Energy in Center of Momentum Frame

Why is the total energy in the center of momentum frame of two particles the rest energies of the particles? I'm imagining the specific example of two identical particles with rest mass $m$, one at rest and one moving in the $+x$ direction. If we switch to the CoM frame, we should have one particle moving with $+\beta$ and one moving with $-\beta$ (both will have the same $\gamma$) so that the sum of the 3-momentum goes to 0. If I try to calculate the total energy, it is then $\gamma mc^2$ for each particle, so the total energy should be $2\gamma mc^2$. This is different than the sum of the rest energies, $2mc^2$, which is the energy that the CoM frame is supposed to have. Where did I go wrong?

• Do you have a reference for the claim in your first sentence? – Alfred Centauri Nov 24 '16 at 0:26

The total relativistic 4 vector momentum squared is $E_{\rm total}^2 - p_{\rm total}^2 = m_{\rm total}^2$, with $E_{\rm total}$ being the total energy, $m_{\rm total}^2$ is the square of the total mass, $p_{\rm total}^2$ is the square of the total 3 momentum and units $c=1$. This 4-momentum squared value is an invariant (a constant) in every possible reference frame, even though the total 4-momentum vector varies from frame to frame.
In the CoM frame, $p_{\rm total}^2 = 0$. So $E^2 = m_{\rm total}^2$. For energy and momentum we can use simple addition $p_{\rm total}=p_1+p_2$ and $E_{\rm total}=E_1+E_2$, where subscript 1 and 2 indicate the quantity corresponds to particle 1 or 2. But for the mass $m_{\rm total}=m_1+m_2$ does not always apply. This may seem counter-intuitive, but as you will realize from other examples our intuitions aren't always that reliable when it comes to relativity. The theoretical reason is that you can add four-momenta where the four-momenta have components of energy in the time term and 3-momentum in the space terms. But as the mass is equal to the Minkowski metric length of the four-momentum vector, the masses don't always add for individual particles. The one special case they do add is when the particle are not moving in the CoM frame.
Your answer $2\gamma mc^2$is correct, with $m$ the rest mass of each particle. $2\gamma mc^2$ is the total energy of the 2 particles in the CM frame for your example. It's the effective rest mass of the system in the CM frame.