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Scalar field energy for two delta function sources

  1. Aug 19, 2010 #1
    I'm trying to evaluate the energy shift in a scalar field described by the Klein-Gordon equation caused by adding two time-independent point sources. In Zee's Quantum Field Theory in a Nutshell, he shows the derivation for a (3+1)-dimensional universe, and I'm trying to do the same for an (N+1)-dimensional universe.

    At some point I obtain the integral

    [tex]
    E = -\int \frac{d^N \vec{k}}{(2 \pi)^N} \frac{e^{i \vec{k} \cdot (\vec{x}_1 - \vec{x}_2)}}{\vec{k}^2 + m^2}
    [/tex]

    where the two [itex]x[/itex] vectors are the locations of the sources. This integral isn't hard to evaluate for 3+1 dimensions, it can be done by going to spherical coordinates, where you get an integration over [itex]d(\cos{\theta})[/itex], making the exponential doable. However, in N+1 dimensions, going to hyperspherical coordinates, I can't perform any such simplification because the volume element contains not a simple sine, but a [itex]sin^{N-2}[/itex]. I tried this road and it gives me some hypergeometric function. Can anyone see how I might proceed with this integral?
     
  2. jcsd
  3. Aug 19, 2010 #2
    "It" being Mathematica?

    Some of the hypergeometric functions are bessel functions in disguise if that helps. But in any case, it the integral equals a hypergeometric function, then so be it. They are perfectly well defined functions just like sin and cos and scientific software can evaluate them.
     
  4. Aug 20, 2010 #3

    DrDu

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    Don't know whether this really works: Try to express the denominator as an integral over exp(-r(k^2+m^2) with r=0 to infinity. Then you can do a quadratic substitution and perform integration over shifted k. For the final integral over r you probably will have to substitute y=1/r.
     
  5. Aug 20, 2010 #4
    "It" being the integral, but yes, I had a look at what Mathematica had to say about it. Let me clarify. The full integral doesn't yield a hypergeometric function, just the angular integral, so I'm left with an integral over [itex]k[/itex] with the integrand consisting, among other things, of this hypergeometric function. I'm aware that the hypergeometric functions are well-defined and that they have specific special functions as special cases. The reason I'm not happy with it appearing is that the final result should be simpler, since I should be able to derive from it the analog of the inverse-square law in N dimensions. This leads me to think there might be a simpler way to do this integral, perhaps by choosing different coordinates or representing the dot product differently. I'm really just hoping someone has seen this integral before, since it's an exercise in Zee.

    I'm not really sure right now how that will work, but I'll have a look at it. Thanks.
     
  6. Aug 25, 2010 #5
    I tried DrDu's suggested approach and I got to an answer, but there's a problem. If I choose N=3 in the final result, things don't reduce to what I know the N=3 case should be. I haven't been able to find my mistake, if anyone would take a look I'd appreciate it. Here is what I did.

    Writing

    [tex]
    \frac{1}{k^2+m^2} = \int_{-\infty}^{0} dr e^{r(k^2+m^2)}
    [/tex]

    the integral becomes

    [tex]
    \int_{-\infty}^{0} dr \int \frac{d^N \vec{k}}{(2 \pi)^N} e^{r k^2 + i k(x1-x2) +r m^2}.
    [/tex]

    Rewriting the exponential I get

    [tex]
    \int_{-\infty}^{0} dr \int \frac{d^N \vec{k}}{(2 \pi)^N} e^{r (k + \frac{i (x1-x2)}{2 r})^2 + \frac{(x1-x2)^2}{4r} + r m^2}.
    [/tex]

    I substitute [itex]u = (k + \frac{i (x1-x2)}{2 r}[/itex] to get

    [tex]
    \int_{-\infty}^{0} dr \int \frac{d^N \vec{u}}{(2 \pi)^N} e^{r u^2 + \frac{(x1-x2)^2}{4r} + r m^2}.
    [/tex]

    Pulling out the exponentials without dependence on [itex]u[/itex] and performing the integral over [itex]u[/itex] eventually gives me.

    [tex]
    \int_{-\infty}^{0} dr e^{\frac{(x1-x2)^2}{4r} + r m^2} 2^{-N} (-r \pi)^{-N/2}.
    [/tex]

    I didn't know how to do this integral, but apparently the resulting function goes like

    [tex]
    |x1-x2|^{\frac{2+N}{2}} K_{-\frac{2+N}{2}}(\sqrt{m} |x1-x2|),
    [/tex]

    where [itex]K_a[/itex] is the modified Bessel function of the second kind. If I substitute N = 3, I get an exponential multiplied with [itex]|x1-x2|[/itex] to powers 0, 1 and 2. The actual solution, however, is an exponential (this part is correct) multiplied by [itex]|x1-x2|[/itex] to the power of -1.

    If anyone can spot a mistake in what I did, I'd appreciate it. Perhaps my substitution [itex]u[/itex] isn't valid? I'm not sure that it's okay for it to be a function of [itex]r[/itex].
     
  7. Aug 26, 2010 #6

    DrDu

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    Good boy! The substitution is ok. Wolfram alpha didn't do the integral for me. I suspected it to be some Bessel function as the exponent looks like the generating function. I don't know where the error comes from but the result you are looking for would be obtained if (N+2)/2 is replaced by (N-2)/2.
     
  8. Aug 26, 2010 #7

    DrDu

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  9. Aug 26, 2010 #8
    Thanks, that helped me spot my error. I had lost a minus sign (but somehow added it back when I typed the LaTeX code). How did you happen to notice this?

    Now back to special functions to figure out how I do this by hand...
     
  10. Aug 26, 2010 #9
    Thanks for the link. I had never actually encountered these specific functions, but the expression you referenced helped me figure out the integral. I'm still looking at one more problem, though. I've obtained the expression

    [tex]
    E = (2 \pi)^{-\frac{N}{2}} m^{\frac{N-2}{2}} |x1-x2|^{\frac{2-N}{2}} K_{\frac{N-2}{2}}(m |x1-x2|).
    [/tex]

    Now I need to take the limit of [itex]m[/itex] goes to zero. I had a look at some of the individual limits for different [itex]N[/itex], and it seems that

    [tex]
    \lim_{m \rightarrow 0} m^{\frac{N-2}{2}} K_{\frac{N-2}{2}}(m |x1-x2|) = |x1-x2|^{\frac{2-N}{2}}.
    [/tex]

    Is this a known result or does anyone see how I might prove this limit for general [itex]N[/itex]?
     
  11. Aug 26, 2010 #10

    DrDu

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  12. Aug 26, 2010 #11
    I'm afraid I've never seen that book before in my life. We used Riley, Hobson & Bence. It looks like it contains a lot of information, perhaps I should get a copy (or bookmark that site, at least).

    Any way, thanks again. I can finally cross this pesky integral off my list.
     
  13. Aug 27, 2010 #12

    DrDu

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    I just wanted to point out that the massless case can be quite easily derived from the integral representation of the Bessel function you were deriving.
     
  14. Aug 27, 2010 #13
    Yes, you're right. I haven't tried, but setting mass to zero I could have just partially integrated [itex]N/2[/itex] times to get to my final result, I think. Is that what you meant?
     
  15. Aug 27, 2010 #14

    DrDu

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    I thought in something like introducing a new variable of integration q=-(Delta x)^2/4r, then you get hopefully the pre-factor (Delta x)^(2-N) and the integral over q is the defining function of the Gamma function.
     
  16. Aug 27, 2010 #15
    Sounds about right. I looked up the standard integral

    [tex]
    \int_0^\infty x^n e^{-a x} dx = \frac{\Gamma(n+1)}{a^{n+1}}
    [/tex]

    and applied that after substituting [itex]y = 1/r[/itex], that worked perfectly. Thanks for the tip.
     
  17. Aug 27, 2010 #16
    The volume element in N-dimensions is given by:

    [tex]
    dV_{N} = r^{N - 1} \, dr \, d\Omega_{N}
    [/tex]

    where the solid angle is:

    [tex]
    d\Omega_{N} = \frac{2 \, \left[\Gamma(\frac{1}{2})\right]^{\frac{N - 1}{2}}}{\Gamma\left(\frac{N - 1}{2}\right)} \, \sin^{N - 2}{\theta} \, d\theta, \; 0 \le \theta \le \pi
    [/tex]
     
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