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Inverse Muon Decay differential cross section

  1. Dec 6, 2011 #1
    Why is it that the differential cross section for processes like Inverse Muon Decay (IMD) are always given in the CM frame? Every paper I have seen that discusses Inverse Muon Decay gives the differential cross section in CM frame. Is it very hard to calculate the differential cross section in the lab frame where the electron is at rest?

    To be clear, IMD is this process:

    [itex]\nu[/itex][itex]\mu[/itex] + e [itex]\rightarrow[/itex] [itex]\nu[/itex]e + [itex]\mu[/itex]
  2. jcsd
  3. Dec 7, 2011 #2
    It is typically easier to do cross section calculations in the CM frame. You just then do a lorentz transformation to the Lab frame. Or you use a lorentz invariant form of the differential cross section.
  4. Dec 7, 2011 #3
    Thanks for the reply! That was kind of what I was thinking. But the differential cross section isn't lorentz invariant (at least I would think it isn't). So the best you could do is find some relationship between the CM frame components and the Lab frame components and dump those in. But the differential cross section will still be from the CM point of view, regardless of the fact that you could calculate it with measured lab frame quantities.
  5. Dec 7, 2011 #4
    For a 2 -> 2 process (like you talked about above) the spectral (or equivalently the angular) differential cross section is not lorentz invariant. But for other types of interactions, for instance the A + B -> C + X where A, B, C are defined particles and X is any other combination of particles, the so-called inclusive differential cross section can be written in a lorentz invariant way. Usually called the lorentz invariant differential cross section (LIDCS).

    You can have a look at this paper (section 4): http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20080043627_2008043417.pdf

    Or this one: http://www-zeuthen.desy.de/~pohlmadq/teach/582/ch2.pdf [Broken] For more information.
    Last edited by a moderator: May 5, 2017
  6. Dec 8, 2011 #5
    Thanks again for the reply. I believe that answers everything I wanted to know.

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