Name _________________
Solutions to Test 1
September 21, 2012
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
 
u pc

c E
f 
E     E  vpx 
x    x  vt 
px    px  vE c
t     t  vx c 2 
py  p y
y  y,
z  z
f0
 1  v cos  c 
pz  pz
1   
n
2

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 
 1  n  12 n  n  1  2  
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. To make an electron move in a circle, the easiest way to
A) Use a magnetic field perpendicular to its motion
B) Use a magnetic field parallel to its motion
C) Use an electric field perpendicular to its motion
D) Use an electric field parallel to its motion
E) Strap tiny wheels on it and give it a circular racetrack to go on
2. The four dimensional distance formula is
2
2
2
2
A) s 2   x    y    z    t 
B) s 2   x    y    z    t 
2
2
2
2
C) s 2   x    y    z   c 2  t 
2
D) s 2   x    y    z   c 2  t 
2
2
2
2
2
2
2
E) None of the above
3. Suppose I am running forward at speed 0.75c towards a light beam that is going
perpendicular to me at speed c. How fast, according to me, will the light beam be
going?
B) 0.75c
C) 1.25c
D) 1.75c
E) c
A) 0.25c
4. Suppose you have a stretched spring. Suppose you now allow the spring to relax to
its unstretched length. According to relativity, does the mass change?
A) Yes, it increases
B) Yes, it decreases
C) Yes, it will definitely change, but there is insufficient information to tell how
D) Maybe, it depends on additional details, like how fast it changes
E) No
5. If I am standing still and a rocket goes past me, I would say that the clocks on the
rocket go ________, and the person on the rocket would say my clocks go ________.
A) slower, faster
B) faster, slower
C) slower, slower
D) faster, faster
E) Insufficient information
6. If you have a collection of many objects moving at different speeds in different
direction, how would you find the effective mass of the whole collection?
A) Add up the energies of all the particles, then divide by c2
B) Add up the energies of all the particles, then divide by  c 2

C) Find the total momentum and energy, then use M 2 c 4  E 2  c 2 p 2
D) Add up the total momentum, then divide by the average velocity
E) Add up the masses of all the pieces
7. Another way to think of the Lorentz transformation (called Lorentz boost in class) is
A) It’s like a rotation in space-time, mixing time with space
B) It is a formula for how coordinates appear to transform, because the objects that
measure them are all Lorentz contracted
C) Spacetime, like all objects in special relativity, is somewhat compressible, which
leads to the distortions
D) It is a formula that has to do with clocks that run on light waves; if we used
ordinary pendulum clocks, the formulas would be different
E) It is caused by the changing length of objects when you accelerate them
8. If a particle of mass m approaches the speed of light, what would the limit of the
momentum and energy be?
A) p  mc , E  12 mc 2
B)
C)
D)
E)
p  mc , E  mc 2
p  , E  mc 2
p  mc , E  
p, E 
9. Which of the following statements is false in special relativity?
A) Energy is conserved
B) Mass is conserved
C) Momentum is conserved
D) There are no rigid objects
E) It is generally impossible for different observers to agree on simultaneity
10. What fundamentally makes equations like F  kqq r 2 for the force between two
charges unacceptable in special relativity?
A) Lorentz contraction causes the charges to change size, and hence get bigger or
smaller, and this isn’t concluded
B) The distance r will be different as viewed by different observers
C) Since the formula F = ma is false in relativity, there is no point in calculating the
force
D) This formula assumes instantaneous action at a distance, which is impossible
in special relativity
E) This is a force of electromagnetism, which is not compatible with relativity
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 that according to one observer, event A occurs before event B. Under
what circumstance, if any, will all observers agree that event A occurs before
event B?
This will be guaranteed to be the same as viewed by all observers only if B is in
the absolute future or on the future light cone, which implies that the separation is
2
2
2
2
timelike or lightlike, so that s 2   x    y    z   c 2  t   0 .
A
12. Two stars are in orbit around each other, and each
To
produces a radio wave (a kind of light) with a
Earth
frequency of 1420.45 MHz, as viewed by an observer
B
standing on the star (if that’s possible). Explain how
this frequency would change as viewed from Earth. Predict which (if either)
would have an increase/decrease in the frequency.
The Doppler shift formula gives the frequency for each of these. Since star A is
moving towards us, it would experience a blue shift, shifting it to a higher frequency.
Since star B is moving away from us, it would experience a lower frequency.
13. In a homework problem, you found that a “bunch” of protons, when it passes
through the Atlas detector, as viewed by us is much shorter than the Atlas
detector, but if you were moving along with them it is much longer than the
Atlas detector. But surely the protons are either all in the detector at the same
time or not. Resolve this apparent paradox.
The key to understanding this paradox is the phrase “all in the detector at the
same time.” The phrase “at the same time” only makes sense if everyone can agree at
what the same time is, and simultaneous has a different meaning in the lab frame and in
the frame that is commoving with the protons.
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
14. The probability of a particle lasting a proper time  is given by P  e  0 , where
 0 is the mean lifetime. It is observed that for a group of B+ mesons moving at
0.99630c, after traveling a distance d = 0.680 cm, only 30.3% of them still exist.
(a) How long did they last as viewed in our frame?
Time is just distance divided by velocity, so
t 
d
0.00680 m

 2.277 1011 s  22.77 ps .
8
v 0.9963  2.998 10 m/s 
(b) How long would they have lasted as viewed by someone moving with them?
Since it is moving relativistically, t   , so

t

 t 1  v 2 c 2   22.77 ps  1  0.99632  1.957 ps .
(c) What is the mean lifetime  0 of the B+ meson?
We need to solve P  e  0 for  0 . We have
P  e   0 ,
ln P     0 ,
 0 ln P   ,
1.957 ps 1.957 ps



 1.639 ps .
0  
ln P
ln  0.303
1.194
15. A pair of spherical spacecraft are each passing me at high
velocity in opposite directions. They are actually spherical, but
because of their high speed, they appear as ovals 40 m  32 m in
diameter.
(a) Which direction (horizontal or vertical) are they actually
moving? What would be their diameter if someone were
moving along with them?
Since moving objects are Lorenz contracted, they must be
moving in the vertical direction. The transverse or horizontal direction
is not affected by Lorentz contraction, so the horizontal size is the actual diameter, or 40
m.
(b) How fast are each of these spacecraft moving, according to me, as a fraction
of c?
The apparent length is related to the actual (proper) length by L  L p  . We
therefore have
L  Lp 1  v 2 c 2 ,
L2
1 v c  2 ,
Lp
2
2
2
v2
L2
2
 32 m 
 1 2  1 
  1   0.8   1  0.64  0.36 ,
2
c
Lp
 40 m 
v c  0.36  0.60 .
(c) How fast is one of these spacecraft moving, according to the other spacecraft,
as a fraction of c?
We can calculate it from either perspective. We can call the vertical direction x,
then we can, for example, examine the sphere moving at speed u x  0.60c from the
perspective of the observer moving at v  0.60c . Since all the action is in the xdirection, we use only the velocity in this direction formula to obtain
ux 
0.6c   0.6c 
ux  v
1.2c
1.2c



 0.882c .
2
2
2
1  vu x c 1   0.6c  0.6c  c 1  0.6 1.36
16. An X particle slams into a proton at rest (m = 938
MeV/c2), causing it to annihilate with it to make two
(massless) photons, each with energy E = 1876 MeV,
and moving perpendicular to each other, as
sketched at right.
(a) What is the momentum of each of the final state
photons, in MeV/c? What is the total
momentum?
photon
photon
proton
X
Because the photon is massless, it satisfies E  pc , so p  E c  1876 MeV/c .
However, momentum is a vector, so we must also include an indication of the direction.
If we make to the right the +x direction, and upwards is the +y direction, then
p1  1876ˆi MeV/c , p 2  1876ˆj MeV/c , and ptot  1876ˆi  1876ˆj MeV/c .




We could also work out the magnitude of ptot  1876ˆi  1876ˆj MeV/c , which is
2653MeV/c .
(b) What is the momentum and energy of the initial X particle?
The final momentum must equal the initial momentum, and the same is true of the
energy. The proton has momentum zero, so we have
p X  p p  p1  p 2 ,


p X  1876ˆi  1876ˆj MeV/c
Energy is also conserved. The proton starts with energy E0  m p c 2  938 MeV . We
therefore have
E X  E0  E1  E2 ,
E X  E1  E2  E0  1876  1876  938  MeV = 2814 MeV
(c) What is the mass (in MeV/c2) of the X?
We use the equation
m 2 c 4  E 2   cp    2814 MeV   1876 MeV   1876 MeV   8.798 105 MeV ,
2
2
2
2
mc 2  938 MeV .
It isn’t surprising it came out the same as the proton mass, since it is actually an antiproton, which has the same mass as the proton.
17. A rocket with mass M = 1000.0 kg is being sent to a distant star. It is supposed
to be accelerated from rest up to v  2.80 108 m/s by firing its rockets at a
steady rate for one year ( t  3.156 107 s )
(a) What is the initial energy and momentum of the rocket ship? What is the
final energy and momentum of the rocket ship?
The rocket is initially at rest, so
p0  0 ,
E0  mc 2  103 kg  2.998 108 m/s   8.99 1019 J.
2
The final momentum and energy are
p   vm 
E   mc 
2
10
3
kg  2.80  108 m/s 
1   2.80 108 m/s 2.998 108 m/s 
10
3
kg  2.998  108 m/s 
2
 7.83 1011 kg  m/s ,
2
1   2.80 108 m/s 2.998 108 m/s 
2
 2.515  1020 J .
(b) What force (in N) is required to achieve this final speed in one year?
We use the fact that F  dp dt , but since it is implied that the force is constant,
we have F  p t , so
F
p 7.83  1011 kg  m/s

 24,800 kg  m/s 2  24,800 N .
7
t
3.156 10 s
(c) How far, in light-years, does the rocket travel in this first year?
( 1 ly  c  y  9.47 1015 m )
We now use the idea that E  W  Fd , so we have
W E 2.515 1020 J  8.99  1019 J
1 ly
d 

  6.511015 m 
 0.687 ly .
F
F
24,800 N
9.47  1015 m