Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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 '13 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 '13 at 3:58
1  
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 '13 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 '13 at 5:29

Well, Einstein discovered that objects moving at great speed will appear to slow down. For example, if you could observe the passengers on a train moving close to the speed of light, they would appear to be going about their business-sending a message, reading a book, talking on the phone-much more slowly than if they had been at rest. This effect is negligible for common velocities, of course. So yes, your clock will show an earlier time on the outer ring than than the one on the center, but for most purposes, this difference will be negligible as well.

share|improve this answer
    
But the person on the ring would observe the person at the center of the ring to be moving, not him and so the person at the center's clock would be slower? Is my statement contradicted by the fact that the person on the ring knows he is moving due to the felt gravitational field/ acceleration? –  Keir Simmons Mar 28 at 5:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.