About the definition of Born rigidity

[cianfa72]

TL;DR

On the definition of Born rigidity. Is the notion of proper distance well-defined not just locally ?

I'd ask for clarification on the definition of Born rigidity, see for instance Born Rigidity.

In the context of SR (flat spacetime) consider a ruler moving through spacetime. Its points define a timelike congruence in the region of flat spacetime it occupies.

The general definition of Born rigidity involves the notion of proper distance (length) between neighboring congruence's worldlines. Basically, at any point/event along a timelike congruence's wordline, one takes the spacelike hyperplane orthogonal to the 4-velocity at that point. In general such an hyperplane won't be orthogonal to all the congruence's members (the congruence may not be irrotational, i.e. have not zero vorticity).

Therefore the notion of proper distance is well-defined only locally in an (open) neighborhood of the point along the chosen worldline. In other words the notion of proper distance (length) isn't well defined globally from a general point of view.

Does it makes sense ? Thanks.

 

[Ibix]

You can define proper distance along an arbitrary spacelike path parameterised by $\lambda$ as $\int\sqrt{|g_{ab}\frac{dx^a}{d\lambda}\frac{dx^b}{d\lambda}|}d\lambda$, analogous to proper time along a timelike path. I would think that to call it "proper length" the path needs to be orthogonal to some timelike congruence representing fixed points on the object. So I don't think proper length is only defined locally, no.

 

[cianfa72]

You can define proper distance along an arbitrary spacelike path parameterised by $\lambda$ as $\int\sqrt{|g_{ab}\frac{dx^a}{d\lambda}\frac{dx^b}{d\lambda}|}d\lambda$, analogous to proper time along a timelike path. I would think that to call it "proper length" the path needs to be orthogonal to some timelike congruence representing fixed points on the object. So I don't think proper length is only defined locally, no.

But, what if the timelike congruence representing fixed points on the object (say a ruler) hasn't zero vorticity ? In that case there isn't a spacelike hypersurface orthogonal to the congruence's worldlines. So there isn't any such spacelike path you were talking about.

 

[Ibix]

But, what if the timelike congruence representing fixed points on the object (say a ruler) hasn't zero vorticity ? In that case there isn't a spacelike hypersurface orthogonal to the congruence's worldlines. So there isn't any such spacelike path you were talking about.

The hypersurface doesn't matter. You're only drawing a 1d line, not extending it into a plane. You can always draw a line segment perpendicular to one member of a congruence and extend it to a nearby one, then start again with a new perpendicular line.

For example, consider a concentric circle on the rotating disc. You an draw a line in the "world cylinder" thus defined that is everywhere perpendicular to the worldline of the point it's at - it forms a shallow helix in a (2+1)d Minkowski diagram. It doesn't close, which is a problem if you try to construct a plane out of it, but you don't need to form the plane to be able to make (spatial) arc-length measurements.

 

[cianfa72]

The hypersurface doesn't matter. You're only drawing a 1d line, not extending it into a plane. You can always draw a line segment perpendicular to one member of a congruence and extend it to a nearby one, then start again with a new perpendicular line.

Ah ok, basically your point is that if we look at 1d lines, then for any timelike congruence there always exists such an orthogonal (spacelike) line (this boils down to the fact that when looking at 1d submanifolds, the Frobenius integrability condition is always met - basically what is required to do is integrate a spacelike vector field orthogonal to the congruence).

 

[Ibix]

Yes.

 

[pervect]

The two formal defintions of born rigidity I'm aware of apply to timelike congruence which I will denote as $u^a$. I believe the definitions are equivalent. The one I use the most is that the shear and expansion tensor of the congruence $u^a$ must vanish, I think this is the easiest to understand if one is familiar with how to compute the shear and expansion of a vector field. The other definition, which I think is Born's original definition, is that the Lie derivative of the spatial metric vanishes along the congruence $u^a$. The spatial metric is a 3-metric, $h_{ab}$, computed via projection from the 4-meteric $g_{ab}$ by $h_{ab} = g_{ab} + {u_a}{u_b}$ for a (-1,+1,+1,+1) sign convention. Formally we'd write $\mathcal{L}_u h_{ab} = 0$.

I don't use the Lie derivative form much, but it may be mathematically simpler. What the notation is saying again is that we take the Lie derivative of $h_{ab}$, the spatial projection of the metric tensor, which is a rank 2 tensor, along the vector fields u^a of our congruence.

To justify my informal statement, at any point along the congruence, you can consider the space with the metric $h_{ab}$ which is orthogonal to the congruence. In general, the local "space" orthogonal to the congruence may be different at every point , only when the vorticity is zero can you construct a global space orthogonal to the congruence. But locally , you can use this notion of space to determine distances.

 

[cianfa72]

The other definition, which I think is Born's original definition, is that the Lie derivative of the spatial metric vanishes along the congruence $u^a$. The spatial metric is a 3-metric, $h_{ab}$, computed via projection from the 4-meteric $g_{ab}$ by $h_{ab} = g_{ab} + {u_a}{u_b}$ for a (-1,+1,+1,+1) sign convention. Formally we'd write $\mathcal{L}_u h_{ab} = 0$.

You said $h_{ab}$ is a spatial 3-metric tensor. Your definition of it is $h_{ab} = g_{ab} + {u_a}{u_b}$. However the indices $a,b$ still run from 0 to 3.

To justify my informal statement, at any point along the congruence, you can consider the space with the metric $h_{ab}$ which is orthogonal to the congruence. In general, the local "space" orthogonal to the congruence may be different at every point, only when the vorticity is zero can you construct a global space orthogonal to the congruence.

Yes, since zero vorticity implies (iff) hypersurface orthogonality.

But locally, you can use this notion of space to determine distances.

Ok.

 

[pervect]

I had to look this up. My source is not the best, but it says the indices in h_ab do run from 0..3. I'm not sure if I actually said otherwise, but I was sure thinking it. Mea culpa.

The point is that $h_{ab} u^a = 0$. So $h_{ab}$ is a linear operator that projects away the vector u^a to a zero vector.

The following proof was offered:

because we are using a -+++ metric signature $u^a u_a$ = -1. Then

$$ (g_{ab} + u_a u_b) u^a = (g_{ab} u^a) +u_a u_b u^a = u_b + (u_a u^a) u_b = u_b + -u_b = 0$$

 

[cianfa72]

Quoting Wikipedia - kinematical decomposition

$h_{ab} = g_{ab} + X_aX_b$
for the projection tensor which projects tensors into their transverse parts; for example, the transverse part of a vector is the part orthogonal to $X^a$. This tensor can be seen as the metric tensor of the hypersurface whose tangent vectors are orthogonal to $X^a$.

In the notation used there, $X^a_{;b}$ should be the (1,1) tensor $\nabla_b X^a$ written in abstract index notation (note the Latin indices).

Sometimes I've seen $\nabla_b X^a := (\nabla X)^{a}{}_{b}$ rather than $\nabla_b X^a := (\nabla X)_{b}{}^{a}$.

 

[cianfa72]

For example, consider a concentric circle on the rotating disc. You an draw a line in the "world cylinder" thus defined that is everywhere perpendicular to the worldline of the point it's at - it forms a shallow helix in a (2+1)d Minkowski diagram. It doesn't close, which is a problem if you try to construct a plane out of it, but you don't need to form the plane to be able to make (spatial) arc-length measurements.

Back to your example, consider now in (2+1)d Minkowski spacetime a piece of a concentric annulus on the rotating disk. At any point P on the timelike congruence's members it defines, pick in the tangent space an element (vector) orthogonal to the 4-velocity of the congruence's worldline passing through it. There are infinitely many ways to pick this bunch of vectors. By restricing this choice to give a smooth vector field, then its (spacelike) integral curves (orbits) define paths to integrate over to get the proper length/distance.

So, as expected, in this case we get infinitely many answers for the proper lenght of the 2D object. Like the case of 1D object, I believe one can define a notion of "proper area" which in this case turns out to be unique.

 

[Ibix]

So, as expected, in this case we get infinitely many answers for the proper lenght of the 2D object.

In a cylindrical coordinate system where the disc is rotating with angular velocity $\omega$ about the origin, the line element is $ds^2=dt^2-(dr^2+r^2d\phi^2)$, let the drawn circle have radius $R$. The congruence thus defined is a one parameter family parameterised by $\phi$ and formed from the integral curves of the unit vector field $(\gamma,0,\gamma\omega)$ where $\gamma$ is the Lorentz gamma factor associated with speed $\omega R$.

There is only one smooth unit vector field orthogonal to that congruence and lying in the circle, $\frac 1A(\omega R^2,0,1)$, where $A$ is a normalisation factor I haven't bothered to calculate. The proper length of any arc along the circle is $\int \sqrt{|ds^2|}$ along an integral curve of that field. What's not unique about that?

Generally, you can draw any arbitrary path on the disc and this picks out a one-parameter family of the congruence defining the disc. You need to do a bit more work to derive the appropriate spacelike vector field from it, but then you can integrate along its integral curves.

Physically, you can build a strong arbitrarily curved wall on the surface of the disc and glue a piece of string to it while the disc is rotating, so the string is unstressed (supported against centrifugal motion by the wall) and marking out some arbitrary path. Marks on the string show its proper length, and will correspond to locally measured lengths. Thus there must be some global measure of length.

 

[cianfa72]

In a cylindrical coordinate system where the disc is rotating with angular velocity $\omega$ about the origin, the line element is $ds^2=dt^2-(dr^2+r^2d\phi^2)$, let the drawn circle have radius $R$. The congruence thus defined is a one parameter family parameterised by $\phi$ and formed from the integral curves of the unit vector field $(\gamma,0,\gamma\omega)$ where $\gamma$ is the Lorentz gamma factor associated with speed $\omega R$.

Yes, in cylindrical coordinates $(t,r,\phi)$ for (2+1)d Minkowski spacetime, the points on the concentric circle drawn on the rotating disk are described from helices.

There is only one smooth unit vector field orthogonal to that congruence and lying in the circle, $\frac 1A(\omega R^2,0,1)$, where $A$ is a normalisation factor I haven't bothered to calculate. The proper length of any arc along the circle is $\int \sqrt{|ds^2|}$ along an integral curve of that field. What's not unique about that?

Yes definitely. My point was to considerer a 2D object on the rotating disk, for instance a piece of a concentric 2D annulus.

Generally, you can draw any arbitrary path on the disc and this picks out a one-parameter family of the congruence defining the disc. You need to do a bit more work to derive the appropriate spacelike vector field from it, but then you can integrate along its integral curves.

Yes.

 

[A.T.]

Marks on the string show its proper length, and will correspond to locally measured lengths. Thus there must be some global measure of length.

Yes, the string can be a closed loop going all around the rotating disk, thus measuring it's proper circumference. That global geometry will be non-Euclidean though, with circumference / diameter > pi.

 

[cianfa72]

Marks on the string show its proper length, and will correspond to locally measured lengths. Thus there must be some global measure of length.

As locally measured lenghts, you mean the lenght between marks on the string as measured from an inertial momentarily comoving frame.

 

[Ibix]

My point was to considerer a 2D object on the rotating disk, for instance a piece of a concentric 2D annulus.

Ok, but isn't that just a general fact about shapes in more than 1d? "Length" isn't really well defined for a 2d shape even in Euclidean geometry unless you add some rule like "length of the longest straight line contained within the shape" or "length of longest edge".

As locally measured lenghts, you mean the lenght between marks on the string as measured from an inertial momentarily comoving frame.

Yes.

 

[cianfa72]

Ok, but isn't that just a general fact about shapes in more than 1d? "Length" isn't really well defined for a 2d shape even in Euclidean geometry unless you add some rule like "length of the longest straight line contained within the shape" or "length of longest edge".

Yes, definitely. Indeed I was wondering whether one could define a sort of "proper area" for 2D object/shapes.

 

[A.T.]

Yes, definitely. Indeed I was wondering whether one could define a sort of "proper area" for 2D object/shapes.

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.

 

[cianfa72]

Only later I realized the following: take a set of Langevin observers at fixed radius on the rotating disc (let's call it $S$). This is a set of timelike worldlines out of the Langevin congruence that describes all the points on the disc.

The attempt to synchronize standard clocks carried by each member of $S$ using Einstein synchronization procedure, basically defines an helix curve in cylindrical coordinates for Minkowski spacetime with the property of being orthogonal (w.r.t the Minkowski flat metric) to any member of $S$. This is the blue helix in the picture

Since the Langevin congruence has zero expansion and shear it is Born rigid, so is $S$ too. As far as I can tell, the latter means that the round trip time for light signals exchanged back and forth for any pair of members in $S$ stays constant.

 

[Ibix]

Yes. It's obvious from symmetry - the congruence is identical under the combination of time translation and then rotation by $\Delta t$ and $-\omega\Delta t$ respectively, where $\Delta t$ is arbitrary. So any experiment carried out between Langevin observers now will work identically later.

 

[A.T.]

As far as I can tell, the latter means that the round trip time for light signals exchanged back and forth for any pair of members in $S$ stays constant.

Sure, why would one expect otherwise, if the disc spins at a constant rate?

However, the time for a trip around the entire circumference (back to the same point on the disc) depends on the direction of the trip. Light propagation speed is not isotropic in a rotating frame, so using Einstein synchronization for the disc clocks on the circumference is maybe not the best choice.

 

[cianfa72]

However, the time for a trip around the entire circumference (back to the same point on the disc) depends on the direction of the trip. Light propagation speed is not isotropic in a rotating frame, so using Einstein synchronization for the disc clocks on the circumference is maybe not the best choice.

I'm trying to understand your claim from Minkowski spacetime in standard cylindrical coordinates.

Consider the (subset) of Langevin observers $S$ on a fixed circumference on the rotating disc. Start from event $A_1$ on the worldline of the Langevin observer $A$ in $S$ and draw the light cone in those coordinates. It will intersect the neighboring Langevin worldline $B$ at the event $B_1$. Proceed this way from $B_1$ up to reach the $A$'s worldline again at event $A_2$.

One can do the light trip along the circumference in two ways/directions. So, are you claiming that from the point of view of Langevin observers in $S$ the light propagation process isn't isotropic ?

 

[Ibix]

One can do the light trip along the circumference in both ways. So, are you claiming that from the point of view of the Langevin observers in S the light propagation process isn't isotropic ?

Locally it is, because the coordinate system is orthogonal. But @A.T. is talking about light pulses emitted by one observer and travelling completely around the circle, perhaps bouncing off many mirrors in a many-sided polygon. Seen from the Minkowski rest frame of the disc the light propagates at $c$ in either direction, but the emitter/receiver moves towards one pulse and away from the other so the reception times are not equal if the emission times were. Seen from the rotating frame, light takes longer to orbit in one direction than the other. The diagram you posted in #19 explains how the local isotropy leads to a global anisotropy - if you Einstein synchronise successive pairs of clocks around the circle, you will define the blue helix and get back to your start point out of sync with the start clock by the proper time along a Langevin worldline connecting "levels" of the blue helix (the orange line).

 

[A.T.]

I'm trying to understand your claim from Minkowski spacetime in standard cylindrical coordinates.

Consider the (subset) of Langevin observers $S$ on a fixed circumference on the rotating disc.

For what I described, you only need one "observer" (emitter and detector) and some mirrors, all fixed to the disc circumference.

See also:
https://en.wikipedia.org/wiki/Sagnac_effect
https://en.wikipedia.org/wiki/Ring_laser_gyroscope
https://en.wikipedia.org/wiki/Fibre-optic_gyroscope

 

[cianfa72]

To me this topic of rotating disc is a bit of a headache.

The diagram you posted in #19 explains how the local isotropy leads to a global anisotropy - if you Einstein synchronise successive pairs of clocks around the circle, you will define the blue helix and get back to your start point out of sync with the start clock by the proper time along a Langevin worldline connecting "levels" of the blue helix (the orange line).

The spacelike blue helix curve in #19 (defined by using Einstein synchronization) results to be orthogonal to the worldlines of the (subset) of Langevin congruence representing points/observers on the circumference at fixed radius on the rotating disc (I named $S$ this set). We can calculate its proper lenght by integrating along its path in flat spacetime. We can do the same along the subset (say $C$) of the Langevin congruence's worldlines restricted to points on a diameter of the cirumference as seen in the interial rest frame of the center of the disc. The ratio between them is greather than $\pi$. So far so good.

Now to business. Langevin observers in $S$ lay down their own standard rulers to measure the lenght of the circumference. Say the number of rulers they must juxtapose is $N$. Same procedure for the Langevin observers in $C$ along the diameter, say they must juxtapose $M$ rulers.

Does it follow that the ratio $N/M > \pi$ ?

 

[Ibix]

Does it follow that the ratio $N/M > \pi$ ?

Yes.

 

[A.T.]

To me this topic of rotating disc is a bit of a headache.

Now consider this:

You use many of your standard rulers to build 2 rings of the same diameter, placed co-axially, one just slightly above the other (almost co-located):
O_i : is at rest in the inertial frame
O_r : is rotating in the inertial frame, such that the rulers are contracted to half their proper length: The rotating rulers are of course supported, so they are under no tangential stresses.

Consequently O_r contains twice the number of the rulers that O_i contains. This makes prefect sense in the inertial frame, where the moving rulers in O_r are half as long as the static rulers in O_i.

In the rotating frame the rings are still of the same diameter, co-axial, one right above the other (almost co-located), so they should have the same circumference. But here the static rulers in O_r should be twice as long (and still twice as many) as the moving rulers in O_i.

So how can the rings still be close to each other everywhere along their perimeter (be almost co-located) in the rotating frame, if O_r contains twice as many rulers, each of twice the length, compared to those in O_i?

 

[cianfa72]

In the rotating frame the rings are still of the same diameter, co-axial, one right above the other, so they should have the same circumference. But here the rulers in O_r are twice as long (and still twice as many) as the rulers in O_i.

I'm confused about this. We can look at the worldsheets describing the standard rulers held by the Langevin observers riding on the rotating circumference. Say there are N of them along it, so the proper lenght of the blue helix in #19 is N times the proper lenght of such a ruler (at rest in the rotating circumference/Born coordinates). These worldsheets intersect the plane $t=0$ (inertial rest frame of the disc's center in cylindrical coordinates) in N pieces of the circumference. Look now at the worldsheets of the rulers that intersect the plane $t=0$ along the diameter of the circumference in M pieces.

The ratio N/M therefore should be $N/M = \pi$.

What I am missing?

 

[A.T.]

The ratio N/M therefore should be $N/M = \pi$.

If $N_i$ is the number of rulers along the non-rotating (static in the inertial frame) ring $O_i$ with $M$ rulers across its diameter, then obviously:

$N_i/M = \pi$

But since the rulers in the spinning ring $O_r$ are length contracted (in the inertial frame), you can fit more of them ($N_r$) on a circle of the same diameter (still $M$ rulers across) so:

$N_r > N_i$

and thus:

$N_r/M > \pi$

What I am missing?

I think mostly these parts:

Seen from the rotating frame, light takes longer to orbit in one direction than the other. The diagram you posted in #19 explains how the local isotropy leads to a global anisotropy - if you Einstein synchronise successive pairs of clocks around the circle, you will define the blue helix and get back to your start point out of sync with the start clock

Light propagation speed is not isotropic in a rotating frame, so using Einstein synchronization for the disc clocks on the circumference is maybe not the best choice.

See also:

The Einstein synchronisation looks this natural only in inertial frames. One can easily forget that it is only a convention. In rotating frames, even in special relativity, the non-transitivity of Einstein synchronisation diminishes its usefulness. If clock 1 and clock 2 are not synchronised directly, but by using a chain of intermediate clocks, the synchronisation depends on the path chosen. Synchronisation around the circumference of a rotating disk gives a non-vanishing time difference that depends on the direction used. This is important in the Sagnac effect and the Ehrenfest paradox. The Global Positioning System accounts for this effect.

 

[Sagittarius A-Star]

In the rotating frame the rings are still of the same diameter, co-axial, one right above the other (almost co-located), so they should have the same circumference. But here the static rulers in O_r should be twice as long (and still twice as many) as the moving rulers in O_i.

So how can the rings still be close to each other everywhere along their perimeter (be almost co-located) in the rotating frame, if O_r contains twice as many rulers, each of twice the length, compared to those in O_i?

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. But you can do this only for a section of the rim of the disk, which is smaller than the circumference to avoid a jump in O_r coordinate-time.