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Message from discussion SR treatment of arbitrarily accelerated motion
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Bill Hobba  
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 More options 24 Aug 2003, 03:43
Newsgroups: sci.physics.relativity
From: "Bill Hobba" <bho...@iprimus.com.au>
Date: Sun, 24 Aug 2003 12:46:26 +1000
Local: Sun 24 Aug 2003 03:46
Subject: Re: SR treatment of arbitrarily accelerated motion
Dirk Van de moortel wrote:
> Suppose I is an inertial observer and A is an arbitrarily accelerated
> observer that constantly monitors the acceleration he feels as a
> function a(T) of his own proper time T (this is 'tau').
> At each time T, observer A can write the amount with which his
> velocity has changed during a small (infinitesimal) proper time
> interval dT during which the acceleration does not change as
>     dV(T) = a(T) dT
> This change of velocity is to be regarded with respect to the
> instantaneously comoving inertial frame at time T.

> Observer I can parametrize the worldline of A with the same
> proper time T, so he will see infinitesimally consecutive velocities
> of A as v(T) and v(T+dT).

> Since v(T+dT) is the standard SR composition ('addition') of the
> velocities v(T) and dV(T), we can write (using c=1):
>     v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
> which, using the expression for dV(T), becomes:
>     v(T+dT) = [ v(T) + a(T) dT ] / [ 1 + v(T) a(T) dT ]

> To calculate dv(T)/dT we use
>     [ v(T+dT) - v(T) ] / dT = a(T) [1-v^2(T)] / [ 1 + v(T) a(T) dT ]
> Take the limit dT --> 0
>     dv(T)/dT = a(T) [ 1 - v^2(T) ]

> Rearrange:
>     dv(T) / [ 1 - v^2(T) ] = a(T) dT
> Integrating between 0 and T
>     argtanh(v(T)) - argtanh(v(0)) = Int{ 0 to T; a(T') dT' }
> and using the abbreviation
>     A(T) = Int{ 0 to T; a(T') dT' },
> we get
>     argtanh(v(T)) - argtanh(v(0)) = A(T)

> Inverting to
>     [ v(T) - v(0) ] / [ 1 - v(T) v(0) ] = tanh(A(T))
> and isolating v(T) gives:
>     v(T) = [ v(0) + tanh(A(T)) ] / [ 1 + v(0) tanh(A(T)) ]
> or simply, if v(0) = 0:
>     v(T) = tanh(A(T))

> Use the standard SR Lorentz transformation (with c=1) between
> the frame I and the instantaneously comoving inertial frame of A
> at time T:
>     dt = gamma(T) ( dT + v(T) dX )
>     dx = gamma(T) ( dX + v(T) dT )
> Since we are working on the worldline of I where X=0 and thus
> dX=0, we get:
>     dt/dT = gamma(T)
>           = 1 / sqrt( 1-v^2(T) )
>           = 1 / sqrt( 1-tanh^2(A(T)) )
>           = cosh(A(T))
> and
>     dx/dT = gamma(T) * v(T)
>           = v(T) / sqrt( 1-v^2(T) )
>           = tanh(A(T)) / sqrt( 1-tanh^2(A(T)) )
>           = sinh(A(T))

> Integrate beteen 0 and T:
>     x(T) = Int{ 0 to T; sinh(A(T')) dT' }
>     t(T) = Int{ 0 to T; cosh(A(T')) dT' }

> If possible, eliminate T to find the equation of the worldline x(t)
> of the accelerated observer A in the frame of the intertial
> observer I:
>     x(t) = ...

> If possible, invert the expression for t(T) to find the proper time
> of A as a function of the coordinate time t of I:
>     T(t) = ...
> and use it to find the velocity v(t) as a function of coordinate time t:
>     v(t) = tanh(A(T(t)))

> Summary:
> ========
>     x and t = coordinates of object as seen by inertial frame.
>     a(T) = felt proper acceleration as function of proper time T.
>     A(T) = Int{ 0 to T; a(T') dT' }
>     v(T) = tanh(A(T))       if v(0) = 0
>     dt/dT = cosh(A(T))
>     dx/dT = sinh(A(T))
>     x(T) = Int{ 0 to T; sinh(A(T')) dT' }
>     t(T) = Int{ 0 to T; cosh(A(T')) dT' }
>     eliminate T to find worldline equation x(t)

> Example:
> ========
> The rocket with constant acceleration of the FAQ:

http://hermes.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/rocket...

> Take
>    a(T) = a = contant
> So
>    A(T) = Int{ 0 to T; a(T') dT' }
>           = a T
> So
>    v(T) = tanh(A(T))
>           = tanh(a T)
> and
>    x(T) = Int{ 0 to T; sinh(a T') dT' }
>           = 1/a * ( cosh(a T) - 1 )
> and
>    t(T) = Int{ 0 to T; cosh(A(T')) dT' }
>           = 1/a * sinh(a T)

> Eliminate T:
>    (x+1/a)^2 - t^2 = 1/a^2
> giving the hyperbola
>    x(t) = 1/a [ sqrt( 1 + (a t)^2 ) - 1 ]

> Also proper time as a function of coordinate time:
>    T(t) = 1/a * argsinh(a t)
> so
>    v(t) = tanh( a T(t) )
>          = tanh( argsinh(a t) )
>          = a t / sqrt[ 1 + (a t)^2 ]

> Re-introduce c and we find:
>    v(T) = c tanh(aT/c)
>    x(T) = c^2/a [ cosh(aT/c) - 1 ]
>    t(T) = c/a sinh(aT/c)
>    gamma(T) = cosh(aT/c)
> and as functions of t:
>    T(t) = c/a argsinh(at/c)
> and
>    x(t) = c^2/a sqrt( 1 + (at/c)^2 )      (the hyperbola)
>    v(t) = at / sqrt( 1 + (at/c)^2 )
>    gamma(t) = sqrt( 1 + (at/c)^2 )
> which are the equations of the FAQ entry.

Nice to see the full details Dirk  But at a logical level all you really
need to remember is that during an infinitesimal time period an accelerated
system can be considered to be traveling and constant velocity so SR
applies.

Thanks
Bill


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