SR treatment of arbitrarily accelerated motion

Dirk Van de moortel

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Aug 23, 2003, 11:50:30 AM8/23/03

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FWIW, I did this as a little exercise a few weeks ago on a vacation
day when it was too hot to do anything else.
Since there are a few active threads on accelerated motion and
whether it can or cannot be treated by special relativity, I decided
to post this for those who might be interested.
Everything is pretty straighforward, so I have not attached numbers
to the equations.
Feel free to point out the typo's and errors.

Once we know the acceleration that is "felt" by an observer as a
function of his own proper time (called the proper acceleration),
which can be very easily measured, this function can be used to
calculate every aspect of his motion as seen by any interested
inertial observer.

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.html
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.

Dirk Vdm

Bill Hobba

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Aug 23, 2003, 10:46:26 PM8/23/03

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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

Dirk Van de moortel

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Aug 24, 2003, 4:14:36 AM8/24/03

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"Bill Hobba" <bho...@iprimus.com.au> wrote in message news:3f482...@news.iprimus.com.au...

[snip]

> 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

Yep :-)

Dirk Vdm