Geodesic applied to twins paradox

[Garth]

Let me give a realistic gedanken example, which is realistic in the sense that it can be constructed in standard GR cosmology without any exotic topologies, worm holes etc.

Consider a model with greater than closure density, $\Omega_{Total}$> 1, in the slowly contracting phase of its history. (You may have to introduce enough DE so the universe is small enough to circumnavigate yet only slowly contracting.)

Consider deep intergalactic space far away from local gravitational fields where the typical density is that of the cosmological average so only cosmological curvature is significant.

Twin inertial observers, Alice and Bob, pass close by each other at high mutual velocity, each thinking they are at rest and the other is traveling fast. They set their (identical) clocks at the first encounter.

Their paths cross again at a second close encounter after one of them has circumnavigated the universe, but which one; is it Alice or is it Bob?

The topological answer is the one that circumnavigates the universe is the one with the greater ‘winding number’, but both Alice and Bob think their winding number is zero and the other’s is one.

The only way to work out which one is actually circumnavigating the universe and will actually experience the greater lapse of proper time between encounters is by referring to the distribution of matter and its average momentum in the rest of the universe - a Machian-type resolution.
Garth

 

[robphy]

Let me give a realistic gedanken example, which is realistic in the sense that it can be constructed in standard GR cosmology without any exotic topologies, worm holes etc.

Consider a model with greater than closure density, $\Omega_{Total}$> 1, in the slowly contracting phase of its history. (You may have to introduce enough DE so the universe is small enough to circumnavigate yet only slowly contracting.)

Consider deep intergalactic space far away from local gravitational fields where the typical density is that of the cosmological average so only cosmological curvature is significant.

Twin inertial observers, Alice and Bob, pass close by each other at high mutual velocity, each thinking they are at rest and the other is traveling fast. They set their (identical) clocks at the first encounter.

Their paths cross again at a second close encounter after one of them has circumnavigated the universe, but which one; is it Alice or is it Bob?

The topological answer is the one that circumnavigates the universe is the one with the greater ‘winding number’, but both Alice and Bob think their winding number is zero and the other’s is one.

Why would "both Alice and Bob think their winding number is zero and the other’s is one"?
In an earlier thread (mention by pervect in https://www.physicsforums.com/showthread.php?p=1535007#post1535007), they can exchange light signals and (after waiting sufficiently long , possibly before they meet) distinguish themselves based on the reception of light signals. https://www.physicsforums.com/showthread.php?p=367371#post367371

The only way to work out which one is actually circumnavigating the universe and will actually experience the greater lapse of proper time between encounters is by referring to the distribution of matter and its average momentum in the rest of the universe - a Machian-type resolution.

Garth

Well... the simplest and most direct measurement is that each can look at their wristwatches at the meeting events... i.e. the proper time elapsed on their respective worldlines. No matter sources needed.

 

[Garth]

Why would "both Alice and Bob think their winding number is zero and the other’s is one"?

Because without reference to the rest of the universe each thinks that they are stationary and the other moving.

In an earlier thread (mention by pervect in https://www.physicsforums.com/showthread.php?p=1535007#post1535007), they can exchange light signals and (after waiting sufficiently long , possibly before they meet) distinguish themselves based on the reception of light signals. https://www.physicsforums.com/showthread.php?p=367371#post367371

Measuring the time taken for their light beam to circumnavigate the universe and return to them is one way of probing the mass and momentum distribution of the rest of the universe, because the topology that determines the outcome of the experiment is itself determined by that distribution.

Well... the simplest and most direct measurement is that each can look at their wristwatches at the meeting events... i.e. the proper time elapsed on their respective worldlines. No matter sources needed.

Of course only one observer will measure the greatest time elapse, but the only way to determine beforehand which one that will be is to look at the rest of the universe.

See Uzan et al. paper
Twin paradox and space topology

Thus in Friedmann–Lemaıtre universes, (i) the expansion of the universe and (ii) the existence of a non–trivial topology for the constant time hypersurfaces both break the Poincare invariance and single out the same “privileged” inertial observer who will age more quickly than any other twin: the one comoving with the cosmic fluid – although aging more quickly than all her traveling sisters may be not a real privilege!

(emphasis mine)

To distinguish the "privileged" observer requires knowing the 4-vector of the cosmic fluid.

Garth

 

[robphy]

Why would "both Alice and Bob think their winding number is zero and the other’s is one"?

Because without reference to the rest of the universe each thinks that they are stationary and the other moving.

I don't think either of them can say anything about their winding number...without global information about the spacetime.

Measuring the time taken for their light beam to circumnavigate the universe and return to them is one way of probing the mass and momentum distribution of the rest of the universe, because the topology that determines the outcome of the experiment is itself determined by that distribution.

Two vacuum (i.e. zero stress energy) spacetimes can have different topologies... e.g. Minkowski and the cylindrical-Minkowski. So, matter plays no role in determining the topologies of these spacetimes.

 

[Garth]

I don't think either of them can say anything about their winding number...without global information about the spacetime.

If each thinks they are stationary then each believes that they are not the one circumnavigating the universe.

Two vacuum (i.e. zero stress energy) spacetimes can have different topologies... e.g. Minkowski and the cylindrical-Minkowski. So, matter plays no role in determining the topologies of these spacetimes.

It is the cylindrical-Minkowski that I would consider to be unphysical, i.e. not realisable in any "real" universe, because of this paradox.

Garth

 

[lightarrow]

Sorry if I "dare" expressing my opinion about such difficult subject; I want to remark it's just an opinion.

Maybe, the cosmological Twin Paradox could be resolved if it would be possible to modify SR, so that time dilation formula could become "anysotropic": time of an approaching ref. frame pass faster, time of a receding one passes slower. At the beginning, before reaching their maximum relative distance, both starships observe the other's time going slower, but when they start to approach in the second-half of the trip, they see the opposite, so when they meet again, they have the same age.

I have this idea from doppler effect.

 

[robphy]

Sorry if I "dare" expressing my opinion about such difficult subject; I want to remark it's just an opinion.

Maybe, the cosmological Twin Paradox could be resolved if it would be possible to modify SR, so that time dilation formula could become "anysotropic": time of an approaching ref. frame pass faster, time of a receding one passes slower. At the beginning, before reaching their maximum relative distance, both starships observe the other's time going slower, but when they start to approach in the second-half of the trip, they see the opposite, so when they meet again, they have the same age.

I have this idea from doppler effect.

If they meet again with the same age, that effectively reduces the situation to the non-relativistic Galilean case, where the elapsed time between two events is independent of the spacetime-path.

(There may also be a problem with the time-reversed scenario.)

 

[lightarrow]

If they meet again with the same age, that effectively reduces the situation to the non-relativistic Galilean case, where the elapsed time between two events is independent of the spacetime-path.

Can't understand why. In the "vanilla" twin paradox there would be however an asymmetry and so dependence on the spacetime-path.

In the cosmological TP instead, the two starship seems to me in a complete simmetrical situation.

 

[Garth]

In the cosmological TP instead, the two starship seems to me in a complete simmetrical situation.

Which is precisely the problem, 'blind' twins are in a symmetrical situation at each close encounter.

NB: 'Blind: refers to the fact that cannot see out of their own 'small enough' laboratories. The close encounters goes through both such laboratories and they both remain at all times in inertial frames of reference.

The resolution of the paradox is to break the symmetry by observing, one way or another, the distribution of matter in motion, and therefore topology, of the rest of the universe.

Garth

 

[robphy]

Can't understand why. In the "vanilla" twin paradox there would be however an asymmetry and so dependence on the spacetime-path.

In the cosmological TP instead, the two starship seems to me in a complete simmetrical situation.

Here's is the spacetime diagram I drew in https://www.physicsforums.com/showthread.php?p=367371#post367371

$$\] \begin{picture}(140.00,150.00)(0,0) \multiput(0.00,100.00)(0.12,-0.12){833}{\line(1,0){0.12}} \multiput(0.00,100.00)(0.12,0.12){417}{\line(1,0){0.12}} \multiput(0.00,140.00)(0.12,-0.12){1167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,-0.12){167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,0.12){1083}{\line(1,0){0.12}} \multiput(0.00,30.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(0.00,60.00)(0.12,-0.12){500}{\line(1,0){0.12}} \multiput(0.00,60.00)(0.12,0.12){750}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.12){333}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.18){333}{\line(0,1){0.18}} \multiput(110.00,150.00)(0.12,-0.12){250}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.12){1000}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.18){833}{\line(0,1){0.18}} \multiput(30.00,150.00)(0.12,-0.12){917}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.12){667}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(70.00,150.00)(0.12,-0.12){583}{\line(1,0){0.12}} \put(20.00,0.00){\line(0,1){150.00}} \put(60.00,0.00){\line(0,1){150.00}} \put(100.00,0.00){\line(0,1){150.00}} \put(140.00,0.00){\line(0,1){150.00}} \put(35.99,24.05){\circle{2.00}} \put(51.97,48.03){\circle{2.00}} \put(67.89,72.11){\circle{2.00}} \put(83.95,96.05){\circle{2.00}} \put(100.26,119.47){\circle{2.00}} \end{picture} \[$$Although the two observers appear to be identical at the start,
after some time before they reunite, they will receive different patterns of light signals sent by the other.
So, they are not completely symmetrical.

 

[lightarrow]

Here's is the spacetime diagram I drew in https://www.physicsforums.com/showthread.php?p=367371#post367371

$$\] \begin{picture}(140.00,150.00)(0,0) \multiput(0.00,100.00)(0.12,-0.12){833}{\line(1,0){0.12}} \multiput(0.00,100.00)(0.12,0.12){417}{\line(1,0){0.12}} \multiput(0.00,140.00)(0.12,-0.12){1167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,-0.12){167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,0.12){1083}{\line(1,0){0.12}} \multiput(0.00,30.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(0.00,60.00)(0.12,-0.12){500}{\line(1,0){0.12}} \multiput(0.00,60.00)(0.12,0.12){750}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.12){333}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.18){333}{\line(0,1){0.18}} \multiput(110.00,150.00)(0.12,-0.12){250}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.12){1000}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.18){833}{\line(0,1){0.18}} \multiput(30.00,150.00)(0.12,-0.12){917}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.12){667}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(70.00,150.00)(0.12,-0.12){583}{\line(1,0){0.12}} \put(20.00,0.00){\line(0,1){150.00}} \put(60.00,0.00){\line(0,1){150.00}} \put(100.00,0.00){\line(0,1){150.00}} \put(140.00,0.00){\line(0,1){150.00}} \put(35.99,24.05){\circle{2.00}} \put(51.97,48.03){\circle{2.00}} \put(67.89,72.11){\circle{2.00}} \put(83.95,96.05){\circle{2.00}} \put(100.26,119.47){\circle{2.00}} \end{picture} \[$$


Although the two observers appear to be identical at the start,
after some time before they reunite, they will receive different patterns of light signals sent by the other.
So, they are not completely symmetrical.

Ok. Now forgive me if I keep saying strange ideas: in my very personal opinion, that fact, maybe, proves there is something to change in our descriptions with SR, because it would be more intuitive to me if there was complete symmetry between the two observers (and so no age difference every time they reunite). Even if there was a preferred ref frame in such a universe, as Garth propose, it seems to me we could always think of the two observers moving in opposite directions with respect to that frame so making the situation symmetric. But probably I'm saying just a lot of stupid things.

 

[Garth]

Here's is the spacetime diagram I drew in https://www.physicsforums.com/showthread.php?p=367371#post367371

$$\] \begin{picture}(140.00,150.00)(0,0) \multiput(0.00,100.00)(0.12,-0.12){833}{\line(1,0){0.12}} \multiput(0.00,100.00)(0.12,0.12){417}{\line(1,0){0.12}} \multiput(0.00,140.00)(0.12,-0.12){1167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,-0.12){167}{\line(1,0){0.12}} \multiput(0.00,20.00)(0.12,0.12){1083}{\line(1,0){0.12}} \multiput(0.00,30.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(0.00,60.00)(0.12,-0.12){500}{\line(1,0){0.12}} \multiput(0.00,60.00)(0.12,0.12){750}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.12){333}{\line(1,0){0.12}} \multiput(100.00,0.00)(0.12,0.18){333}{\line(0,1){0.18}} \multiput(110.00,150.00)(0.12,-0.12){250}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.12){1000}{\line(1,0){0.12}} \multiput(20.00,0.00)(0.12,0.18){833}{\line(0,1){0.18}} \multiput(30.00,150.00)(0.12,-0.12){917}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.12){667}{\line(1,0){0.12}} \multiput(60.00,0.00)(0.12,0.18){667}{\line(0,1){0.18}} \multiput(70.00,150.00)(0.12,-0.12){583}{\line(1,0){0.12}} \put(20.00,0.00){\line(0,1){150.00}} \put(60.00,0.00){\line(0,1){150.00}} \put(100.00,0.00){\line(0,1){150.00}} \put(140.00,0.00){\line(0,1){150.00}} \put(35.99,24.05){\circle{2.00}} \put(51.97,48.03){\circle{2.00}} \put(67.89,72.11){\circle{2.00}} \put(83.95,96.05){\circle{2.00}} \put(100.26,119.47){\circle{2.00}} \end{picture} \[$$Although the two observers appear to be identical at the start,
after some time before they reunite, they will receive different patterns of light signals sent by the other.
So, they are not completely symmetrical.

The two observers obviously are not symmetrical, but they don't know it without 'looking outside'.

As we have discussed in the thread you linked to, the paradox probes our understanding of the Equivalence Principle. i.e. How does the external field differentiate between local inertial frames of reference that are momentarily 'close enough' for the EEP to apply?

In your experiment the sending out of light signals probes the topology of the rest of the universe, and that topology I would argue is determined by the distribution of the matter and energy within it. (I know that in the hypothetical case you are considering a featureless compact cylindrical space, but I consider that not to be a realistic scenario. But then again that is just the Machian in me rising to the surface!)

Garth

 

[robphy]

Ok. Now forgive me if I keep saying strange ideas: in my very personal opinion, that fact, maybe, proves there is something to change in our descriptions with SR, because it would be more intuitive to me if there was complete symmetry between the two observers (and so no age difference every time they reunite). Even if there was a preferred ref frame in such a universe, as Garth propose, it seems to me we could always think of the two observers moving in opposite directions with respect to that frame so making the situation symmetric. But probably I'm saying just a lot of stupid things.

Let's not forget that, in addition to the "principle of relativity" (due to Galileo), there is the "constancy of the speed of light" (due to Einstein). In the diagram above, the slant of the worldlines of light-rays have to be maintained in any attempt to "twist" the cylinder to make the slanted worldline of the observer with winding number 1 into a worldline that is vertical (like the worldline of the observer with winding number 0)... ... but that can't happen.