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Time dilation & symmetrical twin paradox variant

  1. Apr 12, 2010 #1
    I'm still trying to understand time dilation, the twin paradox, and the effect of acceleration.

    Yes, I've read the twin paradox FAQ, but it only gets me so far.

    Here's a hypo I came up with. I've tried to simplify it by (a) making the observers symmetrical and (b) eliminating the distance between them. Any help in working through this would be greatly appreciated:

    Outside any gravitational field is a long, narrow train platform. There are two trains, one on each side of the platform, on parallel tracks. Each train is ten light-years long. The trains are initially at rest with respect to the platform and each other. The trains have clocks on the outside and a set of red and green lights, visible from the other train. All the clocks are initially synchronized. The trains are filled with observers.

    The trains spend one day accelerating in opposite directions. At the end of the day, they've each reached 1/2 the speed of light, with respect to the platform, and stop accelerating. At this time each each train flashes its green lights. The trains continue traveling at 1/2 the speed of light, with respect to the platform. After X days have passed (measured internally, on board each train) the trains then flash their red lights and spend one day "decelerating" (with respect to the platform) and come to rest (with respect to the platform and each other). Thus, the only difference between the two trains is that they traveled in opposite directions, with respect to the platform. Otherwise, they are symmetrical (unlike the classic twin paradox).

    I assume that, during the experiment, observers on one train, looking just across the platform, will observe the clocks on the other train to run at a slower pace than the clocks on their own train, thus creating a perceived "lag" that grows with time. I also assume that, at the end of the experiment, both sets of clocks will display the same time, regardless of how long X is (because, after all, the trains are symmetrical in all except direction).

    So how is the paradox resolved here? In other words, if the clocks of the "other" train appear to run slower to each set of observers during the experiment, how can they be synchronized at the end? Would the observers on one train see the other train's green lights (signaling the end of acceleration) before, after, or at the same time as their own train stopped accelerating? What about the red light (signaling the beginning of "deceleration")? I assume the observers would perceive the other train to come to a stop (with respect to the platform) at the same time as their own train came to a stop (or would they)?

    Also, I didn't say how long X was. If the trains spent 5 years traveling 1/2 the speed of light with respect to the platform, then I assume the the perceived time "lag," prior to "deceleration" would be greater than if they spent only 5 days traveling at that speed. But the value of X should have no effect on acceleration (or whatever effect - simultaneity, "frame jumping," etc., one uses to explain how the clocks end up synchronized). How is it that whatever effect precisely "cancels out" the perceived velocity-related time "lag" manages to do so regardless of X (and, therefore, regardless of the size of the "lag")?

    Thanks in advance.

    - Tom
     
  2. jcsd
  3. Apr 13, 2010 #2

    George Jones

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    This is a common misconception. In reality, the phrase "a moving clock runs slow" does not necessarily mean "a moving clock is visually seen to run slow."

    Consider the following simpler example.

    Assume that Alice is moving with constant speed directly towards Ted. When Ted uses his telescope to watch Alice's wristwatch, he sees her watch running at a faster rate than his watch. Ted sees Alice's moving watch running fast, not slow! Ted sees this because of the Doppler shift. Because Alice moves towrds Ted, the light that Ted sees from her watch is Doppler-shifted to a higher frequency. But the rate at which a clock or watch runs is like frequency, i.e., a second-hand revolves at a certain frequency, and all frequencies are Doppler-Shifted.

    To explain what "A moving clock runs slow." means, I first have to explain how Ted (with helpers Bob and Carol) establishes his frame of reference.

    Starting from Ted, a series of metre sticks, all at rest with respect to Ted, are laid end-to-end along the straight line oining Alice and Ted. At each joint between two consecutive metre sticks, a small clock is placed. The metre sticks and clocks are at rest with respect to Ted. Initially none of the clocks are running; before turning them on, the clocks have to be synchronized. To do this, Ted directs a laser pointer along the line joining Ted and Alice, and then sends a flash of light. Each clock is turned on when the flash of light reaches it. The speed of light is not infinite, so the time taken for the light to travel from Ted to each clock has to be taken into account. To do this, the clocks' hands are set initially as follows. The clock one metre away from Ted is set to the time taken for light to travel one metre; the clock two metres away from the tower is set to the time taken for light to travel two metres; ... .

    This whole setup of metre sticks and clocks establishes Ted's reference frame.

    Now, As Alice moves toward Ted, Ted uses his telescope to watch Alice's wristwatch, and to watch his clocks. First, he watches one of the distant clocks in his reference frame. The time he sees on the clock is the time at which the light he views set out, so Ted sees an earlier time on the distant clock than he sees on his wristwatch. He does, however, see the distant clock running at the same rate as his watch. Similarly, Ted's sees all the clocks in his frame running at the same rate as his watch.

    As Alice approaches Ted, she whizzes by clock after clock of Ted's reference frame. Using his telescope, Ted sees that Alice is beside a particular clock, and he notes the time on her watch and the time on the clock adjacent to her. Some time later, Ted sees Alice beside a different clock, and he again notes the time on her watch and the time on the clock adjacent to her.

    Ted checks his notes, and he finds that the time that elapsed on Alice's watch as she moved between these two clocks of his frame is less than the difference of the readings of the two clocks. This what is meant by "A moving clock runs slow."

    For a quantitative example, have a look at

    https://www.physicsforums.com/showthread.php?p=2186296#post2186296.

    Can you qualitatively apply this to your train example?
     
  4. Apr 13, 2010 #3
    George,

    Thanks for your reply. I was hoping to take the doppler effect out of the equation in my hypo, by having the observers look straight out their train windows, not ahead or toward the rear, so that at any given instant, the clock they were observing was traveling laterally, not advancing or receding.

    But I like your hypo too, so I hope you don't mind if I piggy back off of it:

    Ok, I think I get what Ted sees: Alice's watch moving fast, according to him, but according to his notes, still slower than his own clocks.

    But let's say that Alice decides to take her own notes. At the exact moment she whizzes by the first of Ted's clocks, she notes the time on Ted's clock and the time on her own watch. She does this for each subsequent clock. When she reaches Ted, the two compare notes.

    Will Alice's notes correspond with Ted's? In other words, will Alice's notes indicate that Ted's clocks appeared to be running faster than her watch? If the answer is "yes," that would seem to violate the principle that "a moving clock runs slow," since Ted's clocks are moving with respect to Alice. But I have a hard time understanding how the answer could be "no." Both made observations about a clock and the watch of a person who was, at that time, directly next to the clock. While the light from the image took time (and progressively less time) to reach Ted, that shouldn't affect the content of the image.

    Am I missing something here?
     
  5. Apr 13, 2010 #4
    You will still detect a Doppler effect, it is called "Transverse Doppler Effect".
     
  6. Apr 13, 2010 #5
    Okay, so in terms of the train hypo, the Transverse Doppler effect would just seem to add to paradox. It's a reason for the clocks to appear to run even slower ("redder") than they would under time dilation alone. Or is "Transverse Doppler Effect" just another way of saying "pure time dilation, uncomplicated by any Classical Doppler Effect?"

    Either way, I still have the problem of not understanding what causes the clocks to regain apparent synchronicity by the end of the experiment. Assuming that the "other" trains clocks appear to run slower during the unaccelerated portion, due to time dilation, transverse doppler effect, or both, and assuming that they clocks all display the same time at the end of the experiment, because the trains are symmetrical, then something must cause the clocks to run faster in the accelerated portion, and to a sufficient degree to offset them running slower during the unaccelerated portion. But what causes the clocks to appear to run faster? And how does it succeed in offsetting the slowing effects of time dilation/transverse doppler effect, regardless of the length of time the clocks were subject to this apparent slowing?
     
    Last edited: Apr 13, 2010
  7. Apr 13, 2010 #6

    Janus

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    So if I'm reading this right, you are talking about the observer looking out and reading the clock that is directly opposite them in the other train at that instant. In that case, he will always note that the opposite clock reads exactly the same time as his own.

    How is possible when we know that the observers is each train will determine that the clocks in the other train run slow?

    There is another effect that has to be accounted for; the Relativity of Simultaneity. What this effect means is that, while the train's are moving relative to each other, each train will determine that the clocks in the other train are not synchronized. The clocks in the tail of the other train will be ahead of the clocks at the head of the train.

    What this means is that the clocks "ahead" of you in the other train all show times that are ahead of your own. However, they run slower than yours, so that when they reach a point just opposite of your own, they read the same as yours. As the difference in the relative speeds of your train decreases to zero, the other clocks drift back into sync, until at zero relative velocity, they are all in sync again and read the same as yours.
     
  8. Apr 14, 2010 #7

    Ich

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    No, the opposite clocks go "faster", i.e. read increasingly more time than the own clock.
    That's exactly what "moving clocks run slow" means: a moving clock shows less time than the clocks that show coordinate time in the "stationary" frame. (I think that's also what George Jones said.)
    The effect is reciprocal.
    If you know the Lorentz transformations (the basic ingredient of SR, time dilation is just a derived effect under some strictly defined circumstances): It's the difference between constant x (measurement a Ted's position) or constant x' (measurement at Alice's position).
     
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