Special Rel. 4 momenta, particle decay.

[Awesomesauce]

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$^{2}$
E=mc$^{2}$+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$^{2}$=Ea+Eb. Ea=Eb does it not? In that case
Ea+Eb=E=γ(m$_{x}$/2)c$^{2}$
because Ea = E=γ(m$_{x}$/4)c$^{2}$.
I think I have some bad logic somewhere. I get γ=2 from this, so β=v/c=$\sqrt{3}$/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 :)

 

[TSny]

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?

Yes. Good.

I am having trouble finding the energy and speed of the particles in the rest frame.
I think I have mxc$^{2}$=Ea+Eb. Ea=Eb does it not?

Yes, but you shouldn't be hesitant on this point! Remember, 3-momentum must also be conserved and the created particles have equal mass.

In that case
Ea+Eb=E=γ(m$_{x}$/2)c$^{2}$
because Ea = E=γ(m$_{x}$/4)c$^{2}$.
I think I have some bad logic somewhere.

I'd say your logic is great here!

I get γ=2 from this, so β=v/c=$\sqrt{3}$/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.

Each final particle has β=$\frac{\sqrt{3}}{2}$ in this frame of reference.