Solutions for chapter 7
Page 461, problem 17:
(a) Plugging in, we find √
1 −w
1 +w
=
√
1 −u
1 +u√
1 −v
1 +v.
(b) First let’s simplify by squaring both sides.
1 −w
1 +w=
1 −u
1 +u·
1 −v
1 +v.
For convenience, let’s writeAfor the right-hand side of this equation. We then have
1 −w
1 +w=A
1 −w=A+Aw.Solving forw,w=1 −A
1 +A
=
(1 +u)(1 +v)−(1−u)(1−v)
(1 +u)(1 +v) + (1−u)(1−v)=2(u+v)
2(1 +uv)
=u+v
1 +uv(c) This is all in units wherec= 1. The correspondence principle says that we should get
w≈u+vwhen bothuandvare small compared to 1. Under those circumstances,uvis the
product of two very small numbers, which makes it very, very small. Neglecting this term in
the denominator, we recover the nonrelativistic result.
Page 461, problem 18:
Among the spacelike vectors,aandeare clearly congruent, because they’re the same except
for a rotation in space; this is the same as the definition of congruence in ordinary Euclidean
geometry, where rotation doesn’t matter. Vectorbis also congruent to these, since it represents
an interval 3^2 − 52 =− 42 , just like the other two.
The lightlike vectorscanddboth represent intervals of zero, so they’re congruent, even
thoughcis a double-scale version ofd.
The timelike vectorsf andgare not congruent to each other or to any of the others; f
represents an interval of 2^2 , whileg’s interval is 4^2.
Page 462, problem 22:
At the center of each of the large triangle’s sides, the angles add up to 180◦because they form
a straight line. Therefore 4s=S+ 3× 180 ◦, soS− 180 ◦= 4(s− 180 ◦).
Page 463, problem 28:
By the equivalence principle, we can adopt a frame tied to the tossed clock, B, and in this frame
there is no gravitational field. We see a desk and clock A go by. The desk applies a force to clock
A, decelerating it and then reaccelerating it so that it comes back. We’ve already established
that the effect of motion is to slow down time, so clock A reads a smaller time interval.