Name _________________
Solutions to Test 1
September 20, 2013
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:
F  qE  qu  B
p  qRB
E     E  vpx 
t     t  vx c 2 
y  y,
z  z
u pc

c E
f 
x    x  vt 
f0
 1  v cos  c 
1   
px    px  vE c 2 
py  p y
pz  pz
n
 1  n  12 n  n  1  2 
ux 
uy 
uz 
ux  v
1  vux c 2
uy
 1  vux c 2 
uz
 1  vux c 2 
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. For small velocities v, we can approximate 1  v2 c2 as
A) 1  v c
B) 1  v c
C) 1  iv c
D) 1  v2 2c2
E) 1  v2 2c2
2. If I see a clock going by at sufficient speed, it will look like it is slowed down by a
factor of two. If someone moving with the clock looks at my clock, he will say my
clock
A) Is slowed down by a factor of two
B) Is speeded up by a factor of two
C) Is slowed down, but by a factor that is generally different from two
D) Is speeded up, but by a factor that is generally different from two
E) May be speeded up or slowed down, depending on which of us is actually moving
3. As the velocity of a massive particle approach c, the momentum approaches ___ and
the energy approaches ___
A) mc, 12 mc 2
B) mc, mc2 C) mc, 
D) , mc2
E) , 
4. Suppose a spacecraft is moving relative to us, and is emitting electromagnetic
radiation with frequency f0. In which circumstances will the detected frequency f be
lower than f0?
A) If it is moving directly away from us (only)
B) If it is moving directly towards us (only)
C) If it is moving perpendicular to our line of sight (only)
D) A and C are true, but not B
E) B and C are true, but not A
5. Which of the following equations is still valid in special relativity?
A) p = mv
B) F = ma C) W = Fd D) E = ½mv2 E) none of these are valid
6. Suppose I try to communicate over long distances instantaneously by making a long
rigid rod and moving one end, with my partner noticing that his end moves in
response. This would fail because
A) No matter how slowly I move it, it will Lorentz contract, counteracting my effects
B) There is no such thing as a rigid rod in special relativity
C) Time dilation will ensure that there is a delay before it reaches the other end
D) The rod will distort spacetime, creating a black hole
E) Actually, this method would work
7. Ultimately, the Galilean transformation has to be replaced by the Lorentz boost
formula because the Galilean transformation does not
A) Include the effects of length contraction
B) Include the effects of time dilation
C) Preserve the four-dimensional distance formula
D) Allow for any kind of Doppler shift
E) Include the effects of the binomial expansion
8. The fourth component of space is time. What is the fourth component of momentum?
A) Force
B) Energy C) Distance D) Mass
E) Angular momentum
9. Which quantity below is not always conserved, according to special relativity?
A) Mass (only)
B) Momentum (only)
C) Energy (only)
D) Mass and momentum
E) Actually, all of these are always conserved
10. For known particles, the quantity E2 – c2p2 is always
A) Positive, but never negative or zero
B) Negative, but never zero or positive
C) Zero, but never negative or positive
D) Negative or zero, but never positive
E) Positive or zero, but never negative
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences)
(10 points each).
11. We wish to make a negatively charged electron go in a clockwise
circle, as sketched at right. What sort of field, and which direction
should we have it point, to accomplish this?
A particle moving in a circle must have a centripetal force acting on it, pulling it
towards the center. A magnetic field can achieve this, with a force given by F  qu  B .
Consider, for example, when the electron is at the top of the circle, and moving to the
right. If we have B pointing into the paper, then u  B will point straight up, but when
you multiply by the negative charge q this will reverse and point inwards. So we need B
pointing into the paper.
12. Suppose I have 1 kg of some substance, and I now add energy to it by one of the
following: (a) heating it, (b) moving it; adding kinetic energy to it, (c) having it
undergo a chemical reaction that adds energy to it, (d) adding nuclear energy to
it, (e) compressing it to add mechanical energy. In which of these cases, if any,
does its mass increase?
For a particle at rest, E0 = mc2, so an increase in the energy will increase the mass.
Hence (a), (c), (d), and (e) all increase the mass. Kinetic energy, in contrast, does not
increase the energy, but rather changes the energy to E = mc2, so (b) doesn’t increase the
mass.
13. If you push on an object of mass m with constant force F, it should undergo
constant acceleration, a, and hence it will eventually exceed the speed of light c.
Explain either why this reasoning is incorrect, or why the result does not, in fact,
violate relativity. What does happen if you push on an object with a constant
force for a long time?
The argument assumes F = ma, which is false, so in fact constant force does not
equate to constant acceleration. Instead, F = dp/dt, so the momentum will increase
linearly to infinity, but this corresponds only to a velocity very close to (but still less than)
c, the speed of light.
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
14. In the year 2100, a pair of spacecraft are each launched towards the star Sirius,
9 ly away (ly = cy). Spacecraft A travels directly there at a steady speed of v =
0.25c. Spacecraft B, due to engine trouble, is stuck at rest for 18 years, but then
they get their engines working and travel thereafter at v = 0.5c, so that both
arrive in 2136.
(a) What is the total time experienced by each of the travelers, from 2100–2136?
First, we should check that both spacecraft do, indeed, arrive in 2136. Spacecraft
A takes time t  d v   9c  y   0.25c   36 y , so that works out. Spacecraft B took 18
years to do nothing, and then an additional t  d v   9c  y   0.5c   18 y to travel
there, so indeed, they will both get there in 2136. So far so good.
To get the proper time, in each case we use the formula t   , or rewrite this as
  t   t 1  v 2 c 2 . For spaceship A, this works out to
 A  t 1  v2 c2  36 y  1   0.25c  c2  36 y  1  0.252  34.857 y
2
For spaceship B, there are two stages, and we find
 B  t1 1  v12 c2  t2 1  v22 c2  18 y   18 y  1   0.5c  c 2
2
 18 y   18 y  1  0.52  33.589 y
(b) Which of the two spacecraft experienced less total proper time? Explain
qualitatively how you could have known the answer to this without doing
part (a).
Obviously, spaceship B has less proper time, with its astronauts arriving more
than a year younger. We know that the longest proper time path between any two events
is always the one that doesn’t accelerate, and since spaceship A moved at constant
velocity, it must have the longer proper time.
(c) As spacecraft A approaches Sirius, how quickly do they see spacecraft B
catching up to them?
This is a straightforward subtraction of velocities formula, with v = 0.25c as the
speed of the observing spacecraft and u = 0.5c the speed of the observed spacecraft. We
therefore have
ux 
ux  v
0.5c  0.25c
0.25c


 0.2857c  8.566 107 m/s .
2
1  ux v c 1   0.25 0.5 0.875
15. The Super Duper Hyperloop passenger
signal
transportation system consists of cylindrical cars
u
that (when at rest) are 20.0 m long, and 3.0 m in
diameter. They are then accelerated to a speed of u = 1.00×108 m/s before being
injected into the transport tunnel.
(a) What are the dimensions of the cars, as viewed by a stationary observer?
The cars Lorentz contract only in the direction of motion, so the diameter does not
change at all. The length changes to
2
 1.00 108 m/s 
2
1 
   20.0 m  1  0.334
8
 2.998 10 m/s 
L  LP   LP 1  u c   20.0 m 
2
2
 18.86 m .
(b) A businesswoman boards the cars when her watch says (correctly) the time is
9:20, and gets off when her watch says the time is 9:53. What time is it really
when she gets off? Ignore time zones.
The watch presumably measures proper time , but external clocks will measure
the time t   . We therefore have
t   

1 u c
2
2

33 min
1  0.3342
 35.00 min .
The time will therefore be about 35 minutes later, or 9:55.
(c) As the train approaches a signal light, a red light with frequency 407 THz
shines signals the driver that the train needs to stop. At what frequency does
the driver of the train see the light?
We use the relativistic Doppler shift equation, with angle  = 0. So we have
f 
f 1  v2 c2  407 THz  1  0.3342
f0


 576 THz .
 1  v cos c 
1 v c
1  0.334
Unfortunately, this is right in the green part of the spectrum, so we can only hope the
driver realizes the problem and stops the train anyway.
16. The BaBar experiment produces the (4s)
e–
e+
particle by colliding electrons with energy E1 =
9.0 GeV and positrons with energy E2 = 3.1
GeV head on, as sketched at right. The electron and positron are essentially
massless.
(a) What is the momentum (in GeV/c) of the initial electron and positron? Don’t
forget that momentum is a vector!
Because the electron and positron are close to massless, they obey E  p c .
Keeping in mind that momentum is a vector, we then have
p1  9.0ˆi GeV / c and
p2  3.1ˆi GeV / c .
We could leave out the unit vectors if we wanted to, since it is one dimensional, but it is
important to note that the positron is moving to the left.
(b) What is the momentum and energy of the (4s) particles that are produced?
By conservation of momentum, the momentum and energy of the (4s) must be
the sum of the momenta and energies of the initial particles. Hence
E  E1  E2   9.0 GeV    3.1 GeV   12.1 GeV ,

 

p  9.0ˆi GeV / c  3.1ˆi GeV / c  5.9ˆi GeV / c .
(c) What is the mass (in GeV/c2) and velocity (as a fraction of c) of the resulting
(4s) particles?
We find the mass using
 mc 
2 2


2
 E 2  c2 p 2  12.1 GeV   5.9ˆi GeV  111.6 GeV2 ,
mc2  111.6 GeV  10.56 GeV .
So the mass is 10.56 GeV/c2. The velocity is given by
u pc 5.9ˆi GeV


 0.488ˆi .
c E 12.1 GeV
So it was moving at 0.488c to the right.
2
17. A proton with mass m = 1.673×10-27 kg and charge e = +1.602×10-19 C is initially
moving to the left at velocity ui = –2.40×108 m/s. Suddenly an electric field is
turned on pointing to the right of magnitude E = 3.00×108 V/m. After a short
time, it is found that the proton is now moving to the right at uf = +2.40×108 m/s
(a) What is the initial and final momentum of the proton, in kgm/s? What is the
initial and final energy of the proton, in J? How much did each of them
change?
The magnitude of the initial and final momenta are each
p   mu 
mu
1 u c
2
2
1.67310

27
kg  2.40 108 m/s 
 2.40 10 m/s 
1 

8
 2.998 10 m/s 
8
2
 6.70 1019 kg  m/s
However, the directions are opposite, so p f  6.70 1019 kg  m/s and
pi  6.70 1019 kg  m/s . The change in momentum is p  1.340 1018 kg  m/s
The energy of both the initial particle and the final particle are each given by
E   mc 
2
mc 2
1 u c
2
2
1.67310

27
kg  2.998 108 m/s 
 2.40 108 m/s 
1 

8
 2.998 10 m/s 
2
2
 2.509 1010 kg  m2 /s 2 .
This works for both, so Ei  E f  2.509 1010 J . There is no change in energy.
(b) What is the force on the proton, in N (note: CV = Nm)?



This is trivial: F  eE  1.602 1019 C 3.00 108 V/m  4.80 1011 N .
(c) How long (in s) would it take for this change in direction to occur? What is
the total displacement that the particle has moved during this process?
Fortunately, the force is constant, so we may use the formulas
F
p
and W  Fd .
t
Solving for the time difference, we have
t 
p 1.340 1018 kg  m/s

 2.79 108 s  27.9 ns .
F
4.80 1011 N
However, there was no work done on the proton, since its final energy matched its initial
energy. Hence the displacement is zero, and the proton is exactly where it started.