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.
-
1Each one will see the other twin dancing in slow motion – Jan 25 '17 at 21:38
-
Indeed, dancing, clapping your hands or blinking your eyes are various means of measuring time (though not the most precise). – Jan 25 '17 at 21:40
-
1We appreciate that this seems weird to you. It seems weird to everybody. But ... it seems weird to everybody, which means that questions about it get asked over and over and over again. Start with http://physics.stackexchange.com/q/241772/ and possibly also some of the links in the sidebar. – dmckee --- ex-moderator kitten Jan 25 '17 at 22:18
-
What do you think? – sammy gerbil Jan 25 '17 at 22:28
-
My thought was - since the stationary twin will be older than the other twin when he/she returns, the stationary twin must have danced more steps and therefore danced faster as viewed by the travelling twin - assuming both twins continually view the other throughout the rocket travel – user309371 Jan 25 '17 at 22:47
-
1@dmckee: I don't think the question you linked to is terribly relevant, as it concerns only time dilation, whereas this question asks about the combined effects of time dilation and the time it takes for a signal to travel from one twin to the other. In particular, the effects in the linked question are independent of whether Alice is moving toward or away from Bob, whereas that matters very much for the question at hand. – WillO Jan 26 '17 at 05:48
-
Voted to reopen, because the alleged duplicate addresses only half the question, and the less interesting half at that. The answer to the present question depends on whether the travelers are moving toward or away from each other; the linked "duplicate" addresses only the (easier) case where they're moving away. – WillO Jan 26 '17 at 15:59
2 Answers
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.)
- 15,072
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
- 239
-
3This is wrong. The OP did not ask what Bob's dance looks like in Alice's coordinates; he asked what Alice sees through her telescope, which is a different thing. – WillO Jan 26 '17 at 02:26
-
1@WillO Oh no, sorry. Could you explain in more detail the differences between the two cases? – smaude Jan 26 '17 at 03:09
-
2At any given moment, the Lorentz Transformation (which is embodied in your formula) tells us what Alice thinks Bob is doing right now. But what she sees through her telescope is not what she thinks Bob is doing right now; it's what she thinks Bob did quite some time ago (just as when you look at the sun through your telescope, you don't think you're looking at the sun right now, but the sun as it was eight minutes ago). You are answering "What does Alice think Bob is doing now?". The OP asked "what is Alice seeing now?".. Those are different. – WillO Jan 26 '17 at 03:22
-
1Thanks, that makes a lot of sense. I'd downvote my answer if I had more reputation:) – smaude Jan 26 '17 at 03:28
-
1Thanks for thinking this through and for your general good nature. :) You'd think it would go without saying that we all mistakes and we say "oops" when those mistakes are pointed out. Unfortunately, not all users are quite so mature. Thanks for being one of the grown-ups. – WillO Jan 26 '17 at 05:25
-
1
-
@WillO, what you tell smaude sounds a little pedantic... same as when saying "You didn't specify whether the twins were in the same location when they were put to sleep". What's the actual difference between the two cases? – Helen Apr 16 '17 at 15:01
-
@Helen: The difference is that if they started out in one place/time, then we know they are moving apart from each other and have only one case to consider. Otherwise, we have to consider two cases, depending on whether the motion is toward or away from each other. – WillO Apr 16 '17 at 15:18
-
@WillO, (I would still call it pedantic but the point is) would smaude's answer be different in each of the two cases? – Helen Apr 19 '17 at 17:00
-