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Close to C orbit around Earth and how time would be affected

  1. Mar 27, 2010 #1
    There is a nuance of special relativity that I cannot seem to grasp. This may be very similar to the twin paradox, so I am trying to find out what the critical points are.

    If I hop into my spaceship and orbit Earth at 0.995C for, say, 100 years, I will see Earth's clock progress at an extremely slow rate, correct? By the same measure, an observer from Earth would notice the same about me.

    So what happens if I negate all acceleration effects and snap immediately back into Earth's inertial reference frame at the end of the 100 years? Will Earth's clock read what I read it to read while in flight or will it "catch up" in that sudden stop? Does the twin paradox actually resolve? Does stopping in Earth's frame make me subject to its time?

    I heard the statement that going faster than light would violate causality, and I accept that, but I also heard that traveling faster than light would hypothetically cause the traveler to move backwards in time... WHAT? I do not understand this statement.

    If the traveler goes backwards through time, does this mean that the universe around him will move backwards, and that he, upon stopping, would find himself in the past... or would the universe hurtle itself at an impossible speed beyond conceptual infinity into the future?

    The reason I ask such a strange question is because I cannot seem to be able to tell the difference between the rocket jock and the rest of the cosmos.

    To go backwards in time in the classical sci-fi novel sense generally meant that the traveler found himself in a world that went backwards... My reference frames seem to be intertwined.

    I am sorry if none of this makes sense, but this how confused I am when trying to deal with the time discrepancies.
  2. jcsd
  3. Mar 27, 2010 #2


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    If you are in orbit around the earth, that is a priori impossible.
    You would need to constantly accelerate to stay in orbit (the whole F = mv2/r thingy) as with any circular motion, hence your rocket frame is not equivalent to that of the observer on earth.
  4. Mar 27, 2010 #3
    I did not give the best example, but let us imagine whatever it would take to remove general relativity, orbital trajectory, and everything that would detract from the point. I wish I could isolate only frames, time, length, speed, etc. but I'm not that creative.

    Let us assume there was a way to read Earth's clock as it appears from a .995C frame of reference without accelerating and/or experiencing gravitational effects. Let us assume I am in this special reference frame and Earth moving relative to me in the same way I am moving relative to it. No doppler effects or anything of the sort.

    If it is impossible to "snap" into the earth's reference frame, then let us simply say that negative acceleration occurred very quickly. I only want to compare the time in flight to the time when I am back in Earth's frame.

    I just do not understand how two objects moving relative to each other can cause a difference in the final time since they both disagree on which clock was slow.

    The only thing I can see that causes a difference is that my rocket would now be in Earth's frame of reference, making me subject to the observations of a new frame. I do not know how to explain this, really...

    Let's say (in the twin paradox) twin A flew past Earth and twin B was waiting in his rocket to accelerate into the same reference frame as twin A... Would the results be opposite of the classic twin paradox?

    I am wondering if the final time discrepancy depends upon which frame we end up in.
    Last edited: Mar 27, 2010
  5. Mar 27, 2010 #4

    Filip Larsen

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    If you consider the "classical" twin paradox where A first fly away from B with constant speed [itex]v = \beta c[/itex] and then instantaneously turned around and flew back towards B with speed [itex]-\beta[/itex], then the image of B's clock would arrive at A showing B's clock rate to be

    [tex] dt = d\tau \sqrt{\frac{1-\beta}{1+\beta}} [/tex]

    where [itex]d\tau[/itex] is the proper time of the clocks. If you integrating the rate of B's clock image at A for the complete trip out and back, you will end up with

    [tex] t = \tau \frac{1}{\sqrt{1-\beta^2}} = \tau \gamma [/tex]

    which is the same time dilation you will find if you analyse what happens using, say, event intervals.

    For instance, if [itex] \beta = 0.999 [/itex], then [itex]dt/d\tau \approx 1/44[/itex] when heading out and [itex] dt/d\tau \approx 44 [/itex] when heading home, resulting in a total of [itex] t/\tau \approx 22 [/itex].
    Last edited: Mar 27, 2010
  6. Mar 27, 2010 #5


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    You can't ignore the very concept that is the answer to the question!
  7. Mar 27, 2010 #6
    Let me assist you with your creativity :wink:.

    To orbit at 0.995c will require a central massive body of much greater mass than that of the Earth, so let us consider Earth2 that has the required mass. To remove differential ageing due to gravity we need to place the stationary observer (A) at the same height as the orbiting observer (B) so we need a tower or a very low surface skimming orbit. Neither is impossible in principle. Now all differential time dilation effects between A and B are due to velocity time dilation only. Now at 0.995c the gamma factor is about 10, so if for example B orbits for 1 year according to his own clock, then 10 years passes for A. As orbiting observer B passes very close by observer A, B measures A's clock to be ticking 10 times slower than his own clock and A measures B's clock to tick 10 times slower than his own clock, just as in Special Relativity. Doppler effects make the clock of B appear to be tick faster as he approaches A and slower as he recedes from A (from the point of view of A) but we are ignoring the visual Doppler effect here.

    During the rapid negative de-acceleration of B to when he comes to rest in the frame of A, we can ignore the effect of acceleration on the clock of B, because time dilation is only a function of instantaneous relative velocity and not affected by acceleration. During the slow down process B is travelling faster than A at all times until the moment B is at rest alongside A. Let us say that during this slowdown process B ages by 1 day and A ages by 5 days. This difference of 4 days is independent of how long B has been orbiting. So:

    If B orbits for 1 year and then slows down, (1 year + 1 day) will have passed for B and (10 years +5 days) will have passed for A.

    If B orbits for 1 day and then slows down, 2 days will have passed for B and 15 days will have passed for A.

    As you can see, the slowdown process can not cancel out the differential ageing during the orbiting phase and in fact only adds to the differential.

    In the example I gave above of B ageing (1 year +1 day) and A ageing (10 years +5 days) the times are proper times (i.e. measured by clocks that are carried by the observers) and so all observers will have to agree on the discrepancy, whatever frame they are in, because proper times are not frame dependent.

    You can do a linear version of this thought experiment. Consider a a very long row of clocks (A) that all synchronised with each other and another very long row of clocks (B) parallel to the A row, that are also synchronised with each other and initially also synchronised with all the A clocks. Now row B accelerates and cruises for a period at a velocity v relative to the row of A clocks and then finally slows down to come to rest again with the row of A clocks. All the B clocks will show less elapsed time than the A clocks, but when they come to rest with the A clocks they will of course, once again be ticking at the same rate as the A clocks. While the B clocks are cruising relative to the A clocks, the A observers will say the B clocks are ticking slower than the A clocks and the B observers will say the A clocks are ticking slower than the B clocks.
    Last edited: Mar 27, 2010
  8. Mar 27, 2010 #7

    You nailed the point I was getting at. Thanks! I think we are almost there.

    The final elapsed time is my major concern here. I am trying to figure out if it is frame-dependent.

    I do not know how to use LaTeX, nor have I analyzed the maths, so I will try to do the best I can without them.

    I will use Kev's row of clocks, since it is a far simplified version of what I was getting at.

    Are these the correct postulates?


    1. If Row A is moving relative to Row B and then comes to rest, Row A will show a final elapsed time less than Row B.

    (state 1, from Frame B's reference)

    A --------->
    B ----------

    2. According to Row A's frame of reference during motion, Row B is moving relative to Row A, the only difference being the vector of motion (opposite of A).

    (state 1, from Frame A's reference)

    A ---------
    B <--------

    3. Row A must show less elapsed time than Row B because Row A is found to have conformed to Row B's frame of reference for the final reading.

    (state 2, from Frame B's original reference, now occupied by both rows)

    A --------- (less time elapsed in final state)
    B ---------

    4. If Row B had sped up to conform to Row A's frame of reference, Row B must show less elapsed time than Row A.

    (state 2, from Frame A's original reference, now occupied by both rows)

    A ---------
    B --------- (less time elapsed in final state)


    The reason I think it must be this way is because there is no way to tell the difference between who is at rest and who is moving... It's all relative. Looking at the problem from different views should yield different results. The only thing that changes in the final state is which Row conforms to the other.

    Please tell me if the postulates are correct.
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