Relativity Problem Sheet.pdf - Of the Clux

University of Warwick
Department of Physics
Relativity – PX 109
WEEK 1
1. By considering the length of a rod in two reference frames S and S′ (where S′ is
moving with speed v in the x-direction with respect to S) show explicitly that under a
Galilean transformation length does not change. Let the co-ordinates of the ends of
the rod be x
A
and x
B
in frames S in which the rod is at rest. (Note that a meaningful
measurement of length requires the measurement of the co-ordinates of each end of
the rod at the same instant in time).
2. To demonstrate the Galilean invariance of a conservation law consider the following:
* A particle of mass m 1 = 3 kg, moving at a velocity of u 1 = +4 ms -1 along the x-axis of
an inertial frame S, approaches a second particle of mass, m 2 = l kg, moving at u 1 = 3
ms -1 along this axis. After a head-on collision, it is found that m 2 has a velocity U 2 = +
3 ms -1 along the x-axis.
i) Using the conservation of linear momentum calculate the velocity, U 1 of m l after
the collision.
ii) The collision is seen by an observer in another inertial frame S which moves with
a (uniform) speed of + 2m/s relative to S along the x-direction (the standard
configuration).
(a) Calculate the velocities u l ', u 2 ', U l ' and U 2 relative to S.
(b) Calculate the initial and final values of the total linear momentum (p i and p f )
relative to S.
(c) Hence show that the total linear momentum is conserved in S (as it is in S)
[the conservation of linear momentum is a Galilean invariant law] while the
initial and final values of total linear momentum in S are different from the
values in S.
3 (a) Consider two frames of reference S and S' in standard configuration with S'
moving at 10 miles per hour relative to S in the direction of the +ve x-axis. At 10.00
am an explosion occurs at a point in space whose coordinates in S' are x' = 4 miles, y'
= 1 mile, z' = 2 miles. Given that the origins of S and S' were coincident at 9.00 am,
find the coordinates of the explosion in the frame S, assuming Galilean relativity.
(b) In a frame of reference S two events occur simultaneously at two different
points which are separated by 310 8 m. Using Special Relativity find (as a fraction of
the speed of light c = 310 8 ms -l ) the velocities of the inertial frames S' (in standard
configuration relative to S) in which the two events are separated by (i) 0.1 s, (ii) 1 s,
and (iii) 10 s.
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SOLUTION
(i).
y
y'
10 miles per hour
z
o
z '
0'
4 miles
Explosion at 10am
[1]
x
x'
The standard Galilean transformation is:
x xvt
y y
z z
t t
the absolute time
- the same in S and S
We have been given the information with respect to S′ but we require it with respect to S.
[The frames are coincident at 9am hence at 10 am 00′ = 10 miles]
Invert the equation
x
x
vt
y y, z z,
t t
x 4 10 1 14 miles
y= 1 mile, z
2 miles
14,1, 2 are the co-ordinates in S
(ii)
According to Special Relativity, the time intervals measured by different inertial
observers no longer agree (as they did in Galilean Relativity).
If t is the time inertial frame S and
t
that in frame S′ then
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Lorentz transformation gives
t
v
t
x
2
c
2 2
1 v / c
We are told that the two events occur simultaneously (therefore
t
0
) and at a
separation
x
8
310 m
Hence,
vx
c c v/
c
t
1 v / c 1 v / c
2 2 2 2
Rearranging gives
Hence the velocities
v/
c
are:
v/
c
t
t 2
1
(i) 0.099 for t = 0.1s
(ii) 0.707 for t = 1s
(iii) 0.995 for t = 10s
4. (a) The factor
2
1 enters into the relativistic expression for length contraction and
1 1
2
occurs in the expression for relativistic time dilation. To gain some experience
in the magnitude of these two factors as a function of
v
c calculate their values for
0.100,0.300, 0.600, 0.800, 0.900, 0.950
and 0.990 , and plot your results as a
function of .
(b) Up to what value of are the binomial expansions of these
2
1
and
1 1
2
acceptable approximations
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WEEK 2
5. Using the Lorentz transformation for space and time co-ordinates, verify, explicitly
the (Lorentz) covariance of the following equation:
2 2 2 2 2
x y z c t 0
[Hint: Write this equation in terms of x, y, z and t and substituting from the Lorentz
transformations show that the resulting equation is of the same form as the one above]
6. The space and time coordinates of two events as measured in a frame S are as follows:
Event 1: x
1
x0, t1
x0
/ c,
y1
0, z1
0
Event 2: x x , t x / 2c,
y
0, z 0
2
2
0 2 0
2 2
Show that there exists an inertial frame in which these events occur at the same time,
and find the velocity of this frame relative to S. At what value of time in the new
frame do both events occur
[Answer: 1 2 ,
x 0
t 3 ]
c
7. Show that two successive Lorentz transformations with velocities v 1 and v 2 in the
same direction are equivalent to a single Lorentz transformation with velocity
2
v v1 v2
1 v1
v2
c .
SOLUTION
We need to make two successive transformations, so for the first frame
x
2
x
v1t
v
1x
t
v1x
c
x v t
t t
1
2
1
1 v c
and the second frame
x
2
c
2
2
1
1 v c
x
v2t
x
v t
2
v2x
t
v2x
c
t t
2
2
2
1
v c
The two transformations together give
2
c
2
2
2
2
1
v c
2
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x
x
v t
2
2
x
v t t
v x c
1
v
2
1
1
c
2
v
1
v
2
2
2
c
2
1
1
v
2
1
c
2
x
2
1
v v c
tv
v
1
2 1
2
v1
c
2
1
v
1
2
2
c
2
2
x
2
x
v t
2
2
x t v1
v2
1
v1
v2
c
2 2 2 2
2
1
v c 1
v c 1
v v c
1
2
2
x t v1
v2
1
v1
v2
c
2 2 2 2
2
1
v c 1
v c 1
v v c
1
2
1
x x
v t
[The top is in the correct form]
1
2
2
Look carefully at denominator
2 2 2 2
2 2 2 2
2
1
v
2
1 2
2
1 c 1
v
1
1
1 1
2 c
v c v c v v c
1
v 2
1
v2
c
2 2 2 2
1
( v1
c v2
c )
2
1
v v c 2
Look for the trick!
1
v
2
1
v
2
2
4
1
2
c 2v
v
1
2
v
c
2 2
1
v2
2
c
( v
4
2
1
v v c 2
1
2
1
2
c
2
v
2
2
2
c ) 2v
v
1
2
c
2
1
v1
v
2
c
2
2
1
v v c 2
1
2
v
2
1
v
c
2
2
1 2
1
v v
1
v1
v2
c
2
c
2
So now we find that
2
v
v 1
v v c
x t
x
1 2
v1
v2
1
1
v1
v2
c
2
1
2
c
2
x vt
1 v c
2
x vt
2
where v v
v v v
1 2
1
1 2
c
which is of the form required. Thus two successive Lorentz transformations do indeed yield
the same as one using the transformation of velocity formula.
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8. The following table shows the time of flight (over 10 m) for electrons that had
*
previously been accelerated through a potential to give them the specified kinetic
energy (KE). The KE is specified in units of MeV.
[1MeV = 10 6 eV and 1 eV = 1.60210 -19 J]
KE
(MeV)
Time of Flight
over 10 m
(S)
0.0051 0.237
0.051 0.080
0.255 0.045
2.55 0.034
5.1 0.033
25.5 0.033
Velocity v
(ms -1 )
v 2
(m 2 s -2 )
v
2KE(
J
m
2 )
0e
v
2
c
2
KE(
MeV )
1
1
0.51
2
(a) Using the data provided calculate the values of v and v 2 .
(b) Plot the value of v 2 (m 2 s -2 ) against the KE (MeV) on a log 10 vs log 10 scale.
Velocity Squared (m^2s^-2)
1.0E+19
1.0E+18
1.0E+17
1.0E+16
1.0E+15
0.001 0.01 0.1 1 10 100
Energy (MeV)
(c)
(d)
2
In Newtonian mechanics ( )
m 0e
KE J
v where m 0e = 9.109410 -31 kg. Use this
2
expression to calculate a value for v 2 (remembering to convert the KE into the
correct units) and plot this on the same graph as used in part (b). Why does
Newtonian theory not agree with the experimental data over the whole range of
energies
The relativistic expression for the kinetic energy of an electron travelling at
velocity v is
0.5
KE ( MeV ) 0.51 1
v 1
2 c 2 . Re-arrange this expression and
show that the calculated values of v 2 agree with experiment.
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WEEK 3
9. Proxima Centauri, the star nearest our own, is some 4.2 light-years away. (a) If a
spaceship could travel at a speed of 0.80c, how long would it take to reach the star,
according to the spaceship’s pilot (b) What would someone in the frame that moves
with the spaceship measure as the distance to Proxima Centauri
[Ans: (a) 3.1 yr, (b) 2.5 ly]
SOLUTION
Proxima Centauri, the star nearest our own, is some 4.2 ly away. (a)
v ship
0. 8c
, hence
1.67 and 0. 8 . In the spaceship frame, from the Lorentz transformations we have, with
origin of space and time in both frames at the departure point of the spaceship,
xEarth
tship 1 .67
tEarth
0. 8
c
and
t
Earth
4.2
0.8 5.25 years
hence
t
Earth
5.25
0.8
4.2 3. 15
1.67
years
or from the invariance of the interval,
giving
Ship
2 2 2
5.25
4.2 tShip
t 3.15 yrs as before.
(b) Here we need simply to multiply
[Ans: (a) 3.15 yrs, (b) 2.5 ly]
t Ship
by the velocity of the ship (0.8c) to give 2.52 ly.
10. Rigel is 900 light years from Earth. (a) What is the minimum Earth time a spacecraft
could take to make a return trip to Rigel (b) How fast should the spacecraft travel to
complete the journey in just 30 astronaut years (c) What is the distance from Earth to
Rigel measured by astronauts on their 30 year round trip
11. Our galaxy is 10 5 light years across. How long does it take to cross the galaxy in a
spacecraft travelling at 0.995c
12. A new Klingon battleship races at a top speed of 0.2c away from the planet XG4C.
The star ship Enterprise follows at a speed of 0.25c relative to the Klingon ship. With
what speed does the Enterprise appear to catch up with the Klingon “wessel”,
according to Mr. Chekov on the planet [Ans: 0.23c]
13. A new Borg battleship has a proper length of 217 m and travels at a speed of 0.20c
with respect to planet XG12 (away from the planet). The Borg prepare to battle the
Enterprise, which is moving at the same speed with respect to the same planet
(towards the planet). If the Borg are heading straight at the Enterprise, what is the
length of the Borg ship as measured by Captain Jean Luke Pickard
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[Ans: 200 m]
14. The proper mean lifetime of
*
+ mesons is 26 ns.
(a) What is the mean lifetime of a burst of + mesons travelling with = v/c = 0.73
(b) What distance is travelled at this velocity during one mean lifetime
(c) What distance would be travelled without the effect of time dilation
[Ans: (a) 38 ns, (b) 8.32 m, (c) 5.69 m.]
15. A burst of muons is produced by a cosmic ray interacting in the outer reaches of the
upper atmosphere. They travel towards the Earth’s surface with an average speed of
0.99c. If 1% of the burst survives to reach ground level, estimate the height of the
burst and calculate the distance as measured in the muon’s frame. The muon mean
lifetime is 2.2 sec.
[Ans: 2.1310 4 m, 3.0110 3 m.]
SOLUTON
The muon mean (rest) lifetime is 2.2 sec. The average speed of the muons in the Earth
frame is 0.99c. Only 1% reach the ground, so if Earth is the apparent lifetime in the earth
frame, then
I I 0 exp t Earth
or I 1
ln
ln
t
Earth
I
0 100
1
v
1
c
2
2
1
1
0.99
2
7.09
6
5
Earth
Rest
7.09 2.2 10
1.56 10
s
So in the Earth’s frame
ln
1
100
4.61
t
t 4.611.56
10
5
1.56 10
7.19 10
8
5
5
5
d 0.99
2.99810
7.19 10
2.1310
s
4
m
and using the Lorentz contraction we have in the muon’s rest frame
d
d
2.1310
7.09
4
3.0110
3
m
[Ans: 2.1310 4 m, 3.01103 m.]
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16. A beam of particles travelling at 99% the speed of light passes through two detectors
60 m apart. The number of particles per second recorded passing the second detector
is only one quarter of the number of particles per second recorded passing the first.
What is the half-life of the particle in its rest frame
WEEK 4
17. An absorption line in the hydrogen spectrum from a distant galaxy are red-shifted
from =394 nm to =500 nm. Assuming that this is due to the recession of the
galaxy calculate its velocity relative to the Earth.
SOLUTION
Using the relativistic Doppler formula (where =v/c):
0
1
1
0.5
500
1.27
394
1 .61
1
1
1 .611.61 1
2.61 0.61
v
c
0.61
0.234
2.61
So the galaxy is moving away from the Earth at a velocity v = 0.234c
18. By 2025 a spaceship shuttle service operates between the Earth and Mars. Each
spaceship is equipped with identical monochromatic head and taillights, and their
cruising speed v relative to earth, is such that the headlight of a homeward-bound
spaceship appears green ( = 500 nm) and the taillight of a departing spaceship
appears red ( = 600 nm).
(i) What is v/c
(ii) What is the wavelength of the headlight of an earth-bound spaceship is observed
from an outward-bound one
[Ans: (i) 1/11, (ii) 457 nm.]
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19. A driver was caught running a red light. His defence is that he saw the light as green,
as a result of the Doppler shift. He is arrested. What for Estimate the seriousness of
his transgression if he is fined £1 for each m.p.h. above the speed limit of 70 m.p.h.
You may assume red light is 620 nm, and green light 530 nm.
20. Fresnel suggested that the movement of material bodies through the Ether might drag
the ether along with it and affect the speed of light relative to an external laboratory.
In 1861 Fizeau tested this idea and found that the speed of light in water relative to the
laboratory was affected by the velocity of flow of the water. If the ether is dragged
along at the velocity of the medium the velocity of light relative to the laboratory
would have this flow velocity v (Galilean Transformation), added to it so that the
velocity relative to the laboratory is about:
c L
c
v where n is the refractive index of the medium.
n
In 1886 Michelson and Morley carried out an accurate measurement of this effect and
found that the velocity relative to the laboratory is about:
c L
c
0 . 453
v
n
Show that this result is consistent with relativity and does not need the ether
hypothesis.
SOLUTION
If the flow of water is parallel to the direction of the light beam the velocity c L relative to the
laboratory is found from the relativistic addition of the flow velocity v and velocity of light
relative to the medium c
c n
ux
v c
v c n v v
ux
cL
1
2
c n v
1 vu
c vc
vc
1 1
nc
x
2
2
c nc
v
Since 1 we can expand the second term using the binomial expansion
nc
2
v v
v
c L
c n v 1 c n v
2
nc nc n
1
2
v
since 0
nc
so
c 1 c 1 c 1 c
c L v 1
v 1
v 1 0. 435v
n
2
n n
2
n n
2
1.33 n
c
c L
0. 435v
n
So relativity saves the day yet again!
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21. (a) An electron has a kinetic energy of 2.53 MeV; calculate
*
(i) its total relativistic energy, E.
(ii) the magnitude, p, of its relativistic linear momentum in units of MeV/c.
(iii)its relativistic mass m(u)
2
[ m 0 c for the electron is 0.511 MeV]
(b)
A proton is accelerated to 1.08 TeV in the Fermilab accelerator; calculate
(i) its speed as a fraction of c.
(ii) how much slower than c is this high-energy proton moving.
2
[ m 0 c for the proton is 983.3 MeV]
22. The Galaxy is about 10 5 light years across and the most energetic cosmic rays known
have energies of the order of 10 19 eV. How long would it take a proton with this
energy to cross the Galaxy as measured in the rest frame of (i) the Galaxy and (ii) the
proton The rest-mass of the proton is 938 MeV/c 2 .
[Ans: (i) 10 5 years, (ii) 4.96 min.]
23. (a) What is the ratio of total mass to rest mass for a 500 GeV proton (i.e. a proton that
has been accelerated through a potential difference of 500 GV). The rest mass of a
proton is 1.6710 -27 kg.
SOLUTION
(b) What is the velocity of a particle whose total mass is double its rest mass
(c) What is the rest mass, total mass, energy and momentum of an electron accelerated
through 1.5 MV (Rest mass of an electron is 9.110 -31 kg)
(a) 500 GeV is the kinetic energy of the proton, T. The total energy is the sum of the kinetic
energy and rest energy.
KE m
2
0c
1
Total Energy E m
2
0
m0c
2
0c
KE E0
2 2 0.
5
E 1
v c
E
10
2
27
16
10
0
m0c
1.627 10
9 10
1.510
eV
19
E 500 0.94 500.94 GeV
1.510
J
1.602 10
0.94 GeV
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E 500.94
Note 533 . This corresponds to a proton velocity of about 0.999998c. The
E0
0.94
KE in this case is so large that the rest mass energy makes a negligible contribution to the
total energy.
(b)
m
m 0
2m 0
so
2 0.5
1
v c 2
2
2 2
2 2
3
1 4 1 v c v c 3 4 v c
2
(c)
14
2
31
16
14
8.19 10
E
0
m0c
9.110
9 10
8.19 10
J
eV
19
1.602 10
0.51 MeV
Kinetic Energy KE = 1.5 MeV
Total Energy
E 1.5
0.51
2.01MeV
2
E m0c
mc
E
Total Mass 01
2
c
2. MeV/c 2
E
2
2
2 2 2
E
E0
E0
p c so momentum p
1. 9 MeV/c
c
2
2
0.5
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WEEK 5
24. Inertial frame S is moving at a speed 0.6c in the x (x parallel to x) direction with
*
respect to inertial frame S. At time t = t = 0, x = x = 0.
(a) Draw the Minkowski space-time diagram for S and S showing the x- and ct-axes,
as well as the x- and ct-axes. The x-axis should span the range from 0 to 5 m, and
the ct-axis should span the range from 0 to 5 m. Remember everywhere on the xaxis
ct = 0 and likewise on the ct-axis x=0
x
vx
x
vt t t
2
c
1
1 v
2
c
2
2
2
(b) Using the relations x ( ct)
1calibrate the x- and ct-axes.
(c) Indicate the events x = 2.5 m, ct = 1.5 m, and x = 4 m, ct = 4 m on the diagram.
(d) What are the co-ordinates of the above events in the S frame
(e) Hence use the diagram to demonstrate that “moving clocks run slow”.
25. Two rockets A and B depart from Earth at steady speeds of 0.6c in opposite
directions, having synchronised clocks with each other and with Earth at departure.
After one year measured in the Earths reference frame, rocket B emits a light signal.
At what times, in the reference frames of rockets A and B, does rocket A receive the
signal [Hint: It is very helpful to draw the Space-time diagram(s)!]
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Mark Newton 2003