# 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?

• Where does the half in the second expression come from? May 13, 2020 at 20:14
• Sorry, My mistake. The 1/2 is wrong May 14, 2020 at 10:22

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.

• To clarify, the center of mass energy wouldn't be $(m_1+m_2)c^2$ unless both the particles are also at rest in the COM frame. I guess this is nonetheless correct for the case OP is interested in, assuming they mean the two particles are at rest with respect to each other when they say "two particles are moving together". It's a bit unclear to me what the OP meant with that phrase, thus the comment.
– user87745
May 13, 2020 at 23:14
• Yes, what I wrote was for the case where the particles are moving together. Thank you for clarifying. May 14, 2020 at 1:39
• How do I then transform the energy back to lab frame? May 14, 2020 at 10:26
• I edited this answer in order to make it more compete and thus less misleading. May 14, 2020 at 16:29

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 $$E_{lab}^2-p_{lab}^2 c^2=E_A^2-p_A^2 c^2 =8.1667 m^2 c^4$$ .

• Wow thank you!! May 14, 2020 at 20:09
• To improve your answer, can you answer the question directly before providing a numerical example (which contains the answer... but one has to go through it to see or else know enough to recognize it)? May 14, 2020 at 20:19