1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Bessel Function identity

  1. Nov 8, 2009 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    By appropriate limiting procedures prove the following expansion:

    [tex]J_0 (k\sqrt {\rho ^2 + \rho '^2 - 2\rho \rho '\cos (\phi )} ) = \sum\limits_{m = - \infty }^\infty {e^{im\phi } J_m (k\rho )J_m (k\rho ')} [/tex]

    2. Relevant equations

    \over x} - \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over x} '|}} = \frac{1}{{\sqrt {\rho ^2 + \rho '^2 - 2\rho \rho '\cos (\phi - \phi ') + (z - z')^2 } }}[/tex]

    = \int\limits_0^\infty {\sum\limits_{m = - \infty }^\infty {J_m (k\rho )J_m (k\rho ')} e^{im(\phi - \phi ')} e^{ - k(z_ > - z_ < )} dk}


    3. The attempt at a solution

    I can't even really figure out what is going on at this part. Apparently something has rid the expansion of the integral, and the only thing that could do that is a delta function in k. I don't under what situation we are going to just have a 0th order bessel function either. This question is quite the stumper for me... any help would be appreciated :)
  2. jcsd
  3. Nov 8, 2009 #2


    User Avatar
    Gold Member

    Actually ok nevermind, I solved it but what in the world is this used for? It's letting z' = 0 and [tex]\phi '[/tex] = 0 but ... what is the interpretation of this?
  4. Nov 9, 2009 #3
    I don't have an answer but I think I am able to express this identity in descriptive terms. I don't know why it should hold but it goes like this:

    Take three points in the plane: A, B and C. The 0-order Bessel function of C with respect to A is the left hand side of the identity. We consider this as a kind of quantum-mechanical amplitude for a particle to get from A to C. The right hand side says:

    You take the Bessel function as being the amplitude to get from A to B and another one to get from from B to C. The amplitude to get from A to C is the product of the two amplitudes. But now you have to add up the amplitudes for every possible Bessel function, from 0th order, 1st order, 2nd order, all the way to infinity. The sum of those amplitudes is simply equal to the direct amplitude of the zeroth order Bessel function going directly from A to C.

    I think I have correctly put this identity into words, although I don't know why it should hold. The exp(i*m*theta) terms in the right hand side are because the vector angle between AB and BC is associated with a variation in the phase of the Bessel functions of higher order: you need to have J_n(r, theta) = J_n(r)*exp(2i*n*theta).

    I hope someone can see how this all fits together.
  5. Nov 9, 2009 #4

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    I assume this is from Jackson (an equation number would help). Bessel functions turn up in cylindrical problems. This formula is what is called a "sum rule". A similar thing happens with spherical harmonics.

    The point is this: The most symmetrical situation in cylindrical coordinates is to have a point charge at the origin. Then you have only J_0 in your expression for the Green's function. But what if we move the source away from the origin? This equation says that we can still write down the Green's function from the point of view of the origin, using an infinite sum of Bessel functions.

    On the left you have J_0 a function of radial distance from the source point, rather than from the origin. On the right, it is re-expressed as a sum of Bessel functions of distances 1) from the origin to the source point, and 2) from the origin to the observation point. This is good, because generically speaking, the source will not be at the origin. This new sum allows us to easily account for a variety of sources in cylindrical coordinates that are not maximally symmetrical.

    By the way, on the left I think you should have [itex]\cos \gamma = \cos (\theta - \theta')[/itex]. (Or, I suppose Jackson has set [itex]\theta' = 0[/itex] for some reason).
  6. Nov 10, 2009 #5
    OK. I like Ben's interpretation better than my original suggestion. I think I was misled by seeing what looked like the product of two Bessel functions on the RHS. I think a good way to interpret this is to special case it down to a simple shift of coordinates.

    Let's set k equal to 1, theta to zero, and think of rho-prime as a constant, not a variable. Now all we're doing is taking a zero-order Bessel function shifted a distance rho-prime to the left of the origin and re-expressing it as the sum of nth-order Bessel functions. Since the set of nth-order Bessel functions is a complete basis on the two-dimensional plane, this should be a straighforward exercise. The only problem is determining the coefficients of each term. We naturally do this by taking the dot-product of the target function (the displaced zeroth-order Bessel function) with the sequential basis elements (the nth-order Bessel functions).

    The given identity then tells us the result of this decomposition. Specifically, it tells us that the nth-term coefficient is J_m(rho-prime). In other words, the dot-product of J_m with the displaced J_0 is simply J_m evaluated at the distance of separation. It's a very simple result for what looks like a fairly difficult integration. I wonder if there is an easy way to see why it must be so.
  7. Nov 10, 2009 #6


    User Avatar
    Gold Member

    Ah cool, sounds good :) I asked my professor and got some similar, less detailed explanations that seem to have helped me put all the pieces together as to what's going on. Thanks you two!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook