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Time dilation in the period of a pendulum

  1. Sep 4, 2012 #1
    i just started studying special theory of relativity and am confused a lot. its first consequence,that is moving clocks run slow confuses me a lot. if two events happen then the time difference between them as measured by a clock in a frame moving at .9c would be lesser than the time measured by a similar clock which is in a frame that is at rest.
    now should i first specify here that which is the proper time? the proper time,as i understand it, would be the time measured by the clock in whose frame the two events occured at the same position. right?
    lets say that the events were stationery with respect to the frame which i see to be moving at .9c. (i am at rest.) now since i am at rest and we know that moving clocks run slow then i will say that the time measured in the frame of refrence of the events is less than the actual time(as measured by the clock in my frame.) right? so the moving clock measured 5s but mine measured 10s.
    but then a thought occured to me that to someone who is in the moving frame(that is who is at rest in the 0.9c frame) i will look like moving with -.9c. and he has equal right in saying that my clock is slow(the one who seems to be moving at -.9c) and his is normal. so if his measures 5s then mine measures less that is 2.5s.
    now one of the situations is obviously wrong. but i cant understand which one? kindly explain how to tell proper time from "improper" time and which situation is correct?

    actually this doubt arose because of a numerical in one book i was reading. here is it:


    Question :
    The period of a pendulum is measured to be 3.00 s in the reference frame of the pendulum. What is the period when
    measured by an observer moving at a speed of 0.960c relative to the pendulum?

    SOLUTION :

    Concept: Let’s change frames of reference. Instead of the observer moving at 0.960c, we can take the equivalent point of view that the observer is at rest and the pendulum is moving at 0.960c past the stationary observer. Hence, the pendulum is an example of a clock moving at high speed with respect to an observer.

    Analyze : The proper time interval, measured in the rest frame of the pendulum, is Δtp = 3.00 s.
    Use time-dilation equation to find the dilated time interval: Δt = Δtp*omega
    omega = 1/((1-(v^2)/(c^2))^0.5)

    the time interval after calculation comes to be 9.57s.

    now i say that instead of the observer which was moving, his time is slow and his clock must measure less than 3.00s but the book says otherwise.
    where do i go wrong? do explain.
    thanks in advance
     
  2. jcsd
  3. Sep 4, 2012 #2

    Doc Al

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    Think of proper time between events (such as the ticking of the pendulum clock) as the time measured on a clock that is collocated with those events.

    So, a clock in the rest frame of the pendulum would measure the proper time between those events. But in the frame of the moving 'observer', the ticks of the pendulum clock take place at different locations. So the time between those events according to that observer would not be the proper time. (That observer could effectively be thought of as using multiple clocks to measure the time interval between events.)
     
  4. Sep 4, 2012 #3
    okay i understand what you mean by proper time. you mean to say that i am sitting at the origin of the frame and the pendulum starts its oscillation from my right and then ends it there that is at the same position.
    but i still do not understand the situation that i posted.
    i mean that i being in the pendulum frame measure its time as 5 s. someone else sees me moving at 0.9c so he says that my clock is running slow and so he measures 10s.
    but do i not have the same liberty of saying that i am not moving at 0.9c. instead the 'someone else' is moivng in the opposite direction at 0.9c. so his clock must be slow not mine.
    so his will measure time 2.5s.
    now i do not understand where i go wrong? both situations seem fine to me
     
  5. Sep 4, 2012 #4

    Doc Al

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    OK.
    Sure, according to you it is his clocks that are running slowly.
    Realize that you will disagree that the other frame's clocks elapsed 10s between the events. (You will claim that his clocks are not synchronized.) According to you, his clocks only elapse 2.5s during that time, which is perfectly consistent with the time dilation formula.
     
  6. Sep 4, 2012 #5
    ok i get your point. both situations are correct but only for me and the other for the other frame.
    but then why in the question about the pendulum the book takes the pendulum to be moving with .96c and the observer to be at rest.?
    i could have done the same question by keeping the observer moving at .96c and myself stationery. then i would have found out that the time measured by the moving observer is less than 3.00s.
    by the way in that case the dilation equation would have been:
    Δtp = omega*Δt . right?
    Δtp is the proper time
     
  7. Sep 4, 2012 #6

    ghwellsjr

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    Every clock keeps track of proper time. It doesn't matter what happens to it, it ticks away its own proper time.

    When you're talking about the time difference between two events, you could be talking about two different kinds of thing. One of them is the time-like spacetime interval between those two events. This is a measurement that is not dependent on the definitions of any particular Frame of Reference. All that is required is that an inertial (non-accelerated) clock be present at both events. If a second clock which is present at the first event has to accelerate in order to arrive at the location of the second event prior to or simultaneously with the time of the second event, then it will measure a lesser time than the first clock which did not accelerate. Now, obviously, unless you know ahead of time the location and time of the second event, it's going to be impossible to provide for a clock to arrive there without accelerating. So this is where you calculate the spacetime interval between the two events using whatever multiple clocks you have previously synchronized in any arbitrary Frame of Reference. You can look up spacetime interval for a timelike pair of events to see how this is done.

    Now the second thing you could be talking about is the coordinate time difference between the two events which is not the same in every Frame of Reference. For this you have to understand how a Frame of Reference is defined. It takes many clocks, previously synchronized so that at every location, there is a clock. If you want to know the time of an event, you just look at the clock at the location and time of the event. Each different Frame of Reference has its own set of clocks.

    So when we talk about a moving clock running slow in a particular Frame of Reference, we are comparing the time on that one moving clock in relation to whatever clock it is located at that was previously synchronized in that Frame of Reference.

    Maybe now you can understand how two clocks moving with respect to each other determine that the other clock is running slow--because you're comparing one clock to a bunch of other clocks.
    First off, if you want to get a 2 to 1 relationship the speed involved is 0.866c, not 0.9c. But otherwise everything you say is correct. What you are calling "improper" time is normally called "coordinate" time. The reason you go from 10s in your rest frame to 5s in the moving frame is because the events which are at the same location in your frame are not at the same location in the moving frame. An observer at rest in that frame would pick two events that are at rest with respect to him that are 10s apart and determine that they are 5s apart in your frame.
    Your book appears mixed up. I can't figure out how they did that calculation and the fact that the call omega what everyone else calls gamma is suspect.
     
  8. Sep 4, 2012 #7

    Doc Al

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    The two frames are in relative motion. Each one views itself as being at rest and the other moving.
    I'm not sure what you mean. Knowing relativity, you can certainly calculate what the other frame would measure as the time interval between ticks of the pendulum clock. They would measure a greater time. The proper time between events is always the shortest time interval.
    No. The time dilation formula says that 'moving clocks run slow' by a factor of gamma. So, Δt = gamma*Δtp. The other frame sees your clock as moving, and during the time that you measure to be 3 s (which is the proper time between the events in question), they will measure a greater time on their clocks.
     
  9. Sep 4, 2012 #8
    what i meant was that the book considered the observer to be stationery and the pendulum moving and found out that the moving time (3.0s) was less than that measured by the stationery observer(9.57s). now i am saying that the motion between the two is relative. what if i say that i want to see it this way that the pendulum is stationery and the observer moving. then the observer should have measured less time than me beacuse i am stationery.
    but you are saying that the proper time is the shortest time between two events. does not this put a restriction on the formula that always the frame which measures the proper time will have the shortest time. then i will not have the liberty of saying to the other person that motion is relative and your clock seems slower to me. instead my time will always be the shortest no matter what.
     
  10. Sep 4, 2012 #9

    Doc Al

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    The observer always considers themselves to be at rest (with respect to themselves) and the other frame as moving. For clarity, lets call the two frames A (the rest frame of the pendulum) and B (the other frame). These two frames are in relative motion. Either one is perfectly permitted to make observations about things going on in the other frame.

    Viewed from B, the pendulum clock at rest in frame A is moving and thus runs slow. When the pendulum clock read a 3 second interval, B frame clocks will read a 9.57s interval.

    Of course it works both ways. If you in the A frame are observing a clock in the B frame read a 3 second interval, then you would measure on your A frame clocks a 9.57s interval.

    But realize that the measurements made in frame B of the time between the ticking of the pendulum clock at rest in A do not represent the proper time between those events.
    For events that take place at a single location in one frame, the proper time between those events is measured by a clock in that frame.
    It works both ways. Say you both had pendulum clocks. You both measure the period of your own clock to be 3 s. Frame A will measure the period of the B pendulum to be 9.57s. And frame B will measure the period of the A pendulum to be 9.57s.

    (FYI: In order to fully understand how things work out, you need to understand the relativity of simultaneity as well as time dilation. When frame A measures the time between ticks on B's pendulum clock, he is essentially using two clocks to make the measurement. One A clock located where the pendulum is at the beginning of its interval and another A clock located where it is at the end of the interval. A considers his two clocks to be synchronized, but B does not.)
     
  11. Sep 4, 2012 #10

    ghwellsjr

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    Doc_Al, can you show the calculation that leads to 9.57s. I can't figure it out.
     
  12. Sep 4, 2012 #11

    Doc Al

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    I was just using the numbers quoted by khurram. I didn't bother to check or correct them. (As the point did not depend on the details of the speed or gamma.) I probably should have, though.

    He had several speeds quoted through this thread, none of which quite match the gamma factors he was using. (As I'm sure you realized.)

    I have no idea where he got the 9.57s from, as he quotes a speed of 0.96c. That would give a gamma = 3.57, and thus a time of 3.57*3 = 10.7 s. (Edit: A speed of 0.95c would give you a gamma = 3.2 and thus a time of around 9.6 s. Maybe that's what he meant.)

    He also used a speed of .9c at one point with a gamma of 2; it would have been closer to gamma = 2.3.
     
    Last edited: Sep 4, 2012
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