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 $\nabla_b X^a := (\nabla X)_{b}{}^{a}$.