In Section 2.8 we discussed what happened when Mary traveled on a spaceship away from her twin brother Frank, who remained on Earth. Analyze the light signals sent out by Frank and Mary by using the relativistic Doppler effect. Strategy We will use Equation (2.34) for both the outbound and return

Question

In Section 2.8 we discussed what happened when Mary traveled on a spaceship away from her twin brother Frank, who remained on Earth. Analyze the light signals sent out by Frank and Mary by using the relativistic Doppler effect.

Strategy We will use Equation (2.34) for both the outbound and return trip to analyze the frequency of the light signals sent and received. During the outbound trip the source (Frank) and receiver (Mary) are receding so that β = -0.8. For the return trip, we have β = +0.8. The frequency [latex]f_{0}[/latex] will be the signals that Frank sends; the frequency f will be those that Mary receives.

[latex]f=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}} f_{0} \quad ext { Relativistic Doppler effect }[/latex] (2.34)

Accepted Answer

First, we analyze the frequency of the light signals that Mary receives from Frank. Equation (2.34) gives

[latex]f=\frac{\sqrt{1+(-0.8)}}{\sqrt{1-(-0.8)}} f_{0}=\frac{f_{0}}{3}[/latex]

Because Frank sends out signals annually, Mary will receive the signals only every 3 years. Therefore during the 6-year trip in Mary’s system to the star system, she will receive only 2 signals.

During the return trip, β = 0.8 and Equation (2.34) gives

[latex]f=\frac{\sqrt{1+0.8}}{\sqrt{1-0.8}} f_{0}=3 f_{0}[/latex]

so that Mary receives 3 signals each year for a total of 18 signals during the return trip. Mary receives a total of 20 annual light signals from Frank, and she concludes that Frank has aged 20 years during her trip.

Now let’s analyze the light signals that Mary sends Frank. During the outbound trip the frequency at which Frank receives signals from Mary will also be [latex]f_{0} / 3[/latex]. During the 10 years that it takes Mary to reach the star system on his clock, he will receive 10/3 signals—3 signals plus 1/3 of the time to the next one. Frank continues to receive Mary’s signals at the rate [latex]f_{0} / 3[/latex] for another 8 years, because that is how long it takes the sixth signal she sent him to reach Earth. Therefore, for the first 18 years of her journey, according to his own clock he receives 18/3 = 6 signals. Frank has no way of knowing that Mary has turned around and is coming back until he starts receiving signals at frequency [latex]3 f_{0}[/latex]. During Mary’s return trip Frank will receive signals at the frequency [latex]3 f_{0}[/latex] or 3 per year. However, in his system, Mary returns 2 years after he has received her sixth signal and turned around to come back. During this 2-year period he will receive 6 more signals, so he concludes she has aged a total of only 12 years.

Notice that this analysis is in total agreement with the spacetime diagram of Figure 2.26 and is somewhat easier to obtain. Although geometrical constructions like spacetime diagrams are sometimes useful, an analytical calculation is usually easier

Question 2.8: In Section 2.8 we discussed what happened when Mary traveled...

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