I Einstein's elevator: gravity without curvature?

[pervect]

This is a rather old issue, but one that has recently been on my mind.

We often say that gravity is the curvature of space-time, with good reason. At the same time, we also talk about the "gravity" in Einstein's elevator, as an example of the equivalence principle. This is also with good reason, and is historically important. But Einstein's elevator is set in flat space-time. There is no curvature. So if we talk about gravity in Einstein's elevator, we're talking about gravity without curvature. But we have perhaps just previously said - or some other poster has previously said - that gravity is due to the curvature of space-time. Thus we are left with what appears to be a confused message.

What's the best way of resolving this issue? Are we forced to talk about connections and Christoffel symbols to adequately define what we mean by gravity in Einstein's elevator? Is there a better way of doing this, preferably one that is accessible to B and I level readers? I don't really regard either connections or Christoffel symbols as being I level, and definitely not B level.

On the flips side, can we modify or qualify saying that gravity is due to the curvature of space-time , in a way that will not confuse the presentation, but also allow us to talk about the "gravity" in Einstein's elevator?

 

[David Lewis]

In his thought experiment, by "gravity", Einstein meant Newtonian gravity, not GR gravity.

 

[Dale]

@pervect

I agree with your point. In Einstein's historical papers he seemed to (inconsistently but usually) use "gravity" to refer to the Christoffel symbols. That is also what is closest to the Newtonian force of gravity.

Perhaps we should use "gravitation" to refer to the general phenomenon which is modeled using curved spacetime in GR and use "gravity" to refer specifically to the Christoffel symbols.

 

[Ibix]

Perhaps we should use "gravitation" to refer to the general phenomenon which is modeled using curved spacetime in GR and use "gravity" to refer specifically to the Christoffel symbols.

I'd be wary of that particular choice leading to the gravity wave/gravitational wave thing again.

 

[Nugatory]

So if we talk about gravity in Einstein's elevator...

That's always felt backwards to me. The equivalence principle doesn't quite say that acceleration IS gravity, so it doesn't say that what's going on in the elevator is gravity. Instead, it says that (the local effects that we ascribe to) gravity can be understood as acceleration effects, like those we observe in the elevator. The elevator tells us a lot about how gravity works, but gravity tells us nothing about the elevator.

 

[PeterDonis]

What's the best way of resolving this issue?

To be clear that the term "gravity" does not have a single meaning; it can refer to different things, so for clarity, if you use it, you have to make clear which meaning you are using. For example, if I want to use "gravity" to mean "spacetime curvature", for clarity I will say specifically "tidal gravity". Or if I want to use "gravity" to mean what is observed in an "Einstein elevator" scenario, I will say "acceleration due to gravity".

 

[Ibix]

The equivalence principle doesn't quite say that acceleration IS gravity, so it doesn't say that what's going on in the elevator is gravity. Instead, it says that (the local effects that we ascribe to) gravity can be understood as acceleration effects, like those we observe in the elevator.

Agreed. Newton says that there's a fundamental difference between the elevator under thrust and sitting on a planet, but there's no way to tell which is which. The elevator is Einstein saying "here are two things that have webbed feet and go quack - maybe we should stop calling one of them a cat".

So I think the point is that if we're trapped in a box and the floor is pushing up on our feet, we really ought to limit ourselves to observing that we're using an accelerating frame of reference. Only when we've detected tidal effects, or managed to spot some stars with changing redshift or something should we come down one way or another and say "this is gravity" or "this is acceleration".

 

[SlowThinker]

Perhaps I'm just illustrating the confusion you're talking about, but

We often say that gravity is the curvature of space-time, with good reason. At the same time, we also talk about the "gravity" in Einstein's elevator, as an example of the equivalence principle. This is also with good reason, and is historically important. But Einstein's elevator is set in flat space-time. There is no curvature. So if we talk about gravity in Einstein's elevator, we're talking about gravity without curvature.

The clock at the top of the elevator run faster than the clock at its bottom, don't they? So we do have a curved spacetime here, don't we?
Maybe it's not "curved" in the mathematical sense, but it certainly isn't Minkowski spacetime.

 

[PeterDonis]

The clock at the top of the elevator run faster than the clock at its bottom, don't they?

Yes.

So we do have a curved spacetime here, don't we?

No. The difference in clock rates from bottom to top of the elevator does not, in and of itself, mean that spacetime is curved.

Maybe it's not "curved" in the mathematical sense, but it certainly isn't Minkowski spacetime.

Yes, it is. Google "Rindler coordinates". These are the coordinates in which observers at rest in the elevator are at rest. You will find that in these coordinates, the "rate of time flow" of an observer at rest depends on their "height" (position in the direction from bottom to top of the elevator), and that these coordinates are just a different set of coordinates on Minkowski spacetime.

 

[Ibix]

Maybe it's not "curved" in the mathematical sense, but it certainly isn't Minkowski spacetime.

It is Minkowski spacetime. How could me sitting on top of a rocket change the shape of spacetime? It's just that the natural coordinates to use in an accelerating elevator are curved ones that don't have the same notion of simultaneity as the usual inertial coordinates. And therefore objects that are at rest in this coordinate system are not following inertial paths and don't have the same simultaneity relationship that you would expect from inertial clocks.

 

[DrGreg]

The clock at the top of the elevator run faster than the clock at its bottom, don't they? So we do have a curved spacetime here, don't we?
Maybe it's not "curved" in the mathematical sense, but it certainly isn't Minkowski spacetime.

The coordinates in which the elevator is at rest are not Minkowski coordinates (and so non-inertial coordinates), but the spacetime itself is independent of your choice of coordinates and is still flat and is still Minkowski spacetime.

 

[SlowThinker]

No. The difference in clock rates from bottom to top of the elevator does not, in and of itself, mean that spacetime is curved.

So maybe the curvature isn't the most important part of gravity?
Or, are you saying that gravity is the difference between an accelerating elevator, and a stationary elevator near a massive object?

That would be quite an important problem with terminology, because to most people, gravity is the thing pulling everyone down, not the minuscule effects that can't be directly measured until you actually orbit the Earth in a rocket.

 

[SlowThinker]

How could me sitting on top of a rocket change the shape of spacetime?

Well I never quite understood how you can "transform away" acceleration. No matter how hard I'm imagining space and time axes around me, I still can't fly. So something must be wrong with the spacetime around here.

 

[PeterDonis]

So maybe the curvature isn't the most important part of gravity?

No, it means that curvature is not the same as gravitational time dilation. Spacetime curvature is tidal gravity.

are you saying that gravity is the difference between an accelerating elevator, and a stationary elevator near a massive object?

The difference in the geometry of spacetime, and therefore in the global "shape" of the path through spacetime of the elevator, yes. See below.

to most people, gravity is the thing pulling everyone down, not the minuscule effects that can't be directly measured until you actually orbit the Earth in a rocket.

That's why, when we want to be precise, we say that spacetime curvature is tidal gravity, specifically, instead of just "gravity".

I never quite understood how you can "transform away" acceleration.

You can't. The acceleration (meaning proper acceleration--the acceleration that is felt by the person inside the elevator) is exactly what remains the same in the flat spacetime vs. the curved spacetime cases. What is different is the global "shape" of the path through spacetime of the person inside the elevator; but there is no way for that person, purely on the basis of measurements made inside the elevator, to know what the global "shape" of his path through spacetime is. To know that he has to look outside, at distant objects.

 

[Ibix]

Well I never quite understood how you can "transform away" acceleration.

You can transform away coordinate acceleration. Just drop a set-square, and use it to define your coordinate system while it is free-falling, rather than while it is sitting on the surface of the Earth. In this description of the world the freely moving set-square is at rest, and the floor is accelerating upwards towards it.

No matter how hard I'm imagining space and time axes around me, I still can't fly.

No, but you can free-fall. And you can choose whether to treat the floor as coming up to meet you or yourself as falling towards the floor. The former is the natural description in Newtonian physics; the latter in general relativity (at least at the local level).

The point of the elevator/surface of the Earth is that, in both cases, the floor is not moving inertially and a dropped ball is moving inertially. So the explanation for the ball falling is, in both cases, better phrased as "the floor is coming up to meet it". What's different is why the floor is non-inertial. In the elevator it's because the spacetime is flat and the floor is being accelerated through it. In the room it's because spacetime is curved but the "bottom" of the curved region is filled with matter which pushes up on the layer above it, which pushes up on the layer above it until you get to the surface, which pushes upwards on your feet, but not the ball because it's not in contact with it.

 

[RockyMarciano]

AFAIK Einstein claims that there is no physical way to tell apart the effects of gravity from the effects of acceleration in the elevator. And not just locally but for as long as the elevator accelerates. And as DrGreg says the use of noinertial coordinates shouldn't affect the fact the elevator is in flat Minkowski spacetime. So I can se pervect's point that this thought experiment seems to blur the distinction between flat and curved spacetime as pertains to gravity physical effects. On the other hand there's the fact that the gravity effects that are usually associated to curvature in the curved spacetime setting are tidal effects that seem different from the effect obtained from acceleration in flat spacetime, but then one has to wonder how exactly then the principle of equivalence links acceleration and "actual" gravitational tidal effects.

 

[PeterDonis]

The difference in clock rates from bottom to top of the elevator does not, in and of itself, mean that spacetime is curved.

I realized after posting this that this does raise a question about an argument that appears in the literature. I have posted a separate thread about it (note that this thread is an "A" level thread, since it requires considerable technical background to fully understand the argument in the literature that I am referring to):

https://www.physicsforums.com/threa...me-dilation-imply-spacetime-curvature.919181/

 

[PeterDonis]

not just locally but for as long as the elevator accelerates

No, this is not the case. Einstein's argument is limited to a short enough interval of time as well as a small enough elevator in spatial extent.

one has to wonder how exactly then the principle of equivalence links acceleration and "actual" gravitational tidal effects

It doesn't. Investigating whether, and to what extent, tidal gravity (spacetime curvature) is present goes beyond the EP.

 

[RockyMarciano]

No, this is not the case. Einstein's argument is limited to a short enough interval of time as well as a small enough elevator in spatial extent.

Sure, I was referring to the "gravity" effect, feeling weight.

It doesn't. Investigating whether, and to what extent, tidal gravity (spacetime curvature) is present goes beyond the EP.

Yes. But then to what purely gravitational effect does the equivalence principle link acceleration in flat spacetime? Or you mean the EP is not related to GR?

 

[SlowThinker]

(Feel free to move this to the other thread)
So if there is an infinitely wide and finitely deep flat slab, is the spacetime above it curved or not? Do I feel gravity or not? What is attracting me to it? And most importantly, how is this different from the accelerating elevator?

 

[PeterDonis]

to what purely gravitational effect does the equivalence principle link acceleration in flat spacetime?

None. It links proper acceleration in flat spacetime to proper acceleration in curved spacetime.

 

[PeterDonis]

(Feel free to move this to the other thread)

It doesn't belong there. If anything, it belongs in a separate thread of its own, at least if you want to dig into it deeply enough. But I'll give a brief response here.

an infinitely wide and finitely deep flat slab

It turns out that there is not a single consistent solution to the Einstein Field Equation that meets this description. (There is in Newtonian gravity, but we're not talking about Newtonian gravity here.) There are different solutions that come somewhat close, but each one has properties that are different from the Newtonian solution, and which will create intuitive difficulties if you try to interpret it the way the Newtonian solution is interpreted.

Do I feel gravity or not?

You never feel gravity. You feel proper acceleration. If you put yourself in a rocket and turn on its engine, you will feel proper acceleration, which you can adjust to any desired value (such as 1 g) by adjusting the thrust of the engine.

What your global path through spacetime looks like when you do that will depend on the global spacetime geometry--it will be different in flat spacetime vs. Schwarzschild spacetime vs. whatever "infinitely wide flat slab" solution you adopt for purposes of discussion--but none of that has anything to do with the equivalence principle or Einstein's "elevator" argument. As far as you can tell from measurements made inside the elevator, all of those cases will be indistinguishable.

What is attracting me to it?

Um, the stress-energy of the slab, and its effect on the spacetime geometry? Remember that in GR, gravity is not a force, so thinking of yourself as being "attracted" to the slab isn't the best way to think of it--that's Newtonian thinking, and we're not talking about Newtonian gravity.

how is this different from the accelerating elevator?

As far as measurements made inside the elevator are concerned, it isn't. See above.

 

[SlowThinker]

It turns out that there is not a single consistent solution to the Einstein Field Equation that meets this description.

Really? What about a globally (horizontally) closed space, so that the total mass would be finite?
My point is, the space above the slab will likely have no curvature, since it is the same everywhere in space, and one dimension (time) is not enough to create a curvature.
So we have a flat spacetime, but I can still walk on the slab. And there is a "gravitational" time dilation. Sounds like gravity to me.

Edit:

What is attracting me to it?

Um, the stress-energy of the slab, and its effect on the spacetime geometry?

What effect really? If it's not curvature, then there must be some intermediate spacetime that fits between a flat and a curved one?

Remember that in GR, gravity is not a force, so thinking of yourself as being "attracted" to the slab isn't the best way to think of it--that's Newtonian thinking, and we're not talking about Newtonian gravity.

Hmmm... (thinks a lot)... So what is "accelerating" me towards the slab?

 

[pervect]

So maybe the curvature isn't the most important part of gravity?
Or, are you saying that gravity is the difference between an accelerating elevator, and a stationary elevator near a massive object?

That would be quite an important problem with terminology, because to most people, gravity is the thing pulling everyone down, not the minuscule effects that can't be directly measured until you actually orbit the Earth in a rocket.

I'd tend to agree that most people think of gravity as a "thing that pulls everyone down". Henceforth "TTPED". And this does raises some important issues of terminology, which is one of the issues I wanted to discuss - what is a good terminology to use, one that ideally will reach a broad audience, from the most to the least sophisticated?

In Newtonian physics, we can consider two cases. One is "real gravity" due to some nearby massive body. In Newtonian physics, "TTTPED" is in this case considered to be a real force. Another case we can consider is an accelerating elevator, Einstein's Elevator. In this case, "TTTPED" is not a real force at all, but a pseudo-force, or an inertial force. The easiest way to see this is to analyze EInstein's elevator in an inertial frame, where we can directly apply Newton's laws, and note that the only real force on an object on the floor of the elevator is the floor pushing up on the object, causing it to accelerate. Most I (intermediate) level and above readers will have been exposed to this concept, and perhaps some B-level readers as well, so in many cases it should (hopefully) just be a matter of refreshing their memory.

Where do we get the idea that there is a "TTPED" in the elevator? This happens when we switch to a non-inertial frame of reference. To do this, we have to modify Newton's laws, they no longer directly apply in a non-inertial frame of reference. This is perhaps not a terribly hard thing to do, but the details of doing it are somewhat involved, and I think there are a lot of readers who just accept the result, because they've heard it before, without going through the calculations, or knowing the details of the calculations. Also, calculations of this sort would be at I-level, I think, and doing them would leave out all the B-level readers. But the result is what's important, in Newtonian physics, when we switch to an accelerated frame of rerference, we use laws that are formally similar to Newton's laws, but we add in these "fictitious forces" that are not real forces. And it is these fictitious forces that we feel on our feet when we stand in Einstein's elevator. (add). To be a bit pendantic, and try to be as precise as possible, I would say that what we actually feel on the soles of our feet is pressure, and that when we integrate the normal component of the pressure over the area of the "footprint", we get the total force, or weight.

I'll touch lightly on another issue here that is that we expect there to be a lot of similarities between the elevator rider and "real gravity". It appears though that this point occasionally needs more discussion - perhaps not in this thread, though. We can be a bit more specific here - we can use a scale to measure the weight (as a force) of an object on the elevator, via the technique mentioned above (integrating the normal component of the pressure over the footprint), and that an identical scale in the "real" gravitational field will yield an identical reading for the weight (force) on the scale. In Newtonian theory this is a happy accident, GR provides a theoretical basis for the two scales to read the same number.

The above discussion was Newtonian, things start to get even more interesting when we try to extend the calculations for special relativity. What happens is that in an accelerated frame, we not only have a "TTPED", but we also have other effects. One of the best-known of these effects is what can be loosely termed "gravitational time dilation". When we look at how we express the laws of physics in an accelerating frame in special relativity, we wind up having to do more than just add in a fictitious force, as we used to be able to do in Newtonian physics.

I would say at the A-level that the best mathematical representation of the "TTPED" s the Christoffel symbol. (At least it's the best I can think of). Specifically we consider a free particle, and write the geodesic equations of motion for said free particle via the Christoffel symbols. We compare the resulting equations of motoins to those we'd get via "Newton's laws" to gain insight into how the particle moves. And we can write out the metric for the accelerated frame, and discuss the aspects of "gravitational time dilation". Unfortunately, I don't see these observations as being too helpful to the vast majority of readers at the B and I levels But I hope I can get across the point that going from an inertial frame to an accelerated frame is no longer a matter of adding in a fictitious force - we need to do more. And that this starts us down the path (though it doesn't take us all the way to the end of the path) in thinking that gravity in GR isn't just a force, at least not a "real" force.

Well I never quite understood how you can "transform away" acceleration. No matter how hard I'm imagining space and time axes around me, I still can't fly. So something must be wrong with the spacetime around here.

I don't think this is all that hard to understand, personally. If you take a ride on the "Vomit comet", or imagine doing so, you have a physical example of what it means to "transform away acceleration". You can fly - you just need a plane. When you do fly, you see that spacetime isn't so very different. Specifically, riding in a plane doesn't change the nature of space-time, but it can (with the right trajectory) get rid of the "TTPED", at least for a while.

 

[A.T.]

What's the best way of resolving this issue?

As Peter said, distinguish between "attraction" and "tidal effects". DrGreg made a great diagram (see below). Explain that attraction needs no curvature locally, but is globally the only way to combine all those local non-inertial coordinate charts, that are at rest relative to each other.

This is my own non-animated way of looking at it:

• A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[SlowThinker]

So one thing that might help is renaming gravitational time dilation to accelerational time dilation. That gets it out of the way.

It seems to me that the infinite slab example is really crucial. It's a direct extension of the Einstein's Elevator, and while I can't speak for others, to me it looks like a very good approximation of the Earth's surface, unlike a sphere, which would be the next level.
If you can explain the infinite slab, wrapping it around Earth should be easy.
I'd say that the river model can do that.

If you don't want to use the river model, you'll probably have to go backwards: curvature allows acceleration without (spatial) movement, and Earth's acceleration is TTPED.
But this approach leaves a question: if I have a finite but large slab, how is the curvature far away at the rim relevant to the gravity felt around its center, where the spacetime is nearly flat?
You can't have gravity without curvature popping up somewhere, but saying that gravity is curvature is completely counterintuitive.

 

[pervect]

As Peter said, distinguish between "attraction" and "tidal effects". DrGreg made a great diagram (see below). Explain that attraction needs no curvature locally, but is globally the only way to combine all those local non-inertial coordinate charts, that are at rest relative to each other.

I tend to agree, though I have been remiss in saying so. I like the idea of talking about "the acceleration due to gravity" as a way to disambiguate it from tidal gravity, aka the Riemann curvature tensor.

 

[PeterDonis]

What about a globally (horizontally) closed space, so that the total mass would be finite?

I have no idea what you are talking about here. I suggest that, rather than waving your hands, you actually take some time to see if you can find references discussing this type of solution in GR. (You might also search PF: IIRC we have had past threads on this topic.)

the space above the slab will likely have no curvature, since it is the same everywhere in space, and one dimension (time) is not enough to create a curvature.

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.

 

[timmdeeg]

Some questions.

Is it true that tidal gravity manifests itself by nearby geodesics showing relative acceleration which is not frame dependent?

If yes consider the elevator accelerating in flat spacetime on top of which two marbles are being dropped one after the other. Wouldn't their geodesics accelerate away from each other and thus proof tidal gravity inside the elevator? Or is the term tidal gravity restricted to the case where the relative acceleration of geodesics doesn't happen in only one direction, as in the elevator.

Is it legitimate to consider an infinite large massive plate such that the gravitational field lines are parallel? Then relative acceleration of geodesics of freely falling objects should happen in only one direction (vertical, like in the accelerating elevator). Is the answer whether or not the term tidal gravity is correct in this case the same as for the accelerating elevator?

EDIT I think the infinite plate isn't allowed as a thought experiment because it would probably collapse.

 

[PeterDonis]

Is it true that tidal gravity manifests itself by nearby geodesics showing relative acceleration which is not frame dependent?

Tidal gravity is relative acceleration of nearby geodesics, and yes, it is not frame dependent.

consider the elevator accelerating in flat spacetime on top of which two marbles are being dropped one after the other. Wouldn't their geodesics accelerate away from each other

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).

Is it legitimate to consider an infinite large massive plate such that the gravitational field lines are parallel?

Only if you can find a solution of the Einstein Field Equation that has this property.

 

[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.