# Lorentz transformation of delta function

1. ### Chenkb

30
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!

Last edited by a moderator: Feb 8, 2014
2. ### maajdl

379
What do you mean by "which can be described by a delta function" ?

3. ### Chenkb

30
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^*}|)##

4. ### Bill_K

4,159
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.

5. ### maajdl

379
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?