Name _________________
Solutions to Test 1
September 18, 2015
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered.
Possibly useful formulas:


 
E     E  vpx 
x    x  vt 
F  qE  qu  B
p  qRB
px    px  vE c 2 
t     t  vx c 2 
 
u pc
py  p y
y  y,

c E
z  z
pz  pz
f0
n
f 
1     1  n  12 n  n  1  2 
 1  v cos  c 
u x 
ux  v
1  vu x c 2
uy
u y 
 1  vu x c 2 
u z 
uz
 1  vu x c 2 
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. A fast-moving clock will run _____ and be _____ than the same clock at rest
A) Slower, longer
B) Faster, longer
C) Slower, shorter
D) Faster, shorter
E) None of the above
2. Which particles are massless?
A) Protons
B) Neutrons
C) Electrons D) Photons
E) None of these
3. Suppose I take some object, and add some energy to it. Which types of energy will
cause a change in the object’s mass m (also known as rest-mass)
A) Any type of energy, including potential or kinetic energy
B) Potential energy, like nuclear, chemical or mechanical energy, but not kinetic
energy
C) Nuclear or chemical energy, but not mechanical energy
D) Nuclear energy, but not chemical or mechanical energy
E) No type of energy changes an object’s rest-mass
4. Which of the following formulas can be used for massless particles?
2

A) E 2  c 2 p 2   mc 2  (only)
B)
C)
D)
E)
E   mc 2 (only)


p   mu (only)
All of the above
None of the above
5. As an electron approaches the speed of light, v  c, the momentum approaches ___
and the energy approaches ____ .
B) mc, 
C) , ½mc2 D) , mc2
E) , 
A) mc, ½mc2
6. In the twin paradox, one twin stays at rest on Earth, while the other travels quickly
and ages less. But from the perspective of the moving twin, the one on Earth should
age less. Why are these two perspectives not truly interchangeable?
A) The one on Earth is actually at rest, and the other is truly moving
B) Lorentz contraction means that the distances travelled are not truly
interchangeable
C) Because all objects are not rigid in special relativity, time itself can be stretched,
making such comparisons meaningless
D) The traveling twin did not travel at constant velocity. He changed speed
when he turned around to come back
E) Actually, they are interchangeable. The moving twin will return to discover the
stationary twin is younger
7. Which of the following is a pretty good approximation for  for small velocities v?
A)   1 
v2
v2
B)


1

c2
2c 2
C)   1 
v2
c2
D)   1 
v2
2c 2
E)   1 
c2
2v 2
8. Which of the formulas below is still valid in special relativity?
 dp


mv 2
A) F 
B) F = ma
C) p  mv
D) E  12 mv 2 E) F 
r
dt
9. In special relativity, why are the Galilean transformations ( x  x  vt , t   t ) not
“good”?
A) They imply the wrong formula for conservation of energy
B) They imply the wrong formula for conservation of momentum
C) They don’t preserve the 4D-distance formula
D) They are for physical objects, not light
E) They are for rotating coordinates, not inertial coordinates
10. Since spacetime is four dimensional, all vectors should have four components. The
fourth component of momentum is
A) Time
B) Energy C) Mass
D) Force
E) Work
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences)
(10 points each).
11. Suppose I build a launcher that launches a space ship at half the speed of light.
On this ship, I put a second launcher that launches a smaller ship, also at half of
the speed of light. Would this be a way to make a space ship that goes at the
speed of light?
It is impossible to make physical objects go at the speed of light. The naïve
assumption is the addition of velocity formula, which is incorrect. As viewed by the first
ship, I would be moving backwards with velocity  12 c while the small ship would be
moving forwards at velocity  12 c . It is not hard to use the subtraction of velocity
formula then to work out that the actual velocity of the small ship as viewed by me would
be 54 c .
12. A very simple device for communicating faster than light would be to have a
long pole from point A to point B. The person at point A moves one end of the
pole back and forth, perhaps in Morse code, and the person at point B measures
and interprets the message. Explain why this would or would not work.
In general, it is impossible to send any signals faster than light, so clearly this is
impossible. The naïve device describes fails because adjustments at point A do not move
instantaneously, but instead are transmitted at the speed of sound, which is much lower
than the speed of light c. This is a very slow form of communication.
13. Particle detectors typically have strong magnetic fields. Explain how tracking a
particle in the presence of a magnetic field could allow one to determine the sign
of the charge of a particle, as well as some other quantity. At least one equation
is necessary.
In the presence of a magnetic field, a charged particle will bend and move in a

 
circular motion. The formula F  qu  B describes this force. If the particle accelerates
 
in the direction of u  B , then it must be positively charged, while if it accelerates in the
opposite direction, it must be negatively charged (and neutral if it goes straight). If the
radius of curvature is measured, its momentum can be deduced from the formula
p  qRB , assuming the charge is known (usually of magnitude e).
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated calculations (20
points each)
14. Consider the three events A, B, and C, which in one frame have coordinates
 x, y, z, ct  given by
A   0, 0, 0, 0  , B   5.0 m,12.0 m, 0,13.0 m  , C   5.0 m,5.0 m, 0,5.0 m 
(a) For each pair (AB, AC, and BC) find s2 and determine how the pair is related
to each other (absolute future/past, future/past light cone, or elsewhere)
We calculate the proper distance between each pair, and then when necessary just
notice which one has a smaller time coordinate to figure out future vs. past. For A vs. B:
2
s AB
  5.0 m   12.0 m   02  13.0 m    25  144  169  m 2  0.
2
2
2
So B is on the future light cone of A, or A is on the past light cone of B. For A vs. C:
2
s AC
  5.0 m    5.0 m   02   5.0 m    25  25  25  m 2  25 m 2 .
2
2
2
Hence A and C are elsewhere compared to each other. For B vs. C:
2
sBC
 02   7.0 m   02   8.0 m    49  64  m 2  15 m 2 .
2
2
So C is in the absolute past of B, or B is in the absolute future of C.
(b) If another observer were moving at a different velocity, the coordinates of
the three events would change. How could this affect the answers to part (a)?
The distance squared is invariant, and, indeed, the categorizations are invariant,
so nothing would change.
(c) In which pair of events would it be possible for a massive object to move
from one of these to the other? How about for a massless object? How about
for a tachyon?
Massive objects move slower than the speed of light, and from the past to the
future, so such an object could go from C to B. Massless objects move at the speed of
light, following lightlike paths, so such an object could go from A to B. Tachyons always
move faster than the speed of light, and therefore could go from A to C or from C to A.
Indeed, since it is ambiguous which order these two events occurred in, they can go
either way.
150 m
0.6c
150 m
120 m
Earth
15. An alien spaceship has a shape (as viewed by us) of a square-based pyramid with
square side 150.0 m, and height 120.0 m is travelling point first directly towards
us at v  0.600c . It is sending us signals at a frequency that we measure as 800.0
MHz.
(a) What are the dimensions of the actual ship, as viewed by aliens aboard the
ship?
We need the factor of , which is given by

1
1 v c
2
2

1
1  0.600
2

1
1
1


 1.250.
1  0.3600
0.6400 0.800
Since the actual height of the spacecraft is the proper length, we use the formula
L  L p  to find
L p   L  1.250 120.0 m   150.0 m.
The other dimensions of the pyramid are unaffected by Lorentz contraction, so they are
actually 150 m. Hence all three dimensions are actually 150 m.
(b) What is the actual frequency at which the spacecraft is broadcasting?
Rearranging the Doppler shift formula, f 
f0
, and since it is
 1  v cos  c 
coming towards us, it now has   0 , so we have
f 0  f  1  v cos  c    800.0 MHz 1.25  1  0.600 cos  0  
 1000.0 MHz  0.400   400.0 MHz .
(c) It continues broadcasting at the same frequency. What frequency do we
detect at closest approach? After it has passed us and is heading away from
us?
The actual frequency f0 remains unchanged, but we have   12  at closest approach
and    when heading away, so
f 
f0
400.0 MHz
400.0 MHz


 320.0 MHz ,
1
1.250
 1  v cos  c  1.250 1  0.600 cos  2   
f 
f0
400.0 MHz
400.0 MHz


 200.0 MHz .
 1  v cos  c  1.250 1  0.600 cos   1.250 1.600 
16. A charged pion with mass m = 139.6 MeV/c2 is travelling in a circle of radius
1.50 m. It completes one circle in a time of 40.0 ns (= 40.0  10-9 s) before
decaying
(a) How long, if we were traveling along with the pion, would the pion appear to
last?
The velocity is the distance over time. For a circle, this is
v
2 r 2 1.50 m 

 2.36 108 m/s.
t
40.0 109 s
We therefore have a Lorentz factor of

1
1  v2 c2
1

1   2.36 10 m/s   2.998  10 m/s  
8
8
2

1
1  0.78592
We then calculate the proper time  using t   , so
  t    40.0 ns  1.617  24.7 ns .
(b) What is the energy of the pion, in MeV?
This is computed using
E   mc 2  1.617 139.6 MeV/c 2  c 2  225.7 MeV .
(c) What is the momentum of the pion in MeV/c?
This is computed using
p   mu  1.617 139.6 MeV/c 2  2.36 108 m/s 

1.617 139.6 MeV/c   2.36 108 m/s 
2.998 108 m/s
 177.7 MeV/c
 1.617 .
17. Dr. Carlson has a rest mass of m = 78.5 kg. He is travelling in his spacecraft at
v  2.70 108 m/s when the galactic traffic light turns red, at which point his
brakes turn on, applying a force of F  1220 N to attempt to stop him.
(a) What is the initial and final energy of Dr. Carlson (in J)?
The initial Lorentz factor is given by
i 
1
1  v2 c2

1
1   2.70  10 m/s 
8
2
 2.998 10
8
m/s 
2

1
1  0.90062
 2.301.
The final Lorentz factor is just  f  1 . The initial and final energies are therefore
Ei   i mc 2  2.301 78.5 kg   2.998  108 m/s   1.623  1019 J,
2
E f   f mc 2  1 78.5 kg   2.998 108 m/s   7.056 1018 J .
2
(b) What is the initial and final momentum (in kgm/s) of Dr. Carlson?
The final momentum is zero, since he stopped. The initial momentum is
pi   i mu  2.301 78.5 kg   2.70 108 m/s   4.876 1010 kg  m/s,
2
p f  0.
(c) What is the stopping distance at this speed, in light years (1 ly = 9.458  1015
m)?
We use the work formula, W = Fd, and use the fact that the change in energy is
the work. We therefore have
18
19
W E E f  Ei  7.056 10 J   1.623 10 J  9.174  1018 J
d 



F
F
F
1220 N
1220 N
15
7.520 10 m

 0.795 ly .
9.458  1015 m/ly
(d) How long does it take to come to a stop, in years (1 y = 3.156 107 s)?
We use the formula F  dp dt , which for constant force is F  p t , to obtain
t 
3.997 107 s
p p f  pi 4.876  1010 kg  m/s



 1.266 y .
3.156  107 s/y
F
F
1220 N