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Very simple thought experiment re: time dilation

  1. Apr 15, 2010 #1
    Very simple thought experiment about time dilation:

    Let's say I have two clocks that are a substantial distance (light years) apart, at rest with respect to each other. The clocks are Einstein synchronized. Also, the clocks are specially designed to withstand any force or collision. However, if subject to a collision, they are designed to stop ticking, and will be "frozen" at the moment of the collision. Then I travel, in a large spaceship, past one of the clocks so that I and the two clocks are all in a line (Me, clock 1, clock 2). I then accelerate toward the clocks. I cease accelerating before reaching clock 1, and travel at a constant speed, a substantial fraction of c, directly toward both clocks. I get out my stopwatch. When my spaceship strikes clock 1, I start the my stopwatch, and maintain my velocity toward clock 2. Years later, when my spaceship strikes clock 2, I stop my stopwatch. I then pull my spaceship over (or not), get out, and peel both clocks off the front of my spaceship. Is the difference in time displayed on the clocks greater, less than, or the same as the time elapsed on my stopwatch between collisions?

    If they are unequal, then what about the principle of symmetry? To me, the clocks are moving. To the clocks, I am moving.

    If they are equal, then is "time dilation" just a visual illusion?
     
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  3. Apr 15, 2010 #2

    JesseM

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    The difference between time on the clocks is greater than the time elapsed on your stopwatch.
    Once you have stopped accelerating and are moving at constant velocity, consider how things look in your inertial rest frame. The clocks are both ticking slow relative to your stopwatch in this frame until the moment you bump them, but if they were synchronized using the Einstein convention in their own frame, they will be out-of-sync in your rest frame due to the relativity of simultaneity. Whatever time clock 1 reads at the moment you bump it, then according to the definition of simultaneity in your rest frame, clock 2 reads a time significantly ahead of clock 1 at that moment (whereas according to the definition of simultaneity in the clocks' own rest frame, clock 2 reads exactly the same time as clock 1 at the moment you bump it). This explains why the difference between clock 1 and clock 2 can be greater than the elapsed time on your stopwatch despite the fact that both clocks were ticking slower than your stopwatch from the perspective of your rest frame.

    Here's a simple numerical example. Suppose the two clocks are 20 light years apart in their own rest frame, and you are traveling at 0.8c in their frame. So, in the clocks' rest frame you will take 20/0.8 = 25 years to travel between clock 1 and clock 2, so if clock 1 reads t=0 years when you hit it, then since they were synchronized and ticking normally in this frame, clock 2 must read t=25 years when you hit it. Meanwhile in this frame your stopwatch was running slow by a factor of [tex]\sqrt{1 - 0.8^2}[/tex] = 0.6 due to time dilation, so your stopwatch only elapsed T = 25*0.6 = 15 years between hitting clock 1 and hitting clock 2.

    Now analyze the situation from the perspective of your rest frame. If the two clocks were 20 light-years apart in their rest frame, then according to the length contraction equation, they are only 20*0.6=12 light-years apart in your frame. And the relativity of simultaneity tells us that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at speed v, they will be out-of-sync by vL/c^2, with the trailing clock showing a time that is ahead of the leading clock. So, since in your frame the clocks are coming at you at v=0.8c, clock 2 must be ahead of clock 1 by (0.8c)*(20 light years)/c^2 = 16 years. Thus at the moment clock 1 reads t=0 years, clock 2 must already read t=16 years in your frame. Since the distance between them is 12 light-years in your frame, it will take 12/0.8 = 15 years for clock 2 to reach you, and your stopwatch is ticking normally so that's how much time elapses on your stopwatch. But clock 2 is ticking slow by a factor of 0.6 in your frame due to time dilation, so it only ticks forward by 15*0.6 = 9 years in this time. But since it started out reading t=16 years, by the time clock 2 reaches you it will read 16 + 9 = 25 years. So this frame predicts the same thing, that your stopwatch elapses 20 years while the difference between clock 1 and clock 2 once they are stopped is 25 years, in spite of the fact that clock 1 and clock 2 were ticking slower than your stopwatch in this frame.
     
  4. Apr 15, 2010 #3

    Doc Al

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    Greater than, by a factor of gamma.

    The "time dilation" effect is perfectly symmetric, but the set up is not. You are comparing your single clock to two spatially separated clocks. Nonetheless, you see each moving clock as running slow and as showing different times, since they are not synchronized in your frame. Observers with the two clocks see your clock as running slow.
     
  5. Apr 15, 2010 #4

    phyzguy

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    A very clear explanation JesseM. Thanks,
     
  6. Apr 15, 2010 #5
    Jesse,

    Thanks so much for your response. It made a lot of sense. I wonder if I could use your numbers to pose another question I have about simultaneity.

    So at the time I strike clock 1, if I look ahead to clock 2, I should see clock 2 display a time of 4 years. (16 years, the actual time it's currently displaying in my frame, minus 12 years, to account for the time it takes light to travel the distance between the two clocks, as that distance is measured in my frame). But if an observer who is next to clock 1, and stationary with respect to both the clocks, looks at clock 2, he will see clock 2 display a time of -20 years (0 years, the actual time it's displaying in his frame, minus 20 years, to account for the time it takes light to travel the distance between the two clocks, as that distance is measured in his frame). So even though the observer and I are at the same place (the location of clock 1), at the same time (when I strike clock 1), we each see clock 2 as displaying vastly different times.

    If I'm calculating this correctly, does that mean that, if the observer jumped on board my spaceship (please overlook the extreme g-forces involved), then that observer would instantly go from seeing clock 2 read -20 years to seeing it read 4 years? And if so, wouldn't the reverse hold true, that if, at the time I hit the first clock, I quickly decelerated my spaceship by 0.8c, then, during that deceleration, I would see clock 2 moving backwards from 4 years to -20 years? If that's true, wouldn't it be possible for an observer at clock 1's location to just keep accelerating and decelerating repeatedly, and thereby see the same event (e.g., clock 2 displays 1 year), multiple times, in forward motion and reverse motion?
     
    Last edited: Apr 15, 2010
  7. Apr 15, 2010 #6

    JesseM

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    No, that would only work if clock 2 was at rest 12 light-years away in your frame, since it's moving in your frame you'll see something different because the clock was farther away in the past when the light hitting your eyes now was originally emitted. If you define the position in your frame where you meet clock 1 as x=0 and the time in your frame when you meet clock 1 as t=0, then clock 2's position as a function of time is given by x(t) = 12 - 0.8c*t (so at t=0 clock 2 is at x=12, then 10 years later it's 8 light-years closer at x=4, etc.) Meanwhile if a light beam is coming in the -x direction and reaches x=0 at t=0, then its position as a function of time must be x(t) = -1c*t. So, the light beam reaching your eyes at t=0 must have intersected the worldline of clock 2 when 12 - 0.8c*t = -1c*t, so adding 0.8c*t to both sides gives 12 = -0.2c*t, and dividing both sides by -0.2 gives t = -60. At this moment clock 2 would have been at a distance of 12 + 0.8*60 = 12 + 48 = 60 light years away from you, so it makes sense that the light from clock 2 at that time would just be reaching you now. Since clock 2 is running slow by a factor of 0.6 due to time dilation, over the past 60 years it has only ticked forward by 0.6*60 = 36 years, so if it reads a time of 16 years now, back then it must have read a time of 16 - 36 = -20 years. So, at the moment you hit clock 1, you should be seeing an image of clock 2 reading a time of -20 years.

    What if we try to figure things out in the clocks' rest frame? In this frame the clocks are at a distance of 20 light years apart, and both are at rest so the light reaching clock 1 now must have been emitted by clock 2 a time of 20 years earlier. Since the clocks are synchronized and ticking normally in this frame, if clock 1 and 2 read a time of 0 years now, 20 years earlier clocks 1 and 2 both read a time of -20 years, so again it makes sense that right now the image of clock 2 seen by clock 1 will show a time of -20 years.
     
  8. Apr 15, 2010 #7

    Mentz114

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    No, there's no way to that except travel backwards in time. The most extreme thing that can happen is that the other clocks appear to be very slow or very fast.

    In the diagram I've attached, which is the basic traveller scenario, you can see that any worldline that goes from event 1 to event 2 must receive the same number of synch pulses from the base clock. A realistic worldline must lean less than 45 deg from the time axis at all times. Try sketching a worldline that breaks that rule.

    I don't think there's any way to estimate the base-clock's rate except by using synch pulses, because seeing a clock running is hard thing to do if the clock is a long way off.

    I can't see what you're trying to demonstrate with this thought experiment, I hope you can be convinced that the weird things you suggest can't happen.
     

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  9. Apr 16, 2010 #8
    Oh man! You mean the Barnard Slingshot-Telescope I've spent all this time constructing is NOT going revolutionize the field of astronomy?!?!?

    Okay, in all seriousness, thanks for your replies. They really helped.

    I wasn't attempting to demonstrate anything in particular, just trying to determine if the weird things I predicted were possible, and if they were not, then where exactly I was going wrong. I see now it's in calculating the time seen by an observer of an advancing clock. That's what was causing me to over-estimate the effects of shifting simultaneity on what an accelerating observer actually sees in far distances. Thanks for putting your finger on that error in my chain of reasoning, and bringing me one step closer to actually understanding SR.

    And I'm glad to see that while weird things undoubtedly happen in SR (clocks can move slower), there's a limit to the weirdness (they can't move backwards).
     
    Last edited: Apr 16, 2010
  10. Apr 17, 2010 #9
    So, I'm still trying to put all this together. Please let me know if I've got anything wrong. In the twin paradox, would the attached diagram, in which the "traveling twin" occupies the vertical axis, accurately depict the transmission of light pulses from the stationary twin to the traveling twin, with the following three things happening as a result of the "turnaround," all from viewpoint of the traveling twin:

    1. The incoming distance traveled by the incoming beams "jumps" higher;
    2. The time it takes for the incoming beams to reach the traveling twin "jumps" higher;
    3. "Present" or "now" at the stationary twin's location "jumps" forward.

    And these effects, in a sense, cancel each other out, preserving the continuity (but not the rate), of the incoming light pulses. In other words, despite the fact that these three things "jump" at turnaround, the stationary twin's clock, as observed by the traveling twin, would not "jump" at all, but merely go from appearing to tick slowly to appearing to tick quickly.
     

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  11. Apr 17, 2010 #10
    p.s. twinparadox2 includes my guess as to what the traveling twin's lines of simultaneity might look like, before and after the turnaround (in blue).
     

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  12. Apr 17, 2010 #11

    Mentz114

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    Hi,
    I looked at the picture, but one twin travels backwards in time so it's not possible. Apart from that, you've got the right idea.
    (unless I misunderstood the diagram ).
     

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  13. Apr 17, 2010 #12
    Here's my understanding: to the the traveling twin, the stationary twin does move back in time. This is just like in the example that Jesse explained above - If I suddenly decelerated when I struck clock 1, then clock 2 would move back in time (from year 16 to year 0).

    However, this takes place outside my lightcone, so it doesn't violate the laws of SR. The traveling twin doesn't actually see the stationary twin move back in time (he doesn't see the stationary twin's clocks move in reverse), because of the corresponding increase in the distance that light has traveled.

    Let me know if you think this is wrong, and why.
     
  14. Apr 17, 2010 #13

    JesseM

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    The problem is that the phrase "to the traveling twin" isn't very well-defined--as you pointed out, you're not talking about what the traveling twin actually sees visually. Are you talking about what happens in a non-inertial frame where the traveling twin remains at a constant position both before and after acceleration? Unlike with inertial frames, there isn't a single unique way to construct a rest frame for a non-inertial observer, you could construct many different types of coordinate systems in which this observer was at rest with different opinions about simultaneity, none would be preferred over any other by the laws of physics. If you specify that you want a non-inertial frame where the definition of simultaneity and distance at each point on the non-inertial observer's worldline should match up with the definition of simultaneity and distance in his instantaneous co-moving inertial frame at that moment, then it's true that if you decelerated when you struck clock 1, clock 2 would move backwards from 16 to 0. Still this is basically an arbitrary matter of choice, there's no physical reason that compels you to define things this way.

    And if you do want to give the traveling twin in the twin paradox a non-inertial frame whose definition of simultaneity always matches up with his instantaneous co-moving inertial frame, then in the standard version of the twin paradox he should actually see the inertial twin's clock jump forward, not back. The last section of this twin paradox FAQ has a diagram:

    gr.gif
     
  15. Apr 18, 2010 #14

    Mentz114

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    I've said my piece - your space-time diagram has one backwards in time segment. The observer on this leg will see other clocks running backwards.
     
  16. Apr 18, 2010 #15
    I think it's consistent. When I decelerate after striking clock 1, I'm accelerating away from clock 2. The traveling twin, during "turnaround" is accelerating toward the stationary twin. So I think it makes sense that, to me, time at clock 2 would move backward, and, to the traveling twin, time at the stationary twin's location would move forward.

    As far giving an accelerating observer a non-inertial frame whose definition of simultaneity always matches up with his instantaneous co-moving inertial frame, I think that's what happening in the animation below (from http://en.wikipedia.org/wiki/Lorentz_transformation). To me, it looks like it is possible for time to move "backwards" (i.e., in the diagram, events outside the lightcone move "up," from the past to the future). However, no event passes the lightcone boundary twice (it's not possible to "see" a clock move backward).
     

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  17. Apr 18, 2010 #16
    That's a fair criticism. By "To X," I mean just, "In X's reality." If X is inertial, that's the same as saying "In X's frame." If X is accelerating, then, yes, we'd be talking about a range of frames.
     
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