I recently read about the Maxwell's laws of Electromagnetic Waves and I found that Light is made up of both Electric and magnetic fields.

So now if i pass the light through a capacitor such that the plates are parallel to the light will the light be deflected?

If it is deflected then what about the particle nature of light in which the photons are neutral without any charge(as far as i know charge do not exist without mass).

If not why are the electric and magnetic fields not affecting the light in the wave nature if it is solely due to the wave nature why are electrons being deflected in the external fields.

I hope someone give me a clear idea of what is wrong with my idea.

  • $\begingroup$ Nope, light isn't. $\endgroup$ – Kunal Pawar Apr 13 '17 at 6:18
  • $\begingroup$ The photon will not be deflected by the plates. It doesnt carry a charge. $\endgroup$ – jaromrax Apr 13 '17 at 6:45

Looking at the classical electromagnetic wave with E and B fields propagating perpendicularly to each other and the direction of motion. These fields do not carry charge, and it is only charge that is deflected/senses electric and magnetic fields. Classically, there can be superposition of two waves which show interference patterns, but superposition is not interaction. The classical electromagnetic wave interacts only when it sees charges (and,possibly, magnetic dipoles). Another way of looking at it is that the dielectric constant of the space between the capacitor plates does not change whether the capacitor is charged or not.

If one goes to the photon level, firing individual photons through the opening of the capacitor plates: photons as neutral will change direction only if they interact. There exists a small probability that a photon meeting an electric field may interact with the electromagnetic interaction, but the probability is tiny as the relevant Feynman diagram has at least five coupling constants. The four of this diagram + one extra because the interaction with the field is with a virtual photon which means an extra vertex.

  • $\begingroup$ If an em wave hits a charge that is free to move (initial static) then should we expect a second em wave from the charge ? Is the first wave vanish or continue to travel after it has interact with the charge ? $\endgroup$ – ado sar Apr 29 at 12:46
  • $\begingroup$ @ado In classical elecromagnetism, the wave has a momentum. If it hits a charge, it may be absorbed, partially absorbed and reflected or completely reflected. . Analogously, part or all of the momentum may be transferred to the charge. When a photon hits an electron, it may be absorbed and rescatered, depending on the particlular boundary conditions. en.wikipedia.org/wiki/Compton_scattering $\endgroup$ – anna v Apr 29 at 13:04
  • $\begingroup$ But if electron gets momentum then from Newton second law it feels accelaration so shouldnt radiate? $\endgroup$ – ado sar Apr 29 at 17:52
  • $\begingroup$ @adosar Newton's law and Maxwell's do not work the same way in the quantum frame. , where one talks of interactions and feynman diagrams. Yes, there will be reatiation or scattering depending on quantum mechanical probabilities calculable from the appropriate feyinman diagram $\endgroup$ – anna v Apr 29 at 19:34
  • $\begingroup$ Thanks for the answer. $\endgroup$ – ado sar May 1 at 19:54

Light is not deflected when it passes through a capacitor.

Light is classicly an electromagnetic wave, and visualizing it as the water waves in a pond might help you solve this cognitive dissonance.

In the finite region in which there is a electric field generated by the capacitor, the electric field of the wave and the previous one interfere through superposition (because, as you might have studied, Maxwell equations are linear), but after that, since the wave hasn't got charge anywhere in it, there is nothing the Lorentz force can interact with, and thus nothing is accelerated, nothing is deflected, and the wave exists the capacitor as it entered it.

This is of course consisten with the quantum mechanical notion that light is made of photons which have no charge, and therefore cant be deflected by electric fields.

  • $\begingroup$ I'm assuming a normal, domestic range of energies for the electromagnetic fields. I'm not sure how this will work with sufficiently high energies. $\endgroup$ – Mario Apr 13 '17 at 10:19

All electromagnetic waves have the electric and magnetic fields vibrating perpendicular to each other. But this does not mean that they are affected by electric and magnetic fields.

The quantum theory shows that light is made up of photons which do not carry charge, during to which it is unaffected by electric or magnetic fields.


Depends how precise you want to be. Classical theory says light wouldn't be affected. If you go to quantum level you then find effects like photon-photon scattering and this means there will be some effect on passing light. Mind that this effect is very weak and also non-linear.

  • $\begingroup$ What do you mean by non-linear (with respect to what quantities) and photon-photon scattering? $\endgroup$ – Yashas Apr 13 '17 at 11:26
  • $\begingroup$ Non-linear here means that superposition principle doesn't apply anymore. I.e., in linear case when field's strengh increases twice, effect also increases twice. If effect is non-linear you can have 0.1 times increase in effect. Regarding photon-photon scattering you can have a look at this wikipedia article: en.wikipedia.org/wiki/Two-photon_physics It's about interaction between photons in vaccum. This is pure QED effect which doesn't exist in classical electrodynamics. $\endgroup$ – MariusM Apr 14 '17 at 8:23

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