Simple electron recoil question

1. Oct 27, 2013

johne1618

Imagine two electrons $A$ and $B$ at rest.

Electron $B$ is at a vertical distance $r$ above electron $A$.

Let us assume that the electrons are constrained to move on horizontal rails.

At time $t=0$ I give electron $A$ a horizontal acceleration $a$ for a time interval $\Delta t$.

At time $t=r/c$, due to the impulse provided by the classical Lienard-Weichert radiation field of electron $A$, electron $B$ gains horizontal momentum:

$$\Delta p = F\Delta t=-\frac{e^2a\Delta t}{4\pi\epsilon_0c^2r}$$

In terms of quantum electrodynamics a photon is exchanged between electron $A$ and electron $B$.

Therefore if electron $B$ receives a momentum $\Delta p$ from the photon does that mean that electron $A$ must recoil with momentum $-\Delta p$ as it emits the photon?

2. Oct 27, 2013

nikol

Yes, that what will happen. The momentum conservation is valid at every vertex(vertex is the point where the interaction happens). So in your case the moving electron A with a momentum $p_{A}$ emits a photon with momentum let's say $q$ and continues propagating now with momentum $(p_{A}-q)$. The electron B which in your case has initially momentum $p_{B}=0$ absorbs the photon and then continues moving with momentum $p_{B}=q$. Only I would not use recoil, since the same process cannot be distinguished from the one in which the electron B is emitting a photon and "recoiling" and the electron A receiving a "boost", a better description would be to say that they exchanged photon with momentum $q$ ($\Delta p$ in your case)

3. Oct 28, 2013

johne1618

So the classical (retarded) Lienard-Weichert interaction theory describes (later) momentum transfer to the static charge $B$ due to the (earlier) accelerating charge $A$ but it neglects the fact that charge $A$ receives (earlier) recoil momentum during the process.

Maybe this recoil momentum is described by a time-reversed advanced Lienard-Weichert interaction in which the earlier momentum transfer to the accelerating charge $A$ is due to the later interaction with the static charge $B$?

Last edited: Oct 28, 2013
4. Oct 28, 2013

nikol

What I wanted to say is(but was not very clear) - for me your description is equivalent to the situation electron A emits a photon and electron B absorbs it. But QFT treats the 2 processes as one, you do not care if A emits and B absorbs or the opposite (B emits and A absorbs), only that they exchange a photon, So I was thinking that if we change the reference frame on your setup such as electron A to be at rest we will get the other part of the process (but I did not check it, sorry).

5. Oct 28, 2013

johne1618

Ok but I don't think special relativity will allow one to change the reference frame to electron A as it is accelerating.

6. Oct 28, 2013

The_Duck

QFT and photons are probably a distraction here. Let's speak purely classically. When you accelerate electron A, you disturb the electromagnetic field. Electromagnetic waves are emitted in all directions. The field ends up carrying some momentum; the momentum density is $\vec{E} \times \vec{B}$ in some units. So not only does the particle gain momentum, so does the field. This means that the force required to accelerate the electron is larger than it would be for an uncharged particle, since the change in total momentum (electron + field) is larger. This effect is known as the "radiation reaction force"; it effectively feels like there is an extra force resisting the acceleration of charged particles. This radiation reaction is the recoil you are looking for.

Finally, when the electromagnetic waves reach electron B, they accelerate it. Some of the momentum stored in the field is transferred to electron B. But there is still a lot of momentum in the field. The radiation reaction recoil is much larger than the change in electron B's momentum, because there is still a lot of momentum left in the electromagnetic field.

7. Oct 28, 2013

johne1618

But the radiation reaction force depends on the rate of change of acceleration and I am assuming electron $A$ has a constant acceleration.

8. Oct 28, 2013

nikol

I was thinking this way - you have your electron A that got accelerated for the time interval $\Delta t$. If you imagine this interval to be very small $\Delta t->0$ so you can say that in words of QFT it emitted a photon, the photon carried away the change in the momentum $\Delta p$ and then at some later time$t=r/c$ was absorbed. The change of frames I meant was after the acceleration boost, so they would be inertial. I was trying to imagine how the situation can be understood as a exchange of photons, but may be this is not a good way to approach this problem and we should look at it from the classical point of view, like the other guy suggests.