The last 25 years I thought that time dilation, mass increase and length contraction is "scaled" that way that on the first glance the crew of an accelerating starship could think everything is newtionain. I thought that when a starship accelerates constantly with 1g for 10 years (from the crews pov) it reaches roughly the speed that gives them a time dilation, mass increase and length contraction of factor 10, so they expierience a 10 light years journey (from stationary pov) in just one year of their personal time. From stationary pov they need 10 years and time ist slowed down to 0.1, from ships pov they need one year because the distance is contracted by 0.1. So I thought that you have to accelerate for 100 years with 1g (ships pov) to get a time dilation of factor 100.

Yesterday I read that article.


which states:

At a constant acceleration of 1g, a rocket could travel the diameter of our galaxy in about 12 years; if the last half of the trip involves deceleration at 1g, the trip would take about 24 years.

1g constant acceleration for 12 years could only be meant from the ships pov.

But on the other hand it says:

A half-myth: It gets harder to push a ship faster as it gets closer to the speed of light This is a half-myth because it depends on the frame of reference. It is true for those watching from the planetary reference frame. For those experiencing the journey (in the ship's reference frame) it is not true. For both the planetary frame and the ship's reference frame, the ship will change speed in a Newtonian way—push it a little and it speeds up a little, push it a lot and it speeds up a lot. However, in the planetary frame the ship will appear to be gaining mass due to its high kinetic energy, and the mass-energy equivalence principle. Should the engines be giving a constant thrust, this will result in progressively smaller acceleration due to the higher mass it is required to accelerate. From the ship's frame, the acceleration would continue at the same rate.

which seems to backup my thoughts. But apparently I was wrong that the scaling of the effect acts the way I thought.

Am I wrong, or the article?

  • $\begingroup$ If somebody can come up with a better title of the question that comes more to the point, please edit it. $\endgroup$
    – Hothie
    Commented Oct 12, 2015 at 15:38

1 Answer 1


Ok. I think I found the answer here:


Which is from "John Baez's Stuff". Afaik he is an expert on this topic. :-)

First of all we need to be clear what we mean by continuous acceleration at 1g. The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket (see relativity FAQ on accelerating clocks). This acceleration will be denoted by a. The proper time as measured by the crew of the rocket (i.e. how much they age) will be denoted by T, and the time as measured in the non-accelerating frame of reference in which they started (e.g. Earth) will be denoted by t. We assume that the stars are essentially at rest in this frame. The distance covered as measured in this frame of reference will be denoted by d and the final speed v. The time dilation or length contraction factor at any instant is the gamma factor γ.


T          t         d          v                γ
1 year     1.19 yrs  0.56 lyrs  0.77c            1.58  
2          3.75      2.90       0.97             3.99
5          83.7      82.7       0.99993          86.2
8          1,840     1,839      0.9999998        1,895
12         113,243   113,242    0.99999999996    116,641


In the rocket, you can make measurements of the world around you. One thing you might do is ask how the distance to an interesting star you are headed towards changes with T, the time on your clock. At blast-off (t=T=0) the rocket is at rest, so this distance initially equals the distance D to the star in the non-accelerating frame. But once you are moving, however you choose to measure this distance, it will be reduced by your current distance d travelled in the non-accelerating frame, as well as the whole lot contracted by a factor of γ, your Lorentz factor at time T.


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