Genady

I don't know what is unclear. Clarify. Restate the unintentionally struck question, please.

The physics is clear then.

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.

What happens if the sources are everywhere?

Yep, it was obvious indeed.

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

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

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

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

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

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

Only perhaps if it's preexisting / primordial. Got it. Sure*. In case the mass we have is zero, the geometry is not necessarily flat. * Depends, not uniquely determined.

I don't know what you mean. Does mass cause geometry to exist?

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.

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.

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.

All solutions obey this equation. The mass is not at the same location where the Schwarzschild metric is. 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.

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

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

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

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.

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

Sorry, I guess it makes sense to you, but it doesn't make sense to me. Just words. Got a model?

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

Just replace "non-physical" with "physical" and "mind" with "brain": Science (as concept) is physical. Without this physical brain you could not read these words here on the page or any other book for that matter. Take the physical brain away from the world, and science ceases to exist. Yes.