# Simultaneity, rotation and gravity.

1. Mar 11, 2014

### yuiop

We have had a number of threads on how to synchronise clocks around a rotating ring. One method of doing this is to start all the clocks on the ring via a signal from the centre of the ring. This method has the advantage of being transitive, but has the disadvantage that the local one-way speed of light depends on which way it is going around the ring. Einstein synchronisation has the advantage that the local one way speed of light is independent of direction. I think it was Pervect that pointed out that Einstein synchronisation also has the advantage that locally things still work according to our Newtonian expectations. For example two equal masses sent simultaneously with equal speed in opposite directions around the ring, would collide and come to rest with the ring at the mid point, if the collision is inelastic. This would not happen with the transitive synchronisation.

I was wondering what would happen if we use gravity to define simultaneity. Consider a balance with equal length horizontal arms. I would expect that if I place equal weights at the ends of the arms simultaneously, the balance would not rotate, but which definition of simultaneity applies to gravity if the balance is attached to a rotating ring? My initial thoughts are that for the balance to 'balance', the weights would have to attached simultaneously according to the Einstein synchronisation convention. If the weights were attached simultaneously according to the transitive clocks, the leading weight would appear to be attached to the balance before the trailing weight (according to the Einstein clocks) and the balance would rotate about its fulcrum in the same direction as the rotation of the ring in the time interval between the first and second weights being attached. It seems as far as gravity is concerned, Einstein synchronisation is the 'natural' method. Agree?

Last edited: Mar 11, 2014
2. Mar 11, 2014

### bcrowell

Staff Emeritus
The arm of the balance can't be perfectly rigid. Transverse waves will travel along it at some speed v<c. So I think this scheme amounts to doing something like Einstein synchronization, but carrying it out using mechanical waves rather than light waves.

3. Mar 11, 2014

### yuiop

On a related subject, consider geodesic orbits in the Kerr metric. The coordinate geodesic orbital velocity is given by:

$\frac{r}{a\pm \sqrt{r^3/m}}$

The coordinate speed of light is given by:

$\frac{2 m a \pm r \sqrt{r^2+a^2-2mr}}{(r^2+a^2+2ma^2/r)}$

where in both cases, the positive sign is for prograde.

Taking the ratio, the geodesic orbital velocity relative to the speed of light is:

$\frac{r(r^2+a^2+2ma^2/r)}{(a\pm \sqrt{r^3/m})(2 m a \pm r \sqrt{r^2+a^2-2mr})}$

The implication is that even a local stationary observer will see satellites at the same altitude orbiting at different velocities, with the retrograde satellites orbiting faster. Things get worse from the point of view of a ZAMO observer. However, all the above is assuming that observers on a ring of radius (r) synchronise there clocks in a transitive manner. The question is, if we use Einstein synchronisation, would the orbits have equal speeds in both directions according to the stationary or ZAMO observers? If there a unique rotation speed for a ring that gives equal orbital speeds? Is there anything else special about the frame in which the orbital velocities are equal?