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Velocity in terms of proper time

  1. Mar 25, 2014 #1
    Two questions, based on the same situation: in
    http://physics.stackexchange.com/questions/34204/relativistic-acceleration-equation
    (question A) it is mentioned that, for an object with a constant acceleration g[itex]_{M}[/itex], and with [itex]\tau[/itex][itex]_{0}[/itex] =1/g[itex]_{M}[/itex] , after proper time [itex]\tau[/itex], the coordinates are
    x= cosh([itex]\tau[/itex]/[itex]\tau[/itex][itex]_{0}[/itex])
    t = sinh([itex]\tau[/itex]/[itex]\tau[/itex][itex]_{0}[/itex])
    and that therefore
    v = tanh([itex]\tau[/itex]/[itex]\tau[/itex][itex]_{0}[/itex])
    My first reaction was that this should be coth (position/time, cosh/sinh), but then I figured that tanh comes from v=dx/dt =d(cosh ...)/d(sinh ...) = sinh.../cosh..... Is this correct?
    (question B) However, this is the end velocity after d[itex]\tau[/itex]. So, I presume one would need to call this v= dv. If we want the end velocity after a finite amount of time, I presume integration would be in order, but since it is the rapidities that add rather than the velocities, I am not sure how this integration would look. Or perhaps there is a simpler method to find the coordinate velocity after finite proper time [itex]\tau[/itex] with constant acceleration g[itex]_{M}[/itex]? (Starting at (0,0).)
    I would be grateful for anyone who can untangle me from this mess.
     
  2. jcsd
  3. Mar 25, 2014 #2

    Bill_K

    User Avatar
    Science Advisor

    Yes

    No, v = dx/dt, as you've calculated it, is the desired instantaneous coordinate velocity at time τ.
     
  4. Mar 25, 2014 #3
    Thanks, Bill_K
     
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