Solving Relativistic Collision using Four Momentum Vector

[darkmatter820]

1. A particle of mass M decays from rest into two particles. One particle has mass m and the other particle is massless. The momentum of the massless particle is...
2. Ei = Ef, Pi= Pf
3. This is a GRE practice problem. I can solve this problem using the old method as listed in the step 2, but I just learned about four momentum yesterday and it seems to be the fastest way so far to solve these kind of problems; but I'm struggling to solve this problem with this method.

My attempt: P1 =(Mc^2/c,0), P2= (mc^2/c, pm), P3 = (E/c, p*); P1 is the initial energy-momentum vector, P2 is the final energy momentum for the particle with mass m, and P3 is the vector for massless particle. I also assumed the massless particle is moving at c. Then,
P1= P2+P3,
P1°P1 = (P2+P3)°(P2+P3), p* = E, c =1.

This is as far as I can go. I might have made some mistakes somewhere, so please be specific when you response so I correct them next time. Thanks in advance! Cheers

 

[vela]

The four-momentum for particle 2 isn't correct. Its energy isn't simply its rest energy. Also, use the fact that the two particles have to have opposite but equal three-momenta to write
\begin{eqnarray}
p_1 &= (Mc, 0) \\
p_2 &= (E_2/c, p) \\
p_3 &= (p, -p)
\end{eqnarray}

 

[darkmatter820]

Hi Vela, with the correction I got (M^2 -m^2)/2 = p^2 + p(p^2+m^2)^(1/2), with E = (p^2+m^2)^(1/2), and c = 1, which is kind of nasty (for me at least). So is four- momentum always the quickest way or just depends on the problem? Thanks

 

[vela]

Try using conservation of energy to solve for E₂ instead, then you'll have a comparatively simple expression from which you can isolate p.

Occasionally, there are problems where not using four-vectors is an easier route, but I've found that more often than not using four-vectors is much simpler. The thing with SR is that if you take a suboptimal approach, you can sometimes go in circles with the algebra. You learn how to avoid that with practice.