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.

 

[PFfan01]

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.

The paper referenced in the OP is also cited in wiki, saying:
It is generally argued that Maxwell equations are manifestly Lorentz covariant while the EM stress-energy tensor follows from the Maxwell equations; thus the EM momentum defined from the EM tensor certainly respects the principle of relativity. However a recent study indicates that “such an argument is based on an incomplete understanding of the relativity principle”, and states that the EM stress-energy tensor is not sufficient to define EM momentum correctly. https://en.wikipedia.org/wiki/Abraham–Minkowski_controversy

 

[PeterDonis]

The paper referenced in the OP is also cited in wiki

And the words "reactionless drive" at the top of that wiki page are an indication that what is being discussed there is not mainstream science. So it's off topic 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 $c$.

A photon always moves at c?

From Einstein’s special relativity on down, the invariance of the speed of light in free space has been a central tenet of physics. Now, in a clever set of experiments, scientists in the United Kingdom have demonstrated that, in certain conditions, individual photons in free space can be slowed down to speeds measurably below the supposedly invariant light speed (Science, doi: 10.1126/science.aaa3035).

Spatially structured photons that travel in free space slower than the speed of light

Published Online January 22 2015
Science 20 February 2015:
Vol. 347 no. 6224 pp. 857-860
DOI: 10.1126/science.aaa3035

I think Science must have reflected the mainstream science.
PS: I don’t understand what the “spatially structured” means. I never see any reports on structures within a photon.

 

[PWiz]

@physicsforum01 Theoretically, a photon has an ill-defined 4-velocity, as PeterDonis has reaffirmed in post #27. I don't know what the experimental setup of those scientists was as I can't read beyond the abstract, so I will refrain from commenting on that.

I never see any reports on structures within a photon.

Yes, a photon has no internal structure (it's not even a particle in the ordinary sense). I don't they're talking about that here though.

Btw, it would be a good idea to not to type your entire post in bold style with a large font size. The post will then look a bit neater.

 

[Ibix]

This has been discussed here before. You can take a collimated light beam, pass it through a lenses to expand the beam and then condense it back down again. Light takes longer to run through this system than just going straight through the same length of free space - the edges of the beam move on a diagonal during the expansion and condensation of the beam, and the center of the beam moves through a long length of glass, rather than the whole lot traveling in a straight line straight down the middle in free space.

It turns out that if you use diffractive optics instead of lenses, you can do the same thing with a single photon, and it takes longer to traverse with the optics in place than without. So a spherically expanding photon travels more slowly than a plane wave photon.

I'm simplifying somewhat, but the paper notes that (at least for the cases they talk about) the geometric optics approximation I've laid out is pretty good. I've no idea whether one can assign a four velocity to a photon in such a state in a coherent manner.

 

[PWiz]

You can take a collimated light beam, pass it through a lenses to expand the beam and then condense it back down again.

But is light really only moving through empty space in this setup? Won't the material the lenses are made of "delay" the photon in a manner previously discussed in this thread when the photon passes through them?

 

[Ibix]

But is light really only moving through empty space in this setup? Won't the material the lenses are made of "delay" the photon in a manner previously discussed in this thread when the photon passes through them?

With lenses, yes. But note that different parts of the wave-front experience different thicknesses of glass and different path lengths in free space (summing to the same optical path length). With diffractive optics (i.e. diffraction gratings), I don't think the answer is quite so obvious, since these can be plane structures. Although there's still some interaction with matter.

As I recall, the above linked paper by Giovanni et al describes a simpler setup than the one I described (fiber, free space, difraction grating, free space, fiber, from memory) and concludes that their experiment shows that the group velocity in free space is less than c. They provide the geometric argument ("some bits of the photon travel on a diagonal path") as a visualisation of why that's the case, and maths to show that the description matches their results. I suspect that whether the photon is "actually spreading out" or "actually has a lower group velocity when prepared this way" is a matter of interpretation - although I'm happy to be corrected on that.

 

[PFfan01]

As I recall, the above linked paper by Giovanni et al describes a simpler setup than the one I described (fiber, free space, difraction grating, free space, fiber, from memory) and concludes that their experiment shows that the group velocity in free space is less than c. They provide the geometric argument ("some bits of the photon travel on a diagonal path") as a visualisation of why that's the case, and maths to show that the description matches their results. I suspect that whether the photon is "actually spreading out" or "actually has a lower group velocity when prepared this way" is a matter of interpretation - although I'm happy to be corrected on that.

Constancy of photon speed in free space. According to the principle of relativity, Einstein light-quantum hypothesis, momentum-energy conservation law, and Maxwell equations are equally valid in all inertial frames. Thus as the carriers of light energy and momentum, any photons in free space keep moving uniformly after they leave a source observed in any inertial frames. On the other hand, observed far away from the source (especially at the infinity, which was used as an assumption to derive Doppler effect in the 1905 paper by Einstein), the light wave behaves as a (local) plane wave, while the photons for a plane wave move at the light speed in all inertial frames due to the invariance of Maxwell equations. From this it follows that the photons in free space move at the light speed in all directions independently of the motion of the source or the observer, which is the direct result from the principle of relativity.

 

[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

This paper is conceptually a complete mess. It is a classical paper on classical light, but repeatedly talks about photons and gives them classical attributes. A lot of the justification of their concepts is based on this mostly-classical (and therefore mostly-wrong) picture of a photon.

 

[PFfan01]

This paper is conceptually a complete mess. It is a classical paper on classical light, but repeatedly talks about photons and gives them classical attributes. A lot of the justification of their concepts is based on this mostly-classical (and therefore mostly-wrong) picture of a photon.

From my understanding, the photon concept was introduced by Einstein’s light-quantum hypothesis, thus the photon energy cannot be solved by quantum theory because the whole quantum theory is developed based on the fundamental assumption: the Planck constant is a Lorentz invariant constant (Dirac) and the photon energy is equal to the Planck constant multiplied by frequency while the frequency is a pure classical concept. Just like the photon energy, the photon momentum in free space cannot be solved by quantum theory, because it is the direct result of the principle of relativity and Einstein’s light-quantum hypothesis. Thus I don’t think that paper “is conceptually a complete mess”.

PS: As we know, there is a serious contradiction between nonlocal indeterminacy of quantum theory and local reality of special relativity, specifically reflected in the superluminal propagation of quantum states of an entangled electron pair. In fact, there is another serious self-contradiction in quantum assumptions: As we know, the canonical momentum for an electron in a uniform magnetic field is not unique (indeterminacy), but its canonical momentum operator should correspond to an observable quantity according to the quantum assumptions. Let us forget those controversies.

 

[Dale]

Thus I don’t think that paper “is conceptually a complete mess”.

I do. I am still wading through it, but that is my opinion this far.

A single author is generally a bad sign for the quality of a paper. Occasional good papers are solo-authored, but more commonly such papers are of low quality. Usually it means that the author has never had their ideas seriously challenged and the writing has had insufficient internal review before submission.

PS, the rest of your post shows a lack of understanding of modern QFT, but seems off topic for this thread.

 

[PeterDonis]

I don’t understand what the “spatially structured” means.

It means that a "photon" is not what you think it is. See below.

a photon has no internal structure (it's not even a particle in the ordinary sense). I don't they're talking about that here though.

They are talking about the fact that in their experiment, the waves associated with the "photons" are not plane waves: as the abstract says:

"light beams have finite transverse size, which leads to a modification of their wave vectors resulting in a change to their phase and group velocities."

In other words, the word "photon" as they are using it does not mean "a particle of light"; it means "a wave packet of light whose properties are such that its group velocity is lower than c".

(Note that I can only read the abstract, as the full paper is behind a paywall, so I don't know if they are using classical wave optics or quantum wave optics to do their analysis. Either way, however, what I said above applies; in the quantum case, the "group velocity" would just be an expectation value instead of a classically calculated number.)

 

[PeterDonis]

From my understanding, the photon concept was introduced by Einstein’s light-quantum hypothesis

The original light-quantum hypothesis was Planck's, not Einstein's; he introduced it in order to derive a formula for black-body radiation that matched experimental data, which the classical Rayleigh-Jeans formula did not (the failure of the latter to do so was called the "ultraviolet catastrophe").

the whole quantum theory is developed based on the fundamental assumption: the Planck constant is a Lorentz invariant constant (Dirac) and the photon energy is equal to the Planck constant multiplied by frequency while the frequency is a pure classical concept.

No, you have it backwards. In quantum electrodynamics, the "frequency" of the photon is the energy. More precisely, the energy is the fundamental concept; the "frequency" is just an interpretation that arises when we take the classical limit. The same goes for momentum vs. wave number.

Let us forget those controversies.

These opinions about quantum theory are off topic for this thread.

 

[PWiz]

EDIT: Nevermind.

 

[Nugatory]

This thread started off on slightly shaky ground:

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...I wonder who's right?

The answer to this question is that they're both right, and the apparent contradiction appears because a photon isn't what you think it is.

We often use the word "photon" in relativity discussions when we really mean "a pulse of light that we've localized to within the precision of our thought experiment so that we can speak as if it is at a single point instead of spread out through a region of space like any real pulse of light" - we do this because nobody wants to say, write, or read forty-three words when one word is available and will get the message across.

However, this convenient oversimplification leads to confusion and apparent contradiction when we come across something that is true of a pulse of light but not true of a photon, and that's what's happening here. A photon does not have a four-velocity, but there is a way of associating a worldline and a four-velocity to the light in Fizeau's experiment. If a photon were that pulse of light we'd have a contradiction, but it isn't. Now look at the informal blurb about that paper in Science:

A photon always moves at c?
From Einstein’s special relativity on down, the invariance of the speed of light in free space has been a central tenet of physics. Now, in a clever set of experiments, scientists in the United Kingdom have demonstrated that, in certain conditions, individual photons in free space can be slowed down to speeds measurably below the supposedly invariant light speed .

You'll see the same confusion there, shifting smoothly from the behavior of photons to the "supposedly invariant light speed" (and note that the abstract of the paper is more precise than the informal blurb and does not hint that the invariance of $c$ is only "supposed").

From my understanding, the photon concept was introduced by Einstein’s light-quantum hypothesis, thus the photon energy cannot be solved by quantum theory

The concept introduced by Einstein's light quantum hypothesis bears very little resemblance to the modern understanding of what photon is (for example, Einstein would not have hesitated to assign positions and velocities to his hypothetical light quanta). It is not altogether lacking in irony that Einstein's Nobel was awarded for the piece of his anno mirabilis work that has stood up least well to the test of the time.

It's easier to say what a photon is not than what it is, but that's a better discussion for the QM forum.

 

[Nugatory]

This seems like an abuse of terminology, does it not?

There is abuse going on, but it's the phrase "particle of light" that is abusive (or at least routinely misinterpreted).

The basic terminology problem here comes from the way that the word "particle" is used in quantum field theories for historical reasons (and because no one wants to be saying "quantized excitation of the <whatever> field" all the time). That meaning is so different from the standard English-language meaning of the word "particle" that confusion is almost inevitable when someone hears the term "particle of light" outside of a QFT textbook.

 

[PeterDonis]

This seems like an abuse of terminology, does it not?

How so? A wave packet is a perfectly well-defined notion, as is its group velocity, and it is what the term "photon" as it is used in the paper, as far as I can tell, is being used to refer to.

 

[PFfan01]

... In other words, the word "photon" as they are using it does not mean "a particle of light"; it means "a wave packet of light whose properties are such that its group velocity is lower than c".

(Note that I can only read the abstract, as the full paper is behind a paywall, so I don't know if they are using classical wave optics or quantum wave optics to do their analysis. Either way, however, what I said above applies; in the quantum case, the "group velocity" would just be an expectation value instead of a classically calculated number.)

Here there is a paper that presents Padgett-team experimental work in popular words:
“Photon Footrace: Slowing Down Light in Free Space”, by Stewart Wills, http://www.osa-opn.org/home/newsroom/2015/january/photon_footrace_slowing_down_light_in_free_space/#.Vk4HF2zovIW It says: “Now, in a clever set of experiments, scientists in the United Kingdom have demonstrated that, in certain conditions, individual photons in free space can be slowed down to speeds measurably below the supposedly invariant light speed (Science, doi: 10.1126/science.aaa3035).”

From the following link, you have free-access to Padgett-team original experimental report:
https://www.researchgate.net/profile/Daniel_Giovannini

 

[Ibix]

(Note that I can only read the abstract, as the full paper is behind a paywall

Arxiv: http://arxiv.org/abs/1411.3987

 

[PeterDonis]

n certain conditions, individual photons in free space can be slowed down to speeds measurably below the supposedly invariant light speed

And what do they mean by "individual photons"? It certainly isn't "a particle of light".

Furthermore, to the extent they use a 4-vector (the "wave vector") to describe the light, the vector is always null; that is obvious from their equation

$$
k_z^2 + k_x^2 + k_y^2 = k_0^2
$$

So there is no way of associating their "photon" with a 4-velocity, since that would be converting a null vector into a unit vector, which is impossible, as has already been pointed out in this thread. Of course, since the "photon" is moving in free space, not a medium, we would expect its wave vector to be null no matter what we think might happen in a medium. In short, this paper, interesting as it is, appears to be irrelevant to the topic of this thread.

(But if the wave vector is always null, what is all this about the "speed" of the photon being slower than c? That's because they are using "speed" to mean $k_z / k_0$, i.e., the longitudinal component of the wave vector divided by the timelike component. Or, to put it another way, they are using "speed" to mean the momentum of the wave along its direction of propagation divided by its energy. Because the wave is finite in spatial extent, i.e., it has non-zero transverse components to its wave vector, its "speed" defined in this way will be less than c even though the wave vector as a whole is null. But this "speed" is not the speed of a "particle" in any case.)