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

Physics Asked by user133382 on January 20, 2021

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?

One Answer

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.

Answered by mathquest on January 20, 2021

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