About the definition of Born rigidity

[A.T.]

Regarding "But here the static rulers in O_r should be twice as long": This is only true, if you apply the standard Lorentz transformation. You can do this only if you define, that the one-way speed of light with reference to O_r is the same clockwise and counterclockwise.

Yes, this is the flaw in the reasoning. The usual length contraction factor is isotropic: It depends only on the speed, not on the direction. The moving ruler (aligned with direction of motion) is always shorter than the resting ruler of the same proper length.

But this isotropic length contraction factor is derived using the assumption that light propagation is isotropic. Which is not valid in the rotating frame along the circumference. In the example I gave, the rulers forming O_i are not at half their proper length in the rotating rest frame of O_r, but rather must be at twice their proper length, in order to span the circumference measured in the rotating frame by the resting rulers forming O_r. So in rotating frames, you can have not only kinematic length contraction, but also kinematic length elongation, depending on the direction of motion.

 

[Sagittarius A-Star]

assumption that light propagation is isotropic. Which is not valid in the rotating frame along the circumference.

I think for clarificaion one should distinguish between one-way- and two-way speeds.

For applying the LT for a segment of the rim of the disk, one defines (not assumes) that the one-way speed of light is isotropic for synchronizing clocks, fixed to the rim of the disk. For "measuring" a one-way-speed, one needs always 2 synchonized clocks.

If you use only 1 clock, fixed to the rim of the disk, you will measure, that a two-way-speed is anisotropic around the complete rim of a rotating disk: The clockwise two-way-speed is different from the counterclockwise two-way speed. Two-way-speeds have a physical effect (i.e. Sagnac-effect). What one measures with only 1 clock should not be called a one-way-speed.

 

[cianfa72]

Yes, this is the flaw in the reasoning. The usual length contraction factor is isotropic: It depends only on the speed, not on the direction. The moving ruler (aligned with direction of motion) is always shorter than the resting ruler of the same proper length.

Yes, this turns out by looking at the worldsheet of the moving ruler described in the inertial rest frame of the center's disc (cut the cylinder in #19 along a vertical side and unroll it on the plane using the induced Minkowski metric).

My confusion is about which circumference Langevin observers actually measure. Do they measure (by using their standard rulers) the proper lenght $2\pi R$ where R is the proper lenght of the radius using rulers at rest in inertial rest frame of the rotating circumference?

 

[Sagittarius A-Star]

My confusion is about what circumference Langevin observers actually measure. Do they measure (by using their standard rulers) the proper lenght $2\pi R$ where R is the proper lenght of the radius using rulers at rest in inertial rest frame of the rotating circumference?

The circumference will be measured in the non-Euclidean geometry of the rotating disk as
$U=\gamma 2\pi R$ with $\gamma=1/\sqrt{1-\omega^2R^2/c^2}$.

Source:
https://en.wikipedia.org/wiki/Ehrenfest_paradox#Einstein_and_general_relativity

 

[Ibix]

My confusion is about which circumference Langevin observers actually measure. Do they measure (by using their standard rulers) the proper lenght $2\pi R$ where R is the proper lenght of the radius using rulers at rest in inertial rest frame of the rotating circumference?

A (small) ruler they carry with them will measure the same distances that an instantaneously co-moving inertial ruler would measure. So it'll measure (small) distances along the blue line because its tangent is one of the spatial basis vectors of the instantaneously co-moving frame.

One way to see it is to imagine starting with the disc at rest and laying a chain of small rulers around a circle, touching end-to-end. Hold each one in place with a single nail in the middle (and maybe supported against a rim so they won't bend due to centrifugal force when we start the disc rotating). Then start it rotating. Seen from the initial inertial frame, each ruler will length contract, so will open a small space between the ends. Thus (after spin up, at steady speed) a disc-riding Langevin observer must measure the circumference lengthened compared to before the spin up.

 

[pervect]

I suggest thinking of a "standard ruler" as a pair of worldlines, one for each end. On a space-time diagram, a "standard ruler" is a filled rectangular block.

Length contraction means that the "ordinary" length of a ruler changes when you change your frame of referece. But there is a sort of length to a ruler that is invariant, the so-called "proper length".

So, to measure the circumference of a rotating disk with rulers, draw the worldlines of the end of the rulers (which you've already done, it's the Langevian congruence). And then find the proper length of each ruler, which is a time and observer invariant quantity. And then, add them together. The proper length of a ruler is also the length of the ruler it's own frame of reference.

The ruler as a pair of worldines exists through all time, and the rulers fill all of the space-time of the circumference of the disk. It does this without needing any notion of simultaneity. Simultaneity is only needed when you want to use a ruler to measure the length of something that's moving relative to the ruler, but it's not needed when rulers fill a co-moving geometry.

Alternatively, you can interpret the rotating disk as a quotient manifold, where you associate each worldline in the 4-d space-time congruence with some point in an abstract space 3d "space" of the quotient manifold. A paper that takes this approach (which is an approach I like personally) is https://arxiv.org/abs/gr-qc/0207104. "Space geometry of rotating platforms", by Rizzi and Ruggerio. They authors go so far as to apply the SI defintion of length to basically come up with the induced metric of the quotient manifold, the metric induced on the abstract 3d space by the Lorentzian space-time of the 4d disk.

 

[cianfa72]

Back to the Langevin congruence: it is a timelike stationary congruence.

Which is the definition of timelike stationary congruence ? The definition I have is a timelike congruence such that the vector field defined by tangent vectors to its worldlines is parallel to a timelike Killing Vector field (KVF) of the underlying spacetime metric.

It turns out that the definition above implies zero expansion and shear tensors for the kinematic decomposition of the congruence.

 

[A.T.]

Simultaneity is only needed when you want to use a ruler to measure the length of something that's moving relative to the ruler, but it's not needed when rulers fill a co-moving geometry.

This is the key insight, that helps to avoid lots of confusion.

Alternatively, you can interpret the rotating disk as a quotient manifold, where you associate each worldline in the 4-d space-time congruence with some point in an abstract space 3d "space" of the quotient manifold. A paper that takes this approach (which is an approach I like personally) is https://arxiv.org/abs/gr-qc/0207104. "Space geometry of rotating platforms", by Rizzi and Ruggerio. They authors go so far as to apply the SI defintion of length to basically come up with the induced metric of the quotient manifold, the metric induced on the abstract 3d space by the Lorentzian space-time of the 4d disk.

If by "abstract 3d space" you mean what the authors call "relative space", then I would note that it's not that abstract. It is what actually quantifies the amount of physical material needed to build the disc, and/or it's physical strain. As the authors write:

This important property suggests a procedure to define an extended 3-space, which we shall call ‘relative space’: it will be recognized as the only space having an actual physical meaning from an operational point of view, and it will be identified as the ’physical space of the rotating platform’.

 

[Sagittarius A-Star]

If you want to know how much paint you will need to paint that rotating disc, then you will have to use the non-Euclidean geometry, which is measured by rulers or tape-measures attached to the disc, to figure out its proper area.

With $\gamma=1/\sqrt{1-\omega^2r^2/c^2}$
$dA = \gamma r dr d\phi$

The proper area is
$A(R)= 2\pi \int_0^R \gamma r \, dr= {2 \pi c^2 \over \omega^2}(1-\sqrt{1-\omega^2R^2/c^2})$

Calculation (using a substitution):
https://www.wolframalpha.com/input?i2d=true&i=2πIntegrate[Divide[Power[c,2]β,Power[ω,2]Sqrt[1-Power[β,2]]],{β,0,B}]