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Velocity of accelerated observer

  1. Oct 28, 2014 #1
    Hey,

    I have a question regarding accelerated motion in special relativity. Suppose an observer has a constant proper acceleration α, his velocity in an inertial frame of a distant observer will then be given [itex]v = c tanh(ατ/c)[/itex], where τ is the proper time of the accelerated observer. Since every quantity in the argument of the function is lorentzinvariant, v also should be an invariant quantity. But that would imply, that every observer agrees on the velocity of the accelerated observer at a given moment, which doesn't fit in my knowledge of special relativity so far... is there a fault in my reasoning or am I missing something else?

    Thanks in advance!
     
  2. jcsd
  3. Oct 28, 2014 #2

    WannabeNewton

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    Recall that in order to derive ##v = \tanh(\alpha \tau)## one must explicitly use the fact that the worldline of the observer relative to a fixed inertial frame is ##t(\tau) = \sinh \alpha\tau## and ##x(\tau) = \cosh \alpha \tau##. Clearly under a boost to another inertial frame or any arbitrary coordinate transformation the worldline of the observer will not preserve this exact form because ##t,x## change under the transformation e.g. if I boost then ##t\rightarrow \gamma(t - vx), x\rightarrow \gamma(x - vt)## while the ##\sinh \alpha\tau,\cosh \alpha\tau## remain the same so if I express everything explicitly in terms of the new ##t,x## in this boosted frame then I won't have the same simple form as before for the coordinates of the accelerated observer.

    In other words the equality ##v = \tanh \alpha \tau## is not a Lorentz invariant equality because it is only valid in a specific background Lorentz frame.
     
    Last edited: Oct 28, 2014
  4. Oct 28, 2014 #3

    robphy

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    Hmm...
    In ##v=c\tanh(\alpha \tau/c)##, the quantity ##\alpha \tau/c## is like
    an angle between spacelike radial directions (analogous to the rapidity as the angle between timelike-worldlines from an event).
    The origin of these radial directions is the event whose lightcone forms the apparent horizon for the family of concentric accelerating observers.

    A boost will essentially add a constant to this angle, which alters your velocity in that boosted frame.

    Maybe the best way to think about what is going on is that
    if you set your clock to read ##\tau=0##, when you are at rest in that frame,
    after a time ##\tau## on your clock, your velocity will be ##v=c\tanh(\alpha \tau/c)## in that frame.
     
  5. Oct 28, 2014 #4

    mfb

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    The assumptions behind the formula itself are not lorentz invariant - it assumes the spaceship starts with a velocity of zero at time τ=0.

    Every observer will agree that your distant observer will observe the velocity of v for the spaceship.
     
  6. Oct 29, 2014 #5
    Thanks for all your answers, I got it now :)
     
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