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.

 

[paranormal]

I can say that your approach using four momentum vectors is correct. The four momentum vector is a useful tool in solving relativistic collision problems as it takes into account both energy and momentum conservation. In your attempt, you correctly identified the initial and final energy-momentum vectors for each particle and used the fact that the total initial four momentum must equal the total final four momentum. You also correctly assumed that the massless particle is moving at the speed of light, which means its momentum is equal to its energy. Your final equation, P1°P1 = (P2+P3)°(P2+P3), is a valid equation for solving the problem.

However, if you are struggling to solve the problem using this method, I suggest practicing more problems and familiarizing yourself with the concept of four momentum vectors. It is a powerful tool in solving relativistic problems and with more practice, you will become more comfortable with it. Also, make sure to double check your calculations and units to avoid any mistakes. Good luck!