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Module 12 Paradoxes
How does one test a scientific theory? Experimental testing is certainly the preferred method. Perform an experiment and
show that the results are consistent with the predictions made by the theory. But one can also test a theory by trying to find
conceptual flaws in it. One such conceptual flaw is a paradox, a situation where it seems that the theory gives two possible,
conflicting results. If a paradox cannot be resolved, then the foundation of the theory develops a crack.
Relativity theory has been tested experimentally and it has passed every test. It has also been subjected to many apparent
paradoxes, perhaps more than most theories because it is a theory of relativity (it deals with the observations of two
observers) and because many of its predictions are, for lack of a better word, unusual. So far, all of the paradoxes have been
resolved, although some people may not be satisfied with some of the resolutions. But that is also the nature of scientific
theories; you can’t satisfy everybody.
We’ll look at a few apparent paradoxes and discuss their conventional resolutions.
The Twins
The twin paradox is arguably the most popular paradox of special relativity theory. The usual statement of the paradox goes
something like this.
There are two twins, A and B. B leaves Earth in a space ship, travels near the speed of
light, and returns to Earth. The time for the roundtrip measured by B’s clock is a proper
time interval. Thus, A’s clock will measure a longer time for the trip. So A will be older
than B when B steps out of the ship. But from B’s point of view, B is stationary and the
Earth and A are traveling. A’s clock will measure a proper time interval and B’s clock will
measure a nonproper interval. Thus, B will be older than A when B returns.
(The fact that A and B are twins isn’t important at all. It just adds dramatic effect to the scenario.)
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The reason that this paradox is so popular isn’t so much because of its resolution but rather because of the difference in
aging of the twins. This aging difference is time-travel into the future for one of the twins. We will explore this time-travel
in detail because it is extremely interesting and it is supported by the theory. We also will briefly discuss the rather
anticlimactic resolution of the paradox.
Let’s say that B travels at 0.8c in his ship, which corresponds to a gamma factor of 1.67 between A and B. Because we are
dealing with the special theory of relativity, we can’t readily handle accelerations. So we will make B go in a straight path
from Earth and return along the same trajectory. We will neglect the acceleration from rest to 0.8c, the acceleration during
the turning around, and the deceleration during the landing. Let’s say that the trip out takes 3 years measured by B on the
ship’s clock. The return trip, then, also takes 3 years. So B measures 6 years for the total trip.
If we take the stance that this is a proper time interval, then A will measure a dilated, nonproper time interval of 1.67 times 6
years or 10 years. Thus, when the ship returns, A will be 10 years older than when B left, B will be 6 years older, and B will
have traveled 4 years into Earth’s future as far as B is concerned. It should be noted at this point that even though the
conclusion of 6 years for B being 10 years for A is correct, using the principle of time dilation the way that was just done is
really not appropriate. The principle strictly applies to the case where one clock, moving in a straight line in the same
direction, records the proper interval while a stationary observer measures a nonproper time interval between two events
that occur at different positions. In our scenario, the time measured by A is not a valid nonproper time interval. First, B
changes directions during the trip and, second, A sees the two events occur at the same position. We can break the trip up
into two parts and use time dilation, however. From Earth to the turn-around point takes 3 years on B’s clock. This is
straight-line motion so A measures 1.67 times 3 years or 5 years for the flight. The return flight back to Earth takes another
3 years for B so another 5 years for A. We reach the same conclusion that the trip takes 6 years for B and 10 years for A.
We have analyzed the trip with A at rest in frame S and B in frame S’. But can’t we say that B is in frame S and A is in S’
so that B ages 10 years and A ages 6 years? This is the apparent paradox. The answer is no, this symmetry does not
exist. What breaks the symmetry? The answer commonly proffered is that B must accelerate to return to Earth. A does
not accelerate. It is this acceleration which breaks the symmetry and resolves the paradox. You might argue, however,
that we could think of B as stationary and the Earth with A on it accelerating as it goes out and returns to B’s ship. Thus,
who can really say which twin is accelerating? But this can be determined experimentally. Give each twin an
accelerometer before launch. Only one will record an acceleration and that will be the one on B’s ship. You can’t have it
both ways here; B will be younger when the ship lands.
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Okay, that’s it, the twin paradox is solved. But wait…there is another way to investigate this scenario that leads to the same
final result but it gives us more details about how this time-travel pans out as seen by the twins if we allow them to
communicate with transmissions that travel at the speed of light. And the results of the investigation are strange indeed!
What we do is give A and B lasers and have each of them send a pulse of light to the other at a rate of one pulse per year as
measured on their respective clocks. We now want to figure out the rate at which the signals are detected.
To do this, we use the relativistic Doppler shift results of Module 6. The frequencies that we are interested in comparing
here are not the frequencies of the electromagnetic field oscillations (although those are shifted as well) but rather the pulse
transmission and reception frequencies.
For example, let’s look at twin B. B sends a pulse of light once each year. If we say that B is in frame S’, then the
transmission frequency is f ‘= 1 pulse per year or 1 ppy. Now we find the Doppler-shifted reception frequency f to see when
A receives these pulses in frame S.
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On the trip out, A sees the source of the frequency moving away. We can therefore use the longitudinal Doppler shift
expression [Eq. (6.27)] to find
f  f'
cv
c  0.8c 1
 (1 ppy)
 ppy.
cv
c  0.8c 3
We interpret this result as meaning that A receives the first pulse form B after 3 years on A’s clock, the second pulse after
another 3 years, and the third pulse after another 3 years. Recall that time dilation says that 3 years for B takes 5 years on
A’s clock. So when B sends the third pulse, 5 years have elapsed on A’s clock but that pulse isn’t received by A until
another 4 years as measured by A!
On the trip back, A sees the source of the frequency moving towards Earth. We use the pertinent longitudinal Doppler shift
expression [Eq. (6.25)] to find
f  f'
cv
c  0.8c
 (1 ppy)
 3 ppy.
cv
c  0.8c
We interpret this result as meaning that A receives the remaining 3 pulses form B during a single year, the tenth year on A’s
clock. Again, the 3-year trip back for B is 5 years for A. However, A receives all three of the return pulses during the tenth
year on A’s clock!
We see right away that the 6 years for B is 10 years for A, just like the time-dilation result. But we also see that the
reception of the pulses by A is not uniform nor symmetric. Based on received pulses, A measures the majority of the total
trip time to be for the trip out. In fact, the return trip is measured by A to take only 1 year, shorter than the 3 years on B’s
clock.
If you are having a bit of trouble sorting this out, don’t feel too bad. What might help make things clearer is a pictorial
representation of what is going on here. Maybe a space-time diagram will help.
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Check out the space-time diagram in Figure 12-1. The
diagram is drawn such that A is seen to be at rest at the
origin of frame S. Thus, we see the world line of A going
right on top of the ct axis. Each marked division represents
one year on A’s clock. We have B at rest at the origin of
frame S’, with this origin moving to the right during the trip
out at a constant speed of 0.8c. Thus, B’s outgoing world
line corresponds to the ct’ axis which is tilted clockwise at
38.7 from the vertical. A division of 1 year on B’s world
line is 2.13 times longer than a division of 1 year on the
world line of A. (You should confirm this angle of 38.7 
and this factor of 2.13 using what you learned in Module 8!)
All we need to do to make B’s world line for the return trip
is to draw a straight line tilted 38.7  counterclockwise from
vertical.
The orange lines are the world lines of the laser pulses
sent out by B. These lines are all directed towards A and
tilted 45 from vertical since they are light world lines.
We see that B sends a pulse after each year of the ship’s
clock. We see that the first pulse is received by A after 3
years on the Earth clock, just as we deduced earlier. And,
sure enough, the second pulse is received after 6 years, the
third after 9 years, and the final 3 pulses all arrive during
the last year. The last pulse is received by A as B lands.
Also note the black dashed line that verifies that 3 years
for B is 5 years for A.
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Paradoxes
ct (c-years)
10 6
9
5
8
7
4
returning
world line
of B
38.7
6
3 turnaround
point
5
4
2
3
outgoing
world line
of B
2
1
1
38.7
x
Fig. 12-1
5
You might be thinking that we could also show what
happens to the signals received by B on a space-time
diagram. You would be right! Look at Figure 12-2. Now
the light world lines are drawn emanating from A after
each year on A’s clock. We see that the first pulse is not
received until 3 years after launch on B’s clock, at the
turnaround point. On the return trip, B receives nine
pulses at a constant frequency with the tenth pulse
received as B lands. This is consistent with the Doppler
shifted pulse frequencies observed by B. On the trip out,
the observed frequency is 1/3 ppy, and on the trip back it
is 3 ppy. So a three-year trip out means B receives one
pulse and a three-year trip back means nine more pulses
are received.
ct (c-years)
10 6
9
5
8
7
4
returning
world line
of B
6
3 turnaround
point
5
4
2
3
outgoing
world line
of B
2
1
1
x
Fig. 12-2
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As if things aren’t strange enough, let’s look at how each twin
observes the aging of the other to occur based on the reception of
the light pulses. Each knows that the other is sending one pulse
each year so each can track the apparent age of the other by
counting the received pulses. We look at Figure 12-1 for A’s
perspective and Figure 12-2 for B’s perspective. Let’s say that the
ship leaves when the twins are 20 years old. Table 12.1
summarizes the observed ages.
Notice that the aging looks identical for both twins on the trip out.
Each says on their twenty-third birthdays, “I’m 23 and my twin is
21.” It is only after the turn back home that the aging starts to
look different for each twin. Both turn 24 on the same day if you
ask B, but A would say that B is still 21 on A’s twenty-fourth
birthday! However, both agree that A is 30 and B is 26 when the
ship lands, and that B has traveled four years into Earth’s future.
Table 12.1. Observed Ages of Twins Based on
Reception of Light Pulses
Twin A
My Age B’s Age
20
20
Twin B
My Age A’s Age
20
21
21
22
22
23
23
21
24
24
24
25
25
27
26
30
26
21
20
22
27
What an interesting and bizarre trip this has turned out to be!
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29
23
30
26
7
Faster Than Light
Many paradoxes put forth involve scenarios where it looks like something travels faster than light. We have seen that
relativity theory predicts that superluminal speeds are not possible, hence the paradox. In all of these cases one can show
that no object is traveling faster than light and that no information is traveling faster than light. We’ll take a look at a few
scenarios.
Scenario 1
v
Let’s take a rod of length r and rotate it at an angular speed of w as
shown in Figure 12-3. The piece of the rod at its end has a tangential
velocity of
v  wr
(12.1)
r
We can make this speed be faster than that of light if
w
c
r
(12.2)
w
To do this at a reasonable angular speed would take a very long rod but the math says it
can be done. But the physics does not! The rod is made of matter. As the chunk of rod
at the end speeds up, its inertial mass will increase and its angular acceleration will
decrease. The rod will warp and bend as it spins and will definitely break.
Fig. 12-3
Wait, you say, what if in the future we could find a miracle material that would not
break? It still won’t work. As the rod spins faster, the moment of inertia increases due
to the increased relativistic mass of each chunk of rod. The moment of inertia
approaches infinity as the rod spins faster. Thus, it would require an infinite torque and
an infinite amount of work to reach the speed of light.
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Scenario 2
The previous paradox was simple to resolve. Matter can not reach a speed of c, period. But what if we replace the
rod with a beam of light? Consider Figure 12-4. Here we have a laser rotating at angular speed w. The beam
travels a distance of r whereupon it hits a curved wall. The tangential speed of the spot on the wall is
v  wr
(12.3)
We can make this speed be faster than that of light if the beam length satisfies
c
r
w
B
A
(12.4)
So, for instance, if the laser rotates at 5000 rpm, w = 523 rad/s, and r must be
greater than 574 km. Again, that’s a long distance, but it should work, right?
The laser spot should move from A to B at a speed faster than c, right? It
seems so. Since this is a beam of light and not a rod, we don’t have any
inertial mass to prevent this from being so.
Is this paradox irresolvable? Is the theory of special relativity threatened?
No. Look closely at what is happening. The photons travel from the laser to
point A at c. They travel from the laser to point B at c. They are not moving
from A to B along the wall. It’s okay if this spot moves faster than c; we’re
not violating relativity here. But doesn’t that mean we could somehow send a
signal from A to B at a speed faster than light? Nope. Remember that the
light is coming from the laser and it is confined to speed c. Any information
must be encoded in the laser light and it simply cannot travel faster than light.
(You’re welcome to try to think of a way that you could use this set-up to send
information from A to B with the laser spot. If you come up with a way that
works, contact the Nobel Prize headquarters.)
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Paradoxes
r
w
Fig. 12-4
9
Scenario 3
Let’s try a rod again to send information faster than light. This time, though, we’ll keep the rod’s speed under c. Suppose
we want to transmit information along the x-axis by means of a long rod as shown in Figure 12-5. The rod is tilted at angle
q with respect to the x-axis. The intersection point of the rod and the axis is labeled P. Notice that this intersection point
will move to the right if the rod is moved downward. Thus, we can use the intersection point to send a signal along the xaxis. Point P begins at the transmitter. All we have to do is to set our detector farther out along on the axis and, at some
later time, point P will hit the detector. Let’s now figure out how fast we can move point P.
y
t=0
Dx
t>0
Transmitter
q
Dy
Receiver
x
P
uy
Fig. 12-5
Suppose the rod moves downward at speed uy. During a time of Dt point P moves to the right a distance of Dx.
Thus, its velocity along the x-axis is
v
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Paradoxes
Dx
Dt
(12.5)
10
During this same time the rod moves downward a distance of
Dy  u y Dt
(12.6)
We can relate the ratio of the distances to angle q with
tan q 
Dy
Dx
(12.7)
Combining the previous three equations gives the speed of the intersection point P as
v
uy
tan q
(12.8)
Interesting…the rod can move less than the speed of light but we can easily obtain v > c by choosing an appropriate
angle. For instance, if uy = 0.6c and q = 25, then v = 1.3c. It seems that we can transmit our signal faster than light
can travel.
Hold on. Alas, we cannot. While it is true that this mathematical intersection point can travel faster than c, there is
no way to transmit information with it. For example, suppose you want to signal the detector when a light goes on
at the transmitter. When the light turns on, the rod begins to move. It must accelerate to reach its speed and this
takes time. Point P will not beat the light signal to the receiver. Good try, but this paradox has been resolved.
You may want to try and replacing the rod with a laser beam as we did in Scenario 2. The beam would form the
angle q with the x-axis and a downward movement of the laser would result in the laser spot on the x-axis moving
to the right faster than light. But just as in Scenario 2, there is no way to send information with this laser spot. In
fact, all we have done here is to change from the circular geometry of Scenarios 1 and 2 to a linear geometry.
Scenarios can be concocted where points of intersection of light with walls or axes can move faster than c but there
is no way to send information with these intersection points.
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The conclusion that information cannot travel faster than light is commonly accepted. However, quantum theory
does predict what may be a violation of this rule. When certain particles that have the same possible quantum
states interact and then separate, they seem to be able to transmit information about their states instantaneously
over long distances. These particles exhibit what is called quantum entanglement. This principle is key to the
design of proposed quantum computers where information about many different quantum states is transmitted
extremely fast. Is information really traveling faster than light? Some would say yes. Others would say no; rather
space-time is being warped by quantum effects and that distances between these particles are being shrunk so that it
just appears that information is traveling faster than c. In the end, it may not matter which view is correct. What
we do know is that without quantum effects, information has a speed limit and that limit is c.
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