# Generating artificial gravity by using rotation

If I want to design for futuristic space stations and I want to use rotation to produce artificial gravity. One of such designs consists of a giant ring that is rotated about its centre. If it were rotating fast enough, inhabitants in the ring will feel a ‘gravitational’ force just like on Earth.

Suppose there are two clocks in the space station, one at the centre and the other attached to the rotating ring. Which clock would be slower?

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The clock placed on the rotating ring would tick slower by a factor of $\sqrt{1-\omega^2 r^2/c^2}$, where $\omega$ is the angular velocity of the ring, $r$ is the radius of the ring and $c$ is the speed of light.

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In other words it is only the speed of the clock that makes it slower. The artificial "gravity" ie acceleration, has no effect. –  Martin Beckett Mar 19 at 3:00
so I can interpret this effect as the acceleration itself has no effect on the speed of the clock, however the speed caused by the artificial "gravity" makes the clock runs slower? –  noPDE Mar 19 at 3:58
This is a rather subtle issue. In General Relativity, observables (like time, distance, etc.) are dependent on the metric tensor field. The value of metric is determined by the energy/momentum distribution in spacetime. It is assumed in your problem that the objects involved have little effect on the metric. The metric of a rotating coordinate system (i.e. the rotating clock) yields the above result. If the ring, etc., were sufficiently massive then they too would have an effect on the metric. –  elfmotat Mar 19 at 4:59
You can view it as either due to the velocity (though you must use instantaneous comoving intertial reference frames instead of the accelerated rest frame of an observer on the ring) or the acceleration (in the rest frame on the ring). Either works due to the equivalence principle. Both are equivalent to the metric construction elfmotat describes. –  Michael Brown Mar 19 at 5:29