I have been taught that in Classical Mechanics, the total energy of a system of two particles in the Centre of Mass Frame is given by
$$
E_\mathrm{total} = \frac{1}{2}MV^2 + \frac{1}{2}\mu v_r^2
$$
where $M$ is the total mass of the system, $\mathbf{V}$ is the velocity of the centre of mass, $\mu$ is the reduced mass and $\mathbf{v_r}$ is the separation velocity of the two particles.
Furthermore, I have been taught that the first term on the right hand side, $\frac{1}{2}MV^2$ represents kinetic energy of the centre of mass and the second term $\frac{1}{2}\mu v_r^2$ is the "energy to do Physics with" – that is, the energy which can be transfered to any particles that would exit a collision of the original particles.
My question is this: Is it possible to find analogous expressions for each of these terms in the relativistic case? According to wikipedia the total energy of the COM frame is given by
$$
E_\mathrm{tot} = Mc^2
$$
but is it possible to find an equivalent to this "energy to do Physics with?"
Best Answer
Particle 1 and particle 2 each have relativistic momentum 4-vectors: $$\pmatrix{\frac{E_1}{c}\\p_{1x}\\p_{1y}\\p_{1z}}\text{ and } \pmatrix{\frac{E_2}{c}\\ p_{2x}\\p_{2y}\\p_{2z}}$$, so the total momentum 4-vector is $$\pmatrix{\frac{E_1+E_2}{c}\\p_{1x}+p_{2x}\\p_{1y}+p_{2y}\\p_{1z}+p_{2z}},$$ where $E_1=m_1c^2+K_1$ and $E_2=m_2c^2+K_2$.
These have squared magnitudes of $E_j^2/c^2-(p_{jx}^2+p_{jy}^2+p_{jz}^2)=(m_jc)^2$, where $j$ represents 1, 2, or the totals.
The squared magnitude of the total is $$(m_{CM}c)^2 = E_1^2/c^2+E_2^2/c^2+2E_1E_2/c^2-[(p_{1x}+p_{2x})^2+(p_{1y}+p_{2y})^2+(p_{1z}+p_{2z})^2]$$
Doing the algebra and substituting the $(m_1c)^2$ and $(m_2c)^2$ terms we get $$m_{CM}^2c^2 = m_1^2c^2 + m_2^2c^2 + 2(\frac{E_1E_2}{c^2}-\vec{p}_1\cdot\vec{p}_2)$$
Notice the mass in the center of mass is not simply the sum of the individual masses. You might think of this mass as "what you can do physics with" although one might interpret it differently.