Quick check: relativity, rockets, clocks and the equivalence principle

One common thought experiment that introduces relativity on gravitational fields is the "clocks on an accelerating rocket":

Paraphrasing Mr. Feynman:

Suppose a rocket, with two clocks, one on each end, accelerates and you (the observer) stands on the rear. Each time the clock on the top ticks, a light signal is sent to the rear clock. The difference in lengths travelled by light (due to acceleration) will be perceived as though the clock on the top had a faster rate.

My question is: To the observer, wouldn't the rocket have always the same length, thus taking light the same time to travel from the top to the bottom?

• Define "length". In special relativity an inertial observer can build a reference frame: a grid of meter sticks with synchronized clocks, as Einstein explained. What do you mean by the length exactly in our case? Commented Nov 6, 2017 at 13:05
• Excellent point. No comments on length but probably it's the key to the solution. However, I believe that for the observer the rocket will always be the same, even when accelerating.
– user133382
Commented Nov 6, 2017 at 13:07
• Causality is independent of the observer. Why are you not satisfied with being an inertial observer? If you want to be the accelerating observer with the accelerating grid of meter sticks, then the coordinate expression of the metric, though still flat, is not going to be simple anymore, and therefore the null geodesics on which light travels will be parametrized differently. Commented Nov 6, 2017 at 13:27
• Because this is not what the thought experiment proposes.
– user133382
Commented Nov 6, 2017 at 15:55
• Commented Jan 7, 2023 at 20:10

"I consider myself to be the finish line, that is always x meters away from the starting line. And I always move towards the place where the starting line was when the light started propagating from the starting line", says the observer.

The observer adds: "The finish line moves, the current starting line moves, the distance between the finish line and the current starting line is not affected by the motion of the two lines, because the two lines move the same way. But the distance between the finish line and the past position of the starting line is affected by the motion of one of the two lines."

To resolve this question from the point of view of an accelerating observer you need to go outside of Special Relativity. In General Relativity frame of references could be very "unphysical", recording "speeds" higher than the speed of light, recording enormous accelerations, etc.

An inertial observer can build a very special (ignoring the three degrees of freedom associated to the orientation of the frame) reference frame as explained in any introduction to SR: a grid of meter sticks with synchronized clocks. These coordinates posses some very nice properties, for example light travels on straight lines with constant speed c. In GR you have infinite degrees of freedom of how you choose your coordinates. In flat GR (assuming no mass or gravitational waves) one can try to use the same natural coordinates, but they will not be as nice:

Assume your spaceship is a meter stick with an observer placed at every millimeter of it with their own clock. They synchronize, and then you start the engine. They record time as the impulse of light travels past them along the rocket. Even though every one of them will think light travels at the speed c (by performing some natural local experiment), when you aggregate their data, you will see that the impulse of light accelerated through the ship. That will happen because the physical clocks the observers at millimeter marks posses will be accumulating delay due to acceleration (similarly to the twin paradox), but they will have accumulated different amounts by the time they need to record the ray of light is passing. You will see signals coming more often from the other end because of that "acceleration" of light.

Alternatively, you can order the observers to keep adjusting their clocks to counteract that. In these coordinates there will be no acceleration of light, and the observer at the end of the rocket will receive equally spaced (with respect to his adjusted clock) signals coming from the adjusted clock at the opposite end. The observer will know, though, that the clocks are adjusted, and are not equally adjusted as it takes time for the signal to travel through the rocket. Then he can predict, that signals from an unadjusted physical clock would come more often than his unadjusted clock ticks.

Finally, the observer at the end of the rocket can order other observers to synchronize their clocks according to how he perceives them. That will probably be the most natural frame of reference for that observer. Light in such coordinates will travel with constant speed, but there is no contradiction because clocks are already adjusted as suggested by Feynman.

To summarize. To perform calculations in accelerating frames you need to specify coordinates. I gave three examples of such coordinate systems but all of them are harder than the inertial frame of reference to work with, and computations anyway rely on one done in an inertial frame to predict how physical clocks behave when accelerated.

I've given this thought experiment quite some attention some time ago as it confused me. I think that some concepts are a bit mixed up in the example. The experiment is meant to get "a feeling".
I think that in the experiment (described by Feynman in his little book "Six not so easy pieces") Feynman sees space-time initially as absolute space-time, with the condition that the speed of light is the same for everyone. The two are not compatible, obviously, but he just wants to highlight one feature of general relativity. This means that the metric of space and the metric of time are not dependent on each other, that they are the same for every observer. Which of course means that the rocket's length is independent of the observer. Observers moving relative to each other will see each other's time pass exactly at the same rate as their own, and distances will be equal. If she finds herself in an inertial frame or a non-inertial one is not yet important. This only becomes important in the context of relativity, about which Feynman wants to explain a still-to-come general relativity feature.
If two light pulses are emitted one second after another, then the observer on the rocket's floor will receive them one second apart if the rocket moves with uniform velocity. The pulses will be received in less than one second if the rocket accelerates. This means that the observer sees things happening faster at the top of the rocket. I'm not sure if Feynman insists that the speed of light must be the same for the observer (from the premises of a constant light speed follows the whole of special relativity). I guess he does because that's the only way that the observer can say that the two pulses were really sent from the top in less than a second apart. Feynman concludes that in a real gravity field time goes slower if you find yourself deeper in the field (closer to the bottom of the rocket) because the accelerated rocket is equivalent to a real gravity field.
When time passes at different rates at different places, then time is curved. This automatically means that space is curved. This curvature of space-time is not yet included in Feynman's thought experiment. You can even ask yourself if the rocket's length will be different in a gravity field, as compared to its length in free space while moving with constant speed.