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Propagator - Hankel function - infinite? - Weinberg

  1. Feb 14, 2008 #1

    tiny-tim

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    I'm completely confused by page 202 of Weinberg's QTF, Vol.I.

    In particular:

    I can't find "Hankel" anywhere on PF, and wikipedia is no real help.

    Surely integral (5.2.8) doesn't converge (the integrand oscillates between ±1)?
    In fact, I think it's -∞. :redface:
    So how can it be a standard function? :confused:
    :frown: [size=-1]What am I missing?[/size] :frown:
     
  2. jcsd
  3. Feb 14, 2008 #2
    Hmm... can't help with that I'm afraid! Have you tried your library for a book of mathematical methods for physicists? Specific functions like Leguerre or Hermite polynomials that often crop up can usually be found in a book like that somewhere.
     
  4. Feb 14, 2008 #3

    dextercioby

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    There are numerous books on the theory of Bessel functions of all possible kinds; I'd reccomend J. Watson 1928 classic "The theory of Bessel functions".
     
  5. Feb 15, 2008 #4
  6. Feb 18, 2008 #5

    tiny-tim

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    good news … and bad news!

    Aha! Someone on another forum has given me a very helpful reference on Wolfram:

    Cool! :cool: Now I see what Weinberg is getting at! :smile:

    Weinberg is saying (puting [tex]X = m\sqrt{x^2}[/tex]):
    [tex]K_1(X) = \int_{0}^{\infty}\frac{u\mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)\, ,[/tex]​
    which, putting u = sinht, is the same as:
    [tex]K_1(X) = \int_{0}^{\infty}\sinh t.\mathrm{d}t.\sin(X\sinh t)\,.[/tex]​
    But there's still a problem -
    Wolfram defines:
    [tex]\large K_n(X) = \int_{0}^{\infty}\sinh nt.\mathrm{d}t.\sin(X\sinh t)\,,[/tex]​
    which would be the same as Weinberg for n = 1,
    EXCEPT THAT WOLFRAM STIPULATES THAT |Re(n)| < 1.

    Changing the variable back from t to u, and with n real, gives:
    [tex]\Large K_n(X) = \int_{0}^{\infty}\frac{\sinh (n.\sinh^{-1} u) \mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)\,,[/tex]​
    and. for large values of u, we can put, approximately:
    [tex]\large\frac{\sinh (n.\sinh^{-1} u)}{\sqrt{u^2 + 1}}\,=\,\frac{e^{n.logu}}{\sqrt{u^2 + 1}}\,=\,\frac{u^n}{\sqrt{u^2 + 1}}\,=\,u^{n\,-\,1}\,,[/tex]​
    which tends to zero,
    and so the integrand tends very comfortably to zero provided n < 1.

    So the Wolfram integral for [tex]\large K_n(X)[/tex] does converge for n < 1, but it doesn't seem to for n = 1 (because the integrand oscillates between ±1), and Wolfram doesn't claim it does!

    :cry:: So surely Weinberg's derivation of (5.2.8) and (5.2.9) is just wrong? :cry:
     
  7. Feb 18, 2008 #6
    Hi,

    I've been working the last few days on this exact same problem (evaluating the Feynman propagator in closed form). I haven't closed the problem out yet, but perhaps I can pass on some things I've found out and I'll be interested to see what else you find.

    I came to the same integral (5.2.8) and concluded that it is indeed divergent, because as you say the integrand oscillates indefinitely. However, within the context of distributions, it can be interpreted in a rigorous and precise fashion.

    In this case, the distribution is u/sqrt(u^2 + 1), which for large u is similar to a Heaviside step function, and the integral (5.2.8) asks for the Fourier transform of this distribution. The step function has a well-defined Fourier transform (there's a delta function in there somewhere - you'll have to look it up), and this is the way forward: evaluate this divergent piece first, and then what's left is finite and integrable: it's the Hankel function referred to by Weinberg.

    More precisely, write

    u/ sqrt(u^2 + 1) = H(u) + ( u/sqrt(u^2 + 1) - 1 )

    H(u) is the step function; it's Fourier transform has a delta function in it somewhere. (Weinberg didn't mention it because it's concentrated on the light cone. It is therefore irrelevant to his purpose of evaluating the propagator in the spacelike region.) The second bit (in parentheses) is O(1/u^2) for large u, so it's absolutely integrable and its Fourier transform is well-defined within the context of standard real analysis. You can transform it to one of the standard integral representations of Bessel functions.

    And that's where I'm up to right now. I also wanted to figure out the form of the propagator in the timelike region, and then I wanted to the whole thing over again for spin-half particles (satisfying the Dirac equation) as well.

    Hope that helps,

    Dave
     
  8. Feb 18, 2008 #7

    tiny-tim

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    I'm not following this - Weinberg only integrates from 0 to ∞, and so isn't interested in u < 0, so you might as well use the constant function 1 as the step function H.

    And doesn't the Fourier transform integrate from -∞ to ∞ anyway? So his integral
    [tex]K_1(X) = \int_{0}^{\infty}\frac{u\mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)[/tex]​
    is half of the Fourier transform
    [tex]\int_{-\infty}^{\infty}\frac{u\mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)\,,[/tex]​
    in which the distribution is certainly not similar to a step function?

    I agree that your second bit (in the parentheses) is O(1/u^2) for large u, but that still leaves the integral
    [tex]\int_{0}^{\infty}H(u)\mathrm{d}u\sin(Xu)\,,\,=\,\int_{0}^{\infty}\mathrm{d}u\sin(Xu)\,,[/tex]​
    which doesn't converge.

    And where does the light-cone come into it? Both u and x are one-dimensional variables. :confused:
     
  9. Feb 18, 2008 #8
    Ah, but integral
    \int H(u)sin(Xu) du
    is actually referring to (the imaginary part of) the Fourier transform of the step function, which is well defined:

    http://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html

    It's a distribution, not a function, so its fourier transform is not necessarily defined in terms of a convergent integral.

    In the context of this problem, it brings in a contribution \delta(X), where your notation X stands for m\sqrt(x^2) in your original posting. That's what I was referring to when I meant the delta function was concentrated on the light cone, because the light cone is precisely the region X=0.

    Dave
     
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