# Einstein definition of simultaneity for Langevin observers

• I
cianfa72
TL;DR Summary
Einstein definition of simultaneity for Langevin observers
Hi,
reading this old thread Second postulate of SR quiz question I'd like to ask for a clarification on the following:
For example, consider the family of Langevin observers, who are all moving in circular trajectories about a common origin, with the same angular velocity . In the accelerated coordinates in which these observers are at rest (we use cylindrical coordinates here to make things look as simple as possible), the metric is

$$ds^2 = - \left( 1 - \omega^2 r^2 \right) dt^2 + 2 \omega r^2 dt d\phi + dz^2 + dr^2 + r^2 d\phi^2$$

If we look at a spacelike slice of constant coordinate time in this metric, we find something unexpected: it is Euclidean! The metric of such a slice is simply ##dz^2 + dr^2 + r^2 d\phi^2##, which is the metric of Euclidean 3-space in cylindrical coordinates. Why, then, is it always said that "space" is not Euclidean for such observers?

The answer is that, although the observers are at rest (constant spatial coordinates ##z,r,\phi##) in this chart, the spacelike slices of constant coordinate time ##t## are not simultaneous spaces for those observers. That is, the set of events all sharing a given coordinate time ##t## are not all simultaneous (by the Einstein definition of simultaneity) for the observers. In fact, the set of events which are simultaneous, by the Einstein definition of simultaneity, to a given event on a given observer's worldline do not even form a well-defined spacelike hypersurface at all. So we can't even use that obvious definition of "space" for such observers.
Here the Einstein definition of simultaneity to a given event on the Langevin observer's worldline locally means take the events on the 3D spacelike orthogonal complement to the worldline's timelike tangent vector at that point. Since the Langevin congruence does rotate (i.e. its vorticity is not null) then it is not hypersurface orthogonal, hence there is not a spacelike foliation of Minkowski spacetime orthogonal in each point to the Langevin's worldlines.

Is the above correct ? thank you.

2022 Award
Yes. The Euclidean space referred to is actually the space of the inertial frame where the disc is at rest. This is not orthogonal to the Langevin worldlines. But if you imagine a flock of closely spaced clocks following Langevin trajectories and sync each one at radius ##r## to the one in front by the Einstein procedure, you find that you define a family of non-closing helical surfaces. Those surfaces are everywhere orthogonal to the worldlines, but "now" is multiply defined. If you want to close the surfaces so that they are a valid foliation, somewhere they must not be orthogonal to the Langevin congruence.

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• Orodruin
cianfa72
Yes. The Euclidean space referred to is actually the space of the inertial frame where the disc is at rest. This is not orthogonal to the Langevin worldlines.
Sorry, do you mean the global inertial frame for Minkowski spacetime in which the center of the disk is actually at rest ?

2022 Award
Yes.

cianfa72
you find that you define a family of non-closing helical surfaces. Those surfaces are everywhere orthogonal to the worldlines, but "now" is multiply defined.
Ah ok, you mean that such non-closing spacelike helical hypersurfaces are everywhere orthogonal to the timelike worldlines of the Langevin's congruence. However they do not form a foliation since two different timelike separated events actually belong to the same spacelike hypersurface.

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2022 Award
Yes. They are spacelike surfaces but not achronal. The easiest way to see it is to take the cylinder defined by Langevin observers at some fixed radius ##r## and "unwrap it" into a flat sheet. This surface is infinitely long in the timelike direction and ##2\pi r## wide in the spacelike direction and has a (1+1)d Minkowski metric defined on it. You can easily sketch the Langevin worldlines - they are sets of parallel timelike worldlines. You can then easily sketch their orthogonal spacelike surfaces. Then, identifying the left and right edges of the diagram with one another, you can see that the surfaces only close for Langevin observers with ##\omega=0## - i.e. regular Minkowski observers.

You can also see that the spacelike surfaces can be extended both ways across the edge of the diagram and will become another spacelike plane in the future/past of the event where you started drawing. This can be iterated infinitely. If you wrap the plane back up into a cylinder, you are drawing helical lines around it.

cianfa72
They are spacelike surfaces but not achronal. The easiest way to see it is to take the cylinder defined by Langevin observers at some fixed radius ##r## and "unwrap it" into a flat sheet. This surface is infinitely long in the timelike direction and ##2\pi r## wide in the spacelike direction and has a (1+1)d Minkowski metric defined on it.
It should be the induced metric this surface inherit from the 3D Minkowski metric (we can drop a spatial dimension to the extent of describe a 2D disk).

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Gold Member
But if you imagine a flock of closely spaced clocks following Langevin trajectories and sync each one at radius ##r## to the one in front by the Einstein procedure, you find that you define a family of non-closing helical surfaces. Those surfaces are everywhere orthogonal to the worldlines, but "now" is multiply defined. If you want to close the surfaces so that they are a valid foliation, somewhere they must not be orthogonal to the Langevin congruence.
Unfortunately this method doesn't create a surface that is everywhere orthogonal in all directions to the Langevin congruence at all radii. I'm assuming you repeat the procedure described for all ##r##, and then the surface will be orthogonal to the congruence in the tangential direction, but won't be orthogonal in the radial direction.

To see this, picture this in an inertial frame in which the centre is at rest. In the tangential direction the surface is generated by helixes, where the "slope" of each helix is proportional to its radius. But that means radial lines can't be horizontal (except for one) which means in the radial direction they are not orthogonal to the congruence.

My objection wouldn't apply if by "Langevin congruence" you meant only the worldlines that are at a single radius ##r##. In that case you could construct an orthogonal surface.

2022 Award
My objection wouldn't apply if by "Langevin congruence" you meant only the worldlines that are at a single radius . In that case you could construct an orthogonal surface
No, I see what you mean. So is the quotient space Peter refers to in the quote not everywhere orthogonal to the congruence, or is it not quite what I thought it was?

cianfa72
Unfortunately this method doesn't create a surface that is everywhere orthogonal in all directions to the Langevin congruence at all radii.
Btw, since the 'full' Langevin congruence has not zero vorticity then the 2D spacelike distribution (drop a spatial dimension to the extent of analyze this specific problem) is not integrable. So there are not spacelike hypersurfaces othogonal in each direction to the timelike Langevin congruence.

If we restrict to Langevin worldlines at a single radius, then yes such 1D spacelike surfaces exist however, as pointed out before, they do not foliate the 2D spacetime.

Mentor
is the quotient space Peter refers to in the quote not everywhere orthogonal to the congruence
The quotient space is not a subspace of the spacetime, so asking whether it is orthogonal to the congruence makes no sense.

• cianfa72