twin clocks-is it acceleration?

DaleSpam

Quote by A.T.

When you draw a space time diagram or integrate the path element, you are assuming a frame.

I think we are talking about different kinds of explanations. What I would call a geometric explanation wouldn't draw any coordinates.

ghwellsjr

Quote by A.T.

Quote by ghwellsjr

In each of these three cases I provided an inertial spacetime diagram for the stay-at-home twin's inertial rest frame and a non-inertial spacetime diagram for the traveling twin's non-inertial rest frame and yet none of the OP's for which I provided these diagrams responded with any statement to the effect that they understood the diagrams. I just don't know if they are effective.

I don't understand them either.

I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram. If the signals are simply a picture seen through telescope, the angular size of blue twin, as seen by black twin would be increasing after turnaround. But in the diagram the viewing distance is constant for a few years.

I think it is very difficult to make a smooth and correct diagram for the sharp turnaround version. The simpler case might be the one with constant proper acceleration, then you have just one coordinates chart (Rindler) for the rest frame of the accelerating twin, not two that you have to merge somehow.

See Fig 7:
http://cds.cern.ch/record/497203/files/0104077.pdf

See Fig 9 of your above link. Barbara is the "Immediate Turn-around" traveler, Alex stays at home. The diagram is showing the non-inertial rest frame of Barbara.

DaleSpam

Quote by A.T.

I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram.

The diagram uses radar coordinates, the same approach as Einstein's convention, but applied to a non inertial observer. Since the resulting coordinates are non inertial you can get red and blue shifting without relative motion. I.e. The standard inertial frame formulas don't apply.

A.T.

Quote by PAllen

[Edit: On further thought, it appears that Fermi-Normal coordinates as above, for a uniformly accelerating observer would still have the same simultaneity surfaces as Rindler and Radar - just labeled differently. Thus, it seems, you need change in acceleration - e.g. the classic twin with slightly rounded turnaround - to expose the difference between Radar simultaneity and Born Rigid simultaneity (which is what Fermi-Normal uses).]

Thanks for the explanation. So you say for constant g there is no difference. But what would the difference be otherwise? Would you actually get different age differences by integrating the paths, or just different simultaneity in between?

PAllen

Quote by A.T.

Thanks for the explanation. So you say for constant g there is no difference. But what would the difference be otherwise? Would you actually get different age differences by integrating the paths, or just different simultaneity in between?

Just different simultaneity in between. Also different applicability:

- Fermi-Normal simultaneity = Born rigid simultaneity = spacelike geodesic 4-orthogonal to world line tangent = simultaneity of intantaneous comoving inertial frame are all inapplicable to an instant turnaround; they are also inapplicable to a W shaped twin path (the simultaneity lines intersect).

- Radar simultaneity works for instant turnaround and W shaped trajectory with no problems.

However, where they both work, any observable is computed to be the same. Something non-observable, like distant simultaneity between world lines, is different.

A.T.

Quote by PAllen

Just different simultaneity in between. Also different applicability:

- Fermi-Normal simultaneity = Born rigid simultaneity = spacelike geodesic 4-orthogonal to world line tangent = simultaneity of intantaneous comoving inertial frame are all inapplicable to an instant turnaround; they are also inapplicable to a W shaped twin path (the simultaneity lines intersect).

- Radar simultaneity works for instant turnaround and W shaped trajectory with no problems.

Do Fermi-Normal coordinates have the same proper acceleration for every x? Because in Rindler coordinates it varies with x.

PAllen

Quote by A.T.

Do Fermi-Normal coordinates have the same proper acceleration for every x? Because in Rindler coordinates it varies with x.

As I realized when I saw that a trivial coordinate transform went from Rindler to Fermi-Normal:

- their lines of simultaneity are identical
- the feature of varying proper acceleration for world lines of constant x, at different x values, is a feature of both.

phyti

Since the writer of a paper is so familiar with their own material, there is
always the possibility for the reader to misinterpret or not understand some
parts due to insufficient details, excessive complexity, verboseness,
mistakes, etc. Having experienced this as a reader, this is an attempt at an
improved version of the original post.

twin clocks2.doc

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phyti	Since the writer of a paper is so familiar with their own material, there is always the possibility for the reader to misinterpret or not understand some parts due to insufficient details, excessive complexity, verboseness, mistakes, etc. Having experienced this as a reader, this is an attempt at an improved version of the original post. twin clocks2.doc