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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 $c$.