1-15. a.

Let's call the Earth's frame S; the frame of the spaceship on the left is S': y' x' v u x y x
EARTH b.

In the Earth's reference frame, the spaceship on the left has a velocity xcomponent of v = 0.9c, and the spaceship on the right has a velocity x-component of u x
= -0.9c.

We want to know the speed of the spaceship on the right as seen by the spaceship on the left. In other words, we want to know u x
' (or its absolute value).

We simply use the relativistic velocity transformation: u x
'

1 u
 x
 vu
2 v x

1


(
0
0
.
.
9
9 c c

)(
2
0

.
0
9
.
c
9 c )
 
0 .
99 c c c

Since we are asked for a speed (not velocity), the best answer is 0.99c
.

If we had chosen to attach S' to the spaceship on the right, we would have obtained the same magnitude for u x
'.
This time v and –u x
are 30000 m/s, so

u x
'
 u
1
 x
 v vu x c
2

1


30000 m / s

30000 m
( 30000 m

3

/ s )(
10 8

30000 m / s

2 m
/
/ s s )
 
60000 m / s , so the relative speed is 60000 m/s. This is what the Galilean velocity transformation would have given us: u x
– v. Relativistic effects are negligible because the speeds are too small.
1-18. a.

Let's draw this from the point of view of the S' frame. We're told that the light moves in the y' direction with speed c, and we infer that the S frame is moving to the left with speed v: y y' v x c x' b.

The speed of the light is u y
' = c, so all we have to do is use the inverse velocity transformations to find u x
and u y
. (Note: the inverse velocity transformations are used when we transform to the frame that is moving in the negative x direction of the other frame.) u x
 u x
'
 v
1
 vu x
' c
2

0
1

 v
 v
0 c
2
 v u y

 

1 u y
' vu x
' c
2






1
 c v

0 c
2

 c
 c 1
 v 2 c
2
The magnitude of the velocity of the light beam in frame S is

u x
2  u y
2  v
2  c
2

 1
 v
2 c
2


 v
2  c
2  v
2  c .
1-24. a.
The proper (life)time is always shorter by a factor of

than the time measured in any other frame, so the mean lifetime measured in the laboratory frame is a factor of
 greater than the proper lifetime:

2.6

10
-8
s = (1 – v
2
/c
2
)
-1/2 
2.6

10
-8
s =
(1 – 0.9
2
)
-1/2 
2.6

10
-8
= 5.96

10 -8 s . b.
In 5.96

10 -8 s, a laboratory observer sees a pion travel a distance D = vt =
(0.9c)(5.96

10 -8 s) = (0.9

3.0

10 8 m/s)(5.96

10 -8 s) = 16.1 m . c.
If we incorrectly use the proper lifetime in the laboratory frame, we find D = vt =
(0.9c)(2.6

10
-8
s) = (0.9

3.0

10
8
m/s)(2.6

10
-8
s) = 7.02 m . d.

The spacetime interval is independent of reference frame. It's easy to use the laboratory frame since we already know
 x = 16.1 m and
 t = 5.96

10 -8 s.

The spacetime interval is
 s

   
2 

 8 
5 .
96
 
8

2 

16 .
1 m

2
7 .
78 m
3 10 m / s 10 s


Alternative solution: compute the spacetime interval in the pion's reference frame. In the pion's frame,
 x' = 0 because every object is motionless in its own frame.
 t' = 2.6

10
-8
s. So
 s = c
 t' = (3

10
8
m/s)(2.6

10
-8
s) = 7.8 m .
1-49. a.

Since both events (1 and 2) are simultaneous in S', we can write t
1
' = t
2
'.

The Lorentz transformation for time gives us expressions for t
1
' and t
2
': t
1
' =

(t
1
– vx
1
/c 2 ) t
2
' =

(t
2
– vx
2
/c
2
)

a.

Since t
1
' = t
2
', we can set the two preceding equations equal to each other to obtain an expression for v:

(t
1
– vx
1
/c t
1
– vx
1
/c 2 v = c
2
(t
2
– t
2
(x
2
– x
1
)v/c
2
1
) =

(t
)/(x
2
2
– vx
2
/c
= t
2
– vx
2
/c 2
= t
2
– t
1
– x
1
)
2
)

Next we just plug in the given values: x
1
= 1 c
.
y, t
1
= 1 y, x
2
= 2 c
.
y, and t
2
= 0.5 y.

So v = c 2 (0.5 y – 1 y)/(2 c .
y – 1 c .
y) = -0.5c.

So S sees S' travel in the –x direction with speed 0.5c. b.
We can use the expression for either t
1
' or t
2
'. So t
1
' =

(t
1
– vx
1
/c 2 )
=

[1 y – (-0.5c)(1 c
.
y/c
2
)] =

(1.5 y), where

= (1 – .5
2
)
-1/2
= 1.15. Then t both are 1.15(1.5 y)

1.73 y .
1
' and t
2
' d.
1-56.

The square of the spacetime interval is (
 s)
2
= (c
 t)
2
– (
 x)
2
= (-0.5 c
.
y)
2
– (1 c
.
y)
2
= -0.75 (c
.
y)
2
.

(
 s) 2 is negative, so the interval is spacelike .

We simply use the Lorentz transformation for t
1
' and t
2
': t
1
' =

(t
1
– vx
1
/c
2
) t
2
' =

(t
2
– vx
2
/c
2
)

When we subtract t
1
' from t
2
', we immediately obtain the desired result: t
2
' – t
1
' =

(T – vD/c
2
), where T = t
2
– t
1
and D = x
2
– x
1
.

This is a very useful fact: the Lorentz equations take exactly the same form for intervals (
 x,
 t,
 x', and
 t') as they do for coordinates. b.


If the two events are simultaneous in S', then t
2
' – t
1
' = 0.

Then from the preceding equation for t
2
' – t
1
', T = vD/c 2 . Solving for v, v = c
2
T/D, or v = c(cT/D). Since v must be less than c, the factor cT/D must be less than 1. cT/D < 1 can be rearranged to D > cT. c.
Just in case you're curious…

We must show that t
2
' > t
1
' when D < cT.

We must assume that t
2
> t
1
; in other words, T > 0. (We want to show that if event 1 causes event 2, then event 1 precedes event 2 in all reference frames. We must begin with the assumption that event 1 precedes event 2 in the S frame.)

Let's look again at the expression t
2
' – t
1
' =

(T – vD/c and manipulate it:
2
) t
2
' – t
1
' =

(T – vD/c
2
) =

/c(cT – vD/c) =

/c[cT – v(D/c)].
1. a.

The term in the brackets is the difference between two products. The two factors in the first product are always larger than the corresponding factors in the second product: c > v and T > D/c (by the assumption D < cT).

Thus the first product is always larger than the second product and t
2
' – t
1
' is always positive, or t
2
' > t
1
'. Since this is true for all v < c, this is true for all reference frames (moving with any speed v relative to S). d. Just in case you're curious…

We again write t
2
' – t
1
' =

(T – vD/c 2 ).

Plug in D = c'T, where c' > c: t
2
' – t
1
' =

(T – vc'T/c
2
) =

T[1 – (v/c)(c'/c)].
 v/c < 1, but c'/c > 1. As long as v and c' are chosen so that the product of these factors is greater than 1, then t
2
' – t
1
' < 0 and the effect (event 2) precedes the cause (event 1).

Door 1 has coordinate x = 0. The two doors are 960 m apart. So door 2 has coordinate x = 960 m.

c.

The front of the train crosses door 1 at time t = 0. To travel from door 1 to door 2, the front of the train traverses 960 m. The time required is distance/speed =
960 m/(0.8

300 m/
 s) = 4
 s.

Thus the requested coordinates for F2 are (960 m, 4
 s) . b.

In the train's frame, the back of the train has coordinate x' = -960 m.

In the train's frame, the tunnel entrance travels from the front of the train to the back of the train with velocity -0.8c. This requires time = distance/speed
= 960 m/(0.8

300 m/
 s) = 4
 s.

Thus the requested coordinates for B1 are (-960 m, 4
 s) .

The (inverse) Lorentz equations are x =

(x' + vt') and t =

(t' + vx'/c
2
), where

=
(1 – 0.8
2
)
-1/2
= 5/3.

Using these equations to transform (x', t') = (-960 m, 4
 s), we find (x, t)
= (0, 2.4
 s) for B1 in S tunnel
. So B1 precedes F2: the tunnel observer sees the back of the train enter the tunnel before the front of the train leaves the tunnel.
This is expected: the train's length is contracted in S tunnel
, so the entire train fits inside the tunnel with plenty of room to spare.

We could have found the answer without using the Lorentz equations. We immediately know that x coordinate of B1 (the back of the train entering the tunnel) is 0; the entrance to the tunnel is the tunnel's origin. We know that the tunnel observer records the proper time interval between the front of the train entering the tunnel and the rear of the train entering the tunnel. In S train
, the time interval between these two events (4
 s) is dilated, so we divided by

to find the proper time: 4
 s/(5/3) = 2.4
 s. Yet another method would be to use length contraction: the train has length (960 m)/

= 576 m in S traveled in time 576 m/0.8c = 2.4
 s. tunnel
. This distance is d.

The Lorentz equations are x' =

(x - vt) and t' =

(t - vx/c
2
).

Using these equations to transform (x, t) = (960 m, 4
 s), we find (x', t')
= (0, 2.4
 s) for F2 in S train
. So F2 precedes B1: the train observer sees the tunnel exit pass the front of the train before the tunnel entrance reaches the back of the train. This is expected: the tunnel's length is contracted in S train
, so only part of the train can fit inside the tunnel.

In S train
, the train is longer than the tunnel. In S tunnel
, the tunnel is longer than the train. Each observer can explain the other observer's "mistake" by considering that the moving observer's clocks appear unsynchronized. These details are not required for this problem.

e.

Let's start in S tunnel
. The two events are (x
B1
, t
B1
) = (0, 2.4
 s) and (x
F2
, t
F2
)
= (960 m, 4
 s).

In some frame S'', the Lorentz equations yield t
F2
'' – t
B1
''
=

[(t
F2
– t
B1
) – v''(x
F2
– x
B1
)/c
2
], where v'' is the speed of S'' relative to S tunnel
.

Plugging in known values yields t
F2
'' – t
B1
'' =

[1.6
 s – v''(960 m)/c
2
]

If B1 and F2 are simultaneous in S'', then t v'' = (1.6
 s)c
2
/960 m = 0.5c.
F2
'' – t
B1
'' = 0, and we can solve for v'':

So S tunnel
and S'' see each other moving with speed 0.5c. How fast is the train moving in S''? y y'' tunnel x v'' x'' train v

We use the relativistic velocity transformation: u x
'

1 u
 x
 v vu x
, c
2 but we have to be very careful.

As written above, this equation transforms an object's velocity u x
measured in the
S frame to the velocity u x
' measured in the S' frame, where v is the velocity of S' as seen in S. We want to transform the train's velocity v measured in the S frame to the velocity u x
'' measured in the S'' frame, and v'' is the velocity of S'' as seen in
S. (In this paragraph, I was lazy; everywhere I wrote "velocity," I should have written "x-component of the velocity.")

Comparing the two statements above implies that the appropriate statement of the velocity transformation is u x
''
 v
 v ''
1
 vv ''
, c
2 where v = 0.8c and v'' = 0.5c.


So u x
''
 v
 v
1
 vv '' c
2
''

0 .
8 c

0 .
5 c
1

( 0 .
8 c )( 0 .
5 c ) c
2

0 .
3 c
1

0 .
4

0 .
3 c
0 .
6

0 .
5 c .

So both the tunnel and the train move with speed 0.5c (in opposite directions) relative to S''.

So in S'', both the tunnel and the train are subject to exactly the same length contraction. Since S'' sees both tunnel and train as having the same length, the train fits perfectly in the tunnel: the back of the train crosses door 1 exactly as the front crosses door 2.