What is gravity?

[DanMP]

3 minutes ago, Genady said:

Since energy-momentum and curvature are not timelike separated, there is no meaningful causal relation between them.

So the curvature has no cause?

[Genady]

9 minutes ago, DanMP said:

So the curvature has no cause?

Yes, it has not. Every spacetime has a curvature.

[DanMP]

22 minutes ago, Genady said:

Yes, it has not. Every spacetime has a curvature.

Maybe, but the curvature depends on the local mass and/or energy ... You remove the mass/energy and the gravity is gone.

[Genady]

30 minutes ago, DanMP said:

curvature depends on the local mass and/or energy

Not necessarily. For example, the Schwarzschild geometry exists in vacuum.

[swansont]

1 hour ago, Genady said:

Since energy-momentum and curvature are not timelike separated, there is no meaningful causal relation between them.

This implies that lightlike separations are not causal. 

[Genady]

3 minutes ago, swansont said:

This implies that lightlike separations are not causal. 

Thank you for the correction. I should've said, "... are not timelike or lightlike separated ..."

[joigus]

2 hours ago, DanMP said:

Maybe, but the curvature depends on the local mass and/or energy ... You remove the mass/energy and the gravity is gone.

It's a bit more subtle than this, I think. You can have vacuum solutions with curvature. If you think about it, the Schwarzschild solution is a vacuum solution. De Sitter and anti-De Sitter are too. OTOH, the Einstein field equations are nonlinear, so I wouldn't rule out other exotic vacuum solutions with curvature.

[swansont]

1 hour ago, Genady said:

Thank you for the correction. I should've said, "... are not timelike or lightlike separated ..."

Isn’t the separation of momentum-energy and curvature lightlike? The fluctuations (gravitational waves) propagate at c

23 minutes ago, joigus said:

It's a bit more subtle than this, I think. You can have vacuum solutions with curvature. If you think about it, the Schwarzschild solution is a vacuum solution. De Sitter and anti-De Sitter are too. OTOH, the Einstein field equations are nonlinear, so I wouldn't rule out other exotic vacuum solutions with curvature.

What’s the curvature of the Schwarzschild solution?

1 hour ago, Genady said:

Not necessarily. For example, the Schwarzschild geometry exists in vacuum.

I’m confused.

“The Schwarzschild geometry describes the spacetime geometry of empty space surrounding any spherical mass”

https://jila.colorado.edu/~ajsh/bh/schwp.html

[joigus]

5 minutes ago, swansont said:

What’s the curvature of the Schwarzschild solution?

Infinite at one point. Zero everywhere else. But you're right. It's not a good example.

De Sitter is more what I was thinking about.

[Genady]

1 hour ago, swansont said:

Isn’t the separation of momentum-energy and curvature lightlike?

In the Einstein field equation, the curvature on one side and the energy-momentum on the other side are not lightlike separated.

1 hour ago, swansont said:

The Schwarzschild geometry describes the spacetime geometry of empty space surrounding any spherical mass

It is not local, contrary to

4 hours ago, DanMP said:

but the curvature depends on the local mass and/or energy

 

[MigL]

1 hour ago, joigus said:

De Sitter is more what I was thinking about.

You're gonna have to elaborate on that one also.

My understanding is that De Sitter models an essentially flat universe, devoid of ordinary matter, where the dynamics are dominated by the Cosmological Constant/Dark Energy.
IOW, there is a local energy that accounts for any deviation from absolute flatness.
 

[Genady]

2 hours ago, joigus said:

It's a bit more subtle than this, I think. You can have vacuum solutions with curvature. If you think about it, the Schwarzschild solution is a vacuum solution. De Sitter and anti-De Sitter are too. OTOH, the Einstein field equations are nonlinear, so I wouldn't rule out other exotic vacuum solutions with curvature.

Aren't there many examples, at least in principle? In particular:

"Q: The information that gets lost when we go from the Riemann tensor to the Ricci tensor does not affect the energy-momentum tensor nor Einstein’s equations. What is the meaning of this lost information then?

A: It means that for a given source configuration, there can be many solutions to Einstein’s equations. They all have the same right-hand side, namely \(T^{\mu \nu}\). But they simply have different physical properties. For example, the simplest case is to ask: what if this energy-momentum stuff is zero? If it is zero, does it mean that there is no gravitation, no interesting geometry at all? No. It allows gravitational waves."

Susskind, Cabannes. General Relativity: The Theoretical Minimum. 

2 hours ago, joigus said:

Zero everywhere else.

Not according to this: homework and exercises - Non-zero components of the Riemann tensor for the Schwarzschild metric - Physics Stack Exchange

[MigL]

I can't add much to the mathematical discussion, but from a physical standpoint, the fact that gravity gravitates is what causes its non-linearity.
If it were somehow possible to remove the energy-momentum components of an existing gravitational field/geometry, some part of that field/geometry would still exist, because its own energy density causes ( caused ? ) a part of the field/geometry.
But that original gravitational field, or geometry, had to have been caused by a local energy-momentum distribution.

I don't know about mathematical considerations, but physical considerations tell us local curvature has to be caused by something ( and from my understanding of De Sitter, so is global curvature ) because our universe is causal.

[swansont]

3 hours ago, joigus said:

Infinite at one point. Zero everywhere else. But you're right. It's not a good example.

Isn’t the sun’s (or earth’s) field approximately a solution to the Schwarzschild geometry? 

2 hours ago, Genady said:

In the Einstein field equation, the curvature on one side and the energy-momentum on the other side are not lightlike separated.

They are equal, but isn’t that a static solution? And if you perturb the energy-momentum, don’t you get a lightlike fluctuation in the curvature?

2 hours ago, Genady said:

It is not local, contrary to

How is it not local?

1 hour ago, Genady said:

 

"Q: The information that gets lost when we go from the Riemann tensor to the Ricci tensor does not affect the energy-momentum tensor nor Einstein’s equations. What is the meaning of this lost information then?

A: It means that for a given source configuration, there can be many solutions to Einstein’s equations. They all have the same right-hand side, namely Tμν . But they simply have different physical properties. For example, the simplest case is to ask: what if this energy-momentum stuff is zero? If it is zero, does it mean that there is no gravitation, no interesting geometry at all? No. It allows gravitational waves."

Susskind, Cabannes. General Relativity: The Theoretical Minimum. 

Gravitational waves are a dynamic effect, though. What if we limit ourselves to a static configuration?

[Genady]

10 minutes ago, swansont said:

They are equal, but isn’t that a static solution?

All solutions obey this equation.

 

11 minutes ago, swansont said:

How is it not local?

The mass is not at the same location where the Schwarzschild metric is. 

 

13 minutes ago, swansont said:

Gravitational waves are a dynamic effect, though. What if we limit ourselves to a static configuration?

I don't see why it would matter. There are simply more independent variables in the curvature tensor than there are independent equations in the Einstein field equation.

[swansont]

3 minutes ago, Genady said:

The mass is not at the same location where the Schwarzschild metric is. 

Local ≠ co-located

[joigus]

2 hours ago, MigL said:

You're gonna have to elaborate on that one also.

Spatially flat and space-time flat are often conflated in the literature. I would have to review the Riemann coefficients with 0t pairings of indices (a space cannot warp in just one dimension). I'm not sure nor do I have the time (nor the energy) now to review these notions. Maybe someone can do it for all of us. Most likely @Markus Hanke. I'm sure DS space-time is often characterised as having constant curvature*. We're kind of mixing it all together as if the scalar curvature were "the thing" that says whether a manifold is flat or nor. It's  more involved. If just one R_ijkl is non-zero, the manifold is just not flat.

Calabi-Yau manifolds are another example which are Ricci-flat (R=0), but not flat.

2 hours ago, Genady said:

Aren't there many examples, at least in principle? In particular:

"Q: The information that gets lost when we go from the Riemann tensor to the Ricci tensor does not affect the energy-momentum tensor nor Einstein’s equations. What is the meaning of this lost information then?

A: It means that for a given source configuration, there can be many solutions to Einstein’s equations. They all have the same right-hand side, namely Tμν . But they simply have different physical properties. For example, the simplest case is to ask: what if this energy-momentum stuff is zero? If it is zero, does it mean that there is no gravitation, no interesting geometry at all? No. It allows gravitational waves."

Susskind, Cabannes. General Relativity: The Theoretical Minimum. 

Not according to this: homework and exercises - Non-zero components of the Riemann tensor for the Schwarzschild metric - Physics Stack Exchange

Yes. Thank you. Read my comments to @MigL on flat vs spatially flat, Ricci-flat, and so on. They're very much in the direction you're pointing. Right now I'm beat, but I promise to follow up on this.

41 minutes ago, swansont said:

Isn’t the sun’s (or earth’s) field approximately a solution to the Schwarzschild geometry? 

Yes, of course you're right. This theorem due to Birkhoff[?] that the external solution is unique as long as it's static and spherically symmetric. Schwarzschild's solution was just an unfortunate example. I know very little about exact solutions in GR. I just figure there must be solutions with not all curvatures zero with no clearly identifiable matter distribution giving rise to them.

* 

Quote

In mathematical physics, n-dimensional de Sitter space (often abbreviated to dS_n) is a maximally symmetric Lorentzian manifold with constant positive scalar curvature. It is the Lorentzian^{[further explanation needed]} analogue of an n-sphere (with its canonical Riemannian metric).

 

[Markus Hanke]

6 hours ago, joigus said:

Maybe someone can do it for all of us. Most likely @Markus Hanke.

Sorry, I have not been able to keep up with these discussions over the last few days, as I’m busy with a large RL project.

What was the question here?

deSitter spacetime has non-zero Riemann tensor, so it’s not flat.

[DanMP]

17 hours ago, Genady said:

It is not local, contrary to

21 hours ago, DanMP said:

but the curvature depends on the local mass and/or energy

 

When I wrote "local" I meant "in the proximity".

Let's consider a simple example: I am on a spaceship hovering somewhere on the Earth's orbit around the Sun, not exactly on collision course  , but 1 million km closer to the Sun (at 148.6 million km from the Sun, when the Earth is at 149.6). Most of the time/year, in order to stay at that fixed point, I have to use the propulsion to compensate the Sun's "gravitational pull". Once a year, when the Earth is near, I have to orient the ship thrusters towards the Earth, not the Sun, in order to maintain my fixed position in Sun's reference frame.

The question is: why I have yo do it?

1. because of gravitational time dilation changes?
2. because of spacetime curvature changes?
3. or because the presence of Earth (more exactly its mass) is changing "the curvature of spacetime" in my proximity? Another question is how exactly the Earth does that?

[Genady]

5 minutes ago, DanMP said:

When I wrote "local" I meant "in the proximity".

Let's consider a simple example: I am on a spaceship hovering somewhere on the Earth's orbit around the Sun, not exactly on collision course  , but 1 million km closer to the Sun (at 148.6 million km from the Sun, when the Earth is at 149.6). Most of the time/year, in order to stay at that fixed point, I have to use the propulsion to compensate the Sun's "gravitational pull". Once a year, when the Earth is near, I have to orient the ship thrusters towards the Earth, not the Sun, in order to maintain my fixed position in Sun's reference frame.

The question is: why I have yo do it?

1. because of gravitational time dilation changes?
2. because of spacetime curvature changes?
3. or because the presence of Earth (more exactly its mass) is changing "the curvature of spacetime" in my proximity? Another question is how exactly the Earth does that?

1 and 2 (they are different expressions of the same, but 2 is more precise.) You have to do it because what happens at your location at that time. I.e., the spacetime geometry at that spacetime event.

[joigus]

9 hours ago, Markus Hanke said:

deSitter spacetime has non-zero Riemann tensor, so it’s not flat.

Yes, thank you. That's what I thought.

 

[DanMP]

56 minutes ago, Genady said:

1 and 2

And what is wrong with 3? The changes mentioned in 1 and 2 only occur when the Earth is approaching ... Why are you discounting the Earth as a cause?

[Genady]

2 minutes ago, DanMP said:

And what is wrong with 3? The changes mentioned in 1 and 2 only occur when the Earth is approaching ... Why are you discounting the Earth as a cause?

Because the same changes in geometry can occur in other circumstances, e.g., different body or bodies, with different parameters / locations/ movements, but your response will be the same.

[DanMP]

16 minutes ago, Genady said:

the same changes in geometry can occur in other circumstances, e.g., different body or bodies, with different parameters / locations/ movements

Yes, but there is always a mass (or energy) required ... The geometry doesn't change without a change in mass/energy. So I would "blame" the approaching mass, not the geometry that is affected by it.

[Genady]

1 minute ago, DanMP said:

Yes, but there is always a mass (or energy) required ... The geometry doesn't change without a change in mass/energy. So I would "blame" the approaching mass, not the geometry that is affected by it.

But why Earth gets near there? Because it follows a geodesic according to the spacetime geometry.

So, I would "blame" the spacetime geometry again. But you would blame some other mass-energy changes for that geometry. Then I would blame geometry for changes which occur to that mass-energy. Etc.

The point is that geometry and mass-energy "conspire" in such a way that they together obey the Einstein field equation of GR.