0
$\begingroup$

We can all agree empirically that a stream of photons encountering an arrangement of atoms forming a lens bends.

But do the photons actually make contact with the atoms in the process of bending?

I think this question maps neatly onto the similar question, do photons passing through swirling water change direction because of making contact with the water atoms, or because of a quantum field of some sort created by virtual interactions with the water atoms, averaged out?

$\endgroup$
  • $\begingroup$ The answer might depend on what you think "contact" means. When you say that your finger makes "contact" with a tabletop, you are talking about electromagnetic interactions between electrons in the atoms that make up your finger and electrons in atoms that comprise the table top. A photon doesn't have atoms, it doesn't have electrons, and it isn't an electron. Whatever "contact" a photon makes with the atoms of matter, it is not the same phenomenon as in the usual meaning of the word. $\endgroup$ – Solomon Slow Feb 14 '18 at 15:18
  • $\begingroup$ Okay, but there are some interactions between photons and atoms which have a concrete finality to them, right? Such as activating an atom in a cell on a photo detector? Am I right in understanding that what's happening there is that the energy from a photon has been absorbed by an atom and resulted in some change in that atom's state? $\endgroup$ – Ben Wheeler Feb 15 '18 at 14:46
1
$\begingroup$

It depends on what do you understand by it. Interaction via the intermediate of "quantum field of some sort created by virtual interactions with the water atoms" is not a valid photon-water interaction? I really think this is only question about labeling, about human consensus what we call "interaction".

I would certainly say that water and light interact (no interaction means no effect). Maxwell equations of light-in-water can be seen it two equivalent ways:

1) One can describe the effect of the water by the material constants $\epsilon$ and $\mu$ of the water. Then one describes by the equations only light, presence of water is hidden in these constants.

2) One can describe all (light+water) as one system and use vacuum constants $\epsilon_0$ and $\mu_0$.

Electrodynamics inside medium is electrodynamics in vacuum where one takes into account charges and currents of the medium.

This all just to say that I do not think your question is "core" physics question.

$\endgroup$
  • $\begingroup$ Let me ask a slightly related question. Presumably if a photon travels through many miles of water, it eventually interacts with an atom in a more substantial way than just being affected by a field. What is the nature of this collision? Is the photon absorbed by adding energy to the electron state which is then released as another photon when the electron moves back down? If that's what happens, is there any pattern to which direction the new photon is sent out at? $\endgroup$ – Ben Wheeler Feb 15 '18 at 14:41
  • $\begingroup$ What I'm trying to get at is, is the bending of light or the passage of light through a transparent medium at all a matter of being absorbed and reemitted by atoms in the same rough direction that the photon originally traveled? $\endgroup$ – Ben Wheeler Feb 15 '18 at 14:44
  • $\begingroup$ Honestly I do not know. Some qualified physicist might know. Light bending is certainly related to the speed of propagation: more rapid in vaccum then in water. Because of this, the light-front is deviated. How exactly (micorscopically) light is slowed-down by water, I do not know. This is maybe what you look for: askamathematician.com/2011/08/… and en.wikipedia.org/wiki/Refractive_index#Microscopic_explanation $\endgroup$ – F. Jatpil Feb 16 '18 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.