# Time dilation at perceived constant acceleration

Let's say that we have a spaceship. From the beginning, the ship is stationary (relative to me) and not affected by outer forces (like Earth's or Sun's gravity). Then the ship starts to accelerate at $$10\space m/s^2$$, that should be comfortable for the astronaut onboard (artificial gravity near $$g$$). Normally this is an easy problem: after 1 s the speed is 10 ms/s, after 2 s the speed is $$20\space m/s$$, etc. But the perceived acceleration is constant for the astronaut, meaning that we need to account for time dilation (otherwise, after a year the ships speed would be higher than $$c$$).

How do I calculate the acceleration of the ship after $$x$$ seconds? (the ships acceleration from my point of view, not the ships) As I understand the acceleration constantly increases the velocity, which increases the time dilation, which decreases the acceleration.

The acceleration measured in the rest frame of the rocket, i.e. by the people on the rocket, is called the proper acceleration. Constant proper acceleration is a standard problem that you'll find treated in all books on relativity, and you'll find a summary of the results in Phil Gibbs' article on the relativistic rocket.

You ask specifically about the acceleration measured by the observer watching the accelerating rocket, and this is given by:

$$a' = \frac{a}{\gamma^3}$$

where $$a$$ is the proper acceleration of the rocket and $$\gamma$$ is the Lorentz factor. The Lorentz factor is given by:

$$\gamma = \sqrt{1 + (at/c)^2}$$

Combine these to get the acceleration as function of time.

The relativistic rocket article I linked includes lots of other useful equations e.g. for the speed of the rocket observed from Earth, the distance travelled of the rocket, etc. If you're curious to see the derivations see Chapter 6 of the book Gravitation by Misner, Thorne and Wheeler.

• Thnx, it was proper acceleration I was looking for. – user255136 Jul 17 '19 at 15:30