# A Twin Paradox Variant, With a Surprise

The “Twin Paradox” in Special Relativity is well known.  One of  the pair, conventionally Homer, remains on Earth while his twin, Ulysses, takes off in a high-speed rocket for another star, rounds the star and comes home.  When Ulysses emerges from his spacecraft he finds that Homer has aged far faster than he has.  A naive reading of SR holds that Ulysses could have considered himself stationary, while Homer moved at high speed, and therefore Ulysses should have aged faster, and that is the paradox.

The answer is that the roles of the twins are not symmetric: Homer remained in a single frame of reference, while Ulysses changed his frame of reference with at least three accelerations, and the age measurement is taken in Homer’s unchanged frame of reference.  Every detailed calculation shows that the “paradox” disappears when this is taken into account.

But a friend recently came up with a variant on this, where the roles of the twins were symmetric.  Suppose, he said, that Homer and Ulysses blasted off in opposite directions at the same speed, turned around after a set time, and then returned.  As they approach each other on the return, each sees the other’s clock as running more slowly than his own.  So when they pass each other, which is older?

The answer is that each sees the other’s clock as running more quickly than his own, but, amazingly, when they meet both find that they have aged the same amount!  The analysis follows here, and is encapsulated in this Spacetime Diagram

If you aren’t familiar with a spacetime diagram, the x axis is space, the t axis is time.   Lines are the motion of objects, their so-called “worldlines”.  The scale is such that the speed of light is 1; lines which are at 45o (slope = 1) are light rays. Lines with slope $m>1$ are the paths of objects moving at velocity $v=\frac{1}{m}$.

In this diagram, the red lines are the path of Homer’s Rocket, and the green lines are the path of Ulysses’ rocket.   The black lines are light rays sent between the ships.

The diagram is always drawn in some frame of reference.  In this case, I’ve chosen a frame of reference which moves along with Homer’s rocket on its outbound journey.

There are some text notations on the diagram.  Beside each colored line segment is a notation of the form $\tau=kt$, for some $0.  This gives the value of  the proper time  $\tau$ for a ship on that line  as a fraction of the time $t$; the proper time is just the time kept by a clock moving on the ship.

Beside the light rays are the values for the Doppler Shift of the light rays: the difference in frequency from the source to the observer.  When $D(v)=2$, the light rays have twice the frequency at the receiver that they had at the source; when $D(v)=0.5$, they have half the frequency.  It’s important to note that the Doppler shift affects every periodic phenomenon: in particular, it affects clock ticking.  When the Doppler Shift is 2, the observer will see clocks ticking twice as fast as the person at the source sees them.

For completeness, we should give the formulae for these things.  When two objects are approaching with velocity $v$, $D(v)=\sqrt{\frac{1+v}{1-v}}$.  When they are receding with velocity $v$, $D(v)=\sqrt{\frac{1-v}{1+v}}$.  The formula for $\tau$ is also simple: $\tau=(\sqrt{1-v^2})t$.

Now that that’s in place, we can describe the experiment.  Homer and Ulysses blast away from each other at 60% of light speed: $v=0.6$.  After one year of proper time ($\tau$, not $t$), they turn and blast towards each other at $v=0.6$.  At all phases of their journey, they agree to flash lights every second, so each can keep track of the proper time on the other’s ship.  The total number of light flashes each sends is the total proper time on that ship.

Their journey is the red and green parallelogram in the diagram, which is the spacetime diagram as seen from an observer who moves in the direction of Homer’s ship as it moves away from Ulysses’ ship.  When Homer makes the turn, the observer doesn’t.

We start out at the bottom.  From $x=t=0$, the ships blast away from each other.  Since the observer starts out moving with Homer, in the observer’s frame Homer is motionless but Ulysses is blasting away at $v=0.6$, represented by a line with slope $-5/3$.  We can see that Homer’s proper time is just $t$, but Ulysses’ proper time is $0.8t$ — according to the observer, Ulysses’ clock is running slow.  Homer and Ulysses, meanwhile, are exchanging light signals (say one per second) but the Doppler Shift is 0.5 — they are receiving light signals at the rate of one every two seconds.  According to each, the other’s clock is running at half-speed.  At $t=1$, Homer’s $\tau,\tau_H=1$, so Homer makes the turn: Ulysses, however, keeps blasting away until $t=1.25$, at which time Ulysses’ $\tau,\tau_U=1$.  Once this happens, Ulysses is now co-moving with the observer (his velocity is the same as Homer’s original velocity), so from the perspective of the observer Ulysses is now motionless and Homer is blasting towards Ulysses at 60% of the speed of light, so now $\tau_U=t,\tau_H=0.8t$.

For the first part of this epoch, Homer is seeing light from Ulysses that left Ulysses’ ship while it was moving in the same direction that Homer’s ship is now.  As a result, there is no Doppler shift, and Homer and Ulysses both see the other’s clock as keeping perfect time.  This lasts until the light ray that left Ulysses’ ship at Ulysses’ turn reaches Homer’s ship, which is when $t=x+2=-5/3x+1$, or $x=-3/8,t=13/8$.  Homer’s proper time at this point is $12/8$.   The Doppler shift is now 2, so each of  Homer and Ulysses think the other’s clock is moving twice as fast as  his own.  This continues until they pass, at which time each sees the other’s clock is moving half as fast as his own.

However, if they stopped and compared clocks, they would each see that the same proper time had elapsed for both of them; if they counted flashes received vs sent, they would see that they had sent the same number of flashes.  In fact, if they exchanged one flash every tenth of a year (proper time), they would each report that during the first year of their journey, they sent 10 flashes and received 5; over the next six months, each sent 5 and received 5; and during the final six months, each sent 5 and received 10.  Each would report that the proper time of his journey was 2 years, and each sent and received 20 flashes.

Now (for the sake of completeness) let’s consider how an observer in the original frame of reference (the one Ulysses and Homer inhabited before they left) saw both journeys.  If Homer and Ulysses had relative velocity $s$, each had velocity $v$ from the perspective of an observer in the original frame of reference, where $s=\frac{2v}{1-v^2}$.  A little algebra shows $v=\frac{1-\sqrt{1-s^2}}{2s}$, and plugging in $s=0.6$ we get $v=1/3$  So from his perspective, Homer and Ulysses each blasted away at 1/3 of the speed of light for $1/\sqrt{1-(1/3)^2}=1.06$ years, turned, and returned at the same rate.  According to him, both were gone 2.12 years, while Homer and Ulysses each saw the journey take 2 years exactly.