The Counter-Revolving Twins |

The question is real, and so is the effect -- it's not just an illusion. Suppose there's a clock in the nose

What's more, when the ships approach each other, ship A can see that ship B's clocks are

Finally, can the astronauts on ship A ever actually

These questions do not seem to be easy to answer. My first attempts used curved coordinates, much like the ones used in handling the revolving clock. That got me exactly nowhere. The problem is that the question I'm asking requires figuring out exactly where the "other" ship is at each moment

Finally, as far as the questions I've posed here are concerned, the effect of gravity is uninteresting. It slows the clocks on the two ships equally, it slows a clock in the center of the circle more than it slows the clocks on either of the two ships, it makes the computations harder, and it adds nothing to the apparent paradox or the interest its solution holds (for me, at least). So, I have assumed that space is flat, and the two ships are blasting furiously with their engines in order to maintain their circular paths. Of course, their rockets must point directly outward at all times to keep them going in circles (and that's where the logo for this site comes from -- as I'm sure you've already realized).

This is a rather long page, so here's a table of contents:

Discussion of the Method

Coordinate Conventions

Statement of the Problem and Equations of Motion

The Rotated Coordinate System

Finding Ship B at time τ_{A}= 0

Ship B's Clock Rate in A's MCRF

Porthole View: Relative Ship Locations

Porthole View: Relative Clock Rates

Some Graphs of the Results

My first goal on this page, and what seems to be the hardest part of the problem, is to determine the location of Ship B in the momentarily comoving reference frame (MCRF) for Ship A at each moment in Ship A's time. In other words, we need to determine

To simplify the math, the radius of the circle is assumed to be

Given a particular moment in time in Earth's frame, we will rotate the axes to (ξ, ζ) coordinates, so that ship A lies on the (rotated) ζ axis. Then Earth time which corresponds to a particular moment in Ship A's MCRF varies with the (rotated) ξ coordinate. So throughout the Earth frame, we can assign an

This explanation probably does not seem very clear at this point. Hopefully the meaning will become more apparent as we go through the actual math.

At a particular moment in time, we're interested in several specific coordinate values. We need the coordinates of both spaceships in all three frames, and we may need the time coordinate of Earth in all three frames, as well. We'll use two subscripts to represent each value, where each subscript can be A, B, or E. The first subscript will indicate what it's a coordinate of; the second will indicate whose frame of reference it's evaluated in. So we have,

We will adopt the additional convention that the angles θ and φ (see figure 1) are measured only in the

Equations (4b) and (4c) follow from what we already know of revolving clocks.

Our goal is to find φ as a function of θ. Exactly what this function should represent is a little hard to describe in words, but I will try.

Consider the MCRF for Ship A at the moment when Ship A is at angle θ. Suppose this (inertial) frame is populated with observers who fill all space, and that they have all previously synchronized their clocks. At the moment when Ship A is at angle θ, and is comoving with this frame,

In the Earth frame, we observe the flare.

The angle φ which is the location of the flare on the unit circle, as observed from the Earth frame, is the value we are looking for. It represents the location of Ship B at the moment when ship A is at angle θ,

To see this, consider that the angular velocity of each ship, viewed in the Earth frame, is exactly proportional to the rate at which that ship's clock is ticking. But the observers on Ship A saw that Ship B's clock was ticking slowly. Therefore, at the moment when the ships passed, Ship A's observers saw that Ship B

To put this yet a different way, the position of Ship B as viewed from Earth, immediately after the ships pass each other, must represent a

Enough explanation -- let's go on with the derivation.

In the inertial frame of Earth, we have

The MCRF for ship A is moving to the left along the ξ axis at velocity v, so the Earth frame is moving to the right relative to ship A's MCRF, and:

From figure 2, we can find the ξ coordinate of ship B in Earth's FoR as a function of the angles:

But we also know the location of Ship B in Earth's FoR as a function of its Earth time coordinate:

From equations (8a) and (11),

Substituting (9) into (12),

and substituting for δ

and, for small values of v,

Later on I'll plug the exact formula (15) into a program to get some graphs of what the ships are doing. We can see the general form of the behavior by inspecting equation (16), however.

The ships pass when θ = 0 and θ = π, and at both of those points φ = θ, as it must. The two angles are also equal when θ = π/2, and for values between π/2 and π, ship B is

Note that at π/2, the two ships are traveling

First, from (16) we have immediately the approximate solution,

Differentiating (15) we obtain the exact solution as a function of θ and φ:

and finally,

As we might have guessed from the fact that Ship B falls behind initially, but catches up and passes ship A's angular position at the point when they're opposite each other, ship B's clock runs fastest relative to A's at the moment when they are

This is very surprising! At the moment when B's clock is zooming ahead, the relative velocity between A and B is

Intuitively we might expect the clock rates to match at that point, since the ships are relatively stationary. However, intuition fails badly in this case, because both ships are accelerating, and when we try to take the consequential rotation of the reference frames into account we see that Ship B is actually moving very quickly across the "line of simultaneity" between it and ship A. And the consequence is that its time is changing rapidly relative to ship A's time. (But as always, be cautious with all "intuitive" explanations when dealing with relativity!)

Again, I'll plug formula (19) into a program a little later and get some graphs out.

This is much simpler than the previous question, since we can do it entirely in the frame of reference of the Earth (central) observer. All we need to determine is when a pulse of light sent from Ship B will arrive at ship A, or, equivalently, we need to determine at what angle φ the pulse must be emitted by ship B in order to be received at angle θ by ship A.

Ship B sends the pulse at time

and it's observed by Ship A at time

and the total pathlength, by simple trigonometry, must be

Since C=1, the time it takes the photon to traverse the path is P. Therefore, in the Earth frame, the time when the photon reaches A must be later than the time it left B, by P:

Substituting (20) and (21) into (23b) and combining with (22), we get

and finally

or, for small values of v,

It's obvious from these formulas -- and from simply looking at Figure 3 -- that Ship A can

Note that if this were

This is in stark contrast to the MCRF view, in which Ship B's clock does, indeed, get ahead of Ship A's clock. But it happens only when the ships are separated spatially, and by the time the image of the clock can get from B to A, A's clock has caught up. So, again, observers on Ship A can never actually

and finally

or for small values of v,

In contrast to the actual clock readings, the observers on A can indeed see Ship B's clock running fast.

This is hardly surprising -- it's just the blue shift of the signals from Ship B when the ships are moving toward each other.

The program is a trivial Newton's Method solver, in three files. It's hardly a work of art, but if you want to look at it, it's here:

twins.cxxThe plots were done via direct commands issued in Gnuplot.

twin_solve.cxx

twin_solve.h

I plotted data for 0.5C and for 0.88C. The colored labels on the traces were added with Gimp, which was also used to jpg the plots afterward. The line representing "present time" on Ship A's clock was also added with Gimp. No other changes were made. The weird trace labels in the upper right corner will be familiar to anyone who's used Gnuplot.

Both axes on each graph show

As expected, the relative rate of Ship B's clock is at a maximum at the point when the ships are running parallel, and are farthest apart. The line labeled 'Time on "Other" Clock at "Present Time" is the time seen by an observer in the MCRF of Ship A who happens to be standing next to Ship B, and whose clock shows the same reading as Ship A's clock. Whenever that curve is above the gray "Present Time" line, Ship B's clock is

The clock

Finally, note that this is one cycle of a periodic graph. If we glue another copy onto the right-hand edge, we noticed something a little surprising, which is that the line showing the rate as seen in the telescope is discontinuous! At the moment when the ships pass each other, the observed rate drops abruptly, as the blue shift changes to red shift. In practice, unless the ships actually collide there will actually be a smooth transition from blue shift to red shift as they pass "near" each other.

This graph shows nothing new, except for the perhaps unexpectedly sharp peak on the computed relative rate curve (green), and the skyrocketing value of the observed rate curve (magenta).