If I have charged particle in an external electric field such that the particle accelerates, then energy must be radiated according to the Larmor formula. If this motion is non-relativistic, then I can ignore back-reactive forces, so the particle will never lose energy because it never interacts with its own field. Its kinetic and potential energies may change, but the total energy will remain constant. But this seems to contradict the fact that energy is radiated. Does this mean that the non-relativistic treatment violates conservation of energy?

The context for this question is a solution to a problem in Jackson's Electromagnetism. It has a non-relativistic charged particle incident on a potential field and asks for the total radiated energy. The solution treats the energy of the particle as a constant, and uses this to write the potential in terms of the kinetic energy at infinity. But if there's radiation, isn't this untrue?

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    $\begingroup$ If you derive the non-relativistic limit in a rigorous way, then you'll also find that the emitted radiation becomes zero. The fact that radiation is emitted depends on the fact that the fields at some point depend on the position of the charge a time d/c ago where d is the distance at that time. In the limit of c to infinity you recover the electrostatic case where no radiation is emitted. $\endgroup$ – Count Iblis Nov 10 '16 at 1:04
  • $\begingroup$ I think it is a weakness of ignoring the way a particle gets accelerated. Going deep enougth it's always an EM interaction. Try to disprove it. - Accepting this it is obvious that any acceleration is accompanied by radiation of the accelerated particle. The movement of a relativistic particle (movement means no interaction with something) is not accompanied by any radiation. Otherwise it would indeed violate the conservation of energy. $\endgroup$ – HolgerFiedler Nov 10 '16 at 5:00

So here's the answer. Ignoring back-reactive forces is equivalent to assuming that the radiated energy is small compared to the energy of the particle. This is not the same thing as assuming the particle moves non-relativistically. The radiation from a relativistic particle can still be negligible, and non-relativistic particle can radiate a substantial amount of energy. For a trivial example of the first case, just consider a particle moving at a relativistic constant speed. Then there is no radiation at all. An example of the second case is the Bremsstrahlung radiation from a particle moving at a non-relativistic speed that suddenly stops.

We usually assume that the radiated energy is small, and this was an unstated assumption in the problem in Jackson. If we let the particle accelerate in a potential field for long enough, the total radiated energy will be comparable to the energy of the particle, and our treatment of the energy of the particle as a constant is no longer valid.

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A non-relativistic treatment of a moving charged particle only means that its velocity is much smaller than the speed of light. That doesn't affect how the EM radiation emitted is affected by the acceleration of the charged particle.

The general expression for the electric and magnetic fields from a generally moving charge can be resolved into the radiation part that falls off as $1/r$. This is affected by the particle's acceleration regardless of how fast its moving.

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  • $\begingroup$ Ok, but if we only considered the fields in the radiation zone, i.e., the $1/r$ terms, does this lead to a violation of conservation of energy as I explained above? The particle never loses energy, but there is energy radiated out to infinity, so from where does that energy come? $\endgroup$ – Klein Four Nov 10 '16 at 14:22

A charge q moving in vacuum with a constant velocity v has conserved overall energy because the Electric field density on the direction of v increases while the field density in the other direction decreases.

For sake of argument let's consider the following example: particle position is [x, y, z] = [0, 0, 0] at time t1 = 0 v = [vx, vy, vz] = [1, 0, 0] = 1 m/s (moving along x axis)

Electric field along x-axis at time t1
at E(-1, 0, 0) = q / (4π ε x2) = q / (4π ε)
at E( 1, 0, 0) = q / (4π ε x2) = q / (4π ε)

At time time t2 = 0.5s the particle position will become [0.5, 0, 0]
the electric field alonx x-axis is now
at E(-1, 0, 0) = q / (4π ε (-1 - 0.5)2) =q / (4πε 2.25)
at E( 1, 0, 0) = q / (4π ε (1 - 0.5)2) = q / (4πε 0.25)

dE(-1, 0, 0) = (q / (4πε 2.25) - q / (4π ε) ) is negative while
dE( 1, 0, 0) = (q / (4πε 0.25) - q / (4π ε) ) is positive

Moving particle field is conserved. In front of the particle the field increases at every point, while in the back it decreases at every point (with respect to previous position).

The potential energy of the field in front of the particle is increasing and vice verse for the back.
Epot = (1/2) ε * dE2 * dVolume

If one will integrate over all volume the potential field energy is constant, but the energy density will increase along the direction of movement while it will decrease in the opposite direction.

PS: we have a non relativistic velocity v << c, but the change in field will propagate with the speed of light that means that the field density on the direction of the movement in greater than in the back. This is like red/blue shifting. If you are in front of an ray emitting object the light will come blue-ished while if you are in the opposite direction of speed it will come red-ish.

Note: to make my explanation shorter I've only considered electric field along one axis; but you can deduce by your self what happens in all other directions.

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