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Spiral.. Physics
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Modern Physics
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The Special Theory of Relativity:
Kinematics
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Copyright © 2003
Paul D’Alessandris
Spiral Physics
Rochester, NY 14623
All rights reserved. No part of this book may be reproduced or transmitted in any
form or by any means, electronic or mechanical, including photocopying, recording,
or any information storage and retrieval system without permission in writing from
the author.
This project was supported, in part, by the National Science Foundation. Opinions expressed are those of
the author and not necessarily those of the Foundation.
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The Special Theory of Relativity:
Kinematics
Deriving Time Dilation
Einstein developed much of his understanding of relativity through the use of gedanken, or
thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that
could not be performed due to technological limitations, and “perform” the experiment in his
head. By analyzing the results of these experiments, he was lead to a deeper understanding of
his theory.
In developing his understanding of the relativity of time, Einstein imagined what he
considered the simplest possible clock, consisting of a mirror on the floor and a mirror on the
ceiling of a train car, and a light pulse bouncing back and forth between the mirrors. Each
reflection of the light pulse serves as a “tick” of the clock.
Now imagine two observers and the train car traveling to the right at speed v. A passenger on
the train will see the light pulse travel a vertical path from floor to ceiling. An observer at rest
on the earth, however, will see the train displaced to the right during the time it takes the pulse
to travel from floor to ceiling, and hence see the light pulse follow an angled path.
3
Earth frame of reference
Passenger frame of reference
pulse emitted at floor
pulse reflected at ceiling
Let t represent the travel time from floor to ceiling measured on the earth, and t0 represent
the travel time from floor to ceiling measured on the train. Based on these definitions, and the
speed of the train (v), the distances in the diagram can be determined.
Earth frame of reference
Passenger frame of reference
pulse emitted at floor
ct0
pulse reflected at ceiling
ct0
ct
vt
By Pythagoras’ Theorem,
(ct ) 2  (vt ) 2  (ct 0 ) 2
(ct ) 2  (vt ) 2  (ct 0 ) 2
(c 2  v 2 )t 2  c 2 t 0
v2
2
)t 2  t 0
2
c
t 0
t 
v2
1 2
c
(1 
4
2
Therefore, since the denominator is less than 1 for any speed v, the time interval between the
“ticks” of the clock measured on the earth is greater than the time interval measured on the
train. To earth-bound observers, the clock is ticking slower than normal. Einstein proposed
that this is not a property of the clock, but rather a reflection of the fact that time itself passes
slower for an object is in motion.
Using Time Dilation
The half-life of a muon at rest is 2.2 s. One can store muons for a much longer
time (as measured in the laboratory) by accelerating them to a speed very close
to that of light and then keeping them circulating at that speed in an evacuated
ring. Assume that you want to design a ring that can keep muons moving so fast
that they have a laboratory half-life of 20 s. How fast will the muons have to
be moving?
The time between two events, for example the “birth” and “death” of a muon, depends on
who makes the measurements. Within any particular reference system, the familiar results of
classical physics are valid. However, in comparing results between observers in different
reference systems, a method of relating one observer’s measurements to another is needed.
The formula for time dilation, derived above, is typically written in the form
t   (t 0 )
where

t0 is the proper time, the time between two events in the frame of reference in
which both events occur at the same spatial point,

t is the time between the same two events in a different frame, moving at relative
speed v,

and  is the Lorentz factor, given by
 
5
1
v2
1 2
c
In the muon’s frame of reference, the muon’s are at rest and hence have a half-life of 2.2 s.
Moreover, this is a proper time since in this frame of reference both the birth and death of the
muons occur at the same point. (Since the muons aren’t moving, everything that happens to
them happens at the same point.)
 t   ( t 0 )
20 s   (2.2 s )
  9.09
1
 
1
9.09 
v2
c2
1
v2
c2
v2
1 2
1 2  (
)
9.09
c
v2
1 2
 1 (
)
2
9.09
c
v2
 0.9879
c2
v  0.994c
1
Time Dilation at Everyday Speeds
The cruising speed of a jet airplane is approximately 250 m/s relative to the
earth. For each hour of earth-time that passes, how much less time passes for
the passengers on the plane?
The time interval measured by the passengers is the proper time, t0, so
t 
t 0
1
v2
c2
t 0  t 1 
v2
c2
250 2
t 0  (1hr ) 1 
(3 x10 8 ) 2
t 0  (1hr ) 1  6.94 x10 13
6
Unfortunately, at this point my calculator tells me that t0 = 1 hr, which is obviously not true.
I need a better way to deal with the extremely small numbers that sometimes show up in
relativity. One way to do this is with the binomial approximation.
Expressions of the form:
(1  x) n
where x is much less than 1 can be approximated by:
(1  x) n  1  nx
if x  1
Using this approximation allows me to find that:
t 0  (1hr ) 1  6.94 x10 13
t 0  (1hr )(1  6.94 x10 13 )1 / 2
1
t 0  (1hr )(1  ( )6.94 x10 13 )
2
t 0  (1hr )(1  3.47 x10 13 )
t 0  1hr  3.47 x10 13 hr
t 0  1hr  1.25 x10 9 s
Thus, for every hour that passes on earth, 1.25 nanoseconds less time passes on the airplane!
This may seem like an incredibly small amount of time (and it is) but this effect has been
measured in numerous experiments.
Deriving Length Contraction
If the time interval between two events depends on the relative motion of the observer,
Einstein realized that the spatial separation between the events must also be observerdependent.
Consider a hypothetical spaceship journey from earth to a distant star. Assume the star is a
distance L0 from the earth, as measured by stationary earth-bound observers. Therefore, the
elapsed time for the spaceship to reach the star, as measured on earth, is
t 
L0
v
where v is the speed of the spaceship measured on earth.
7
By time dilation, however, the elapsed time for the spaceship to reach the star, as measured on
the spaceship (a proper time), is
t
t 0 

L
t 0  0
v
The distance the spaceship travels (L), as measured on the spaceship, is simply the product of
its speed and the elapsed time measured on the ship
L  vt 0
L  v(
L
L0
)
v
L0

Since gamma is greater than one, the distance between the earth and the star as measured on
the ship is less than the distance as measured on the earth. To moving observers, the distance
to the star shrinks.
Using Length Contraction
The star Vega is approximately 25 light-years from Earth (as measured by observers on
Earth).
a. How fast must a spaceship travel in order to reach Vega in 30 years, as measured on
Earth?
b. How fast must a spaceship travel in order to reach Vega in 30 years, as measured on
the spaceship?
The distance between two events, for example leaving Earth and arriving at Vega, depends on
who makes the measurements. Within any particular reference system, the familiar results of
classical physics are valid. However, in comparing results between observers in different
reference systems, a method of relating one observer’s measurements to another is needed.
8
The formula for length contraction is
L  L0 / 
where

L0 the proper length, the distance between two events in the frame of reference in
which both events are at rest,

and L is the distance between the same two events in a different frame, moving at
relative speed v.
For part a, relativity theory is unnecessary. Both the distance and time measurements are
made from the same frame of reference. Therefore, results from classical physics are valid.
d Earth  vtEarth
25cyr  v(30 yr )
25cyr
v
30 yr
v  0.833c
(Note the use of the speed of light as a unit. Rather than substituting 3.0 x 10 8 m/s for c,
simply leave c as a unit of velocity.)
For part b, the distance is measured in the Earth’s frame while the time is in the frame of the
spaceship. To solve part b, you must either convert the distance into the spaceship frame or
the time into the Earth frame. You can convert the distance into the spaceship frame by
realizing that the distance to Vega as measured on Earth is a proper length. Therefore,
d spaceship  vtspaceship
( L0 /  )  v(30 yr )
(25cyr /  )  v(30 yr )
25cyr
 v
30 yr
v
0.833c 
v2
1 2
c
2
v
0.694c 2 
v2
1 2
c
2
0.694c  0.694v 2  v 2
1.694v 2  0.694c 2
v  0.640c
9
Deriving the Lorentz Transformation
The time dilation and length contraction relationships actually have very limited applicability.
To use the time dilation relationship, one of the two observers must measure a proper time,
where the two events occur at exactly the same spatial point. To use the length contraction
relationship, one of the two observers must measure a proper length, where the two events
must be at rest with respect to the observer. What if you want to compare measurements
concerning more general events? To do this requires the Lorentz Transformation, which
allows you to transform the spacetime coordinates of an event in one inertial reference system
to any other inertial reference system.
To derive the Lorentz transformation, imagine two inertial reference systems, labeled O and
O’. Let the origins of O and O’ overlap at time zero, and allow O’ to move with speed u
relative to O. (Therefore, at a later time t, the origins are separated by a distance ut.) Call the
direction of motion the x-direction.
u
x’
ut
event
x
O’
O
Now imagine an event that occurs somewhere in spacetime. This event is located at position x
relative to the O system, and position x’ relative to the O’ system. How are these two
locations related?
You may be tempted to state that
x  x'ut
however, this can’t be correct because x and x’ are measured in different reference systems.
However, imagine that the event is the tip of a meterstick, fixed in O’, striking some object.
Since x’ is now a proper length in O’, it will appear contacted in O by the gamma factor.
Therefore, the correct relationship between x and x’ is
x
x'

 ut
rearranging yields
x'   ( x  ut )
10
Since there is no relative motion in the y and z directions, these positions are the same in both
coordinate systems
y'  y
z'  z
This completes the spatial part of the Lorentz transformation, but what about the temporal
part? To determine how t and t’ are related, now imagine that the event under investigation is
the result of a light pulse, emitted from the origin when the two origins overlapped at time
zero, striking some detector. Since the speed of light is the same in both systems, the distance
measured in each system must be equal to the product of c and the elapsed time
x'   ( x  ut )
x
ct '   (ct  u )
c
x
t '   (t  u 2 )
c
Using the Lorentz Transformation
Inside of a spaceship zooming past earth at 0.5c, I fire a laser (in the same
direction as the ship’s motion) and let it strike a mirror10 m in front of the
laser.
a. What is the elapsed time measured on the earth between turning on the laser
and the light striking the mirror?
b. How far has the light traveled before hitting the mirror, as measured on
earth?
Since neither the earth’s observers nor the observers on the ship measure a proper time or a
proper length between the two events (turning on the laser and the laser striking the mirror), a
more general method of relating different observers’ measurements is needed. This general
method of relating measurements is the Lorentz Transformation. The Lorentz Transformation
relates the coordinates of a spacetime event, (x, y, z, t), measured in one frame to the
coordinates of the same event in a frame moving with relative velocity u, (x’, y’, z’, t’) as
follows:
x'   ( x  ut )
y'  y
z'  z
ux
t '   (t  2 )
c
11
These equations are written in a form that easily allows the determination of the primed
coordinates from the unprimed. If the situation requires the inverse of this task, the equations
can be easily inverted (by changing the sign of u and flipping the primed and unprimed
notation) to yield
x   ( x'ut ' )
y  y'
z  z'
ux'
t   (t ' 2 )
c
Let the two coordinate systems overlap at the first event (the laser is fired). Thus, the position
and time of the laser’s firing is zero in both coordinate systems. We now must find the
position and time of the second event (the laser strikes the mirror). This is relatively easy to
determine in the frame of the spaceship (the primed frame):
x'  10m
10m
t' 
 3.33x10 8 s
c
Since we know the spacetime location of the event in the primed frame, the Lorentz
Transformation allows us to transform this information into the earth frame. With u = 0.5c
(=1.155),
ux'
)
c2
10 (0.5c)(10)
t  1.155(( ) 
)
c
c2
17.3m
t
 5.78 x10 8 s
c
t   (t '
x   ( x'ut ' )
10
x  1.155(10  (0.5c)( ))
c
x  17.3m
The light travels 17.3 m and takes 5.78 x 10-8 s to strike the mirror in the earth’s frame.
12
Deriving Velocity Addition
Since the Lorentz transformation allows you to relate the position and time of an event in one
coordinate system to the position and time in any other coordinate system, it also allows you
to relate quantities that depend on position and time, like velocity and acceleration. Therefore,
using Lorentz we can derive equations that allow use to transform velocities measured by one
observer to velocities measured by other observers.
Refer back to the Lorentz transformation derivation. This time, imagine that the event of
interest is a particle, launched from the origin when the two origins overlapped at time zero,
striking some detector. Thus,
x  vxt
x'  v x ' t '
Substituting these relationships into the Lorentz transformation yields
x'   ( x  ut )
v x ' t '   (v x t  ut )
v x ' t '   (v x  u )t
and
ux
)
c2
uv t
t '   (t  2x )
c
uv
t '   (1  2x )t
c
t '   (t 
dividing the first equation by the second yields
v x ' t '  (v x  u )t

uv
t'
 (1  2x )t
c
v u
vx '  x
uv
1  2x
c
This directly relates the x-speed of an object in one reference system (vx) to the speed of the
same object measured in a different system (vx’).
13
Although y- and z-positions are not effected by the Lorentz transformation, y- and z-velocities
are different in different systems. Starting with,
y'  y
v y 't'  v yt
divide by the time transformation derived above
v y 't'
vyt
uv
t'
 (1  2x )t
c
vy
vy '
uv
 (1  2x )
c

The same type of transformation holds for z-velocity.
Using Velocity Addition
A spaceship travels at 0.8c with respect to the solar system. An unmanned probe
is ejected at 0.6c at an angle of 300 from the direction of travel of the ship (both
the speed and angle of the probe are measured with respect to the ship). What
are the speed and angle of launch of the probe as measured in the solar system
frame?
Just as distance and time depend on the relative motion between observers, so does velocity.
The following relationships allow you to compare velocity measurements between two
observers in relative motion.
vx  u
uv
1  2x
c
vy
v 'y 
uv
 (1  2x )
c
v x' 
where v’x and v’y are the velocities of an object measured in a frame (the “primed” frame)
moving at speed u relative to the “unprimed” frame (where observers measure vx and vy).
14
These equations are written in a form that easily allows the determination of v’x and v’y if vx
and vy are known. If the situation requires the inverse of this task, the equations can be easily
inverted (by changing the sign of u and flipping the primed and unprimed notation) to yield
vx 
vy 
v x'  u
uv '
1  2x
c
v 'y
 (1 
uv x'
)
c2
Using this form of the equations, with the spaceship the primed frame and u = 0.8c (=1.67),
yields
vx'  u
vx 
uv '
1  2x
c
0.6c(cos(30))  0.8c
vx 
(0.8c)(0.6c(cos(30)))
1
c2
vx  0.932c
vy 
v 'y
uvx'
)
c2
0.6c(sin(30))
vy 
(0.8c)(0.6c(cos(30)))
(1.67)(1 
)
c2
v y  0.127c
 (1 
15
The solar system observers detect the probe’s motion as
v  v x2  v y2
v  (0.932c) 2  (0.127c) 2
v  0.941c
  tan 1 (
vy
  tan 1 (
0.127c
)
0.932c
  7.76 0
16
vx
)
..
..
..
..
..
The Special Theory of Relativity:
Kinematics
Activities
17
As you pass a sign reading “Speed Limit 55 mph”, your speedometer reads 65 mph.
a. How fast are you moving relative to the ground?
b. How fast are you moving relative to someone in your passenger seat?
c. How fast are you moving relative to the law-abiding citizens you are passing?
d. How fast are you moving relative to a Porsche with speedometer reading 75 mph?
A police officer is in pursuit of the Porsche. The officer's speedometer reads 85 mph.
e. How fast are you moving relative to the police car?
The police officer decides to shoot out the tires of the Porsche and fires her revolver. The bullets have a muzzle
velocity of 400 mph. (The muzzle velocity is the velocity of the bullets relative to the gun.)
f. How fast are you moving relative to the bullets?
g. You probably have different answers for questions a, b, c, d, e, and f. Given all these different answers for
your speed, which one is correct?
h. If someone simply asked you, “How fast are you moving?” what would be the correct answer?
18
In the following gedanken experiment, Isaac has created a jetpack that he claims allows him to travel at
1.5c relative to the ground. To test his claim, Albert challenges Isaac to a 100 m dash against a laser. If
Isaac is correct, he should be able to beat the laser in this race.
The starting gun is fired, the jetpack is turned on, and the race begins. During the race, Isaac sneaks a
glance at his “opponent”.
a. How fast does Isaac see the laser beam traveling relative to himself?
b. Based on your answer above, who does Isaac see winning the race?
c. Based on the results of the race, which was traveling faster, Isaac or the laser beam?
Isaac goes back to the design lab and creates a better jetpack. He thinks this jetpack will allow him to go
travel at 2.5c. He again challenges the laser to a race. The starting gun is fired, the jetpack is turned on, and
the race begins. During the race, Isaac sneaks a glance at his “opponent”.
d. How fast does Isaac see the laser beam traveling relative to himself?
e. Based on your answer above, who does Isaac see winning the race?
f. Based on the results of the race, which was traveling faster, Isaac or the laser beam?
Regardless of the power of Isaac’s future jetpacks, the laser always moves past him as if he was stationary. Since
you seldom win races when you are stationary relative to your opponent, Isaac can never beat a laser beam in a race.
If you always lose to light in a race, that’s equivalent to always moving slower than light.
19
In each the following gedanken experiments, Albert is in the center of a glass-sided freight car speeding to
the right at a very high speed relative to you.
I. Albert has a flashlight in each hand, directed at the front and rear panels of the freight car. Albert switches them
on at the same time. Which panel, front or rear, is struck by light first, or are they struck at the same time? Explain.
a. In Albert’s frame of reference:
b. In your frame of reference:
II. Albert has a flashlight in each hand, directed at the front and rear panels of the freight car. The panels are
replaced by mirrors. Albert switches the flashlights on at the same time. Which beam of light returns to Albert first,
the one directed toward the front or the one toward the rear, or do they return at the same time? Explain.
c. In Albert’s frame of reference:
d. In your frame of reference:
III. Two assassins sneak into the freight car with Albert, one against the front panel and one against the rear panel.
They each fire a laser at Albert. The two lasers strike Albert at the same time. Who fired first, the person against
the rear wall or the person against the front wall, or did they fire at the same time? Explain.
e. In Albert’s frame of reference:
f. In your frame of reference:
20
In each the following gedanken experiments, Albert is in the center of a glass-sided freight car speeding to
the right at a very high speed relative to you.
I. Albert has a flashlight in each hand, directed at the ceiling and floor of the freight car. The flashlights are located
midway between the ceiling and the floor. Albert switches them on at the same time. Which is struck by light first,
the ceiling or the floor, or are they struck at the same time? Explain.
a. In Albert’s frame of reference:
b. In your frame of reference:
II. Albert has a flashlight in each hand, directed at the front and rear panels of the freight car. Albert switches the
flashlights on at the same time. Which beam of light travels at a greater speed, the one directed toward the front or
the one toward the rear, or do they travel at the same speed? Explain.
c. In Albert’s frame of reference:
d. In your frame of reference:
III. Two assassins sneak into the freight car with Albert, one against the front panel and one against the rear panel.
Albert sees them before they see him. He quickly pulls out two lasers and fires them at the same time, one aimed
toward the front and one aimed toward the rear. Which assassin bites the dust first, or do they die at the same time?
Explain.
e. In Albert’s frame of reference:
f. In your frame of reference:
21
Two identical spaceships of rest length 94 feet are traveling in the same direction. The two ships, spaceship
A and spaceship B, zoom past an interstellar basketball stadium along the length of the court, just as the
final seconds are ticking off the stadium clock. (The rest length of a basketball court is 94 feet.) When the
stadium clock reads 1.0 s, a last-second shot is made, and as the ball passes through the hoop, the stadium
clock reads 0.0 s. A physicist in the stands at center court measures spaceship A’s speed as 0.5c and
spaceship B’s speed as 0.6c.
I. Comparing Time Intervals
a. From the captain of ship A’s frame of reference, was the shot in the air greater than, less than, or
exactly 1.0 s? Explain.
b. From the captain of ship B’s frame of reference, was the shot in the air greater than, less than, or
exactly 1.0 s? Explain.
c. The captain of ship A and the captain of ship B are in radio contact, and they report to each other
their measurement of the time duration of the shot. Which captain measures the shot as being in the air
longer, or do they both measure the time duration of the shot as equal? Explain.
II. Comparing Lengths
d. From the captain of ship A’s frame of reference, rank the following lengths.
A
B
C
The length of ship A.
The length of ship B.
The length of the basketball court.
Largest 1. _____ 2. _____ 3. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
e. From the captain of ship B’s frame of reference, rank the following lengths.
A
B
C
The length of ship A.
The length of ship B.
The length of the basketball court.
Largest 1. _____ 2. _____ 3. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
f. The captain of ship A and the captain of ship B are in radio contact, and they report to each other
their measurement of the length of the court. Which captain measures the court as longer, or do they
both measure the court as having the same length?
22
I. One consequence of Special Relativity is that no massive object can travel at speed c. Three students
discuss why this is so.
Trina:
What Special Relativity is really about is what we can see, not really about what is possible or
impossible in principle. This means that objects can go faster than light, but since we see
things using light, we would never be able to see an object doing this because the light could
never bounce off the object and return to reach our eyes.
Hans:
For a massive object to travel at c requires an infinite amount of energy. Therefore it is
impossible.
Shaun: If an object traveled at c, in its frame of reference some beams of light would be at rest. Since
this contradicts the initial assumption that the speed of light in vacuum is the same for all
observers, this means that traveling at c is a logical impossibility.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
II. Three students debate whether time really slows down as you travel closer and closer to the speed of
light.
Ahmed: Scientists put an atomic clock on an airplane and another on the ground. The one on the
airplane ticked slower than the one on the ground. End of story. Speed slows time.
Jehona: The amount of time those clocks slowed down by was ridiculously small. They aren’t even
accurate to that small of a time. Maybe at really close to the speed of light time does slow
down, but that’s just a theory since we can’t get clocks going that fast. Basically, we’re not
sure whether time slows down or not.
Kyrai:
Time doesn’t slow down since it’s not a physical thing. Maybe the clock slowed down, but
that’s because of how the clock was built. Clocks don’t really measure time, they measure the
time for gears to turn or springs to expand. Those are mechanical things and time itself is not
mechanical.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
23
I. A muon is created in the upper atmosphere and travels toward the surface of the earth. The muon is
traveling at a speed which is too slow to make it to the surface of the earth during its lifetime, yet it reaches
the surface of the earth before decaying. Three students construct explanations for how this is possible.
Sabina: Since the muon is traveling at close to the speed of light relative to the scientists on earth, its
lifetime is dilated. Thus it “lives” long enough to reach the surface of the earth before
decaying.
Henryk: Actually, since the muon is traveling at close to the speed of light relative to the atmosphere,
the length of the atmosphere contracts so it doesn’t have as long a distance to travel. This is
why it reaches the surface of the earth before decaying.
Izydor: Actually, you both have it backwards. Since the scientists are moving relative to the muon,
their clocks run slow, giving the muon enough time to reach the surface before decaying.
Also, since the muon is moving, it is contracted. Since it is smaller than “normal” it can reach
the earth in a smaller amount of time.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
II. Three students are trying to understand the famous twin paradox, where one twin, Alice, goes on a long
spaceship flight while her twin, Bob, stays home at rest on the earth. The paradox is to decide which twin is
older when Alice returns.
Jake:
In Alice’s frame of reference, Bob is in motion, so Bob ages slower than Alice. In Bob’s
frame, Alice is in motion, so Alice ages slower than Bob. When they get back together, they
each think the other is younger. Alice sees Bob as younger and Bob sees Alice as younger.
This is not a paradox, because in relativity time, and therefore age, are relative to the
observer.
Josh:
Your first two sentences are correct, then you blew it. They don’t each think the other is
younger. They are both aware of both points of view and the two points of view cancel. They
both expect to be the same age when Alice returns and they are the same age when she
returns.
Lukas: As Alice is cruising away from Bob, they do each see the other is younger. However, as she
turns around to come home, Alice changes her reference system (she accelerates). This causes
Alice to age slower than Bob in all reference frames. Thus, when she returns home, they both
agree she is younger than Bob.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
24
Six identical septuplets leave earth when they reach the age of 21, in the year 2121. Each septuplet goes on
a spaceship journey that takes T years, as measured by a clock in each spaceship. During the journey they
travel at a constant speed v, as measured on earth, except during the relatively short acceleration phases of
their journey.
A
B
C
D
E
F
T (yrs)
10
20
10
5
20
10
v
0.8c
0.4c
0.4c
0.2c
0.8c
0.9c
a. Rank these septuplets on the basis of the year on earth when they return from their journey.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
b. Rank these septuplets on the basis of their age when they return from their journey.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
25
Six spaceships with rest lengths L0 zoom past an intergalactic speed trap. The officer on duty records the
speed of each ship, v. (None are going in excess of the stated speed limit of c, so she doesn’t have to pull
anyone over for a ticket.)
A
B
C
D
E
F
L0 (m)
100
200
100
400
200
100
v
0.8c
0.4c
0.4c
0.2c
0.8c
0.9c
a. Rank these spaceships on the basis of their length measured by the police officer.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
b. Rank these spaceships on the basis of their length as measured by their respective captains (i.e., the
length of ship A measured by ship A’s captain, etc.).
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
26
A spaceship, traveling at 0.3c with respect to the solar system, ejects a reconnaissance pod forward at 0.8c
with respect to the ship. The pod, after collecting data, radios the data back to the ship at c, with respect to
the pod. Consider the following six speeds.
A
B
C
D
E
F
The speed of the reconnaissance pod measured by the spaceship.
The speed of the reconnaissance pod measured by observers at rest in the solar system.
The speed of the data measured by the spaceship.
The speed of the data measured by observers at rest in the solar system.
The speed of the spaceship measured by the reconnaissance pod.
The speed of the spaceship measured by the spaceship.
a. Rank these six speeds from largest to smallest.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
As part of Homeplanet Security, a defense spaceship zooms past Earth at 0.5c, as measured on Earth,
monitoring space for incoming alien craft. The spaceship detects an alien craft coming directly toward it at
0.5c, as measured on the spaceship. The spaceship launches a missile at 0.8c, as measured on the spaceship,
directly toward the oncoming alien craft. Consider the following six speeds.
A
B
C
D
E
F
The speed of the spaceship measured on Earth.
The speed of the aliens measured on Earth.
The speed of the missile measured on Earth.
The speed of the missile measured on the spaceship.
The speed of the missile measured by the aliens.
The speed of the spaceship measured by the aliens.
b. Rank these six speeds from largest to smallest.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
27
The nearest star to our Sun is Proxima Centauri, located 4.28 light-years away, as measured by observers
on Earth.
a. I would like to build a spaceship that will allow me to get to Proxima Centauri in 10 years (as
measured by a clock I leave behind in my office at work). What is the minimum speed required to
accomplish this trip?
b. I would like to build a spaceship that will allow me to get to Proxima Centauri in 10 years (as
measured by a clock on the dashboard of my ship, right under the fuzzy dice). What is the minimum
speed required to accomplish this trip?
Mathematical Analysis
28
The center of the Milky Way galaxy is approximately 50,000 lightyears away, as measured by observers on
Earth. (By the way, an enormous black hole with a mass of approximately 3.7 million times the mass of the
sun resides at the center of the Milky Way.)
a. If I travel at 0.9c, how long will it take me (as measured on Earth) to make the trip to the center of the
galaxy?
b. If I travel at 0.9c, how long will it take me (as measured by the wristwatch on my long-dead arm) to
make the trip to the center of the galaxy?
c. If I really want to visit the supermassive black hole at the center of the galaxy and only age 40 years
during the journey which what minimum speed must I travel?
d. Assuming I travel at the above speed, what is my odometer reading when I reach the black hole?
(Assume this is my first journey in my new spaceship.)
Mathematical Analysis
29
I drive approximately 20 miles (as measured by Rand McNally) at 45 mph on my commute to school.
a. How many miles fewer than 20 do I put on my car for each trip to school?
b. Compared to synchronized clocks at my home and at school, how “early” do I arrive each day
according to my wristwatch?
Mathematical Analysis
30
Imagine a new high speed jet that can carry you from New York to Las Vegas (a distance of about 4000
km) in only 1.0 minute (as measured on Earth).
a. What is the approximate speed of the jet?
b. Is this fast enough for you to notice (using a normal wristwatch) the effects of relativity?
Now imagine an even newer jet that can go all the way around the world in 1.0 minute as measured by
clocks at the airport. The radius of the earth is about 6380 km.
c. By how much are the pilots' watches “slow” for a complete circumnavigation of the globe?
d. Is this fast enough for you to notice (using a normal wristwatch) the effects of relativity?
Mathematical Analysis
31
Particle accelerators routinely accelerate particles to close to the speed of light. Imagine an electron
traveling at 0.9995c in a circular accelerator of diameter 3000 m.
a. How long does it take the electron to complete one path around the accelerator, as measured in the
laboratory?
b. How long does it take the electron to complete one path around the accelerator, as measured by the
electron?
c. What is the diameter of the accelerator, as measured by the electron?
Mathematical Analysis
32
The half-life of a neutron at rest is 889 s. A neutron emitted from a particular nuclear reaction is traveling
at 0.78c.
a. What is the half-life of this neutron (as measured in the laboratory)?
b. How far will the neutron travel (as measured in the laboratory), if it decays after 2.0 half-lives?
Mathematical Analysis
33
My new spaceship came with a 10 year, 1016 meter warranty. The small print said that the warranty expires
when either a clock on earth records that 10 years have passed since I bought the spaceship or the
odometer on the spaceship reads 1016 m. I jump in the ship and travel directly away from earth at 0.9 c
until the warranty expires.
a. Where am I, as measured on earth, when the warranty expires?
b. How many years have I been traveling, as measured on the spaceship, when the warranty expires?
Mathematical Analysis
34
Two spaceships traveling in the same direction, spaceship A and spaceship B, zoom past an interstellar
speed trap. The officer on duty measures spaceship A’s speed as 0.5c and spaceship B’s speed as 0.8c.
Complete the following table.
Velocity of
Officer
Velocity of
Ship A
Measured by
Officer
Measured by
Ship A
Measured by
Ship B
Mathematical Analysis
35
Velocity of
Ship B
A spaceship, traveling at 0.8c with respect to the solar system, ejects two reconnaissance pods. Pod A
travels at +0.6c with respect to the ship, and pod B travels at -0.4c with respect to the ship. Both pods
travel along the initial direction of travel of the ship. Complete the following table.
Velocity of
Ship
Velocity of
Pod A
Measured by
Solar System
Measured by
Ship
Measured by
Pod A
Mathematical Analysis
36
Velocity of
Pod B
A spaceship, traveling at 0.3c with respect to the solar system, ejects a reconnaissance pod forward at 0.8c
with respect to the ship. The pod, after collecting data, beams the data back to the ship at c, with respect to
the pod. Complete the following table.
Velocity of
Ship
Velocity of
Pod
Measured by
Solar System
Measured by
Ship
Measured by
Pod
Mathematical Analysis
37
Velocity of
Data Stream
As part of Homeplanet Security, a defense spaceship continuously encircles Earth at 0.5c, as measured on
Earth, monitoring space for incoming alien craft. The spaceship detects an alien craft coming directly
toward it at 0.4c, as measured on the spaceship. The spaceship launches a missile at 0.8c, as measured on
the spaceship, directly toward the oncoming alien craft. Complete the following table.
Velocity of
Spaceship
Velocity of
Aliens
Measured by
Earth
Measured by
Spaceship
Measured by
Aliens
Mathematical Analysis
38
Velocity of
Missile
A spaceship, traveling at 0.8c with respect to the solar system, ejects a reconnaissance pod at 0.6c with
respect to the ship, perpendicular to the direction of motion of the ship.
a. What is the speed of the pod with respect to the solar system?
b. What is the angle, with respect to the solar system, that the path of the pod makes with respect to the
initial direction of motion of the ship?
c. What would this angle be without relativistic effects?
Mathematical Analysis
39
A spaceship, traveling at 0.8c with respect to the solar system, fires a laser perpendicular to the direction
of motion of the ship.
a. What is the speed of the laser beam with respect to the solar system?
b. What is the angle, with respect to the solar system, that the path of the laser makes with respect to the
initial direction of motion of the ship?
c. What would this angle be without relativistic effects?
Mathematical Analysis
40
A spaceship, traveling at 0.4c with respect to the solar system, ejects a reconnaissance pod at 0.9c, 250
from the direction of motion of the ship, all with respect to the ship.
a. What is the speed of the pod with respect to the solar system?
b. What is the angle, with respect to the solar system, that the path of the pod makes with respect to the
initial direction of motion of the ship?
Mathematical Analysis
41
A neutral kaon traveling at 0.85c, in the laboratory frame of reference, decays into a pair of pions. The
pions are emitted at 0.83c perpendicular to the initial direction of motion of the kaon, in the kaon’s frame
of reference. At what angle and speed are the pions detected in the laboratory frame of reference?
Mathematical Analysis
42
I build a jetpack that enables me to travel at 0.9c relative to Earth and challenge a beam of light to a 100 m
(in the earth’s frame of reference) race. Assuming a fair start, complete the following table:
Event
Starting gun
fires
Earth
frame
Position
0m
Time
0s
My
frame
Position
0m
Time
0s
Light reaches
finish line
I reach
finish line
a. In my frame of reference, how far behind the light am I when it crosses the finish line?
b. In my frame of reference, how long after the light crosses the finish line do I cross the finish line?
c. In the Earth’s frame of reference, how far behind the light am I when it crosses the finish line?
d. In the Earth’s frame of reference, how long after the light crosses the finish line do I cross the finish
line?
Mathematical Analysis
43
In the following gedanken experiment, I am in the center of a glass-sided freight car of proper length L0
speeding to the right at speed v relative to you. I have a flashlight in each hand, directed at the front and
rear panels of the freight car. I switch them on at the same time. Complete the following table:
Event
Flashlights are
turned on
My
frame
Position
0m
Time
0s
Earth
frame
Position
0m
Time
0s
Light strikes
front panel
Light strikes
rear panel
a. In the Earth’s frame of reference, construct an expression for the time difference between the two
panels being struck by light.
b. What is this time difference for a 20 m long car traveling at 0.9c?
Mathematical Analysis
44
A train travels along level, straight tracks at 0.9c. Two guns are placed on the ground perpendicular to the
tracks, oriented such that they will shoot holes in the side of the train as it passes. As measured on the
Earth, the guns are 10 m apart and are fired at the same time. Earth-bound observers expect to see two
bullet holes 10 m apart in a contracted train. Thus, when the train stops at the next station and returns to
“normal” length, the bullet holes should be more than 10 m apart. However, the train-bound observers see
the Earth as moving and the distance between the two guns as contracted to less than 10 m. They think the
train is always at its “normal” length and argue that the bullet holes will remain less than 10 m apart
when the train pulls into the next station. How can we resolve this apparent paradox? Maybe we should
start by completing the following table:
Event
Rear gun is
fired
Earth
frame
Position
0m
Time
0s
Train
frame
Position
0m
Time
0s
Front gun is
fired
a. How far apart are the bullet holes when the train stops at the next station?
b. CLEARLY explain what is wrong in the reasoning of either the Earth-bound or train-bound observers
presented in the introductory paragraph.
Mathematical Analysis
45
A farmer, who also knows a little physics, has a 6 m long turbo-tractor and a 5 m long barn. She’d like to
store the tractor in the barn and thinks relativity will allow her to do this. If a friend drives the tractor at
high speed, the farmer knows the tractor will shrink and thus fit into the barn, at least temporarily…
a. In the farmer’s frame of reference, how fast must the tractor be moving in order to fit into the barn? At
this speed, the front of the tractor reaching the rear of the barn, and the rear of the tractor reaching
the front of the barn are two simultaneous events.
However, her friend says that from her point of view, it will be the barn that’s in motion, thus the barn will
shrink and the tractor will not fit. How can we resolve this apparent paradox? Maybe we should start by
completing the following table:
Event
Front of
tractor reaches
rear of barn
Barn
frame
Position
0m
Time
0s
Tractor
frame
Position
0m
Time
0s
Rear of tractor
reaches front of
barn
b. So, does the tractor fit in the barn?
c. CLEARLY resolve the apparent paradox presented in the introductory paragraph.
d. In the tractor frame of reference, how much of the tractor is outside of the barn when the front of the
tractor plows into the rear of the barn?
Mathematical Analysis
46
Two spacecraft (O and O’) of equal rest length 100 m pass very close to each other as they travel in
opposite directions. Both travel at 0.8c (although in opposite directions) as measured on a nearby star. The
captain of O has a laser cannon at the stern of her ship, directed perpendicular to her ship, and intends to
fire it the instant her bow is lined up with the stern of ship O’. Since she sees O’ as Lorentz contracted, she
expects this blast to miss O’ (it’s just a practical joke…). However, since the captain of O’ sees ship O as
Lorentz contracted, the captain of O’ fears the blast will destroy her ship. How can we resolve this
apparent paradox? Maybe we should start by completing the following table:
Event
Bow of O lined
up with stern
of O’
a.
b.
c.
O
frame
Position
0m
Time
0s
O’
frame
Position
0m
Time
0s
Cannon on O
is fired
Does the laser blast from O strike O’?
If not, by how much does the blast miss O’, in both the ship O and ship O’ frame of reference.
CLEARLY resolve the apparent paradox presented in the introductory paragraph.
Mathematical Analysis
47
A Federation space cruiser is floating in Federation territory at rest relative to the border of Klingon
space, which is 6.0 lightminutes away in the +x direction. Suddenly, a Klingon warship flies past the
cruiser in the direction of the border at a speed 0.6c. 5.0 minutes later, according to cruiser clocks, the
Klingons fire a disrupter blast that travels at the speed of light back to the cruiser. The disrupter blast hits
the cruiser and disables it. A bit later (according to cruiser radar measurements) the Klingons cross the
border into Klingon territory. The Klingon-Federation Treaty states that it is illegal for a Klingon ship in
Federation territory to damage Federation property. However, when the case comes up in interstellar
court, the Klingons claim that according to measurements made in their reference frame, the damage to the
cruiser occurred after they had crossed back into Klingon territory. Thus they were not in Federation
territory when the blast hit the cruiser and are not guilty of breaking the treaty. As the judge in this case,
what should you do? Maybe you should start by completing the following table:
Event
Klingon
warship passes
cruiser
Cruiser
frame
Position
0m
Time
0s
Warship
frame
Position
0m
Time
0s
Warship fires
blast
Are the Klingons guilty of breaking the treaty?
Mathematical Analysis
48
Blast hits
cruiser
Warship enters
Klingon
territory