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Suppose we have an electromagnetic wave the propagates through space, in the direction of x-axis. A charged particles lies along the x-axis and at some time $t_0$ the EM wave hits the particle. The electric field exerts a force:

$$\mathbf{F} = \frac{\mathbf{E}}{q}$$

I can't understand the way energy is conserved in this case and the interplay between incident and emitted radiation (we know that light must be emitted because the charged particle is accelerated by the force).

The EM wave before time $t_0$ has an energy $E_{inc}$, that is the energy of the incident wave. When it hits the atom it loses some energy which is converted in kinetic energy of the particle. So at this point we have:

$$E = K + E'_{incident} + E_{emitted}$$

where $E'_{incident}$ denotes the energy of the original wave after interaction with the particle has take place. Is this description correct? I tried to find some classical description of such situation and during searching I found the following:

Optical radiation accelerates charges in a material, and accelerating charges emit light that adds to (or subtracts from) the incident light.

I can't understand the final statement, that is how emitted light adds or subtracts from the incident light if the energy is conserved?

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  • $\begingroup$ Please note that "When it hits the atom" is a point of confusion between classical electromagnetism and quantum electrodynamics. An electromagnetic wave can only interact with an atom quantum mechanically, i.e. a photon that is building up the EM wave will interact with the particle. This does not have to do with the kinematics you are asking about, but if you are a student you have to learn the difference. $\endgroup$
    – anna v
    Commented Apr 12, 2022 at 4:03
  • $\begingroup$ This link may help to understand how energy is carried in classical electromagnetic waves .hyperphysics.phy-astr.gsu.edu/hbase/electric/engfie.html $\endgroup$
    – anna v
    Commented Apr 12, 2022 at 4:09

2 Answers 2

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The EM wave before time $t_0$ has an energy $E_{inc}$, that is the energy of the incident wave. When it hits the atom it loses some energy which is converted in kinetic energy of the particle. So at this point we have:

$$E = K + E'_{incident} + E_{emitted}$$

where $E'_{incident}$ denotes the energy of the original wave after interaction with the particle has take place. Is this description correct?

No.

We can't assign separate energies to "partial waves" and add them up.

In macroscopic EM theory, EM energy is associated with any space region in which there are EM fields, but this energy can't be distributed/assigned in some objective and useful way to individual elementary waves present.

One correct description of the process would be: when the EM wave hits a charged particle initially at rest, the particle accelerates and thus produces secondary wave spreading out from the particle.

In terms of energy, some EM energy (in the space around the particle, due to EM field of both primary wave and secondary wave being present) is being transferred to the particle and this shows up as increased kinetic energy or internal potential energy of the particle ( if the particle has inner degrees of freedom). Some other EM energy is being redirected in direction; before the interaction, EM energy moved in direction of the primary wave, but when interaction is going on, a different energy flow appears, where some energy is redirected to all possible directions (due to presence of the secondary wave).

Optical radiation accelerates charges in a material, and accelerating charges emit light that adds to (or subtracts from) the incident light. -- I can't understand the final statement, that is how emitted light adds or subtracts from the incident light if the energy is conserved?

They mean EM field. Accelerating charge produces its own EM field, which adds up to the field of the primary wave. This does not mean "energies of secondary and primary wave" add up. There are no such separate energies. There is only total EM energy of the resulting EM field.

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  • $\begingroup$ Just to make it clear: the incident radiation (the disturbance of the em field) continues to propagate irrespective of the obstacle (charged particle). Is that correct? $\endgroup$
    – Anton
    Commented Apr 17, 2022 at 16:42
  • $\begingroup$ @Anton yes the primary wave propagates unchanged. Secondary wave is created and together they form total EM field whose spatial pattern is changed by presence of the charged particle. $\endgroup$ Commented Apr 17, 2022 at 17:27
  • $\begingroup$ So there must be destructive interference between the primary and secondary wave that lowers the energy stored in the EM field? I mean if we have mechanical energy stored in the particle's motion then the energy stored in EM field must be lowered (conservation of energy). Lower EM energy can only achieved if the there is some destructive interference between primary and secondary wave. Am I getting it right? $\endgroup$
    – Anton
    Commented Apr 18, 2022 at 20:02
  • $\begingroup$ @Anton yes, the partial EM fields add up to total field that in some regions has lower intensity than that of one partial field. Especially the region behind the charged particle in the wave zone (where particle's field is predominantly due to wave component) is a good example of this cancellation and decrease of EM energy. $\endgroup$ Commented Apr 18, 2022 at 20:45
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The total energy is $E_{incident+emmitted} + E_{mechanical}$, the mechanism by which the "incident" waves energy is reduced, is due to the superposition of the emitted wave and the incident wave. The incident wave doesn't change to ', it stays as it is, and the resultant field energy of the 2 waves decreases.

The "lost" field energy, is made up for in the increase in mechanical energy.

Field energy is not a conserved qauntity, total energy is.

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  • $\begingroup$ Why incident wave doesn't change after interaction? If it is unchanged doesn't this violate energy conservation as we have the same incident plus mechanical energy? $\endgroup$
    – Anton
    Commented Apr 11, 2022 at 20:00
  • $\begingroup$ We have the incident wave, plus the mechanical energy, plus the emitted wave(+original field from the charge). There is destructive interference. $\endgroup$ Commented Apr 11, 2022 at 20:02
  • $\begingroup$ If I understand it correctly, the incident wave just propagates like the particle wasn't there. The particles increases its mechanical energy but this gain is offset by destructive interference between emitted and incident radiation? Can it be shown that there will always be destructive interference? Because the blockquote states that it may adds (constructive interference). $\endgroup$
    – Anton
    Commented Apr 11, 2022 at 20:07

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