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Time Dilation. The faster you travel the longer I have to wait for you to return?

  1. Dec 5, 2012 #1
    Time Dilation problem

    I understand Time Dilation and most of the principals involved. However I am still stuck on this one lingering question that I can’t make sense out of. If anyone could answer this I would really appreciate it.

    Given:
    1. Person A is the traveler

    2. Person B is stationary

    The next two are calculated via the Lorentz Transformation

    3. If Person A travels at 50% the speed of light then one year for Person A is equivalent to 1.15 years for Person B

    4. If Person A travels at 99.99% the speed of light then one year for Person A is equivalent to 70.71 years for Person B

    Here’s the lead up to my question:

    Person A is going to the best burrito shop in the galaxy located on Planet X. This planet is exactly one light year away. He will bring back two burritos for himself and Person B to eat. Let’s calculate how long Person B will be waiting.

    If Person A travels at 50% the speed of light it will take 4 years to return (two years over and two years back). By this time 4.60 years would have passed for Person B (4x1.15)

    If Person A travels at 99.99% the speed of light it will take 2 years to return (It will take a tiny bit longer than two years but this discrepancy is negligible for this problem). In this scenario 141.42 years would have passed for Person B (2x70.71).

    Which brings me to the question:
    If I send someone to get me lunch on another planet, the faster they travel the longer I have to wait?
     
  2. jcsd
  3. Dec 5, 2012 #2

    Ibix

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    When you say "it takes x amount of time for something to happen" you must always ask "x amount of time for who?". That's what has tripped you up here.
    It would take four years by whose watch?

    It would take two years by whose watch?

    What you appear to have done is calculated the travel time as seen by the stay-at-home person B, assumed that that travel time will show on person A's watch when he returns, and then calculated the time that would show on person B's watch if that assumption were correct. It's not correct - you know the time on person B's watch (4 years or 2 years) and you need to calculate the time on person A's watch.

    Both the distance travelled and the time taken are different for person A and person B. According to B, who stays at home, Person A goes out at 0.5c and returns in four years (he had two light years to cross and t=d/v) but Person A's watch will read γt=0.866×4=3.46 years. According to person A, the burrito shop travelled to him at 0.5c, but it only travelled γd=0.866×2=1.73 light years, so he has no problem with his watch coming up at less than 4 years.

    You can do the calculations for 0.9999c. There will be a bigger discrepancy between the watches, but the same thing will happen as above.
     
    Last edited: Dec 5, 2012
  4. Dec 5, 2012 #3
    Thanks for the response IBIX. Sorry, but perhaps I didn't make this one point clear enough. Planet X is Exactly one light year away.


    So if Person A travels 50% the speed of light to a planet that is one light year away it will take him two years to reach the planet and two years to return.

    1. If Person A travels the speed of light then it will take that person one year to reach the planet and one year to return. Same as if Person A was traveling 99.99% the speed of light minus the negligible difference.

    2. Now that I know the exact amount of time that has passed for Person A, I can calculate the amount of time that has passed for stationary Person B

    3. If Person A travels at 50% the speed of light then one year for Person A is equivalent to 1.15 years for Person B

    4. If Person A travels at 99.99% the speed of light then one year for Person A is equivalent to 70.71 years for Person B

    Are any of these four "givens" incorrect?
     
  5. Dec 5, 2012 #4

    Doc Al

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    Careful. It is person B who sees person A as traveling at 50% the speed of light. So, according to person B's clocks the trip takes 4 years. According to person A the trip is shorter.
     
  6. Dec 5, 2012 #5

    Nugatory

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    I've made the necessary additions in boldface.

    You have it backwards. You know the time that passed as measured by person B, and you can calculate what person A, the traveller, experiences.
     
    Last edited: Dec 5, 2012
  7. Dec 5, 2012 #6

    Ibix

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    Number 1 is wrong. If planet X is exactly 1ly away according to the stay-at-home person, it is 0.866ly away according to the traveller - length contraction at work. It is the stay-at-home whose watch shows 4 years when the traveller returns - the traveller's watch shows less time. You have the time dilation factor correct, but are applying it backwards here.

    If I may, I think that you haven't quite grasped special relativity quite as well as you think. You keep saying things like "the distance is x" and "the time is t" without specifying whose distance or whose time. The key point about SR is that two people who are not at rest with respect to each other do not, in general, agree on positions, lengths, times, and a host of other quantities. Saying "planet X"is 1ly away is not enough - you must say 1ly away according to person B (or whoever).

    I have been assuming that you mean planet X is 1ly away according to the stay-at-home person B, because he is the only constant in your problem. But you didn't specify, and it is that lack of clear thinking about who exactly is measuring what that is leading to your confusion.

    Edit: I obviously type too slow. Nugatory and Doc Al have rendered my response... er... nugatory.
     
  8. Dec 5, 2012 #7
    OK Thanks Doc, Ibix and Nugatory. You guys helped clear up my confusion. I understand where I was making the mistake now.
     
  9. Dec 8, 2012 #8
    Btw, the key point why A and B measure different traveling times is not really the speed at which one of them travels (because from A's perspective it's B who is moving and from B's perspective it's A who is moving, so the situation is symmetrical). The crucial difference is the acceleration from one speed (B's) to another (A's first acceleration away from B, then after a time a new acceleration back towards B). The acceleration causes a change in frames of reference.

    (This is the answer to the so-called twin paradox.)
     
  10. Dec 8, 2012 #9

    ghwellsjr

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    It's true that B accelerates but that doesn't mean there must be a change in frames of reference in order to answer the so-called twin paradox. You can use any Inertial Reference Frame (IRF) and get the correct answer. Here are three IRF's based on the OP's first scenario at 50% of light speed to show you what I mean.

    First, in the IRF in which both A and B start out at rest. B (black) travels at 50%c for 1 light-year and then returns. Note that it takes him 21 months according to his clock to get to Planet X and another 21 months to return. Since his speed in this IRF is 0.5c, his time dilation 1.1547 which means that the dots marking off each month are spaced slightly farther apart than the coordinate months. Meanwhile, A (blue) has aged by 48 months. (I have made B travel exactly 21 months to get to Planet X which is a little long and so A has aged a little long also.) B's total aging is 42 months while A's is a little over 48 months.

    attachment.php?attachmentid=53734&stc=1&d=1354960880.png

    That is all that is necessary to explain the scenario. However, we can transform the entire scenario into another IRF and show it all over again. Here is the IRF in which B is stationary during his trip to Planet X. During this time, he is experiencing no time dilation and his months proceed in step with the coordinate time. However, A is traveling away from him at 0.5c according to the IRF so his time is dilated by the factor 1.1547. After 21 months, B turns around and heads for home. However, he now has to travel at 0.8c according to the IRF so his time is now dilated by 1.6667. When he arrives back home, A has aged slightly more than 48 months and B has aged 42 months, exactly like in the first IRF.

    attachment.php?attachmentid=53735&stc=1&d=1354961278.png

    Now we can do another transformation and see what the scenario looks like in the IRF in which B is stationary for his return trip:

    attachment.php?attachmentid=53736&stc=1&d=1354961935.png

    This looks very much like the previous IRF and the numbers apply in a similar manner so I won't go into the details.

    Please note that each IRF is self-contained and explains everything about the scenario. Each IRF can easily handle the three accelerations that B experiences without the need to jump between frames or use a non inertial reference frame. If you want to claim that these IRF's do not explain the OP's scenario (or the so-called twin paradox) then I invite you to describe what you think is wrong with them and then produce your own detailed explanation and hopefully a plot to illustrate your explanation.
     

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  11. Dec 8, 2012 #10

    K^2

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    The point on acceleration is to answer the question, "Why can't we use the frame in which B is stationary all the time?" Answer being that you can, but it's not an inertial frame, so SR is insufficient to describe it. But any inertial frame should certainly give you consistent answers with SR alone.
     
  12. Dec 8, 2012 #11
    It depends on what you call "faster". "Faster" can mean "great speed" or "greater acceleration." So there is a clearer way to present the same idea.

    The more they accelerate, the longer I have to wait.

    Here, acceleration refers to the dynamic acceleration which is force on the traveler divided by mass of the traveler. Also, deceleration is equivalent to acceleration. The issue comes down to the force on the traveler.

    In the case of the twin scenario, it turns out to be the same thing. If the rocket travels with great velocity, then it has have a very large acceleration toward earth in order to turn around.

    For me, the weird thing was this is a nonlocal effect. Weird= antiintuitive. Basically, the person in the rocket ages the one on earth merely by using his rockets to turn around. Although this is weird, it logically works out. It seems a little strange that the twin by running his rockets a large distance away makes the twin on earth instantly age faster. However, this is an illusion in the following sense. The speed of light is the fastest possible speed in the universe. Therefore, the twin in the rocket can't immediately know that the twin aged faster while the rockets were running.

    Basically, the traveler which has the biggest impulse ages the slowest. Impulse equals integral of force with time = change in linear momentum due to force. The earth twin never has a force applied to him, so he ages fastest.

    The Lorentz time dilation formula is expressed in terms of the speed of the traveler. However, hidden in the derivation of this formula is the dynamic acceleration. The "logical adjustments" in the physics occur while the twin in the rocket is turning around.

    Another problem is with the twin scenario itself. To simplify the concepts, the twin conundrum is expressed in the extreme limit of an instantaneous turn around. This makes the math a lot easier, compared to if it was a slow turn around. However, an instantaneous turn around is impossible. An instantaneous turn around would squash the traveling twin into a quark plasma! The acceleration is important but hidden in the set up of the problem.

    The problem with a slow turn around is it takes more time. The traveler feels less force, but he feels it over a much longer time. The two effects cancel out so that the Lorentz contraction still works, even with the slow turn around.

    The physical symmetry is broken by the impulse on the traveler. The observer experiencing the greatest impulse ages slowest.

    There are some physicists who state that there is no acceleration in relativity. However, they mean it in a very technical sense. The Lorentz time dilation formula doesn't explicitly contain acceleration. It contains only speed. However, the dynamic acceleration is subtly implied.
     
  13. Dec 8, 2012 #12

    Ibix

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    I'm almost afraid to start this.
    I'd love to see the maths supporting that statement, because it seems to me to be incorrect. If I send out two people to get burritos, and one accelerates harder than the second then he will always be ahead of the second guy, and will return sooner.

    I'm fairly certain this isn't true, either. Consider the cash going out to the burrito shop floating free, and the burrito coming back the same way. According to me, the cash takes t1=d/v to get to the shop and the burrito takes t2=d/v to get back. According to a clock travelling with the cash, it takes t1'=d/(γv) to get to the shop; according to a clock travelling with the burrito it takes t2''=d/(γv) to get back to me. My total wait is 2d/v; the total travel time of the cash and burrito is 2d/(γv).

    The key point here is that the acceleration phase of the cash and the burrito is completely irrelevant - it happens outside the scope of the experiment and can be as hard or as gently as you like. You can even imagine a universe empty except for me, the burrito shop and some cash and a burrito created (fiat burrito) at the correct velocities. You still find that the total elapsed time for the travellers is less than that for the stay-at-home.

    It isn't the acceleration that "causes" the difference. For a single observer to follow the path of the cash-and-burrito, they do need to accelerate, but the difference in elapsed time is due to the two observers following unequal length paths through spacetime. A good analogy is me driving from A to B in a straight line, while you drive from A to C to B. Certainly you can't get to B without making a turn at C, but that turn doesn't "cause your path to be longer"; it just is longer because it's not the direct route.
     
  14. Dec 8, 2012 #13

    K^2

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    It happens to be wrong in OP's example, but don't just assume things are always going to be so clear cut.

    Imagine that two people on a space station orbiting a planet want to go get burritos that are sitting on a station half a revolution ahead. They both get into their shuttles. One gives a short burn forward to accelerate, the other gives a short burn in reverse to slow down. About half way, they adjust their orbits again to intercept the station with burritos. Which one gets to burritos first? One that initially slowed down. Why? Because he took a shorter, faster path.
     
  15. Dec 8, 2012 #14

    Ibix

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    I wasn't assuming, although I should have explained my reasoning, which is as follows. In a flat space in the low-velocity limit, Darwin123's contention is obviously wrong. As far as I am aware, the acceleration observed in an inertial frame when a body undergoes proper acceleration of a' is a=a'/γ3 - and that doesn't let a slower-accelerating body overtake a faster accelerating one either.

    I know enough orbital mechanics (including that result, as it happens) to know that intuition is risky (at best) there. I dread to think what general relativistic orbital mechanics looks like.
     
  16. Dec 8, 2012 #15
    Round trip time and turn around time are different. Round trip time is the total time the round trip takes. The turn around time is the time the external force is working on the observer. In the twin conundrum, the external force on the traveling twin is the thrust of the rockets.

    The so called paradox is in round trip time. Newtonian physics says that the round trip time is the same for both twins. Relativity says that if the relative speed of the two is ever close to the speed of light, then the round trip time is longer for the twin on earth. This is what we call an asymmetry.

    There is no asymmetry in your burrito example. In your burrito example, at low speed, both observers have to wait the same amount of time during the round trip. The one who went out to get the burrito is just as old as the one who waited for the burrito. Acceleration shortens the round trip time the same way for both observers.

    A higher nonrelativistic velocity may shorten the turn around time for both. However, the time it takes to turn around, which was defined in terms of force, will be longer if the one who goes out is running. Newton's first law says that a body in motion tends to stay in motion unless acted on by an outside force. In order to turn around, he has to accelerate in the opposite direction. He can either turn around suddenly with great force or turn around slowly with less force. The time is the same for both.

    At high speeds close to the speed of light, the round trip time is different for both observers. The time that the rocket engines are on gets longer for both observers if the twin in the rocket starts moving at high speed. However, the asymmetry between observers is made while the rocket engines are on.

    The rocket engine does something in relativity to the clocks and rulers of the accelerating twin that isn't done to the clocks and rulers in Newtonian physics. The rockets make an asymmetry that can only be seen after the round trip.
     
  17. Dec 8, 2012 #16

    K^2

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    Surprisingly benign, so long as you stay above 3rs.
     
  18. Dec 9, 2012 #17

    Ibix

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    Ghwellsjr's spacetime diagrams already demonstrate that this isn't true. Let his blue line be the worldline of the stay-at-home, one of the black legs be the worldline of the outbound cash, and the other black leg be the worldline of the inbound burrito. His diagrams accurately describe my scenario, and the proper time experienced by the cash on its trip plus the proper time experienced by the burrito on its trip total less than the proper time of the stay at home person, with no acceleration involved.

    I note that you haven't posted maths or references to back up your claim that greater acceleration leads to a longer wait time for the stay at home twin. It seems wrong to me for reasons I articulated in my response to K^2. Could you let me know (with maths) what it is that you think happens? What is the parameter that determines when the Newtonian result (higher acceleration gets there in less time) gives way to your regime?
     
  19. Dec 9, 2012 #18

    Mentz114

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    It looks very good. This paper has the maths and some good orbital plots.

    :http://arxiv.org/pdf/1201.5611v1.pdf
     
  20. Dec 9, 2012 #19
    Most of these mathematical proofs using GR rely on one path being a geodesic and the other not a geodesic. The twin on earth is traveling a geodesic, which is the shortest path in space time between two point. The traveling twin is not on a geodesic. His path is not on the shortest path through space time.

    Ghwellsjr's spacetime diagrams would not be satisfying because the diagram couldn't show the symmetry breaking mechanism. This is a case where a word could be worth a thousand pictures. Even if the analysis was mathematical, what would be useful is if the equation with the symmetry breaking mechanism was presented. At least one would get an idea of how the symmetry breaking occurred.

    The problem is that there is no physical hypothesis that distinguishes between trajectories that are geodesics and trajectories that are not geodesics. One can always find a set of variables in which an observer is traveling a geodesic. After all, the physical laws are independent of the path of the observer. "Everything" is relative, or so people have been told.

    What many people are asking is not whether there is mathematics that distinguish one twin from the other. They are asking for a physical hypothesis that distinguishes between the two observers. They want to know the "symmetry breaking" feature in the calculation. "Physical intuition" is not sufficient for distinguishing between a geodesic and a nongeodesic.

    Many books on science for laymen say straight out that it is the rockets that break the symmetry. I am just generalizing what these books in "mainstream science" say. Instead of "rockets", I say "external force". What you seem to be saying is that the rockets have nothing to do with the twin on earth aging faster than the traveling twin. What is asked for is a physical hypothesis, not a mathematical proof. The mathematical proof is certainly worth a discussion on its own, but it is irrelevant here.

    There are many problems in physics where a symmetry is broken. The question of what interaction physically breaks a symmetry is often a valid scientific question, regardless of scientific field.
     
  21. Dec 9, 2012 #20

    K^2

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    Longer path would mean longer proper-time, which would mean that twin that traveled aged more than the twin that remained in one place.

    The fact is, twin that traveled has actually taken a shorter path. How? Geodesics are local minima. In fact, they don't even have to be minima, merely extrema, but I don't know if that's ever relevant in GR. At any rate, there can exist paths that are shorter.

    For a simple analogy, think of an object in a glass of water placed in such a way that you can see the object both through the wall of the glass and through the surface. It looks like there are two copies of the object, one distorted more than the other. I'm sure you've seen this. In optics, light takes the "shortest" path as well. Fact that you can see two images of the same object tells you that there are two "shortest" paths between the object and your eye. Again, the path only needs to be locally shortest. Meaning that any small perturbation of the path has to increase the length.



    Anyways, back to the twins. I'm not entirely sure whether it's fully equivalent to engine thrust, but imagine that the twin that traveled used gravity of a massive object, like a black hole, to turn around. That way, his trajectory is also a geodesic. It is also locally shortest. However, the question of which path is shorter remains.

    We know the answer, of course. Special Relativity tells us that the twin that stayed put aged more. That means, his path was longer.
     
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