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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.

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I have a particle of rest mass $M$, total energy $E$ colliding with a stationery particle of rest mass $m$

The combined total energy $\mathscr E$ of this entire two-particle system (wrt. the "reference lab system" in which the particle of mass $m$ is stationary) is accordingly $$\mathscr E = E + m~c^2,$$

and the combined total momentum

$$\mathscr P = \sqrt{ (E/c)^2 - (M~c)^2 } + 0.$$

I have to show that [...] the total energy, $E'$ in the frame where their centre of mass is at rest is given by: [...]

For this I find helpful to think of the two-particle system described above as being replaced by just one (ficticious) particle of (suitable) mass $\mathscr M$, which carries the same total energy $\mathscr E$ and the same total momentum $\mathscr P$.

Obviously therefore $\mathscr M^2~c^4 = \mathscr E^2 - \mathscr P^2~c^2$.

Now, the point to note is that this (invariant) mass $\mathscr M$ is exactly the equivalent of the " total energy, $E'$ in the frame where their centre of mass is at rest" that you're seeking;

$$E' := \mathscr M~c^2.$$

Or in other words: the combined four-momentum vector of the two-particle system, as expressed in its center-of-mass frame (i.e. where its three-momentum vector is exactly zero) is equivalently

$ \{ \mathscr M~c, 0, 0, 0 \} = \{ E'/c, 0, 0, 0 \} $.

Accordingly

$$ E'^2 = \mathscr E^2 - \mathscr P^2~c^2 = (E + m~c^2)^2 - \left( \sqrt{ (E/c)^2 - (M~c)^2 } \right)^2~c^2,$$

which readily gives the expected expression for $E'$.

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