Lorentz transformation of delta function

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 P: 27 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? regards!
 PF Gold P: 377 What do you mean by "which can be described by a delta function" ?
P: 27
 Quote by maajdl What do you mean by "which can be described by a delta function" ?
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^*}|)##

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Lorentz transformation of delta function

 Quote by Chenkb When we go to lab frame (boost in z direction), what's the Lorentz transformation of the delta function?
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.
 PF Gold P: 377 Chenkb, 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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