# Consider two electrons released from rest at a distance $d$. How much energy is radiated?

Abraham-Lorentz?

Two electrons d apart have potential energy. Release them, they will be repelled according to Coulomb's law.

I could make an assumption about the associated vector potential, but I think that would ignore some important relativistic modifications. I'm only familiar with relativistically correct potentials of a charge with uniform velocity, there will be acceleration here.

But accelerating charges radiate, producing radiation, resulting in breaking action.

How much energy us lost instead of becoming kinetic?

$$\frac{e^2}{4\pi\epsilon_0d}=\frac{e^2}{4\pi\epsilon_0x}+(2)\frac{1}{2}m_e\dot{x}^2+E_\mathrm{rad}$$

This is a straight forward algebra problem if $$E_\mathrm{rad}$$ is assumed to be zero. The potential energy term goes to zero, leaving in expression easy to solve for $$\dot{x}$$.

What's $$E_\mathrm{rad}$$? I'm not sure where to start without inserting assumptions I suspect might be faulty.

At first blush, I'd begin with assuming $$\vec{A}=\frac{\mu_0}{4\pi r}\frac{d\vec{p}}{dt}$$ where $$\vec{A}$$ is the vector potential and $$\vec{p}$$ is the dipole moment. This is my habit for radiation problems. But the dipole moment in this case is $$\vec{0}$$.

What if attention is just placed on one of the electrons?

Then $$\vec{p}=-ex\hat{i}$$ since the electrons move in the + and - x directions \begin{align} \nabla \times \vec{A} & = \vec{B}=\frac{\mu_0}{4\pi}\left[\frac{-\hat{r}}{r^2}\times (-e)\dot{x}\hat{i}+\frac{-\hat{r}}{r^2c}\times (-e)\ddot{x}\hat{i}\right] \\ \vec{E} & = c\vec{B}\times\hat{r} \\ \vec{S} & = \frac{1}{\mu_0}\vec{E}\times\vec{B} \\ E_\mathrm{rad} & = \iiint\vec{S}\cdot\hat{r}R^2\sin(\phi)d\phi d\theta \end{align} I'd flesh the math some more, but I'm pretty confident I made a mistake with my expression for the Vector Potential.

I might try starting with a relativistic expression for a moving charge.

• I think this is interesting enough that I would like to see a solution, so I have voted to reopen. – John Rennie Dec 10 '18 at 15:35
• *This isn't a homework question. If this question is asked as homework, your professor is a sadist. If the radiation were ignored it would be a homework question.This is an invitation to investigate the Abraham-Lorentz force in an elementary context. I see flaw with all three ways I can think of to address the issue. – R. Romero Dec 10 '18 at 15:57
• I think that this is a much harder problem than you're allowing for. As you correctly note, this is an Abraham-Lorentz problem, and it's much harder to get that to work correctly than what naive expectations would set. For starters, I see no reason why energy methods should work - I don't see why anything other than a full solution to the Abraham-Lorentz equation of motion will work, with the energy balance then inferred from it. And then that puts you into the runaway-solutions and third-order-in-time types for problems with Abraham-Lorentz. – Emilio Pisanty Dec 10 '18 at 16:01
• What does this look like if you go in reverse? Start the electrons at infinity. Bring them toward each other at constant velocity until they are d apart. I think this avoids having to tangle with the Abraham-Lorentz force. A force would have to be applied to cancel out the Coulomb repulsion. – R. Romero Dec 10 '18 at 17:00
• Why are you asking for a maximum rather than a definite value? If the charges are released at rest, isn't there a definite answer, not just a maximum? If I do a simple estimate based on the Larmor formula, I get $E_\text{rad}\sim k^{5/2}c^{-3}e^5m^{-3/2}r^{-5/2}$. For $d=0.1$ nm, this comes out to about $10^{-24}$ J, which is about $10^{-7}$ of the energy released. – Ben Crowell Dec 10 '18 at 17:56