(1). A tree and a pole are 30km apart in the Earth’s frame. Lightning strikes the
tree and the pole at t=10ms. A rocket is traveling in the x direction at 0.5c.
(a). What are the spacetime coordinates of the two events in the rocket’s frame?
(b) How far apart in time are the events in the rocket’s frame?
S is the ground’s frame and S′ is the rocket’s frame. S′ moves with velocity v =
0.5c relative to S.
(a) Let the tree be at x=0km and the pole at x=30km in S. We have
[
! = 1 " ( v c)
] = [1 " (0.50 ) ]
1
2
"2
1
2 "2
= 1.155 .
Applying the Lorentz transformations to the lightning strike at x = 0 m and t =
10 ms,
x ! = " ( x # vt ) = (1.155 ) %& 0 m # ( 0.5 ) 3.0 $ 10 8 m/s 1 $ 10 #2 s '( = #1732 km
vx ,
)
t ! = " + t # 2 . = (1.155 ) 1 $ 10 #2 s # 0 s = 11.55 ms
*
c -
(
(
)(
)
)
For the lightning strike at x = 30 km and t = 10 ms,
(
)(
)
m) &
) = 11.49ms
x ! = (1.155 ) $% 3.0 " 10 4 m # ( 0.50 ) 3.0 " 10 8 m/s 1 " 10 #2 s &' = #1697 km
(
)(
$
0.50 ) 3.0 " 10 8 m/s 3.0 " 10 4
(
#2
t ! = (1.155 ) (1 " 10 s #
2
8
(
)
3.0
"
10
m/s
%
'
(b) The events in the rocket’s frame are not simultaneous. The lightning is
observed to strike the pole before the tree by 57.75 µs.
(
)
(2). A cube has a density of 2000kg/m3 when it is at rest in the lab. What is
its density when it is moving at 0.9c in a direction perpendicular to one face?
The length of an object is contracted when it is measured in any reference
frame moving relative to the object. The contraction occurs only along the
direction of motion of the object.
The cube at rest has a density of 2000 kg/m3. As the cube moves with a speed
of 0.9c, its dimension along the direction of motion is contracted according to:
L ! = 1 " ( v c ) L = 1 " ( 0.9 ) L = 0.436L
2
2
Because the other dimensions of the cube are not affected by the cube’s
motion, the new density will be
" ! m / 0.436L3
1
=
=
3
"
m/L
0.436
"! =
2000kg/m 3
= 4600 kg/m 3
0.436
(3). A rocket moving at 0.8c shoots a bullet in the opposite direction at 0.9c
relative to the rocket. What is the bullet’s speed relative to the earth?
The rocket and the earth are inertial frames. Let the earth be frame S and the
rocket be frame S′. S′ moves with v = 0.8c relative to S. The bullet’s velocity
in reference frame S′ is u′ = 0.9c.
Using the Lorentz velocity transformation equation,
u=
u! + v
1 + u ! v c2
=
"0.9c + 0.8c
1 + ( "0.9c)( 0.8c ) c 2
= "0.36 c
The bullet’s speed is 0.36c along the x direction. Note that the velocity
transformations use velocity, which can be negative, and not speed.
(4). Rockets Orion and Sirius have proper lengths of 1000m. Orion, traveling
at 0.8c relative to Earth passes Sirius traveling at 0.6c relative to Earth. How
long does Orion take to pass Sirius as measured by the Sirius crew? After
Orion has passed Sirius and traveled some distance ahead, Sirius sends a
message to Orion encoded in a pulse of light. Will Orion be able to receive the
message? Explain.
Let S be the earth’s reference frame, and S′ be the rocket Sirius’s reference
frame. S′ moves relative to S with a speed of v = 0.6c. In S, rocket Orion’s
speed is u = 0.8c.
Orion’s speed relative to Sirius is
u! =
u"v
1 " uv c2
=
0.8c " 0.6c
1 " ( 0.8c )(0.6c) c 2
= 0.385c
Thus, the length of the Orion rocket as measured in S′ (Sirius rocket) is
L ! = (1000 m ) 1 " (0.385 ) = 923 m
2
Now, Orion is moving at a speed of 0.385c and has to move 1923 m to
completely pass Sirius. Thus, the time taken is
!t =
1923 m
0.385 (300 m 1 µs )
= 16.6 µ s
Orion will eventually get the message since the message is sent using a pulse
of light and Orion cannot travel faster than the speed of light.
(5). Calculate for the judge how fast you were going in miles/hr when you ran
the red light because it seemed green to you. Red light has a wavelength of
650nm and green has a wavelength of 550nm.
For an observer approaching a light source,
# (1 " v c )1 2 &
(!
.
! ob = %
% (1 + v c )1 2 ( source
$
'
Setting
!=
v
c
and after some algebra we find,
" 2source # "2obs (650 nm) # ( 550 nm)
=
= 0.166
"2source + "2obs (650 nm )2 + ( 550 nm) 2
2
!=
(
2
)
v = 0.166c = 4.98 $ 107 m s (2.237 mi h) (m s)
#1
= 1.11 $ 108 mi h .
(6). Alpha and Beta are 100 ly (light years) apart. Beta goes supernova 10
years after Alpha goes supernova.
(a) Could Alpha’s explosion have caused Beta’s explosion?
(b) A spaceship measures the distance between the explosions to be 120ly.
What does a clock on the spaceship measure to be the time between the
explosions?
Let S be the galaxy’s frame and S′ the spaceship’s frame. The spacetime
interval Ds between the two events is invariant in all frames.
(a) The light from Alpha’s explosion will travel 10 ly in 10 years. Since
neither light nor any other signal from Alpha can travel 100 ly in 10 years to
reach Beta, the explosion of Alpha could not cause the explosion of Beta.
(b) Because the spacetime interval Ds between the two events is invariant,
( !s )2 = c 2 ( !t )2 " ( !x )2 = c 2 ( !t # )2 " ( !x# )2
2
% 1 ly (
% 1 ly (
2
2
$'
10
y
"
100
ly
=
(
)
(
)
'& y *)
& y *)
2
( !t # )2 " (120 ly )2
$ (10 y ) " (100 y ) = ( !t # ) " (120 y ) $ !t # = 67.1 years
2
2
2
2