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Hyperbolic motion in special relativity is rectilinear motion with constant proper acceleration. A representative positive-$x$ world line is $x=c{t}_0\cosh (\tau /t_0)$, $t=t_0\sinh (\tau /t_0)$, with proper acceleration $c/t_0$. This describes a hyperbola in spacetime, not a hyperbolic Kepler orbit.

Newtonian mechanics describes motion using forces, masses and acceleration in inertial frames with absolute time. It is the low-speed limit of special relativity; for constant proper acceleration with $|at|\ll c$, relativistic hyperbolic motion reduces to $x\simeq a{t}^2/2$.

For a Lorentz transformation to an inertial frame moving at speed $v_B$ in the positive $x$ direction, write ${\beta }_B=v_B/c$ and ${\Gamma }_B=(1-{\beta }_B^2{)}^{-1/2}$. Any four-vector transforms asThis convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ Minkowski metric as the four-momentum calculation.

The definition of proper time gives $d\tau =dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the four-velocity isSince $\dot{\gamma }={\gamma }^{3}u\dot{u}/c^2$, differentiating with respect to proper time givesHere $1+{\gamma }^2{u}^2/c^2={\gamma }^2$ simplifies the spatial component of the four-acceleration.

At each event, boost with $v_B=u$ to the instantaneous rest inertial frame. Then $A^{\prime 0}=0$ and $A^{\prime 1}=\gamma {\gamma }^4\dot{u}(1-u^2/c^2)={\gamma }^3\dot{u}$. Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar proper acceleration.

If this signed proper acceleration is constant and the particle starts from rest, integration gives $\gamma u=at$, soIntegrating once more and imposing $x(0)=0$ yields the hyperbolic motion worldlineFor $a=0$ the continuous limiting solution is $x=0$. Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than $c$.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

For motion on a fixed axis, the signed spatial four-acceleration in the instantaneous rest inertial frame is $a={\gamma }^3\dot{u}$. Its magnitude is the scalar proper acceleration. Constant signed $a$ from rest gives $\gamma u=at$ and the hyperbolic motion worldline $x=(c^2/a)(\sqrt{1+a^2{t}^2/c^2}-1)$.

A static observer at Schwarzschild radius $r>2M$ must maintain the outward proper accelerationIt approaches the Newtonian value $M/r^2$ at large radius and diverges at the horizon.

A particle released from rest in a constant electric field parallel to its motion follows a spacetime hyperbola and has rapidity proportional to proper time.