More Relativity: The Train and The Twins
Michael Fowler, UVa Physics.
Einstein’s Definition of Common Sense
As you can see from the lectures so far, although Einstein’s Theory of
Special Relativity solves the problem posed by the Michelson-Morley experiment—the
nonexistence of an ether—it is at a price. The simple assertion
that the speed of a flash of light is always c in any inertial frame
leads to consequences that defy common sense. When this was pointed out
somewhat forcefully to Einstein, his response was that common sense is the
layer of prejudices put down before the age of eighteen. All our
intuition about space, time and motion is based on childhood observation of a
world in which no objects move at speeds comparable to that of light. Perhaps
if we had been raised in a civilization zipping around the universe in
spaceships moving at relativistic speeds, Einstein’s assertions about
space and time would just seem to be common sense. The real question, from
a scientific point of view, is not whether Special Relativity defies common
sense, but whether it can be shown to lead to a contradiction. If
that is so, common sense wins. Ever since the theory was published,
people have been writing papers claiming it does lead to contradictions.
The previous lecture, the worked example on time dilation, shows how careful
analysis of an apparent contradiction leads to the conclusion that in fact
there was no contradiction after all. In this lecture, we shall consider
other apparent contradictions and think about how to resolve them. This
is the best way to build up an understanding of Relativity.
Trapping a Train in a Tunnel
One of the first paradoxes to be aired was based on the Fitzgerald
contraction. Recall that any object moving relative to an observer will
be seen by that observer to be contracted, foreshortened in the direction of
motion by the ubiquitous factor . Einstein lived in Switzerland, a very mountainous country where the railroads between towns often go through
tunnels deep in the mountains.
Suppose a train of length L is moving along a straight track at a
relativistic speed and enters a tunnel, also of length L. There
are bandits inhabiting the mountain above the tunnel. They observe a
short train, one of length , so they wait until this short train
is completely inside the tunnel of length L, then they close doors at
the two ends, and the train is trapped fully inside the mountain. Now look
at this same scenario from the point of view of someone on the train. He
sees a train of length L, approaching a tunnel of length , so the
tunnel is not as long as the train from his viewpoint! What does he think
happens when the bandits close both the doors?
The Tunnel Doors are Closed Simultaneously
The key to understanding what is happening here is that we said the bandits
closed the two doors at the ends of the tunnel at the same time. How
could they arrange to do that, since the doors are far apart? They could
use walkie-talkies, which transmit radio waves, or just flash a light down the
tunnel, since it’s long and straight. Remember, though, that the
train is itself going at a speed close to that of light, so they have to be
quite precise about this timing! The simplest way to imagine them
synchronizing the closings of the two doors is to assume they know the train’s
timetable, and at a prearranged appropriate time, a light is flashed halfway
down the tunnel, and the end doors are closed when the flash of light reaches
the ends of the tunnel. Assuming the light was positioned correctly in
the middle of the tunnel, that should ensure that the two doors close
Or are They?
Now consider this door-closing operation from the point of view of someone
on the train. Assume he’s in an observation car and has incredible
eyesight, and there’s a little mist, so he actually sees the light flash,
and the two flashes traveling down the tunnels towards the two end doors.
Of course, the train is a perfectly good inertial frame, so he sees
these two flashes to be traveling in opposite directions, but both at c,
relative to the train. Meanwhile, he sees the tunnel itself to be
moving rapidly relative to the train. Let us say the train enters the
mountain through the “front” door. The observer will see the
door at the other end of the tunnel, the “back” door, to be rushing
towards him, and rushing to meet the flash of light. Meanwhile, once he’s
in the tunnel, the front door is receding rapidly behind him, so the flash of
light making its way to that door has to travel further to catch it. So
the two flashes of light going down the tunnel in opposite directions do not
reach the two doors simultaneously as seen from the train.
The concept of simultaneity, events happening at the same time, is
not invariant as we move from one inertial frame to another. The
man on the train sees the back door close first, and, if it is not quickly
reopened, the front of the train will pile into it before the front door is
closed behind the train.
Does the Fitzgerald Contraction Work Sideways?
The above discussion is based on Einstein’s prediction that objects
moving at relativistic speed appear shrunken in their direction of motion.
How do we know that they’re not shrunken in all three directions, i.e.
moving objects maybe keep the same shape, but just get smaller? This can be
seen not to be the case through a symmetry argument, also due to
Einstein. Suppose two trains traveling at equal and opposite relativistic
speeds, one north, one south, pass on parallel tracks. Suppose two
passengers of equal height, one on each train, are standing leaning slightly
out of open windows so that their noses should very lightly touch as they pass
each other. Now, if N (the northbound passenger) sees S as
shrunken in height, N’s nose will brush against S’s
forehead, say, and N will feel S’s nose brush his chin.
Afterwards, then, N will have a bruised chin (plus nose), S a
bruised forehead (plus nose). But this is a perfectly symmetric problem,
so S would say N had the bruised forehead, etc. They can
both get off their trains at the next stations and get together to check out
bruises. They must certainly be symmetrical! The only consistent
symmetrical solution is given by asserting that neither sees the
other to shrink in height (i.e. in the direction perpendicular to their
relative motion), so that their noses touch each other. Therefore, the
Lorentz contraction only operates in the direction of motion, objects
get squashed but not shrunken.
How to Give Twins Very Different Birthdays
Perhaps the most famous of the paradoxes of special relativity, which was
still being hotly debated in national journals in the fifties, is the twin
paradox. The scenario is as follows. One of two twins—the
sister—is an astronaut. (Flouting tradition, we will take fraternal
rather than identical twins, so that we can use “he” and “she”
to make clear which twin we mean). She sets off in a relativistic spaceship
to alpha-centauri, four light-years away, at a speed of, say, 0.6c.
When she gets there, she immediately turns around and comes back. As seen
by her brother on earth, her clocks ran slowly by the time dilation factor , so
although the round trip took 8/0.6 years = 160 months by earth time, she has
only aged by 4/5 of that, or 128 months. So as she steps down out of the
spaceship, she is 32 months younger than her twin brother.
But wait a minute—how does this look from her point of view? She
sees the earth to be moving at 0.6c, first away from her then towards
her. So she must see her brother’s clock on earth to be running
slow! So doesn’t she expect her brother on earth to be the younger
one after this trip?
The key to this paradox is that this situation is not as symmetrical as it
looks. The two twins have quite different experiences. The one on
the spaceship is not in an inertial frame during the initial
acceleration and the turnaround and braking periods. (To get an idea of
the speeds involved, to get to 0.6c at the acceleration of a falling
stone would take over six months.) Our analysis of how a clock in one
inertial frame looks as viewed from another doesn’t work during times
when one of the frames isn’t inertial—in other words, when one is
The Twins Stay in Touch
To try to see just how the difference in ages might develop, let us imagine
that the twins stay in touch with each other throughout the trip. Each
twin flashes a powerful light once a month, according to their calendars and
clocks, so that by counting the flashes, each one can monitor how fast the
other one is aging.
The questions we must resolve are:
If the brother, on earth, flashes a light once a month, how frequently, as
measured by her clock, does the sister see his light to be flashing as she
moves away from earth at speed 0.6c?
How frequently does she see the flashes as she is returning at 0.6c?
How frequently does the brother on earth see the flashes from the spaceship?
Once we have answered these questions, it will be a matter of simple
bookkeeping to find how much each twin has aged.
Figuring the Observed Time between Flashes
To figure out how frequently each twin observes the other’s flashes to
be, we will use some results from the previous
lecture, on time dilation. In some ways, that was a very small scale
version of the present problem. Recall that we had two “ground”
clocks only one million miles apart. As the astronaut, conveniently
moving at 0.6c, passed the first ground clock, both that clock and her
own clock read zero. As she passed the second ground clock, her own clock
read 8 seconds and the first ground clock, which she photographed at
that instant, she observed to read 4 seconds.
That is to say, after 8 seconds had elapsed on her own clock, constant observation
of the first ground clock would have revealed it to have registered only 4
seconds. (This effect is compounded of time dilation and the fact that as she
moves away, the light from the clock is taking longer and longer to reach her.)
Our twin problem is the same thing, at the same speed, but over a longer
time - we conclude that observation of any earth clock from the receding
spacecraft will reveal it to be running at half speed, so the brother’s
flashes will be seen at the spacecraft to arrive every two months, by
Symmetrically, as long as the brother on earth observes his sister’s
spacecraft to be moving away at 0.6c, he will see light from her flashes to be
arriving at the earth every two months by earth time.
To figure the frequency of her brother’s flashes observed as she
returns towards earth, we have to go back to our previous example and find how
the astronaut traveling at 0.6c observes time to be registered by the second
ground clock, the one she’s approaching.
We know that as she passes that clock, it reads 10 seconds and her own clock
reads 8 seconds. We must figure out what she would have seen that second
ground clock to read had she glanced at it through a telescope as she passed
the first ground clock, at which point both her own clock and the first ground
clock read zero. But at that instant, the reading she would see on the
second ground clock must be the same as would be seen by an observer on the
ground, standing by the first ground clock and observing the second ground
clock through a telescope. Since the ground observer knows both ground
clocks are synchronized, and the first ground clock reads zero, and the second
is 6 light seconds distant, it must read -6 seconds if observed at that instant.
Hence the astronaut will observe the second ground clock to progress from -6
seconds to +10 seconds during the period that her own clock goes from 0 to 8
seconds. In other words, she sees the clock she is approaching at 0.6c
to be running at double speed.
Finally, back to the twins. During her journey back to earth, the
sister will see the brother’s light flashing twice a month. (Evidently,
the time dilation effect does not fully compensate for the fact that each
succeeding flash has less far to go to reach her.)
We are now ready to do the bookkeeping, first, from the sister’s point
What does she see?
At 0.6c, she sees the distance to alpha-centauri to be contracted by
the familiar to a distance of 3.2 light
years, which at 0.6c will take her a time 5.333 years, or, more conveniently,
64 months. During the outward trip, then, she will see 32 flashes from
home, she will see her brother to age by 32 months.
Her return trip will also take 64 months, during which time she will see 128
flashes, so over the whole trip she will see 128 + 32 = 160 flashes, so she
will have seen her brother to age by 160 months or 13 years 4 months.
What does he see?
As he watches for flashes through his telescope, the stay-at-home brother
will see his sister to be aging at half his own rate of aging as long as he
sees her to be moving away from him, then aging at twice his rate as he sees
her coming back. At first glance, this sounds the same as what she sees—but
it isn’t! The important question to ask is when does he see
her turn around? To him, her outward journey of 4 light years’
distance at a speed of 0.6c takes her 4/0.6 years, or 80 months. BUT
he doesn’t see her turn around until 4 years later, because of the
time light takes to get back to earth from alpha-centauri! In other words, he
will actually see her aging at half his rate for 80 + 48 = 128 months, during
which time he will see 64 flashes.
When he sees his sister turn around, she is already more than half
way back! Remember, in his frame the whole trip takes 160 months (8 light years
at 0.6c) so he will only see her aging at twice his rate during the last
160 - 128 = 32 months, during which period he will see all 64 flashes she sent
out on her return trip.
Therefore, by counting the flashes of light she transmitted once a month, he
will conclude she has aged 128 months on the trip, which by his clock and
calendar took 160 months. So when she steps off the spacecraft 32 months
younger than her twin brother, neither of them will be surprised!
The Doppler Effect
The above analysis hinges on the fact that a traveler approaching a flashing
light at 0.6c will see it flashing at double its “natural”
rate—the rate observed by someone standing still with the light—and
a traveler receding at 0.6c from a flashing light will see it to flash
at only half its natural rate.
This is a particular example of the Doppler Effect, first discussed
in 1842 by the German physicist Christian Doppler. There is a Doppler
Effect for sound waves too. Sound is generated by a vibrating object
sending a succession of pressure pulses through the air. These pressure
waves are analogous to the flashes of light. If you are approaching a
sound source you will encounter the pressure waves more frequently than if you
stand still. This means you will hear a higher frequency sound. If
the distance between you and the source of sound is increasing, you will hear a
lower frequency. This is why the note of a jet plane or a siren goes
lower as it passes you. The details of the Doppler Effect for sound are a
little different than those for light, because the speed of sound is not the
same for all observers—it’s 330 meters per second relative to the
An important astronomical application of the Doppler Effect is the red
shift. The light from very distant galaxies is redder than the light
from similar galaxies nearer to us. This is because the further away a
galaxy is, the faster it is moving away from us, as the Universe expands.
The light is redder because red light is low frequency light (blue is high) and
we see low frequency light for the same reason that the astronaut receding from
earth sees flashes less frequently. In fact, the farthest away galaxies
we can see are receding faster than the 0.6c of our astronaut!
In the next lecture, we shall brush up on the pre-relativistic concepts of momentum, work and energy
to be ready for their relativistic generalizations.