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Consider two charged particles initially at rest in the configuration below.

diag

Let us assume the following:

  1. Starting at time $t=0$, we apply a constant force $f$ to the the bottom particle so that it has a constant acceleration $a=f/m$.
  2. The top particle has a large mass $M$.
  3. 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 radiative electric field due to the bottom particle produces a force $F$ on the top particle given by:

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

For simplicity we assume that the mass $M$ of the top particle is so large that its acceleration due to force $F$ is negligible. Thus it does not produce a significant radiative electric field back at the position of the bottom particle.

Now let us calculate the energy $E_{in}$ that we supply to the system.

Let us assume that we apply a force $f$ to the bottom particle for a time interval $\delta t$.

During time interval $\delta t$ the bottom particle travels a distance $d$ so that the energy supplied $E_{in}$ is given by:

$$E_{in} = f \times d$$

The bottom particle has a constant acceleration $a$ so the distance it travels in time interval $\delta t$ is given by:

$$d = \frac{1}{2}a \delta t^2$$

Using the expression for the acceleration of the bottom particle, $a=f/m$, we find from the above two relations that the energy supplied to the system during time interval $\delta t$ is given by:

$$E_{in} = \frac{f^2\delta t^2}{2m}$$

Where has this energy gone?

The kinetic energy, $KE$, of the bottom particle after the time interval $\delta t$ is:

$$KE = \frac{1}{2} m v^2$$

The velocity of the bottom particle after a time interval $\delta t$ is given by:

$$v = a \delta t$$

Since the acceleration $a=f/m$ the above two equations imply that the kinetic energy of the bottom particle is given by:

$$KE = \frac{f^2 \delta t^2}{2 m^2}$$

Therefore, as expected, all the energy $E_{in}$ that we supplied during time $\delta t$ has gone into the kinetic energy $KE$ of the bottom particle.

But, as stated above, since the bottom particle is accelerating, after a slight time delay $t=r/c$, there is a force $F$ acting on the top particle. During the time interval $\delta t$ an energy $E_{top}$ is supplied to the top particle given by:

$$E_{top} = \frac{F^2\delta t^2}{2M}$$

My question is where has this energy come from given that all the energy we supplied, $E_{in}$, is fully accounted for in the kinetic energy of the bottom particle alone?

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  • $\begingroup$ You say "an energy is supplied to the top particle". Only if it moves, which you said it wouldn't by assumption. $\endgroup$ – ACuriousMind Jan 16 '15 at 16:24
  • $\begingroup$ I assume that the heavy top particle accelerates slightly, so that it absorbs some energy, but not enough to produce a significant force back on the light bottom particle. $\endgroup$ – John Eastmond Jan 16 '15 at 23:48
  • $\begingroup$ Yes, and that's basically the limit $M \to \infty$, so that $E_\text{top} = 0$. $\endgroup$ – ACuriousMind Jan 16 '15 at 23:57
  • $\begingroup$ The question does not assume $M\rightarrow \infty$, only that $M$ is great enough so that the action of the secondary radiation on the source can be neglected. Negligible acceleration of $M$ is not the same thing as zero acceleration. Actually, this assumption is not that important for the question. $\endgroup$ – Ján Lalinský Jan 19 '15 at 17:27
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To say that

we assume that the mass $M$ of the top particle is so large that its acceleration due to force $F$ is negligible.

means to take the limit $M \to \infty$ (since $a = \frac{F}{M}$ shall produce $a = 0$), since only infinitely heavy things do not accelerate when a force is applied. Under this assumption,

$$ \lim_{M\to\infty} E_\text{top} = 0$$

hence there is no violation of energy conservation in the case considered.

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  • $\begingroup$ @Jan: If the acceleration $a$ is neglegible, then $E_\text{top} = a\frac{F\delta t}{2}$ is neglegible as well. The question is just inconsistent in the approximations made - if you include the back-reaction of the moving top particle, then the velocity of the small particle will be a bit smaller than calculated here and the "missing" energy is in the movement of the top particle. $\endgroup$ – ACuriousMind Jan 19 '15 at 17:31
  • $\begingroup$ Kinetic energy of the top particle $E_{top}$ is the point of the question, so it is not negligible. The acceleration $A$ of the top particle cannot thus cannot be zero, but it can be small enough in the sense its retarded field action on the bottom particle is negligible. Whatever the magnitudes, they do not change answer to the question - the extra kinetic energy comes as a result of EM energy flux. There is no reason to expect total kinetic energy of the particles equals work done on the bottom particle. $\endgroup$ – Ján Lalinský Jan 19 '15 at 17:50
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My question is where has this energy come from given that all the energy we supplied, $E_{in}$, is fully accounted for in the kinetic energy of the bottom particle alone?

The work done by external force accelerating the particle 1 (bottom) goes into kinetic energy of the particle 1.

The work done by electric force of the particle 1 on the particle 2 (top) goes to kinetic energy of the particle 2.

Work-energy theorem is valid because the kinetic energy of particle 2 increases due to work of electric force of particle 1 on particle 2. In the picture of continuous transfer of energy, energy is being accumulated in the particle 2 while there is flow of EM energy in the region surrounding it and while some EM energy is being sucked up into the particle.

The rate of increase of kinetic energy equals income of EM energy through boundary surface of the region (arbitrary closed surface that encloses the particle 2) minus the rate of increase of EM energy in the region.

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