# Lorentz transformation of delta function

1. Feb 8, 2014

### Chenkb

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. Feb 8, 2014

### maajdl

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

3. Feb 8, 2014

### Chenkb

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. Feb 9, 2014

### Bill_K

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. Feb 10, 2014

### maajdl

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?