I How Does the Twin Paradox Apply to Light and 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.

 

[Herman Trivilino]

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.

 

[Ibix]

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?

As noted by others, it depends. An easy thing to do is switch to a frame in which the A and B observers have equal and opposite velocities. If the grid pitches are equal in that frame then the observers age equally, obvious from symmetry. If the pitches are not equal then the observers in the finer pitched grid will age more than those in the coarser pitched grid. This is easy to see by imagining an extreme case of fine spacing where the coarsely spaced observers meet finely spaced ones almost continuously, but the finely spaced ones only meet coarsely spaced ones occasionally.

Note that the ratio of grid pitches is not frame invariant but the proper times between meetings is, so this analysis only works in this velocity-symmetric frame. In the particular case of symmetric aging, in the rest frame of either grid the other (moving) grid will be length contracted and hence more finely spaced than the rest grid.

 

[stevendaryl]

A cylindrical universe is indistinguishable from a flat universe where all physical quantities are periodic in space.

In terms of inertial coordinates, there is some spacetime vector $V$ with components $V^x, V^y,V^z,V^t$ such that all physical variables (the electromagnetic field, temperature, charge density, matter density, etc.) at a point $X$ with coordinates $x,y,z,t$ are the exact same as those at the point $x+V^x, y+V^y,z+V^z, t+V^t$

So this means that if there is an Earth at the point $X$, there will be an identical Earth at the point $X+V$. In a deterministic world, you can’t distinguish between the case of a cylinder, where $X$ and $X+V$ are the same point, and the case of there being an infinite line of identical copies of Earth, one at $X$, one at $X+nV$ for each integer $n$.

In either case, there is a preferred reference frame, namely the one in which $V^t =0$. This is true in only one reference frame. Since a Lorentz transformation mixes time and space components of 4-vectors.

It’s only in the preferred frame that there would appear to be an infinite line of identical Earths. In every other frame, there would seem to be a line of Earths of different ages. For someone in a rocket traveling from one Earth to the next, it would appear that the Earth he is traveling towards was older than the one he left. So when he got to the second Earth, people will be older than their counterparts on Earth one. If the traveler had a twin, the twin’s counterpart on Earth two would be older than the traveling twin.

 

[Killtech]

[...]

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Thanks for the constructive responses. Finally we have at least established the basics of my original post and seem to agree on this very simple 2 grid scenario.

But the twin paradox is called so for a reason - sure i know it's not really a paradox, but the intent of it is to stress test the assumptions of the theory to check if no contradiction can be constructed... and checking a set of assumptions for contradictions is the 101 for mathematicians. For me of particular interest is to get a better understanding why the symmetry assumption in no circumstance creates one in the open world scenario - and more precisely in the very specific construction i tried.

I hope the discussion provided enough insight why the grids scenario hat it's perks/usefulness. For one it allows comparison with the closed world scenario albeit admittedly within a limited capacity. The other thing is the girds periodic structure does allow to examine further relations which otherwise take up a lot more argumentation to achieve.

And finally we can go back to the comparison i intended with the classical inbound/outbound/stay-at-home open world twins scenario in first post. The core idea was to identify each twin in that experiment with a vertex on one of the grids, but obviously that means I need a 3 grid setup like this:

... ------ A1 ------ A2 ------ A3 ------ A4 ------ ...
... -----> B1 -----> B2 -----> B3 -----> B4 -----> ...
... <----- C1 <----- C2 <----- C3 <----- C4 <----- ...

All 3 grids need to have the same lattice size/spacing in terms of their proper lengths to maintain their symmetry. From A's perspective grids B and C move at the same speed but opposing directions. As in the picture, the initial condition ist that A2,B2,C2 occupy the same point at the start of the experiment.

On the other hand, whenever we ignore one of the grids and just look at any pair of them we should be exactly in the previous simple 2 grid scenario (which we can now use as reduction to a known/discussed problem).

My thought was that i could identify the stay-home twin as A2 and B2 as the outbound sibling. And for that matter also C2 since we can conduct the experiment symmetrically in both directions since we have grids. The turn around point must be chosen when B meets a C vertex but i was thinking it may be more interesting to wait till the next tripple vertex meeting of all grids (the symmetic construction and initial alignment should maintain that those happen periodically every two B-C meetings). That would identify the inbound return trip twin as C4 (and B0).

Now an idea was to now massively use to the symmetry argument in an attempt to yield a different outcome - i.e. a contradiction. So using the previously established symmetry extensively that $\Delta t_{I,J} = \Delta t_{I,J}$ between each pair of grids the idea is to try to conclude a result that doesn't agree with the original twin paradox scenario i.e. no age difference at reunion of the inbound/home twin. What confuses me is that this does seem to work and i don't see where there is an error in the identification of the twins with grid vertices, their age differences with the frame independent clock time intervals $\Delta t_{I,J}$, or the symmetry argumentations. This attempt however conveniently breaks down when trying to translate it into the closed world scenario because there the assumed symmetry never held up to begin with.

And there is also another quite different scenario (a rather mathematical one) which originally lead me to have a closer look at this, because it indicated there might be a hidden asymmetry. So yeah, i just want to understand why there is not - but in a more general setup that is more useful for my understanding/analysis.

 

[Ibix]

As in the picture, the initial condition ist that A2,B2,C2 occupy the same point at the start of the experiment.

Then A1, B1 and C1 will not be in the same position, since the grids have the same pitch measured in their respective rest frames and the two moving grids are length contracted in this frame.

 

[Dale]

What confuses me is that this does seem to work

Can you post your math?

 

[Rene Dekker]

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 $$\frac 1 2 \Delta t_{O,A} + \frac 1 2 \Delta t_{B,A} < \Delta t_{A,O} = \Delta t_{A,B}$$

In your clarifications, it seems clear that you imagine that all the O's meet all the A's simultaneously in A's reference frame (and similarly for the B's). That automatically means that the proper distance between A's is smaller than the proper distance between O's (and the proper distance between B's). After all, if the proper distance between O's is the same as the proper distance between A's, then all O's cannot meet simultaneously with all A's due to length contraction.

That is, in your setup $$\Delta t_{O,A} < \Delta t_{A,O}$$ The situation is not symmetric, because of the way you have defined the experiment.

On top of that, it is not really valid to add $$\frac 1 2 \Delta t_{O,A} + \frac 1 2 \Delta t_{B,A}$$ together and expect it to say something meaningful about the twin paradox. The key to the resolution of the outgoing/incoming twin paradox, is that O and B disagree about what the clock of A indicates. When the hand-over between O and B occurs, the perception of A's clock jumps forward in time. You have not accounted for that jump in perception in your formula.

 

[Killtech]

Then A1, B1 and C1 will not be in the same position, since the grids have the same pitch measured in their respective rest frames and the two moving grids are length contracted in this frame.

Can you post your math?

Well, the event of A1,B1,C1 meeting is not used in determining any of the $\Delta t_{I,J}$. However, I wrongly assumed that B2 will meet the C4 at the point where A3 is. Length contraction will indeed offset that meeting to happen before B2 arrives at A3 (in the scenario of symmetric grids). Hmm, but instead, since the speed of B/C was not specified, i can chose it just so that $\gamma=2$ thus B2 will instead meet C6 exactly when it reaches A3 instead. Therefore:
$$\Delta t_{B,A} + \Delta t_{C,A} = 4 \Delta t_{A,B}$$
left side is adding clock intervals from B2 starting at A2 and passing by 4 C vertices to meet C6 which is the exact same event as its meeting with A3, i.e. $4\Delta t_{B,C} =\Delta t_{B,A}$. The other term is C6 starting at its meeting with B2 going back to A2 which needs $\Delta t_{C,A}$. right side represents A2 starting together with B2 and waiting until 4 B vertices pass by - which exactly coincides with C6 arrival $4 \Delta t_{A,B} = 4 \Delta t_{A,C}$. Both sides should now follow continuous (if not differentiable) paths that start at A2$(t=0)$ and end at the same point A2$(t=T)$.
Now using further symmetries $\Delta t_{B,A} = \Delta t_{C,A}$ i.e. time between meeting of two A vertices is the same for both grids B and C and $\Delta t_{B,A} = \Delta t_{A,B}$ because grids have same proper spacing/are now symmetric. Then i get $2\Delta t_{B,A} = 4\Delta t_{B,A}$, so something went wrong somewhere, because $\Delta t_{B,A}>0$ and finite.

As you can see i am not interested in calculating the exact values of any of the $\Delta t_{I,J}$. Merely interested in finding the relations amongst them.

In your clarifications, it seems clear that you imagine that all the O's meet all the A's simultaneously in A's reference frame (and similarly for the B's). That automatically means that the proper distance between A's is smaller than the proper distance between O's (and the proper distance between B's). After all, if the proper distance between O's is the same as the proper distance between A's, then all O's cannot meet simultaneously with all A's due to length contraction.

That is, in your setup $$\Delta t_{O,A} < \Delta t_{A,O}$$ The situation is not symmetric, because of the way you have defined the experiment.

Thanks for the answer. I noticed my mistake from from the previous responses which hinted at this problem already but i haven't figured out how I want to approach it.

I naively had the picture in my head that all grids could be constructed at rest in A's frame to ensured they are perfectly equal (in particular equally spaced). Only then grid B and C would be steadily accelerated in opposing directions up until they reach the target speed. The idea was that i could do this in both a closed or open world and should end up with comparable setup in different worlds. But to involve acceleration in this is like opening Pandora's box - even if it is only to define the initial setup.

Naively i first though I don't really need any math for it because i could just treating each vertex independently as if they were not connected, then in A's frame each accelerates the same and their distances won't change throughout the speedup phase. But yes indeed, doing so skips the length contraction and that conversely means that in the accelerated frames (and finally frames B and C), the proper length between vertices changes and i end up in a scenario of asymmetric grids - which @PeterDonis correctly described before. In that scenario meeting points are indeed simultaneous from A's perspective.

On the other hand if the grids were rigid connected bodies, then my best guess would be that the body/grid would contract around its center of mass from A's point of view requiring additional acceleration for vertices that are off center. This would imply the acceleration to grow the further off center they are making them go somewhat too fast. I don't see how to make the grid contraction instantaneous so I guess i would have to think of it as a contracting force with the vertices inertia making it act reasonably.

But apparently there is something wrong in both approaches. I noticed that in the case of an unconnected grid, the accelerated frames notices the effects of acceleration differently depending on which frame it starts from and in which direction it accelerates. Accelerating from A to B will make the lattice proper spacing grow but slowing down from B to A will have to make it expand. So an observer on a grid would notice an asymmetry between acceleration into different directions depending on which frame he starts from.

Same would be the case for the rigid grid if it is sufficiently large to measure the contracting/expanding force before the vertices can reach their equilibrium position which the original proper grid spacing.

 

[Ibix]

Therefore:
$$\Delta t_{B,A} + \Delta t_{C,A} = 4 \Delta t_{A,B}$$

$\Delta t_{A,B}$ is the time a clock attached to grid A measures between meeting adjacent B grid elements, right? Then this is clearly wrong since $\Delta t_{A,B}=\Delta t_{B,A}=\Delta t_{A,C}=\Delta t_{C,A}$ from symmetry, so it simplifies to 2=1. Where did you get the idea from?

But apparently there is something wrong in both approaches.

You have not taken into account the relativity of simultaneity (this is the answer to 99% of problems with special relativity). You are apparently assuming that decelerating grid elements will start their deceleration simultaneously in their final rest frame, but accelerating grid elements will start their acceleration simultaneously in their initial rest frame. You should look up Bell's spaceship paradox. In the meantime, I would strongly suggest not introducing acceleration into an already overly complicated scenario.

 

[Killtech]

$\Delta t_{A,B}$ is the time a clock attached to grid A measures between meeting adjacent B grid elements, right? Then this is clearly wrong since $\Delta t_{A,B}=\Delta t_{B,A}=\Delta t_{A,C}=\Delta t_{C,A}$ from symmetry, so it simplifies to 2=1. Where did you get the idea from?

Yeah, that's what $\Delta t_{A,B}$ is supposed to be and i get that it's a contradiction. My idea to use the grids and the $\Delta t_{I,J}$ is just so i can at least mathematically treat the twins frames as equivalent even though the experimental twin setups is not. Only in the close world case it is technically symmetric, but the assumption that inertial frames are equivalent is dropped there anyway (since obviously they are not). The point of the grids is that they offer more relations between quantities and which allows to involve the symmetry of inertial frames assumption which isn't otherwise used in any of the twin paradox scenarios. But without using this assumptions how are you to check if it could cause contradictions - i.e. a real paradox?

As I originally wrote, I stumbled on the problem from the analogy with sound waves mechanics which has an asymmetry just like in the closed world scenario - but for sound waves it must be also present in an open world. My problem was that a theory that has perfectly symmetric inertial frames is actually really tricky. So i figured i could dissect SRT with the $\Delta t_{I,J}$ to find out how it internally works (or if it works).

You have not taken into account the relativity of simultaneity (this is the answer to 99% of problems with special relativity). You are apparently assuming that decelerating grid elements will start their deceleration simultaneously in their final rest frame, but accelerating grid elements will start their acceleration simultaneously in their initial rest frame. You should look up Bell's spaceship paradox. In the meantime, I would strongly suggest not introducing acceleration into an already overly complicated scenario.

Ah, right it's obvious in hindsight. The vertices don't transform by a Lorentz trafo or anything trivial but instead i have to ray trace them along their trajectory to where they were/will be in terms of the new surface of simultaneity of the frame. Yeah, moving time back and forth of course displaces the vertices causing the contraction around the origin of the frame in this simple example - since ray tracing is easy enough for inertial movement.

means that if there is a cat at A3, then for A2 at $t=0$ the cat will be alive, for B2 it won't be born yet and for C2 it will have died already. And if it moved around during it's life I will have to take into account the displacement it had towards A3 at that phase of its life. Albeit not as funny as Schödinger's cat, it still visualizes the issue and i think Physics is more understandable if explained by cats.

As for the acceleration seeming to be asymmetric, well indeed with this explanation it's trivial either. I have chosen for it to be simultanious from A's perspective which is why i build an asymmetry into the system all by myself.

 

[Ibix]

The point of the grids is that they offer more relations between quantities and which allows to involve the symmetry of inertial frames assumption which isn't otherwise used in any of the twin paradox scenarios.

And my point is that I cannot see how you could think that $\Delta t_{B,A}+\Delta t_{C,A}=4\Delta t_{A,B}$ could be plausible. Quite apart from the contradictions, I can't see any motivation at all for that equation - they aren't even times in the same reference frame. Obviously you had some motivation for writing it - can you explain what it is?

 

[Ibix]

Here's a Minkowski diagram of your setup:

The red lines are the worldlines of your stationary grid, A. The green lines are your grid B and the blue lines your grid C. I've added labelled black lines identifying intervals that correspond to the various $\Delta t$ quantities you have defined. Remember that this is in Minkowski space, so the interval along the lines is $\sqrt{\Delta t^2-\Delta x^2}$, not the Euclidean length.

Perhaps now you can see why I'm so confused by your $\Delta t_{B,A}+\Delta t_{C,A}=4\Delta t_{A,B}$. The worldlines these intervals run along aren't parallel, so it isn't at all clear to me that there's any reason to think that they should add up - at the least some geometry or hyperbolic trig would be required to show something like this were reasonable. I'm wondering if you have forgotten that the concept of proper time in relativity is similar to that of length in Euclidean geometry. Adding lengths of non-parallel lines isn't necessarily a meaningful thing to do, and neither is adding proper times.

 

[Killtech]

The red lines are the worldlines of your stationary grid, A. The green lines are your grid B and the blue lines your grid C. I've added labelled black lines identifying intervals that correspond to the various $\Delta t$ quantities you have defined. Remember that this is in Minkowski space, so the interval along the lines is $\sqrt{\Delta t^2-\Delta x^2}$, not the Euclidean length.

Perhaps now you can see why I'm so confused by your $\Delta t_{B,A}+\Delta t_{C,A}=4\Delta t_{A,B}$. The worldlines these intervals run along aren't parallel, so it isn't at all clear to me that there's any reason to think that they should add up - at the least some geometry or hyperbolic trig would be required to show something like this were reasonable. I'm wondering if you have forgotten that the concept of proper time in relativity is similar to that of length in Euclidean geometry. Adding lengths of non-parallel lines isn't necessarily a meaningful thing to do, and neither is adding proper times.

Thank you very much for the effort and yes you are perfectly right. Sorry, i was somewhat occupied with work so didn't find the time to check this forum.

The "equation" i wrote down is just the one from my original post but corrected - and a mistake added. It represents the age relation of the twin at reunion in the outbound/inbound scenario but expressed in terms of grid quantities with the left hand side representing the proper time of the outbound/inbound path with the right being the stay at home path. So the correct relation is therefore:
$$\Delta t_{B,A}+\Delta t_{C,A}<4\Delta t_{A,B}$$
(one twin will be older then the other at reunion) Of course one can calculate the actual left and right hand values, but that was never the point. In this particular choice of $\gamma = 2$ it would seem that $1<2$ i.e. one twin will have twice as aged as the other. Using the symmetry arguments one can now represent the same calculation from different frames but that doesn't change anything and now i understand why.

I think i also know how one could properly proof that the equivalence of inertial frames postulate doesn't cause a contradiction. it boils down to the statement that a difference in the one-way-speed is undetectable using physics based on light (EM-force) - as that is what effectively causes the age differences. Choosing whatever inertial frame just adds a constant offset to the one-way-speed of light and because every closed path integral over a constant vector field yields exactly a zero result, the choice of frame has no influence on the result/age difference at reunion. Of course the situation changes in a topologically different setup like a closed world where closed path integrals around the world yield a value other then zero. Same is true if the one-way-speed of light offset vector field had a non-zero curl somewhere. So it's not entirely pointless to try to measure it.

Anyhow, the discussion here was quite enlightening for me. Now i understand how this is construction works and it made me realized that i made the same errors when i was looking at acoustics. It's not like this math setup is for waves without a medium only, but rather it's generally works for all waves - simply because the medium doesn't matter (as in the one way speed of light offset).

 

[pervect]

A quick comment.

It is assumed that the reader is principle with the SR principle of "maximal aging". See for instance http://www1.kcn.ne.jp/~h-uchii/extrem.aging.html, which references Taylor & Wheeler.

Pick a cylinder in 3d space, with the usual cylindrical coordinates, r, theta, and z, with r set to 1. Consider two points with the same value of theta and different values of z, i.e.geodesics (straight lines) between the points (r, theta,z) and (r,theta,z+X). Informally, we might say that one point is "above" the other on the cylinder. Then there are an infinite number of geodesics (straight lines) that connect the two points, that wind around the cylinder a different number of times. All of the paths are straight, but the path with a winding number of zero is shorter than the others. We can see that in the cylindrical spatial geometry, there are straight lines (geodesics) that are not the shortest distance between two points. One way of describing this is that the principle that the a straight line is the shortest distance in the plane gets slightly modified, to a similar but slightly different principle, the principle of stationary or extremal path length, when we switch to a cylindrical geometry.

The situation is very similar with two twins in a 4d cylindrical spacetime, with coordinates t,r,theta, and z. There will be an infinite number of geodesics that a an observer might follow (timelike geodesics) that lead from a point (t,r,theta,z) to a point (t+X,r,theta,z+Y). But path one will be longer than the other paths, the path of longest proper time having a zero winding number.

Thus, the analogy between "straight lines" (geodesics) in space as the shortest distance between two points, and the principle of "straight lines"(geodesics) on space-time diagrams, as paths of maximum time holds with only a small modification. In both the space and space-time cases there are multiple geodesics for a cylindrical geometry, and a very slight modification of the wording, changing "minimum distance" to "extremal distance" and "maximum aging" to "extremal aging" allows one to recover a very similar result for the cylindrical geometry.

 

[Killtech]

[...] One way of describing this is that the principle that the a straight line is the shortest distance in the plane gets slightly modified, to a similar but slightly different principle, the principle of stationary or extremal path length, when we switch to a cylindrical geometry.
[...]

Yeah, i used exactly that approach to get the acoustic wave equation to conform to the same math as SRT - i.e. i derived the math framework from acoustic geodesics. Given a seemingly analog derivation it wasn't clear to me where the differences stem from which is why i tried to figure that out by expressing the situations where i saw disagreements through various grid setups. But again all those turned out to be just my very own misinterpretations and mistakes (mostly forgetting about mismatching SoS).