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Transforming a 4-vector integration measure

  1. Nov 27, 2012 #1
    Hello,

    I have a particular derivation of a four-vector integration measure, basically changing the measure to some related more useful measure - but I'd like to do this in 3-vector notation. Here it is, from the integral:

    [tex]-i\lambda\int_{-\infty}^{\infty}\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2}[/tex]

    where k is the four vector: [tex]k=(E,\mathbf{p})[/tex]

    We transform the integration measure like so:

    [tex]d^3k=4\pi k^{2}dk\: ,\: \frac{dk^2}{dk}=2k\: ,\: d^3k=2\pi kdk^2[/tex]

    Where the first step is done due to spherical symmetry (can anyone explain why this is spherically symmetric? I'm guessing the 'k' is isotropic in some sense?)

    Anyway we attain that the integration measure is:

    [tex]\frac{d^4k}{(2\pi)^4}=\frac{k^2dk^2}{16\pi^2}[/tex]

    My main question is how to change this derivation to a 3-vector one? I'm guess I pretty much subsitute 'k' for 'p' but I'm not sure.


    Thanks,
    SK
     
  2. jcsd
  3. Nov 27, 2012 #2

    tom.stoer

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    too many identical k's w/o index, vector notation
     
  4. Nov 27, 2012 #3
    Sorry, but I did not quite get what you want to do. You want to turn the given integral in 4-space to an integral in 3-space? Well, you cannot do it in general, unless you have a constrain (like the "mass shell": k2 = m2) . If you do, please let me know so I can tell you what to do in this case. About the spherical symmetry question, the integrand is a function of k2 = k02 - k2 i.e. it depents only of the magnitude of k and not it's direction.
     
  5. Nov 28, 2012 #4
    d4k=dk0d3k,also k2=k02-k2
     
  6. Nov 28, 2012 #5

    tom.stoer

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    again too many k's w/o index ;-)

    k2=k02-k2 → 2k2=k02

    but this is definitly NOT what you mean ...
     
  7. Nov 28, 2012 #6
    ahh yes,so
    k2=k02-[itex]k^2 [/itex]
     
  8. Nov 28, 2012 #7

    tom.stoer

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    but this is what cosmic dust says: you cannot do that in general w/o assuming that there is a constraint; and in the above mentioned integral there isn't such a constraint, so the mass-shell condition does not apply.
     
  9. Nov 28, 2012 #8
    one can use it by simply giving a negative imaginary part to mass,if you wish.
     
  10. Nov 28, 2012 #9
    What I wanted to do was just take this out of 4 vector notation, how would I express this with a scalar part and a vector part.

    I think if k is a four vector then:

    [tex]d^4k=dEd^3\mathbf{p} \: ,\: k=(E,\mathbf{p})[/tex],

    is this right? If so I just want to change the derivation I stated intially but using the integration measure on the RHS above this line.

    Thanks,
    SK
     
  11. Nov 28, 2012 #10

    tom.stoer

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    yes, that's correct
     
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