2
$\begingroup$

This question already has an answer here:

Here I state and try to answer three variations of the twin paradox

1) "Classical" problem, no acceleration, no turn around

Consider the case where there's a stationary planet, and a moving spaceship moving at close to the speed of light, starting at the left going right at constant velocity. Now imagine at some instant, two beings spawn on earth and in the spaceship. They are both aged 0 at that point. Now as the spaceship continues, both frames remain inertial. Which twin is older at any given point?

According to one of the answers here (the 4th answer down to be exact): How is the classical twin paradox resolved?, there is no solution to this paradox. Either A or B will be older depending on who you ask (person A or B or anyone else).

Is this problem reasonable and is the "solution" correct?

2) Normal problem, no acceleration, yes turn around

Everyone knows this one; the paradox is resolved because the person traveling changes direction. http://www.physicsmatt.com/blog/2017/1/18/the-twin-paradox-in-special-and-general-relativity. Been there, done that.

3) Weird problem, no acceleration, no turn around, curved universe?

Now I was thinking, "what if I combine the two?". Imagine the earth is like an asteroids game, or it's a sphere/torus. Going in a certain direction for long enough means that I eventually end up where I started. Consider the case I stated in (1), where two beings aged 0 appear on the earth and on the spaceship. The spaceship proceeds to go at a constant velocity to the right, never changing direction, never under any gravitational field, but due to the nature of the universe they are in, the spaceship twin ends up reaching the earth again. Now at this point, who is older? The solution for (2) won't work because there's no turning point, right? In the link I provided:

You do this by just putting a sheet above the one you started with (or to the side, if moving in that direction). Nothing wrong with that, and no edges to worry about crossing.

The trick is though, when the traveler gets to this new point, you yell "surprise" and identify the new point with the point in the original sheet using the symmetry of the torus.

So the poor sap thinks they're getting away from their twin, only for you to change the nice new spot they picked out for themselves to the place they started from. It's a mathematical sleight-of-hand, but it is the easiest way to see exactly who is younger in the Twin Paradox on the torus.

And my question that I don't feel like the other question sufficiently addresses:

A specification question: I don't quite see how this particular "mathematical sleight of hand" makes it so that the moving twin is aging less. Sorry if I'm being slow, but could anyone explain that part more thoroughly? I still don't see how even in an inertial constant-velocity frames one twin is aging slower.

Feel free to tell me that this problem is nonsensical and stupid--I'm a complete noob so there is a high chance that I have no idea what I'm talking about (please tell me if that's the case)

$\endgroup$

marked as duplicate by Aaron Stevens, WillO, John Rennie special-relativity Mar 18 at 7:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ So you are asking about scenario 3? $\endgroup$ – Aaron Stevens Mar 18 at 1:25
  • $\begingroup$ I'm asking if that answer that was upvoted once compared to twenty times is correct (question 1) and I'm asking about question 3. $\endgroup$ – D.R Mar 18 at 1:27
  • $\begingroup$ The page you have linked in scenario 2 discusses your scenario 3 toward the bottom. $\endgroup$ – Paul T. Mar 18 at 1:27
  • 2
    $\begingroup$ Your (1) is a non-paradox and is explained entirely in terms of the relativity of simultaneity. Learning to believe in that is non-trivial, but it is the answer. And yes, that means that it depends on who you ask. That is the nature of the universe: some questions have no definitive answer. $\endgroup$ – dmckee Mar 18 at 1:30
  • 1
    $\begingroup$ For (3), try working out the simplest example: spacetime is a flat cylinder, parameterized by $\theta$ in the circular direction and $t$ in the "vertical" direction. The metric is $ds^2=dt^2-d\theta^2$. Twin 1 travels along a line $\theta=0$, Twin 2 along a spiral $\theta=2\pi v t$. Compute the lengths of both paths from $t=0$ to $t=1/v$. The path length for either twin tells you how much that twin's clock advances. You'll generally learn a lot more working out simple examples for yourself rather than just trusting other people's answers. $\endgroup$ – WillO Mar 18 at 3:15
1
$\begingroup$

The question is not stupid. So in the normal twin paradox, the traveling twin's path includes acceleration, and hence can be distinguished from the "stationary" twin.

In the toroidal version, both twins are inertial, but the traveling twin's path cannot be continuously deformed to a point. As explained in https://arxiv.org/pdf/0910.5847.pdf, this introduces "new kind of asymmetry between the spacetime paths joining two events, an asymmetry which is not due to acceleration but to the multiply connected topology.... all the inertial frames are not equivalent, and the topology introduces a preferred class of inertial frames."

The authors go onto explain that, "The loss of equivalence between inertial frames is due to the fact that a multiply connected spatial topology globally breaks the Poincaré group."

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.