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Lorentz transformation of delta function

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Chenkb
#1
Feb8-14, 03:11 AM
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?
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maajdl
#2
Feb8-14, 03:10 PM
PF Gold
P: 354
What do you mean by "which can be described by a delta function" ?
Chenkb
#3
Feb8-14, 07:37 PM
P: 27
Quote Quote by maajdl View Post
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^*}|)##

Bill_K
#4
Feb9-14, 07:53 AM
Sci Advisor
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Lorentz transformation of delta function

Quote Quote by Chenkb View Post
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.
maajdl
#5
Feb10-14, 03:35 AM
PF Gold
P: 354
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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