# How can relativistic electron coliding with electron at rest transfer all it's energy?

I was been reading the paper of Lev Okun "The theory of relativity and the Pythagorean theorem" and I came across the following statement.

In order to ‘feel’ how an electron at rest can receive the entire energy of a moving electron it is sufficient to use their center-of-inertia frame to consider scattering by 180 degrees, and then return back to the laboratory frame.

I am going to need some help to figure that out. I imagine the 180 degree scattering in the CM frame as the two electrons just shifting places. In the CM frame the total 3-momentum is zero, therefore after the collision we have $$(E'_1+E'_2,0)=(W,0)$$ Where $W$ can be written as $$W=2p_{cm}^2+2m_e^2+2E_1E_2=4(p_{cm}^2+m_e^2),$$ The expression for the product $E_1E_2=p^2_{CM}+m_e^2$ holds, since in the CM frame momenta differ by direction only. In the lab frame $$(E_{in}+m_e,p).$$ The question is what is the final state in the lab frame. According to Lev Okun we can obtain it by transforming from the CM to the LAB frame. We know that $$\gamma_{CM}=\frac{E_{lab}}{E_{CM}}=\frac{E_{in}+m_e}{E'_1+E'_2}=\frac{E_{in}+m_e}{W},$$ and $$\beta_{CM}=\frac{p_{lab}}{E_{lab}}=\frac{p^*}{E_{in}+m_e}$$ Where the last follows since in the final state, as well as in the initial we have one electron moving and one at rest, I assume that after the collision $E_{lab}=E_{in}+m_e$. Unfortunately, I don't know what the final momentum in the lab frame $p^*$ is. For the transformation back to the lab frame after the collision we have $$p'_{\mu}=(\beta_{CM} W,\beta_{CM}\gamma_{CM} W)$$ So the final momentum in terms of the unknown lab frame momentum and $W$ should be $$p'_{\mu}=\left(\frac{p^*W}{E_{in}+m_e},p^*\right ).$$ But since $p^* \& W$ are unknown, I can't really solve anything. I am not really sure what I am doing is correct. Perhaps I have some misunderstanding. I would appreciate if someone could explain to me what the meaning of Lev Okun's words is.

EDIT: I was made a mistake,by assuming that the two electrons are at rest in the CM frame, which led to the wrong conclusion that $W=2m_e$. Actually the energy of each electron in the CM frame is $E^2=m^2-p_{CM}^2$ and they are not at rest. I corrected the errors. Please see the comment of "dmckee" to this question.

• Neither electron is at rest in the CoM frame, so their energy is $\sqrt{m_e^2 + p_{com}^2}$, not $m_e$. – dmckee --- ex-moderator kitten Sep 8 '16 at 20:24
• I don't understand that. Suppose you have electrons one at rest, the other moving with velocity $v$, and you ask what is the energy to create $e,e,\overline{e},e$, the answer is $7m_e$ and it is solved by using that in the final state CM we have $(4m_e,0)$, why here the CM momentum is zero and in my it is not. I got confused :( – Alexander Cska Sep 8 '16 at 21:05
• Slow down. If you rush ahead from an erroneous starting place you can expect to have trouble. You write *"In the CM frame the total 3-momentum is zero, therefore before an after the collision we have $(m_1+m_2,0)=(2m_e,0).$"*, but this is incorrect. You have to *add the four-vectors*. Not add the three-vectors and then try to deduce the time-like component. Each of the electrons is *moving* in the CoM frame. Call the magnitude of their momenta $p_{com}$, and the total energy-momentum 4-vector is $(2\sqrt{m_e^2 + p_{com}^2},0)$. – dmckee --- ex-moderator kitten Sep 8 '16 at 21:13
• OK and give me some rope here. In the CM we have $\overline{p_1}=\overline{p_1}=\overline{p}$. Using the basic formula for the square of the 4-momentum $E^2-p^2=m^2$ will give us what you wrote. My source of initial confusion is the problem I cited above. Actually it is a different problem, where it is said that the 4 particles are at rest. I got it now. – Alexander Cska Sep 8 '16 at 21:22