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Suppose two twins are put asleep. One is put in a rocket ship and accelerated to 90 percent of the speed of light. They are then both awakened, after which they both train their telescope on each each other. They both start dancing. Will the twin in the rocket ship and the twin on the ground both see the other dancing at the appropriate rhythm; or will the twin in the rocket ship see the other twin dancing at a too fast rhythm and the twin on the ground see the other twin dancing in slow motion.

sammy gerbil
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user309371
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2 Answers2

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This depends on whether the twins are moving toward each other or away from each other.

You didn't specify whether the twins were in the same location when they were put to sleep. If they were far apart, then put to sleep, then Alice was accelerated, she might have been accelerated either toward Bob or away from him.

Now for what she sees in her telescope, there are two effects. Effect One: Time dilation makes Bob's dance appear slower. Effect Two: The distance between Alice and Bob keeps changing, making Bob's dance appear slower if they're getting farther apart, or faster if they're getting closer together.

(Everything we say about what Alice sees applies equally, of course, to what Bob sees, since everything is symmetric.)

So if they're moving apart, the two effects reinforce each other, and Bob certainly appears to be dancing in slow motion. If they're moving toward each other, the effects work in opposite directions, so you've got to do a little algebra to see which wins out.

If you do that algebra, you'll find that Bob's dance appears to be speeded up by a factor of $\sqrt{1+v}/\sqrt{1-v}$. (Hint for the algebra: First do everything in Alice's frame, figuring out when and where the light signal she emits at time $t$ reaches Bob. Then Lorentz-transform that "when and where" to Bob's frame.)

WillO
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Actually, they will both see each other dancing in what appears to be slow motion. It's not super intuitive, but if one of them is accelerating away from earth in a rocket, it's equally valid to say that the one on earth is effectively accelerating away from the rocket. The observed time of another person(t2) will always be less than your own(t1) as you can see in the formula for time dilation;

t2 = t1/sqr(1-(v^2/c^2) (I'm not quite sure how to do formulas in a more professional way.)

This can easily be derived by thinking about how light appears to bounce between mirrors to while accelerating: http://users.sussex.ac.uk/~waa22/relativity/Complete_Derivation_files/derivation.pdf

smaude
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