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Quite a few of the questions given on this site mention a photon in vacuum 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 speed of light in all frames according to special relativity.

Does a photon in vacuum have a rest frame?

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    $\begingroup$ related: physics.stackexchange.com/q/29082 $\endgroup$
    – user4552
    Commented Jul 19, 2013 at 22:40
  • $\begingroup$ It's a basic feature of relativity that inertial frames travel at speed LESS THAN $c$. There is no such thing as a frame of reference with speed $c$ relative to another frame. This is an interesting example of the difference between the surface of a light cone (exactly null) and the interior of a light cone (which is time-like). There is a big difference between zero and not-zero! $\endgroup$ Commented Dec 13, 2023 at 17:07

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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): $$ m^2 \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2 $$

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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    $\begingroup$ 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. $\endgroup$
    – FrankH
    Commented Oct 21, 2011 at 19:58
  • $\begingroup$ And in QM the photon energy is $\hbar\omega$ and $\omega$ in a medium is the same, so $m_{photon}=0$. $\endgroup$ Commented Oct 21, 2011 at 20:41
  • $\begingroup$ I understand how at light speed (i.e. a photon) the problem of no rest frame exists, but what about if we ask ourselves about the limit as something approaches light speed. What does the world look like in the limit? Is that a legitimate question I might ask here do you think? $\endgroup$ Commented Jul 10, 2023 at 14:09
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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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It is not possible to find a frame of reference where a photon is at rest. I will argument in two different ways:

1. Maxwell equations and electromagnetic argument:

From Maxwell it is expected that electromagnetic disturbances propagate in vacuum at a constant speed c~299792458 m/s which is the maximum speed for the propagation of electromagnetic interactions.

If you could find a rest frame for a photon (i.e. a frame of reverence where the speed of photons is zero), then, in this frame of reference any electromagnetic interaction would be impossible (as photons are the carriers of the electromagnetic interaction). For example, the force between two electrons at rest would be $F=0$ for any location of the electrons as the field would not be able to propagate between them. This is absurd, and therefore it is not possible to find a frame of reference where a photon is at rest.

2. Corpuscular nature of photons and Quantum Mechanics:

The energy $E$ of a photon is defined as $E=hf$ where $h$ is Plank's constant and $f$ stands for the photon's frequency but $c = \lambda f$ (with $\lambda$ being the wavelength). This product can be zero in three different ways:

  1. $\lambda = 0$, $f$ finite. In this case, the photon has zero wavelength and therefore infinite momentum and finite energy which is absurd.
  2. $f = 0$, $\lambda$ finite. In this case, the photon has no energy but a finite momentum ($p = h/\lambda$) which is again absurd.
  3. $\lambda = 0$ and $f = 0$. The photon has zero frequency (zero energy) and zero wavelength (infinite momentum) which is double absurd.

Therefore both Classical Electromagnetism and Quantum Theory of Light deny the possibility of a frame of reference where a photon can be found at rest.

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I am going to approach the matter otherwise. I do not dispute the arguments made in answers in [1] or answers in [2]. However, based on my learning style, these arguments do not work for me. This is because in my opinion, there is---as yet no---convincing argument offered that draws a clear line between the energy and momentum of a massless particle and the truth of the proposition, Photons have no rest frame, which is found in [3] among many places. My approach is a logical proof, which may be suitable for people who favor other methods of learning.

My approach is a proof by contradiction [4] that relies on one postulate and one definition.

Postulate The speed of light, $c$, in vacuum is the same for all observers, regardless of the motion of light source or observer [5].

Definition In special relativity, the rest frame of a particle is the frame of reference in which the particle is at rest.

Proof by contradiction

The proposition to be proved is $P$, where $$ P = \text{Photons do not have a rest frame.}$$

We assume $P$ to be false. In other words, we assume $\tilde P$, where $$\tilde P = \text{Photons do have a rest frame.} $$

I must now show that $~\tilde P$ implies falsehood.

By definition, in special relativity, the rest frame of a photon in vacuum is the frame of reference in which the photon is at rest. This is in direct contradiction to the postulate that reads here as, the speed of a photon in vacuum, which is a universal physical constant that is approximately $3\times10^8$ m/s [6], is the same for all observers, regardless of the motion of photon source or observer.

I appeal to the law of non-contradiction. Since assuming that photons do not have a rest frame is false leads to a contradiction, therefore it is concluded that photons do not have a rest frame is, in fact, true.

Q.E.D.

Bibliography

[1] Does a photon in vacuum have a rest frame?

[2] Is the non-existence of a rest frame for photon in vacuum a consequence of the second postulate?

[3] Modern Electrodynamics, Andrew Zangwill, p 838.

[4] https://en.wikipedia.org/wiki/Proof_by_contradiction

[5] https://en.wikipedia.org/wiki/Special_relativity

[6] https://en.wikipedia.org/wiki/Speed_of_light

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For simplicity, assume $(1+1)$ dimensions. Let $M$ be the $2\times 2$ matrix relating two frames. Then (setting $c=1$) the relative velocity of the two frames is the ratio $v=M_{12}/M_{22}$ (or some analogous ratio depending on your preferred ordering for your frame).

Suppose $v=1$. Then $M_{12}=M_{22}$. From this and Lorentz-orthogonality, one easily gets $M_{11}=M_{21}$, whence $M$ is singular, contradiction.

Therefore there cannot be two frames with relative velocity $1$.

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Let me put it this way:

Einstein's special relativity does not give us a way to make a photon's velocity be the frame of reference.

You might find a way do that. There may be some advantage to doing it, but I haven't heard of any advantage. This may be the reason that nobody reputable has bothered to do it, or if they have that it hasn't gotten a lot of attention.

Now humor me for a moment, I have an analogy.

Imagine that people insisted that we have to do our geography like the world is flat. So everybody has a "horizontal" frame. We think that the farther away people are, the more they are tilted. People in australia are upside down, and if somehow they could be immediately translated here they would arrive on their heads. We can do a transformation to get from our frame to somebody else's frame if they're somewhere else. And there are various other adjustments we have to make. The area of a flat circle would be $\pi r^2$, but in the real world circles have less area than that (or maybe more area, depending on how you measure the radius. It's less area than a flat circle with the same radius as the great circle radius). But the difference is only important for great big circles.

We don't do that. We don't worry about how distant things are tilted, and we don't adjust for horizontal frames, we just accept that the world is round.

Maybe there's a way to accept that things that have different velocities just don't work the same, and have one single way to measure it all which fits together.

Wouldn't that be better? Instead of an infinite number of misleading frames to transform among, we'd just have the one description, one way to measure stuff and it would just work. Like using spherical trig in place of flat geometry.

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Not at all. 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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    $\begingroup$ "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. $\endgroup$
    – Gerben
    Commented Oct 22, 2011 at 11:21
  • $\begingroup$ You are absolutely right. I framed it badly and corrected my answer. Thank you. $\endgroup$
    – user1999
    Commented Dec 2, 2021 at 4:25

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