# Momentum conservation in an electromagnetic system?

Suppose that I have two charged particles in the configuration below. Let us assume the following:

1. We apply a constant force $f$ to the the bottom particle so that it has a constant acceleration $a(t)=f/m$.
2. The velocity of the bottom particle is negligible to simplify the calculation of the electric field.
3. The top particle is initially stationary with a large mass $M$.
4. The distance $r$ is large enough so that the Coulomb repulsion between the particles, which is inversely proportional to $r^2$, is negligible.

Under these conditions the Lienard-Wiechert retarded electric field due to the bottom particle, accelerating at time $t=0$, produces a force $F$ on the top particle, at a later time $t=r/c$, given by:

$$F(t=r/c)=\frac{qQa(t=0)}{4\pi\epsilon_0c^2r}.$$

Let us say that in a time interval $\Delta t$ the top particle gains a momentum $F\Delta t$ towards the left.

My question is the following: How is this momentum change balanced?

The conventional answer is to say that the EM field gains an opposite momentum to the right.

But the only way I can see the EM field changing is if the top particle accelerates to the left, under the action of $F$, producing its own counter electric field towards the right. However this won't work as the mass $M$ of the top particle is assumed to be large so that its acceleration, and thus its induced electric field, is negligible.

In summary I can see how the electric field in the vicinity of the top particle can transfer momentum to it but I can't see any mechanism whereby the top particle transfers momentum back to the field if we are free to make the assumption that it is so heavy that it can absorb the momentum without changing its motion appreciably.

P.S. My hypothesis is that a balancing momentum $F\Delta t$ to the right is transmitted backwards in time from the top particle at time $t=r/c$ to the bottom particle at time $t=0$ using an advanced electromagnetic interaction. This momentum then has the effect of reducing the effective mass of the bottom particle (less external force has to be supplied to produce a given acceleration $a$).

Two things.

First of all, in general, if you make simplifying assumptions that aren't actually true, then why would you expect actual momentum conservation? You should only expect your calculation to yield approximate momentum conservation.

Secondly (and more importantly), is that if you are applying a constant force $f$ to this system, then you should be increasing the momentum of this system. Of course, if you were to include in your system that which is producing $f$, then the total momentum would be conserved, but as you are not, you should expect the momentum of the system (these two particles and their E&M field) to be increasing over time in the direction of $f$.

• But I guess I can say that in a time interval $\Delta t$ I would expect a change in momentum of the system to be given by $f \Delta t$ and not $f \Delta t - F \Delta t$. Nov 12 '14 at 18:15
• @JohnEastmond That's right. The momentum of your entire system should change (in magnitude) by $f\Delta t$ over a time period of $\Delta t$. Nov 12 '14 at 18:26
• Then my problem is that I don't understand the mechanism by which the change in momentum of the top particle, $-F \Delta t$, is balanced by a corresponding change in momentum of the EM field, $F \Delta t$. Nov 12 '14 at 18:47
• So I am admittedly trying to avoid details because I believe the actual computation will be at least slightly tedious. Nevertheless, I think it's easy to see that this is possible. The system has three things contributing to its momentum: the mechanical momentum of the top particle, the mechanical momentum of the bottom particle, and the momentum of the E&M field. If we declare that the positive direction is to the right, then the momentum of the bottom particle is increasing over time and the momentum of the top particle is decreasing over time . . . Nov 12 '14 at 20:17
• . . . Then the momentum change of the E&M field would just have to be so that, after you add these three together, you get $f\Delta t$. Nov 12 '14 at 20:17

But the only way I can see the EM field changing is if the top particle accelerates to the left, under the action of F, producing its own counter electric field towards the right. However this won't work as the mass M of the top particle is assumed to be large so that its acceleration, and thus its induced electric field, is negligible.

Density of momentum of electromagnetic field and the stress tensor in macroscopic theory are functions of total electromagnetic fields, not function of their acceleration-dependent parts.

If the particle is very heavy, it will move with weak acceleration and the acceleration-dependent part of its retarded field will be also weak. But there will be also the static electric field of the particle, which is proportional to its charge and does not depend on its mass. The greater the charge, the greater the change in particle momentum and the greater the electric field strength which figures in the EM momentum expressions.

The static field of the upper particle together with the propagating field of the particle down below combine to substantial total field which may then lead to substantial density of momentum of electromagnetic field and the stress tensor near the upper particle.