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Quite a few of the questions given on this site mention a photon having a rest frame such as it having a zero mass in its rest frame. I find this contradictory since photons must travel at the seed of light in all frames according to special relativity

Does a photon have a rest frame?

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Not in vacuum, but the question makes sense in a transparent medium. Experts, what does an observer see in a moving RF in a medium if its velocity is c/n? – Vladimir Kalitvianski Oct 21 '11 at 18:12
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I understand that in a medium you are dealing with a wave packet of photons, and not a single photon. In the lab frame, each photon in the packet moves at the speed c, but the packet's group speed is c/n. If you shot a pulse of light into a medium and then followed it at the speed c/n, you would just see a Lorentz-transformed packet of photons. Each photon will still move at the speed c, but the group speed of the packet would be zero. And no, a photon does not have a rest frame, if special relativity applies. – drlemon Oct 21 '11 at 19:40
A one-photon propagation mode is also possible in a medium, hence a photon is a wave train of a finite length, no? – Vladimir Kalitvianski Oct 21 '11 at 20:43
@drlemon: your comment is not right--- there is no packet--- it happens photon by photon. – Ron Maimon Oct 21 '11 at 20:52
@Ron and Vladimir: In E&M textbooks, the speed of light in a medium is derived for a monochromatic wave. The physical reasoning for slowing down the speed of light in a medium is that atoms polarize and produce their own E/M fields, interfering with the incident wave. The total E/M field is that of a monochromatic wave traveling at the speed c/n. There are no individual photons in this analysis. An individual photon will be scattered by atoms back and forth, and will, in general, have some stochastic transport. On average it will travel at c/n, but it will not just slow down to c/n. – drlemon Oct 21 '11 at 21:39
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3 Answers

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Short answer: no.

Explanation:

Many introductory text books talk about "rest mass" and "relativistic mass" and say that the "rest mass" is the mass measured in the particles rest frame.

That's not wrong, you can do physics in that point of view, but that is not how people talk about and define mass anymore.

In the modern view each particle has one and only one mass defined by the square of it's energy--momentum four vector (which being a Lorentz invariant you can calculate in any inertial frame): $$ m \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2 $$

For a photon this value is zero. In any frame, and that allows people to reasonably say that the photon has zero mass without needing to define a rest frame for it.

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I agree completely with @dmckee and would only add that for any particle the elapsed time experienced by that particle in it's rest frame is called the proper time and can be calculated (in units where $c=1$) by any observer as $$d\tau^2 = dt^2 - d\vec{x}^2$$ and for a photon in a vacuum the proper time is always identically $0$. So photons do not experience any passage of time so in that sense also, they do not have a rest frame. – FrankH Oct 21 '11 at 19:58
And in QM the photon energy is $\hbar\omega$ and $\omega$ in a medium is the same, so $m_{photon}=0$. – Vladimir Kalitvianski Oct 21 '11 at 20:41
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Your answers are right,a solitary photon has no rest frame, nonetheless I find quite interesting to note that a system of massless particles(such as photons) can have a nonzero mass provided that all the momenta are not oriented in the same axis and that for such systems zero momentum frames CAN actually be defined.

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Not at all. Rest frame is a concept that does not exist in nature. Had it would exist, nature wouldn't be causal. A photon propagating through medium does not 'move' in a speed smaller than the speed of light in vacuum. It simply interacts electromagnetically with the medium and these interactions slows down its propagation through the medium.

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"Rest frame is a concept that does not exist in nature." That's a strange way of stating things. If (in SR) in some frame $L$ you observe a (massive) particle moving at a speed $v < c$, you can most definitely pass to some frame $L'$ in which the particular doesn't move. – Gerben Oct 22 '11 at 11:21

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