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.