# Do photons have absolute motion?

Absolute rest is not possible as the concept of motion is relative. But can we assert that photons have absolute motion as the observers in all the frames of reference would agree to the same value of speed of light.

Do we consider frames of references moving at speed of light, if we do than above assertion won't be correct.

• By tagging this with general-relativity you open a whole 'nother can of worms. There are perfectly good ways of looking at things in GR when the speed of light at places distant from you is not constant. – dmckee --- ex-moderator kitten Nov 22 '16 at 5:52
• We cannot consider frames of references moving at speed of light because Lorentz transofmation formulas become undefined at $v=c$. We can only take $v<c$ and try to obtain some limiting results at $v\rightarrow c$, but they should be interpreted with caution. – Alexey Sokolik Nov 22 '16 at 9:25
• – ACuriousMind Nov 22 '16 at 14:01
• Also related: physics.stackexchange.com/q/16018/2451 – Qmechanic Nov 23 '16 at 8:14
• @dmckee---ex-moderatorkitten --Just out of curiosity, would the distant places you're referring to be in causally-separated regions (i.e., in regions of different density than the parts of our observable region that haven't collapsed gravitationally, at least in our times)? (If it's something more subtle than that, don't bother to elaborate, as I might not have the math for it.) – Edouard Jan 26 at 19:18

In a certain sense you are right: there is no inertial frame of reference in which a photon propagating in vacuum is locally at rest.

When measured locally and in an inertial frame, the speed of light is $c$ independently of the inertial frame we choose: this is one of the postulates of special relativity.

Notice anyway the keyword locally: on larger scales, where spacetime cannot be considered flat, we have to use the formalism of general relativity, and things may be very different: see for example this question & answer and also this one.

Also, it is important that we choose an inertial frame: if the frame of reference is not inertial, the speed of light may be different from c.

The similar behaviors found in the presence of gravitational fields and of non-inertial frames is no coincidence: the equivalence principle actually tells us that the forces experienced in a gravitational field are the same as those expereinced in a non-inertial frame of reference.

Consider a photon born on Earth, and traveling to a distant galaxy light years away. In the reference frame of the photon, the distance between Earth and the galaxy is contracted from Earth's $$L_0$$ to

$$L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \longrightarrow 0$$

The time required to get to the distant galaxy is also "contracted," relative to our time $$T_0$$, as

$$T = T_0{\sqrt{1-\frac{v^2}{c^2}}} \longrightarrow 0$$

However, if we compute the limit of $$\frac{L}{T}$$ instead as $$v \to c$$, we get

$$lim_{ v \to c}\frac{L}{T} = lim_{ v \to c} \frac{L_0}{T_0}=c$$

The speed of light in a reference frame moving at the speed of light is still the speed of light, but it becomes rather meaningless since light does not live in the time domain. We often say light lives in only spatial domain, but according to equations it appears it does not travel through space in its own reference frame either. In order to measure speed, we need length and duration - i.e. travel. Since neither length nor time appears to progress from the photon's perspective - i.e. it is born and ceases to exist in the same moment, and in the same place : everything in between is contracted to a single point in the direction of travel - then there is no speed.