Differential cross sections in Peskin & Schroeder

  • #1

Main Question or Discussion Point

I'm a bit confused by something Peskin & Schroeder say about differential cross sections. In my printing, this is on page 101 in the paragraph preceding the one that contains eq. 4.62:

"In the simplest case, where there are only two final-state particles, this leaves only two unconstrained momentum components, usually taken to be the angles [itex]\theta[/itex] and [itex]\phi[/itex] of the momentum of one of the particles. Integrating [itex]d\sigma/(d^3p_1d^3p_2)[/itex] over the four constrained momentum components then leaves us with the usual differential cross section [itex]d\sigma/d\Omega[/itex]."

The second sentence seems like a very odd thing to say. How do you integrate over constrained variables? They have defined the generic differential cross [itex]d\sigma/(d^3p_1...d^3p_n)[/itex] as the quantity that, when integrated over any small region [itex]d^3p_1...d^3p_n[/itex] in final momentum space, gives the cross section for scattering into a state with momenta in that region. So, for two particles, if we specify small ranges for [itex]\theta[/itex] and [itex]\phi[/itex] for one of the final momenta then we have specified the entire region [itex]d^3p_1...d^3p_n[/itex] in final momentum space. We can't integrate over the remaining components because they've already been fixed. So what exactly do P&S mean?
 

Answers and Replies

  • #2
174
5
I think you're already assuming that the four integrations have been done which by the fact that you have an energy-momentum conserving delta function, relates all the other variables to θ and ϕ.
before integrating over the "constrained" variables (or taking into account energy-momentum conservation) specifying θ and ϕ does not specify the final momenta completely.
 
  • #3
So would the unintegrated form contain delta functions? Otherwise, it still wouldn't make sense. As [itex]d\sigma/(d^3p_1...d^3p_n)[/itex] is associated with the probability for scattering into a particular momentum state, it must be identically zero when evaluated at any [itex](\vec{p}_1,...,\vec{p}_n)[/itex] that don't satisfy 4-momentum conservation. I.e., to be consistent with the definition of the cross section, it must vanish for any impossible final momentum states—which include those don't satisfy 4-momentum conservation. Only a delta function does that.
 
  • #4
1,023
30
Does not your transition matrix contains terms dependent on momentum variables.Also there is a 4 delta function for conservation of energy and momentum which is always multiplied by this transition matrix square to get cross section.you can use this delta function to reduce the variables.
 
  • #5
174
5
yes there will be an energy momentum conserving delta function.
look at equation 4.73 (pg 104), when he defines the invariant matrix element.
 

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