SR treatment of arbitrarily accelerated motion

I posted this a while ago on Usenet in group sci.physics.relativity.
See the thread: a3M1b.82977$F92.8834@afrodite.telenet-ops.be ( V1.7 - Most recent additions and modifications: 6-May-2013 ) This is the MathJax version. Click here for the original ASCII text version. FWIW, I did this as a little exercise a few weeks ago on a vacation day when it was too hot to do anything else. And no swimming pool anywhere near. Since there are a few active threads on accelerated motion and whether it can or cannot be treated by special relativity, I decided to post this for those who might be interested. Everything is pretty straighforward, so I have not attached numbers to the equations. Feel free to point out the typos and errors. See also http://hermes.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/acceleration.html http://groups.google.co.uk/groups?&threadm=bi75vd$qbm\$1@glue.ucr.edu

Time rate of arbitrarily accelerated clock observed by arbitrarily accelerated observer:

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($$\tau$$)$ of his own proper time $\tau$ (this is "tau").

At each time $\tau$, observer A can write the amount with which his
velocity has changed during a small (infinitesimal) proper time
interval $d \tau$ during which the acceleration does not change as
$dV(\tau)$ $= a(\tau) d \tau$
This change of velocity is to be regarded with respect to the
instantaneously comoving inertial frame at time $\tau$.

Observer I can parametrize the worldline of A with the same
proper time $\tau$, so he will see infinitesimally consecutive velocities
of A as $v(\tau)$ and $v(\tau+d \tau)$. Since $v(\tau+d \tau)$ is the standard SR composition ('addition') of the
velocities $v(\tau)$ and $dV(\tau)$, we can write (using $c=1$): $$v(\tau+d \tau) = \frac{ v(\tau) + dV(\tau) }{ 1 + v(\tau) dV(\tau) }$$
 This part was added for the benefit of Mr. "Mike" aka "Bill Smith" aka "Undeniable" aka "Eleatis". See also the threads "An interesting SR puzzle" and "The Bell Tolls for Dinky-Donkey" You know that SR says that     "When an object P has a relative velocity u w.r.t.      to an inertial frame Q , of which the observer itself      has a relative velocity v w.r.t. an inertial observer R ,     then the object P has a relative velocity          w = (u+v)/(1+u v)     w.r.t. the observer R." Now replace     P => observer A at proper time T+dT     Q => instantaneously comoving inertial frame of observer             at proper time T of observer     R => initial rest frame I     u => dV(T)     v => v(T)     w => v(T+dT) We must do this because it must be valid for *all* values of T, and v(T) becomes quite large, so using the Galilean transformation here would be totally wrong. Together with the fact that dV(T) can be written as         dV(T) = a(T) dT because dT is infinitesimal, that is the very essence of the derivation.

which, using the expression for $dV(\tau)$, becomes $$v(\tau+d \tau) = \frac{ v(\tau) + a(\tau) d \tau }{ 1 + v(\tau) a(\tau) d \tau }.$$ To calculate $\frac{dv}{d \tau}(\tau)$ we write $$\frac{ v(\tau+d \tau) - v(\tau) }{d \tau} = a(\tau) \frac{1-v^2(\tau)}{ 1 + v(\tau) a(\tau) d \tau }$$ and take the limit $d \tau \rightarrow 0$, giving $$\frac{dv}{d \tau}(\tau) = a(\tau) ( 1 - v^2(\tau) ).$$ Rearranging to $$\frac{dv(\tau)}{ 1 - v^2(\tau) } = a(\tau) d \tau$$ and integrating between $\tau_0$ and $\tau$, writing $v_0 = v(\tau_0)$, this becomes $$\text{artanh}(v(\tau)) - \text{artanh}(v_0) = \int_{\tau_0}^\tau a(\tau') d \tau',$$ which, using the abbreviation $$A(\tau) = \int_{\tau_0}^\tau a(\tau') d \tau',$$ produces $$\text{artanh}(v(\tau)) - \text{artanh}(v_0) = A(\tau).$$ Inverting to $$\frac{ v(\tau) - v_0 }{ 1 - v(\tau) v_0 } = \tanh(A(\tau))$$ and isolating $v(\tau)$ gives $$v(\tau) = \frac{ v_0 + \tanh(A(\tau)) }{ 1 + v_0 \tanh(A(\tau)) },$$ or, written more tersely: $$v(\tau) = \tanh( A(\tau) + \text{artanh}(v_0) ),$$ which, with $v_0 = 0$, reduces to $$v(\tau) = \tanh(A(\tau)).$$ Use the standard SR Lorentz transformation (with $c=1$) between
the frame I and the instantaneously comoving inertial frame of A at time $\tau$: $$dt = \gamma(\tau) ( d \tau + v(\tau) d \xi )$$ $$dx = \gamma(\tau) ( d \xi + v(\tau) d \tau ).$$ Since we are working on the worldline of I where $\xi=0$ and thus $d \xi=0$, we get: \begin{align} \frac{dt}{d \tau} &= \gamma(\tau) \\ &= \frac{1} { \sqrt{1-v^2(\tau) } } \\ &= \frac{1} { \sqrt{1-\tanh^2( A(\tau)+\text{artanh}(v_0) ) } } \\ &= \cosh( A(\tau)+\text{artanh}(v_0) ) \end{align} and \begin{align} \frac{dx}{d \tau} &= \gamma(\tau) v(\tau) \\ &= \frac{v(\tau)}{ \sqrt{ 1-v^2(\tau) } } \\ &= \frac{\tanh( A(\tau)+\text{artanh}(v_0) )}{ \sqrt{ 1-\tanh^2( A(\tau)+\text{artanh}(v_0) ) } } \\ &= \sinh( A(\tau)+\text{artanh}(v_0) ) \end{align}
Integrated between $\tau_0$ and $\tau$, and with $x_0 = x(\tau_0)$ and $t_0 = t(\tau_0)$, this results in $$x(\tau) = x_0 + \int_{\tau_0}^\tau \sinh( A(\tau')+\text{artanh}(v_0) ) d \tau'$$ $$t(\tau) = t_0 + \int_{\tau_0}^\tau \cosh( A(\tau')+\text{artanh}(v_0) ) d \tau'.$$
If possible, eliminate $\tau$ to find the equation of the worldline $x(t)$
of the accelerated observer A in the frame of the inertial observer I: $$x(t) = \dots$$ If possible, invert the expression for $t(\tau)$ to find the proper time
of A as a function of the coordinate time $t$ of I: $$\tau(t) = \dots$$ and use it to find the velocity $v(t)$ as a function of coordinate time $t$: $$v(t) = \tanh( A(\tau(t))+\text{artanh}(v_0) ).$$
Summary:

$x$ and $t$: coordinates of object as seen in inertial frame

$a(\tau)$: felt proper acceleration as function of proper time $\tau$
$$A(\tau) = \int_{\tau_0}^\tau a(\tau') d \tau'$$ $$v(\tau) = \tanh( A(\tau)+\text{artanh}(v_0) )$$ $$\frac{dt}{d \tau} = \cosh( A(\tau)+\text{artanh}(v_0) )$$ $$\frac{dx}{d \tau} = \sinh( A(\tau)+\text{artanh}(v_0) )$$ $$t(\tau) = t_0 + \int_{\tau_0}^\tau \cosh( A(\tau')+\text{artanh}(v_0) ) d \tau'$$ $$x(\tau) = x_0 + \int_{\tau_0}^\tau \sinh( A(\tau')+\text{artanh}(v_0) ) d \tau'$$         Eliminate $\tau$ to find the worldline equation $x(t)$.

Note: These equations are derived in a different way in the article
http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf
See equations (3), (4), (5) on page 3.

Special case

When $\tau_0 = t_0 = x_0 = v_0 = 0$,
$$v(\tau) = \tanh(A(\tau))$$ $$\frac{dt}{d \tau} = \cosh(A(\tau))$$ $$\frac{dx}{d \tau} = \sinh(A(\tau))$$ $$t(\tau) = \int_0^\tau \cosh(A(\tau')) d \tau'$$ $$x(\tau) = \int_0^\tau \sinh(A(\tau')) d \tau'$$ Example:

The rocket with constant acceleration of the FAQ:
Take $$\tau_0 = t_0 = x_0 = v_0 = 0$$ and $$a(\tau) = a = \text{constant}.$$ So $$A(\tau) = \int_0^\tau a(\tau') d \tau' = a \tau$$ and $$v(\tau) = \tanh(A(\tau)) = \tanh(a \tau)$$ $$t(\tau) = \int_0^\tau \cosh(a \tau') d \tau' = \frac{1}{a} \sinh(a \tau).$$ $$x(\tau) = \int_0^\tau \sinh(a \tau') d \tau' = \frac{1}{a} ( \cosh(a \tau) - 1 )$$ Eliminate $\tau$:
$$\left(x+\frac{1}{a}\right)^2 - t^2 = \frac{1}{a^2}$$ giving the hyperbola $$x(t) = \frac{1}{a} \left( \sqrt{ 1 + (a t)^2 } - 1 \right).$$
Proper time as a function of coordinate time:
$$\tau(t) = \frac{1}{a} \text{arsinh}(a t)$$ so
\begin{align} v(t) &= \tanh( a \tau(t) ) \\ &= \tanh( \text{arsinh}(a t) ) \\ &= \frac{a t}{ \sqrt{ 1 + (a t)^2 } }. \end{align}
Re-introduce c and we find as functions of proper time $\tau$:

$$v(\tau) = c \tanh\left(\frac{a \tau}{c}\right)$$ $$\gamma(\tau) = \cosh\left(\frac{a \tau}{c}\right)$$ $$t(\tau) = \frac{c}{a} \sinh\left(\frac{a \tau}{c}\right)$$ $$x(\tau) = \frac{c^2}{a} \left( \cosh\left(\frac{a \tau}{c}\right) - 1 \right)$$ and as functions of coordinate time t:

$$v(t) = \frac{a t}{ \sqrt{ 1 + \left(\frac{a t}{c}\right)^2 } }$$ $$\gamma(t) = \sqrt{ 1 + \left(\frac{a t}{c}\right)^2 }$$ $$\tau(t) = \frac{c}{a} \text{arsinh}\left(\frac{a t}{c}\right)$$ $$x(t) = \frac{c^2}{a} \left( \sqrt{ 1 + \left(\frac{a t}{c}\right)^2 } -1 \right) \qquad (\text{the hyperbola})$$ which are the equations of the FAQ entry.

Furthermore we have the momentum given by $$p(t) = m \gamma(t) v(t) = m a t \qquad (\text{linear})$$ and the force $$F(t) = \frac{dp}{dt}(t) = m a \qquad (\text{constant})$$ Dirk Vdm

Much more straightforward and concise, using 4-vector formalism

Posted on Usenet:

Putting $c = 1$ and using accent notation for the derivative w.r.t. proper time $\frac{d}{d \tau}$:

Velocity 4-vector $$V = ( t', x', 0, 0 ) = ( \frac{dt}{d \tau}, \frac{dx}{d \tau}, 0, 0 ).$$ Acceleration 4-vector $$V' = ( t'', x'', 0, 0 ) = ( \frac{d^2t}{d \tau^2}, \frac{d^2x}{d \tau^2}, 0, 0 )$$ with unknown functions $t(\tau)$ and $x(\tau)$.

You have $V.V = 1$, giving $$\left(\frac{dt}{d \tau}\right)^2 - \left(\frac{dx}{d \tau}\right)^2 = 1.$$ Since $\frac{dV}{d \tau}.\frac{dV}{d \tau}$ is invariant, it must have the value it has in the comoving inertial frame, giving $$\left(\frac{d^2t}{d \tau^2}\right)^2 - \left(\frac{d^2x}{d \tau^2}\right)^2 = 0 - a^2.$$ These 2 differential equations are pretty common and easily solved. With the boundary conditions $t(0) = 0$ and $x(0) = 0$: $$t(\tau) = \frac{1}{a} \sinh( a \tau )$$ $$x(\tau) = \frac{1}{a} \cosh( a \tau ) - \frac{1}{a}$$ from which you can eliminate the proper time $\tau$ and find the hyperbola $$x(t) = \frac{1}{a} \left( \sqrt{ 1 + (a t)^2 } -1 \right)$$