how the delta transforms between coordinate systems.

    Furthermore, using a coordinate system which is singular in the point where you have your delta is not the best idea, in particular since this puts your delta on the boundary of the coordinate range. You really want to avoid these things.
     
  4. Feb 11, 2016 #3

    Orodruin

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    You may also want to check with your class mates who were at the lecture the day before yesterday. We covered a very similar example.
     
  5. Feb 11, 2016 #4
    Where did I write ##\delta(r)##? I know that the delta function is given by ##\delta(\bar{\textbf{r}}-\bar{\textbf{r}}_0)= \frac{1}{r}\delta(r-r_0)\delta(\varphi-\varphi_0)## in polar coordinates. My idea is to use this coordinate transformation I did so I can get the delta function well defined or how to speak.
     
  6. Feb 11, 2016 #5

    Orodruin

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    Yes, but you still have the problem that your delta is now at the boundary of your coordinate region, which makes it rather fishy to perform the integral in polar coordinates. For obvious reasons I cannot comment directly on whether your "trick" is correct or not, I am just going to say that obviously you can compute the integral in whatever coordinate system you want to as long as you transform the integrand accordingly.
     
  7. Feb 11, 2016 #6
    I didn't even notice it was you Professor! :) Please notice how I pointed out that I didn't want someone to solve it for me, only some consulting and guidance on this specific part towards the solution. I understand you points and I think I know where I went wrong here.
     
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