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Twin paradox

  1. May 16, 2008 #1
    suppose we start with a long line of stationary, evenly spaced and perfectly synchronized clocks along the x axis. if the stationary twin is at the origin and at t=0 the other twin passes the origin moving at relativistic speed with gamma=2 then from the point of view of the stationary twin the moving twin is length contracted to 1/2 his normal length. but despite this moving twin, due to loss of simultaneity, will see the line of clocks as no longer synchronized and therefore as being contracted to 1/2 its normal length. thats obvious. presumably, due to loss of simultaneity, each individual clock will seem to be running at 1/2 its normal speed even though the time read off as each individual clock passes would be running at twice normal speed. i think thats right.

    but what happens when the moving twin suddenly stops? now all the clocks are synchronized and there is no length or time contraction. in particular, from the point of view of the moving twin, what time does the non-moving twins clock show just before and just after the moving twin suddenly stops?

    what happens then when the moving twin begins moving back toward the origin? what does he see?
     
    Last edited: May 17, 2008
  2. jcsd
  3. May 17, 2008 #2
    Look at this example - (My value for gamma differs from yours a little bit):

    Let us assume three clocks A, B, and C. A and B are at rest in S and C is at rest in S', while the relative velocity between S and S' is 0.5c.

    In S:
    A and B are synchronous and the distance between them is 1 light-second.
    At t=0 clock C starts from A and arrives at t=2 at B. In S, clock C is time-dilated and therefore shows the t'=1,73..s. The same is happening when C travels back from B to A. So t=4s and t'=3,46..s..

    In S':
    We are no changing the frame and are looking into S'. In that frame, A and B are not synchronous, becaus of relativity of simultaneity B starts 0,577..s before A. Due to Length-contraction the distance between A and B is 0,866..Light seconds.
    When A starts at t=0, B is showing t= 0,5s (Not 0,577..s because B is time dilated).

    Now we can answer your question:
    When B is just before C, C shows t'=1.73..s. Both A and B are time-dilated, so A shows t=1.5s and B shows t=2s.
    When B stops at C, A jumps forward in time to be synchronous with B, so A and B show t=2s.
    When B is just after C, A jumps forward in time again and shows t= 2.5..s.
    When A stops at C, C shows t'=3,46..s. Both A and B are time-dilated, so A shows t=4s and B shows t=4.5s. So when A and C are coming togeter, in both frames A shows 4s and C shows 3.46..s - so A is older. :smile:
     
    Last edited: May 17, 2008
  4. May 17, 2008 #3
    thank you. I figured it was something like that.

    now I presume that this shift in A's time from B's point of view as B accelerates has something to do with general relativity. (of course, the time on A's clock doesnt really change when B accelerates. it just appears to B to do so). I suggest another thought experiment.

    suppose the length between A and B is some absurdly large distance (jillions of light years) and C, having already stopped at B, instead of changing speed instantly, accelerates at 1 g for some length of time (about a year) till it reaches relativistic speed. the time spent accelerating is such a small part of the total transit time that it can be ignored. during this acceleration the time on A's clock continually shifts (from B's point of view). how is this equivalent to gravity?

    I'm just trying to understand general relativity here. I dont need a detailed mathematical explanation.
     
    Last edited: May 17, 2008
  5. May 17, 2008 #4
    That would be better stated as {each individual clock will seem to be running at 1/2 its normal speed even though the time read off as each individual clock passes would show an elapsed time that is twice that shown on his own clock.}

    When he is moving the line of clocks is no longer synchronised and he sees each clock as having a positive offset right from the outset and the further away each clock is the greater the offset proportional to Do*v/c^2 where Do is the proper distance measured in the rest frame of the line of clocks. Thereafter he see the line clocks advancing at 1/2 the rate of his own clock but the elapsed time on each clock that he passes is twice that shown on his own clock because as far as he is concerned they had a head start.

    Lets say he is looking through a telescope at the non moving twins clock and it shows 2.00.00 PM just before he stops. 1f he stops within one second then he will see a time of somewhere between 2.00.00 PM and 2.00.01 PM on the non moving twin's clock. In other words, he will notice very little change in elapsed time of the non moving twin's clock. What he will notice, is that just before he stopped the non moving twins clock was advancing at rate of 1/2 a second to every second that passes on his clock and just after he stops the non moving twin's clock has speeded up to the same rate as his own clock. When he has stopped he will notice that the clock that he has stopped next to, shows an elapsed time that is twice that of is own clock. If he allows for light travel times, he will notice that all the clocks in the line of clocks show twice the elapsed time of his own clock.

    For the sake of argument let's say the clock that he stopped next to, shows an elapsed time of two hours while his own clock shows a journey time of one hour. If he moved back to the non movig twin so slowly that there was no further time dilation of his own clock, then when he got back his clock would be one hour behind the non moving twins clock when they are standing next to each other once again. However, no matter how slow he moves, there will be some time dilation and the slower he moves the longer that additional time dilation will be over so that in practice his clock will be slightly more than a hour behind the non moving twin's clock.

    [EDIT] When I stated that when the moving twin looks at the non moving twins clock through a telescope I should have added that he would have to allow for a classic doppler shift in his calculations.
     
    Last edited: May 17, 2008
  6. May 17, 2008 #5
    Time dilation is not necessarily due to acceleration. I read an analogy in the context of "the clock postulate" that went something like this. Imagine a cyclist on a cold but calm day. When he accelerates from rest to a velocity relative to the air he feels a wind chill factor that cools him down more than when he is at rest. It is not the acceleration that causes the wid chill, but the motion relative to the air. The acceleration simply changes his motion relative to the air. When he stops accelerating and sentles down to a cruising speed he continues to be cooled at greater rate by the wind chill factor even though there is no longer any acceleration. For wind chill factor you could substitute time dilation gamma factor.
     
  7. May 17, 2008 #6
    i know that velocity results in time dilation. that isnt what i was referring to. when B stops then the line of clocks are all synchronous. before that they were out of synch. that is a time shift.
     
    Last edited: May 17, 2008
  8. May 17, 2008 #7
    Ok, let's try and create the equivalent situation in a gravitational context. C is sitting on the surface of the Earth experiencing a constant acceleration of 1g. A is high up and his clock rate is faster than that of C because he experiences less gravitational time dilation. To make the situation completely equivalent C observes A to moving away from him at a progressively greater rate in your kinetic example so we will have to do the same in the gravitational example and have A accelerating upwards. A's upward acceleration will cause his clock to kinetically time dilate and classic doppler shift due to motion going away from C will make his clock appear to be slowing even more as far as C is concerned. As you can see the situation, is is not simple for your example and proportional effects of kinetic time dilation have to be weighed against the gravitational time dilation. Without a detailed mathematical numerical analysis it would be difficult to conclude anything.
     
  9. May 17, 2008 #8
    classic doppler shift due to motion going away from C will make his clock appear to be slowing even more as far as C is concerned.

    this is factored out when talking about time dilation (and loss of simultaneity) as is light travel time.
     
    Last edited: May 17, 2008
  10. May 17, 2008 #9

    Janus

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    The thing to keep in mind is that gravitational time dilation is due to a difference in potential rather than difference in local g force. Thus a clock runs faster at the top of a mountain than at sea level not because gravity is weaker, but because it is higher in Earth's field. If we were to replace Earth's gravity field with an uniform one (one that didn't fall off with increased distance from the center of the Earth), the Clock on the mountain would still run faster, even though it experiences the same gravitational force as one at sea level. In fact, it the difference in clock rates will actually be greater, for the very reason that Earth's gravity in the first example falls off with distance, which leads to a smaller difference in potential than the uniform field does.

    When B accelerates at one g towards A, this is the equivalent of a uniform gravity field of 1 g strength extending from B to A, from B's point of view. Since A is vastly higher than B in this "field" from B's point of view, A is at a vastly greater potential and runs vastly faster than B. The greater the distance from A to B, the faster A runs according to B.
     
  11. May 17, 2008 #10
    When the moving twin accelerates, he sees a linear spatial gradient in the speed of light, and all things under a pseudo-force equal to the force per unit mass he experiences times the energy of the particle under examination. The linear spatial gradient in the speed of light is equal to the force per unit mass he experiences. Note that I am using a slightly different definition of force, since the velocities in the momentum formulas are normalized by light speed. Since the pseudo-force is proportional to energy, it is like gravity. I have posted a calculation on my hi5 account, under the name johnwilliams22. I hope this helps.
     
  12. May 17, 2008 #11
    Hello dude222

    Quote

    ------When the moving twin accelerates, he sees a linear spatial Gradient in the speed of light, and all things under a pseudo-force equal to the force per unit mass he experiences times the energy of the particle under examination. The linear spatial gradient in the speed of light is equal to the force per unit mass he experiences. Note that I am using a slightly different definition of force, since the velocities in the Momentum formulas are normalized by light speed. Since the pseudo-force is proportional to energy, it is like gravity. I have posted a calculation on my hi5 account, under the name johnwilliams22. I hope this helps.-------

    What is that in layman's terms.

    Matheinste.
     
  13. May 19, 2008 #12
    Janus:
    wikipedia backs you up. you are 100% right.
     
  14. May 20, 2008 #13
    http://www.sysmatrix.net/~kavs/kjs/addend4.html

    The standard version of the twin paradox has an astronaut twin – call her Stella – accelerating away from her Earthbound twin sister, Terra. Stella rockets through the cosmos at relativistic speeds, then later turns around and flies home, eventually landing back on Earth.

    But in order to eliminate pesky accelerations, we'll say that the astronaut twin, Stella, was already in motion as she passes over Earthbound twin Terra. At the pass-by, their clocks each start at zero. Then, instead of later turning around, Stella eventually passes another astronaut, Alf, moving equally fast in the opposite direction, who adopts Stella's clock reading and continues back toward Earth. When Alf arrives at Earth, he passes over homebound Terra without slowing, at which time their clock readings are compared. Thus there are no accelerations and the scenario can be examined with the simplest (SR) arithmetic, like what's been shown in diagrams thus far.
     
  15. May 20, 2008 #14
    Hello again.

    I thought that the twins (non)paradox required them to be in the same frame as each other at the start and end of the travelling twins journey.

    Matheinste.
     
  16. May 20, 2008 #15
    tehy have to be at the same place at the same time.
     
  17. May 20, 2008 #16
    Hello again

    In your scenario above the traveller never returns.

    Matheinste.
     
  18. May 20, 2008 #17
    so??


    btw, the total elapsed time is exactly the same.
     
  19. May 20, 2008 #18
    Hello again.

    For who.

    Matheinste
     
  20. May 20, 2008 #19
    stella + alf
     
  21. May 20, 2008 #20
    Hello again.

    I think i'll pass on this discusion and leave it to those who understand these things.

    Goodbye .

    Matheinste.
     
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