Light speed and reflection

  • #1

Main Question or Discussion Point

i have a somewhat elusive question (for me) about light. when you drop a bouncy ball, after it hits the ground, right before it bounces back up, its velocity is zero. correct? it becomes troubling when the ball is replaced with light. when light is reflected straight back to its source, it cannot stop or even slow down because of the absolute speed of light. how does it accomplish this? i think ive read its massless but that doesn't seem to help me.
 

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  • #2
chroot
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The photon coming back at you from the mirror is not actually the same photon as the one that hit the mirror. The photons are not just "turned around," they are actually absorbed by atoms in the mirror, then re-emitted. The directions involved are governed by quantum electrodynamics, which can be used to derive, among aother things, the angle of incidence = angle of reflection feature of reflection.

- Warren
 
  • #3
so mirror atoms contain photons and the energy from the photons hitting them release them? does everything contain photons? (should i post this in the classical physics section?)
 
  • #4
chroot
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No, mirrors contain atoms. All atoms have the ability to absorb and create photons. Indeed, all charged objects can absorb and create photons.

- Warren
 
  • #5
so when the light hits the mirror, more photons are ejected simultaneously so that the speed of light is never broken?
 
  • #6
chroot
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Photons are first absorbed by the atoms, then new photons are emitted by the same atoms.

- Warren
 
  • #7
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Question, i remember in phys 120 one of the bonus questions was to prove, using energy, that a free electron cannot absorb (or maybe it was emit, i can't remember) a photon. So what happens when a photon collides with an electron in that case??? Is there merely an elastic collision? and in that case is the acceleration of the photon infinite? (wrt velocity rather than speed?)
 
  • #8
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That would be something like to compton scattering, wouldn't it?
 
  • #9
ZapperZ
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Here's a hint: The most reflective surface tends to be a metal (electrical conductor) rather than an insulator.

What is involved here is the conduction electrons within the metal. Under the first approximation, these electrons can be considered to be a free electron gas. When photons hit the surface, these electrons absorb and oscillate (within a certain freq. range) at the same frequency as the E-field vector of the photons. An oscillating bunch of electrons can generate EM radiation. Thus, these conduction electrons retransmit the same frequency of light.

The reason why the reflection laws are there is because of conservation of momentum parallel to the surface. The photon's initial parallel momentum is retransmitted back by the electrons because the conduction electrons can make the same "crystal momentum" transition. The perpendicular momentum, while not exactly conserved, is "reflected" back, and this requires the presence of the metal's heavier ions to absorb the recoil momentum.

[This is where it gets a bit confusing, because a "pure" free electron gas will not work since, by definition, the electrons are not coupled to anything else. The requirement for the ions to be able to absorb the recoil momentum means there must be at least a very weak coupling or interaction between the free electron gas and the lattice ions].

Also note that if you study in detail the response of the conduction electron to the photon's E-field, you'll see that it will oscillate and thus radiate pi out of phase with the incoming photons. And that's what you also get upon reflection, that the reflected light has a phase that has been shifted by pi (180 degrees).

Zz.
 
  • #10
russ_watters
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relativelyslow said:
so when the light hits the mirror, more photons are ejected simultaneously so that the speed of light is never broken?
Photons are first absorbed by the atoms, then new photons are emitted by the same atoms.
...and there is an associated time delay.
 
  • #11
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Ah yess, the 180' phase shift. I remember about that, but how come that phase shift isn't random is really thick mediums.

Also, was what i stated earlier possible?
 
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  • #12
ZapperZ
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Goalie_Ca said:
Ah yess, the 180' phase shift. I remember about that, but how come that phase shift isn't random is really thick mediums.

Also, was what i stated earlier possible?
The "thickness" has nothing to do with the phase shift. As long as the metal is thicker than the skin depth, the phase shift is independent of the thickness.

I'm not sure which one you are refering to as "state earlier". You were talking about "photon acceleration", which isn't possible since photons do not change speed. You may want to read up on things like compton scattering if you want to know more about photon interaction with free particles such as electrons.

Zz.
 
  • #13
Here's a hint: The most reflective surface tends to be a metal (electrical conductor) rather than an insulator.
What about water? Its surface can be pretty good mirror. Thus distilled water has no conductivity, i.e. free electrons.
 
  • #14
ZapperZ
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Michael F. Dmitriyev said:
What about water? Its surface can be pretty good mirror. Thus distilled water has no conductivity, i.e. free electrons.
Hint: when one studies optics, there is such a thing as "Brewster angle" when light moves from one medium to another that have different index of refraction. Although not the case here, there is also such a thing as a "total internal reflection". You should also check if the "reflectivity" off a water surface is of the same order as reflectivity off a metallic surface.

Note that I said that the MOST reflective surface TENDS to be a metal. I didn't say ALL reflective surface are metals. When high reflectivity is of importance (such as in a reflecting telescope or other optics applications), thin film metallic surfaces are always used.

Zz.
 
  • #15
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Acceleration of a photon

Wouldn't a photon have to acceleration up to the speed of light when being emitted?

Nate
 
  • #16
chroot
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Photons do not accelerate; they always travel at the speed of light.

- Warren
 
  • #17
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The photon coming back at you from the mirror is not actually the same photon as the one that hit the mirror. The photons are not just "turned around," they are actually absorbed by atoms in the mirror, then re-emitted. The directions involved are governed by quantum electrodynamics, which can be used to derive, among aother things, the angle of incidence = angle of reflection feature of reflection.
Arguably, in quantum electrodynamics, the photon you shot at the mirror is not the same one that hit it.

In the end, photons are not bouncy balls.
 
  • #18
chroot
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FZ+,

It would seem to me that it cannot be the same photon -- since making a photon do an about face would require it to accelerate in a way that would cause it to slow down, and even be at rest at some point during the "transition." Photons can't be steered by electric or magnetic fields either, which are the only "tools" the atomic electrons could "use" to make a photon turn around. It seems to me that it simply must be absorbed and a new photon emitted.

- Warren
 
  • #19
FZ+
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You misunderstand me - of course I agree. It can't be the same photon. I was just nudging in the way of virtual particles, and suchlike. My understanding of QED is kinda foggy. IIRC, it's about the probabilities of events in certain situations. So I wonder, if you fire a photon at a detector, and subsequently detect a photon there, is it the 'same' photon you fired, or just a different one with the same properties? How can you tell? And does it matter?
 
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  • #20
chroot
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Just making sure we're on the same page. :smile:

- Warren
 
  • #21
ZapperZ
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FZ+ said:
So I wonder, if you fire a photon at a detector, and subsequently detect a photon there, is it the 'same' photon you fired, or just a different one with the same properties? How can you tell? And does it matter?
You need to remember that in a photo detector, the photon is converted either to an electrical signal, or a cascade of electrons as in a photomultiplier. So if you are asking of the electrical signal or impulses that you receive is due to "that" photon, the answer is yes. But this only goes as far as indicating that there is a "cause and effect", meaning that if no photon is present, then you get no signal. However, the reverse isn't necessarily true always, i.e. if you get no signal signal, it doesn't mean a photon isn't there. This is because we do not have a photodetector that is 100% efficient (or else all those EPR-type experiments will be devoid of all detection loopholes).

Zz.
 

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