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Differential cross sections in Peskin & Schroeder

  1. Jan 2, 2013 #1
    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?
  2. jcsd
  3. Jan 2, 2013 #2
    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.
  4. Jan 3, 2013 #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.
  5. Jan 3, 2013 #4
    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.
  6. Jan 3, 2013 #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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