v1.6.1 / chapter 2 of 4 / 01 apr 07 / greg goebel / public domain
* The previous chapter outlined the relativistic concepts of time dilation, length contraction, and relativistic simultaneity. All these are somewhat counterintuitive notions, and more to the point they seem to lead to paradoxes. As it turns out, if the basic concepts of special relativity are understood, the paradoxes evaporate on closer inspection.
* It might seem that a consideration of relativistic time dilation and length
contraction leads to completely absurd contradictions. Imagine Alice is
flying a starship of length L. If Alice's starship is moving at Cf, then it
is length-contracted by the factor:
Lc1 = L * SQRT( 1 - Cf^2 )
For example, if L is 100 meters and Cf is 0.8 C, then Lc1 is 60 meters.
Now let's imagine that the starship passes through a tube floating in space,
at rest relative to Bob, whose length Bob sees as exactly Lc1, or 60 meters
in this case. Bob will be able to observe that at one instant the starship
will fit exactly into the tube. However, from the point of view of Alice on
the starship, it is the tube that is length-contracted, by the factor:
Lc2 = Lc1 * SQRT( 1 - Cf^2 ) = L * ( 1 - Cf^2 )
If L = 100 and Cf = 0.8 C, then Lc2 = 36 meters. There's no way Alice's
starship will fit in the tube -- is there?
* From a simple consideration of length contraction, we have just painted ourselves into a corner, but there's a way out, through one of the other basic elements of Special Relativity: relativistic simultaneity. Bob can see both ends of Alice's starship fit inside the tube while Alice doesn't, because as far as he is concerned time in the rear of her starship is running ahead of time in the front.
Let's imagine that Alice has a clock at each end of her starship, and that she has synchronized the clocks. Alice, who perceives her starship at rest and the tube in motion, sees the Lc2 = 36 meter tube approaching her L = 100 meter starship at Cf = 0.8 C. The entrance of the tube swallows up the front of her starship; when exit of the tube passes the front of her starship, the tube is still traversing the rest of her starship. More specifically, when the exit of the tube reaches the front of the starship, then the rear of her starship will still be L - Lc2 meters, or 64 meters, from the entrance of the tube.
Alice will measure a short time interval from the time the nose of her starship reaches the exit of the tube to the time the entrance of the tube reaches the rear of the starship. Of course, we know that Bob sees the front of Alice's starship reach the exit of the tube at exactly the same instant that the rear of her starship reaches the entrance of the tube. Bob knows he sees Alice's entire starship in the tube at a single instant because the clock in the rear her starship is running ahead of the clock in the front. The short time interval that Alice measures will give the advance of the clock in the rear as seen by Bob.
Doing the algebra, the entrance of the tube will reach the rear of her
starship after a time interval of:
Td = ( L - Lc2 ) / ( Cf * C )
= ( L - L * ( 1 - Cf^2 ) ) / ( Cf * C )
= L * ( 1 - ( 1 - Cf^2 ) ) / ( Cf * C )
= L * ( 1 - 1 + Cf^2 ) / ( Cf * C )
= L * Cf^2 / ( Cf * C )
= L * Cf / C
If L = 100 and Cf = 0.8 C, then Td = 267 nanoseconds (nsec), or in other
words Bob sees the count on the clock at the rear of Alice's starship running
267 nsec ahead of a clock at the front, while Alice sees the clocks as
synchronized. This is a worthwhile little formula to remember: it provides
a key for unlocking a number of seeming paradoxes.
There really is no paradox here. When the exit of the tube reaches the front of Alice's starship, both Bob and Alice agree on that event. However, Bob also sees the rear of the starship enter the entrance of the tube at what amounts to 267 nsec in the future for Alice. Alice will not see the entrance of the tube reach the rear of her starship until a (very) short time later.
* The faster Alice's starship goes, the shorter it becomes, and the more time advances in the rear of her starship. A cartoonish way to visualize length contraction is to imagine a centipede in which each successive set of legs is marching a bit ahead of the legs in front of it, forcing the centipede to pile up on itself and become shorter.
Let's take a closer look at this notion by considering light propagating from the clock in the rear of Alice's 100 meter long starship to the clock in the front. From Alice's point of view, this is a straightforward event, with the light propagating from front to back in 100 / 300,000,000 = 333 nsec.
From Bob's point of view, Alice's starship is 60 meters long and moving at
80% of the speed of light, V = Cf * C = 240,000,000 meters per second. Since
Bob sees Alice's starship as in motion, the front clock will move away while
the light from the rear clock is propagating forward, and so the time T Bob
measures for propagation of light from the clock in the rear to the clock in
the front is given by:
C * T = 60 + V * T
C * T - V * T = 60
T * ( C - V ) = 60
T = 60 / ( C - V ) = 60 / ( 300,000,000 - 240,000,000 )
= 60 / 60,000,000 = 1 microsecond
Bob knows that Alice's clock is only running at 60% of his, so he calculates
that in her frame of reference it takes 600 nsec for light to get from the
rear of the ship to the front. Since Alice knows it only takes 333 nsec that
would seem to be a contradiction, but Bob sees the clock in the rear as
running 267 nsec ahead; 267 + 333 = 600 nsec, and the contradiction goes
away.
No matter how fast Alice is going relative to Bob, as far as she is concerned
light will always take 333 nsec to propagate from the front to the rear.
However, as her relative velocity increases, Bob will calculate an ever
increasing discrepancy, and so the clock in the rear of her ship must run
increasingly ahead to maintain the equivalence of the two frames of
reference. More formally, Alice sees the time for light to propagate from
front to rear of her starship as:
Trfa = L / C
Since the observed time discrepancy of the clock in the rear at a velocity Cf
is:
Td = L * Cf / C
-- then Bob observes that the difference in time value of the clock in the
rear of Alice's starship when the pulse is emitted and the time value of the
clock in the front of Alice's starship when the pulse is detected as:
Trfb = Trfa - Td = ( L / C ) - ( L * Cf / C)
= ( L / C ) * ( 1 - Cf )
As Cf increases, the delay Trfb will become smaller and smaller, though since
Cf is always less than 1, there will always be a delay: the time value on
the rear clock received at the front clock will approach that on the front
clock, but it will never reach it, much less exceed it.
* Another interesting paradox often cited in books on relativity is the "running man" paradox. Suppose Alice owns a barn with doors on both ends that is 10 meters long. Now suppose Bob -- who is actually the secret identity of the Flash, the fastest man alive -- is running towards the barn at 80% of the speed of light, carrying a pole 10 meters long, while Alice stands outside the barn and watches him zip past.
At Bob's speed, Alice observes the pole as being length-contracted by 60%, reducing its length to 6 meters. Bob enters the barn through an open door, while the door on the opposite side through which he intends to exit is closed. The entry door is closed automatically when the tail end of the pole clears it, and then the exit door is opened in sequence quickly enough to allow him to pass through without obstruction. This scenario makes perfect sense. Alice observes a 6 meter pole inside a barn 10 meters long, and it obviously fits inside with both the doors closed.
Now let's look at it from Bob's point of view. He can consider himself standing still in his frame of reference, with Alice's barn approaching him at 80% of the speed of light. That means that the barn is length-contracted to 6 meters, while his pole still remains 10 meters long. How can it possibly fit inside the barn with both doors closed?
It doesn't take much thought to realize this is almost exactly the same
scenario as Alice's starship passing through a tube, just restated slightly.
The trick is that the doors are closed and opened in synchronization in the
barn's frame of reference, but Bob does not see the doors as synchronized in
this way. Going back to the little equation derived above:
Td = L * Cf / C
-- then if L, the length of the barn or the pole, is 10 meters, and Cf is 0.8
C, the value of Td is 10 * 0.8 / 300,000,000 = 26.7 nsec.
From Bob's point of view, when he runs through the barn and the front of his pole is about to reach the exit door, he still has 4 meters of pole sticking out of the entry door. However, he knows that the time at the exit door is running 26.7 nsec ahead of the time at the entry door. Since he knows time in Alice's frame of reference is running 60% slower, that means that as far as he's concerned the discrepancy is actually 26.7 / 0.6 = 44.4 nsec. At Cf = 0.8 C, the 4 meters of pole still outside the entry door will pass through in 16.7 nsec. He has plenty of time to clear the entry door before it closes.
By the way, the fact that pole actually fits into the barn in the barn's rest frame demonstrates that length contraction is not an optical illusion: the pole really is shorter in Alice's frame of reference. There's also the interesting question of what happens if the Flash stops inside the barn, but this leads to a really ugly consideration of subtleties that are too complicated to detail in this short document.
* Here's another puzzle that is particularly confusing. Let's go back to Alice and her 100 meter long starship. Suppose she is moving at 80% of the speed of light past a little space buoy that Bob has set up to track her passage. Alice's spaceship is fitted with a little module on the front and the rear, and the buoy has one of the modules as well. These modules can detect the proximity of each other, using microwave beams or lasers or whatever; the specifics are irrelevant, all we have to do is assume the modules detect each other at a short enough range and take little enough time so that the details of how it is done can be ignored. When the module on the buoy detects the module on the front of Alice's starship flying by, it starts a timer, and when it detects the module on the rear of Alice's starship flying by, it stops the timer.
Let's see what happens in Bob's frame of reference. Alice's starship zips past him at 80% the speed of light, 240,000,000 meters per second. At this speed, the starship is contracted to 60 meters, and the buoy clocks the transit of the starship as 60 / 240,000,000 = 250 nsec.
Since Alice's clocks are running 60% slower, Bob will calculate that the transit takes 250 / 0.6 = 417 nsec in Alice's frame of reference. In Alice's frame of reference, the buoy zips past her at 80% the speed of light. Since the length contraction of Bob's space buoy is not an issue here, as far as she's concerned, the transit takes 100 / 240,000,000 = 417 nsec, and both are in agreement.
The problem arises when Alice factors in the 60% time dilation in Bob's frame of reference. Alice thinks that Bob must have measured the transit as taking 417 / 0.6 = 694 nsec -- but as shown, he actually measured 250 nsec. How can this be?
The trick is, as before, Bob sees the clock in the rear of Alice's starship as advanced by 267 nsec. When the rear of her starship passes the buoy, the clock in the rear does read 417 nsec, and if it had been reading zero when the timer started counting, Bob would indeed have to conclude that 694 nsec passed on Alice's starship. However, the clock in the rear of the starship was already at 267 nsec when the timer started counting, and it has only really advanced 417 - 267 = 150 nsec. Bob factors in the 60% time dilation to get 150 / 0.6 = 250 nsec, exactly what the timer measured.
By the way, this puzzle isn't so different from the one involving Alice's starship zipping through a tube. The elements of the puzzle would be entirely the same if the tube was cut down in length to a ring. It's just that the props are different.
* Finally, let's pretend both Bob and Alice are flying in 100 meter starships with modules on each end, and pass each other at a relative speed of 80% of the speed of light. The modules work just as they did in the previous example. Bob starts counting time when the module on the front of his starship encounters the module on the front of Alice's starship, and then stops counting when the module on the rear of his starship encounters the module on the rear of Alice's starship.
In Bob's frame of reference, Alice's starship travels 160 meters during that interval, which gives a transit time of 160 / 240,000,000 = 667 nsec. Bob knows that Alice's clock is running slow by 60%, and so this amounts to 667 / 0.6 = 400 nsec in Alice's frame of reference. But Alice sees exactly the same situation in reverse, believing that the transit takes 667 nsec in her frame of reference and 400 nsec in Bob's.
Of course, we know the answer by now. We've already calculated that the discrepancy in clock synchronization for a 100 meter starship moving at 80% of the speed of light is 267 nsec. 400 + 267 = 667 nsec, and the paradox goes away. Once again, by the way, this scenario isn't so different from the puzzle involving Alice's starship going down a tube, it's just that in this case the length of the tube as seen by Bob is 160 meters; only the props are different.
* There are a lot of variations on these puzzles -- a moving block falling through a hole is another common one -- but after performing enough of them they start to seem alike. Sometimes people dream up puzzles along such lines and think for a while they have refuted special relativity, but it turns out they didn't take relativistic simultaneity into account. Einstein was a step ahead of them all along.
* The section above demonstrated that length contraction isn't an illusion. Time dilation isn't an illusion either, and that can be shown by the most famous of all relativistic paradoxes. Suppose Alice and Bob are twins. Alice flies to the stars in a starship moving at a good percentage of the speed of light ("relativistic" speed) while Bob stays at home. When Alice comes back to Earth, she will be younger than Bob.
For example, suppose that Alice and Bob are both 20 years old when Alice sets
out in a starship to travel to the planet Minbar, 10 light-years away, at
half the speed of light. She spends a negligible amount of time at Minbar,
and then comes back to Earth at half the speed of light. Her time-dilation
factor would be:
1 / SQRT( 1 - Cf^2 ) = 1 / SQRT( 1 - 0.5^2 ) = 1.15
Her trip would take 40 years as far as Bob back on Earth was concerned, but
only 40 / 1.15 = 34.6 years as far as Alice was concerned. Bob would be 60
years old; Alice would not quite be 55 years old.
This is known as the "Twin Paradox", and it certainly seems contradictory. Alice's starship can be just as validly regarded as being at rest and the Earth as being in motion. This leads to the first objection: if Bob sees Alice's clock as running slow and Alice sees Bob's clock as running slow, then how can Alice age less than Bob?
The answer is that the two scenarios, though equivalent in the strict sense, are not evenly balanced. To Bob back on Earth, Alice's starship is moving at half the speed of light, and only her starship is length-contracted. To Alice, the rest of the Universe is moving at half the speed of light, and the rest of the Universe is length-contracted: the distance to Minbar is only 10 / 1.15 = 8.66 light-years.
This is the critical point of the Twin Paradox. Alice is taking a trip through a Universe that appears length-contracted to her, and so a trip at half the speed of light takes a shorter time. In fact, it was the necessity to ensure that the Alice and Bob's clock could balance out that was used in the previous chapter to introduce the notion of length contraction in the first place.
* This still doesn't show how Alice and Bob can see each other's clock running slow and come up with different times. In fact, Alice and Bob do not always see the other's clock running slow. This is because of the fact that although time dilation is not an optical illusion, optical illusions are also involved in the twin paradox.
The phenomenon of relativistic time dilation was derived in an earlier section in the case of a starship moving across our line of sight, in which the clock appears to run more slowly. However, that does not mean that the clock will be seen to run slower in all circumstances, which is a common misconception about Special Relativity. If Alice's starship is moving toward Bob, her clock will seem to run faster than his. If Alice's starship is moving away from Bob, her clock will seem to run even more slowly than it would if her starship were running across his line of sight.
Discussion of this phenomenon requires an understanding of the "Doppler
shift", which was well understood by classical physicists at the beginning of
the 20th century. It is the change in pitch caused by the motion of an
object. If a train approaching at high speed blows a whistle, the pitch of
the whistle is higher than it would be if the train were at rest. Similarly,
if the train is moving away, the pitch of the whistle is lower. If the time
of the period of the whistle's wavelength when the train is at rest is T and
the period when the train is moving at velocity V is Tm, then if the speed of
sound is given by S, the ratio of the change in period due to the Doppler
shift is given by:
Tm / T = ( 1 - V/S )
-- if the train is approaching. If the train is moving away, the ratio is
given by:
Tm / T = ( 1 + V/S )
The classical Doppler shift can be redefined to apply to light with a minor
change in variable definitions:
Tm / T = ( 1 - Cf ) ! if approaching
Tm / T = ( 1 + Cf ) ! if receding
The classic Doppler shift is not correct at relativistic speeds, since the
time dilation factor applies and has to be also multiplied in. This gives:
Tm / T = ( 1 - Cf ) ) / SQRT( 1 - Cf^2 )
= SQRT( ( 1 - Cf )^2 / ( 1 - Cf^2 ) )
= SQRT( ( 1 - Cf ) * ( 1 - Cf ) / ( 1 - Cf ) * ( 1 + Cf) )
= SQRT( ( 1 - Cf ) / ( 1 + Cf ) )
So, for an approaching starship, the relativistic Doppler shift is:
Tm / T = SQRT(( 1 - Cf ) / ( 1 + Cf ))
For a receding starship, the relativistic Doppler shift is:
Tm / T = SQRT(( 1 + Cf ) / ( 1 - Cf ))
If we have a clock that emits a pulse of light on every tick, the pulse /
tick period will decrease if the clock is approaching and increase if it is
receding. More generally, the Doppler effect causes the light of objects
approaching us to be reduced in period, or equivalently increased in
frequency, becoming bluer and more energetic, or "blueshifted". Similarly,
the light from objects moving away from is increased in period, or
equivalently reduced in frequency, becoming redder and less energetic, or
"redshifted". We can use the terms "blueshift" and "redshift" as a
convenient shorthand for describing the two Doppler effects.
Now let's suppose that Bob's clock on Earth and a clock on Minbar are synchronized with each other by transmissions of coded timing signals. Although events that are synchronized in one frame of reference are not necessarily synchronized in another, the clocks can still be synchronized with each other in the same frame of reference. Assuming that Earth and Minbar are exactly ten light-years apart, each will receive clock signals that should be exactly ten years old and can adjust their clocks accordingly.
Just before Alice leaves the Earth at some arbitrary "time zero" Tz, she will read the signal from the clock on Minbar and find that it gives a value of Tz - 10 years. She knows that it will take 20 years in Bob's frame of reference for her to reach Minbar, so she knows that when she arrives at Minbar the clock there will read Tz + 20 years. She will observe the clock on Minbar count through 30 years during the course of her journey, while her own clock only counts through 17.3 years.
This means that she sees the clock on Minbar "blueshifted" by the Doppler
effect, running fast by a factor of Tm / T = 17.3 / 30 = 0.577. This is
exactly what would be predicted by the relativistic Doppler effect:
Tm / T = SQRT(( 1 - Cf ) / ( 1 + Cf ))
= SQRT(( 1 - 0.5 ) / ( 1 + 0.5))
= 0.577
When Alice ends her brief stay on Minbar, she turns around and heads back
home. By exactly the same logic, the signal from Bob's clock arriving on
Minbar from Earth will be 10 years out of date, and she knows Bob's clock
will advance 30 years while she flies home at half the speed of light. This
is exactly the same scenario as occurred on the outbound leg of the journey
and it works out in exactly the same way.
Of course, while Alice was traveling away from the Earth on the outbound leg
of the voyage to Minbar, she was picking up the signals from Bob's clock.
When she left, the signal from Bob's clock was reading a value of Tz, and
when she arrived at Minbar, the signal was reading as Tz + 10. As she went
from Earth to Minbar, she observed Bob's clock as "redshifted" by the Doppler
effect by a factor of Tm / T = 17.3 / 10 = 1.73. Again, this is exactly what
would be predicted by the relativistic Doppler effect:
Tm / T = SQRT(( 1 + Cf ) / ( 1 - Cf ))
= SQRT(( 1 + 0.5 ) / ( 1 - 0.5))
= 1.73
So ... during the course of her journey, Alice reads Bob's clock counting
over 10 years at a slow rate and over 30 years at a fast rate, and she
sees that Bob's clock has indeed counted through 40 years while hers has only
counted through 34.6 years.
Incidentally, this is a somewhat artificial scenario, since it assumes that Minbar and Earth are motionless relative to each other and that clocks between the two planets can be effectively synchronized, but given the fact that the relative motions of star systems and planets within such a cosmically short distance are far below the speed of light, the error factor in the analysis remains insignificant.
* OK, that shows how the scenario works from Alice's point of view. Now we need to consider how the same scenario appears from Bob's point of view.
Let's suppose that Alice's starship also has a communications system to allow it to relay its current clock time back to Bob. As Alice flies to Minbar, Bob observes her "redshifted" clock running slow by a factor of 1.73, and when she turns around and comes back, he observes her "blueshifted" clock running fast by a factor of 0.577.
As noted, however, he will not observe Alice's clock as "redshifted" and "blueshifted" for the same amount of time:
In sum, the observations between the clocks in the two frames of reference work out because Bob, at the start and end point of Alice's round trip, sees Alice's clock running slow 75% of the time and running fast 25% of the time, while for Alice the proportions are exactly 50% each.
* A graph known as a "spacetime diagram" or "Minkowski diagram" -- after Hermann Minkowski (1864:1909), a Russian expatriate who had been one of Einstein's instructors at the Swiss Federal Institute of Technology (ETH in its French acronym) in Zurich -- can be used to help visualize the scenario. It's just a graph with time on the vertical axis and space on the horizontal axis, with the time and space defined for one observer's frame of reference, in this case the Earth's.
If the time axis is measured in years, the space axis is measured in light-years, and the two axes have equal increments, then light always travels at a 45 degree path on the diagram. An object at rest in the Earth's frame of reference follows a vertical path, since it passes through time but not space, while an object moving closer and closer to the speed of light approaches the 45 degree angle. Minkowski diagrams are often used in texts on relativistic physics.
* There's one more problem, a particularly sticky one, with the twin paradox that has to be resolved. During the trip to Minbar, Alice, by her own clock, determines that 17.3 years have passed. However, she knows that, in spite of the fact that "blueshifting" makes the clock on Minbar appear to run fast, it is actually running slow due to time dilation by the factor of 1.15, and so it only counts 17.3 / 1.15 = 15 years during the course of her outbound journey. Since the clocks on Earth and Minbar are synchronized in their frame of reference using coded signals, then when Alice departed from Earth the clock on Minbar read Tz. When she arrived at Minbar, it read Tz + 20, not Tz + 15.
The answer should be apparent, though maybe not obvious, if you remember how
relativistic simultaneity works, and noticed the qualification that the
clocks on Earth and Minbar are synchronized in their own frame of
reference. They are not synchronized in Alice's frame of reference,
with the difference in synchronization is given by, as before:
Td = L * Cf / C
If L is specified in light-years and Td in years, this simplifies to:
Td = L * Cf
Given that L is ten light-years and Cf is 0.5 C, this means that as far as
Alice is concerned the clock in Minbar is running 5 years ahead through the
entire outbound journey. It times through 15 years during her trip, and ends
up at a value of 20 years.
By the same logic, while Alice spends 34.6 years on her interstellar round trip, she knows that Bob's clock is running slow by a factor of 1.15 and so only counts through 34.6 / 1.15 = 30 years. However, Alice got five years out of synch with the rest of the local Universe, and when she turns around she gets out of synch with it by another five years, for a total of ten years, which means that when she arrives back home Bob's clock has counted 40 years since her departure.
* The twin paradox sounds so dodgy that it's worthwhile to summarize how it works. A "question and answer" format works well in this case:
A: Suppose Bob and Alice are twins. Bob stays on Earth while Alice takes a round-trip in a starship to Minbar, 10 light-years away, at half the speed of light. Einstein's equation show that Alice's clock will run slow by a factor of 1.15. When Alice returns to Earth, Bob has aged 40 years, while Alice has aged only 34.6 years.
A: Actually, they don't see each other's clock running slow all the time. Due to the optical illusion created by the Doppler shift, when they are flying apart, their clocks seem to be running slower than they actually are; when they are coming together, their clocks seem to be running faster than they are. Alice will see Bob's clocks running slow for half her journey and running fast for half her journey. However, since Bob won't see Alice's clock running fast until the light from her "U-turn" at Minbar finally reaches Earth, he will see Alice's clock running slow for 75% her journey and running fast for 25% of her journey. The accounting adds up.
A: They add up because the frames of reference of Alice and Bob are, though equivalent in the strict physical sense, not perfectly balanced. Alice sees herself as standing still in a Universe that is moving at half the speed of light: by the Lorenz-Fitzgerald contraction, the whole Universe is length-contracted, and her voyage is shortened as a result to 8.66 light-years each way.
In addition, although the clocks on Earth and Minbar are synchronized as far as Bob's concerned, they aren't as far as Alice is concerned: on her way to Minbar, she knows the clock on Minbar is running five years ahead, and on her way back to Earth, she knows the clock on Earth is running five years ahead.
Bob sees himself as standing still and Alice as moving at half the speed of light. He sees her ship as length contracted, and sees the clocks at the ends of her ship as out of synchronization; neither of these phenomena are a factor in crunching the numbers for the twin paradox. Once again, the accounting adds up.
The twin paradox is really only due to time dilation: if Alice went into orbit and spun dizzily around the Earth at half the speed of light for 40 years by Bob's clock, she will have only aged 34.6 years at the end. It is easy to demonstrate this: if a set of space stations were placed in a ring around the Earth, motionless relative to the Earth and with their clocks synchronized to Bob's Earth clock, then Alice could take her ship from station to station in a straight line until she traversed the entire ring. The scenario for each jump from station to station would be the same, except for distances, as it would be if she flew from Earth to Minbar.
The same sort of reasoning could be used for any flight path of any configuration, with Alice performing straight-line flights from station to station conveniently arranged along the flight path. Accounting for the equivalent-but-unbalanced readings of Bob's clock and Alice's clock is trickier, but the accountings will always have to add up.
* Incidentally, some sources tend to give the impression, maybe not deliberately, that the twin paradox is due to Alice flying away and coming back while Bob stays put, but that's not completely the case: the twin paradox is due to the fact that Alice is flying through a length-contracted Universe, while Bob simply sees that Alice is flying in a length-contracted ship. The fact that Alice flies away and comes back while Bob stays put merely explains how their accountings of clock ticks balance out.
Suppose that instead of flying ten light-years to Minbar, Alice flies twenty light-years at half the speed of light to the planet Zahadum, where she plans to settle with no thought of coming back home to Earth. When Alice departs, Bob sends a message to Zahadum telling the planet's high council that Alice is 20 years old and that she will arrive in 40 years' time -- by a remarkable coincidence a year on Zahadum is the same length as a year on Earth. The high council receives the message 20 years later and realizes that Alice will arrive in 20 more years.
Their local customs require that Alice be greeted by someone her own age, and so they select Dot, who happens to be 40 years old. Twenty years later, Alice arrives to meet Dot, who is now 60 -- but Alice is still not quite 55 years old. Alice's trip was one-way, but the twin paradox applied just as much as if she had gone to Minbar and then returned to Earth.
* Sources also sometimes say that the twin paradox is due to the fact that Alice undergoes acceleration while Bob does not. This is not exactly wrong, since Alice has to accelerate to reach half the speed of light, but it's not particularly insightful. Once again, the twin paradox is really only due to the fact that she is traveling at half the speed of light relative to Bob and the rest of Bob's frame of reference.
Let's go back to the example of Alice's ship passing a space buoy. Suppose Bob has a space buoy near Earth and Dot has a space buoy near Zahadum. Alice comes zipping past Bob's space buoy at half the speed of light, and when she does so she relays her current age of 20 to the buoy. Bob then relays Alice's age on to Zahadum, and 40 years later (as Bob and Dot read the clock) Alice zips by Dot's space buoy, once again relaying her age. Although Dot is 60, Alice will report her own age as under 55, as before. No accelerations were directly involved in this scenario.
* There's another way to look at the equivalence principle involved in the Twin Paradox by considering what would have to be done to reverse the scenario so that Bob ages less than Alice instead of the other way around. As described above, Alice goes to Minbar 10 light-years away at 50% of the speed of light, a journey that takes 40 years or 20 years each way as far as Bob is concerned, but only 34.6 years or 17.3 years each way as far as Alice is concerned.
Now suppose Alice's starship flies at 50% of the speed of light for 40 years by her clock, without stopping and turning around. Suppose that after 17.3 years on Earth, Bob decides to use an improved, faster starship to catch up with Alice and rendezvous after 40 years of flight by Alice's clock.
Bob has to do some calculation to figure out how fast he needs to go to make
the rendezvous. From his point of view, Alice's clocks are running slow by a
factor of 1.15. This means her trip actually takes 46.2 years from the
Earth's frame of reference, and Alice will be 23.1 light-years away in the
Earth's frame of reference at the time of the rendezvous. Since Bob takes
off 17.3 years after Alice's departure, he calculates that he will have to
catch up with Alice in:
46.2 - 17.3 = 28.9 years
-- in the Earth's frame of reference. Since Alice will be 23.1 light-years
away when they meet, that means that Bob must fly at:
23.1 / 28.9 = 0.8 C
-- or 80% of the speed of light, to make the rendezvous.
The 28.9 year flight-time was, as mentioned, given for the Earth's frame of
reference. However, once Bob takes off in his starship at 80% of the speed
of light, time slows down by a factor of 0.6, and so Bob actually only takes:
28.9 * 0.6 = 17.3 years
-- to catch up by his own clock. When the two rendezvous, Bob has only aged
34.6 years and Alice has aged 40 years.
When Bob sets off in his starship, he immediately sees the clock on the Alice's starship change from ticking slower to ticking faster due to the Doppler shift. You can do the math if you like and see that, due to the time lag of the speed of light, Alice won't see Bob depart Earth until 30 years after her own departure by the starship's clock, and so will only see Bob's clock ticking faster due to the Doppler shift for ten years.
Similarly, Alice will measure the velocity of Bob's starship as 50% of the speed of light and the distance of its travel as 10 light-years. Once again, the equivalence principle is preserved. Alice now sees Bob as traveling ten light-years away from her at half the speed of light, then turning around and coming back to her at half the speed of light, with all elements of the scenario calculating out to exactly the same values as the original twin paradox.
* Consideration of this scenario leads to another seeming paradox: from Alice's point of view she's standing still, while the Earth is flying away at half the speed of light and Bob is approaching at half the speed of light. From Alice's point of view, she might think that the speed of Bob relative to the Earth is the speed of light -- when it's actually only 0.8 C.
In fact, this general scenario holds no matter how high a fraction of the speed of light the velocities are. Suppose the twin paradox scenario is changed to Alice flying to 10 light-years to Minbar at Cf = 0.9 C. That means that Alice's round trip time would be 20 / 0.9 = 22.2 years as far as Bob in concerned, though only 9.69 years as far as Alice is concerned. Now shift the scenario around as above, with Bob leaving at 9.69 / 2 = 4.84 years in the Earth's frame of reference and trying to catch up with Alice after a total of 22.2 years of flight in her frame of reference.
Her clocks are running slow relative to an Earth clock by a factor of 2.29, so the actual rendezvous time will be 2.29 * 22.2 = 51 years after Alice's departure by the Earth clock, by which time she will have traveled 51 * 0.9 = 45.9 light-years. Bob has to catch up in 51 - 4.84 = 46.1 years by the Earth's clock, meaning he has to travel at 45.9 / 46.1 = 0.994 C to make the rendezvous. Since his own clock is running slow by a factor of 9.53, that means it only takes him 4.84 years by his own clock to make the rendezvous.
Once again, though Alice would see the Earth and Bob moving in opposite directions at 0.9 C, their relative velocity is only 0.994 C, not 1.8 C as Alice might think. The same exercise can be performed for any other values of Cf; Bob will always end up flying slower than the speed of light relative to Earth.
* This can be thought of as the "three-way paradox", involving three frames of reference, not just two, with the relative velocities of two frames of reference as observed by a third not adding up to what they actually are and even seeming to exceed the speed of light. If Alice observes two objects flying away from each other at a speed of 0.5 C relative to her in each direction, as far as they are concerned they are only flying away from each other at 0.8 C, not the speed of light. If Alice observes two objects flying away from each other at 0.9 C, their relative velocity as far as they are concerned is only 0.994 C, not 1.8 C. Of course, the same relative velocities are observed if the two objects are approaching each other, since the only change has been in their positions.
In more general terms, consider two objects moving at Cf1 and Cf2 as observed
from a third frame of reference. There is actually a simple equation to give
the relative velocity Cf12 as observed in the frames of reference of the two
objects; it can be derived with a little calculus or some ugly and tiresome
algebra, but in the interests of simplicity the details can be ignored to
just give the result:
Cf12 = ( Cf2 + Cf1 ) / ( 1 + Cf2 * Cf1 )
-- if they are moving in opposite directions. Of course, it's also nice to
be able to figure out the relative velocity of the two objects if they are
traveling in the same direction -- for example, to allow an Earth observer
tracking Bob in his effort to catch up with Alice to determine the velocity
of Bob and Alice relative to each other. Not surprisingly, that equation
just involves changes in signs:
Cf12 = ( Cf2 - Cf1 ) / ( 1 - Cf2 * Cf1 )
For example, if Alice observes the Earth and Bob moving in opposite
directions at Cf1 = Cf2 = 0.5 C, then the equation for motion in opposite
directions gives their relative velocity as:
Cf12 = ( 0.5 + 0.5 ) / ( 1 + 0.5 * 0.5 )
= 1 / 1.25
= 0.8 C
-- as expected. Push the relative velocity up to 0.9 C and the equation
gives 0.994 C. If an Earth observer watches Bob at Cf2 = 0.8 C trying to
catch up with Alice at Cf1 = 0.5 C, then the equation for motion in the same
direction gives the relative velocity as observed by the two spacefarers as:
Cf12 = ( 0.8 - 0.5 ) / ( 1 - 0.8 * 0.5 )
= 0.3 / ( 1 - 0.4 )
= 0.3 / 0.6
= 0.5 C
-- again, as expected. Change the scenario to Bob flying at 0.994 C to catch
up with Alice flying at 0.9 C, and the result is 0.9 C. These equations are
strictly one-dimensional, working along a straight line, but equations can be
expanded to two or three dimensions using the same principles -- though that
is also a bit beyond the scope of this document.
* As mentioned, although time dilation and length contraction are not optical illusions, the optical illusions created by the Doppler shift do have to be taken into consideration in examples of special relativity. There is another optical illusion that may be involved as well, which for a lack of a better term might be called "relativistic perspective". It wasn't mentioned above because it would have complicated the discussion without really changing anything.
Consider the examples of Alice flying at Cf = 0.8 in her 100 meter long starship. As discussed, Bob will see Alice's starship as 60 meters long, and with the clock in the rear 267 nsec ahead of the clock in the front. However, this is only strictly true if the starship crosses Bob's line of sight at a right angle.
Suppose that Alice is flying straight towards Bob at relativistic speeds, and that her starship has a clock mounted on an arm at a right angle to the nose, with a second clock mounted on an arm on the opposite side on the rear. Suppose further that Bob takes a snapshot of the starship; it will actually show the clock in the rear being behind the clock in the front, simply because the light from the back of the ship had to travel both the length of the ship and the distance the ship moved during the time of transit of the light to the nose of the ship.
Given that the starship has a length L and a velocity of V = Cf * C, then as
far as Bob's concerned, the time T it takes for the light from the rear to
reach the nose can be derived as follows:
T = ( L * SQRT( 1 - Cf^2 ) + T * V ) / C
C * T = L * SQRT( 1 - Cf^2 ) + T * V
C * T - T * V = L * SQRT( 1 - Cf^2 )
T * ( C - V ) = L * SQRT( 1 - Cf^2 )
T = L * SQRT( 1 - Cf^2 ) / ( C - V )
For L = 100 meters and Cf = 0.8, this gives a value of 1,000 nsec. Since the
clock in the rear of the starship is running ahead by 267 nsec, this means
the snapshot will show the clock in the back running 1,000 - 267 = 633 nsec
behind of the clock in the front. The trick of relativistic perspective will
also imply that the starship seems much longer than it does from the side.
If Alice's starship is flying away from Bob, the snapshot will show the clock
in the rear being well ahead of the clock in the front, since the light from
the clock in the front has to travel to the rear of the starship as the
starship moves forward. Much the same sort of analysis, involving simple
changes of sign, gives:
T = L * SQRT( 1 - Cf^2 ) / ( C + V )
For L = 100 meters and Cf = 0.8, this gives a value of 112 nsec; the snapshot
will show the clock in the back of the starship running 112 + 267 = 379 nsec
ahead of the clock in the front. From this relativistic perspective, the
starship will seem shorter to Bob than it does from the side.
Once again, the idea of relativistic perspective wasn't discussed in the previous sections because it would have complicated matters without really providing much useful insight. Although the perceived length of Alice's starship will change depending on the direction of the starship relative to Bob's line of sight, at Cf = 0.8 the starship's actual relativistic length will be 60 meters, no matter what its direction is.
The concept of relativistic perspective is a nice if not really essential thing to know. There are also some crackpots who claim that there is really no time dilation or length contraction, it's all just a trick of relativistic perspective -- but the notion wasn't even figured in to the derivations of time dilation and length contraction given in the first chapter.
