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When light interacts with matter, it is (most often) the electric field that is absorbed. So what happens with the magnetic field when the electric field interacts with, say, an electron? Assuming that the electron absorbs the light and oscillates at the frequency as the incident wave, and then re-emits this light in all directions, the magnetic wave has to "follow" the electric component in order for light to be re-emitted from the electron?

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    $\begingroup$ Separating the electric field from the magnetic field in a photon is not possible - they are one and the same. $\endgroup$ – Jon Custer Sep 12 '16 at 13:43
  • $\begingroup$ to follow unless if it is the same object seen from 2 points of view $\endgroup$ – user46925 Sep 12 '16 at 14:03
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    $\begingroup$ Its a pity whoever down voted my answer didn't give a reason, but live and learn. Best of luck with a proper answer though. $\endgroup$ – user108787 Sep 12 '16 at 14:04
  • $\begingroup$ @CountTo10 Your answer was down voted because it was incorrect. You deleted it, which is your choice, but in doing so you cut off any chance of hearing an explanation of why it was wrong. Down voting is a way of expressing the fact that an answer has problems. It would be nice to have an explanation of the problem, but that takes time. A simple down vote is an imperfect way of saying "watch out for this one". $\endgroup$ – garyp Sep 12 '16 at 14:20
  • $\begingroup$ @garyp hi Gary, I would be a total hypocrite if I moaned too much about it, I hope I didn't, as I have done exactly the same myself to other people :) I have it as a favorite, so I will watch it. Thanks for taking the time to reply, much appreciated. $\endgroup$ – user108787 Sep 12 '16 at 14:41
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In an electromagnetic wave the electric and magnetic fields are always intrinsically linked. In a plane transverse electromagnetic wave the amplitude of the transverse magnetic field H (which oscillates in a plane normal to that of the electric field) is given by H=E/Z, where Z is a constant called wave impedance, which is about 377 Ohms in free space. When light interacts with matter via the electric field both the electric and the magnetic field are reduced simultaneously.

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In the classical scenario you pose, no work is done by the electromagnetic field - the dot product of the E-field and current is zero. Thonson scattering is an elastic interaction.

The scattered waves have exactly the same relationships between the E- and B-field as any other vacuum wave solution of Maxwell's equations.

Things change if the wave interacts dissipatively - for example at an interface with a conductor. Here there are Ohmic losses.

In vacuum, an EM wave has equal energy densities in the electric and magnetic field and the E- and B-fields oscillate in phase. When you use the appropriate boundary conditions with Maxwell's equations, you find that the ratio of E- to B-field is dramatically reduced in the conductor. Most of the energy density is now in the magnetic component and the E- and B-fields are roughly 45 degrees out of phase.

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    $\begingroup$ If you don't mind me asking: why exactly does the B-field lag behind the E field by roughly 45 degrees in a good conductor? I have come across this fact in Griffiths but have never found an intuitive physical explanation for it $\endgroup$ – SalahTheGoat Apr 21 at 11:38

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