B Sign of Expansion Scalar in Expanding FLRW Universe

[Onyx]

Yes.Indeed.

Yes.Indeed.

Yes, you're probably right that I need more of a foundation with some of these concepts. Just to be absolutely clear, you're right that I didn't focus on the extrinsic curvature tensor that much in my last thread. I think I was coming from very intuition-based thinking where I figured that if the expansion rate of points on the hypersurface is visualized in the top picture (with $\rho=sqrt(y^2+z^2)$) then the volume at a single moment would look like a sort of upside-down $f$ function plus a constant. I have found several papers that diagonalize the metric and seem to back this up. In this paper https://arxiv.org/pdf/gr-qc/9707024.pdf (starting on page 4) there is a transformation to $r=x-vt$, and then a new time coordinate which kills the diagonal. The graph of the determinant (if replacing $1$ on the bottom with $c$, I think, and provided $v<c$) looks like a bowl. However, If I kill the diagonal instead by $dt=d\tau-\frac{-vf}{1-v^2f^2}dx$ https://physics.stackexchange.com/q...ing-proper-volume-in-the-alcubierre-spacetime, the determinant looks like more of a hat, which would make more sense to me if there was expansion in the front and then contraction in the back. All I was wondering was just which one of these approaches is more correct.

 

[PeterDonis]

the expansion rate of points on the hypersurface

Might well have no physical meaning whatever. For it to have any physical meaning, the points and the hypersurface have to be chosen very carefully. And in cases where such a choice is even possible, there are better ways of understanding what is going on than looking at what you are looking at.

For example, in FRW spacetime, the hypersurface has to be chosen to be orthogonal to the worldlines of comoving observers everywhere. But this is only possible in the first place because there are comoving observers everywhere, i.e., observers who always see the universe as homogeneous and isotropic, and because the worldlines of such observers "fit" into the overall spacetime geometry in just the right way that we can choose a family of hypersurfaces that are everywhere orthogonal to those worldlines, and call those hypersurfaces "space" at different "times", and then say that "space" is "expanding". But even given all that, it's simpler to just look at the comoving observers themselves and say that they are always moving apart from each other--which is what the expansion scalar of the congruence of comoving worldlines represents. Then you don't have to worry about what "space" even means. You can just focus on a family of observers (or objects like galaxies or galaxy clusters) and their (average) motion. Much more concrete and much easier to grasp.

A somewhat similar general approach can be taken in Alcubierre spacetime. Here there is no single family of observers that matches up with a symmetry of the spacetime, but there are two obvious kinds of observers to look at: observers inside the warp bubble and at rest relative to it; and observers outside the warp bubble and at rest in some chosen inertial frame of the background flat spacetime (and in which the coordinates in which the metric is written down are defined). Then the "velocity" $v$ of the warp bubble is the effective speed at which the first set of observers moves relative to the second. Simple and easy to grasp (although of course if this effective speed is greater than the speed of light there are many possible complications that need to be dealt with). And you don't need to worry about "the shape of space" or anything like that. Which is good because in this spacetime, unlike Alcubierre spacetime, there is no choice of hypersurfaces that makes everything look simple.

All I was wondering was just which one of these approaches is more correct.

None of them are "more correct". There is no such thing as "more correct" in terms of mathematical descriptions; they're all "correct" (at least as long as they are the result of valid mathematical operations). But some are much easier to interpret physically than others.

 

[Onyx]

Might well have no physical meaning whatever. For it to have any physical meaning, the points and the hypersurface have to be chosen very carefully. And in cases where such a choice is even possible, there are better ways of understanding what is going on than looking at what you are looking at.

For example, in FRW spacetime, the hypersurface has to be chosen to be orthogonal to the worldlines of comoving observers everywhere. But this is only possible in the first place because there are comoving observers everywhere, i.e., observers who always see the universe as homogeneous and isotropic, and because the worldlines of such observers "fit" into the overall spacetime geometry in just the right way that we can choose a family of hypersurfaces that are everywhere orthogonal to those worldlines, and call those hypersurfaces "space" at different "times", and then say that "space" is "expanding". But even given all that, it's simpler to just look at the comoving observers themselves and say that they are always moving apart from each other--which is what the expansion scalar of the congruence of comoving worldlines represents. Then you don't have to worry about what "space" even means. You can just focus on a family of observers (or objects like galaxies or galaxy clusters) and their (average) motion. Much more concrete and much easier to grasp.

A somewhat similar general approach can be taken in Alcubierre spacetime. Here there is no single family of observers that matches up with a symmetry of the spacetime, but there are two obvious kinds of observers to look at: observers inside the warp bubble and at rest relative to it; and observers outside the warp bubble and at rest in some chosen inertial frame of the background flat spacetime (and in which the coordinates in which the metric is written down are defined). Then the "velocity" $v$ of the warp bubble is the effective speed at which the first set of observers moves relative to the second. Simple and easy to grasp (although of course if this effective speed is greater than the speed of light there are many possible complications that need to be dealt with). And you don't need to worry about "the shape of space" or anything like that. Which is good because in this spacetime, unlike Alcubierre spacetime, there is no choice of hypersurfaces that makes everything look simple.None of them are "more correct". There is no such thing as "more correct" in terms of mathematical descriptions; they're all "correct" (at least as long as they are the result of valid mathematical operations). But some are much easier to interpret physically than others.

So then the extrinsic curvature tensor is not any more associated with a static frame $x^i=x$ than it is with a comoving frame $x^i=x-vt$? It is just an entirely different object not relevant to the choice of coordinates (I'm not asking sarcastically or anything, just curious)?

 

[Onyx]

but there are two obvious kinds of observers to look at: observers inside the warp bubble and at rest relative to it; and observers outside the warp bubble and at rest in some chosen inertial frame of the background flat spacetime (and in which the coordinates in which the metric is written down are defined).

Thank you, I find this part very helpful.

 

[PeterDonis]

So then the extrinsic curvature tensor is not any more associated with a static frame $x^i=x$ than it is with a comoving frame $x^i=x-vt$?

There is no "the" extrinsic curvature tensor. There are different ones for different choices of spacelike hypersurfaces. Typically defining a frame also defines a choice of spacelike hypersurfaces, but the hypersurfaces don't necessarily look very simple or have any direct physical meaning.

 

[Onyx]

There is no "the" extrinsic curvature tensor. There are different ones for different choices of spacelike hypersurfaces. Typically defining a frame also defines a choice of spacelike hypersurfaces, but the hypersurfaces don't necessarily look very simple or have any direct physical meaning.

Oh. That's true there, there would be different ones.