I Einstein's elevator: gravity without curvature?

[timmdeeg]

No. The marbles accelerate relative to the elevator, but they do not accelerate relative to each other. They have nonzero velocity relative to each other (because they are released at different times in an elevator that is accelerating in flat spacetime), but their relative velocity does not change with time (because they are just two inertially moving objects in flat spacetime).

Ah I see, thanks for correcting.

 

[SlowThinker]

I have no idea what you are talking about here.

Imagine that we are flat people, and the universe is an infinitely long cylinder. I can paint a closed line around it. To the flat us, the line will appear as an endless line. This line can be extended to a 3D closed universe as a finite but endless sheet. I'm pretty sure everyone here, including you, can understand this, so I didn't explain it in so much detail.
I didn't know what's the difficulty with solving the Einstein's field equation above the infinite sheet, so I was trying to come up with another problem that has the same solution.

Again, instead of waving your hands, you should try to actually look up some solutions. (Hint: you will find that their Riemann tensor is not zero, indicating that spacetime is curved above the slab.)

This is a case where you really, really need to do the math instead of trusting your intuition.

So I did try to look up some solutions.
One introduces a fine tuned cosmological constant and horizontal tension inside the slab, both of which are, to me, obvious nonsense. They are the only ones with a curved spacetime above the sheet.
Mathpages, and http://www.physicspages.com/2014/03/14/riemann-tensor-for-an-infinite-plane-of-mass/ and one other paper conclude that the metric has no curvature.
Personally, I don't see how the metric could be anything else than Rindler metric, but my math is not strong enough to see if these solutions are the Rindler metric or not.
If you have some reference that actually solves the EFE instead of guessing the metric, I'd love to see it.

 

[PeterDonis]

I don't see how the metric could be anything else than Rindler metric

The Rindler metric is a vacuum solution (obviously, since it is just a different coordinate chart on Minkowski spacetime--but if you want to verify it, just compute its Einstein tensor, you will see that it's zero). So it can't describe a spacetime with stress-energy in it (such as a flat slab). The first paper you reference notes this.

a fine tuned cosmological constant and horizontal tension inside the slab, both of which are, to me, obvious nonsense

The fine tuned cosmological constant in the first solution is unlikely, yes. But as the paper notes, if you want the solution to have certain properties, and if you take proper account of how to match the geometry of the infinite plane itself to the geometry of the vacuum region, you're forced to that assumption (because that's what computing the Einstein tensor of the metric tells you).

I don't see what the problem is with the horizontal tension in the slab in the second solution. It might seem unusual, but so is an infinite slab in the first place.

Mathpages, and http://www.physicspages.com/2014/03/14/riemann-tensor-for-an-infinite-plane-of-mass/ and one other paper conclude that the metric has no curvature.

That's because all of the metrics they write down are just flat Minkowski spacetime in disguise. This is easy to show by finding a coordinate transformation that puts the metric into the standard Minkowski form for each case. But it's also easy to show from the known fact that the only solution to the EFE which has zero Riemann tensor is flat Minkowski spacetime. If you assume Minkowski spacetime from the start, it's no surprise that you get it back at the end. And, as noted above, Minkowski spacetime is vacuum everywhere, so it can't describe a spacetime containing stress-energy like an infinite flat plane.

What none of these references do is actually consider the stress-energy tensor of the infinite flat plane, and its effect on the spacetime geometry at the plane, and how to match that geometry to the geometry of the vacuum region above the plane. The reason the first paper you reference gets curved metrics is that it actually tries to do this. And, as above, the reason the resulting solutions have unusual properties is that those are what it takes to realize, as closely as possible in GR, the intuitive picture of "the gravitational field above an infinite flat plane".

In other words, the first paper actually does the homework. The others just wave their hands.

If you have some reference that actually solves the EFE instead of guessing the metric

"Solving the EFE" is a bit of a misstatement. Mathematically speaking, you can write down any metric you like, compute its Einstein tensor, multiply it by $8 \pi$, and call that the "stress-energy tensor" of your spacetime. The question is whether the stress-energy tensor you get is physically reasonable, which is a matter of judgment and opinion (though there are fairly standard conditions in the literature, such as the energy conditions, that are used to classify solutions). That's basically what the first reference you give is doing: writing down metrics based on some assumptions about what a spacetime with an infinite plane of stress-energy in it would look like, computing their Einstein tensor, and seeing what that implies about the stress-energy in the spacetime.

 

[SlowThinker]

I don't see what the problem is with the horizontal tension in the slab in the second solution. It might seem unusual, but so is an infinite slab in the first place.

The infinite sheet (or the endless sheet in a closed universe) is horizontally symmetric, so there is no reason for it to be under tension or pressure. I guess if we actually built it, we could build it with a tension built in, but that's another parameter of the problem, not a part of the solution.

The Rindler metric is a vacuum solution (obviously, since it is just a different coordinate chart on Minkowski spacetime--but if you want to verify it, just compute its Einstein tensor, you will see that it's zero). So it can't describe a spacetime with stress-energy in it (such as a flat slab). The first paper you reference notes this.

It is supposed to be Rindler only above the sheet, not inside it.

So it seems that my original question, whether curvature is a necessary condition for gravity to exist, is somewhat hard to answer.

Can we at least agree that local curvature has little connection to local TTPED? I'd be happy with formulations like "gravity (here) is a consequence of spacetime curvature (somewhere else)" but saying "spacetime is curved on the surface of Earth, therefore we feel gravity" is correct but somewhat misleading.

 

[A.T.]

I'd be happy with formulations like "gravity (here) is a consequence of spacetime curvature (somewhere else)"

One could say: "Gravity on this side of the Earth being opposite to gravity on the other side of the Earth implies intrinsic space-time curvature."

 

[SlowThinker]

One could say: "Gravity on this side of the Earth being opposite to gravity on the other side of the Earth implies intrinsic space-time curvature."

That is correct, but we (or Pervect, Peter and others) are trying to explain gravity using curvature, not the other way around.

 

[A.T.]

explain gravity using curvature, not the other way around.

Then just reverse the sentence.

 

[Elroch]

Gravity (eg due to a single large mass) is only a second order effect locally. It appears like a first order effect because we often prefer to pick an accelerating frame that is stationary relative to the Earth, or whatever. Globally it is characterised by the consequences of this second order effect, which are that there is a sink (or more than one) towards which which inertial frames accelerate. I see Einstein's elevator as entirely compatible with general relativity (contrary to one view above). The inside of an accelerating elevator is indeed indistinguishable from a small region stationary relative to a massive object (eg on the surface of the Earth) if second order effects are small enough (as they are with large bodies and small regions).

 

[PeterDonis]

The infinite sheet (or the endless sheet in a closed universe) is horizontally symmetric, so there is no reason for it to be under tension or pressure.

Sorry, you're using intuition again instead of looking at the math. The math says there is no solution in GR that describes an infinite sheet with no horizontal tension in it--or at least, nobody has found one. And the two solutions in the first paper you linked to both describe infinite sheets (the stress-energy tensor components are all something times $\delta(z)$, so there is stress-energy in the infinite plane $z = 0$ but nowhere else) that have horizontal tension in them. If that seems counterintuitive to you, welcome to GR.

It is supposed to be Rindler only above the sheet, not inside it.

So where is the description of the sheet itself? And where is the math that shows that the spacetime geometry takes proper account of the presence of the sheet? Answer: nowhere. None of those references even ask this question, let alone try to answer it. The first paper, by contrast, does so: note that each of the two solutions it gives explicitly describes a stress-energy tensor that is nonzero only on the infinite $z = 0$ plane, and gives a spacetime geometry consistent with that.

You can't just wave your hands and say "well, the geometry must be this above the sheet", without actually doing the math to show that this can be done consistently with the geometry of the sheet itself. Those references are just web articles, not textbooks or peer-reviewed papers, so it's to be expected they won't necessarily get an advanced question like this right. But that's why we have rules here at PF about acceptable sources.

Can we at least agree that local curvature has little connection to local TTPED?

If you mean that, locally, we can always attribute the effects of "gravity" to local proper acceleration, yes, that's true. Spacetime curvature only comes in on larger scales, when we try to fit together local observations in different localities into a single consistent global picture.

 

[PAllen]

Imagine that we are flat people, and the universe is an infinitely long cylinder. I can paint a closed line around it. To the flat us, the line will appear as an endless line. This line can be extended to a 3D closed universe as a finite but endless sheet. I'm pretty sure everyone here, including you, can understand this, so I didn't explain it in so much detail.
I didn't know what's the difficulty with solving the Einstein's field equation above the infinite sheet, so I was trying to come up with another problem that has the same solution.

Note that cylinder surface is able to have no curvature because curvature doesn't exist for a 1 sphere; the cylinder is the product space of 1 sphere and line. Any generalization to more dimensions introduces curvature, e.g. the product space of a line with 2 sphere. Any closed simply connected 2 surface has curvature. Note that one can construct a product of circle and plane, but here the plane is not closed.

 

[PeterDonis]

Thread closed for moderation.

Edit: Thread reopened. Some off topic posts have been deleted.

 

[SlowThinker]

Note that cylinder surface is able to have no curvature because curvature doesn't exist for a 1 sphere; the cylinder is the product space of 1 sphere and line. Any generalization to more dimensions introduces curvature, e.g. the product space of a line with 2 sphere. Any closed simply connected 2 surface has curvature. Note that one can construct a product of circle and plane, but here the plane is not closed.

I believe this is only relevant if the simply connected 2 surface has to be embedded smoothly in a higher dimensional space. I don't think GR tries to embed the curved spacetime in a flat, higher dimensional space.

I was imagining the product of
- a finite flat (as in, zero curvature) torus, unfolded into a horizontal square with glued edges,
- an infinite vertical line,
- the time axis.

 

[DrGreg]

I believe this is only relevant if the simply connected 2 surface has to be embedded smoothly in a higher dimensional space. I don't think GR tries to embed the curved spacetime in a flat, higher dimensional space.

I was imagining the product of
- a finite flat (as in, zero curvature) torus, unfolded into a horizontal square with glued edges,
- an infinite vertical line,
- the time axis.

But a torus isn't simply connected.

 

[SlowThinker]

But a torus isn't simply connected.

Is that a problem? Einstein's field equation only looks at a neighborhood of a point, so it can work on any global topology, as long as it's on a smooth manifold.

 

[PAllen]

Is that a problem? Einstein's field equation only looks at a neighborhood of a point, so it can work on any global topology, as long as it's on a smooth manifold.

Yes, GR would include the case of torus cross a line cross time. However, expecting people to see that as natural from your initial descriptions is implausible. Further, you can not simply guess that a thin flat (no spatial curvature) stable matter torus would be a possible solution, or what the Weyl curvature outside the matter would look like. This would be nontrivial project to work out GR, and simple intuitions are unlikely to be valid. Note that the thin matter torus must have Ricci curvature, else it wouldn't be matter.

 

[PAllen]

I believe this is only relevant if the simply connected 2 surface has to be embedded smoothly in a higher dimensional space. I don't think GR tries to embed the curved spacetime in a flat, higher dimensional space.

I was imagining the product of
- a finite flat (as in, zero curvature) torus, unfolded into a horizontal square with glued edges,
- an infinite vertical line,
- the time axis.

Note that a simply connected two surface necessarily having curvature has nothing to do with embedding. It is a basic theorem of differential geometry. Dr. Greg already covered that a torus is not simply connected.

 

[SlowThinker]

Yes, GR would include the case of torus cross a line cross time. However, expecting people to see that as natural from your initial descriptions is implausible.

Sorry about that. It looked obvious to me, so I felt that a detailed explanation would be a waste of reader's time. I'll try to be more precise next time(s).

Further, you can not simply guess that a thin flat (no spatial curvature) stable matter torus would be a possible solution, or what the Weyl curvature outside the matter would look like. This would be nontrivial project to work out GR, and simple intuitions are unlikely to be valid. Note that the thin matter torus must have Ricci curvature, else it wouldn't be matter.

I can't really comment on a GR solution, but it should not be very different from a Newtonian solution. In Newtonian gravity, a (thin or thick) flat sheet is in an unstable balance. Because of the symmetry, it can't do anything. As pointed out before, it can be built with horizontal stress, but it's not necessary.
There will be vertical pressure, just like pressure is higher in the Earth's core, but the energy contained in that pressure is less than 11 orders of magnitude smaller than Earth's mass. So while I can't provide an exact GR solution, I don't understand why the behavior, or the field of the sheet, in GR should be significantly different from the Newtonian solution.
I feel something could go wrong due to time dilation in an infinitely deep field, but it should be fine.

 

[PeterDonis]

I can't really comment on a GR solution, but it should not be very different from a Newtonian solution.

This should be a huge red flag to you that you are waving your hands instead of actually looking at the math.

I don't understand why the behavior, or the field of the sheet, in GR should be significantly different from the Newtonian solution.

Because GR is not Newtonian gravity. The fact that in a particular weak field case--a spherically symmetric mass--Newtonian gravity happens to be a good approximation does not mean it will be a good approximation in other cases, even if the fields are weak. You can't just wave your hands and say "well, it seems like it ought to be similar". You have to actually look at the math and let it tell you when Newtonian gravity will be a good approximation and when it won't.

Or to look at it another way: Newtonian gravity happens to be a good approximation for a particular case because that's the case we actually live in, here on Earth. So we can test Newtonian gravity by experience and see that it works well. But as soon as you go outside of that particular case, all bets are off--you can't assume Newtonai gravity will still be a good approximation, or even that heuristic intuitions that work well in that particular case will work well in other cases. We have no actual experience with gravitational fields produced by infinite flat sheets, or even finite flat sheets. Nor do we have actual experience with flat torus topologies.

The reason we have GR is that Newtonian gravity broke down in a fundamental way: it couldn't be made consistent with relativity. That turned out to require completely reworking the fundamental conceptual scheme we use to model gravity. And once you know that, you know that you can't trust Newtonian concepts any more, except in the one particular case where we already know they are a good approximation.