Can photons be accelated or do they always travel at one speed (speed of light)?

I did a search for optical accelerator but it's not clear to me exactly what is being discussed. If so, has the been an invention to do this?

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 Photons always travel at the speed of light and there is nothing that can change that. However, the speed of light differs in different materials. A as a result, a photon cannot be accelerated.
 Recognitions: Science Advisor While the speed of light is constant, the velocity of a given photon most certainly isn't. Mirrors for example technically act as photon accelerators by changing the direction of propagation of photons. I would venture that a device called an 'optical accelerator' would in fact be using light to accelerate something else. Claude.

Can photons be accelated or do they always travel at one speed (speed of light)?

Maybe you took the term Optical Accelerator too literallly ?

 For photons:$a=\infty$
 What if a photon is traveling from one medium to another, both with different indices of refractions (like from air to water)? The velocity of the photon would change upon entering the second medium, would that not be considered an acceleration? Or am I missing something..

 Quote by pete5383 What if a photon is traveling from one medium to another, both with different indices of refractions (like from air to water)? The velocity of the photon would change upon entering the second medium, would that not be considered an acceleration? Or am I missing something..
In that case (air to water) it would decelerate, not accelerate since water is denser than air. As I just found out photons have a maximum speed. Think of light traveling from the sun to earth. It takes around 8 minutes for it to get here even though it is moving through a vacuum, it has a maximum speed: 299,792,458 m/s

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 Quote by pete5383 What if a photon is traveling from one medium to another, both with different indices of refractions (like from air to water)? The velocity of the photon would change upon entering the second medium, would that not be considered an acceleration? Or am I missing something..
Perhaps the following post in the "Physics FAQ" which is stickied at the top of this forum, will help:

Do Photons Move Slower in a Solid Medium?

 Recognitions: Gold Member Umm....were a couple of responses deleted here? I'm confused...

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 Quote by ranger Umm....were a couple of responses deleted here? I'm confused...
Yes. A post that creates more confusion than it answers was deleted, including the responses to that post.

Zz.

 Quote by Claude Bile While the speed of light is constant, the velocity of a given photon most certainly isn't. Mirrors for example technically act as photon accelerators by changing the direction of propagation of photons.
Is this true? I thought photons are absorbed and new photons re-emitted, from a mirror.

 While the speed of light is constant, the velocity of a given photon most certainly isn't.
It has been my understanding that photons (I don't like that term after Aspect) always travel at the speed of light, and that the reason light signals travel more slowly inside of materials is because of an absorption-emission processes.

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 Quote by lightarrow Is this true? I thought photons are absorbed and new photons re-emitted, from a mirror.
Yes indeed. A photon's interaction with matter can be described in terms of absorption and emission.

Photons do not slow down. As per my understanding, it is the group velocity that changes as light moves between different mediums.

 I wasn't implying that the constant speed of light changes depending on the reference frame. When I was creating the post, I accidently deleted half of it and the beginning of the last sentence ended up being something about reference frames, when it was actually supposed to be the end of a paragraph which pertained to an aside comment that I was trying to make. I apologize for the confusion, it was a matter of misclicking buttons and not a misrepresentation of facts. My fault, yo.

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 Quote by lightarrow Is this true? I thought photons are absorbed and new photons re-emitted, from a mirror.
I think that some confusion about "the speed of photons" comes about because we forget how photons are defined, and start thinking of it as little "light bullets", or "wave packets" of classical EM radiation.

In free space, photons are defined as the different eigenstates of the energy and momentum operators of the EM field. It turns out that they correspond to harmonic oscillator solutions for each and every mode of the classical EM field, which, in free space, comes down to plane harmonic waves. So to every classical plane harmonic wave corresponds a quantum-mechanical harmonic oscillator (you know, with energy levels E_n = (n + 1/2) hbar omega), and we call the steps between these energy levels: photons.

If there is a dielectric, we can treat the situation in two different ways. We can have a kind of semiclassical approach, where we look first at the *classical* modes of this setup, and then assign again quantum mechanical oscillators to each of these classical modes, and call them "photons". Or we can treat this dielectric as a quantum-mechanical system which *couples* to the free EM field.

In the first case, you understand that the "photons" of this semiclassical system are ENTIRELY DIFFERENT states than the photons of the free EM field. They are eigenstates of a totally different problem, and hence the "free EM photons" have not much to do with the "semiclassical dielectric photons".
However, because the dielectric interaction has been taken into account (classically), there is no "interaction of these photons with the dielectric": it is already included in the classical field solutions that were to be quantized.

In the second case, the interaction terms between the free EM field and the quantum mechanical dielectric system, which induces transitions between the "free photon states", because of the couplings (perturbations) introduced by the dielectric (seen as a quantum system that introduces interaction terms in the EM hamiltonian).

 Quote by vanesch I think that some confusion about "the speed of photons" comes about because we forget how photons are defined, and start thinking of it as little "light bullets", or "wave packets" of classical EM radiation. In free space, photons are defined as the different eigenstates of the energy and momentum operators of the EM field. It turns out that they correspond to harmonic oscillator solutions for each and every mode of the classical EM field, which, in free space, comes down to plane harmonic waves. So to every classical plane harmonic wave corresponds a quantum-mechanical harmonic oscillator (you know, with energy levels E_n = (n + 1/2) hbar omega), and we call the steps between these energy levels: photons. If there is a dielectric, we can treat the situation in two different ways. We can have a kind of semiclassical approach, where we look first at the *classical* modes of this setup, and then assign again quantum mechanical oscillators to each of these classical modes, and call them "photons". Or we can treat this dielectric as a quantum-mechanical system which *couples* to the free EM field. In the first case, you understand that the "photons" of this semiclassical system are ENTIRELY DIFFERENT states than the photons of the free EM field. They are eigenstates of a totally different problem, and hence the "free EM photons" have not much to do with the "semiclassical dielectric photons". However, because the dielectric interaction has been taken into account (classically), there is no "interaction of these photons with the dielectric": it is already included in the classical field solutions that were to be quantized. In the second case, the interaction terms between the free EM field and the quantum mechanical dielectric system, which induces transitions between the "free photon states", because of the couplings (perturbations) introduced by the dielectric (seen as a quantum system that introduces interaction terms in the EM hamiltonian).
Ok. But I haven't completely understood what happens.
Let's say a single photon is involved in the process. How could you explain it in simple terms (if it's possible)?

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 Quote by lightarrow Ok. But I haven't completely understood what happens. Let's say a single photon is involved in the process. How could you explain it in simple terms (if it's possible)?
You mean, with the mirror ?

Well, first solve for the classical modes, WITH the boundary conditions such as the mirror and all that. A single mode (a harmonic solution to the maxwell equations) will probably be some kind of TEM wave, including the reflexion on the mirror, because that's now included in the boundary conditions of the classical problem. The harmonic oscillator corresponding to that mode will then be quantized, and the energy steps of this oscillator will be "photons" of this system. It is part of the "propagation of this photon" to be reflected on that mirror. You cannot analyze this reflection in more detail, because it was introduced as a boundary condition, and not as a physical interaction.
(a bit in the same way as you cannot analyze the specific nature of the interaction of a constraint in Lagrangian mechanics, if you have imposed a holonomic constraint on the generalized coordinates: you can only do that if you introduce an explicit binding potential and don't introduce a holonomic constraint).

When you quantize this classical system (EM field + perfect mirror as boundary condition), a single photon remains a single photon "as it reflects off the mirror", in the same way as, in free space, a single photon remains itself when it propagates through space, because the "mirror" became kind of part of its "free propagation".

However, this photon (remember that photons are nothing else but steps between stationary states) is a different kind of photon from the photon in the free field solution (without mirror). We can both bring into agreement, by introducing an interaction term in the free field solution, which represents the overall action of the mirror on the photon. It will (in its most rudimentary form) be simply a destruction and a creation operator, which destroys the "incoming photon" and creates the "reflected photon" in the free EM field photon basis.