Velocity in Proper Time: Relativistic Acceleration Equation

[nomadreid]

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_{M}, and with \tau_{0} =1/g_{M} , after proper time \tau, the coordinates are
x= cosh(\tau/\tau_{0})
t = sinh(\tau/\tau_{0})
and that therefore
v = tanh(\tau/\tau_{0})
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\tau. 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 \tau with constant acceleration g_{M}? (Starting at (0,0).)
I would be grateful for anyone who can untangle me from this mess.

 

[Bill_K]

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?

Yes

(question B) However, this is the end velocity after d\tau. 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.

No, v = dx/dt, as you've calculated it, is the desired instantaneous coordinate velocity at time τ.

 

[nomadreid]

Thanks, Bill_K