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Decay of a moving Kaon in to two charged Pions

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A Kaon (k0) decays in to two charged pions (π+ & π-), what is the minimum momentum required for the kaon so that none of the pions move backwards (in the opposite direction of the kaon) in the laboratory frame.

    2. Relevant equations

    Conservation of momentum:
    [itex]\gamma_{k^{0}} m_{k^{0}}v_{k^{0}} = \gamma_{π^{+}} m_{π^{+}}v_{π^{+}} + \gamma_{π^{-}} m_{π^{-}}v_{π^{-}}[/itex]

    Conservation of energy:
    [itex]\gamma_{k^{0}} m_{k^{0}}= \gamma_{π^{+}} m_{π^{+}} + \gamma_{π^{-}} m_{π^{-}}[/itex]

    Mass of the different pions are the same:
    [itex]m_{π^{+}} = m_{π^{-}} = m_{π}[/itex]

    3. The attempt at a solution

    Working in the C.O.M. of the kaon frame:
    [itex]\gamma_{k^{0}} m_{k^{0}}v_{k^{0}} = 0 = \gamma_{π^{+}} m_{π}v_{π^{+}} + \gamma_{π^{-}} m_{π}v_{π^{-}}[/itex]

    thus:

    [itex]\gamma_{π^{+}} m_{π}v_{π^{+}} = -\gamma_{π^{-}} m_{π}v_{π^{-}} [/itex]

    which gives: [itex]v_{π^{+}} = -v_{π^{-}}[/itex]

    Substituting this in to the conservation of energy:

    [itex]\gamma_{k^{0}} m_{k^{0}}= m_{k^{0}} = 2 \gamma_{π^{+}} m_{π}[/itex] and using the fact that [itex]\gamma_{k^{0}} = 1 [/itex] in our reference frame, I can solve for [itex]v_{π^{+}}[/itex]:

    [itex]v_{π^{+}} = \pm c \sqrt{1-\frac{2m_{π}}{m_{k^{0}}}} = 0.828c [/itex]

    My question is now, how do I move from the COM reference frame to the lab reference frame. Can I just say that the minimum velocity of the kaon in the lab system has to be equal to 0.828c? Or would I have to look at the increased mass of the kaon due to the fact it is moving at such a high velocity?
     
  2. jcsd
  3. Dec 10, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Sure. If the lab is moving at .828c in the kaon frame (so pion+lab move at the same speed), the kaon is moving at .828c in the lab (ignoring the direction here).

    The mass is mk and independent of the system. The increased energy will make the analysis look different in the lab frame, but there you didn't calculate it so you don't have to worry about that.
     
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