Re: 1 g acceleration?

Hal Finney (
Sun, 5 Jul 1998 10:29:23 -0700

>> I guess my real question is, according to the people on the
>> ship, does it appear to take more and more energy to maintain a
>> constant 1 g acceleration experience or is this only true in the
>> space/time frame of the people on earth? So, according to the people
>> on the ship, you can accelerate at 1 g forever right?

By the principle of relativity (which says that you can't tell if you're moving, i.e. all motion is relative), it can't make any difference how fast someone else thinks you are going. So it will take the same amount of power, from your point of view, to accelerate at 1 g, no matter how fast someone else thinks you are going.

As you go faster relative to the universe, it looks to you like the universe (stars, galaxies, etc.) is going faster relative to you. As the stars approach the speed of light, they don't speed up as much as you would expect given the 1 g acceleration that you feel. They approach the speed of light slower and slower, but do not exceed it. What happens instead is that they undergo Lorentz contraction, flattening out in the direction of motion.

The net result is that you go past stars, galaxies, etc., faster and faster, not because they are moving past you more quickly, but because they are becoming thinner and flatter, so even though they never move faster than the speed of light, you move past more of them in a given amount of time.

In relativity, given constant acceleration "a", the equations of motion are as follows:

Let t be the time measured by the moving observer. Then the distance he has travelled as measured by a stationary observer is:

X = 1/g * sinh (g*t).

The time it has taken for him to travel this far, as measured by a stationary observer, is:

T = 1/g * cosh (g*t).

sinh and cosh are "hyperbolic sine" and "hyperbolic cosine", defined as

sinh(x) == 1/2 * (e^(x) - e^(-x))
cosh(x) == 1/2 * (e^(x) + e^(-x))

Here, x, t, and a must be measured in relativistic units where c = 1 e.g. x is in light years, t is in years. In these units, 1 g is approximately 1/light-year. So after 1 subjective year at 1 g, t = 1


X = 1/2 * (e - 1/e) = 1.18 light years T = 1/2 * (e + 1/e) = 1.54 years

After a large number of subjective years at 1 g, the equation can be simplified to approximately X = T = e^t. You are moving at close to the speed of light, hence X = T in these units (e.g. it takes 100 years for you to travel 100 light years as measured by a stationary observer). But with time as measured by you, the distance you cross in terms of galaxy units is exponential.