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Why to change from momentum space integrals to spherical coordinate ones?

  1. May 16, 2015 #1
    So I was asked to compute loop contributions to the Higgs and compute the integrals in spherical coordinates, I gave a look to Halzen book but did not found anything. Why, when and how to make that change?
     
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  3. May 16, 2015 #2

    Orodruin

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    I suspect they mean spherical coordinates in momentum space ...
     
  4. May 16, 2015 #3
    "[...]Change the momentum integrals to spherical coordinates[...]"
     
  5. May 16, 2015 #4

    Orodruin

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    So what is the problem? Going to spherical coordinates is a standard procedure from multivariable calculus.
     
  6. May 16, 2015 #5

    ChrisVer

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    [itex] d^3 p \rightarrow p^2 dp d \Omega[/itex]
    is quite useful ...
     
  7. May 16, 2015 #6
    But why? because I solved loop integrals using dim regularization and other tricks and never used the change of variables. That is why I asked when to use the change of variables instead to keep using my previous methods (Used in many loop computations by some textboks like Peskin's)
     
  8. May 16, 2015 #7

    Orodruin

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    Did you perhaps use the standard forms of the integrals that you end up with in dim-reg which you can find, eg, in the end of Peskin? How do you think these integrals are evaluated?
     
  9. May 16, 2015 #8
    oh!

    On the unit sphere area?
     
  10. May 16, 2015 #9

    ChrisVer

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    why "unit" sphere area?
     
  11. May 16, 2015 #10
    I am still wondering haha

    Do not know for now... tip for clairvoyance?
     
  12. May 16, 2015 #11

    Orodruin

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    The entire point with dim reg is to rewrite the integral in 4-d as a limit of an integral in d "dimensions". Interpreting it as the area of a d-1-dimensional unit sphere and an integral where ##p^{d-1}\, dp## has replaced the volume ##d^dp##.
     
  13. May 16, 2015 #12

    ChrisVer

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    Let's try it this way... How would you solve this integral:
    [itex] \int \frac{d^3 p}{(2 \pi)^3} \frac{a }{|p|^2+m^2} e^{-i\vec{p} \cdot \vec{x}}[/itex]
    ?
     
  14. May 16, 2015 #13

    ChrisVer

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    Why saying a unit sphere? This doesn't make sense, exactly because the ##d^d p## is a volume. The [itex]d^d p = |p|^{d-1} dp d\Omega_{d-1}[/itex] is a volume element of an "hypersphere" between radius ##p## and ##p+dp##.

    edit: I think I figured it out
     
  15. May 16, 2015 #14

    Orodruin

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    Yes, exactly. I took one step extra and integrated the ##d\Omega##. This gives you the area of the d-1-dimensional unit sphere.
     
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