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I had received a homework problem by my professor. Please consider part (b) of the problem.

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Can part (b) be solved only using special relativity? And how can you solve this? I think that Alice will be younger but I am not really sure.

Timaeus
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rahulgarg12342
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2 Answers2

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It would certainly seem that Alice is younger, assuming that Bob is the one in the inertial reference frame described at the beginning of the problem. The logic would bethat as measured in Bob's frame, $\Delta x_B = 0$, $\Delta x_A = R$ ($R$ here being the "radius" of the Universe; in other words, Alice has "come back" to Bob's location), and $\Delta t_A = \Delta t_B$. All these together would imply that $\tau_A^2 = \tau_B^2 - R^2/c^2 < \tau_B^2$, where $\tau_A$ and $\tau_B$ are the respective proper times.

The really interesting part of this problem is the last question: "Is your answer consistent with Lorentz invariance? Describe in detail what is going on." Alice sees Bob traveling around the Universe in the opposite direction; and so she could say that Bob should be younger by exactly the same logic. (Just like she could say in part (a).) So either our logic in the previous paragraph concerning the proper times is wrong, or Lorentz invariance (i.e., symmetry between inertial reference frames) doesn't hold here for some reason. Since this is a homework problem, I won't tell you which one — but feel free to post your thoughts in the comments and I'll respond to them.

Michael Seifert
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  • I think that our logic for part (a) cannot be wrong. But I am still very confused. Is there some kind of symmetry breaking? – rahulgarg12342 Jul 27 '15 at 21:35
  • The symmetry of the system is broken because there is a preferred inertial frame. You can identify it by shining light in both directions at the same time. In the preferred frame, they arrive back at the source at the same time. Every other frame will have one arrive before the other. On the other hand, you don't know which will arrive younger since your prof didn't specify which Cosmonaut started out in that frame. – John M. Cavallo Jul 27 '15 at 22:44
  • @JohnM.Cavallo The problem never said what Bob did in either part, but Alice goes all the way around in part b. – Timaeus Jul 27 '15 at 23:29
  • @Timaeus All it really says is that they separate and, without either accelerating, they meet up again. – John M. Cavallo Jul 28 '15 at 00:59
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Twin paradox modified for a periodic universe

I do wish people would keep speculations out of physics lessons. The Planck mission found no evidence of any kind of toroidal topology. There's no evidence at all that we live in "spatially periodic" universe. Besides, the old game was Asteroids, not Pac Man.

Can part (b) be solved only using special relativity? And how can you solve this?

Yes, easily. It's easy when you understand it. See the simple inference of time dilation on Wikipedia. It uses parallel-mirror light clocks. The elapsed time is the number of reflections. That's all it is. There is no actual time passing, time is just a cumulative measure of motion. And whatever happens to the parallel-mirror light clock happens to you too, because of the wave nature of matter. When you move fast relative to me, you experience less time than me. You age less than me. The twins paradox is where we can't say who's really moving, and I say your clock goes slower than mine and you say mine goes slower than yours. That occurs when we're separated by relative velocity. But it's no more of a paradox than when we're separated by distance, when I say you look smaller than me and you say I look smaller than you. And besides, we know who moved, because one of us felt the acceleration.

I think that Alice will be younger but I am not really sure.

Alice will be younger. Once you "get under the maths" and understand it, you will be sure.

John Duffield
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