Answers to Chapter Nine Questions

9.1) C & D travel 4 l-yrs at 3/5 the speed of light, 4/(3/5) = 20/3 = 6 years.

9.2) C & D’s watches ticked off only 80% as much as the two synchronized Earth watches, 0.8*(20/3) = 16/3 = 5 years.

9.3) My answers are consistent with the figures.

9.4) T_B = 5 years

9.5) t_B = (4/5)*(16/3) = 64/15 = 4 years

9.6) Yes

9.7) Two times 6 years = 13 years

9.8) Two times 5 years = 10 years

9.9) (4/5)*(40/3) = 160/15 = 32/3 = 10 years! Yes they agree.

9.10) Earth Years = Rocket Years/ = 1/ = 7071 years!!

9.11) I convinced myself.

9.12) f_received = f_{sent → =} = 2

9.13) t₂

t₁

B

A

The diagram below is analogous to figure 9.3 except the rocket, red worldline, is now traveling away from Earth. The two dashed blue lines represent the two messages sent to Earth. The relationship between the sent and received frequencies is essentially the same as that presented in the text.

ΔT_sent = T_B – T_A and Δt_received = t₂ – t₁

f_sent = 1/ ΔT_sent and f_received = 1/ Δt_received

The first message starts at Earth time zero reaches Earth at t₁ = D/c, where D is the Earth measured distance between event A, the sending of the first message and Earth.

The Earth observers see the rocket watch run slow so that the time between the sending of the two messages, according to Earth observers is larger by the shrinkage factor. According to Earth observers the time between messages is . During that time, the rocket moves an extra distance d = v away from Earth. Therefore the second message has to travel a distance D + d to reach Earth. That message starts at Earth time and takes (D + d)/c to reach Earth, therefore

t₂ = + .

Δt_received = t₂ – t₁ = (1 + ) =

Changing frequencies

f_received = f_sent .

9.14) Yes my answer for 9.13) is equivalent to changing v to –v in equation 9.10.