The twin paradox

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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
Code:
... ------ 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.
 
  • #27
Dale
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Fine, basic question then: in the following setup in a flat open world
Code:
... ------ 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.
 
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  • #28
Mister T
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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.
 
  • #29
PeterDonis
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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.
 
  • #30
PeterDonis
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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.)
 
  • #31
Dale
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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?
 
  • #32
PeterDonis
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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.
 
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  • #33
PeterDonis
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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.)
 
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  • #34
Dale
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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.
 
  • #35
PeterDonis
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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.
 
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  • #36
Ibix
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Fine, basic question then: in the following setup in a flat open world
Code:
... ------ 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.
 
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  • #37
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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.
 
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  • #38
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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:

Code:
... ------ 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.
 
  • #39
Ibix
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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.
 
  • #40
Dale
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What confuses me is that this does seem to work
Can you post your math?
 
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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.
 
  • #42
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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.
 
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  • #43
Ibix
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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.
 
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  • #44
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##\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.
 
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  • #45
Ibix
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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?
 
  • #46
Ibix
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Here's a Minkowski diagram of your setup:
1623957410961.png

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.
 
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  • #47
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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).
 
  • #48
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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.
 
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[...] 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.
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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).
 

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