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

## Main Question or Discussion Point

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.

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. I would venture that a device called an 'optical accelerator' would in fact be using light to accelerate something else.

Claude.

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..

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

jtbell
Mentor
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:

https://www.physicsforums.com/showpost.php?p=899393&postcount=4 [Broken]

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

ZapperZ
Staff Emeritus
2018 Award
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.

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.

ranger
Gold Member
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.

vanesch
Staff Emeritus
Gold Member
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).

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)?

vanesch
Staff Emeritus
Gold Member
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.

Claude Bile
Is this true? I thought photons are absorbed and new photons re-emitted, from a mirror.
Reflection is not an absorption/emission process, nor is transmission through a transparent medium.

vanesch
Staff Emeritus
Gold Member
Reflection is not an absorption/emission process, nor is transmission through a transparent medium.
It really depends on your description. I think you are referring to incoherent absorption and re-emission. That's true, it is not an incoherent absorption and re-emission. However, the process can be modeled by a destruction operator of the incoming beam and a creation operator in the outgoing beam, when you consider the beams as those of the free EM field (without mirror).

Pls anyone make it clear that is there any difference between speed of light and photon if yes, what?
Also explain the difference between propogation of light in vaccum and in any other medium with photon in the pricture and also relate the dual nature of wave with it.

I would say 'light' = photons, therefore they are one and the same and yes they travel at the same speed, the speed of light/photons.

Light in a vacuum is the one or stream of photons of light, they travel until they strike some mass, they are absorbed into that mass. If that mass is made of certain stuff, when the light is absorbed, at almost the same time another photon of light is emitted.

But that takes a finite amout of time, ie, photon hits electron, electron goes to a higher state, then the electron falls back to a lower energy state and emits a photon.

That sequence of steps takes more time that if the photon was not absorbed and was free to travel in its own right.. in a vacuum.

If you could see the light travelling between the electrons in a solid, they would travel at the speed of light in a vacuum, but for a short distance until they interact with another electron and raise it to a higher energy state, causing that electron to emit a photon.

The dual nature of light I do not think has much or anything to do with its speed of travel, because mass, atoms are particles with mass, that are not going at the speed of light also exibit a dual nature just as light does.

To explain the dual nature of light and matter, well hopefully someone smarter than I will take that one up. :)

ZapperZ
Staff Emeritus
2018 Award
We need to be careful here.

In many cases, when people talk about "speed of light" as applied to some measurement, it is typically the speed of the group velocity of light, as opposed to either the phase velocity or even photon velocity.

Zz.

Thanks you all

Light in a vacuum is the one or stream of photons of light, they travel until they strike some mass, they are absorbed into that mass. If that mass is made of certain stuff, when the light is absorbed, at almost the same time another photon of light is emitted.

How does the new photon "know" in which direction to go in order to be properly reflected (or, sent) in an optical instument?

OF

"How does the new photon "know" in which direction to go in order to be properly reflected (or, sent) in an optical instument"

Im not sure of the offical or correct theory of light transmission through matter, or reflection.

But I can imagine, (in my world), that what happens when a light photon is absorbed by 'stuff' is that the photon strikes an electron, that is in orbit arount the nuclious.

When the photon strikes the electron that electron if it gains enough energy will move into a higher energy orbit, and in a finite time, will drop back down to a lower orbit and in that process release a photon.

If that electron has enough time to do half an orbit before it drop back to the lower orbit you can imagin that the photon will travel away from the atom at the opposite side than that it arrived from.

If the electron has enough time to do a full orbit before it drops down and releases a photon then the photon will be released in the direction that it arrived.

So mabey the reflective and transmissive index of a material is how fast the electronics orbit the atoms, and how long in time those electronics take to fall back to a lower orbit and release a photon.

Half an orbit and the photon will emerge from the other side of the atom, a full orbit will result in the photon being created in the direction of the excitation photon.

Its only a guess, and probably wrong. But it works for me :)