Inextendible timelike worldline with finite proper time in Minkowski space? would anybody be able to give me an example of a timelike worldline in Minkowski spacetime which is inextendible (so will run off to infinity) and has a finite length? I'm thinking it will be something that looks like hyperbolic motion, but that doesn't have a fixed acceleration?
 A: Here's a simple example : take a timelike curve with unbounded acceleration in finite time. To do this, consider a curve with the acceleration
$$\|\ddot\gamma(\tau)\| = \alpha(\tau)^2$$
But still timelike : 
$$\|\dot\gamma(\tau)\| = -1$$
So the equations for it are 
\begin{eqnarray}
\ddot{x}^2(\tau) - \ddot{t}^2(\tau) &=& \alpha(\tau)^2\\
\dot{x}^2(\tau) - \dot{t}^2(\tau) &=& -1
\end{eqnarray}
Switching to null coordinates, this gives us
\begin{eqnarray}
\ddot{u}(\tau) \ddot{v}(\tau) &=& \alpha(\tau)^2\\
\dot{u}(\tau)\dot{v}(\tau)  &=& -1
\end{eqnarray}
This means that $\dot{u} = -\dot{v}^{-1}$, which gives us the relation
$$\left(\frac{\ddot{v}}{\dot{v}}\right)^2 = \alpha(\tau)^2$$
There's a few possible solutions to this (they'll correspond to various time and space orientation of the curve), but we'll pick the one where everything is positive, 
$$\ddot{v}  - \alpha(\tau)\dot{v} = 0$$
This is a fairly simple system to solve, with solution
\begin{eqnarray}
\dot v(\tau) &=& A \exp({\int_1^{\tau} \alpha(\xi) d\xi})\\
\dot u(\tau) &=& A^{-1} \exp({-\int_1^{\tau} \alpha(\xi) d\xi})
\end{eqnarray}
or, in Cartesian coordinates, 
\begin{eqnarray}
\dot t(\tau) &=& \sinh(\int_1^{\tau} \alpha(\xi) d\xi)\\
\dot x(\tau) &=& \cosh(\int_1^{\tau} \alpha(\xi) d\xi)
\end{eqnarray}
A very simple unbounded acceleration in finite time is to simply pick $\alpha = \tan(\tau)$, which will diverge for $\tau = \pi/2$. This gives us 
\begin{eqnarray}
\dot t(\tau) &=& \sinh(-\ln(\cos(\tau)))\\
\dot x(\tau) &=& \cosh(-\ln(\cos(\tau)))
\end{eqnarray}
The time coordinate can be calculated explicitely : 
\begin{eqnarray}
t(\tau) &=& \int \sinh(-\ln(\cos(\tau))) d\tau \\
&=& \frac 12 (- \sin (\tau) - \ln( \cos(t/2) - \sin(t/2) )) + \ln( \cos(t/2) + \sin(t/2) )
\end{eqnarray}
which tends to $-\ln(0) / 2$ at $\tau = \pm \pi/2$. 
This curve has no endpoint. It can be shown by considering that if it had an endpoint ending at $\tau_b$, there would be a neighbourgood where $\gamma(\tau_b + d\tau)$ belongs to. Just make some arbitrarily small neighbourhood around that point $(t_a, t_b) \times (x_a, x_b)$. Since $t(\tau_b)$ is strictly increasing and divergent, there can be no extension of that curve. 
The total proper time of the curve is just $\pi$ then.
