Genady

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.

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

I did, and my answer was that \(\mathcal L\) is invariant under translation.

No.

The Lagrangian, \[\mathcal L(x)= \frac 1 2 \partial^{\mu} \phi (x) \partial_{\mu} \phi (x) - \frac 1 2 m^2 \phi (x)^2\] for a scalar field \(\phi (x)\) is said to be "Lorentz invariant and transforms covariantly under translation." What does it mean that it transforms covariantly under translation?

You also need the length of the pellet.

Isn't it rather 6000? 😄

Another ecological effect will be on organisms which use Moon for timing, for example, coral spawning.

It will stop slowing down (almost).

This ^^^^, I understand. This ^^^^, is only a distraction to me. (But generally, more often than not, I dislike analogies. Perhaps something individual.)

Yes, it is, in this sense: "This is extremely general. In any kind of gravitational field, as long as it is more or less constant with time, and not doing anything too radically relativistic, the coefficient in front of \(dt^2\) in the metric is always one plus twice the gravitational potential." Susskind, Cabannes. General Relativity: The Theoretical Minimum (p. 155).

I'm at ease with the Schwartz's notation by now. It is as simple as mentally substituting \(A_{\mu} g^{\mu \alpha} B_{\alpha}\) every time he writes \(A_{\mu} B_{\mu}\).