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Two body decay of the k meson

  1. Aug 2, 2015 #1
    Hi everyone,
    i was reading the article of Cristenson,Cronin,Fitch and Turlay of 1964, the one that discovered CP violation in the system of neutral K mesons. There is one point in it that i don't get it. They say "the angle should be zero for two body decay and is, in general, different from zero for three body decay". Based to what i know, it must be conserved the longitudinal and transversal component of the momentum, so if the boost is high, the angle should be small. It's this the right explanation for the very small angle of decay observed?

    I really appreciate any help you can provide.
     
  2. jcsd
  3. Aug 2, 2015 #2

    mfb

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    Which angle? I don't find the quoted text in the original work.

    Edit: Oops, typo, found it.

    The angle is small for highly boosted kaons, but still notable in most (edit: three-body!) decays - the kaon energy was not that high in their beam.
    The observed peak at 1 in the right mass range (and only there) is a clear evidence for two-body decays.
     
    Last edited: Aug 2, 2015
  4. Aug 2, 2015 #3
    I was thinking the same thing. That's why i don't understand why the angle must be so near to zero.
     
  5. Aug 2, 2015 #4

    mfb

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    For a two-body decay, it must be zero (just smeared out by resolution effects). This is just momentum conservation - momentum before is the same as momentum after the decay.

    For a three-body decay, the angle can be different.
     
  6. Aug 2, 2015 #5
    Thanks, reading the article better i have seen that the angle is between the vector sum of the momentum and the direction of the k meson. I thinked it was between the momentum of the single pion and original direction of the kaon
     
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