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Two twins, Nadia and Aidan, decide to have an adventure when they turn 21. Aidan chooses to travel to a distant star 10 light years away at a speed of 0.8c. Nadia decides to travel to a closer star, which is 8 light years away. How fast must Nadia travel to and from the closer star so that she is the same biological age as Aidan once they both return to Earth?

My attempt: In Aidan's frame of reference, the perceived distance of 10 light years is contracted: $L'= L/\gamma$, where $L$ = proper length and $\gamma = 1/\sqrt{1-v^2/c^2}$

So $L' = 10 \times \sqrt{1-0.64} = 6$. The total distance Aidan travels is 12m. So when he returns back on Earth, he will be $12/0.8 = 15$ years older.

For Nadia, the total distance she travels is: $2 \times 8 \times \sqrt{1-v^2/c^2} = 16 \times \sqrt{1-v^2/c^2}$

In order to return to earth 15 years older, her speed has to be: $16 \times \sqrt{1-v^2/c^2}/v = 15$

Solving for v gives $16c/\sqrt{225c^2-256}$

Is this the correct approach to solving the problem?

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closed as off-topic by user10851, Danu, Kyle Kanos, Ali, Brandon Enright Sep 16 '14 at 3:47

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    $\begingroup$ well, your answer adds something that has units $c^{2}$ to something that does not (256), so there is probably a problem. $\endgroup$ – Jerry Schirmer Sep 15 '14 at 23:25
  • $\begingroup$ There's a bit of a trick to this question, hidden in the words "once they both return to Earth." 25 years will pass on Earth during Aidan's trip. The same will happen with Nadia's trip if she travels at $\frac{16}{25} c$. However, this is not the answer as this will have Nadia aging 19.2 years compared to Aidan's 15. One of the two has to come back to Earth first and then wait for the other to return. During that wait time, the first to return will age at an undilated rate. You need to account for this wait. $\endgroup$ – David Hammen Sep 16 '14 at 1:27
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Is this the correct approach to solving the problem?

The result you give for $v$ can't be correct since, within the parenthesis, you're subtracting a number from a speed squared.

There's a much cleaner approach to the problem that does not require length contraction, time dilation, or the Lorentz factor.

The best and most correct approach, in my opinion, is to use the invariant interval.

Assuming that, for either Aidan or Nadia, the outbound and inbound speed are identical, the proper time for either is simply twice the proper time for the outbound trip.

Thus, Aidan ages (we use units where $c = 1$)

$$\Delta \tau_A = 2 \sqrt{(\Delta t_A)^2 - (\Delta x_A)^2} = 2\sqrt{\left(\frac{10}{0.8}\right)^2 - 10^2}$$

and Nadia ages

$$\Delta \tau_N = 2 \sqrt{(\Delta t_N)^2 - (\Delta x_N)^2} = 2\sqrt{\left(\frac{8}{v_N}\right)^2 - 8^2}$$

Setting these equal so that both age the same amount of time yields a straightforward equation for Nadia's speed $v_N$

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  • $\begingroup$ This may not be the right answer, depending on how one interprets "once they both return to Earth". This is the speed at which Nadia needs to go to age 15 years during her trip. Assuming Aidan and Nadia start their journeys at the same time, Nadia will return about 3 years before Aidan, and hence will be about 3 years older than Aidan once they both return to Earth. $\endgroup$ – David Hammen Sep 16 '14 at 0:33
  • $\begingroup$ @DavidHammen, I agree that there is some ambiguity. I've answered while interpreting the question as the OP evidently has. Perhaps the OP will clarify. $\endgroup$ – Alfred Centauri Sep 16 '14 at 0:58

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