Thought experiment: Elevator going up at an extreme acceleration, pulse of light bouncing up, and down between mirrors on the floor, and the ceiling. Won't it take light longer to travel from the floor to the ceiling, than from the ceiling to the floor? If so,then based on the Equivalence Principle, doesn't this mean that light will move slower from floor to ceiling in an Equivalent gravitational field?

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    $\begingroup$ My answer here might be relevant: physics.stackexchange.com/a/337952/20427 $\endgroup$ – Dvij D.C. Sep 2 '19 at 7:46
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    $\begingroup$ Related: physics.stackexchange.com/q/77227/2451 , physics.stackexchange.com/q/297468/2451 and links therein. $\endgroup$ – Qmechanic Sep 3 '19 at 8:13
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    $\begingroup$ Where is the observer? Standing on the floor of the elevator? And how does the observer define the time-of-flight of the light pulse, given that the observer can't observe the departure or arrival of the light pulse unless some other signal propagates from those events to the observer? Here's the point: the thought experiment is undefined. This is a common situation when learning physics: 50% of the insight comes from learning why the question is undefined, 40% comes from learning how to make it well-defined, and the remaining 10% comes from answering the well-defined question. $\endgroup$ – Chiral Anomaly Sep 7 '19 at 20:50
  • $\begingroup$ The thought experiment implies two frames of references: the accelerated frame of reference, and an observer witnessing the acceleration. The witness will see that the light is traveling farther when going from floor to ceiling, and therefore takes longer, than the return trip; and the observer in the accelerated frame of reference, will see light taking longer from floor to ceiling and will conclude that light is traveling slower, in that direction. From one, the speed of light is a constant; but from the other is the distance traveled is a constant; $\endgroup$ – Steven Flowers Sep 8 '19 at 21:25

Your reading of the thought experiment is not correct.

The equivalence principle implies that locally the laws of physics are described by special relativity, hence locally the speed of light is $c$.

The thought experiment means the ceiling (receiver) is moving away from the floor (source) and thus measures a lower frequency of the light. Conversely the floor is measuring a higher frequency. It is the relativistic Doppler effect in special relativity.

Applying the equivalence to a gravitational field, in the former that would show as the gravitational redshift (light moving away from a massive body) and in the latter as the gravitational blueshift (light approaching a massive body).

  • $\begingroup$ I understand that the frequency of the light will be red shifted going up, and blue shifted coming down, and that the time is slowed down because of acceleration, and that to an observer would see the distance between floor and ceiling will look shorter; but the light clearly travels further from floor to ceiling, than from ceiling to floor. It has to take longer go up than it does to come down, for all observers. $\endgroup$ – Steven Flowers Sep 3 '19 at 18:53
  • $\begingroup$ Consider a first inertial reference frame at rest vs. the elevator at the instant the light is bouncing up from the floor and a second inertial reference frame at rest vs. the elevator at the instant the light is bouncing down from the ceiling. The latter is in relative motion against the former. In both frames according to special relativity the light travels at speed $c$. Or, if you measure the speed of light as $c$ in one of the reference frames and calculate the speed in the other using the Lorentz composition of velocities you get $c$ again. $\endgroup$ – Michele Grosso Sep 5 '19 at 7:17
  • $\begingroup$ You have to reason in the momentarily comoving reference frames, not in the elevator accelerated frame for which you need general relativity. $\endgroup$ – Michele Grosso Sep 5 '19 at 7:19

Light is ALWAYS traveling in the same speed of $c$ in ALL reference frames. This was confirmed by various experiments such as Michelson & Morley experiment and the others. The only thing that changes - is the light frequency,- if light looses energy somehow then it's frequency is red-shifted, but speed is the same $c$. Of course if photon is traveling in vacuum. If photon is traveling in medium with refractive index of $n > 1$, then phase velocity of light is $v < c$. The only reasonable explanation if light speed is the same everywhere - is that time flow changes and is dependent on reference frame (this was solved by Einstein). I suggest you first to read about special relativity, because Einstein has developed it first. Then study general relativity because it is much much more complex that comes after.

  • $\begingroup$ The Michelson Morley experiment was designed to measure the speed of light in two directions from a constant velocity frame of reference. The solution for the speed of light being the same at right angles is Special Relativity, the results have nothing to say about the thought experiment I have presented with an accelerated frame of reference. Isn't this why light can't escape a black hole? $\endgroup$ – Steven Flowers Sep 4 '19 at 2:15
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    $\begingroup$ @Steven Flowers. The reason why light crossing the horizon of a black hole can not escape is because of the curvature of spacetime in the interior region and not because of the speed. $\endgroup$ – Michele Grosso Sep 7 '19 at 10:04
  • $\begingroup$ How does an observer witnessing the accelerated frame, not see that the travel time is longer in one direction than the other? $\endgroup$ – Steven Flowers Sep 8 '19 at 21:36
  • $\begingroup$ Not in ALL reference frames, only inertial ones. $\endgroup$ – Ruslan Jun 5 '20 at 19:37
  • $\begingroup$ @Ruslan Wiki gives about speed of light: "In non-inertial frames of reference (gravitationally curved spacetime or accelerated reference frames), the local speed of light is constant and equal to c, but the speed of light along a trajectory of finite length can differ from c, depending on how distances and times are defined." So local speed is same $c$ also in non-inertial reference frames. $\endgroup$ – Agnius Vasiliauskas Jun 8 '20 at 6:21

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