This is from a discussion on (gasp) another message board that I was hoping to get some expert SD insight on…
When light is reflected off a mirror (or whatever), is it slowed & stopped for an instant or is the reflection instantaneous? (i.e., is there an instant of zero velocity when the velocity vector is reversed?)
Also, how does the reflection actually work? Do the quanta of photons bounce or are they absorbed & re-emitted?
If you look at it as a wave phenomenon then a beam of light falling on a surface is, in general, split into two waves – a transmitted one and a reflected one. The wave “conversion” is simultaneous, and a every instant in time (to within the uncertainty relationship, anyway) energy, momentum, etc. are conserved. In the real world waves extend through space, so you’ll have a part where you still have an incoming wavefront, while elsewhere you will have a reflected and transmitted beam.
On a photon level yo can either be reflected or transmitted, but not both. In either case your wave vector is going to be changing, and this has to happen by interaction with the medium. My guess is that, like any interaction, it must have a characteristic time, during which one could argue that the photon had “stopped”
This sort of thing wasn’t covered in my own optics curses (and I’ve had a LOT of them). I recall one professor making the argument that you get zero reflection along one polarization at Brewster’s angle because you are looking directly along the direction of the oscillating dipole, so of course there is no resulting wave (which seems to imply a sort of absorption). Another professor said that this explanation was hogwash. Certainly photons are not absorbed in the usual sense of the word – the substances usually have negligible absorption bands at the wavelength of reflected light. But that doesn’t mean that you can’t consider the photon as briefly absorbed then re-emitted.
I’m with Cal on this. The photons definitely interact with the reflecting medium, and that interaction can be characterized either as a collision or a quick absorption and re-emission. After all, the reason that mirror materials don’t have absorption bands at visible frequencies is that the resulting orbits for the excited electrons aren’t even a little bit stable, so while it’s not possible for the photon to get captured, there’s no reason that a quick interaction resembling a capture-emission with no intermediate stable state couldn’t occur.
A nice feature of the quick absorption-reemission model is that it doesn’t require the photon to change speed, since for the instant it would have been stopped, it instead disappears entirely. To reverse the velocity vector through a collision without passing through zero would require an infinite impulse (although this is stretching Newtonian mechanics way beyond its domain).
A good analogy would be to tie a rope to a hook in a solid wall. Holding the rope taught, briefly and quickly move your end of the rope up then back down, and stop. You should see a hump in the rope travel towards the wall, then reflect back towards you. For a time when the hump is reflecting, you have both a forward and a backward travelling wave.
Does the hump “stop” briefly? I guess it depends on your definition of “stop,” but I would say no.
What if you have two rope humps, or two photons, meeting head-on? Do they go through each other, or do they reflect off each other? Is one description more or less valid than the other? Not really. To me, it’s just semantics.
As far as how the reflection of light really works, it’s reflecting off the “sea” of electrons in the metal. Within the metal, the electrons aren’t bound to individual electrons. When they are illuminated with photons, they redistribute themselves to cancel out the electric field at the surface, which prevents the photons from entering the metal or being absorbed. You could also argue that the electrons absorb the incident photons by being accelerated, and immediately re-radiate the energy, but this seems more compicated, and doesn’t really buy you anything.
I would argue that the absorption and reemission model does gain you something – if only because the original question was couched in quantum terms. The wave reflection you’re talking about is a perfectly valid model for E-M waves, but if you’re going to regard light in particle form, you need a particular (heh) explanation for the phenomenon, like collision or absorption. Since light is not capable of a continuum of speeds, collision is difficult to conceive, so the absorption-reemission model is appropriate.
Slightly off topic: assume we’re using your classical wave reflection model, and we send a beam of light at a highly (near 100%) reflective interface, normal to the surface. Is there any mechanism for the wave to transfer its forward momentum to the mirror, since all the fields are perpendicular to the surface?
PaulT:
Look up the classical theories of radiation pressure. Before they had the idea of discrete photons carrying momentum to a surface they needed way to explain how light could exert pressure on a surface. (That light COULD exert pressure on a surface was provable using Thermodynamics, of all things.The Crookes radiometer was supposed to demonstrate radiation pressure, but we all know that his explanation was wrong. Radiation pressure WAS successfully demonstrated with what was essentially a higher-tech Crookes radiometer by the 1930s.)
So how does classical radiation pressure work? The EM wave of the light beam sets up eddy currents in the medium. These interact with the magnetic field associated with the wave, causing a force on the mirror or dielectric. You can find an explanation of this in George Bekefi’s book on EM Waves. The original theory is by Peter DeBye.
I’d just call it “interaction” and leave it at that. Even if you’re looking at a photon being absorbed by a single atom, can you really talk about what’s going on during the interaction? I never saw this in quantum mechanics classes, and I don’t think is such a description. Just before and after. (But I’d love to see a description, if there is one!)
The momentum of an electromagnetic field is proportional to E cross B, which will be normal to the surface. The force due to a magnetic field on electrons of charge q is proportional to q * V cross B where V is the velocity. The elctrons are moving in the, say, X direction due to acceleration from the electric field, and B is in the y direction, so V cross B is is the Z direction, normal to the surface.
I dig. Hence the eddy currents Cal mentioned.
[offtopic]
Reminds me of something I read on one of those crackpot ‘free energy’ sites (I think it was josephnewman.com) about how he was the first person EVER to discover that copper and even aluminum are attracted by permanent magnets! Want to see it? Just put a magnet on top of a stack of aluinum plates, and pull it away quickly – and the top plate will be pulled away from the stack!
(pause for awed audience response)
But lest you think he was mistaking Lenz’s Law for ferromagnetism, he has a videotaped demonstration where he shows that magnets also REPEL hanging pieces of aluminum and copper, when pushed toward them – these metals are both attracted AND repelled by magnets! Truly the man is a genius…
[/offtopic]
Cal, in your first post, you referred to your “optics curses”. Freudian slip?
thanks for the input…at the very least, I have some topics I can chase further in my physics textbooks
Optics curses
I wouldn’t mind if it was a Freudian slip, but it’s really a slip of the finger. I find that I often don’t hit a key quite hard enough to make it register. If you look at my past posts you’ll se a lot of this. Then I either don’t preview, or I miss it on the preview.
Don’t give me any more of your Lippmann
Optics Curses