Does gravity as a fictitious force do work? (GR's free-falling frame POV)?

[Austin0]

Yes they do.

Remember that in the accelerating frame, the momentum and kinetic energy added to the rocket's exhaust is *larger* than it is in the freely falling frame. So the energy burned by the fuel *is* entirely taken up by the exhaust.

SO this does seem in contradiction of the 3rd law if this is the case.
If the total energy is accounted for by the exhaust then where does the momentum/energy to provide the acceleration registered by the accelerometer come from?

Also remember that the "inertial force" of gravity on the rocket is balanced by an equal and opposite force of the rocket on the Earth. You have to include the Earth in the 3rd law analysis for everything to balance out.

i assumed we were talking flat spacetime but could you elaborate on this concept?
How is the thrust of the rocket transmitted to the earth, through the air do you mean??

The answer is no. If the answer were yes, you would need to hook up electrical power to a kitchen magnet to keep it stuck to your refrigerator.

Obviously a permanent magnet is a different story. I would imagine the energy in that case is stored potential from the energy/work required to organize the structure.
If that is not so then the conservation of energy regarding such a magnet is a complete mystery as they seem to be able to do an unlimited (timewise) amount of work with no apparent source of energy.
As I understand an electromagnet it does require power to create the field, DO work.
So the refrigerator analogy may not apply.
Not that your answer might not be correct, I obviously don't know, hence the question.

 

[PeterDonis]

If the total energy is accounted for by the exhaust then where does the momentum/energy to provide the acceleration registered by the accelerometer come from?

Don't get your frames confused. In the accelerated frame, the momentum and energy of the rocket don't change. The fact that the accelerometer reads something other than zero is irrelevant; under Newtonian gravity accelerometer readings are not how "acceleration" is defined. (Yes, I know that seems paradoxical; one of the advantages of GR as a theory, IMO, is that it *does* define acceleration, in the invariant sense, by accelerometer readings, i.e., by a direct physical observable instead of a coordinate-dependent definition.)

i assumed we were talking flat spacetime

You can't be if the scenario includes gravity. (If you are talking about Newtonian gravity, technically "spacetime" is not a valid concept at all; but in so far as the term applies, spacetime is not flat in Newtonian gravity any more than it is in GR.)

How is the thrust of the rocket transmitted to the earth, through the air do you mean??

I mean that the "inertial force" of gravity of Earth on the rocket--the force that, in the accelerated frame, is opposed by the rocket's thrust so that the net motion of the rocket is zero in that frame--is balanced, by Newton's 3rd law, by an equal and opposite force of the gravity of the rocket on the Earth. Similarly, the upward thrust of the rocket exhaust on the rocket is balanced, by Newton's 3rd law, by an equal and opposite downward force of the rocket on the exhaust.

Obviously a permanent magnet is a different story. I would imagine the energy in that case is stored potential from the energy/work required to organize the structure.
If that is not so then the conservation of energy regarding such a magnet is a complete mystery as they seem to be able to do an unlimited (timewise) amount of work with no apparent source of energy.
As I understand an electromagnet it does require power to create the field, DO work.
So the refrigerator analogy may not apply.
Not that your answer might not be correct, I obviously don't know, hence the question.

A permanent magnet stuck to your refrigerator does no work because there's no relative motion between the two. So there's no need for energy to be expended to hold it there. If there were, as I said, you would need to have a power source hooked up to it; otherwise the magnet would be able, as you say, to do an unlimited amount of work with no apparent source of energy. (There was some work done when the permanent magnet was originally created, to align the atoms' magnetic moments; but that stored energy is not available to do work, because releasing it would remove the magnet's magnetism.)

An electromagnet does require some power to create the field; but your own statement of the scenario already allows for that, in the magnet's "base electric draw without the mass". You were asking whether, once the magnet has something stuck to it and there is no further relative motion, any *additional* energy has to be expended beyond that base electric draw. The answer to that is no.

 

[Austin0]

Don't get your frames confused. In the accelerated frame, the momentum and energy of the rocket don't change. The fact that the accelerometer reads something other than zero is irrelevant; under Newtonian gravity accelerometer readings are not how "acceleration" is defined. (Yes, I know that seems paradoxical; one of the advantages of GR as a theory, IMO, is that it *does* define acceleration, in the invariant sense, by accelerometer readings, i.e., by a direct physical observable instead of a coordinate-dependent definition.)

There is confusion. If you look at my post #45 stevendaryl was talking about Rindler observers in explicitly flat spacetime. As were my questions.
So i think our discussion is a bit disjointed as we have been talking about different conditions

You can't be if the scenario includes gravity. (If you are talking about Newtonian gravity, technically "spacetime" is not a valid concept at all; but in so far as the term applies, spacetime is not flat in Newtonian gravity any more than it is in GR.)

I mean that the "inertial force" of gravity of Earth on the rocket--the force that, in the accelerated frame, is opposed by the rocket's thrust so that the net motion of the rocket is zero in that frame--is balanced, by Newton's 3rd law, by an equal and opposite force of the gravity of the rocket on the Earth. Similarly, the upward thrust of the rocket exhaust on the rocket is balanced, by Newton's 3rd law, by an equal and opposite downward force of the rocket on the exhaust.

A permanent magnet stuck to your refrigerator does no work because there's no relative motion between the two. So there's no need for energy to be expended to hold it there. If there were, as I said, you would need to have a power source hooked up to it; otherwise the magnet would be able, as you say, to do an unlimited amount of work with no apparent source of energy. (There was some work done when the permanent magnet was originally created, to align the atoms' magnetic moments; but that stored energy is not available to do work, because releasing it would remove the magnet's magnetism.)

`
An electromagnet does require some power to create the field; but your own statement of the scenario already allows for that, in the magnet's "base electric draw without the mass". You were asking whether, once the magnet has something stuck to it and there is no further relative motion, any *additional* energy has to be expended beyond that base electric draw. The answer to that is no.

 

[TrickyDicky]

Then local energy isn't physics since the choice of coordinates does change energy.

What happens to be conserved is the total energy, a local change of coordinates doesn't change energy globally in the system, nor affects anything locally, it is like changing the units from joules to ergs, however precisely to conserve energy globally different observers in different frames must observe different energies and frequencies for matter/radiation, it is in this context that local energy is frame dependent, it must be in order to preserve global energy conservation.

This might explain yet another disagreement we had when you were saying energy was frame dependent and I said total energy is not because energy is globally conserved (always referring to classical mechanics or GR static solutions scenarios). It is obvious that local energy is frame dependent just by looking at the gravitational redshift example.

 

[TrickyDicky]

Well, the discussion has wandered all over the place, but I thought that the issue (or one of the issues) was that from the point of view of a local free-falling frame, the normal force on the table is pushing the book upward, and so by the usual definition of "work", the normal force is doing work on the book.

But the local free-falling frame is not that of the table, the table is non-inertial. Work is computed by an observer considered at rest in a free-falling frame, like for instance one free-falling towards the book and table that sees them moving upwards. Remember coordinate systems are not exactly the same thing as frames.
Wikipedia:
"A frame of reference in physics, may refer to a coordinate system or set of axes within which to measure the position, orientation, and other properties of objects in it, or it may refer to an observational reference frame tied to the state of motion of an observer. It may also refer to both an observational reference frame and an attached coordinate system as a unit."

 

[Dale]

What happens to be conserved is the total energy, a local change of coordinates doesn't change energy globally in the system, nor affects anything locally

Locally energy is the timelike component of the four momentum or the time time component of the stress energy tensor. Both of which do, in fact, change under a change of coordinates. They are conserved, but coordinate dependent.

I am not saying that it is wrong to consider coordinate dependent things like components of tensors to be non physical, just that if you do so then local energy is non physical.

 

[Ben Niehoff]

(There was some work done when the permanent magnet was originally created, to align the atoms' magnetic moments; but that stored energy is not available to do work, because releasing it would remove the magnet's magnetism.)

I'm not going to comment on the relativity discussion in this thread. But you've got this backwards! The state with all magnetic moments aligned is the ground state, and raising the temperature (and hence internal energy) will introduce disorder and demagnetize the magnet.

In order to magnetize a lump of iron, you can place it in a constant magnetic field and lower its temperature. Misaligned magnetic moments will then align with the ambient field, releasing energy over time.

But it does cost energy to generate the ambient magnetic field (e.g. by passing current through coils). As usual, in order to lower entropy in one place, you must increase it somewhere else.

 

[Dale]

Yes they do.

Consider some arbitrary scenario in inertial coordinates in flat spacetime. Every force has a 3rd law reaction force, and these are all real forces. Now, transform that scenario to a reference frame accelerating to the right. All of the real forces still exist, but now each object has an additional inertial force pointing to the left. None of the inertial forces point to the right, so none of them are 3rd law pairs with each other, and all of the real forces are already in 3rd law pairs with other real forces. Therefore inertial forces do not always obey Newtons 3rd law.

 

[TrickyDicky]

Locally energy is the timelike component of the four momentum or the time time component of the stress energy tensor. Both of which do, in fact, change under a change of coordinates. They are conserved, but coordinate dependent.

I am not saying that it is wrong to consider coordinate dependent things like components of tensors to be non physical, just that if you do so then local energy is non physical.

Perhaps a clarification of what is usually meant by "non-physicality of coordinate dependent components" is in order.
IMO it usually means that their change due to a coordinate transformation doesn't imply a change in the physics of the situation, because as long as we are dealing with tensors that coordinate transformation implies the corresponding change in other component that compensates it(or the change of basis if it is a vector). That is the reason we use geoemtrical objects like tensor that are invariant to coordinate system transformations.
There's nothing more to it.
Energy has that issue in GR unlike in classical mechanics: It is well defined only locally (as the tt component of the enegy-stress tensor) in all instances where there's no timelike KV.

 

[A.T.]

Also remember that the "inertial force" of gravity on the rocket is balanced by an equal and opposite force of the rocket on the Earth. You have to include the Earth in the 3rd law analysis for everything to balance out.

Doesn't Newtons 3rd imply instantaneous action at a distance? Even EM-forces don't satisfy Newtons 3rd, but they do satisfy the idea behind it (momentum conservation) by assigning a momentum to the field.

But how is it in GR? Are the inertial gravitational forces by two masses on each other always equal and opposite, in every frame which includes both masses?

 

[PeterDonis]

The state with all magnetic moments aligned is the ground state, and raising the temperature (and hence internal energy) will introduce disorder and demagnetize the magnet.

In order to magnetize a lump of iron, you can place it in a constant magnetic field and lower its temperature. Misaligned magnetic moments will then align with the ambient field, releasing energy over time.

But it does cost energy to generate the ambient magnetic field (e.g. by passing current through coils). As usual, in order to lower entropy in one place, you must increase it somewhere else.

Ben, thanks for the clarification. You're right, I was misdescribing where the energy has to be expended to create a permanent magnet.

 

[PeterDonis]

There is confusion. If you look at my post #45 stevendaryl was talking about Rindler observers in explicitly flat spacetime. As were my questions.

Consider some arbitrary scenario in inertial coordinates in flat spacetime. Every force has a 3rd law reaction force, and these are all real forces. Now, transform that scenario to a reference frame accelerating to the right. All of the real forces still exist, but now each object has an additional inertial force pointing to the left. None of the inertial forces point to the right, so none of them are 3rd law pairs with each other, and all of the real forces are already in 3rd law pairs with other real forces. Therefore inertial forces do not always obey Newtons 3rd law.

Ah, sorry, I was mistaken about the scenario. Trying to post when too tired.

 

[PeterDonis]

Doesn't Newtons 3rd imply instantaneous action at a distance?

In the case of Newtonian gravity (meaning Newton's law of gravity, *not* GR--see below), yes. I don't know that it always does.

Even EM-forces don't satisfy Newtons 3rd, but they do satisfy the idea behind it (momentum conservation) by assigning a momentum to the field.

But assigning momentum to the field means that the "force" is now not directly between two charged objects; it's between the first charged object and the field, and then between the field and the second charged object. Momentum is still conserved, so can't you still pick out matched pairs of forces for Newton's 3rd law that way? Or would you say that momentum conservation and Newton's 3rd law are not equivalent?

But how is it in GR? Are the inertial gravitational forces by two masses on each other always equal and opposite, in every frame which includes both masses?

In GR gravity is not a force, so the question doesn't arise. The curvature of spacetime in GR does not propagate instantaneously; it propagates at the speed of light.

 

[Dale]

Ah, sorry, I was mistaken about the scenario. Trying to post when too tired.

No problem, there have been a lot of scenarios and even more sets of coordinates.

 

[Dale]

But assigning momentum to the field means that the "force" is now not directly between two charged objects; it's between the first charged object and the field, and then between the field and the second charged object. Momentum is still conserved, so can't you still pick out matched pairs of forces for Newton's 3rd law that way?

That is certainly compatible with the idea of generalized forces in the Lagrangian formulation of classical mechanics.

 

[TrickyDicky]

I guess the statement that I wanted to say that only in flat spacetime is it true that every inertial observer can view himself at rest in a coordinate system with time-independent metric components.

I missed this.
This statement doesn't seem right. In a stationary curved spacetime every inertial observer (that is a geodesic observer) can see himself at rest in a coordinate system with time-independent metric components.

 

[PeterDonis]

In a stationary curved spacetime every inertial observer (that is a geodesic observer) can see himself at rest in a coordinate system with time-independent metric components.

This is not correct. Inertial observers in a stationary curved spacetime (e.g., Schwarzschild spacetime) see a time-varying metric; the invariant way of expressing this is that inertial observers in a stationary curved spacetime do not follow orbits of the timelike Killing vector field. Observers who follow orbits of the timelike Killing vector field are not inertial; they experience a nonzero proper acceleration that varies with radius.

 

[TrickyDicky]

This is not correct. Inertial observers in a stationary curved spacetime (e.g., Schwarzschild spacetime) see a time-varying metric; the invariant way of expressing this is that inertial observers in a stationary curved spacetime do not follow orbits of the timelike Killing vector field. Observers who follow orbits of the timelike Killing vector field are not inertial; they experience a nonzero proper acceleration that varies with radius.

I think you should make clear exactly what it means (physically) to "see a time-varying metric" for an observer (I had never heard about "seeing metrics" in that sense, is it a standard expression?) and how is that incompatible with a time-independent metric, specifically how that changes the fact that all time derivatives of the metric tensor are vanishing.
Inertial observers in stationary spacetimes (free-falling observers) follow timelike orbits. As you say observers who follow orbits of the timelike Killing field experience a nonzero proper acceleration(although strictly speaking in a Schwarzschild spacetime being a vacuum we only have test particles following geodesics, no timelike KV fields orbits there then) and are time symmetric, how does that imply that the timelike geodesic orbits aren't, can't they be considered following infinitesimal killing orbits?

 

[Dale]

Perhaps a clarification of what is usually meant by "non-physicality of coordinate dependent components" is in order.
IMO it usually means that their change due to a coordinate transformation doesn't imply a change in the physics of the situation, because as long as we are dealing with tensors that coordinate transformation implies the corresponding change in other component that compensates it(or the change of basis if it is a vector). That is the reason we use geoemtrical objects like tensor that are invariant to coordinate system transformations.
There's nothing more to it.
Energy has that issue in GR unlike in classical mechanics: It is well defined only locally (as the tt component of the enegy-stress tensor) in all instances where there's no timelike KV.

That is fine. Certainly, changing coordinates will not affect the result of any measurement. Some people (including you apparently) restrict "the physics of the situation" to such things, and exclude intermediate values and components. Others think that things like (local) energy and momentum are part of the physics even though they are coordinate dependent. Personally, I am ambivalent, I kind of see both sides on this topic.

 

[PeterDonis]

I think you should make clear exactly what it means (physically) to "see a time-varying metric" for an observer (I had never heard about "seeing metrics" in that sense, is it a standard expression?)

You're right, my terminology was ambiguous. A better way of saying what I was trying to say is that only observers who follow orbits of a timelike Killing vector field will see an unchanging spacetime curvature at every event on their worldlines.

and how is that incompatible with a time-independent metric, specifically how that changes the fact that all time derivatives of the metric tensor are vanishing.

This statement is only true in a restricted sense--actually, one of two senses, depending on what you mean. If you mean "time derivatives" in a coordinate sense, it's only true for a coordinate chart whose time coordinate t is such that \partial / \partial t is a timelike Killing vector field. If you mean "time derivatives" with respect to proper time along a particular worldline, or set of worldlines, the statement is only true for observers whose worldlines are orbits of a timelike Killing vector field, i.e., the tangent vector to each worldline at every event on that worldline is a timelike Killing vector.

Inertial observers in stationary spacetimes (free-falling observers) follow timelike orbits.

Sure, but that's true of any observer (inertial or not) in any spacetime (stationary or not), so it's not saying very much. The question is what specific vector field their worldlines are orbits of.

As you say observers who follow orbits of the timelike Killing field experience a nonzero proper acceleration

Yes. At least, they do in Schwarzschild spacetime, and more generally in any Kerr-Newman spacetime. I'm not sure if it's been proven that this must be true in *any* stationary spacetime, although it seems to me that it ought to be true.

(although strictly speaking in a Schwarzschild spacetime being a vacuum we only have test particles following geodesics, no timelike KV fields orbits there then) and are time symmetric

Not sure what this means or what time symmetry has to do with it.

how does that imply that the timelike geodesic orbits aren't, can't they be considered following infinitesimal killing orbits?

"Following an orbit" means having some property at *every* event on a worldline, not just one. At least, that's my understanding of standard usage.

Moreover, it's hard to see the point of letting "following an orbit" apply only at a single event, because the whole point of picking out observers who follow orbits of a timelike Killing vector field is that only those observers see unchanging spacetime curvature at every event on their worldlines. And in a coordinate chart whose time coordinate is as above (\partial / \partial t is a Killing vector field at every event), only those observers will see unchanging metric coefficients at every event on their worldlines ("unchanging" in the sense of the actual numbers, not the line element formula; obviously the formula is the same everywhere, but the actual numbers can depend on the coordinates). This is a key physical property of these observers, which inertial observers in stationary spacetimes (at least the ones we've discussed--as I said above, I'm not positive that it applies to *every* stationary spacetime, but it seems like it should) do *not have.

 

[Austin0]

Originally Posted by TrickyDicky

In a stationary curved spacetime every inertial observer (that is a geodesic observer) can see himself at rest in a coordinate system with time-independent metric components.

This is not correct. Inertial observers in a stationary curved spacetime (e.g., Schwarzschild spacetime) see a time-varying metric; the invariant way of expressing this is that inertial observers in a stationary curved spacetime do not follow orbits of the timelike Killing vector field. Observers who follow orbits of the timelike Killing vector field are not inertial; they experience a nonzero proper acceleration that varies with radius.

As i am just trying to get a handle on Killing vectors so could you explain this in more fundamental terms.
In a free falling frame what internal experiments would produce different results over time?
How could they determine a time dependent metric?
it is easy to see that relative to flat space inertial observers or static Schwarzschild observers they would have a dynamic metric but I assume that is not what you are talking about.

 

[PeterDonis]

In a free falling frame what internal experiments would produce different results over time?

Let me first restate your question a bit: "What experiments done by a freely falling observer in a stationary curved spacetime--for concreteness, we'll use Schwarzschild spacetime as our example--involving only the properties of spacetime, i.e., gravity, would give different results over time?"

The reason I am restating the question is that "in a free falling frame" is ambiguous. In a *local* freely falling frame, no experiments can show effects of curvature, since that's excluded by the definition of a local freely falling frame. If by "a freely falling frame" you mean "a frame in which a freely falling observer is at rest for his entire fall", none of the standard coordinate charts on Schwarzschild spacetime meet that definition, so I wouldn't know what chart to use to answer your question. In any case, the real question of physics is what actual observations would vary with time for a freely falling observer; which chart (if any) we use to describe them is irrelevant.

The simplest such experiments I can think of that a freely falling observer could do would be ones directly showing tidal gravity. Objects slightly below or slightly above the freely falling observer, also freely falling (accelerometers could be used to ensure this), would slowly move away from the observer. Objects at the same radius (above the central mass) but slightly to one side or the other, also freely falling, would slowly move towards the observer.

This in itself would not necessarily indicate a time-varying spacetime curvature; a static observer (one who stays at the same radius forever) could run similar experiments on bodies freely falling past him and would see the same type of tidal effects. But if the freely falling observer starts such experiments at different events on his worldline, each with the same initial conditions (objects released into free fall, initially at rest relative to him, and at the same distance from him as measured by rulers traveling with him), the experiments will show the objects moving away from or towards him at different *rates*--more precisely, with different "tidal accelerations" (these are coordinate accelerations relative to the observer, not proper accelerations; all objects are freely falling). The variation in tidal accelerations *does* indicate a change in spacetime curvature, and would *not* be seen by a static observer.

 

[TrickyDicky]

Peter, I would say that according to what you explain "seeing a time-varying metric" is just what I supposed, the ability of an observer to choose coordinate systems in order to ascertain the coordinate acceleration of other objects specified by the time coordinate. All this variation is purely coordinate-dependent (even if it can be motivated by tidal accelerations).

The kind of experiment you mention can be performed by any observer regardless if it is inertial or not. Those non-inertial observers like the "static observer" you referred to can do that experiment wrt other objects that are not the one wrt which it keeps constant radius due to its proper acceleration, and see a time-varying metric.
So I would say the possibility of doing those experiments is orthogonal to the existence or not of timelike killng vector fields or whether the the spacetime is static and therefore time-independent or not.
The detection of tidal variations is the common feature of gravity and any curved spacetime and it is coordinate independent while the ability to see a time-varying metric is purely coordinate dependent and not related to flatness or curvature either.

 

[PeterDonis]

Peter, I would say that according to what you explain "seeing a time-varying metric" is just what I supposed, the ability of an observer to choose coordinate systems in order to ascertain the coordinate acceleration of other objects specified by the time coordinate. All this variation is purely coordinate-dependent (even if it can be motivated by tidal accelerations).

This isn't what I meant by "seeing a time-varying metric". I explicitly said that I meant "seeing a changing strength of tidal gravity with respect to proper time", which is *not* coordinate-dependent. I didn't say anything about coordinates. I was talking strictly about actual physical observables. An inertial observer in a curved stationary spacetime sees the strength of tidal gravity in his vicinity change with respect to his proper time. An observer following an orbit of the timelike Killing vector field (who must be accelerated) sees the strength of tidal gravity in his vicinity remain constant with respect to his proper time. This is an observable physical difference.

The kind of experiment you mention can be performed by any observer regardless if it is inertial or not.

Well, of course. I specifically said a static observer (accelerated, staying at constant radius) could perform it.

Those non-inertial observers like the "static observer" you referred to can do that experiment wrt other objects that are not the one wrt which it keeps constant radius due to its proper acceleration, and see a time-varying metric.

The experiment I described has to be done locally. Please describe how a static observer at radius r1 can do an experiment that directly measures the tidal gravity in the vicinity of an inertial observer at radius r2 which is different from r1. Of course the static observer at r1 can receive information *from* the inertial observer at r2, via radio messages, say, communicating the results of the inertial observer's experiments, but that doesn't seem to be what you are talking about.

So I would say the possibility of doing those experiments is orthogonal to the existence or not of timelike killng vector fields or whether the the spacetime is static and therefore time-independent or not.

Yes, of course. You can do the experiment I described to measure the strength of tidal gravity in your vicnity in any spacetime whatsoever. But in a spacetime without a timelike Killing vector field, the results of the experiment will change with respect to your proper time, no matter *what* worldline you follow.

The detection of tidal variations is the common feature of gravity and any curved spacetime and it is coordinate independent while the ability to see a time-varying metric is purely coordinate dependent and not related to flatness or curvature either.

Only with a definition of "time-varying metric" different from the one I gave. Obviously if you *define* "time-varying" as "changing with respect to coordinate time", then a time-varying metric is coordinate-dependent. But I don't care about definitions; if you don't like my usage of the term "time-varying metric", then just read "changing strength of gravity with respect to proper time" in all my posts instead, since that's what I meant. I am trying to talk about the actual physical observables, not coordinates.

 

[TrickyDicky]

This isn't what I meant by "seeing a time-varying metric". I explicitly said that I meant "seeing a changing strength of tidal gravity with respect to proper time", which is *not* coordinate-dependent. I didn't say anything about coordinates. I was talking strictly about actual physical observables. An inertial observer in a curved stationary spacetime sees the strength of tidal gravity in his vicinity change with respect to his proper time. An observer following an orbit of the timelike Killing vector field (who must be accelerated) sees the strength of tidal gravity in his vicinity remain constant with respect to his proper time. This is an observable physical difference.



Well, of course. I specifically said a static observer (accelerated, staying at constant radius) could perform it.



The experiment I described has to be done locally. Please describe how a static observer at radius r1 can do an experiment that directly measures the tidal gravity in the vicinity of an inertial observer at radius r2 which is different from r1. Of course the static observer at r1 can receive information *from* the inertial observer at r2, via radio messages, say, communicating the results of the inertial observer's experiments, but that doesn't seem to be what you are talking about.



Yes, of course. You can do the experiment I described to measure the strength of tidal gravity in your vicnity in any spacetime whatsoever. But in a spacetime without a timelike Killing vector field, the results of the experiment will change with respect to your proper time, no matter *what* worldline you follow.



Only with a definition of "time-varying metric" different from the one I gave. Obviously if you *define* "time-varying" as "changing with respect to coordinate time", then a time-varying metric is coordinate-dependent. But I don't care about definitions; if you don't like my usage of the term "time-varying metric", then just read "changing strength of gravity with respect to proper time" in all my posts instead, since that's what I meant. I am trying to talk about the actual physical observables, not coordinates.

I think we are simply having semantic problems here because I agree with most of what you say.
It is true that one thing that distinguishes a time-independent (like Schwarzschild's) from a time dependent (like FRW) spacetime is precisely the fact that in the time-independent one can define a "static observer".

 

[TrickyDicky]

Y
... only in flat spacetime is it true that every inertial observer can view himself at rest in a coordinate system with time-independent metric components.

After clarifying what it means, certainly different to my initial interpretation, I agree with this statement that it is actually simply the fact that only in curved spacetime can one measure tidal acceleration.
This is due to the non-uniform nature of gravitationl fields. That is why usually the Equivalence principle stresses the fact that the equivalence is local: In GR spacetime is equivalent to flat spacetime only locally (infinitesimally), evidently.

 

[PeterDonis]

It is true that one thing that distinguishes a time-independent (like Schwarzschild's) from a time dependent (like FRW) spacetime is precisely the fact that in the time-independent one can define a "static observer".

Yes, although instead of "time-independent spacetime" I would say "spacetime with a timelike Killing vector field". That's the key difference between the two, and it's a coordinate-free statement.