# 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? – mathquest Nov 6 '17 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 Nov 6 '17 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. – mathquest Nov 6 '17 at 13:27
• Because this is not what the thought experiment proposes. – user133382 Nov 6 '17 at 15:55