Imagine that you are in an elevator that is moving downward at a constant speed that is near the speed of light (let's say close to the surface of the earth: toy problem, needless to say). Now say that you jump. Then (neglecting any general relativistic effects that hopefully you can explain to me) would it be fair to say that the only relevant acceleration after your feet leave the elevator would be -9.8... and your initial velocity when you jump would be the initial velocity you would have provided for yourself on the sidewalk minus the speed of the elevator? That's my first question (asked just as a sanity check, I think the answer is yes). My second and main question is: are there indeed any relativistic effects? Another way of asking my question is: does spacetime curvature change in general relativity for objects traveling at high speeds, and what does this say about the experience you would have when doing the above jump?
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1$\begingroup$ The acceleration won't stay constant if you're moving vertically at that speed. The Earth's diameter is about 0.021 light-seconds. $\endgroup$– PM 2RingCommented Sep 7, 2018 at 3:50
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$\begingroup$ @PM 2Ring I agree that over any appreciable interval of time the -9.8 will not be applicable. But I believe the reason you have in mind for this is the fact that the distance from the center of the earth will change extremely rapidly. If I'm correct about your thought then I do agree with you. Nonetheless, I'm curious about the limiting case where the interval of time is approaching zero (this interval I have in mind is right as the clock just started ticking post-jump). Can gravity "transmit the -9.8 acceleration" even as the initial velocity has such an overwhelmingly large magnitude? $\endgroup$– okcappCommented Sep 7, 2018 at 5:11
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$\begingroup$ The spacetime curvature doesn't change, per se, but you do need to take time dilation & length contraction into account when comparing what the curvature "looks" like to different observers. Is that what you're asking about? $\endgroup$– PM 2RingCommented Sep 7, 2018 at 15:40
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$\begingroup$ @PM 2Ring Yes it is! Thanks for the clear question. I suppose if I could rewrite my question I would ask how spacetime curvature would appear to a person jumping from the sidewalk, the person jumping from the constant velocity elevator (moving at a speed close to the speed of light), and to an observer sitting on the sidewalk watch all of the activity. $\endgroup$– okcappCommented Sep 7, 2018 at 20:25
2 Answers
The key word here is relative.
You are in the elevator and whatever the elevator and you are accelerating relative to doesn't matter to you. Relative to the elevator your are have zero velocity and zero acceleration.
If you jump, you're jumping relative to the elevator. Again the elevator's velocity relative to something else doesn't matter, just the relative motion of you and the elevator.
And from that point of view you can happily view this as an entirely non-relativistic event of you jumping up and falling down ...
... unless ...
If the elevator and you were physically connected (you were e.g. holding on to a rail) or you're both in the same field and affected the same way then the acceleration you feel keeps your motion relative and all is good.
But if you letting go of that rail you're holding on to also means you're no longer being accelerated relative to whatever the elevator is accelerating, then you're going to be left behind (or crash into the roof !).
A simpler way to see this is to use a rocket in space. You're an astronaut. The rocket is accelerating using it's own rocket engine. When you're attached you experience that just as expected - there's a down and an up defined by the pseudo gravity of the rocket's acceleration.
Now go for a spacewalk.
Again while holding on you experience the same acceleration.
Now let go.
The ship starts accelerating away at the suddenly not very trivial rate of $9.8 ms^{-2}$ and you find yourself looking at the rapidly disappearing ship while you float in space.
You maintain the velocity both you and the ship had (relative to something) when you let go, but after that, it's accelerating and you (unhappily) are not.
There is no relativistic effect - the velocity you share with the ship relative to something else doesn't matter - zero or 99% the speed of light produce the same result from your point of view.
Assuming a constant gravitational field, you will perceive your acceleration as 9.81m/s^2 relative to the escalator. However, a 'stationary' observer will measure your acceleration differently, due to length contraction$$L=L_0\sqrt{1-\frac{v^2}{c^2}}$$ So if the escalator was moving at 99% speed of light, the stationary observer will see that you are 14.1% of your height, falling at 14.1% of your perceived acceleration. So to answer the question, yes there are relativistic effects. There always will be relativistic effects if observers have different velocities
