Lorentz transformation of delta function

  1. For two body decay, in CM frame, we know that the magnitude of the final particle momentum is a constant, which can be described by a delta function, ##\delta(|\vec{p^*}|-|\vec{p_0^*}|)##, ##|\vec{p_0^*}|## is a constant.
    When we go to lab frame (boost in z direction), what's the Lorentz transformation of the delta function?
    Last edited by a moderator: Feb 8, 2014
  2. jcsd
  3. maajdl

    maajdl 379
    Gold Member

    What do you mean by "which can be described by a delta function" ?
  4. I mean that we can use a delta function to fix the momentum i.e. p=p0*.
    Maybe my example of two body decay is not so suitable, but my question is just for mathematics, that is the Lorentz transformation of ##\delta(|\vec{p}|-|\vec{p_0^*}|)##
  5. Bill_K

    Bill_K 4,157
    Science Advisor

    One of the basic properties of the delta function is that ∫δ3(x) d3x = 1. So write down how the volume element transforms under a Lorentz transformation (hint: x is Lorentz contracted) and you will have it.
  6. maajdl

    maajdl 379
    Gold Member


    Are talking about a distribution function in the momentum space,
    and about how this function might evolve with an interaction?
    Are you considering 3-momentum or 4-momentum?
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