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Accelerating twin.

  1. Aug 19, 2009 #1
    Assume that a rocket ship leaves the earth in the year 2100. One of a set of twins born in 2080 remains on the earth (inertial frame); the other rides in the rocket. The rocket ship is so constructed that is has an acceleration g in its own rest frame (this makes the occupants feel at home). It accelerates in a straight-line path for 5 years (by its own clocks), decelerates at the same rate for 5 more years, turns around, accelerates for 5 years, decelerates for 5 years, and lands on earth. The twin in the rocket is 40 years old.

    (a) What year is it on earth?

    (b) How far away from the earth did the rocket ship travel?

    I managed to find solutions for this problem...but they seemed paradoxical, not for the reason that the moving observer appears to time travel - that is perfectly legitimate. While in the frame of the earth, the quotient of the outward distance and time was less than c - no problems there. However, I then asked myself "what if the observer during the accelerating portion of the journey decides to stop accelerating and continue in an intertial reference frame?". Both observers are now intertial, and thus must agree on the distance between them: my results say otherwise. In fact they show that the distance between them, divided by the time on the moving observors clock can be greater than c.

    I think the paradox may arise at my assumption that we can see the outward journey as a series of interial rest frames of the rocket. Or perhaps my calulations are correct, up until the point where the rocket stops accelerating: it could be that the deceleration must occur in finite time...

    Anyway, I will not post my solutions (yet), so that anyone who wants to can have a go at the problem themselves. If you want to, you are very welcome to find how the earth appears to the observer in the rocket.
  2. jcsd
  3. Aug 19, 2009 #2


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    Not true, they'd only agree on the distance if they were at rest relative to one another--inertial observers in motion relative to one another can disagree on the answer to questions like "how far apart were we at the moment observer #1 turned 30". You can think of this in terms of the fact that each uses a ruler-clock system at rest relative to themselves to measure distance, and each observer thinks the other observer's ruler is shrunk due to Lorentz contraction, and the other observer's clocks are out-of-sync due to the relativity of simultaneity (you need synchronized clocks to measure distance at a single moment).
  4. Aug 19, 2009 #3
    Yes, I see. Although this only partially resolves my paradox. I am still wondering why my rocket observer looks back and measures a distance from the earth that when divided by the time on his clock is greater than c...
  5. Aug 19, 2009 #4

    George Jones

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    How is this any different from the standard twin paradox?
  6. Aug 19, 2009 #5
    I see. I think what I meant is that each observer records the same distance between them at the time when he notices the other observer stops accelerating. But, I see this may not be true, since the two observers are not in a symmetrical setup. One cannot just suppose that the rocket is inertial and take the frame of the observer on earth as accelerating. The observer certainly feels the acceleration, and walks as if on earth. I thought that I had resolved this by letting the rocket observer continue in an inertial frame, but I suppose that an amount of antisymmetry has now built up.
  7. Aug 19, 2009 #6
    It is the standard twin paradox. But these "paradoxes" aren't really paradoxes...and are usually down to some non-relativistic assumption that has been made - but perhaps hidden. It is rather like the "paradox" that a man can manage to fit a 10 foot ladder into a 5 foot long garage by travelling fast enough into the garage. However, taking the frame of the man...he sees himself trying to fit his ladder into a contracted garage. This isn't a paradox...but one has to really probe why it isn't...which I will leave you to do if you haven't met the problem before...
  8. Aug 21, 2009 #7


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    Velocity is defined as distance divided by time, each measured in the same inertial frame:
    What you are doing here is basically measuring distance in the original inertial frame, and dividing by the proper time of the accelerating twin. For short timscales:
    [tex]u=\frac{dx}{d\tau}=\gamma \frac{dx}{dt}[/tex]
    u is also a useful quantity, for example it is a component of the four velocity, but it is not "velocity" or "speed".
    If you ever heard about galaxies receding "faster than the speed of light": that's a similar definition.
  9. Aug 21, 2009 #8
    I don't think that is what I was doing.

    For the accelerating observer, I was calulating the distance from the earth by using an integral over the instantaneous rest frame differentials...

    For the earth observer I used his own personal time, not the proper time, in my calculations...

    It is probably ssmething fishy with my maths...
  10. Aug 21, 2009 #9


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    Proper time is own personal time.
    Whatever, the "paradox" persists if you wait until the accelerating twin is at rest wrt the earth (at the turnaround point), and also if you neglect acceleration and simly move away with constant speed. That's what I am referring to.
  11. Aug 21, 2009 #10
    There are two "proper times" in this question...I was referring to the proper time of the earth observer, trying to distinguish it from the proper time of the rocket observer.

    I was approximating the outward journey into an "infinite" number of rest frames of the rocket. In each infinitesimal frame, I would spend an "infinitesimal" amount of proper time. I would then caclulate the infinitesimal distance that the earth appeared to recede during this time. I then summed over these differentials using an integral.

    I know what I have done wrong. This integral does not correspond to anything physically useful. I have failed to account for the Lorentz contraction of the distance that had already been travelled. But I could get around this, if I multiply each of the differential elements of distance by an appropriate gamma factor. This gamma factor would have to be different for each differential length, since each instantaneous frame would be travelling at a different speed relative to the frame in which I want to find this distance...
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