Light catching up with a uniformly accelerating object Is there a formula for how long it takes light to catch up with a constantly accelerating object given initial distance?
 A: Geometrically, restricting to 1+1 Minkowski spacetime, your problem asks for the intersection of a hyperbola with timelike tangents (and lightlike asymptotes) and a light signal (a lightlike line). Because of the Rindler horizon, there are some starting events [on the lightlike line] that cannot reach the hyperbola.... so “never” (which is not infinity) would be the answer in those cases. In cases where there is an intersection, it sounds like it should be doable.
(The problem might be easier if you work in light cone coordinates.) In addition to the event, the Spatial direction (forward or backward) of the light signal should be specified and you may wish to allow intersections in the past.
UPDATE: Possibly useful
https://math.stackexchange.com/questions/2676200/prove-all-lines-parallel-to-an-asymptote-of-a-hyperbola-intersect-the-hyperbola 
Of course, a uniformly accelerated object traces out only one branch of the hyperbola.
A: Yes. I follow Thorne, Misler, and Wheeler's derivation of the formula for a uniformly accelerating object, and then my own quick work for the formula for the light catching up. Work is in natural units.

Consider a 1+1 Minkowski space. The four velocity has constant magnitude $-1$, so the 4-acceleration is orthogonal to the velocity vector:
$$u^\mu a_\mu=\frac{1}{2}\frac{d}{d\tau}(u^\mu u_\mu)=0$$
Therefore, when the object is at rest initially, the time component of the 4-acceleration is $0$, so $a^\mu=(0,a)=\left(0,\frac{d^2x}{dt^2}\right)$ in the rest frame. Therefore, in all frames, $$a^\mu a_\mu=\left(\frac{d^2x}{dt^2}\right)^2$$
Now, consider a particle undergoing uniform acceleration $g$ in the $x$ direction. We immediately have
$$\frac{dt}{d\tau}=u_0,\;\;\frac{dx}{d\tau}=u_1, \;\; \frac{du_0}{d\tau}=a_0,\;\;\frac{du_1}{d\tau}=a_1$$
We also have three algebraic relations:
$$ \begin{align}
u^\mu u_\mu &= -1 \\
u^\mu a_\mu &= -u^0 a^0 +u^1a^1=0 \\
a^\mu a_\mu &= g^2 \\
\end{align}
$$
One can solve these for the acceleration to find:
$$a_0=\frac{du_0}{d\tau}=gu^1, \;\; a_1=\frac{du_1}{d\tau}=gu^0$$
This are fairly trivial to solve, and combined with a suitable choice of origin, one finds $$gt=\sinh g\tau, \;\; gx = \cosh g\tau$$ from which we find $$x^2-t^2=\frac{1}{g^2}$$

Now, to solve for the light catching up. We will assume both $x>0$ and $t>0$. One sees that the asymptote is $t= x$. By setting $t=0$, one immediately sees that the x-intercept is $x=\frac{1}{g}$. Since light travels on lines of $t = \pm x - b$, the sign on the $x$ must be positive (or it goes the opposite way), and $0<b\leq \frac{1}{g}$. Solving for $t$, we get $$\bbox[5px,border:2px solid black]{t=\frac{1}{2bg^2}-\frac{b}{2},\;\;0<b\leq\frac{1}{g}}$$
Converting out of natural units, we have $$t=\frac{c^3}{2bg^2}-\frac{b}{2c},\;\;0<b\leq\frac{c^2}{g}$$
Note that the answer is for the rest frame, and not for the accelerating object's frame.
