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When an EM wave is created, the source emits Photons with Energy X. Are these photons the only traveling particles driving the energy (Electron Excitation) in the EM field? -- OR -- the photons created from the resting electron also drives the wave?

Which photons are responsible for driving the wave? The photons produced at source or those produced by the electrons in the $E$ Field?

I ask because if there was superposition - And Electron would have higher energy and thus a more energetic Photon created and absorbed by neighbouring electrons... which would in essence, change the entire wave upon one superposition?

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When an EM wave is created,

The electromagnetic wave is a classical wave, a solution of classical maxwell's equations, it has a frequency nu and propagates as varying electric and magnetic fields

em wave

Electromagnetic waves

Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. Note that the electric and magnetic fields in such a wave are in-phase with each other, reaching minima and maxima together

Then you say:

the source emits Photons with Energy X.

Photons are quantum mechanical entities, each described by a wavefunction which is the solution of a quantized maxwell's equation .

This image can give an intuition on how the photons, which are elementary particles and are characterized only by their energy=h*nu and their spin, add up in great numbers to build up the classical wave.

photspin

The photon wavefunction, a complex function, has the E and B information, and the superposition of the wavefunctions builds up the classical wave. This can be seen mathematically in this link, but it needs familiarity with quantum field theory.

The photons and the beam they build up does not depend on a medium unless in transparent matter where a beam can propagate, when it will be affected by the lattice fields by the interaction of the photons with the lattice.

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  • $\begingroup$ Thank you Anna. So can we say that the photons created at source are the sole photons responsible for the wave propagation, and not the decay photons emitted? $\endgroup$ – Rain Feb 9 '17 at 18:42
  • $\begingroup$ there are no decay photons,unless the beam is in a medium, when there could be scattered ( not decayed) photons and absorption and reemission photons. Sources in vacuum can be accelerating charges, as in a spark, or excitations and deexcitations as an incandescent lamp or fluerescent lamps which might be considered to have excitations and decays as . Once they leave the source the beam is self propelled. $\endgroup$ – anna v Feb 9 '17 at 19:17
  • $\begingroup$ This is in the case of a Radio EM wave source with a charge producing Photons to propagate the wave in free space. would the emission of photons from resting electrons play a role in the propagation, or it's always the same finite set of Photons produced at Source that travel X distance exciting the electrons of the E Field in the wave Path? $\endgroup$ – Rain Feb 9 '17 at 19:20
  • $\begingroup$ it cannot be a finite set, it is a continuous emmission from the antenna source electrons . In the animation above, think of an antenna at z. $\endgroup$ – anna v Feb 10 '17 at 5:19
  • $\begingroup$ Thank you Anna. I've been reading your other answers around antennas and were also very helpful. $\endgroup$ – Rain Feb 10 '17 at 5:27
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A photon will travel through a vaccuum, i.e with no intermediate electrons to help it along its way. However when a photon travels through glass it becomes absorbed and reemitted many times by each successive atom, thus it could by thought of as many photons but we only perceive it as one ray going in and out. So yes the resting electrons are important as they become excited one after the other in glass.

Superposition is possible at the electron if 2 photons arrive almost simultaneously to excite the electron to 2x the energy. Once the electron is excited it can reemit a higher energy photon or it could even emit the 2 lower energy photons depending on circumstances.

An advanced understanding is really that a photon is a wave function, and a wave function is a predetermined path for the photon to take , I believe it requires a starting electron and a finishing point also an electron. Photons never superimpose but for example the electron in the nerve cell in the eye will not see 2 photons if they are out of phase with each other (180 deg), these photon will get absorbed as heat in deeper tissues.

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  • $\begingroup$ " Photons never superimpose" is wrong. Their wavefunction superposition builds up the electric and magnetic fields of classical electromagnetism. $\endgroup$ – anna v Feb 8 '17 at 9:02
  • $\begingroup$ Hi Anna, classically I agree ... but if 2 photons are 180 deg out of phase they still do have energy and can act independently. Superposition implies cancelation at 180 deg phase delta. $\endgroup$ – PhysicsDave Feb 9 '17 at 4:08

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