Outrunning a light-speed signal for a limited amount of time by accelerating and then coasting

Question

I'm not a physicist but I was attempting to write a story that includes some special relativity shenanigans that I hoped someone could verify.

The gist of it is that people in a spaceship with a 60 day head start need to outrun a signal sent from their starting point, by accelerating for $t_1$ time (as perceived from within the ship) at $5g$ and then coasting at the resulting speed so that they are only reached by the signal after 80 years (as perceived from within the ship).

I found this answer very helpful: What is the proper way to explain the twin paradox? but I believe those trajectories are measured from a rest frame outside of the ship.

It seems to me that you could only start coasting once you were $c*(80*365*24*60*60)$ meters away from the source of the signal, a distance which I believe I can calculate from the above linked answer, but would always take longer than 80 years, from the frame of reference of the source of the signal.

But I'm struggling to understand how long it would take to get that far away when measured from within the ship.

Is it possible to outrun a light-speed signal if you have a head start, accelerate for a while and then coast? Or do you have to keep accelerating, never coasting, to that the signal never reaches you?

Answer 1

The problem doesn't have a solution with the given parameters. The reason is that acceleration at $5g$ has a curvature radius in spacetime of $c^2/5g \approx 71\text{ days}$, which is larger than the 60 day separation between the light and the ship. So the light will reach the ship while it's accelerating, in a lot less than 80 years.

If you increase the acceleration or the head start a bit then the problem has a solution, though it will require accelerating to an implausibly high speed to outrun the light for 80 years.

Say the light's path is $x=ct$, and the ship starts at rest at $x(0)=x_0,\,t(0)=0$, accelerates at $a$ for a proper time $τ_1$, then coasts.

After the acceleration, the ship's rapidity is $α = aτ_1/c$, its position is $x(τ_1) = x_0 + (c^2/a)(\cosh α - 1)$, $t(τ_1) = (c/a)\sinh α$, and its velocity is $x'(τ) = c \sinh α$, $t'(τ) = \cosh α$. Its position at any later time is therefore

$$\begin{eqnarray} x(τ) &=& x_0 + (c^2/a)(\cosh α - 1) + c(τ-τ_1)\sinh α \\ t(τ) &=& (c/a)\sinh α + (τ-τ_1)\cosh α \end{eqnarray}$$

Setting $x(τ) = c t(τ)$ and some manipulation gives

$$x_0 + (c^2/a)(e^{-aτ_1/c} - 1) = c(τ-τ_1)e^{-aτ_1/c}$$

or (solving for $τ_1$)

$$τ_1 = τ - \frac{c}{a} \left( 1 + W(e^{a τ/c - 1} (a x_0/c^2 - 1)) \right)$$

(where $W$ is the Lambert W function).

As an example, with $x_0 = 60\text{ days}$, $a = 6g$, $τ = 80\text{ years}$, I get $τ_1 \approx 603\text{ days}$. In that time, they'll accelerate to a rapidity of about $10.2$, which is about $0.999999997c$.