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simple question regarding the twin paradox |
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| May27-12, 01:21 PM | #1 |
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simple question regarding the twin paradox
Hi,
I am a little confused with this paradox. I asked my professor about it and he didnt really give a convincing answer. So the scenario basically seems to be some twins on earth(or anywhere) at rest, and then one leaves at relativistic speed for some time then comes back to see that his/her twin is much older than them. My question is, how come you can tell which one would age more? Why couldn't it just as well be the one on the ship? From the twin on earths reference frame, they are at rest and then the rocket flies away from them, while in the rocket frame it is at rest and the earth flies away from them. From each of the twins perspective the other one moves and they are stationary in their own frames. How come the same thing wouldn't happen to the twin on earth and find the rocket twin older when the earth arrived back at the rocket? I just took this paradox for granted for a long time but now I seem to be confused. Thanks for reading. |
| May27-12, 01:41 PM | #2 |
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In order for twin on the rocket to return they must undergo acceleration, and therefore do not have an inertial frame of reference during the entire round trip.
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| May27-12, 02:03 PM | #3 |
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You need to pick one inertial frame and stick with it from start to finish. And you need to understand that the faster you go in that frame, the slower your clock ticks.
Now can you see that from the earth's frame, only the rocket twin's clock will run slow? And can you see that if you use the rocket's frame during the first half of the trip, only the earth's clock will run slow but during the last half of the trip, the rocket has to travel much faster than the earth in order to catch up with it and so its clock has to run even slower such that it ends up with less time on it when it gets back to earth? |
| May27-12, 02:08 PM | #4 |
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Blog Entries: 6
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simple question regarding the twin paradox
If you are at rest in a given inertial reference frame then you do not feel any proper acceleration. When the travelling twin turns around to head back he experiences proper acceleration and he can now be certain that he is not at rest in inertial reference frame. The stay at home twin does not feel proper acceleration during the turn around event, so the situation is not symmetrical.
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| May27-12, 02:45 PM | #5 |
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Recognitions:
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There is a simple way to understand it based on the Lorentz transformation. Say the traveler goes to a place ten light years away from earth. He quickly speeds up to almost the speed of light. The distance in his frame to his destination is now foreshortened and the time to get there is is greatly reduced. Once he gets there, stops and turns around to go back to earth, the same foreshortening of distance and time takes place. So in his frames the total time would be a lot less than twenty years, while the earth bound twin would age more than twenty years.
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| May27-12, 09:35 PM | #6 |
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Visceral, you might want to read the Usenet Physics FAQ entry on the Twin Paradox:
http://math.ucr.edu/home/baez/physic...n_paradox.html There are a number of different ways of understanding what's going on in this scenario (most of which have been mentioned in this thread), but the FAQ entry ties them all together. My personal preference is to look at everything using a spacetime diagram (the FAQ entry discusses this in some detail). Looking at it this way makes the solution obvious, in my view: you have two twins, each of which takes a different path through spacetime between the same pair of events (the event where they part, and the event where they come back together). The two paths they take have different lengths, so they experience different amounts of proper time passing between those two events (since the "length" of a worldline is just the proper time elapsed for someone following that worldline). Working out the actual math tells you that the stay-at-home twin's path is longer, so he ages more and is older when the two meet up again. It's no different in principle than the fact that two paths through Euclidean space that start and end at the same point can have different lengths; it's just geometry. |
| May28-12, 01:19 PM | #7 |
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| May28-12, 02:01 PM | #8 |
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| May28-12, 02:26 PM | #9 |
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Well, actually, the word "paradox" has multiple meanings including:
Source: Collins Concise Dictionary, 4th Ed 1999 |
| May28-12, 02:48 PM | #10 |
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Well, the OP said he suddenly thought to himself, why isn't this story symmetrical -- why is one twin different from the other?? But that's exactly the point of the story, isn't it? That's why it is the 'twin paradox' not the 'twin effect' or some such. As DrG points out, it seems contradictory, but it isn't.
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| May28-12, 04:11 PM | #11 |
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If the traveler twin, on both his outgoing trip and on his returning trip, says that his home twin is aging slower than he is, then how can he find the home twin to be older when he finally gets back?
I think the answer is that the traveler twin says his home twin ages a lot during his turnaround. |
| May28-12, 04:50 PM | #12 |
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| May28-12, 05:06 PM | #13 |
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| May28-12, 07:10 PM | #14 |
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http://math.ucr.edu/home/baez/physic.../twin_gap.html ...which says The key is that there is not a single unique "right answer" for most of these questions; how you answer them depends on how you interpret various observer-dependent quantities. The only question that has to have a unique answer is, how do the two twins' ages compare *when they meet again*. That answer is unique because both twins are at the same location at the same time, so none of the ambiguities in interpretation come into play. |
| May29-12, 02:44 PM | #15 |
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| May29-12, 10:52 PM | #16 |
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; but the FAQ really means "error" as in "something that has to be compensated for by adding in an additional term". In this case, what has to be compensated for is the difference in the zero of time between the two frames, i.e., the "error" is a "correction" you have to apply to convert one frame's time into the other's. The term "accounting error" is an unfortunate choice of words; the FAQ is not trying to say that the 13 year 8 month difference is an incorrect value; all 3 parts, as given in the FAQ, are correct values, and are added together correctly to give the correct final answer. It's just that one of them, the 13 years 8 months, doesn't correspond (in the Time Gap version, where the turnaround is instantaneous) to anything "observed" by Stella; it's just a change in zero of time that she has to apply when she changes inertial frames.
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| May30-12, 08:27 AM | #17 |
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so has anyone actually flown a clock in space at relativistic speeds(or really fast speeds at least - I know clocks can be very accurate tools nowadays) and then collected it to see if it's fast or slow?
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