Evaluating Integral with Spherical Coordinates Using 4-Vectors

In summary, the conversation discusses evaluating a complicated integral involving time-like four-vectors and using spherical coordinates. The speaker is unsure of how to proceed and asks for help. They mention that the integral is related to the Fourier transform of a complex conjugate expression involving Feynman propagators, but are unsure of how to compute it explicitly. The other person suggests trying different approaches, but the speaker is still unsure of how to handle the spherical coordinates.
  • #1
parton
83
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}))
\end{split}
[/tex].

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?
 
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  • #2
ewww...

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.
 
  • #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?
 
  • #4
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.
 

Related to Evaluating Integral with Spherical Coordinates Using 4-Vectors

What is the purpose of evaluating integrals with spherical coordinates using 4-vectors?

The purpose of evaluating integrals with spherical coordinates using 4-vectors is to solve for the volume or surface area of a three-dimensional region in space. This method is particularly useful for problems involving spherical symmetry, as it allows for a more simplified and efficient calculation.

What is the difference between traditional spherical coordinates and 4-vector spherical coordinates?

Traditional spherical coordinates use three parameters (radius, inclination angle, and azimuth angle) to describe a point in space, while 4-vector spherical coordinates use four parameters (radius, three angles) to describe a point in space-time. This allows for a more comprehensive representation of the point's position and can be useful in solving certain problems in physics and engineering.

What are the advantages of using 4-vectors in evaluating integrals with spherical coordinates?

Using 4-vectors in evaluating integrals with spherical coordinates allows for a more concise and elegant solution to problems involving spherical symmetry. It also allows for a more accurate representation of the problem in space-time, which can be useful in fields such as relativity and cosmology.

What are some common applications of evaluating integrals with spherical coordinates using 4-vectors?

Some common applications of evaluating integrals with spherical coordinates using 4-vectors include calculating the electric or gravitational potential of a spherical distribution of charge or mass, finding the flux of a vector field through a spherical surface, and solving problems involving spherical symmetry in quantum mechanics, electromagnetism, and fluid mechanics.

What are some tips for effectively using 4-vectors in evaluating integrals with spherical coordinates?

It is important to have a good understanding of spherical coordinates and their relationship to 4-vectors before attempting to use them in integrals. It can also be helpful to visualize the problem in three-dimensional space and to carefully choose the orientation of the 4-vector coordinates to simplify the calculation. Practice and familiarity with the method can also improve efficiency and accuracy.

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