Notes below outline (but do not complete) a novel solution to this problem that makes use of Galilean chase-plane time T, and Galilean-kinematic velocity V = dx/dT. These obey Galileo's equations for constant (proper) acceleration at any speed, even the classical expression for kinetic energy K=½mV2. A solution in terms of the more widely used (and useful) coordinate-velocity v = dx/dt may instead be found here. In both cases, we also find it useful to define from Minkowski's spacetime Pythagoras' theorem, i.e. the metric equation: (c dτ)2 = (c dt)2 - (dx)2, the proper or traveler time τ and the proper velocity w = dx/dτ.
where the speed of light c is 3x10^8[m/s2] or 1[ly/yr], proper acceleration α is 9.8[m/s2] or 1.03[ly/yr2], and wf & wo denote final & initial proper velocities (w = dx/dτ) in units of "distance traveled per unit of traveler time". Relativists will recognize proper velocity w as c Sqrt[γ2-1], where γ = E/mc2. Proper velocities here can be figured by the conversion
from "Galilean velocities" Vf and Vo which obey the classical equations for constant acceleration. The standard equation
(where dx is distance traveled) in particular should do the job. Given initial/final proper velocities, the maptime dt elapsed is simply (wf-wo)/α, and coordinate velocity v=dx/dt is w/γ. Can you show that finite proper velocities w require that v is always less than c?
Note: The "Galilean velocity" V is the velocity familiar from introductory physics books extended to all velocities as Galileo might have presumed through the kinetic energy equation K = (1/2)mV2. It thus becomes the derivative of x with respect to a "Galilean chase-plane time" T which can be used to track events during 1D acceleration, but which at high speed follows neither traveler nor earth based clocks. At high speeds Galilean velocity V is not the Lorentz/Minkowski coordinate velocity v = dx/dt.
* The Andromeda Galaxy is one of the most distant objects visible to the naked eye. Total distance traveled is 2.2x106 [ly] x 9.46x1015 [m/ly] = 2.08x1022 [meters].
(Thanks. /philf :)