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Imagine a lone charged particle. It emits an EM field which propagates outwards at the speed of light. But since there is no other charged particle to interact with, there is no repulsion or atraction, and the particle will not be acelereated by the EM field: so there is no EM potential energy.

Now, let's say I hit the particle (with a neutral uncharged baseball bat) causing it to accelerate. Its EM field will move with it, overwriting the old one, creating a ripple in the field with propagates outward at the speed of light. An EM wave, also called light.

My question is, how can this wave carry energy?

Specially considering the particle itself didn't even have EM energy, since there is no other charge that could be affected by the light.

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  • $\begingroup$ see en.wikipedia.org/wiki/Quantum_electrodynamics $\endgroup$ – pawel_winzig Nov 19 '17 at 13:05
  • $\begingroup$ Are you familiar with the Abraham-Lorentz force? $\endgroup$ – Alfred Centauri Nov 19 '17 at 13:11
  • $\begingroup$ @pawel_winzig I was hoping for an answer that predated QED, something more intuitive like how the EM wave can be simply though of a ripple in the field. -AlfredCentauri No, I was not. Read the article but I didn't really understand it. But it means that there is an explanation that doesn't need QED, right? I was looking for something more classical... $\endgroup$ – Juan Perez Nov 19 '17 at 13:22
  • $\begingroup$ @JuanPerez: As explained in the wiki article mentioned by Alfred Centauri: The classical description fails, renormalization is necessary. $\endgroup$ – pawel_winzig Nov 19 '17 at 13:32
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    $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/Waves/emwv.html on how an em field carries energy. That accelerated charges radiate is an experimental fact. Before maxwell' unified electricity and magnetism equations it was not known that light was electromagnetism. see also cv.nrao.edu/course/astr534/PDFnewfiles/LarmorRad.pdf $\endgroup$ – anna v Nov 19 '17 at 13:41
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Imagine a lone charged particle. It emits an EM field which propagates outwards at the speed of light.

If a charge would emit EM radiation, it would lose energy. In reality a charge obey a magnetic dipole moment and an electric field. Both are intrinsic (existing independently from outer circumstances) properties. They are static and nothing propagates outwards.

Now, let's say I hit the particle (with a neutral uncharged baseball bat) causing it to accelerate.

The baseball bat may be uncharged, but you over-give kinetic energy - usually associated with virtual photons - from the outer electrons of the bat to the charge. Since you are accelerate the charge, photons are emitted. After acceleration, a free charge moves without acceleration and doesn’t radiate. It has still its magnetic and its electric field.

Its EM field will move with it, overwriting the old one, creating a ripple in the field with propagates outward at the speed of light. An EM wave, also called light.

What are moving are the charges electric and the magnetic fields together with the electron. You can have this imagination, that during accelerations ripples occur. But the EM radiation always happens in quant, called photons. Under strong accelerations (braking is an acceleration too), you get “harder” photons like X-rays. Under soft acceleration you get photons more in the infrared or visible light.

My question is, how can this wave carry energy?

Simple the decrise in kinetic energy is accompanied by photon emission. QM treated a overall existing EM field in which photons are disturbances. But once again, the photons carry energy away.

Specially considering the particle itself didn't even have EM energy, since there is no other charge that could be affected by the light.

An acceleration is the reason for a charge to radiate energy. An acceleration could happens only in two cases. Or under the influence of other particles, like electrons, protons or even neutrons. Or under the influence of a magnetic field, called the Lorentz force.

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  • $\begingroup$ If it was wrong to say that an accelerating charge could be lone, and that there had to be another thing suceptible to the photon, then I can see how energy can be lost to radiation. $\endgroup$ – Juan Perez Nov 19 '17 at 14:25

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