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Spherical coordinates

  1. Dec 29, 2009 #1
    I want to evaluate the following integral:

    [tex] I(p_{1}, p_{2}, p_{3}) = \int \mathrm{d}^{4} q \mathrm{d}^{4}p \, \dfrac{1}{\left[ p_{2} + q \right]^{2} - i0} \dfrac{1}{\left[ p_{1} - q - p \right]^{2} + i0} \Theta(q^{0}) \delta(q^{2}) \Theta(-p_{2}^{0} -p_{3}^{0} - q^{0} -p^{0}) \delta(\left[p_{2} + p_{3} + q + p \right]^{2}) \Theta(p^{0}) \delta(p^{2}) [/tex].

    [tex] p_{1}, p_{2}, p_{3} [/tex] are time-like four-vectors, so e.g. [tex] p_{1}^{2} > 0 [/tex]

    After some work like exploiting the step- and delta-functions [tex] \Theta(q^{0}) \delta(q^{2}) \Theta(p^{0}) \delta(p^{2}) [/tex] and by choosing a special frame with
    [tex] p_{2} + p_{3} = (p_{2}^{0} + p_{3}^{0}, \vec{0}) [/tex] I arrived at:

    [tex] I(p_{1}, p_{2}, p_{3}) = \int \dfrac{\mathrm{d}^{3}q \, \mathrm{d}^{3}p}{4 \vert \vec{q} \vert \cdot \vert \vec{p} \vert} \, \dfrac{1}{p_{2}^{2} + p_{2}^{0} \vert \vec{q} \vert + \vert \vec{p}_{2} \vert \cdot \vert \vec{q} \vert \cos \theta_{1} - i0} \, \dfrac{1}{p_{1}^{2} - 2p_{1}^{0} \vert \vec{q} \vert + 2 \vert \vec{p}_{1} \vert \cdot \vert \vec{q} \vert \cos \theta_{2} - 2 p_{1}^{0} \vert \vec{p} \vert + 2 \vert \vec{p}_{1} \vert \cdot \vert \vec{p} \vert \cos \theta_{3} + 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert - 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert \cos \eta +i0} [/tex]

    [tex] \times \delta((p_{2}^{0}+p_{3}^{0})^{2} + 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert - 2 \vert \vec{q} \vert \cdot \vert \vec{p} \vert \cos \eta + 2 (p_{2}^{0}+p_{3}^{0}))

    So I used spherical coordinates, but I don't know how to integrate that thing. I just know that if [tex] \eta [/tex] is the angle between the vectors [tex] \vec{p} [/tex] and [tex] \vec{q} [/tex] than we must have: [tex] \theta_{3} = \eta - \theta_{2} [/tex].

    But how do I continue? I think I must somehow specify the elevation and azimuth angles in a special way, but I don't know how to do that.

    Could anyone help me please?
  2. jcsd
  3. Dec 31, 2009 #2


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    I have to wonder, how did you get that integral? (Obviously from a Feynman diagram, but which one?) Something about it looks off to me.
  4. Jan 1, 2010 #3
    The integral is just the Fourier transform of [tex] G_{F}^{*} G_{F} G G [/tex]

    (of course with some arguments), where [tex] * [/tex] is complex conjugation, [tex] G_{F} [/tex] is the (massless) Feynman-Propagator in p-space and [tex] G(p) = \dfrac{i}{2 \pi} \Theta(p^{0}) \delta(p^{2})[/tex].

    The first expression of the integral above is correct, but I just don't know how to compute it explicitly. And now there is the problem with spherical coordinates, but I don't know how to continue. Any ideas?
  5. Jan 1, 2010 #4


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    Not really... I tried fiddling with it a bit but I couldn't get it much simpler than you did. Though I'm not sure that picking a specific reference frame is the way to go about it... whenever I've done these propagator integrals, there hasn't been any need to specialize to a particular reference frame. But on the other hand, I typically had expressions like [itex]\delta^{(4)}(p_1^\mu + p_2^\mu)[/itex] instead of your [itex]G(p)[/itex]. A delta function of a momentum four-vector eliminates four degrees of freedom from the integral, but your [itex]\delta(p^2)[/itex] only eliminates one, which means I'd expect your integral to be rather messier than anything I'm used to dealing with.
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