(NOTE: I am an 8th grader, so I may not be capable of perfectly elaborating my point by scientific measures. Also, English is not my first language)

I have read that the speed of light is constant, and does not decrease unless under certain circumstances (such as going through dense and/or thick material.)

When a ball hits the wall and rebounds, it has momentum, except for the moment that it hits the wall, since the speed would be 0 there.

How about light? How can light maintain a constant speed upon hitting something? Shouldn't the speed of the photons be zero when they are neither moving forward nor backward when they have hit the wall?

  • 2
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/35177/2451 , physics.stackexchange.com/q/83105/2451 and links therein. $\endgroup$ – Qmechanic Jul 21 '16 at 9:11
  • 1
    $\begingroup$ While the explanations you were given are not bad, one can point out that, in reality, light does penetrate a reflecting material a little and in that layer the waves will have a different velocity, however, that layer is usually very thin for metallic mirrors (on the order of a few nm to a few ten nm), i.e. it is much thinner than the wavelength. In dielectric mirrors it's a whole different ballgame, again, but the main takeaway is... it depends on how detailed we want to look at the phenomenon of reflection and on what system we are looking at it. $\endgroup$ – CuriousOne Jul 21 '16 at 9:47
  • $\begingroup$ I'm not sure whether explanation in this site is correct circlon-theory.com/HTML/reflection.html $\endgroup$ – Vishnu JK Jul 21 '16 at 10:24
  • $\begingroup$ As you think about this problem, please keep in mind that these models are very crude, and are not good representations of the true nature of light. We all do the best we can to create mental pictures of things in physics, but light is a tough one. In particular, thinking of a photon as a particle that can bounce like a ball is a poor model. $\endgroup$ – garyp Jul 21 '16 at 10:56
  • 1
    $\begingroup$ The light coming out of a reflection is not the same as the light that went in to the reflection. There is a moment when the light is absorbed by the reflective material, just before its energy is then re-emitted by the material. $\endgroup$ – RBarryYoung Jul 21 '16 at 15:32

In this case, it would be useful to not consider light in its particle form as photons, but instead to consider it as a wave - see this wikipedia page. Then, the wave is simply reflected from the surface, without us having to consider the kinetics of any particle. The wave, in a vacuum, would continue to propagate at the speed of light, regardless of the surface it is reflected off of.

However, it is possible to carry on considering the light as photons. These photos are absorbed by the material, which then instantaneously emits a new photon with the same velocity as the incident photon. This doesn't involve the photon being reflected by the material, so avoids the case in which a photon would have a velocity of 0.

Thanks to TheGhostOfPerdition for pointing out that I had missed that second part.

  • 4
    $\begingroup$ You could perfectly explain this considering light in its particle form, A photon is not reflected, A new photon is returned instead. $\endgroup$ – Courage Jul 21 '16 at 8:43
  • 2
    $\begingroup$ The creation of the "new photon" is not instantaneous. The photon mixes with the atoms in the material and lingers for a short period of time before the new photon is emitted. $\endgroup$ – garyp Jul 21 '16 at 10:52
  • 1
    $\begingroup$ @JanDvorak I'm happy to make the answer community wiki for you to include that, but you've lost me! $\endgroup$ – Noah P Jul 21 '16 at 13:35
  • 1
    $\begingroup$ instantaneously that part puzzles me. My understanding is that the photon gives energy to the electron could of an atom, which releases a new photon when the energy decays due to this state being less stable. How can that be instantaneous? (and at the same time, the wave model makes it clear that it should be) $\endgroup$ – njzk2 Jul 21 '16 at 15:33
  • 1
    $\begingroup$ @njzk2 Even in the wave model it is not instantaneous. The incident wave excites a polarization wave in the material. The polarization wave is the source of the reflected wave. But the reflected wave will be phase shifted due to the interaction with the medium. Different media, i.e. different refractive indicies, will produce different phase shifts. In the wave picture, a phase shift is the same as a time delay. $\endgroup$ – garyp Jul 21 '16 at 18:32

Let me explain in purely classical terms (not the description of reality, but easy to imagine).

You realized that when a ball bounces off the wall, at a certain point, it has no momentum. However, it must still have all the energy of the movement (neglecting losses to environment) - where did that energy go?

The ball is formed of many discrete atoms, bound together by electromagnetic forces. As the ball starts hitting the wall, it compresses - the atoms are pushed closer together than they would without the pressure. As the back of the ball approaches the front of the ball, the kinetic energy is converted into that compression, and the momentum is transferred to the wall (and through it, the Earth, accelerating its rotation by a miniscule amount). At some point, the kinetic energy is all gone - into the compressed ball, the sound, slight heating of both the ball and the wall etc. But the ball is still compressed, and the atoms don't want to be so close together - so they push on each other, and on the atoms in the wall. The stored compression energy is converted back into kinetic energy, and momentum is "stolen back" from the wall (again, accelerating the Earth's rotation by a tiny amount). The ball picks up speed, and when all the compression is released, it travels at the same speed as when it hit (ignoring energy losses to the environment).

Photon isn't a composite particle. There's nothing to "compress" to store the kinetic energy. Compare how a rubber ball bounces to the way a billiard ball bounces. In the first case, a lot of the energy is stored in the rubber ball's compression. But a billiard ball is a lot harder, and almost no energy is stored in its compression - the time the ball spends in the "no momentum" stage is much shorter than for the rubber ball. You can consider a photon to be infinitely hard in this picture - it doesn't spend any time at all in the "compressed" phase - it just bounces right off, just like balls in a perfect newton's craddle.

However, I have to note again that this is not how light actually really works. In reality, the photon doesn't really bounce off at all - rather, it is absorbed, and after a bit of time, another photon is emitted in the right direction, as if the photon bounced off.

And the nice thing about physics is that as you go deeper, you'll see a lot of "well, that's not quite the case". All of those behaviours you were taught (and will be taught in the future) are in our models of reality, not reality itself. They are useful models for modelling certain scenarios, and useless for others. Need to calculate where a beam of light will hit if you shine it on a mirror? You can already do that, just with the simple Brewster's angle model. But reality is more complex than that - for example, if you grind off pieces of the mirror at precisely calculated places, you'll have a mirror that reflects light at a different angle (look up "diffraction mirror" if this sounds interesting :)).

To the best of our knowledge, even photons aren't really real. Right now, the best model we have of reality has no balls bouncing around, no photons sometimes behaving like little balls and sometimes like waves in a pool, there's just a quantum field that is neither a little ball nor a wave in a pool. But we only use that model when it's absolutely necessary - because it's very difficult to use. But we know that's closest to reality than anything else we have, because it explains (and predicts) things that no other theory does, and we already depend on the theory in real practical devices in everyday use (I assume you have a personal computer, since you're posting questions on StackExchange :P). Someday, a better theory might gain ground, and we'll be saying "Well, we thought light is really a surface phenomenon of an underlying electromagnetic quantum field, but it so happens that was a silly thing to think! Instead, bananas are really the fundamental thing in the universe, obviously." :)

  • 1
    $\begingroup$ 42 bananas, to be precise. $\endgroup$ – AgapwIesu Jul 21 '16 at 18:59
  • $\begingroup$ The "photon" or plane wave accelerates electrons in a conducting medium and therefore ceases to exist for that instant, however an accelerating electron radiates so generating new EM radiation or a "photon" at a slightly lower frequency if any momentum is exchanged to the medium or there are any I^2*R losses in the medium due to slight resistance in the medium. $\endgroup$ – barry May 2 at 23:33

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