# 9  More On the Twin "Paradox"

### BUT WAIT! (3)

"But wait!  If Larry sees only 2.5 years of Moe's clock on the way out (from 0 to 2.5 years) and 2.5 years on the way back (from 17.5 to 20 years) then that means that 15 years of Moe's time are completely missing from Larry's vision.  That means that if Larry had a big enough super-Hubble telescope that he saw Moe's clock jump from 2.5 to 17.5 years in one swell foop!!  How could that happen?"

### BUT WAIT! (4)

"But wait!  Shouldn't Moe see Larry's clock running slower on the outward part of the trip and faster on the return part?  Doesn't that make the total elapsed time 'even out' so that at the end when Larry is back at Moe's planet the total elapsed time is the same on both clocks?"

### A Closer Look at the Twin Paradox

These two objections are common to someone encounter the twin paradox for the first time.  We'll deal with them in reverse order by analyzing the twin paradox in some detail.  First let's look at "But wait! (4)"

If Larry is traveling away from Moe then each successive "tick" of Larry's clock will occur further away from Moe than did the preceding one.  If Moe is watching Larry's clock through a powerful telescope then it will take longer for each successive tick to be seen by Moe.  This will be due to the extra time that it takes for the light signal from the "tick" to travel the extra distance.  Each successive "tick" will take longer and longer to reach Moe.  This will make the clock appear to be running slower to Moe (more time between ticks)  than it actually is running.

On the other hand when Larry is traveling toward Moe each successive "tick" will take place closer to Moe than the previous "tick."  Since the light signal from later "ticks" will have less distance to travel than previous "ticks" Moe will see them closer together in time than they were actually produced.  This will make the clock appear to be running faster to Moe (less time between ticks) than it is actually running.

This is what is known as the Doppler effect.  It is what makes the sound of a train or truck horn change pitch as it passes you -- from higher pitch as it approaches to lower pitch as it goes away.  If you let the vibrations of the horn be the "ticks" then the horn is kind of a clock "ticking" so many times per second.  As the train or truck approaches each succeeding vibration of the horn has less distance to travel to reach you and therefore is heard by you sooner after the previous vibration than it was emitted by the horn.  This makes the pitch of the horn higher to your ears than it is to the driver's ears.  When the train or truck is going away from you the opposite happens and the frequency sounds lower in pitch to your ears than it does to the driver's ears.

This is NOT what accounts for time dilation.

Moe is no stooge!  he is fully aware of the Doppler effect and can easily take it into account in determining how fast Larry's clock is running.  As he looks through his telescope he can see exactly how far away each "tick" is emitted.  Since all the light signals reach him at speed of c (299,796,458 meters per second -- about 186,000 miles per second) and he can see how much closer each tick occurred than the previously emitted one then he can calculate exactly how long ago each tick was emitted and figure out how long it REALLY took between "ticks" (well, in his frame at least.)

This taking into account of how long the signal takes is not really any new or different idea.  It is done by every successful football quarterback who throws a pass intended for a running receiver.  The quarterback has to take into account the time it will take for the ball to travel, how fast the receiver is running and aim the ball for where he best thinks the receiver will be when the ball finally comes down.  Moe, being a big football fan, does the reverse of this and takes into account where and when each "tick" was emitted as well as how long the signal was on the way.

It is after the Doppler effect has been factored out that relativistic time dilation will remain.  After Moe has received the signals, done his figuring and factored out the Doppler effect he will notice that Larry's clock is running at half the speed of his clock.  This is true for both the receding and approaching parts of Larry's trip.  Remember, in Moe's frame the light pulse on Larry's light clock will be traveling a greater distance than the corresponding light pulse on Moe's light clock.  Since Light travels at the same speed for all inertial observers the only conclusion that doesn't violate one of the two postulates is that Larry's clock is running slower than Moe's (when viewed in Moe's frame of course.  In Larry's frame it is Moe's clock that is moving and therefore running slower)

Now let's look at "But wait (3)"

There is a big difference between what is happening and what Larry sees.  Because of the time that it takes for a light signal to travel great distances Larry must interpret what he sees before he knows what is happening and when it happened on Moe's planet.

Let's consider the following situation:  Before Larry leaves Moe's planet and heads out towards Shemp's planet they agree to send out a radio signal exactly once each year when their clocks say a new year has elapsed.  (Notice that this is the same as looking through a powerful "Super-Hubble telescope and watching the clock's time.  In one case the light from the clock is received by the other and an image formed with the telescope.  In the other the radio signal [a form of "light"] is sent and received.  The difference is that no clear visual image need be formed with the radio signal) Let's ask the following question:  "how many signals will Moe and Larry receive from each other during Larry's trip?"

The answer is that Moe will get send out (and Larry will receive) 20 signals and Larry will send out (and Moe will receive) 10 signals.  This fits in with the idea that Moe is 10 years older than Larry at the end of the trip.  Notice here that Larry will receive all 20 of Moe's signals.   Let's look at it in detail and see exactly when the signals are being sent and received.  It will give us some understanding as to what's going on.

## Larry's Trip with Yearly Signals Exchanged

We'll do a little calculating here to find out exactly when and where Larry is receiving Moe's yearly radio signals.  Pull out the old pocket calculator and dust off your gray cells.

I've chosen the speed of  .866 c [or 86.6% the speed of light] for Larry's speed [as measured in Moe's frame -- or Moe's speed as measured in Larry's frame.  Gotta specify the frame!] because at that speed the time dilation factor is 1/2 which is the same as the length contraction factor.  This means that if I am at rest and I observe a space train zipping by me at 86.6% of the speed of light then I will measure its clocks to be going at half the rate of my clock and it's length to be half the length that it was when it was at rest in my frame.  1/2 is a nice "even" number to work with.

[Actually I've rounded off.  The actual speed is closer to .866025403784..... times c, which is half the square root of 3 times the speed of light.]

Now in analyzing things there are two frame's to do things in.  We can do the calculations in Moe's frame or Larry's frame.  We will choose the frame that both fits the question and is easiest to do the calculations in.  Let's begin.

The first question to look at is what time is on Larry's clock when he receives Moe's first signal?  The answer to this question doesn't depend on whose frame we are in when we make our observations.  Since Larry's clock is in the same place (Larry's ship) as Larry's receiving Moe's signal then we don't have to worry about the relative nature of simultaneity.  This means we are free to choose either Moe's frame or Larry's frame to do the calculations.  Which one will be easier?

Let's try in Moe's frame.  Larry and Moe synchronize clocks at 0.  Larry heads out at .866 c.  One year later Moe sends a signal out at speed c.  We have to find when the signal catches up with Larry, add that time to the year that passed before it was even sent out and then take into account the time dilation effect that has Larry's clock going at half the rate of Moe's clock.  Not too hard, it's a lot like the classic word problem that has a train leaving Cincinnati at noon with a faster train leaving at 1:00 PM and wants us to find where and when the second train overtakes the first train.  This is certainly a do-able problem at the level of 6th grade math.

However...

The answer can be worked out easier if we do it in Larry's frame.  In Larry's frame the clocks were synchronized at 0 and Moe is the one that is traveling at .866 c.  Therefore it is Moe's clock that is running at half the rate of Larry's clock.  If Moe sends out his signal after one year has elapsed on his clock then that will happen after two years have elapsed on Larry's clock.  In Larry's frame this will happen with Moe 1.732 light years away. (Moe is traveling for two years at .866 the speed of light means that in one year he will go .866 light years and in two years he will have gone 1.732 light years)  So a light signal 1.732 light years away is emitted when Larry's clock reads 2 years.  It will take 1.732 years to reach Larry and so Larry's clock will read 3.732 years when he receives Moe's first signal.  [Note the second postulate at work here.  We don't have to make any adjustments for anyone's speed when we use a light signal.  It always travel's at c in anyone's inertial frame.]

Moe's second signal will be sent out when Moe's clock reads 2 years.  In Larry's frame this event is simultaneous with Larry's clock reading 4 years.  Also this event will have happened when Moe has been traveling (we're in Larry's frame now) for 4 years at speed .866c so Moe will be 3.464 light years away (.866 c times 4 years).  So the signal will take 3.464 years to reach Larry and will catch up with him when his clock reads 7.464 years.   This is not correct, however since Larry will have turned around when his clock read 5 years.  So he won't be in the same frame that we are doing our calculations in.  this gives us our 1st result:

Larry will receive one signal during the outward part of his trip when his clock reads 3.732 years.
If Larry were to continue outward he would receive a signal every 3.732 years.  Two years of that time would be because the signals are being emitted every two years (in Larry's frame) because of time dilation of Moe's clock and 1.732 years would be due to the extra distance that the signal had to travel due to Moe's speed of .866 c away from Larry (the Doppler effect!).

As Larry gets the first signal he says to himself (he has lots of spare time to kill), "This signal was emitted 1.732 years ago when Moe was 1.732 light years away.  Since Moe is traveling at .866 c Moe is currently further away than 1.732 light years.  In fact since Moe has been traveling at .866 c for 3.732 years he is currently .866 c times 3.732 years = 3.232 light years away."

[WARNING:  The distance between Moe and Larry when Larry received the first signal was calculated in Larry's frame.  It will be a different distance when calculated in Moe's frame.  What is happening at Moe's planet when Larry receives a signal involves the relativity of simultaneity.  The word "when" makes no sense to separated events unless a frame is specified.]

The important thing to note, is that in Larry's frame Moe is 1.732 light years further away than the "point in space" from where the signal was originally sent when Larry finally gets the signal.

Now for the return part of the trip.

When Shemp's planet reaches Larry (remember, we're looking at it now in Larry's frame) Shemp's clock reads 10 years.  According to Larry's calculations and the signal that Larry got this also happens when (that relative word again!) Moe's clock reads 2.5 years.  This is all in accordance with the relative nature of simultaneity.  Moe's and Shemp's clocks are synchronized in their frame but since they are moving at .866 c in Larry's frame Moe's clock (the front clock) is behind Shemp's clock by 7.5 years.

To avoid having to consider any details of the turn around (how long did it take?? How much coffee got sloshed on Larry's shirt) I will have the part of the returning Larry taken by Curly.  Curly is traveling toward Moe's planet at .866 c (when measured in Moe's frame) and passes Larry just as Larry is passing Shemp's planet.  In Curly's frame (which is the frame Larry would be in after he "turns around") Moe is traveling toward Curly at .866 c rather than away from him as was the case in Larry's frame.  In Curly's frame Moe's planet is the rear planet  and will be 7.5 years ahead of Shemp's planet which is, in Curly's frame, the front planet..

Furthermore, any radio signals that Curly receives will have come from a "point in space" that is further away from Curly than Moe is when Curly receives Moe's signal.  When Curly intercepts Moe's second signal (shortly after passing, or should I say being passed by, Shemp's planet) it will be interpreted as having come from quite some distance away -- much farther away than Moe is "now" in Curly's frame.  Remember, in Larry's frame the signals were interpreted as coming from a "point in space" that were closer than Moe was "now."  This was because Moe was going away when viewed in Larry's frame.  In Curly's frame Moe is coming toward Curly.

So when Curly gets Moe's second signal he will surmise that it came from a much longer distance away than Moe currently is.  He will surmise that Moe's clock read two years when the signal was sent but since it was sent from so far away Moe's clock must read much later when the signal is received.  In fact Curly will calculate that Moe's clock reads a little past the 17.5 years that it read when Shemp's clock read 10 years (all "when's" referring to simultaneity as measured in Curly's frame).

Now with Moe traveling toward Curly at .866 c and sending a signal every two years (remember time dilation) each signal will be sent 1.732 light years closer than the previous signal.  That means that in Curly's frame they will be received .268 years apart.  From the time Curly (or home-bound Larry) receives the 2nd signal until he receives the 20th signal (18 between-signal intervals) a total of 4.824 years will elapse on Curly's clock.  Curly will receive Moe's second signal .176 years after Shemp's planet has passed him and will receive the 20th signal just upon the arrival of Moe's planet.

It takes Larry 3.732 years to receive the first signal.  If Larry "turns around" at Shemp's planet and joins Curly for the return trip then he will receive Moe's 2nd signal when his clock reads 5.176 years.  The next 18 signals will take 4.824 years to arrive.  When Moe's 20th signal is received Larry's clock will read 5.176 years + 4.824 years = 10 years.

Now let's look at it from Moe's perspective.  He will get 5 signals that were sent on the outward part of Larry's trip and 5 signals that were sent on the inward part of Larry's (Curly's) trip.  The 5 signals that were sent on the outward part of the trip will take 3.732 years between signals for a total of  18.66 years.  The 5 signals that were sent on the inward part of the trip will arrive .268 years apart for a total of 1.34 years.  18.66 years + 1.34 years = 20 years.

Moe will spend 20 years receiving Larry's 10 signals while Larry will spend 10 years receiving Moe's 20 signals.

Each twin will receive all of the other twins signals and Larry, the twin who changed frames, will be 10 years younger than Moe the twin that didn't change frames.

Notice how the relative nature of simultaneity fixes it all up.  Larry turns around and joins Curly when (in their frame) Moe's clock reads 2.5 years.  So in Larry's outbound frame the only signal that is received was sent from the outbound Moe.  The Doppler effect made that  signal be received after a lag of 1.732 years.  On the inbound part of Larry's (and Curly's) trip 19 signals were received.  In that frame Moe is traveling toward them so the 19 signals are received only .268 years apart due to the Doppler effect.  Larry and Curly receive 19 of the 20 signals with the time between signals shortened by the doppler effect.  Therefore they are able to receive more signals in less time.

Moe, on the other hand receives 5 signals from outbound Larry with the "long" Doppler effect time of 3.732 years between them and 5 signals with the "short" Doppler effect time of .268 years between them.  Therefore it takes Moe a longer time to receive fewer signals.

Any way you look at it 20 years elapse on Moe's planet between Larry's departure and arrival but only 10 years elapse in Larry's space ship.  Larry can see everything that happens on Moe's planet through Larry's on-board Super-Hubble telescope.  He won't miss a thing.  However for 15 years of what he sees (the time from Moe's clock reading 2.5 years until it read 17.5 years) Larry will have no good answer to the question, "What was I doing when that happened?"

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