# How does gravitational time dilation work in artificial gravity made by rotating a cylinder? [duplicate]

Concerning gravitational time dilation in artificial gravity (made by a rotating torus like in many sci-fi movies) how would you go about calculating the effect?

## marked as duplicate by ACuriousMind♦, Kyle Kanos, John Rennie, RedGrittyBrick, Carl WitthoftJul 31 '15 at 11:45

• Hi Joe. If you look at my answer to the question I've linked I explain how to calculate the time dilation for rotational motion. It's for a centrifuge rather than a space station, but the calculation is exactly the same. As Ed says in his answer, the time dilation is the same as time dilation for straight line motion at the same speed. – John Rennie Jul 31 '15 at 6:20

The time dilation from the motion will be equivalent to a gravitational time dilation with effective mass of $M_{eff} = \frac{v^2r^2}{G}$ (where r is the radius of the rotating torus). And there is a major difference between this time dilation and the usual special relativity time dilation with inertial frames. Since anyone or thing in the torus is undergoing constant acceleration it is constantly changing inertial frames. This motion is therefore not relative and the clocks in the non-rotating frame appear to the entities in the rotating frame to be going faster. Thus, when a twin leaves the rotating frame, he/she will actually be younger than the stay at home twin.