- #1
Awesomesauce
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Homework Statement
A particle X with rest mass mx is traveling with speed vx = 0.8c along
the x-axis in the lab frame.
(a) Write down the four-vector momentum of the particle in the lab frame
in terms of vx and mx.
The particle decays to two particles, A and B, each with mass mx/4.
Calculate the energy of these particles, in terms of mx, in the rest frame of
X. Calculate their speed in this frame.
Homework Equations
E=γmc[itex]^{2}[/itex]
E=mc[itex]^{2}[/itex]+pc
note: four mom. vector = [px,py,pz,E/c] here.
The Attempt at a Solution
Alright, so I have the 4 momenta which is p = [E=γmx0.8c, 0, 0, E=γmxc]. so γ=5/3 using the velocity given?
I am having trouble finding the energy and speed of the particles in the rest frame.
I think I have mxc[itex]^{2}[/itex]=Ea+Eb. Ea=Eb does it not? In that case
Ea+Eb=E=γ(m[itex]_{x}[/itex]/2)c[itex]^{2}[/itex]
because Ea = E=γ(m[itex]_{x}[/itex]/4)c[itex]^{2}[/itex].
I think I have some bad logic somewhere. I get γ=2 from this, so β=v/c=[itex]\sqrt{3}[/itex]/2?
If this is correct, is this v/c value for both particles? As they are traveling in opposite directions. It's mostly the intuition which confuses me.
Thanks :)