Genady

It is curious that the same mechanism gives both the attractive effect of dark matter and the repulsive effect of dark energy.

The hypotheses above may be complementary rather than contrasting because they seem to relate to different aspects of the "four why's of animal behavior" of Nikolaas (Niko) Tinbergen, i.e., causation, development, adaptation, and phylogeny.

2 minutes ago, DanMP said:

actually

What does it mean?

The model tells us how it happens. Some future model might derive this model, i.e., the equations, from more generic principles.

The previous model, aka Newtonian gravity, included a notion of mass acting on the ship. This is an approximation, which is not included in the current best model.

3 minutes ago, DanMP said:

I fixed it:

Can you answer this question please?

The spacetime geometry obeys certain differential equations. The solution of these equations depends on initial and boundary conditions. The mentioned mass is a part of these conditions.

1 minute ago, DanMP said:

Maybe, but your answer is not.

The strike line was unintentional. I try to correct it.

I don't know what is unclear. Clarify.

Restate the unintentionally struck question, please.

16 minutes ago, DanMP said:

Who said that my ship has to be directly affected by the incoming mass? 

The physics is clear then.

12 hours ago, KJW said:

In another thread, I showed that the flat-space Friedmann–Lemaître–Robertson–Walker (FLRW) metric can be coordinate-transformed to a manifestly conformally flat metric. This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar).

 

 

This is true for homogeneously and isotropically distributed sources. What about other "smeared" sources? The Weyl tensor is not necessarily zero, but the curvature is not away from the source.

30 minutes ago, KJW said:

One thing I should mention: I distinguish between the curvature directly associated with the energy-momentum and the curvature that is away from its source energy-momentum, and tend not to use the term "gravity" to describe the curvature directly associated with energy-momentum.

The curvature associated with gravity is called the Weyl conformal tensor. It has the same algebraic structure as the Riemann tensor, but its contraction is zero (—> zero Einstein tensor). In the Cαβγδ form, it is invariant to conformal transformations:

gμν⟶φgμν  for arbitrary scalar function φ

 

 

What happens if the sources are everywhere?

4 minutes ago, KJW said:

grr=1grr

Yep, it was obvious indeed.

2 hours ago, KJW said:

ar=−c221grr gttdgttdr

Perhaps it's obvious but I don't see where \(g_{rr}\) in the denominator comes from, here: \[a^r = -\dfrac{c^2}{2} \dfrac{1}{g_{rr}\ g_{tt}} \dfrac{dg_{tt}}{dr}\]

1 hour ago, KJW said:

For a stationary object in the Schwarzschild metric

What is \(T\) in \[a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}\]

2 hours ago, DanMP said:

You would stil "blame" the affected geometry or some other geometry? Why?

Because Earth or other body you mention are far away and your ship cannot be affected directly by them without "an action at a distance." What it can be affected by directly is the geometry of spacetime at that same point where and when it is "here" and "now".

2 hours ago, DanMP said:

And I repeat the question:

I still don't know what you mean in that question. Does what?

14 minutes ago, swansont said:

I recall a seminar (on Lie algebra, IIRC) in grad school where the prof was asked what the difference was between contravariant and covariant, and the response was something like “contravariant means the indices are along the top” which is true but doesn’t do anything to advance anyone’s understanding.

The following story from Zee, A. Einstein Gravity in a Nutshell (p. 52) is in the same line:

Long ago, an undergrad who later became a distinguished condensed matter physicist came to me after a class on group theory and asked me, “What exactly is a tensor?” I told him that a tensor is something that transforms like a tensor. When I ran into him many years later, he regaled me with the following story. At his graduation, his father, perhaps still smarting from the hefty sum he had paid to the prestigious private university his son attended, asked him what was the most memorable piece of knowledge he acquired during his four years in college. He replied, “A tensor is something that transforms like a tensor.”

😉

17 minutes ago, swansont said:

I’ve read that GR reduces to Newtonian gravity, but have never been able to find a worked example of that, because nothing is ever presented but the tensor mathematics.

Here is one, from MTW's Gravitation. All variables are real numbers. The most important one, \(T_{00}\), is energy density:

14 minutes ago, swansont said:

And what are the boundary conditions of those sets?

Any geometry with zero Ricci tensor and non-zero Riemann tensor.

1 minute ago, swansont said:

How? Usually there’s only one solution for a given set of boundary conditions. 

It can be a non-flat solution for some sets.

2 hours ago, swansont said:

How would one get a gravitational wave absent energy-momentum?

Only perhaps if it's preexisting / primordial.

2 hours ago, swansont said:

The scenario has two parts, the earth influencing geometry and the location. You only addressed the latter.

Got it.

2 hours ago, swansont said:

The geometry you have depends on whatever mass (as a first-order approximation) you have.

Sure*. In case the mass we have is zero, the geometry is not necessarily flat.

* Depends, not uniquely determined.

2 minutes ago, swansont said:

does it cause the mass to exist?

I don't know what you mean.

Does mass cause geometry to exist?

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.

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.

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.

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.

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

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

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 ..."

43 minutes ago, Maartenn100 said:

It's idealism: everything is mind, and fysicality is the appearance of the mind of nature to a localised mind.

That's not my model, that's the model or hypothesis of philosopher and computerscientist Bernardo Kastrup actually. It's called analytic idealism.  We are like whirlpools in a river. Localised minds in the Mind of Nature. When we cease to exist we go back into the broader stream of Consciousness, just like the whirlpool ceases to exist and becomes back part of the greater stream. How other 'whirlpools of localised minds' look like for 'one localised whirlpool of mind' is like a brain and body. A metabolising organism. That's an appearance in the mind of a neuroscientist, a localised mind. A dissociated alter according to Bernardo Kastrup. We are dissociated alters from the broader Mind of Nature. We perceive a shared dream, like  the dissociated alters of someone with dissociated identity disorder in the dream of this person, having a shared dream. When the alters cease to exist, the Mind awakens and recoginises: it was me all along.

I see. It gets worse. Now, it is boring. Regurgitating of age-old philosophies. Hundreds or thousands of years old. In the recent decades, marketed by Deepak Chopra. I am out.