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Killtech

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- 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.

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.