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
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
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
to integrate -- the integral of the proper length of the velocity of
the object over (coordinate) time gives the elapsed proper time for the
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
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
stationary clock from the PoV of the revolving clock, however, that
proves difficult. The problem arises when we try to imagine
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
integral over the path using the MCRF for each piece is not at all
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
stick with that coordinate system for the duration of the
long as the metric is properly transformed, the result will be the same.
An Intuitive Discussion of What's Going
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
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
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
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
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:
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
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.