I'm working through a problem in a special relativity textbook (Woodhouse) and I'm having some difficulty.
I have to show that if I have a particle of rest mass $M$, total energy $E$ colliding with a stationery particle of rest mass $m$ then the total energy, $E'$ in the frame where their centre of mass is at rest is given by:
$$E'^2 = (M^2 + m^2)c^4 + 2Emc^2.$$
Now I understand that the 4 velocity of the centre of mass frame is $V$, and is given by $V = (c,0,0,0)$, I'll call the momenta $P$ and $Q$ (of $M$ and $m$ respectively).
Now in the solutions it says that the centre of mass frame is defined so that it's 4 velocity is proportional to the sum of the momenta of the two particles (Say $V = k(P+Q)$) it then goes on to say that hence the total momentum in the centre of mass frame is $(E'/c,0)$ but how can this be, because the particles are still moving in the centre of mass frame?
I can see that if this were the case then $g(V,P+Q) = E'$ and then if I could work out the form of $V$ I'm pretty sure I could calculate $E'$ okay.