# Another technical post – on time and constant acceleration

For those of you who missed it, a week ago I wrote a post in which I discussed how long it might take to get to the newly discovered exoplanet Gliese 581g.  There was a ton of interest from other sites, and a number of people have posted in their blogs and comments that they might be interested in seeing the details behind the relativistic travel time calculations to other stars.   I’ve already done a post on the energy requirements of matter-antimatter drives a couple of days ago.  Today, I’d like to talk about the time dilation effects, since I don’t know that I’ve seen a web version of this calculation.

Note: As in my last entry, this is a very technical discussion, so if you don’t have some calculus and a bit of relativity under your belt, you’re going to have a tough time following.  For those who are new readers to the blog, I normally keep things at a much lower level (just avoid the “technical” tag).  And, just so you know, there is only one equation in our book.

Okay, enough caveats.  Let’s get started.

Suppose you’re pushing a ship of mass, $m$, with Force, $F=mg$, by whatever mechanism you like.  The idea is that inside the ship, you feel an acceleration, $g$, toward the back of the ship.  It’s like you have artificial gravity! As I mentioned before, if you want an artificial gravity of earth normal, you’ll want:

$g=9.8 m/s^2=1.03 ly/yr^2$

The basic idea is that you accelerate during the first half of your trip and decelerate during the second half.  I’m only going to do the acceleration part here, but the travel times just double when you include the second leg.

Now suppose you’ve traveled a distance, $x$, while accelerating.  The total amount of work applied to your ship is:

$W=Fx=mgx$

For a relativistically moving ship, the energy is:

$E=mc^2gamma$

where, as always, the Lorentz factor is:

$gamma=frac{1}{sqrt{1-v^2/c^2}}$

Thus means that since the ship started with energy $E=mc^2$, the work needed to accelerate a distance, $x$, is equivalent to:

$Delta E=mc^2(gamma-1)=mgx$

or juggling things around:

$gamma=frac{gx}{c^2}+1$

We’re already in a position to figure out how fast the ship is moving once it’s a distance, $x$ from earth, just by inverting the $gamma$ relation:

$v=csqrt{1-frac{1}{gamma^2}}=csqrt{1-frac{1}{(gx/c^2+1)^2}}$

Just to put things in perspective, if you plug in some numbers, after you’ve traveled for 1 lightyear at Earth-normal acceleration, you’ll have reached a speed of 87% the speed of light.

The trip from earth perspective

How long does it take to travel this far? From the earth’s perspective, we start with the simple relation:

$v=frac{Delta x}{Delta t}$

$dt=frac{dx}{v}$

for every step in the trip.  Thus:

$t=int dt=int frac{dx}{csqrt{1-frac{1}{(gx/c^2+1)^2}}}$

this looks like a giant mess, but it simplifies and integrating from $x=0$ to \$d\$ yields:

$t=sqrt{frac{d(gd/c^2+2)}{g}} (1)$

Just plug in your acceleration, how far you want to go, and you’re all set.  For instance, if you plug in a distance of $10.25ly$ (half the trip to Gliese), $g=1.03 ly/yr^2$ (and remember that $c=1 ly/yr$, you get 11.2 years, half the value of the total trip.  Plotting the function:

Notice how this looks. For the first year or so, the curve appears to be a parabola. That’s another way of saying that you get the normal relation you learn in your first term of classical mechanics:

$d=frac{1}{2}at^2$

But after that, the curve looks like a straight line.  In other words, the ship is traveling at close to the speed of light, so approximately constant distance per unit time.

The trip from the ship’s perspective

Time seems to run slower from the ship’s point of view.  In general:

$dt'=frac{dt}{gamma}$

Since $gamma ge 1$, time will always run slower on the ship than back on earth.

To figure out how long the trip will take in all, we simply compute the integral:

$t'=int dt'=int frac{dx}{gamma v}$

which, after substituting in yields:

$t'=frac{clnleft(1+gd/c^2+sqrt{gd/c^2(gd/c^2+2)}right)}{g} (2)$

which is a) atrociously ugly, and was b) tough to compute.  I wussed out and used Maple.

Plugging in for $d=10.25ly$ again, we get: 3.05 years, which means the entire trip to Gliese 581g will take 6.1 years.  More generally:

The ship goes faster and faster, so time slows and slows very little time appears to pass for the person on the ship, despite traveling an enormous distance. Perhaps this plot will be more informative:

In this case, the axes are switched.  I’m showing how long it takes to go various distances (again accelerating at earth-normal gravity) according to earth (solid) and the ship (dashed).  At first, when the ship is traveling much less than the speed of light, the curves overlap.  Of course this is the case, there is very little time dilation.
However, after a while (about 6 months or so), they start to diverge, and the ship time is notably shorter than the earth time.
Feel free to use these calculations to amuse and amaze your friends.  Compute the travel time to any system!
-Dave

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### One Response to Another technical post – on time and constant acceleration

1. Pat Luther says:

Awesome!

This is exactly what I was trying to come up with a few years ago, and various times since, but failed due to my lack of remembering much calculus.