Relativistic rocket

Well, not exactly. I am not accounting for loss of mass for propulsion. Here I am just writing down the relativistic motion with constant proper acceleration. So Alice is riding a rocket that has some kind of propulsion such that she experiences a constant force. Her acceleration is: $\alpha^{\prime\nu}=(0,\alpha)$, where $\alpha$ that does not depends on proper time $\tau$, which is a faithful parameter for the motion in all reference frame.

Let $\beta$ be Alice's instantaneous velocity with respect to an observer Bob. The acceleration in Bob's frame is:

$$\alpha^{\mu} = \Lambda_{\nu}^{\mu} \alpha^{\prime \nu} = \gamma \begin{pmatrix} 1 & \beta \\ \beta & 1 \end{pmatrix} \begin{pmatrix} 0 \\ \alpha \end{pmatrix} =\begin{pmatrix} \alpha \sinh w \\ \alpha \cosh w \end{pmatrix}$$

where $\beta = \tanh w$ and $w$ is the rapidity.

Alice's four-velocity in Bob's frame is:

$$U^{\mu} = \begin{pmatrix} \cosh w \\ \sinh w \end{pmatrix}$$

From the definition, $\alpha^{\mu} = \frac{dU^{\mu}}{d\tau}$, we get $w = \alpha \tau$. This is the reason why rapidity is a better quantity to use to describe Alice's motion than her velocity. Integrating the four-velocity with respect to proper time to get Alice's coordinate:

$$x^{\mu} = \int_{0}^{\tau} U^{\mu} (\tau^{\prime}) d\tau^{\prime} = \frac{1}{\alpha} \int_{0}^{w} U^{\mu}(w^{\prime}) dw^{\prime} = \frac{1}{\alpha} \begin{pmatrix} \sinh w \\ \cosh w - 1 \end{pmatrix}$$

where we applied the initial condition that Alice started her accelerating motion at the same position as Bob.

Plotting parametrically $x^0$ and $x^1$, one sees that the trajectory of Alice in Bob's frame asymptotically approaches a line of slope 1, reflecting the fact that Alice will get arbitrarily close to the speed of light. Indeed, the asymptote of the trajectory is the line $x^0 = x^1 + 1$. Therefore any signal that Bob sent to Alice after $x^0 = 1$ will never reach Alice because the signal will have to be superluminal. From Alice's point of view, as she accelerates away from Bob, there will be a horizon that forms behind her such that any signal from the other side can never reach her.