The Revolving Clock |

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.

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

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.

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,

As we consider points in the moving frame which have

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

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 that leads us to conclude that we may expect to see

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.

The stationary clock is at rest in an inertial frame so its proper time at coordinate time

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

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

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

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.