ALL 12 CHAPTERS COVERED
, Chapter 1
1 3.7×10−24 kg m−1 ; 3.5×10−43 kg m; 10−7 ; 10−4 kg; 1.1×10−12 kg m−3 ;
103 kg m−3 ; 3.7 × 10−16 kg m−3 .
2 3×106 m s−1 ; 9×1035 N m−2 ; 3.3×109 s; 9×1016 J m−3 ; 9×1017 m s−2 .
3 (a)–(k) See the top panel in Figure 1. Note that most of the items
continue into other quadrants (not shown). (l) See the bottom panel
in the same figure. Note that light always travels at speed 1 in this
frame even after reflecting off of a moving mirror.
(c) (g)
(k)
(b)(a) (i)
3
tO
tO
(m) (h)
2
(d) (c)
(e)
xO
(f )
1
(f )
(j)
1 2 x (m) 3
3
tO
tO
(l)
(m)
2
absorption
xO
1
mirror
reflection 1 2 xO (m) 3
detector
Figure 1: Solution to Ex. 3 of Chapter 1. See the solution text for explana-
tion.
2
, 5 (a) See the top panel in Figure 2. The particles (short dashed world
lines) are emitted at event A and reach the detectors at events B and C.
The detectors send out their signals (more short dashed world lines)
at events D and E, which arrive back at the spatial origin at event
F. The line BC joining the two reception events (long dashed) at the
detectors is parallel to the x=axis in our diagram, which means that
they occur at the same time in this frame. Note that the lines DF
and EF are tilted over more than the lines AB and AC because the
returning signals go at speed 0.75 while the particles travel only at
speed 0.5.
(a) 6
F
4
tO
D (m) E
2 C
B
-4 -2 2 4
xO (m)
-2 A
tO
(c) 6 F
E
4 xO
C
tO (m)
2
D
B
-4 -2 2 4
xO (m)
-2
A
Figure 2: Solution to Ex. 5 of Chapter 1. See the solution text for explana-
tion.
3
, (b) The experimenter knows that the detectors are equidistant from
x = 0, and that the signals they send out travel at equal speeds.
Therefore they have equal travel times in this frame. Since they arrive
at the same time, they must have been sent out at the same time.
This conclusion depends on observer-dependent things, such as the
fact that the signals travel at equal speeds. Therefore the conclusion,
while valid in this particular frame, might not be valid in others.
(c) See the bottom panel in Figure 2. This is drawn using the axes of
frame Ō, which we draw in the usual horizontal and vertical directions,
since any observer would normally draw his/her axes this way. In this
frame, the frame O moves forwards at speed 0.75. These axes are
drawn in gray, and calibrated using invariant hyperbolae (not shown),
just as in the solution to Exercise 1.3. Then the events are located
according to their coordinate locations on the axes of O. Note that
to do this, lines of constant tO must be drawn parallel to the xO axis,
and lines of constant xO must be drawn parallel to the tO axis. One
such line is the long-dashed line BC. The detectors (heavy lines) must
pass through the points xO = ±2 m on the xO -axis. The two signal-
emission events D and E are clearly not simultaneous in this frame,
although they are simultaneous in O: D occurs much earlier than E.
Note also that the signal sent at the event E and received at F remains
at rest in Ō, since it was sent backwards at speed 0.75 in frame O,
exactly the same speed as the frame Ō.
(d) The interval is easily computed in frame O because the events D
and E have zero separation in time and are separated by 4 m in x.
So the squared interval is 16 m2 . To compute it in frame Ō, measure
as carefully as you can in the diagram the coordinates tŌ and xŌ for
both emission events. Given the thickness of the lines representing the
detectors, you will not get exactly 16, but you should come close.
6 Write out all the terms.
7 M00 = µ2 − α2 , M01 = µν − αβ, M11 = ν 2 − β 2 , M22 = a2 , M33 =
b2 , M02 = M03 = M12 = M13 = M23 = 0.
8 (c) Use various specific choices of ∆xi ; e.g. ∆x = 1, ∆y = 0, ∆z =
0 ⇒ M11 = −M00 .
10 Null; spacelike; timelike; null.
4
, Chapter 1
1 3.7×10−24 kg m−1 ; 3.5×10−43 kg m; 10−7 ; 10−4 kg; 1.1×10−12 kg m−3 ;
103 kg m−3 ; 3.7 × 10−16 kg m−3 .
2 3×106 m s−1 ; 9×1035 N m−2 ; 3.3×109 s; 9×1016 J m−3 ; 9×1017 m s−2 .
3 (a)–(k) See the top panel in Figure 1. Note that most of the items
continue into other quadrants (not shown). (l) See the bottom panel
in the same figure. Note that light always travels at speed 1 in this
frame even after reflecting off of a moving mirror.
(c) (g)
(k)
(b)(a) (i)
3
tO
tO
(m) (h)
2
(d) (c)
(e)
xO
(f )
1
(f )
(j)
1 2 x (m) 3
3
tO
tO
(l)
(m)
2
absorption
xO
1
mirror
reflection 1 2 xO (m) 3
detector
Figure 1: Solution to Ex. 3 of Chapter 1. See the solution text for explana-
tion.
2
, 5 (a) See the top panel in Figure 2. The particles (short dashed world
lines) are emitted at event A and reach the detectors at events B and C.
The detectors send out their signals (more short dashed world lines)
at events D and E, which arrive back at the spatial origin at event
F. The line BC joining the two reception events (long dashed) at the
detectors is parallel to the x=axis in our diagram, which means that
they occur at the same time in this frame. Note that the lines DF
and EF are tilted over more than the lines AB and AC because the
returning signals go at speed 0.75 while the particles travel only at
speed 0.5.
(a) 6
F
4
tO
D (m) E
2 C
B
-4 -2 2 4
xO (m)
-2 A
tO
(c) 6 F
E
4 xO
C
tO (m)
2
D
B
-4 -2 2 4
xO (m)
-2
A
Figure 2: Solution to Ex. 5 of Chapter 1. See the solution text for explana-
tion.
3
, (b) The experimenter knows that the detectors are equidistant from
x = 0, and that the signals they send out travel at equal speeds.
Therefore they have equal travel times in this frame. Since they arrive
at the same time, they must have been sent out at the same time.
This conclusion depends on observer-dependent things, such as the
fact that the signals travel at equal speeds. Therefore the conclusion,
while valid in this particular frame, might not be valid in others.
(c) See the bottom panel in Figure 2. This is drawn using the axes of
frame Ō, which we draw in the usual horizontal and vertical directions,
since any observer would normally draw his/her axes this way. In this
frame, the frame O moves forwards at speed 0.75. These axes are
drawn in gray, and calibrated using invariant hyperbolae (not shown),
just as in the solution to Exercise 1.3. Then the events are located
according to their coordinate locations on the axes of O. Note that
to do this, lines of constant tO must be drawn parallel to the xO axis,
and lines of constant xO must be drawn parallel to the tO axis. One
such line is the long-dashed line BC. The detectors (heavy lines) must
pass through the points xO = ±2 m on the xO -axis. The two signal-
emission events D and E are clearly not simultaneous in this frame,
although they are simultaneous in O: D occurs much earlier than E.
Note also that the signal sent at the event E and received at F remains
at rest in Ō, since it was sent backwards at speed 0.75 in frame O,
exactly the same speed as the frame Ō.
(d) The interval is easily computed in frame O because the events D
and E have zero separation in time and are separated by 4 m in x.
So the squared interval is 16 m2 . To compute it in frame Ō, measure
as carefully as you can in the diagram the coordinates tŌ and xŌ for
both emission events. Given the thickness of the lines representing the
detectors, you will not get exactly 16, but you should come close.
6 Write out all the terms.
7 M00 = µ2 − α2 , M01 = µν − αβ, M11 = ν 2 − β 2 , M22 = a2 , M33 =
b2 , M02 = M03 = M12 = M13 = M23 = 0.
8 (c) Use various specific choices of ∆xi ; e.g. ∆x = 1, ∆y = 0, ∆z =
0 ⇒ M11 = −M00 .
10 Null; spacelike; timelike; null.
4
