Signals sent during relativistic space travel

[Oijl]

Homework Statement

Suppose rocket traveler Amelia has a clock made on Earth. She flies to and back from a planet 12 light-years away (as measured from rest with respect to Earth) from Earth at a speed of 0.6c. Every year she sends a signal to Earth. How many signals does Earth receive by the time she gets back?

Homework Equations

f = f'*[sqrt((1-u/c)/(1+u/c))]
(relativistic Doppler shift)
L = Lo*sqrt(1-u^2/c^2)
(relativistic length contraction)

The Attempt at a Solution

The distance to the planet for Amelia is shorter than 12 light-years. It is 9.6 light-years. So she takes 16 years (as she measures them) to get there and 16 to get back.

On the way there, by the relativistic Doppler equation I know that Earth receives her signals at a rate of 0.5 signals every Amelia-year, and on her way back Earth receives her signals at a rate of 2 signals every Amelia-year.

There are 16 Amelia-years in the trip to the planet, and 16 in the trip from the planet. So the total number of signals is

16*0.5 + 16*2 = 8 + 32 = 40.

But that is wrong, because it should be 32 signals, if she sent one every year, since her trip lasted 32 years to her.

How am I doing this wrong?
(Thanks.)

 

[HallsofIvy]

You calculated the distance in Amelia's frame of reference but did not account for the slowing of time in her frame.

 

[Oijl]

Ta-Da! Thank you, Halls of Ivy, that was kind of you. That's a load off my mind.

 

[thelahana]

I'm dabbling over the same question but how would that time dilation effect be? Would it be affecting the 32 years time? or, since the Earth has a relativistic speed of 0.6c relative to Amelia, should we consider the effect on Earth's time?

Thanks

 

[rwisz]

It seems that you have made a mistake in your calculations for the relativistic Doppler shift. The correct equation to use in this situation is f' = f*sqrt((1+u/c)/(1-u/c)), where f is the frequency of the signal as measured by Amelia and f' is the frequency of the signal as measured by Earth. This means that on her way to the planet, Earth would receive her signals at a rate of 2 signals every Amelia-year, and on her way back, Earth would receive her signals at a rate of 0.5 signals every Amelia-year. This would result in a total of 32 signals received by Earth by the time Amelia returns.

Additionally, it is important to note that time dilation and length contraction only affect the measurements of time and distance for an observer moving relative to another. In this case, Amelia would experience time passing normally and would measure the distance to the planet as 12 light-years, while an observer on Earth would measure her trip as lasting 32 years and the distance to the planet as 9.6 light-years. However, this does not change the fact that Amelia would still send one signal every year according to her own clock, resulting in a total of 32 signals received by Earth.

I hope this helps clarify your understanding of relativistic effects in space travel. Keep up the good work!