In a flat space, where special relativity works, a travelling body can only return to the same point if we apply some kind of acceleration to the body.

So twin paradox is not a paradox because a travelling body that returns to the same point where it starts is not an inertial reference.

But then we have general relativity, that states that mass (energy) curves space-time. So when a photon changes its trajectory passing nearby sun, it's actually moving in a straight line, but in a curved space-time.

Now we can create a new version of twin paradox, where the spaceship that carries one of the twins uses the curvature of some astronomical body (like Jupiter, sun or a black hole) to return to the same point?

In that case which twin will be older?

How to solve the paradox in this context?


marked as duplicate by WillO, Jon Custer, David Z Nov 16 '18 at 7:24

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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Nov 16 '18 at 7:26
  • $\begingroup$ Can anyone explain me why this is duplicate? The first question is talking about closed universe which leads to a completely different answer, and the second is absolutely a mess... very difficult to understand moderator decision $\endgroup$ – Marco Nov 19 '18 at 19:22

In that case which twin will be older?

Each twin experiences a proper time $\int ds$, where the integral is taken along their world-line. In general, this is all we can really say. However, in the case of a static spacetime, you can define a gravitational potential and then analyze the proper time in terms of two terms, a kinetic term (special-relativistic $\gamma$) and a gravitational one (proportional to the potential).

How to solve the paradox in this context?

The SR paradox occurs if we assume, erroneously, that there is symmetry between the twins. The SR paradox is resolved because the world-lines are different. The symmetry fails because they're distinguishable: only one of them is inertial.

The GR version you've posed is resolved in the same way: the world-lines are distinguishable (although both are inertial), so integrating $\int ds$ along them gives different answers. For instance, one could orbit the earth 47 times in an elliptical orbit, while the other orbits it 10 times in a circular orbit. The orbits intersect at the beginning and end.

  • $\begingroup$ Can't both be inertial, like in two orbits that cross each other? $\endgroup$ – Wolphram jonny Nov 15 '18 at 16:33
  • $\begingroup$ @Wolphramjonny: Yes. I've added a little more explanation to try to make this clearer. $\endgroup$ – Ben Crowell Nov 15 '18 at 16:43
  • $\begingroup$ That now both are inertial, will each other see the other's clock running slow, or is that no longer valid? $\endgroup$ – Wolphram jonny Nov 15 '18 at 16:43
  • $\begingroup$ @Wolphramjonny: I realized after the fact what you meant. Let me know if my edit helps. $\endgroup$ – Ben Crowell Nov 15 '18 at 16:45
  • 3
    $\begingroup$ suppose they are in the same circular orbit, just moving in opposite directions. Both the potential term and the kinetic term are the same, both should see each other's clock running slow, and if not, why not? where is the symmetry broken? there is no difference in gravitational potential, both are in inertial frames, and none of them switches inertial frames. $\endgroup$ – Wolphram jonny Nov 15 '18 at 17:51

First, there is nothing surprising or paradoxical about two geodesic paths from $A$ to $B$ having different lengths. From a point on the equator, you can travel 1/4 of the way around the world to the North Pole while your twin travels 3/4 of the way around the world to the North Pole. Even though you both start and stop at the same places, your odometers show different lengths for your journey.

In the same way, if two travelers take different geodesic paths from one event to another, there's no reason those two paths should have the same length. A clock is a spacetime odometer, so there's no reason their clocks should show the same elapsed time. Which twin is older? The one who followed the longer path through spacetime. Still no paradox.

It's been suggested in the comments that there might be a paradox in the "fact" that Bob's clock always runs slow in Alice's (fixed) reference frame and vice versa. But we're in a curved spacetime, so there are no global reference frames.

So I'm left wondering where the alleged "paradox" is supposed to be.

  • $\begingroup$ Thanks a lot for your answer now I understood very well what you mean with geodesic paths $\endgroup$ – Marco Nov 15 '18 at 20:50
  • $\begingroup$ Sorry for my confusion... hopefully someone will clarify some concepts like I did $\endgroup$ – Marco Nov 15 '18 at 20:50
  • $\begingroup$ Perfect example of the odometer... ;) $\endgroup$ – Marco Nov 15 '18 at 21:14
  • $\begingroup$ Aren't geodesics generically unique though? A perfect sphere isn't a very realistic example. Your answer isn't wrong, but it seems to me to be missing the point, which is that there very rarely are two different geodesics paths from one event to another. $\endgroup$ – tparker Nov 16 '18 at 4:30

Paradox, noun: a seemingly absurd or self-contradictory statement or proposition that when investigated or explained may prove to be well founded or true.

The Twin Paradox is most certainly that.

Before bringing general relativity (GR) into the mix, it's important to understand exactly what the paradox is, as it is much more than just an asymmetry between the twins' world-lines.

In the simplest formulation of the problem (and since it's a thought experiment: why complicate it?), twin A stays at home in his stationary pool--training, while twin B sets off running at near the speed of light. (I'm using triathletes because of the OP's profile).

During the marathon journey, twin A sees twin B aging more slowly. Meanwhile twin B sees twin A aging more slowly. This is already a paradox for some, but easily resolved with the Lorentz transformation.

Now comes the all important "transition" portion of the training: run-to-bike. Twin B is good, he can do it in 0 seconds. Just like that he is headed home on his Zipp, at near the speed of light. Note that twin A looks out and agrees: wow: that exchange took 0 seconds--in fact all observers agree, it was instantaneous.

On the ride home, twin A sees twin B aging more slowly, and vice versa; yet when they meet up, twin B is much younger.

Now that is a paradox: if they always see the other aging more slowly:

(1) how can that be in any case?

(2) how come twin B aged less?

Well, people say it's an asymmetry, or there's acceleration. All true: but the elapsed transition time was 0 seconds in each frame. How can zero seconds in each frame account for years of difference between the frames?

The resolution comes in the relativity of simultaneity: when twin B transitions, his definition of "now" on Earth makes a great leap forward--years. Since all this is occurring outside his light cone, it has no effect on him--he turns around and "computes" that far outside his light cone, his brother is now much older, and when he gets home, he discovers he was right.

In summary: A and B always see each other aging more slowly. When B turns around, A says his clock advances 0 seconds, and he says B's clock advances 0 seconds (though B is far outside his light cone, and he only learns of this later). Meanwhile B says his clock advances zero seconds, but he computes that A's clock has advanced years. Now for B this may seem entirely as an abstraction, much like the Andromeda paradox: when you hang a U-turn on the autobahn, "now" on Andromeda can change years--but what does that mean to you: nothing.

Using curved space time do reduce g-forces during the run-to-bike transition has no effect on the paradox.

  • 2
    $\begingroup$ I think the idea was that in curved spacetime they can return to each other without changing reference frames , each observer staying in its own inertial frame. That adds a twist because you cannot solve the paradox using a switch in reference frames $\endgroup$ – Wolphram jonny Nov 15 '18 at 17:42
  • $\begingroup$ @Wolphramjonny I think it either obfuscates a great thought experiment, or there is no paradox, since a priori parallel transport is path dependent in curved geometries--so the observation doesn't seem to contradict anything. $\endgroup$ – JEB Nov 15 '18 at 18:46
  • $\begingroup$ I consider it better to actually use transition at high but finite acceleration. Because special relativity does define how time dilation behaves in that case, and due to the way general relativity is derived from it, it translates into gravitational time dilation there. $\endgroup$ – Jan Hudec Nov 15 '18 at 21:00
  • $\begingroup$ @JanHudec It doesn't matter because all observers agree the turn around takes 0 seconds. If A says it takes 0 seconds and B (and C, D, E, and F) say it takes 0 seconds: how can it account for 5 missing years, whether it's dilated or not: $ 0 = \gamma 0 = 0/\gamma $. $\endgroup$ – JEB Nov 16 '18 at 3:26
  • $\begingroup$ @JEB, if it takes 0 seconds, then there is discontinuity in the observed time at distance. Which follows from the rules, but is rather contrived, because a realistic transition can't take 0 seconds. My point is that when considering a more realistic transition that takes more than 0 seconds the time dilation is there and makes the transformation fit—and that it translates well to the GR case. $\endgroup$ – Jan Hudec Nov 16 '18 at 19:42

Lets consider slightly more realistic version where the course reversal happens at high but finite acceleration to avoid any unphysical discontinuities.

So how is the course reversal achieved by the moving brother? There is really just one option: he flies a parabolic semi-orbit around extremely heavy body.

That means he descends into a gravitational well near the periapsis, and due to gravitational time dilation his time slows down and he sees the his static brother age fast!

Note that in special relativity when accelerating, you see time flowing faster ahead of you in the direction of acceleration, which is how the moving brother sees the static brother age during the course reversal. And since general relativity is defined from special relativity by equivalence between uniform gravitational field and acceleration of the reference frame, it is logical to expect the time dilation be the same in both cases.


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