Maxwell's equations model EM radiation as propagating away from an accelerating charge. Suppose instead the propagation of this EM radiation is reversed and presented as a source-free boundary condition so that it converges upon a source-less point: How would Maxwell's equations model this?

In particular: would the EM fields converge and then propagate away from the point, or would the temporary EM singularity at the point invalidate modelling after this time?

  • $\begingroup$ What is "the temporary EM singularity"? $\endgroup$ – user195162 May 12 '19 at 1:01
  • $\begingroup$ @Quantumness The EM field converges into a volume getting smaller with time, centered at the point, leading to an increase of the EM fields and energy-momentum density within this volume. This blows up to infinity when all the EM fields have converged to the same point AFAIK. $\endgroup$ – Larry Harson May 12 '19 at 1:48
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    $\begingroup$ > "so that it converges upon a source-less point" if you take a field of a charged particle, run it in time backwards but remove the particle, the resulting process violates Maxwell's equations. There has to be a charged particle at the point of convergence at the time of the convergence. More specifically, there has to be accelerating electric charge where the ordinary radiation field has singularity. $\endgroup$ – Ján Lalinský May 12 '19 at 23:23
  • $\begingroup$ @JánLalinský yes, but my question is asking about radiation which is generally taken to be irreversible via Lamor's formula and hence unbounded to sources thus satisfying $\Box A_{\mu} = 0$ using the Lorenz gauge. $\endgroup$ – Larry Harson May 14 '19 at 1:22
  • $\begingroup$ @LarryHarson the Liénard-Wiechert retarded field of a point charge obeys the wave equation you mention only at points different from the point where the charge is located. That is, that standard radiation field does not obey this equation everywhere. There may be a field that obeys this equation everywhere but then this field does not have to be connected to particle's charge, position or motion in any way - it is an independent background field that obeys Maxwell's equations without sources. $\endgroup$ – Ján Lalinský May 14 '19 at 11:52

Nice question.

The answer I think is that you have to compare like with like.

Let's take field propagating from the accelerated charge at the origin. Once the field is in the vacuum, the Maxwell's equations are time-reversible, so you could simply 'play the video backwards' and see the waves converging into a point. BUT, once the waves get to the origin, in your video playback, there must be a charge there. If there is one, good, the charge will simply absorb the power (where would the power go? no-where. you did not specify how you made the charge move in the first place - that is the fudge-factor).

What if there is no-charge at the origin? What if you built a spherical distribution of antennas aiming to launch the field back into the origin with the same distribution as that from an accelerated charge, but there is no charge there? My intuition tells me that this would not be possible, i.e. you would find it impossible to launch just that field and nothing else. Because that field would not be a solution to Maxwell's equations.

To go any further, I think you need to specify more precisely what field would you launch.

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  • $\begingroup$ But still, the total field can be decomposed into a converging radiation field plus what was needed to create it. These fields can be modelled by Maxwell's equations independently of one another. $\endgroup$ – Larry Harson May 14 '19 at 1:27
  • $\begingroup$ If you start splitting fields into linear consituents, you have to be careful to keep them solutions to Maxwell's equations. Otherwise the whole thing becomes very hard to interpret. Lets say you decompose your 'convering field without the charge' into vector spherical harmonics. The radial part will be given by spherical bessel functions of the first kind. These functions are either zero or finite at the origin. So you would not get diverging fields at the origin unless you start off with infinite energy (IMHO). $\endgroup$ – Cryo May 14 '19 at 10:14
  • $\begingroup$ ... and the spherical waves would continue propagating after converging. i.e. they will 'cohere' into a high-intensity region at the origin, and then decohere out. $\endgroup$ – Cryo May 14 '19 at 10:22

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