The Distant Revolving Astronaut: When Time Runs
Imagine a spaceship in orbit around a distant planet. Imagine
that the Earth lies in the plane of the orbit.
At one point in the orbit, the ship will be traveling directly toward
the Earth. At that moment, in the momentarily comoving reference
frame of the ship, the Earth lies far ahead of the ship. Since
the ship is moving toward Earth, that means that time on the Earth must
appear far ahead of "local" time on the planet below. Looking at
the Lorentz transform, if ship's time is zero, and velocity is
positive, then Earth's velocity relative to the spaceship is negative,
and current time on Earth will be given by v*gamma*distance.
Again, the clocks on Earth seem to the astronauts to be set far in the
At another point in the orbit, the ship will be traveling directly awayfrom the Earth. The sign on the velocity flips, and now it
seems to the astronauts that clocks on Earth are set far in the past,
as the Earth time is now given by -v*gamma*distance.
If the Earth clocks read in the future at one moment, and in the past
at a later moment, then at some point in between they must have run backwards.
When does that happen? Let's find the actual time on Earth clocks
in the MCRF of the orbiting ship, and
see. As we have in the
previous "revolving" cases, we will neglect gravity, as its effect adds
a lot of complexity to the problem without affecting the basic
questions I'm asking.
Distances are measured in lightyears, time in years, and C is 1.
Ship's MCRF coordinates will be (τ,
ξ, ζ). The coordinates based
on the planet the ship is circling will be (t,x,y). Letting the
ship's current location be (ts, xs, ys),
we can immediately write the equations that describe the motion of the
ship from the point of view of the planet.
We can also immediately write the formula for the ship's time as a
function of the planet's time (see the revolving
clock for the derivation).
The "primed" coordinates are rotated and translated relative to the
unprimed frame, but they're still "planet coordinates".
In the primed coordinate system, the ship's coordinates are
and the coordinates of Earth are
The ship's MCRF is moving to the left at velocity v
relative to the primed frame. Therefore the Lorentz transform
from the (similarly rotated and translated) MCRF, represented as a
primed Greek frame, to the primed Latin frame is
All events which are simultaneous in the ship's MCRF, at time ts,
have time coordinate τ' = 0. We know that the Earth's ξ'
coordinate must be contracted by a factor of 1/γ relative to its x'
coordinate, and we know that its ζ' coordinate will match its y'
coordinate. So, the coordinates of Earth in the primed MCRF frame
Applying (5) to (6) we obtain the coordinates of Earth in the primed
Latin frame which correspond to a moment which is simultaneous with the
ship in the ship's MCRF,
The x and y coordinates match the space coordinates of Earth, which is
good -- it means we're probably not totally lost! The time in the
primed frame differs from the time in the unprimed frame by ts,
so the actual time on Earth clocks, as seen from the ship's MCRF at
ship's time ts/γ, must be
Differentiating with respect to τ, we finally get,
This formula deserves some comment. Note what happens when θ =
0: Earth time proceeds γ times faster than ship's
time. But at that moment, from the ship's point of view, Earth is
moving toward the ship, and time on Earth should dilated relative to
ship time. A comoving ship, which was not in orbit, would
see Earth time moving 1/γ times slower than ship's clock
time. And the ship's acceleration along a line connecting it to
Earth is zero at that moment. So what's going on?
The answer is in the geometry. Time dilation is a consequence of
the motion of the "dilated" clock through the "stationary" frame of
reference. If we work out how fast Earth is actually moving
toward the ship when θ = 0, we find that it's not moving at all
along the line of the ship's motion. Along the (rotating) line of
motion, the distance to Earth reaches a maximum when θ = 0.
Plot of the Results
The effects we see depend on the distance from Earth to the planet, the
radius of the orbit, and the velocity. Let's put the ship in an
orbit far out from Alpha Centaurus, going very fast: We'll say xe
= 4, r = 0.1, and v = 0.5 -- that is, the planet the ship is orbiting
is 4 lightyears from Earth, the ship is 1/10 lightyear from the planet,
and it's moving at 1/2 C.
The data are plotted against θ, the angle of the ship's position with
the Y axis (see figure 1). The thin blue
line labeled ve(q) is the ship's velocity in the direction of Earth,
which is v*cos(θ). Velocity units are lightyears/year. The
thin magenta line labeled ts(q) is the current time, in the FoR of the
planet, with units of years.
The red line, te(q), is the time on an Earth clock, in years, as seen
by an observer flying past Earth in the MCRF of the ship. Earth
time starts out at +2 years, when ship's time is at zero.
However, the time line for Earth starts falling immediately, and we see
that it actually follows a rising sine curve. It's trending
upward, but with a peak each time the ship's velocity toward
Earth reaches a maximum, and a trough each time the ship's velocity away
from Earth reaches a maximum. At the points where the ship's
velocity toward/away from Earth is crossing zero, the time line for
Earth is at an intermediate value, and in fact it crosses the local
time line at those points. Note well: The points
where the Earth time line are crossing the local time line are also the
points where Earth time relative to ship time is changing most rapidly.
And that brings us to the Big One: The green line, dte_dtau(q),
is the rate of a clock on Earth, as viewed from the
(continuously changing!) frame of reference of the ship. It
starts out at 1.15 (= γ), when the ship is moving directly toward Earth
accelerating "sideways", but as the ship's orbit bends and the ship
starts to accelerate away from Earth, the rate of the Earth clock drops
... and drops! As the ship's velocity in the direction of Earth
drops to zero (blue line crosses the axis) and it reaches the point in
its orbit closest to Earth, the Earth clocks are running
backwardsten times faster than the ship's clock is
running forward. As the ship eventually starts to accelerate
toward Earth, in the more distant portion of its orbit, the Earth
clocks again run forward at a rapidly increasing rate. At the
point where the ship is farthest from Earth, when its velocity toward
Earth again drops to zero, the Earth clocks are zooming ahead more than
12 times as fast as the ship's clock.
But What Does This Really Mean?
Much, and yet, not so very much.
This effect is important, because it is at the very heart of the "twin
paradox". The traveling twin cannot return to Earth
without accelerating, because he must reverse his direction to get
there. And when he accelerates, very strange things seem to
happen to time. In particular, accelerating toward an
object causes the clock on that object to seem to move faster.
When the twin is first launched, he accelerates toward Alpha Centaurus,
and as he does so the clocks on Alpha Centaurus seem to zoom
ahead. When he gets there, he finds that a great deal of time has
indeed already passed at that location.
When he turns around to come home, the clocks on Earth, which had been
loafing along due to their time dilation relative to the
traveler, suddenly seem to zoom ahead. And when he gets home he
does indeed find that lots of time has passed for the homebound twin.
The issue is really "relativity of simultaneity". What's
simultaneous with what depends on what (inertial!) frame of reference
you happen to be in. When the traveler accelerates, he changes
his inertial frame of reference, and his notion of what's simultaneous
But it's possible to use lots of words in an effort to explain this,
without ever really improving on what the equations already say on
At the same time, keep in mind that this effect, dramatic as it seems,
is completely undetectable. It cannot be seen, it cannot
be measured. It can only be deduced. And so, it doesn't
really mean so very much, after all. Consider, in a little more
detail, what we mean by the "time in the MCRF of the ship":
At a particular moment, and just at that moment, t, the
ship is traveling in a particular direction at a particular
speed. Pretend that all of space is filled with observers who are
not accelerating, but who also happen to be traveling at that
exact speed, in the same direction. Assume, furthermore, that
these observers all have clocks, and that they all have previously
synchronized their clocks. Now, each observer checks his
clock, and at the moment when each observer's clock reads t, he
looks around himself. One of these observers finds he's standing
right next to Earth -- or, rather, he's zooming past Earth. He
looks at an Earth clock, and notes the time he sees on that clock at
the moment when his clock reads t.
That is the Earth time in the MCRF of the traveler. There
is no simple way to actually measure it -- it can only easily be
determined by modeling it. If you were to try to measure it, for
one particular time, you would still need to do it by taking several
measurements at widely different places and then combining the results
afterward -- no single measurement will every capture it, because it
requires comparing times at distant locations.
Now, consider what is meant by the rate of change of Earth's
time in the MCRF of the traveler. To measure the change in
Earth's time, we must repeat the above measurement at time t + dt.
To do that, we need to use a different set of observers.
These observers are also not accelerating -- an important point! -- and
they have also previously synchronized their clocks, and they all
happen to be moving in the same direction at the same speed as the ship
is moving at time t + dt.
So the difference in Earth time we compute is taken between
measurements done by two different sets of observers. When the
clocks are "moving backwards", it's because the member of this second
set of observers who recorded Earth's time did it at an earlier moment
in Earth's history than the observer from the "earlier" MCRF of the
ship. The folks on Earth would disagree with these observers and
with the astronauts about which event "came first", of course!
There is absolutely no way for a single observer to ever see
the Earth clocks moving backwards.
One thing nearly everyone agrees on: This effect is confusing.
What About the Porthole View?
The "porthole view" -- what the astronauts see looking out the porthole
-- does not reflect any of this craziness. Looking
through a telescope at Earth, the astronauts will see the effect of
Doppler shift. They won't see anything more.
As they accelerate toward Earth, the Doppler shift they see will be, as
always, a combination of the relative motion of Earth toward them, and
the apparent slowing of Earth's clocks due to time dilation. The
distance to Earth will seem to decrease due to Fitzgerald contraction,
and the time they compute for the length of time which has
passed since the light they are seeing was emitted will vary as a
result of the change in distance. But then we're once again
wandering off into what they may compute, as opposed to what they
The point to carry away from this is that the MCRF view, though it
contains the key to the twin paradox, can only be computed.
It can't be directly observed.