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Calculating with m/s$^2$ is very helpful when dealing with acceleration on the human range, as accelerating from rest at 4 m/s$^2$ for 3 seconds will give a velocity of 12 m/s. g is also a very helpful referent, as it describes the acceleration we are used to feeling.

For interstellar trips, the obvious unit choices would be light-years (ly) for distance and years (yr) for time, which would mean that acceleration would be in ly/yr$^2$. 1 ly/yr$^2$ is conveniently close to 1 g, and both are close to 10 m/s$^2$, which is nice. However, accelerating at 1 ly/yr$^2$ for 2 years does not result in a velocity of 2 ly/yr (as that would be faster than the speed of light, which is 1 ly/yr). While there are relatively uncomplicated formulae that result in the correct answer, the format of l y/yr$^2$ seems intuitively incorrect if it cannot be easily manipulated (indeed m/s$^2$ suffers from the same issue at such scales as well).

Is there an alternative way of expressing or formatting acceleration that, while perhaps appearing more cumbersome, better captures the effects of prolonged acceleration on velocities approaching the speed of light, and allows manipulation? I could not think of one off-hand, and have no reason to think there is, but I figured no harm in asking.

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    $\begingroup$ With higher speeds & velocities, it is often the case that one uses fractions of the speed of light (i.e., velocities in "units of $c$"). In this case, accelerations would have units of inverse time (e.g., 1/s). $\endgroup$ – Kyle Kanos Sep 15 '14 at 17:13
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Measuring acceleration in $ly/yr^2$ and speed in $ly/yr$ does seem like a convenient choice of units.

If you want to be able to express how fast the ship is going as a trivial function of acceleration and time, a way to do that that works correctly is to express how fast the ship is going by giving the ship's rapidity, instead of the ship's speed. In units in which the speed of light is 1, at low speeds rapidity is basically the same thing as speed. For example, a rapidity of $0.001$ is very nearly the same as a speed of $0.001 ly/yr$.

My point is that rapidity is given very simply in terms of acceleration and time. In units in which the speed of light is 1, the change in rapidity $\Delta\phi$ resulting from applying a constant proper acceleration $a$ for a duration $\Delta\tau$ of proper time is given by

$$\Delta\phi=a \Delta\tau\ ,$$

which at low speeds means the same thing as

$$\Delta v=a \Delta t$$

in Newtonian physics. To use the same numbers as you gave in your example, a proper acceleration of $2 ly/yr^2$ for 2 years of proper time does result in a rapidity of 2.

And if you do want to know the speed instead of the rapidity, that's given simply by $v=\tanh \phi$ in these units.

As a caveat, the above assumes a one-dimensional travel problem. Rapidity doesn't work as a vector the same way that velocity does.

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