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I'm trying to wrap my mind around the twin paradox, but I can't figure out this one problem from my textbook. It uses the relativistic Doppler effect to demonstrate how the paradox works. The first part of the problem is as follows:

Amelia and Casper are twins. Amelia is going on a trip to a distant planet while Casper remains on Earth. The planet is about $12$ light-years from Earth, traveling at $0.6c$. Therefore from Casper's perspective his sister's trip lasts 40 years (20 years outbound and 20 years returning). From Amelia's perspective the trip only lasts 32 years (16 years outbound and 16 years returning). Casper sends a message to Amelia once a year on his birthday. The frequency of these messages will be Doppler shifted as follows:

$$\text{outbound trip:} \\ 0.5yr^{-1}=1yr^{-1}\sqrt{\frac{1-u/c}{1+u/c}}$$

$$\text{return trip:} \\ 2yr^{-1}=1yr^{-1}\sqrt{\frac{1+u/c}{1-u/c}}$$

So Amelia will receive 8 messages on her outgoing trip ($0.5yr^{-1}*16yr)$ and 32 messages on her return trip ($2yr^{-1}*16yr$), totalling to the 40 messages that were sent by Casper.

Now the question asks me to calculate how many messages Casper would receive if Amelia sent a message to him on each of her birthdays. And my calculations are as follows: on the outgoing trip, Casper will receive $10$ messages on Amelia's outgoing trip($0.5yr^{-1}*20yr$), and $40$ messages on her return trip ($2yr^{-1}*20yr$). But that totals to $50$ messages, which is more than what Amelia sent out. Why are my calculations coming out this way?

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1 Answer 1

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There are a couple of solutions based on how we interpret the sentence "Amelia sent a message to him on each ofhisbirthday".

1. If Amelia wanted to make sure that Casper receives the messages on HIS 40 birthdays :
She must do some calculations by using her special-and-general-relativity knowledge. First, she calculates his total age throughout the entire journey and it turns out to be 40 (as in the question). Now She does the same for her age and finds it going to be 32. She understands the symmetry is broken here because this is not special relativity anymore! At half way of her journey, she has to brake out of her inertial frame and accelerate back. OK, so she knows all these stuff and then does the ingenious task of finding out his birthdays while sitting in her own high-speed frame. This is how she does it:

  • I must send messages at different rates for my away journey and return journey, otherwise Doppler shift might mess things up!
  • During my away journey his 1 year is my 0.8 years
    $t_{J} = 1year\times \sqrt{1-0.36}=0.8years$
    So if I sent him messages for 16 years he must get a total of 20 messages ($\frac{16} {0.8}$)

  • But there is going to be a Doppler shift which changes my message-rate of 1msg every 0.8 years to
    $\frac{1 msg}{0.8 yr} \sqrt{\frac{1-0.6}{1+0.6}} = \frac{0.625msg}{yr}$

  • This means he will get a message every 1.6 years (after his birthday! boo)
    $\frac{0.625msg}{yr} = \frac{1msg}{1.6 yr}$

  • So, I must increase my message frequency so that he gets it in time, $F_{new}=\frac{F_{correct}}{\sqrt{\frac{1-0.6}{1+0.6}}}=\frac{1}{0.5}= \frac{2msg}{yr}$

    Now he will get 1 msg every year, yay!
    $\frac{2 msg}{yr} \sqrt{\frac{1-0.6}{1+0.6}} = \frac{1msg}{yr}$
    Note: She will send 32 msgs in 16 years ($16\times2$) and 8 more in the return journey. You can easily change the equations above for a blue-shift and workout the frequency. Otherwise, it might elongate the answer (you can guess it's going to be a small frequency since only 8 msgs are pending and msgs travel faster now)

2. If Amelia did the same as Casper did (How you solved the problem):
She sends messages every year (his 1.25 years). He get's them after a red shift of:
$$\frac{1 msg}{yr} \sqrt{\frac{1-0.6}{1+0.6}} = \frac{0.5msg}{yr}$$
She sends 16 messages in the first 16 years (he receives only 8) and then on her way back she repeats it (but now with a blue-shift):
$$\frac{1 msg}{yr} \sqrt{\frac{1+0.6}{1-0.6}} = \frac{2msg}{yr}$$
Thus, in that 20 years period (her return journey of 16 years) he gets 16 such messages along with those remaining to reach him from the away-trip. This adds up to:
$8\ delayed\ msgs + 8\ more\ delayed\ msgs + 16\ really\ fast\ msgs = 32\ total\ msgs$
Note: Interestingly, this explains that classical who aged more "paradox"! The switching of frames by Amelia is the reason why her brother aged (he did not accelerate, she did). If she was to go on forever with out stopping, both of them would think the other one is younger. But she did stop and came back, thus entering a new frame, which treated her brother as old.
Also, the first 8 messages and the next 8 ones were send to him in a red shift journey. Meanwhile she is on her return journey. 16 messages send during this blue-shift journey will reach him really fast and obviously all of them (32) will be messed up and not synchronous at all! Chances are none of these messages from his sister will actually reach him on his real birthday!

You got the wrong answer because you simply did this:
$\frac{40}{32}=1.25$
$1.25\sqrt{\frac{1-0.6}{1+0.6}} = 0.625$
$0.625\times16=10$
$1.25\sqrt{\frac{1+0.6}{1-0.6}} = 2.5$
$2.5\times16=40$
$10+40 = 50$
1.25 is his years, not hers. Applying it in Doppler shift gives her receiving rate on the R.H.S.

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Brilliant, thanks! My writing "his" was a typo. It was supposed to say "her". I edited. –  Ataraxia Oct 14 '13 at 15:35
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Things get complicated in relativity because we have to keep track of more than one velocities, frames, times etc. In this case there is an additional burden of Doppler-shift! I hope I made myself clear in that answer. I could update it if you need to clear any more doubts. –  user26988 Oct 14 '13 at 15:54

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