Exploring the Twin Paradox for Light & Sound Waves

[Killtech]

TL;DR Summary

How does SRT fix the asymmetries between inertial frames, in particular in the twin paradox?

Originally i just wanted to look at how much analogy can be made between light and sound waves using all that math has to offer to depict them in most similar framework possible - just so as to have a different perspective to understand some things better. Anyhow, no matter how well one tries to hide the (sound) medium (and that can be done pretty well), it will always be there and not allow for a perfect equivalence of inertial frames.

In order to make comparisons with light/SRT of some specific aspects i was curious about i went back to the twin paradox and looked as some special modifications to make them stand out. In particular the twin paradox in a closed world (cylinder manifold which is geometrically perfectly flat) and luckily the internet already had something on it: https://physics.stackexchange.com/questions/353216/twin-paradox-in-closed-universe
I'm not sure if the explanation is fully correct, but it surprised me as it breaks the equivalence of frames (i picked the example specifically to understand how SRT maintains the equivalence under non trivial circumstance).

The problem can be traced back to the issue that in a closed world the one way speed of light seems to be partially measurable: send two signals around the world in opposed directions and if they don't come back at the same time there is a difference in the one way speed of light along that axis. In the example in the link this causes Betties plane of const time to twist such that in her frame her other instances are at a different age on her plane of const time (due to clock synch convention).

And I don't understand how to fix this, because all waves travel at a fixed speed i.e. independently from the source they were emitted from. So if Betty and Albert emitted signals to measure the one way speed of light from Alberts place they will travel together and an asymmetry becomes visible making Alberts frame to stand out (unless a many worlds approach is taken). So Albert's frame seems to be more at rest then Betty's.

Anyhow, so i thought maybe this is a issue unique to a closed world setup. But in all versions of the twin paradox there is a age asymmetry between the "traveling" and "home" twins. I looked at the the "out and back" twin resolution at Wikipedia: https://en.wikipedia.org/wiki/Twin_paradox#A_non_space-time_approach. It needs the 3 twins A (Albert), O (Outgoing) and B (Back) i.e. trins. So to make the example easier let's copy/clone each of the trins indefinitely and arrange them in evenly spaced grids A, O, B. Now whenever a trin passes by another the times of all clocks can be recorded. In particular of interest are the time intervals $\Delta t_{I,J}$ measured by a clock at grid $I$ for the time between two meetings with a trin from grid $J$. Now let all trins be at that at $x,t=0$ point our initial condition. Furthermore the start of the return trip can be set to when O meets B the first time since $t=0$. Then I gather that the resolution implies $$\frac 1 2 \Delta t_{O,A} + \frac 1 2 \Delta t_{B,A} < \Delta t_{A,O} = \Delta t_{A,B}$$
So that would seem to imply that frames O and B are still not fully equivalent to A since $\Delta t_{A,B} \neq \Delta t_{B,A}$ even without the closed world assumption. So it would seem clocks tick indeed faster in one grid then in the other but in an absolute sense (after all the $\Delta t_{I,J}$ just compare two times of the same inertial frame clock between events at its exact location - i.e. independent of any clock synch and whatever). I also considered viewing grid O as the "stay home trin" and A as the outgoing one. That needs introducing another grid C that serves the travel back trin role for O. But doing so is a bad idea because that seems to shatter logic consistency a bit (so far i could not resolve the contradictions between the $\Delta t_{I,J}$ relations of all 4 grids).

Anyhow, the closed world case provides a mean to associated the $\Delta t_{I,J}$ asymmetry with one way speed of light difference and having that, one can add more grids moving at different velocities just to probe the asymmetry. But does that not give a means of deriving the one speed of light in all directions and deduct a frame specific offset velocity vector?

So as you can see I am at a loss now since the one way speed of light is not allowed to have any measurable effect therefore all the asymmetries cannot exist. Or to put it more provocatively: a detectible offset velocity in the one way speed of light is basically an aether wind - so i expected these asymmetries only to pop up for my sound waves analogon but not for light. I don't get where the mistake in all this is and the more I try to understand SRT the less I actually do.

 

[robphy]

Summary:: How does SRT fix the asymmetries between inertial frames, in particular in the twin paradox?

Originally i just wanted to look at how much analogy can be made between light and sound waves using all that math has to offer to depict them in most similar framework possible - just so as to have a different perspective to understand some things better. Anyhow, no matter how well one tries to hide the (sound) medium (and that can be done pretty well), it will always be there and not allow for a perfect equivalence of inertial frames.

In order to make comparisons with light/SRT of some specific aspects i was curious about i went back to the twin paradox and looked as some special modifications to make them stand out. In particular the twin paradox in a closed world (cylinder manifold which is geometrically perfectly flat) and luckily the internet already had something on it: https://physics.stackexchange.com/questions/353216/twin-paradox-in-closed-universe
I'm not sure if the explanation is fully correct, but it surprised me as it breaks the equivalence of frames (i picked the example specifically to understand how SRT maintains the equivalence under non trivial circumstance).

Here's a very old thread on the closed universe case:
https://www.physicsforums.com/threads/the-cosmological-twin-paradox.51197/post-361621

 

[PeterDonis]

The problem can be traced back to the issue that in a closed world the one way speed of light seems to be partially measurable

The root cause is actually that in the closed world, it is no longer the case that all inertial frames are equivalent globally. One particular inertial frame has the property that its surfaces of simultaneity are closed; in all other inertial frames, the surfaces of simultaneity are not closed, and how "not closed" they are varies from frame to frame. (Roughly speaking, in one particular inertial frame, in the 1x1 dimensional case--one dimension of space and one of time--the surfaces of simultaneity are circles, while in all other inertial frames, the surfaces of simultaneity are helixes; they don't close back up on themselves, and by how much they fail to do that is a finite number that varies from frame to frame.) So there is a global geometric asymmetry between inertial frames that exactly correlates to the asymmetry in aging for twins.

i thought maybe this is a issue unique to a closed world setup.

That particular issue is unique to the closed world setup, yes.

But in all versions of the twin paradox there is a age asymmetry between the "traveling" and "home" twins

That is because in all versions of the twin paradox, there is some asymmetry between the twins. It just isn't always the same asymmetry. Different versions have different asymmetries between the twins. This can cause a great deal of confusion among people who think there should be just one "trick" that resolves all possible twin paradoxes. There isn't--at least, not unless you count the general statement that does cover all cases, which is "look at the actual curves in spacetime followed by each twin and compare their lengths", as a "trick". (Most people don't appear to want to do that because the general method looks way too much like actual work instead of a "trick" or "shortcut". )

 

[PeterDonis]

I am at a loss now

My standard recommendation for twin paradox questions is to read the Usenet Physics FAQ article:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Then rethink your analysis in the light of what you've read.

Different versions have different asymmetries between the twins.

Note, btw, that one of the things the Usenet Physics FAQ article will make you realize is that, not only are there different asymmetries between the twins in different twin paradox scenarios, one can even find different asymmetries in the same scenario. As the article shows, there are several different ways of analyzing the standard twin paradox that make it seem like the asymmetry is different things.

 

[pervect]

I think that it's a dead end to expect any close analogy between sound and light in SR. My understanding is that the group structures are different, and that this shows up in the speed of light being frame independent, while the speed of sound is n ot.

However, I'm not familiar enough with the ins and outs of group theory to offer a more rigorous proof that the groups are different.

 

[Dale]

That is because in all versions of the twin paradox, there is some asymmetry between the twins. It just isn't always the same asymmetry. Different versions have different asymmetries between the twins.

That is a good point. In the case of the "repeating" universe there is an asymmetry with respect to the global rest frame.

 

[Killtech]

The root cause is actually that in the closed world, it is no longer the case that all inertial frames are equivalent globally. [...]

So this is indeed considered correct. I understood the original explanation but I was not sure if it was really correct given the loss of equivalence.

That is a good point. In the case of the "repeating" universe there is an asymmetry with respect to the global rest frame.

That particular issue is unique to the closed world setup, yes.

Can you point me to a proof of that is issue is specific to a closed universe? I mean the closed world makes the asymmetry quite obvious, but this does not automatically imply it's not there without it. proof by example does not count for me as a mathematician.

That is because in all versions of the twin paradox, there is some asymmetry between the twins. It just isn't always the same asymmetry. [...]

Yeah, I know some of the asymmetries and looked at the various versions of the twin paradox just to find they did not look at what i was interested in. I understand most of the usual explanations but am not sure how to translate it into my problem. In particular: is it possible to derive whether the SoS is closed purely from local measurements (and if so how is that proofed)?

I am looking at very particular asymmerty that i cannot name any better them $\Delta t_{B,A}<\Delta t_{A,B}$ for two frame independent scalar values. In my case I am not exactly looking at a classic twin paradox, since these quantities technically only make sense using evenly spaced grids which is not the case in any version of the paradox i know of. It's derived from a version though and most of the calculations can be taken from there for a shortcut...

My standard recommendation for twin paradox questions is to read the Usenet Physics FAQ article:
https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
Then rethink your analysis in the light of what you've read.

It's based on the outbound/inbound twin example for a start. But there is no extension of the scenario having enough grid points for the events to occur that define $\Delta t_{I,J}$. These values are however tied to the times the clocks in the original paradox show during certain events, so some of them can be taken from the original calculation.

 

[Ibix]

I think that the simplest explanation of the twin paradox is to note that the elapsed time along a path through spacetime is analogous to the length of a path through Euclidean space. So asking "why is the inertial twin oldest" is analogous to asking "why is a straight line the shortest distance between two points". I usually ask what answer you would give to that question.

Can you point me to a proof of that is issue is specific to a closed universe?

In an open universe you don't loop round and return to your starting place, so there's no frame picked out by that process.

 

[Killtech]

I think that the simplest explanation of the twin paradox is to note that the elapsed time along a path through spacetime is analogous to the length of a path through Euclidean space. So asking "why is the inertial twin oldest" is analogous to asking "why is a straight line the shortest distance between two points". I usually ask what answer you would give to that question.

except that every twin is inertial in my example and the $\Delta t_{I,J}$ treat each as such so by your words finding the oldest would already get quite tricky.

In an open universe you don't loop round and return to your starting place, so there's no frame picked out by that process.

Do you even know what the word "proof" means?

 

[Dale]

Can you point me to a proof of that is issue is specific to a closed universe?

I can’t even write down a general expression for what this “issue” is. Can you? Without that I don’t even know what it is that you want a proof of.

 

[Killtech]

I can’t even write down a general expression for what “this issue” is. Can you?

Sure: apparently in a closed world each frame has a characteristic quantity: the delay between two signals sent out in opposing directions coming back around the world to their source. Unlike the closed/open SoS this one distinguishes each frame in a much finer fashion. Show that no such characteristic quantity can be constructed locally / without the usage of the closed world property.

 

[PeterDonis]

Can you point me to a proof of that is issue is specific to a closed universe?

It follows from the topology of the closed universe: in the 1-1 dimensional case the topology of the closed universe is $S^1 \times R^1$, whereas the topology of standard 1-1 dimensional Minkowski spacetime is $R^2$. The issue I described is only present in topologies that have compact components; $S^1$ is compact, but $R^n$ is not.

If you want to try to visualize this, think of a cylinder vs. a plane. On a plane, you can draw two mutually perpendicular axes in any orientation you want and they will all be equivalent. But on a cylinder, only one orientation of the axes will result in the "horizontal" axis being a closed circle (and the "vertical" axis in this case will go straight up the cylinder without winding around it). In every other orientation, the "horizontal axis" will be a helix, winding up and down the cylinder (and the "vertical" axis will also wind around the cylinder), and the "degree of winding" will be a number that varies continuously with the orientation.

The only difference between the case I just described and the spacetime case is that "rotating" the axes is done hyperbolically, so the definition of "orthogonal" changes (roughly speaking, the "rotated" axes become a narrower and narrower rhombus instead of staying square). But everything else is the same.

in a closed world each frame has a characteristic quantity: the delay between two signals sent out in opposing directions coming back around the world to their source.

This is a side effect (and not the only one) of the underlying topology difference I described above. The underlying topology difference is the fundamental property from which all others come.

 

[PeterDonis]

is it possible to derive whether the SoS is closed purely from local measurements

No. It is a global property, not a local property. More generally, the underlying topology of the spacetime is a global property, not a local property.

I know some of the asymmetries and looked at the various versions of the twin paradox just to find they did not look at what i was interested in.

The FAQ article I linked to does not discuss the "closed universe" case, that's true. So you won't find anything specific to that issue in the article.

 

[PeterDonis]

every twin is inertial in my example

In the closed universe example, yes, all the twins can be "inertial" (as in, always at rest in the same inertial frame), but still meet more than once. That is another side effect of the underlying topology of the closed universe; in standard Minkowski spacetime with underlying topology $R^n$, it is impossible for any two observers who are inertial in this sense to meet more than once.

 

[PeterDonis]

there is no extension of the scenario having enough grid points for the events to occur that define $\Delta t_{I,J}$. These values are however tied to the times the clocks in the original paradox show during certain events, so some of them can be taken from the original calculation.

I don't know what you mean by this. Do you just mean that the article doesn't cover the closed universe scenario (which, as I posted previously, I agree it does not)? Or something else?

 

[Ibix]

except that every twin is inertial in my example and the $\Delta t_{I,J}$ treat each as such so by your words finding the oldest would already get quite tricky.

One pair of twins meet half way through your experiment. Adding the elapsed time between meetups for these two will be less than the elapsed time between meetups for the third. This is just the Minkowski equivalent of the triangle inequality. Again, I'd ask you what would you consider an acceptable response to "why is the triangle inequality true in Euclidean space"?

Sure: apparently in a closed world each frame has a characteristic quantity: the delay between two signals sent out in opposing directions coming back around the world to their source. Unlike the closed/open SoS this one distinguishes each frame in a much finer fashion. Show that no such characteristic quantity can be constructed locally / without the usage of the closed world property.

A flat Minkowski space is symmetric under four translations, three rotations, and three boosts. The "cylindrical" Minkowski spacetime is not symmetric under boost in the tangential direction (not globally, anyway). That lack of symmetry is the source of your "characteristic quantity". So if you want to find such a quantity in flat spacetime you need to find a way in which flat Minkowski spacetime violates the symmetries that define it.

Presuming that SoS is surface of simultaneity, I don't see that your characteristic quantity is any different to the failure of the planes to close. In both cases, the size of the quantity increased with increasing speed relative to the preferred frame.

 

[Dale]

the delay between two signals sent out in opposing directions coming back around the world to their source

Sure. That is easy, the two signals never return to their source in an open universe by definition. QED

I thought you wanted a general proof of some unspecified issue.

 

[Killtech]

I don't know what you mean by this. Do you just mean that the article doesn't cover the closed universe scenario (which, as I posted previously, I agree it does not)? Or something else?

Well, two similar cases actually: yes, the closed world example is one but the grid example is another.

See, I wanted to carry the closed world example to the open world scenario an make them as indistinguishable as possible. So instead of going around the world and coming back from the other side I just clone each twin and put them where each twin previously saw his own image when looking around the loop. Doing so effectively creates a grid of evenly spaced twins (when iteratively done for each twins clone):

-----> B1 -----> B2 -----> B3 ----->
------ A1 ------ A2 ------ A3 ------

Now while the outgoing twin B2 never gets to it's original home he will meet a clone of that home A3 instead. As for the twin A2 staying at home, he will never see his twin B2 again, but he will see the clone of him B1 instead. This case is different from the normal twin paradox as no twin will ever get home exactly yet the age differences between reunions $\Delta t_{I,J}$ are still available - and from the perspectives of the twins itself it's not clear if they would be able to distinguish if they are in a closed or open world within the scope of the experiment.

But if both twins frame are really equivalent, then both the outgoing twin and the home twin must measure the exact same time difference between reunions with their own clocks i.e. $\Delta t_{A,B} = \Delta t_{B,A}$ since there is no way to say which one is outgoing and which one stays. On the other hand the scenario in the closed world yields $\Delta t_{A,B} > \Delta t_{B,A}$

In my original post I used this setup, so i can take some calculations from the inbound/outbound twin variant:

-----> O1 -----> O2 -----> O3 ----->
------ A1 ------ A2 ------ A3 ------
<----- B1 <----- B2 <----- B3 <-----

 

[PeterDonis]

instead of going around the world and coming back from the other side I just clone each twin and put them where each twin previously saw his own image when looking around the loop.

This is just "unrolling the cylinder" onto a flat plane. It doesn't change the underlying topology of the closed world example, since you are treating each clone as if it were the same as the twin it's cloned from, so one twin passing by repeated clones of the other twin counts as the same thing as passing the other twin repeatedly.

from the perspectives of the twins itself it's not clear if they would be able to distinguish if they are in a closed or open world within the scope of the experiment.

If the clones are actually indistinguishable, then the twins will be able to tell they are in a closed world because from their viewpoint they are meeting the same other twin again and again; as noted above, the underlying topology is the same as the closed world so the scenario will have all the same properties, and those properties are testable.

If the clones are not indistinguishable, then your scenario is not equivalent to the closed world scenario any more, so of course it will not have the same properties.

 

[Killtech]

This is just "unrolling the cylinder" onto a flat plane. It doesn't change the underlying topology of the closed world example, since you are treating each clone as if it were the same as the twin it's cloned from, so one twin passing by repeated clones of the other twin counts as the same thing as passing the other twin repeatedly.

Yes, the topologies are clearly different. "Unrolling the cylinder" onto a plane repeatedly by creating a copy of its contents obviously creates a different topology. Take the point of where one twin and its clone is. In the closed world the clone is actually the twin itself and therefore these two points are one and the same, whereas in in the open world topology they are different with quite a distance in between them. Also if there are two spatial dimensions (like on a cylinder) a geodesics will intersect itself while it cannot ever do that in a flat open world.

If the clones are actually indistinguishable, then the twins will be able to tell they are in a closed world because from their viewpoint they are meeting the same other twin again and again; as noted above, the underlying topology is the same as the closed world so the scenario will have all the same properties, and those properties are testable.

If the clones are not indistinguishable, then your scenario is not equivalent to the closed world scenario any more, so of course it will not have the same properties.

The question of indistinguishability is unclear. All the things and events used in calculating the results of the experiments are exactly the same. There are differences, yet none of those seem to be relevant for the calculation done. If the calculations are effectively the same in both cases they cannot lead to different results.

Anyhow, consider the following: for each of the twins it takes a different amount of time to meet their sibling. so the twin that ages less in between meetings will know that the time dilation actually happens on his side and not his twin - even though he is supposedly in a rest frame. This realization however only requires exchanging data on all recorded previous meetings each of the twins had when they meet at the same point. That's the case in the closed world scenario! This is obviously deeply problematic if it happened the same way for the grid twins. So there must be a big difference in the calculation somewhere.

 

[PeterDonis]

"Unrolling the cylinder" onto a plane repeatedly by creating a copy of its contents obviously creates a different topology.

Not if the clones are indistinguishable. Then the unrolled cylinder is indistinguishable from the rolled up cylinder, and has the same topology.

in in the open world topology they are different with quite a distance in between them.

If the clone and the twin are indistinguishable, then this is not correct.

The question of indistinguishability is unclear.

No, it isn't. "Indistinguishable" means just what it says: it means "can't be distinguished". Which in turn means that you are wrong to think of things that you say are "indistinguishable" as still being "different" in some way. You removed all the differences when you said they were indistinguishable.

All the things and events used in calculating the results of the experiments are exactly the same.

In the case where the clones are indistinguishable, yes.

There are differences, yet none of those seem to be relevant for the calculation done.

No, there are no differences; you took them all away when you said the clones were indistinguishable.

If the calculations are effectively the same in both cases they cannot lead to different results.

That's true. But it only applies to the case where you said the clones were indistinguishable. If the clones are not indistinguishable, then it's no longer true that "all the things and events used in calculating the results of the experiments are exactly the same", because being able to tell the difference between one clone and another counts as an experimental result.

for each of the twins it takes a different amount of time to meet their sibling. so the twin that ages less in between meetings will know that the time dilation actually happens on his side and not his twin - even though he is supposedly in a rest frame. This realization however only requires exchanging data on all recorded previous meetings each of the twins had when they meet at the same point. That's the case in the closed world scenario!

Yes, all of this is true for the closed world scenario. Which means it is also true for the "grid scenario" where the clones are indistinguishable, since that scenario is the same scenario as the closed world scenario; you made it that way when you said the clones were indistinguishable. "Indistinguishable" means that it is impossible for any clone of either twin to tell the difference between one meeting with a clone of the other twin and another; to him, they are all meetings with the same other twin. Which means that they are all meetings with the same other twin; you've explicitly ruled out anything that could be used to tell them apart. So you don't actually have two different scenarios; you just have one scenario that you have confused yourself about by describing it in two different ways that make you incorrectly think they're somehow different, when they're actually not.

This is obviously deeply problematic if it happened the same way for the grid twins.

For the case where the grid clones are indistinguishable, no, it's not. See above.

So there must be a big difference in the calculation somewhere.

Not for the scenario where the grid clones are indistinguishable. See above.

I suspect that what is confusing you is that, in your mind, you are not allowing the grid clones to be truly indistinguishable; you are, in your mind, reserving some property as being different between them. For example, you might be thinking of them as being "at different points in space". But if that is actually the case, then there is a way to measure it. "Different points in space" has no meaning if there is literally no observable that can tell them apart. And "indistinguishable" means that there is no observable that can tell them apart. If you are thinking of them as being at "different points in space", you are not thinking of them as being indistinguishable. That is a logical contradiction; you can't have it both ways.

 

[Killtech]

I suspect that what is confusing you is that, in your mind, you are not allowing the grid clones to be truly indistinguishable; you are, in your mind, reserving some property as being different between them. For example, you might be thinking of them as being "at different points in space". But if that is actually the case, then there is a way to measure it. "Different points in space" has no meaning if there is literally no observable that can tell them apart. And "indistinguishable" means that there is no observable that can tell them apart. If you are thinking of them as being at "different points in space", you are not thinking of them as being indistinguishable. That is a logical contradiction; you can't have it both ways.

The twins become indistinguishable if the access to certain information is restricted or rather the twins are forbidden from conducting some other experiments.

Unfortunately math and logic is always quite a bit more tricky when done correctly. Thus to put this into more rigorous mathematical terms: the calculation of the age difference in the closed world scenario does make use of only some postulates/assumptions/properties of that setup, but far from all. This means that the calculation holds true for any setup which satisfies this specific smaller set of assumptions that is actually used. Now this limited set of assumption is the same for the case of the grid, which makes them indistinguishable/equivalent i.e. using these assumptions only will not allow you to tell the scenarios apart (they can indeed be considered as one).

A math analogy: any statement derived without using the AC (axiom of choice) will still hold true when either AC or its negation is added retroactively. So the single scenario where the original conclusion was done becomes two distinct ones which now differ but the conclusion itself remains valid for both. thus indistinguishability is relative to the chosen set of assumptions.

however, when calculating other things which make use of properties of the setups that were previously not used, the differences between these setups show - for example the closed SoS can only happen in the closed world setting for the topological reasons you rightly noted.

More interestingly let's observe a crucial feature of the age calculation: the twins age at their reunion is determined from A's frame. But it could have been done from frame B, too yielding the reverse age difference for the very same event. Although that sounds like a paradox, in the closed world case this is actually okay, because it turns out that by doing so we somehow implicitly make B have a closed SoS, rather then A. So this means that in fact both calculations were made in two different worlds/scenarios: one where A is the preferred frame and in the other it's B.

And it reveals a general problem that this method to determine the outcome of an event (which twin is older at their reunion) actually depends on the choice of frame it is done in! This is more general then the closed world scenario.

Now in the cased of the grid the topology is indeed different as no SoS can ever be closed. So when we switch between the frames A and B there is no difference. mathematically, when the cylinder is cut the border conditions that needed to be satisfied for consistency along the line the cylinder was cut open are no longer required. And therefore the previously two different spaces in the closed worlds are identical and thus one and the same for the grid in the open world scenario.

Yet doing the calculation still yields a frame dependent result for the reunion: we get A will be older then B and B older then A (or equivalently $\Delta t_{A,B} < \Delta t_{B,A}$ and $\Delta t_{A,B} > \Delta t_{B,A}$) - a clean contradiction.

So either two different inertial frames can never be truly equivalent to begin with (even in an open world) or there is a error in the calculation that makes the result depend on the frame.

 

[Dale]

But it could have been done from frame B, too yielding the reverse age difference for the very same event.

Only if you make a mistake in the calculation

And it reveals a general problem that this method to determine the outcome of an event (which twin is older at their reunion) actually depends on the choice of frame it is done in!

This is not correct. The age is always given by the integral of the proper time along the path. The global topology changes the possible paths but not the calculation.

 

[PeterDonis]

The twins become indistinguishable if the access to certain information is restricted or rather the twins are forbidden from conducting some other experiments.

That's not what "indistinguishable" means. Either there are observables that distinguish the clones from each other, or there aren't. Saying that there are, but you won't allow the twins to use them, is not a way of making them indistinguishable; it's just a way of confusing yourself. Stop it.

Unfortunately math and logic is always quite a bit more tricky when done correctly.

Indeed. But that doesn't excuse doing it incorrectly, as you are doing, just because it's easier.

The rest of your post is just you confusing yourself further. You should stop that.

 

[PeterDonis]

Yet doing the calculation still yields a frame dependent result for the reunion

If you're in the open world, there is no reunion. A reunion is only possible in the closed world. @Ibix already pointed this out in post #8 (and I reiterated it in post #14).

This is what comes of inventing elaborate ways of confusing yourself instead of focusing on the basics, namely the difference in topology between the two scenarios and what it implies.

 

[Killtech]

Only if you make a mistake in the calculation

This is what comes of inventing elaborate ways of confusing yourself instead of focusing on the basics

Fine, basic question then: in the following setup in a flat open world

... ------ A1 ------ A2 ------ A3 ------ ...
... -----> B1 -----> B2 -----> B3 -----> ...

which grid-vertex ages more between two meetings with the others grid vertices? those in A or in B?

A2 between meetings with B2 and B1 ($= \Delta t_{A,B}$) or
B2 between meetings with A2 and A3 ($= \Delta t_{B,A}$)?

at each meeting of vertices a picture is taken showing the state of both clocks build into each vertex (during which they are at the exact same point). Age differences can be deducted from reading the clocks in between two recorded consequent pictures.

Unlike all other twin paradox setups this one is perfectly symmetric between frames A and B (at least in an open world) and does include only the most trivial of movement.

 

[Dale]

Fine, basic question then: in the following setup in a flat open world

... ------ A1 ------ A2 ------ A3 ------ ...
... -----> B1 -----> B2 -----> B3 -----> ...

which grid-vertex ages more between two meetings with the others grid vertices? those in A or in B?

This is literally just the Lorentz transform.

A2 between meetings with B2 and B1 (=ΔtA,B) or
B2 between meetings with A2 and A3 (=ΔtB,A)?

$\Delta t_{A,B}=\Delta t’_{B,A}$ (edit: assuming equal proper spacing for the A and B clocks)

Unlike all other twin paradox setups this one is perfectly symmetric between frames A and B (at least in an open world)

Yes, and as such the obvious symmetrical relationship is obtained.

 

[Mister T]

Unlike all other twin paradox setups this one is perfectly symmetric between frames A and B (at least in an open world) and does include only the most trivial of movement.

But there is never a reunion. And hence no way for the twins to compare their ages at a reunion. This is not a twin paradox.

 

[PeterDonis]

which grid-vertex ages more between two meetings with the others grid vertices? those in A or in B?

In ordinary flat Minkowski spacetime with the standard topology, it depends on how you specify the scenario. You can specify it so that the apparent aging is symmetric (which appears to be your intent), or you can specify it so it's not, by any amount you like, in either direction (A's aging less than B's, or the reverse). It all depends on how you specify the A and B observer "grids" to be spaced in spacetime.

Note, by the way, that all of the above is also true in the closed world scenario. You can find a set of "A" and "B" observers that age the same between meetings, or a set that age differently, by any amount you like, in either direction. The only difference in the closed world case is that there is one particular family of specifications that includes an observer or set of observers (which you could label the "A" set or the "B" set, that's just a matter of labeling) whose worldlines are "parallel" to the axis of the cylindrical spacetime, and whose surfaces of simultaneity are closed circles (in the 1x1 dimensional case). (There is a family of such specifications because you still have an infinite range of possibilities for the other family of observers, depending on how much less you want them to age between meetings than the family just described.) And for this particular family of specifications, the set of observers described above are always the ones who age the most between meetings.

 

[PeterDonis]

which grid-vertex ages more between two meetings with the others grid vertices? those in A or in B?

Note, btw, that this scenario you are discussing now has nothing to do with the standard twin paradox, because no observer ever accelerates; all observers are always inertial. The "meetings" are either with successive members of the other set of observers (an A with successive B's, or a B with successive A's) or due to at least one observer going "all the way around the universe" in the closed world case, or a combination of both. (This, btw, is a key difference from the standard twin paradox, since it makes possible scenarios where both sets of observers age the same between meetings, something which is impossible in the standard twin paradox--which seems to be a point that is confusing you. Changing scenarios again and again does not necessarily help you to understand any of them; it is better to get a thorough understanding of just one scenario before trying to ring changes on it.)

 

[Dale]

or you can specify it so it's not, by any amount you like, in either direction (A's aging less than B's, or the reverse)

Hmm, I am not seeing it. What are you thinking of here?

 

[PeterDonis]

I am not seeing it. What are you thinking of here?

Consider the following two examples:

(1) Pick a frame. The A observers all move at speed $v$ to the left, and the B observers all move at speed $v$ to the right, in the chosen frame. Space them out the same distance from each other spatially in the chosen frame. Under these conditions, the A's will age the same between successive B meetings, as the B's age between successive A meetings. (Try it and see, if it's not already obvious from the symmetry in the construction I've just described.)

(2) Pick a frame. This frame is the A rest frame. The A at the spatial origin $x = 0$ meets one B at time $t = 0$. He meets the next B at time $t = T$. Draw the hyperbola of constant proper time $T$ through the $t = T, x = 0$ event. Pick the point on that hyperbola where the worldline of the next A to the right of the A at the spatial origin meets it; call this point P. Draw the worldline of the B observer that passes through the spacetime origin, $t = 0, x = 0$ (and meets the A observer there) so that it meets the next A observer at point P. Then draw the worldline of the B observer next to the left, the one that meets the A observer at $t = T, x = 0$, with the same slope. You now have sufficient information to draw in the entire "grids" of both A and B observers, and by construction, the A's age the same between successive B meetings, as the B's age between successive A meetings.

These sorts of scenarios are not often discussed, but IMO they should be; if people had a better sense of the actual range of possibilities for scenarios like this, it would give them better tools for understanding particular individual scenarios.

 

[PeterDonis]

Consider the following two examples

Oh, and to be clear, both of those constructions were assuming the flat, open world (i.e., standard Minkowski spacetime, 1x1, with topology $R^2$).

In the closed world (flat metric, topology $S^1 \times R^1$), construction #1 can still be done, but construction #2 cannot, at least not globally; it will always break down before it can be extended around the entire cylinder. (This is assuming that the chosen frame for both constructions is the preferred frame of the closed world, i.e., the one whose time axis is parallel to the axis of the cylinder and whose spatial surfaces of simultaneity are closed.)

 

[Dale]

Consider the following two examples:

These two examples were “symmetric”. I understood that. I didn’t see the asymmetric case.

I do see it now. The velocity is not particularly relevant because that is symmetric. However the proper distance between grid clocks can be made asymmetric (in my above post I was assuming equal proper spacing).

If the proper spacing between B clocks is much smaller than between A clocks then A will only age a little between B clocks and B will age a lot between A clocks.

Anyway, assuming that @Killtech was intending equal proper spacing then my comments above hold. And since he didn’t actually state it your comments hold more generally.

 

[PeterDonis]

I do see it now. The velocity is not particularly relevant because that is symmetric. However the proper distance between grid clocks can be made asymmetric (in my above post I was assuming equal proper spacing).

Yes, exactly; that spacing is a free parameter, so we can tailor it however we like to adjust the relative aging.