The key point in the proof is to notice that cos increases (decreases)
monotonically as cos increases (decreases), since
cos =
cos +
1 + cos
This implies
0+
1+0
1+
cos 1+1
cos 1
For the rays in the upper hemisphere ( > 0), this can be inverted to get

1R
1+R
fR
=
fA
For G = 0.6 and =
200km/s
3108 m/s ,
1A
1+A
we have
R = 0.599573162
A = 0.600426496
So
fR
= 0.99866756,
fA
the same as in part a)
[NOTE: This result can be seen in a nice Lorentz invariant way. If
two light pulses are traveling in the same

Problem 5: Spacetime Events and the Invariant
Interval
Because I(BA) = (5 15)2 (10 5)2 = 75 > 0, this is a timelike
interval.
(a) No, events connected by a timelike interval cannot occur simultaneously in any frame.
(b) Yes, there is a frame where events

(c) The error is given by:
1
65mph
uG uE
100% =
) 100%
(65mph
uG
65mph
1 + 60 5(mph2 )/c2
= 6.67 1014 %
E=
(d) Using the Lorentz transformation for the velocity we have:
u=
c/2 + 3c/4
10c
u +v
=
=
1 + u v/c2
1 + 1/2 3/4
11
.
Problem 3
a) Assuming that t

which gives
me
0.51 103
= 3.81
Ee = Ep
GeV = 2 MeV
mp
0.94
Problem 4 - Motion under a constant
force
a) F = dp . Since F is constant, and the particle starts at rest, p = F t.
dt
Using p = m0 v = m0 v 2 , we can solve for the velocity v, which gives
1(v/

(b) High energy regime: In the extreme limit E , cos becomes 1 so
becomes zero. We thus know that will be small in the high energy regime
E
m , making the Taylor approximation
2
cos 1
2
for
1
an appropriate one to use. Taylor-approximating both sides o

d) t = 2.5 5 mins = 12.5 mins. Draw a line parallel to x -axis from
B, and nd the intercept with the ct -axis.
e) x = 1.5 5 light mins = -7.5 light mins. Draw a line parallel to
ct -axis from B, and nd the intercept with the x -axis.
f) t = 4 5 mins = 20

Problem 2: Creation of a Mystery Particle
(pA c)2 + (mA c2 )2 =
(a) EA =
(50 MeV)2 + (140 MeV)2 = 148.7 MeV
(b) In its own frame the particle A is at rest, so the only energy is the rest
mass energy, 140 MeV.
(c) A =
pA
E
=
50 MeV
148.7 MeV
= 0.336, so vA

Rocket 1
Rocket 2
Space Station
(b) The inverse operation entails boosting the velocity u from the S frame
into the S frame. This is the same as the boost we just calculated, except
with the boost speed v pointing in the opposite direction. Thus, all we
h

Homework 7 Solutions
Problem 1: Energy-Momentum 4-vectors
Write down the energy-momentum four-vectors for the following cases. In
each case, the particle has (rest) mass m0 .
We note that, using (E/c, px , py , pz ) as the Energy-momentum four vector, the