# In relativity, how does a traveller "catch up" in time after coming to a rest with respect to a stationary observer?

Please forgive the lack of rigour in this intuition-based question:

Alice and Bob are the twins from the infamous twin experiment. Alice boards a spaceship and accelerates away from Bob, turns around half way, and comes back. Let's let Alice accelerate at precisely the rate required so that for her entire trip her perceived time is 1 day, while Bob's perceived time is 1 year (let's also assume that she has a really good inertial dampening system so that she isn't flattened by the incredible accelerations she'd have to endure).

Now, according to Alice, the day is, let's say, January 17, 2017, while according to Bob, the day is January 17, 2018. Once Alice comes back, after a slight period of adjusting her clocks, Alice agrees with Bob, doesn't she? The entire world agrees with Bob as well. So how did Alice suddenly get here? How did her timeline "accelerate" to match up with Bob's?

• What am I missing here? Once Alice comes back To get back she has to slow down ( negative acceleration), otherwise she will hit Earth at high speed, so during that period she more and more closely matches Earth time. Have I misunderstood, I usually do :)
– user140606
Jan 16, 2017 at 16:12
• Oh, it's probably me explaining badly. I agree that as she slows down her perception of an interval of time, say 1 second, will more and more closely match with Bob's perception of the same interval. The question is more so: once they start to agree on their perception of time, why do they also agree on what time it is? Jan 16, 2017 at 16:17
• Alice's clock does not agree with Bob's. There is no catching up. She has to reset her clock to the proper Earth time. To see this, consider Alice herself as a biological clock: if she had to catch up with Bob, then there would be no more relativistic twin "paradox": she would end up with the same age as Bob. Jan 16, 2017 at 21:10
• All that stuff is very well explained in physics.stackexchange.com/questions/242043/… Jan 11, 2020 at 23:01

In your scenario the traveler cannot "catch up" but in a way the situation can be equalized. Alice travels very quickly for a day and when she returns she finds that a year has passed for Bob. There is no way back from that situation unless Bob travels a similar journey. Then two years would have passed (on Earth) but each of them would only have gained one year (by arriving a year later but only being a day older). If both of them agreed before the journey that they would put a raisin in a barrel each day they were apart then when Alice returns she will have deposited one raisin while Bob will have deposited 365. Since though theoretically she has gained an extra year of life (compared to what she would have had if she hadn't traveled) she can thus add the other raisins with that year to make the same amount as Bob. I think I would be sad though if I came back after a day to find all my friends were a year older.

N.B. It might be possible to substitute sultanas for raisins without affecting overall result

The basic answer here is that then Alice is looking at Bob she will perceive his time as going slower than hers, except precisely at the point where Alice turns.

At the turn (assuming she instantly change her direction, not necessary but makes things easier) Alice is changing her frame of reference in such a way that she considers a much oder Bob to the simultaneous bob to compare to. Then when she travels back, that older Bob, now ages slower that her again an the paradox is resolved.

What Alice actually will see (as the light carrying the image of Bob reaches her in space) is that as when leaves earth, Bob is aging incredibly slowly as she is basically trying to outrun the light from earth . Then when she turn around and comes back, she will see Bob aging in an accelerated pace, as she is now meeting up with all the light that is still being emitted from Bob. This accelerated aging is so fast that bu the time Alice returns to earth, Bob will be much older that she is.

This isn't necessarily the answer to the question but possibly a Verification/Clarification of it.

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In the above example in section A, I have three clocks positioned in roughly the same area, and they are all synchronized. In section B you see that two of the clocks have moved off to the sides. In section C, you can see that the two clocks have moved back to the middle position.

So let's imagine that this continues and thus the two clocks keep moving off to the edge, then back to the middle, then off to the edge, and then back to the middle, and so forth. Now due to this ongoing back and forth motion, these two clocks shall be ticking slower than the one lower clock that remained still.

And so we could watch this phenomenon going on and on and on, but over time we would begin to see that the two upper clocks are ticking at a slower speed, and thus do not indicate that the same amount of time has passed as does the lower clock which remained stationary in our observers frame of reference.

So the point here is that despite the fact that these two clocks, which are going off and then coming back, each of them being somewhat like a twin moving off to a star and then coming back, as we watch them these clocks still remain here with us in the present time from our point of view. Meaning there's no separation in time from our point of view.

All three clocks always remain with us in the present time, or the "NOW" time.

And so the question arises, how is it that all 3 clocks managed remain constantly within the "NOW" time, even though 2 of them are moving through time at a different rate than is the remaining clock.