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toreil
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Homework Statement
A Kaon (k0) decays into 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.
Homework 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]
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 into 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?