Does the photon have a 4-velocity in a medium?

[PFfan01]

From classical electrodynamics textbooks, we know that the Fizeau experiment supports relativistic 4-velocity addition rule. But a recently-published paper says that the photon does not have a 4-velocity. See: "Self-consistent theory for a plane wave in a moving medium and light-momentum criterion", http://www.nrcresearchpress.com/doi/10.1139/cjp-2015-0167#.Vki8xGzovIU

I wonder who's right?

 

[PWiz]

But a recently-published paper says that the photon does not have a 4-velocity.

I thought this was a well known fact.

 

[PFfan01]

I thought this was a well known fact.

What do you mean? Do you mean the Fizeau experiment is not a support in relativistic 4-velocity addition rule?

 

[PWiz]

What do you mean? Do you mean the Fizeau experiment is not a support in relativistic 4-velocity addition rule?

No, I'm saying that the four-velocity is not defined for a photon.

 

[PFfan01]

There is a definition of four-velocity of light in the book by W. Pauli, Theory of relativity, (Pergamon Press, London, 1958), Eq. (14), p. 18, Sec. 6. Seems the definition is the same as that for a matter particle.

 

[PWiz]

There is a definition of four-velocity of light in the book by W. Pauli, Theory of relativity, (Pergamon Press, London, 1958), Eq. (14), p. 18, Sec. 6. Seems the definition is the same as that for a matter particle.

I don't have the book with me. Can you post the definition?

If the definition is the same as that for a massive particle, then it can't be right. Photons move on null lines and experience 0 proper time. Since four velocity is the proper time derivative of four position, it is not defined for a photon.

 

[Dale]

I agree with PWiz. Another way to think of it is that the four velocity is the four momentum divided by the mass, which is 0. Or that the four velocity is the unit tangent to the worldline and a null worldline can only have null tangents, not unit tangents.

 

[PFfan01]

I don't have the book with me. Can you post the definition?

If the definition is the same as that for a massive particle, then it can't be right. Photons move on null lines and experience 0 proper time. Since four velocity is the proper time derivative of four position, it is not defined for a photon.

If in free space, you are right. In a medium, the light speed is less than the vacuum light speed. In the book by Pauli, Fizeau running water experiment is used as a support in relativistic 4-velocity addition rule.

 

[PWiz]

In a medium, the light speed is less than the vacuum light speed.

But a photon always moves at $c$ regardless of the medium (it interacts with other particles in the medium and on a macroscopic scale you can say that the average speed of light reduces, but microscopically individual photons always move at the same speed). The photon still experiences 0 proper time, and you still cannot define its four velocity.

 

[Dale]

If in free space, you are right. In a medium, the light speed is less than the vacuum light speed. In the book by Pauli, Fizeau running water experiment is used as a support in relativistic 4-velocity addition rule.

I think that you probably want to ask about classical light waves rather than photons.

In a medium a plane wave will have a phase velocity which is less than c. You can definitely use the relativistic velocity addition formula on the phase velocity, so I assume that you could make a phase four velocity. Although I don't recall seeing anyone do that before.

 

[PWiz]

You can definitely use the relativistic velocity addition formula on the phase velocity, so I assume that you could make a phase four velocity.

But then there is no contradiction. It might be possible to define a four velocity if you deal with light classically in a medium, and still have an undefined four velocity for the photon treatment.

So addressing the OP, I guess both statements can be right. I would still like to see the four velocity definition for the classical treatment of light in a medium though.

 

[Dale]

But then there is no contradiction. It might be possible to define a four velocity if you deal with light classically in a medium, and still have an undefined four velocity for the photon treatment

Yes. I agree.

 

[PFfan01]

But a photon always moves at $c$ regardless of the medium (it interacts with other particles in the medium and on a macroscopic scale you can say that the average speed of light reduces, but microscopically individual photons always move at the same speed).

Very interesting argument, but could you please show any references for your argument? Thanks a lot.

PS: The paper by Leonhardt, Ulf (2006), "Momentum in an uncertain light", Nature 444 (7121): 823, doi:https://dx.doi.org/10.1038%2F444823a , says that the photon in a dielectric medium moves at the dielectric light speed, but the author did not tell why.

 

[PWiz]

Very interesting argument, but could you please show any references for your argument? Thanks a lot.

References for the 2nd postulate of relativity or for atomic spacing?

 

[PFfan01]

References for the 2nd postulate of relativity or for atomic spacing?

The references for your statement that "a photon always moves at c regardless of the medium".

PS: In my understanding, Einstein's second hypothesis is the constancy of light speed in free space.

 

[PWiz]

In my understanding, Einstein's second hypothesis is the constancy of light speed in free space.

So are you saying that there is no atomic spacing in a medium? Think about it. Between electron interactions in a medium, what does a photon move in? Reading post #9 again might help.

 

[PFfan01]

So are you saying that there is no atomic spacing in a medium? Think about it. Between electron interactions in a medium, what does a photon move in? Reading post #9 again might help.

1. I never said "there is no atomic spacing in a medium".
2. I am just asking you to give any references for your statement that "a photon always moves at c regardless of the medium".
3. In fact, it is enough for you to tell me whether your statement is your reasoning from Einstein's second hypothesis or there are any references to support it.

Even if your statement is your reasoning, I am not able to judge whether it is correct or not, because it is far beyond my knowledge.
Sorry.

 

[PWiz]

A photon is always moving through empty space (when a photon is moving in a medium it's actually moving through the atomic spaces, which is nothing but empty space) or interacting with other particles. The 2nd postulate says that light always moves at $c$ in empty space (as measured in an inertial frame).
I just restated two well known facts. From these two facts, it follows that a photon always moves at $c$. It's just that the interactions of the photon with other particles in a medium "delay" the photon, so the effective speed of light seems to reduce in any particular medium compared to a vacuum (but the photon moves at $c$ between the interactions). That's all I'm saying.
P.S. I'm not trying to be confrontational here.

 

[PFfan01]

A photon is always moving through empty space (when a photon is moving in a medium it's actually moving through the atomic spaces, which is nothing but empty space) or interacting with other particles. The 2nd postulate says that light always moves at $c$ in empty space (as measured in an inertial frame).
I just restated two well known facts. From these two facts, it follows that a photon always moves at $c$. It's just that the interactions of the photon with other particles in a medium "delay" the photon, so the effective speed of light seems to reduce in any particular medium compared to a vacuum. That's all I'm saying.
P.S. I'm not trying to be confrontational here.

So your statement that "a photon always moves at c regardless of the medium" is just your reasoning, without any references to support it. Right?

 

[PeterDonis]

when a photon is moving in a medium it's actually moving through the atomic spaces, which is nothing but empty space

Really? There are hard boundaries around the atoms that delimit them from "empty space"? And the photons never cross those boundaries? See further comments below.

I just restated two well known facts.

Not really. You restated a common model for photon propagation in a medium, but that's a lot different from "well-known fact".

In fact, although it's a common model, it's not actually correct. For example:

the interactions of the photon with other particles in a medium "delay" the photon, so the effective speed of light seems to reduce in any particular medium compared to a vacuum (but the photon moves at $c$ between the interactions).

The interactions you are talking about here are the absorption and emission of photons by atoms in the medium. These interactions do not "delay" one photon; they destroy one photon (when it's absorbed) and create a second photon (when it's emitted). (Note, btw, that the absorption and emission is actually done by electrons in the orbitals of the atom, which means that the photons do in fact have to cross the "boundary" of the atom--the electrons aren't all sitting on the boundary, they are in the interior.)

It is true that, in this somewhat more accurate model, the photons move at $c$ between interactions. However, the model is, as I just said, only somewhat more accurate. We don't actually measure the speed of the photon between interactions; we can't. And if we make our model more accurate still, by bringing in more quantum mechanical details, we will find that the concept of the "speed" of the photon between interactions isn't even well-defined; the quantum amplitudes will have contributions from off shell virtual photons.

The moral is to be very careful what you think of as a "well-known fact".

 

[PWiz]

Really? There are hard boundaries around the atoms that delimit them from "empty space"? And the photons never cross those boundaries? See further comments below.

I never said anything of this sort. When I say "atomic spacing", I'm referring to this:

Atomic spacing refers to the distance between the nuclei of atoms in a material.

In fact, although it's a common model, it's not actually correct.

I've made no allusions to the Bohr model (which I believe is what you think I'm talking about) where electrons are moving in fixed circular orbits (occupying an orbit based on their energy level) around the nucleus. I'm well aware that the exact size of an atom is ill-defined (we can always use bond / Waan der Waal radii to obtain a working value, but I think that's about it).

The interactions you are talking about here are the absorption and emission of photons by atoms in the medium.

Somewhat. IIRC, ZapperZ had an FAQ in which he states that in any medium the interactions between atoms/ions of the medium result in some sort of broadening of energy levels and a "collective" behavior, so I don't think we have the simple case of photons being absorbed by the electrons of individual atoms and then being re-emitted after a slight delay. I can't find that FAQ, but here's an old thread at PF which quotes it: https://www.physicsforums.com/threads/faq-do-photons-move-slower-in-a-solid-medium.243463/ .

These interactions do not "delay" one photon; they destroy one photon (when it's absorbed) and create a second photon (when it's emitted).

This is what I mean when I say "A photon is always moving through empty space or interacting with other particles." When I say "interacting with other particles", I mean the destruction of a photon and the creation of another (after a brief interval of time). Put another way, my statement reads "if a photon is not being created or destroyed, it's propagating through space at an invariant speed $c$ m/s ." (Btw, I know that a photon moves at $c$ right after creation and that there is no "acceleration period" so as to say) I can't see what's wrong with this statement.

the electrons aren't all sitting on the boundary

Yes, I know they aren't, because there is a non-zero probability of finding the electron anywhere in the space around the nucleus. (A probability given by the square of the wavefunction of the electron.)

we will find that the concept of the "speed" of the photon between interactions isn't even well-defined

I wanted to minimize quantum mechanical references here in the relativity subforum, but I guess I should have used the term expectation velocity, right?

It is true that, in this somewhat more accurate model, the photons move at cc between interactions.

But we work with a model until we find a more accurate one which can take its place. I don't think we have that kind of replacement yet.

 

[DrDu]

Photons moving in media are quasi-particles, like electrons in semiconductors. So one could make the statement more precise asking about the 4-velocity of quasi-photons.

 

[PFfan01]

... In a medium a plane wave will have a phase velocity which is less than c. You can definitely use the relativistic velocity addition formula on the phase velocity, so I assume that you could make a phase four velocity. Although I don't recall seeing anyone do that before.

Probably there is no phase four velocity, otherwise it would contradict the wave four vector, according to the recently-published paper. http://www.nrcresearchpress.com/doi/10.1139/cjp-2015-0167#.Vki8xGzovIU

 

[Dale]

Probably there is no phase four velocity, otherwise it would contradict the wave four vector, according to the recently-published paper. http://www.nrcresearchpress.com/doi/10.1139/cjp-2015-0167#.Vki8xGzovIU

That is certainly possible. As I mentioned, I have never seen anyone do it before.

I will read the link and see what they say on the topic.

 

[PeterDonis]

When I say "atomic spacing", I'm referring to this:

Ah, ok, that helps. The nuclei don't really have hard boundaries either, but in the regime under discussion they are certainly a lot "harder" than the boundaries of the atom as a whole.

in any medium the interactions between atoms/ions of the medium result in some sort of broadening of energy levels and a "collective" behavior, so I don't think we have the simple case of photons being absorbed by the electrons of individual atoms and then being re-emitted after a slight delay.

Yes, that's why I said the "slight delay" model was only somewhat more accurate. The other things you mention would be part of a more complete quantum mechanical treatment.

Put another way, my statement reads "if a photon is not being created or destroyed, it's propagating through space at an invariant speed $c$ m/s ."

Yes, this is fine at the "somewhat more accurate" level of modeling (the "slight delay" model). But it's not at the more accurate quantum level of modeling; at that level, as I said before, the photons don't have a definite speed, since the amplitudes have off-shell contributions.

I guess I should have used the term expectation velocity, right?

There are issues even with that for photons (Google "Newton-Wigner localization", for example--it works for massive particles but not for massless ones). But at the "somewhat more accurate" level, just saying the photons move at $c$ is fine. It's when we try to include more accurate quantum effects that issues arise.

we work with a model until we find a more accurate one which can take its place. I don't think we have that kind of replacement yet.

Yes, we do; quantum electrodynamics is a very well-developed theory, and it covers a lot of things that aren't included in the "somewhat more accurate" model.

 

[PWiz]

(Google "Newton-Wigner localization", for example--it works for massive particles but not for massless ones)

Okay, I'll look into that, thanks.

Yes, we do; quantum electrodynamics is a very well-developed theory, and it covers a lot of things that aren't included in the "somewhat more accurate" model.

But do we really need to use QED to answer the OP's question? If yes, then I guess the four-velocity will still be undefined as the ordinary velocity (or the expectation velocity for that matter) is not rigorously defined for the photon.

 

[PeterDonis]

do we really need to use QED to answer the OP's question?

I don't think so; the Fizeau experiment should be analyzable using classical electrodynamics, since no quantum effects come into play.

I guess the four-velocity will still be undefined as the ordinary velocity (or the expectation velocity for that matter) is not rigorously defined for the photon.

Classically, it depends on what model we want to use. If we use the geometric optics approximation, which is the only model in which the term "photon" is really appropriate classically, then the photon has a well-defined 4-momentum, and a well-defined ordinary velocity in any inertial frame (obtained by looking at the spatial components of the 4-momentum in that frame, divided by the time component). However, the photon does not have a well-defined "4-velocity", because its 4-momentum is null, so there is no such thing as a unit vector tangent to the photon's worldline, which is how "4-velocity" is defined.

If, OTOH, we use EM wave theory, then there is no such thing as a "photon", and we aren't using a 4-momentum vector to describe the field, we are using an antisymmetric 4-tensor. So "4-velocity" doesn't even come into play. (We could assign an ordinary velocity to wave crests in a particular inertial frame, but doing that wouldn't play any part in the analysis.)

 

[PeterDonis]

the photon has a well-defined 4-momentum, and a well-defined ordinary velocity in any inertial frame (obtained by looking at the spatial components of the 4-momentum in that frame, divided by the time component). However, the photon does not have a well-defined "4-velocity", because its 4-momentum is null, so there is no such thing as a unit vector tangent to the photon's worldline, which is how "4-velocity" is defined.

Just to put one caveat on this, it looks like the paper referenced in the OP might be treating the propagation of light in a medium by assigning a timelike "4-velocity" to the light instead of a null 4-momentum. This would not really be a "photon" model in the usual sense. The paper is paywalled so I can't read anything besides the abstract (and the abstract has some statements that make me a bit skeptical), so I can't tell for sure that this is what it's doing, or if so, what implications it has.

 

[PWiz]

we are using an antisymmetric 4-tensor

The electromagnetic tensor?

 

[PeterDonis]

The electromagnetic tensor?

Yes.