Mitesh Patel
Handed out 11th May 2020
Relativity – Problem Sheet 2
Topics covered: spacetime diagrams, invariant intervals, four-vectors, relativistic energy
and momentum.
Questions to try in your own time
1. In Lecture 3 we looked at the relativity of simultaneity by analysing a situation whereby Lucy,
Matt, and Rick synchronise their clocks in their reference frame. We will now draw out a spacetime
diagram of this situation to illustrate which events happen simultaneously in which reference frame.
(a) Draw a spacetime diagram in which Rick, Lucy, and Matt are stationary, with Lucy at x = 3,
Rick at x = 3, and Matt at x = 0. At t = 0, Matt emits a light pulse to Rick and another to
Lucy. Draw the paths of the light pulses on the diagram. When does the pulse arrive at Rick,
and when at Lucy?
ANSWER: See the spacetime diagram below. By symmetry, the light pulses must arrive at
Lucy and Rick simultaneously and this must be at ct = 3.
(b) Now draw a separate spacetime diagram for a moving frame, in which Rick, Lucy, and Matt
are moving at = 1/3. Include all three world lines, as well as the paths the light pulses
travel to Rick and Lucy. Draw the ‘line of simultaneity’ (i.e. of equal time) in the moving
frame from the event of the light pulse arriving at Lucy. Does the light pulse still arrive at
Rick simultaneously in this frame?
ANSWER: See the spacetime diagram below. The distances between Lucy & Matt and Rick
& Matt are length-contracted to 3/ in the moving frame. We find that the pulse arrive at
Rick and Lucy at 3/ (1
) = 9/2 = 4.5/ and 3/ (1 + ) = 9/4 = 2.25/ respectively
and so are not simultaneous in the moving frame.
(c) Finally, as soon as Lucy and Rick receive Matt’s light pulse, they each sent a light pulse back
to Matt. Draw these light pulses in your spacetime diagram. Does Matt receive the two light
pulses simultaneously?
ANSWER: The second set of light pulses arrive simultaneously at Matt in the moving frame.
Note, by symmetry, they will also arrive simultaneously in the stationary frame. Remember
that events that occur simultaneously at the same point in spacetime in one frame will occur
simultaneously in any frame.
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ANSWER:
2. Five street lamps A, B, C, D, and E are located on a straight line along the x axis at an equal
distance apart as shown in the figure. They turn on at times tA , tB , tC , tD , and tE respectively, in
the frame at rest relative to the ground. These five events are indicated in the spacetime diagram
below, and a person stands at x = 0 to watch them turn on.
A car is moving at velocity v relative to the ground. At t0 = t = 0, it is at x0 = x = 0. The space
and time axes of the moving frame of the car are also shown in the spacetime diagram. Answer the
questions below by drawing the relevant lines and events in the spacetime diagram.
ANSWER: Refer to the annotated spacetime diagram overleaf.
(a) What is the order in which the lamps turn on in the ground rest frame?
ANSWER: The order is A and C simultaneously, then E, then D, and then B.
(b) What is the order in which the lamps turn on in the car’s rest frame?
ANSWER: C and E simultaneously, then A and D simultaneously, and then B.
(c) What is the order in which the light of the lamps reach the person at x = 0?
ANSWER: A, then C, then B and D simultaneously, and then E.
(d) What is the order in which the light from the lamps reach the person driving the car?
ANSWER: A, then C, then B and D simultaneously, and then E.
(e) Where is the car relative to the street lights when the light from street lamp D reaches it?
ANSWER: Between street lamps A and B.
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3. As mentioned in Lecture 6, the quantity s2 = (ct)2 (x2 + y 2 + z 2 ) is called the “invariant interval”
because it is unchanged under Lorentz transformations. In other words, the distance as well as the
time between two events changes when transforming to a di↵erent frame, but s2 does not. Prove
this is true by applying the Lorentz transformations to the components used in s2 , and showing
that s02 = s2 .
ANSWER: s2 = (ct)2 (x2 + y 2 + z 2 ), s02 = (ct0 )2 (x02 + y 02 + z 02 ). Since y = y 0 and
z = z 0 , the problem boils down to showing (ct0 )2 x02 = (ct)2 x2 . Write out the left hand side:
2
(c ⇣(t vx/c2 ))2
(x vt)2 = (ct
x)2 ⌘2 (x ⇣ ct)2 . Write out the squares:
⌘
2
(ct)2
2ct x +
2 2
x
x2 + 2 ctx
2
(ct)2 =
3
2
(ct)2 (1
2
)
x2 (1
2
) . We know that
1
2
= 1/
2
, so we are done: (ct0 )2
x02 = (ct)2
x2 .
4. This question is about space-like and time-like separations. Either answer the following questions
by drawing out a space-time diagram, or solve algebraically using the Lorentz transformations.
(a) Two events are separated by (c t, x, y, z) = (3, 4, 0, 0) in frame S. Is there a reference
frame S 0 in which the events are simultaneous, or at the same position? What is the relative
velocity between S and the relevant frame S 0 ?
ANSWER: The invariant interval is s2 = 9 16 = 7 < 0, so the events are space-like
separated. This means a reference frame can be found in which the events are simultaneous.
For this frame:
t0 = ( t v x/c2 ) = 0. Therefore we find v x/c2 = t and v =
2
c t/ x = 3c/4. There is no frame in which the events are at the same position as x0 would
be zero and the invariant interval could not then be negative
(b) Two events are separated by (c t, x, y, z) = ( 8, 2, 0, 0) in frame S. Again, find the
relative velocity between S and S 0 where S 0 is the frame in which the two events are either
simultaneous or at the same position.
ANSWER: The invariant interval is s2 = 60 > 0, so the events are time-like separated. This
means a reference frame can be found in which the events occur at the same position. For this
frame: x0 = ( x v t) = 0. Therefore we find v t = x and v = x/ t = 2c/8 =
c/4. There is no frame in which the events are simultaneous as t0 would be zero and the
invariant interval could not then be positive
5. The four-velocity
is given by ( u , u ~u ), where ~u is the usual three-velocity of a particle, ~u = ~u/c
q
and u = 1/ 1 | ~u |2 . What is the squared-length of the four-velocity?
Apply a Lorentz transformation to the four-velocity formed from the three-velocity ~u = (u, 0, 0).
From this, retrieve the velocity addition formula:
u
u0 =
1
2 ~ 2
u| u|
2
u
ANSWER: The squared length is given by
v
.
uv/c2
=
2
u (1
| ~u |2 ) = 1.
Applying the Lorentz transformation to the given four-velocity gives
0
u
= (
Hence
0
u
=
0 0
u u
u u ),
u
0 0
u u
0
u
=
(
(
u u
u0 =
0
uc
=
1
u)
u u)
u
which is the velocity transformation in terms of
= (
u.
=
u)
u u
u
1
u
In terms of u
u/c v/c
u v
c=
(v/c)(u/c)
1 vu/c2
which is the desired velocity addition formula.
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6. A particle has rest mass 6.64 ⇥ 10
27
kg and momentum 2.10 ⇥ 10
18
kg ms
1
.
(a) What is the total energy of the particle?
p
ANSWER:
We are given the rest mass and the momentum, so we use E = (mc2 )2 + (pc)2 =
p
(6.64 ⇥ 10 27 c2 )2 + (2.10 ⇥ 10 18 c)2 = 8.67 ⇥ 10 10 J or 5.41 GeV.
(b) What is the kinetic energy of the particle?
ANSWER: Use K = E mc2 = 8.67 ⇥ 10
10
5.97 ⇥ 10
10
= 2.70 ⇥ 10
10
J or 1.69 GeV.
(c) What is the ratio of the kinetic energy to the rest energy of the particle?
ANSWER: 1.69 GeV/5.41 GeV = 0.452.
7. How much energy is required to change a particle with mass m (i) from rest to a speed of 0.09c?
(ii) From 0.90c to 0.99c? Express your answers in terms of mc2 . How do the answers compare?
ANSWER: (i) 0 to 0.09c: 0 = 1 and 0.09 = 1.004. Hence E = ( 0.09 1)mc2 = 0.004mc2 . (ii)
2
0.9c to 0.99c: 0.9 = 2.294 and 0.99 = 7.089. Hence E = ( 0.99
2.294 =
0.9 )mc = 7.089
4.795mc2 (i.e. 1200 times greater than (i)). There is a much larger inertia to accelerate when the
particle is already at 0.9c, as energy as well as mass have inertia.
8. If a particle’s kinetic energy is n times its rest energy, find its speed in terms of n and c.
ANSWER: E = u mc2 , so K = E mc2 = ( u 1)mc2 . We need to solve the equation
( u 1)mc2 = nmc2 , or u 1 = n. Square both sides:
2
u
We find 1/(n + 1)2 = 1
2
u,
=
1
2
u
1
= (n + 1)2 .
or:
2
u
Hence
u=c
r
=
n2 + 2n
.
+ 2n + 1
n2
n2 + 2n
=c
n2 + 2n + 1
r
N
.
N +1
where N = n2 + 2n. The point to take away is that when n (and hence N ) is large, the velocity
changes very slowly with changes in n, and that the velocity is always less than c.
9. The Large Hadron Collider (LHC) at CERN in Geneva accelerates protons to 7 TeV (1 TeV =
1012 eV). What is the proton velocity? What velocity would a tennis ball (m = 60 g) need such
that it would have a kinetic energy equal to that of a 7 TeV proton?
There are around 1012 protons in each beam bunch at the LHC. Estimate the total energy of each
bunch and compare this (to an order of magnitude) with the kinetic energy of a double-decker bus
moving at 30 mph.
ANSWER: Proton velocity:
= E/mc2 = 7463, therefore
= 1 9 ⇥ 10 9 . So 7 TeV is a
very large energy for a single proton! Ignoring the rest mass energy, convert 7 TeV into joules:
7 ⇥ 1012 ⇥ 1.6 ⇥ 10 19 = 1.12 ⇥ 10 6 J. For the velocity of the tennis ball, guessing the tennis ball
is not relativistic, mv 2 /2 = 1.12 ⇥ 10 6 J gives v = 6.1 mm/s.
The bunch has an energy of around 106 J. Estimating the mass of a bus to be around 10 tonnes
= 104 kg and 30 mph is around 50 km/hour or 15 m/s, the kinetic energy of the bus is around
0.5 ⇥ 104 ⇥ 225 ⇡ 106 J, so approximately the same as the LHC beam bunch.
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