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A problem in special relativity with a surprising result

  1. Jan 4, 2005 #1


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    OK, here is the problem formulation, I won't give the "surprsing" answer for a few days.

    In an inertial frame, we have

    an observer at the origin
    a particle moving at a velocity of .9c in the positive x direction at a postion of x = 10 light years.

    We also have an observe acclerating at 1 light year/year^2 (approximately one gravity) at x=1 light year. This accelerating observer is stationary at t=0 in the inertial frame, and is accelerating in the positive x direction.
    The question is:

    In the local coordinate system of the accelerating observer, what is the velocity of the moving particle?

    hint: it may be helpful to know that if [tex]\mbox{(\tau,\xi)}[/tex] are the coordinates of an object in the local frame of the accelerated observer with acceleration 'a' an inertial observer will assign the coordinates (t,x) as follows:

    t = (1/a + \xi) sinh(a \, \tau)
    x = (1/a + \xi) cosh(a \, \tau)

    Note that [tex]\tau=0,\xi=0 -> t=0, x=1/a[/tex]

    In this problem, a=1.
    Last edited: Jan 4, 2005
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  3. Jan 6, 2005 #2


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    No takers? Well, the suprising answer is that the object that was moving at .9c in the inertial frame has a "velocity" (rate of change of position coordiante [tex]\mbox{\xi}}[/tex] with respect to time coordinate [tex]\mbox{\tau}[/tex]) of 9.0*c, nine times the speed of light!

    Here's the details of how this is arrived at

    given that x = a+vt, (for the problem given, a=10, and v=.9), and given the defintion of the local coordinate system of an accelerated observer, we can write [tex]\mbox{\tau \, , \, \xi}[/tex] as functions of t

    [tex]\tau = arctanh(t/(a+v\,t))[/tex]
    [tex]\xi = (a+v\,t)/cosh(\tau) -1 = (a+v\,t)\sqrt{1-(t/(a+v\,t))^2} -1 [/tex]

    we can confirm that

    [tex](1+\xi) sinh(\tau) = t[/tex]
    [tex](1+\xi) cosh(\tau) = a + v\,t[/tex]


    [tex]\frac{d\xi}{d\tau} = (\frac{d \xi}{d t}) / (\frac{d \tau}{dt}) [/tex]

    This turns out to be

    [tex] \frac{d\xi}{d\tau} = {\frac {\sqrt {{a}^{2}+2\,avt+{v}^{2}{t}^{2}-{t}^{2}} \left( {v}^{2}t-
    t+av \right) }{a}}

    substitutingt=0 into this expression, we find that the intial value of the "velocity" is a*v, and in this example a=10 and v=.9c, so the final answer is 9c.

    About the only way to simply see why this result isn't totally insane is to use the "time dilation" viewpoint. In what passes for the coordinate system of the accelerated observer, the clocks "above" him (greater x) all run at a faster rate, about 10 times faster in this particular example.
  4. Nov 23, 2011 #3
    A couple of questions:

    x = a+vt Can you use this in the relativistic limits you describe?

    Also, when you say that the ship moves at 10c, is everything in the universe moving at 10c? Or is there a situation where the surroundings of the accelerated observer are stationary and the ship ahead of him alone moves at 10c? I would also presume that, if the ship "appears" to move at 10c, that no information from the ship to the accelerated observer could be transmitted, is this correct?

    On another note, maybe you can help me with something.

    Suppose you have the ground frame (Earth).
    Earth sees ship1 start at t=0, v=vo1, at x=xo1.
    Earth sees ship2 start at t=0, v=vo2, at x=xo2

    All velocities are near c. All objects travel along the x-axis only.

    Earth sees ship2 begin to accelerate at a constant acceleration A at t=0.

    If ship1 sends a radio signal (light speed) to ship2, at what position and time will the Earth, ship1, and ship 2 say the signal arrives at ship2.

    I'm willing to do the hard work of this myself, but I am having trouble finding the lorentz transformations for this kind of a problem. Can you giude me to a resource or some help with how to get X,t for the different frames for such a situation? I would really appreciate it.
  5. Nov 23, 2011 #4
    I think this situation is even more interesting when we add a third object at x=-10 and v=0.9c relative to the initial frame and consider its coordinates in the accelerating frame.
  6. Nov 23, 2011 #5


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    I guess you will find that initially, when there is a large spatial separation between accelerating observer and object, it moves faster than light according to the accelerating observers local clock. But In his accelerating frame clocks at rest run at different rates along x. A clock at rest in the accelerating frame, but near the moving object will run faster and thus measure less than c.
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