Einstein Definition of Simultaneity for Langevin Observers

In summary, the space referred to is not Euclidean for observers following a Langevin trajectory, which is surprising since the trajectory is at rest in the inertial frame. However, 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.
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cianfa72
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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.
 
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  • #2
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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  • #3
Ibix said:
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 ?
 
  • #5
Ibix said:
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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  • #6
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.
 
  • #7
Ibix said:
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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  • #8
Ibix said:
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.
 
  • #9
DrGreg said:
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?
 
  • #10
DrGreg said:
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.
 
  • #11
Ibix said:
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.
 
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  • #12
cianfa72 said:
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.
Yes, that's all correct. You can go further and extend to a 2D spacelike surface that's orthogonal to (less than a complete revolution of) the Langevin worldlines at a single radius (but not orthogonal to Langevin worldlines at a different radius), but you can't extend it too far otherwise it would meet the same Langevin worldline twice.
 
  • #13
DrGreg said:
You can go further and extend to a 2D spacelike surface that's orthogonal to (less than a complete revolution of) the Langevin worldlines at a single radius (but not orthogonal to Langevin worldlines at a different radius), but you can't extend it too far otherwise it would meet the same Langevin worldline twice.
Ah ok, so starting from any of a such 1D spacelike 'line' we can actually extend it to a 2D spacelike surface up to the limit you pointed out above.
 
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1. What is Einstein's definition of simultaneity for Langevin observers?

Einstein's definition of simultaneity for Langevin observers is a way of determining whether two events occur at the same time, as perceived by two observers who are moving at different speeds relative to each other. It takes into account the fact that time is relative and can appear to pass differently for observers in different reference frames.

2. How does the concept of simultaneity differ for Langevin observers compared to Galilean observers?

For Galilean observers, simultaneity is defined as two events occurring at the same time if they are perceived to happen at the same moment by an observer who is not moving. However, for Langevin observers, simultaneity is relative and depends on the observer's relative motion and the speed of light.

3. Why is the concept of simultaneity important in Einstein's theory of relativity?

The concept of simultaneity is important in Einstein's theory of relativity because it helps to explain how time and space are relative and can appear differently to different observers. It also helps to reconcile the discrepancies between classical mechanics and the observed behavior of objects at high speeds.

4. How does the definition of simultaneity for Langevin observers affect our understanding of causality?

The definition of simultaneity for Langevin observers affects our understanding of causality by showing that the order in which events occur can appear different to observers in different reference frames. This challenges the traditional concept of cause and effect and highlights the relativity of time and space.

5. Can the concept of simultaneity for Langevin observers be observed or measured?

No, the concept of simultaneity for Langevin observers cannot be directly observed or measured. It is a theoretical concept that helps to explain the behavior of objects at high speeds and is supported by experimental evidence, but it cannot be directly observed or measured in the same way as physical quantities like distance or time.

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