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?


Short answer: no.


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 Oct 21 '11 at 19:58
  • $\begingroup$ And in QM the photon energy is $\hbar\omega$ and $\omega$ in a medium is the same, so $m_{photon}=0$. $\endgroup$ – Vladimir Kalitvianski Oct 21 '11 at 20:41

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.


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.


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 $1$.


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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    $\begingroup$ "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 Oct 22 '11 at 11:21

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