 The Revolving Clock
Notice that I did not say "the orbiting clock".  The effects of the gravitational field on an orbiting clock are straightforward and easy to understand, at least at this level:  the deeper in the well you go the slower you go.  But the issue considered on this page is just the effect of the clock's motion, which causes the clock to go more slowly than a stationary clock in the middle of the circle.  So the clock is assumed to be moving in a circle under its own power, or because it's attached to a tether -- no gravitational field is involved, and space is assumed to be flat.

The revolving clock is moving relative to the stationary clock, so it goes more slowly.  But the stationary clock is moving relative to the revolving clock -- so why doesn't it go more slowly, instead?

We'll look at this from the point of view of the stationary clock, which is simple, and the point of view of the revolving clock, which is not so simple.

### Brief Discussion of the Method

As discussed in the Linear Twins, the elapsed time between any two events experienced by a non-accelerating object in flat space is equal to the proper distance between those events.

If the space isn't flat, or if the object is accelerating, then we can't just take the difference in the coordinates of the two events and use that length, because the object didn't travel in a "straight line".  Instead, we need to follow the path the object traced out between the events, finding the infinitesimal proper length of each infinitesimal bit of the path and add them up.  In other words, we need to integrate -- the integral of the proper length of the velocity of the object over (coordinate) time gives the elapsed proper time for the object.

From the point of view of a stationary clock watching a moving clock, this is easy.  We know the velocity, we know the coordinates, we integrate and we're done.  When the frame of reference we're working in is accelerating, however, it's harder.

In many cases we can evaluate what we need in an inertial frame which is momentarily moving at the same rate as the observer:  the Momentarily Comoving Reference Frame.  For integrating the path of the stationary clock from the PoV of the revolving clock, however, that proves difficult.  The problem arises when we try to imagine adding an infinitesimal piece of the path, and then moving on to the next piece -- with each step, we must change to a new MCRF, and the change affects the time as seen by the stationary clock!  So evaluating the integral over the path using the MCRF for each piece is not at all straightforward.

So, when I treat the moving clock's point of view, I just switch to a coordinate system in which the moving clock is "stationary".  I then stick with that coordinate system for the duration of the integral.  As long as the metric is properly transformed, the result will be the same.

### An Intuitive Discussion of What's Going On

As discussed in the Against Gamma, a helpful way to view "time dilation" is to imagine that the moving object is taking a shortcut through the observer's reference frame.  From the point of view of the stationary clock, it's reasonable to imagine that the revolving clock is doing exactly that.

But from the point of view of the revolving clock, something strange is happening.

Before way say any more, consider an inertial frame in motion relative to an observer.  In the observer's coordinates, draw a line through the (spacial) origin perpendicular to the line of motion.  Consider one instant of time, T.  All events on the line we drew which happen at time T in the observer's frame are simultaneous in that frame.  Furthermore, all of those events are simultaneous in the moving frame, as well.  It's a line through space, but we might call it the "line of simultaneity" with the observer at the origin.

As we consider points in the moving frame which have passed the line, we find that the moving clocks are reading progressively earlier.  Turning that around, as points in the moving frame pass the line, they move into the future in the observer's frame.  They are, indeed, taking a "shortcut", moving into the future by moving through space.

Now, consider the revolving clock.  When we consider the stationary clock's path through the revolving clock's space, we see that the stationary clock is on the line of simultaneity.  As it "moves" through the revolving clock's space, the velocity of the moving clock rotates, and the line of simultaneity rotates as well.

So, if we try to integrate the "motion" of the stationary clock through the space of the moving clock, we find something odd going on.  We add one infinitesimal time step, in which the stationary clock moves a small distance in the time dimension and moves a small distance through time as a result of moving through space across the line of simultaneity.  And then we switch to the next MCRF -- and we find that the line of simultaneity has rotated, and the stationary clock has backed up across it, essentially canceling the "spacial" component of its motion through time!

And that leads us to conclude that we may expect to see just the time portion of dτ/dt here -- and indeed, that is exactly what happens.  From the point of view of the revolving clock, the stationary clock undergoes time contraction, not time dilation.

Beware, though -- as with all intuitive "explanations" of relativistic effects, this can be taken too literally and carried too far.  Only the mathematics really captures what's going on, and for that, see the remainder of this page.

### A Solution:  Viewed from a Stationary Clock

The coordinate system is (t, x, y).  The stationary clock is located at (t, 0, 0).  The revolving clock is moving with constant linear velocity v, and its equations of motion are: (NOTE:  This "velocity" is not the "4-velocity" -- it's the coordinate velocity, which is the 4-velocity divided by γ. But since we'll be integrating over dt rather than dτ, because we haven't found dτ yet, it is what we need here.)

The stationary clock is at rest in an inertial frame so its proper time at coordinate time T will just be T.

The squared proper length of the velocity of the revolving clock will be: And the proper time for the moving clock will be the integral of the square root of the negative squared proper length of its velocity: where ### Another Solution:  Viewed from the Revolving Clock

It would be nice to see this from the point of view of the revolving clock in "flat" coordinates, but that seems to get very messy very quickly.  The problem is that the coordinate system for moving clock's frame is, in effect, rotating as well as translating, which makes it hard to see how to say anything about the stationary clock's behavior using the MCRF.

So we will use an accelerated frame, which will allow us to stick with a single FoR throughout the problem.

The ship's coordinate system (the "primed", or "Greek" frame) will use coordinates (τ, ξ, ζ), defined as We need the metric in the primed frame, and one easy way to obtain it is to use the transformation matrix from the primed frame back to the unprimed ("stationary") frame: We need one additional partial before we can easily evaluate the partials in the matrix: Plugging in values for the partial derivatives, we have: We can then compute the metric in the primed frame, using matrix multiplication notation (no implicit summation on tensor indicies here), as or, after working it out, The path of the ship in ship's coordinates is just (τ, 0, 0), with velocity (1, 0, 0).  To find the ship's elapsed proper time, we need to integrate the length of the velocity vector over the path.  By inspection of the metric, we can see that it's always just 1, so the path length from time 0 to time T will just be T: The path of the stationary clock, given by position vector and coordinate velocity, is The squared proper length of the velocity is: And the elapsed (proper) time for the stationary clock at time T is the integral of the square root of the negative squared length, or which agrees with the result calculated in the stationary clock's frame.