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Time dilation and space ships

  1. Apr 15, 2014 #1
    This has bothered me for a long time. I understand that time varies with velocity relative to the observer. But what about this case -- Suppose we have 3 space ships: A, B, and C. They are spaced widely apart with B in the middle and A and C at opposite sides to B. A and C are moving towards each other at relativistic speed (equal but opposite) and B is sitting still. How are the clocks varying? I can see that B sees A'a and C's clock slowing down, but what about A vs C? For example, from A's pov, B and C are moving towards him with B going half as fast as A. So according to A, B and C's clocks are running slow. But C thinks the same thing about A and B. So everybody is thinking every one else's clock is the slow one. How is this resolved?
  2. jcsd
  3. Apr 15, 2014 #2
    You just accept it. :)

    What you observe will depend on your frame of reference and you should not expect observers in different frames to observe the same thing.
  4. Apr 15, 2014 #3
    I should elaborate... Consider this analogy I stole from pop-science,

    If I hold a ruler straight out in front of me vertically, I see its length is 12 inches. If you look at it from an angle below it will not look to be 12 inches, it will appear to be smaller. If you look at it from nearly underneath it will be nearly zero inches in length. The length you observe the ruler to be is dependent on what angle you are looking at it. There is no reason to expect it to appear 12 inches long to everyone at any angle, you would not expect that because you have an intuitive notion of the euclidean geometry involved.

    In relativity we now use a different geometry to describe lengths and times. In this geometry we do not expect lengths and times to be the same for any observer. It depends on your frame of reference. This is non-intuitive, but it is the case. The only situation where we expect different observers to observe the same times is if they are in the same reference frame (analogous to the only case where we expect different observers to observe the same ruler length is if they observe it at the same angle).
  5. Apr 15, 2014 #4


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    Yes, that is correct. The point is that there is no frame-invariant meaning to the rate of a clock. So the statement "Fred's clock is slow" is incomplete and doesn't mean anything as stated. You have to say something like "Fred's clock is slow in Alice's frame".
  6. Apr 15, 2014 #5


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    Let us suppose that all of these inertial observers meet at event O.

    On (say) A's spacetime diagram,
    can you locate the event on each observer's worldline
    that marks "one tick since O" according to that observer's clock?

    Those events lie on a hyperbola with lightlike asymptotes through O (that is, its asymptotes are on the light-cone of O).

    For each observer, through his one-tick event, draw the tangent-line to the hyperbola.
    That tangent line marks "the same instant of time [according to that observer]".
    Later instants of time according to that observer are parallel to and "above" that tangent line [taking time running upwards].

    Now, note that every other observer's one-tick-event lies above that tangent line... meaning that every other observer's clock takes longer to tick once compared to his own clock.

    These conclusions would be obtained by every such inertial observer who met at event O.
  7. Apr 15, 2014 #6


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    This sounds paradoxical, but the paradox goes away if you think carefully about exactly what it means to say that one clock is running slower than another, and remember the relativity of simultaneity.

    Suppose the hands of both clocks point to noon at the same time (and of course we'll call that time "noon"). I watch one of the clocks until its hands point to 1:00 PM, and then ask where the hands of the other clock are pointing at that moment. If the hands of the other clock are pointing to 1:00 PM at the the same time that the hands of my clock are pointing to 1:00 PM, I'll say that the two clocks are running at the same rate. If they point to somewhere before 1:00 PM, I'll say that the other clock is running slow, and if they point somewhere after 1:00 PM I'll say that the other clock is running fast.

    Note the bolded text above - this entire concept of one clock being slower or faster than another depends on everyone agreeing about what is and is not happening at the same time.

    Because of the relativity of simultaneity, events that happen at the same time for one observer do not necessarily happen at the same time for another. In this case, if the two clocks are moving relative to one another, I may say that at the same time that my clock reads 1:00 PM the other clock reads 12:55 PM so is running slow. However, it does not follow that the other observer will say that my clock reads 1:00 PM at the same time that his clock reads 12:55 PM; instead and because of the relativity of simultaneity, he may say that at the same time that his clock reads 12:55 PM my clock reads 12:50 PM and therefore my clock is running slow relative to his.

    We're both right, because things that are "at the same time" for one of us are not necessarily "at the same time" for the other.
    Last edited: Apr 15, 2014
  8. Apr 15, 2014 #7
    OK, so let's simplify this a bit. Let's say there are only 2 ships. They start out next to each other and synchronize their clocks. Then they move apart for some distance at equal speeds, coast for a while (no acceleration), then turn around and come back together and compare clocks. What will the clocks read? The rules being that each experience the same acceleration (but in opposite directions) during the experiment. Once again each ship sees the other moving relative to himself, so he expects their clock to be slow. Each ship starts and ends their trip in the same frame of reference as their partner so that's no longer a consideration.
  9. Apr 15, 2014 #8


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    In this case, the clocks will end up the same. The flaw is that 'a moving clock is running slow' is only true in inertial frames. Neither rocket remains stationary in any given inertial frame over its trajectory. There is no unique, preferred way to set up a non-inertial frame in which one of the rockets remains at rest. However, in any of the many valid ways to do this, the other rocket's clock is not considered to run slow all the time. It will run fast part of the time, and end up matching the clock of the rocket at rest in the non-inertial frame, when they meet again.
  10. Apr 16, 2014 #9


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  11. Apr 16, 2014 #10


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    Hmmm... You're going from two ships moving at constant speeds in straight lines to two ships changing their speeds and directions... This is a very different problem. Are you sure that you're simplifying things? :smile:

    Actually you have made one simplification: because now we only compare clocks when they are at the same location we no longer have to worry about relativity of simultaneity. Instead of asking "what does his clock read at the same time that my clock reads noon?" we ask "what does my clock read and what does his clock read when the two clocks meet again?"

    However this simplification is somewhat offset by the increased complexity that comes with not working in inertial frames. There is no inertial frame in which either ship is at rest, so there is no simple inertial coordinate system that we can use. Instead, we have to go through a more general coordinate-free calculation like the one that tom.stoer pointed you at to calculate the time that passes on each ship between the events "they separate" and "they meet again".

    When we do, we'll find that under the rules that you've set both clocks read the same. That's sort of to be expected because of the symmetry of the situation. One starts out moving left and then right while the other starts out moving right and then left; they're mirror images so it would be surprising if there were a difference.

    If only one ship accelerated out and back while the other remained in free fall or if they did something else asymmetric, then there would be a difference in the clock readings when they meet up, and no debate about which clock would be the slow one. This is the famous "twin paradox", and google will find you much discussion of this class of problems. http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html is good.
    Last edited: Apr 16, 2014
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