Relativity: correct way expressing energy? In relativity, if two particles are moving together, which way is the correct way expressing their total energy:
$$ E=\sqrt {((m_1+m_2)c^2)^2+((p_1+p_2)c)^2}$$
or:
$$E=\sqrt{(m_1c^2)^2+(p_{1}c)^2}+\sqrt{(m_2c^2)^2+(p_{2}c)^2}$$
Or does it depend on frame of reference?
 A: The total energy of the particles will depend on the reference frame. Any particle in its rest frame has only its rest mass energy $E = mc^2$. In any other frame, it will also have energy of motion in addition to rest mass energy. 
In the center of mass frame, then, if neither particle is moving the total energy would be $E_\text{cm} = (m_1+m_2)c^2$. You can then translate this to other frames if you like.
More generally, to get the total energy of a set of particles in any given frame, just add up the energies of the particles individually---the second method proposed by the OP.
A: Let us see an example :
In the lab frame of reference, there are 2 particles A and B of the same mass $m$ approaching to each other with the velocity of $v_{A,lab}=\frac{3}{5}c$ and $v_{B,lab}=-\frac{4}{5}c$, with $c$ is velocity of light.
The total relativistic energy $E$ and momentum $p$ in this lab frame are
$E_{lab}=E_{A,lab}+E_{B,lab}=\gamma _{A,lab}mc^2 + \gamma _{B,lab}mc^2=(1.25\, +1.6667) mc^2 =2.9167 mc^2 $
$p_{lab}=p_{A,lab}+p_{B,lab}=\gamma _{A,lab}mv_{A,lab} + \gamma _{B,lab}mv_{B,lab}=(0.75\, -1.3333) mc =-0.5833 mc $
, with $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ .
In particle A frame of reference, $v_{A,A}=0$ and 
$v_{B,A}=\frac{v_{A,lab}+v_{B,lab}}{1+\frac{v_{A,lab}\times v_{B,lab}}{c^2}}=\frac{35}{37}c$ (use relativistic velocity addition formula).
The total relativistic energy $E$ and momentum $p$ in this A frame are
$E_{A}=E_{A,A}+E_{B,A}=\gamma _{A,A}mc^2 + \gamma _{B,A}mc^2=(1\, +3.0833) mc^2 =4.0833 mc^2 $
$p_{A}=p_{A,A}+p_{B,A}=\gamma _{A,A}mv_{A,A} + \gamma _{B,A}mv_{B,A}=(0\, -2.9167) mc =-2.9167 mc $
We see that $E_{lab}\neq E_A$ nor $p_{lab}\neq p_A$ .
However, the invariant quantity is
$E_{lab}^2-p_{lab}^2 c^2=E_A^2-p_A^2 c^2 =8.1667 m^2 c^4$ .
