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Time dilation misunderstanding

  1. Sep 17, 2004 #1
    Hi,

    I have noticed what appears to be a basic misunderstanding of time dilation. I would invite anyone to reply if they think I am wrong.

    My view is that all the clocks in all the reference frames actually run at the same rate. It is only when an observer at a relative velocity with respect to the clock (or in a different gravitational state) tries to MEASURE the clock from his point of view that the time dilation phenomenon occurs.

    juju
     
  2. jcsd
  3. Sep 17, 2004 #2

    selfAdjoint

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    This is correct, with only one proviso. The clocks in all REST frames run at the same rate (perfect clocks of course). The time told by a clock in a rest frame (and every massive object has a rest frame!) is called PROPER TIME. When an observer in one frame sees a clock in another frame, so that there is a velocity between them, he will see the clock running slow, compared to his own. Notice that this is symmetical; two observers in two different frames will each see the other's clock running slow compared to his own.
     
  4. Sep 18, 2004 #3

    Doesn't this depend on wether the objects are receding or contracting? I thought that when you observe something getting closer, that you will see it's clock running faster.

    Also isn't it nearly impossible to determine two clocks separated by distance to be in the same rest frame? When you say every massive object has a rest frame then we should be able to determine one here on earth, but as time passes the mass of the earth is rearranged somewhat. It seems that even when two people are face to face they are experiencing "time dilation".
     
  5. Sep 18, 2004 #4

    selfAdjoint

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    No, receding or approaching, it's the same. Maybe you're thinking of the doppler effect?
     
  6. Sep 18, 2004 #5
    selF!!!!!! your back
    Thought something happened to you for awhile
    good to see you back
     
  7. Sep 19, 2004 #6

    selfAdjoint

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    I was on a trip west with my daughter. Got back a week ago.
     
  8. Sep 19, 2004 #7
    Twin Paradox

    Hi,

    I have been trying to explain the twin paradox to myself. here is what I came up with.

    If you use a reference clock attached to the distance to be measured, then both twins OBSERVATION of the reference clock will agree with their MEASUREMENT of time in their own reference frame.

    The twin on earth (Twin A) MEASURES the time to be t and the distance to be d with his own instruments. He also OBSERVES the reference clock as having the same value t, since it exists in his reference frame.

    The twin in the spaceship (Twin B) MEASURES the distance to be D and the time to be T with his own instruments.

    Due to length contraction (there is relative motion between Twin B and the distance to be measured) D=d/gamma (where gamma is the relativistic factor and is greater than 1). Since the velocity is v, Twin B MEASURES the time in his reference frame as T=D/v=(d/gamma)/v.

    This turns out to be the same value that Twin B gets if he OBSERVES the clock of Twin A (or the reference clock). The time for Twin A is t=d/v. Twin B OBSERVES this as (d/v)/gamma which is the SAME as his MEASUREMENT of time is his own reference frame.

    So, Twin B gets the same time value for his MEASUREMENT of time in his reference frame and his OBSERVATION of the time in Twin A’s reference frame (or his observation of the reference clock).

    This is also the same time twin A will CALCULATE for his OBSERVATION of time passing for twin B.

    juju
     
  9. Sep 19, 2004 #8

    selfAdjoint

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    This is right, and as long as neither twin accelerates, the situation remains symmetrical: each twin sees time t on his own clock and time T < t on the other twin's clock. But if the travelling twin should turn arround, thus accelerating, when he gets back home the time he has experienced will be less than that of the unacclelerated twin.
     
  10. Sep 20, 2004 #9
    Hi All,

    Just wanted to make one more comment here.

    What is important in this and all situations are the measurements taken by each observer in their own reference frame, using their own instruments, from their own point of view. The rest is just bells and whistles that explain the different measurements and the observation and calculation of one observers instruments by another.

    Here the important point is that because of length contraction Twin B measures less time on his instruments.

    juju
     
  11. Sep 20, 2004 #10
    That's almost 100% correct.
    "here the important point is that because of length contraction Twin B measures less time on his instruments.."
    He measures time in his own frame and it's not just length contraction in the direction of travel but time dilation dependent on his absolute velocity. ......his clock slows down ( relative to twin A ) * but a year to him is still a year. If he had a female companion who he impregnated the baby would be born ( if all goes well ) ~ 9 Months later ship time. He measures time more slowly in RELATION to twin A so depending on the distance traveled ( or the time spent at .99 c ) when they return ,the infant could be held by a very old uncle or a great,great,great grand niece.

    *where [tex]delta[/tex]t = [tex]delta[/tex]t0 / (1 - v2/c2)1/2 as viewed by twin B
    been tryin to figure out how to graph using LaTex.....anyone have any pointers?
     
    Last edited: Sep 20, 2004
  12. Sep 21, 2004 #11
    Hi all,

    Throw out the reference clock. it was a bad idea. The concept on which it was originally based is flawed and invalid.

    Here’s what I believe actually occurs.

    Twin A is on Earth. Twin B is on the spaceship. Gamma is the relativity factor greater than one, and v is the relative velocity between A and B.

    Twin A measures the distance and time to be d and t on his instruments in his reference frame, where v=d/t. There is no relative velocity between twin A and the distance to be measured.

    Twin B measures the distance and time on his instruments in his reference frame to be D and T. Since there is a relative velocity between Twin B and the distance to be measured, Twin B measures this distance as D=d/gamma.
    The time Twin B measures is then given by the equation Time=Distance/Rate as applied to twin B's measurements.
    T=D/v=(d/gamma)/v=(d/v)gamma=t/gamma.

    Therefore, when twin B gets back to Earth he is younger then Twin A having measured less time passing for himself then twin A measures passing for himself.

    One more concept is necessary. Twin B’s clock runs at the same rate relative to twin B as Twin A’s clock runs relative to twin A. This is true since both twins are at rest relative to their own clocks

    juju
     
  13. Sep 25, 2004 #12
    Yes that's pretty much the case. In fact You seemed to have grasped the idea in Your last post , except as You said there is no referential clock.T is plastic as v is, as is a. The trick ( think I posted this on another board) is that c remains constant, so if t becomes slower and length contracts we still have c to measure properties against . It's like the first time you did the " A car a is travailing west to and intersection at 60mph and car b is traveling east at 49 mph when will they meet at intersection route66" problem.
    The intersection is moving in space-time as are the cars but c is a constant - for car a , car b and the intersection , so we DO have an accessible reference point that we can measure things against.
     
  14. Sep 26, 2004 #13
    I don't think that it is the acceleration that breaks the symmetry (I used to).

    I believe that it is the fact that the distance to be measured is embedded in the rest frame of the earth that actually breaks the symmetry.

    juju
     
  15. Sep 26, 2004 #14

    selfAdjoint

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    Sorry but this statement doesn't convey anything to me. Can you elaborate? The acceleration (curved worldline) theory of the twins has a perfectly clear and diagramable explanation, relying on the anti-pythagoras rule of Minkowski; the sum of the squares of two timelike legs with nonzero space components of a spacetime triangle is less than the square of the pure timelike leg. And this results from the basic metric relation [tex]\Delta s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 [/tex].
     
  16. Sep 27, 2004 #15
    Hi SelfAdjoint,

    Yes, I agree thet the Minkowski formulations works. But without symmetry breaking it works for both observers from their own point of view. All the acceleration does is is either seperate the two observers reference frames, or change the velocity relation. between them if they are already separate. At the new velocity, the basic symmetry still holds.

    The only component that can break this symmetry is the space interval that is measured. if this space interval exists in one reference frame or the other, as it does in the twin situation, then one observer will measure it to be smaller. It is this real measurement of the distance that allows for time dilation to emerge.

    If two obervers are moving toward each other, with no distance defined ( the space interval between them is just shrinking), then no time dilation will be seen in the measured values that each observer measure in his/her own reference frame, since the space interval exists in both reference frames, simultaneously.

    juju
     
    Last edited: Sep 27, 2004
  17. Sep 27, 2004 #16

    selfAdjoint

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    What acceleration can do is bring one twin back to the same point (at a later time) as the other. The twin that does this has a bent worldline, the one who doesn't has a straight worldline. The accumulated proper time on the bent worldline is less, in relativistic arithmetic, than the accumulated proper time on the straight worldline.
     
  18. Sep 27, 2004 #17
    Hi SelfAdjoint,

    It seems to me that the worldline is bent (relative to the earth twin's inertial frame) only during the process of acceleration. It would seem not to be bent at any other time during the entire process. This alone would not account for the magnitude of the time difference

    Also, what happens when you draw the world lines in the frame of the moving twin.

    Here's an article I read on the subject, that has a similar point of view to mine.

    http://www.sciam.com/askexpert_question.cfm?articleID=000BA7D8-2FB2-1E6D-A98A809EC5880105&catID=3

    juju
     
  19. Sep 27, 2004 #18

    selfAdjoint

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    I tried to draw the worldlines with dashes and slants but it didn't work. Visualize first a verticle line, this is the worldline of the stayathome twin; it has two points marked on it. Low down is the initial point where the two twins were together before parting, and higher up a terminus point where the two are back together again. The line is vertical because the stayathome twin covers no space.

    Branching out at an angle from the lower point starts the worldline of the traveling twin. It angles up at less than 45o showing that he has a velocity away from home. Halfway up it turns, and still rising, now points toward the terminus point, showint the velocity is now towards home. Eventually the slanting worldline intersects the vertical worldline at the terminus point; the twins are back together.

    Now the two worldlines form a triangle, and the relativistic arithmetic says the proper time accumulated on the space-covering worldline is less then the accumulated proper time on the non-space-covering one.
     
  20. Sep 28, 2004 #19
    Hi SelfAdjoint,

    I did the same visualizations last night to try to understand your points.


    This is just what I am saying from a different point of view. The fact of the traveler traveling a distance embedded in the rest frame of the observer at rest is what causes the difference in time.

    This difference can be calculated by using length contraction and relative velocity. The fact of acceleration has nothing to do with it.

    A modified version of the event would seem in order to explain.

    Consider a distance with fixed endpoints in the rest frame of observer A, with observer A at the midpoint and obsever Z at the far end.. Observer B is traveling parallel to this distance at a fixed velocity relative to the rest frame of observer A and at a known perpendicular distance.

    When observer B crosses the perpendicular to A, they synchronize their clocks via light signals. At the same time observer A sends a light signal to observer Z to synchronize clocks with him.

    When, B crosses the perpendicular to Z, what is the elapsed time relation between their clocks.

    According to my point of view this can be calculated directly from length contraction and velocity. There is no acceleration involved, and no round trip involved.

    juju
     
    Last edited: Sep 28, 2004
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