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

[stevendaryl]

This comes into play in the scenario under discussion because conservation of energy only makes sense for metrics that are time-independent.

Why conservation of energy is relevant here is because

• In the case of a book sitting on a table in an accelerating rocket in flat spacetime, the table is doing work on the book (as viewed in free-fall coordinates) and we can account for where the energy comes from: The energy comes from burning rocket fuel.
  
• In the case of a book sitting on a table on a planet in curved spacetime, the table is doing work on the book (as viewed in free-fall coordinates) but we can't account for where the energy comes from: No fuel is being used up. But in this case, there is a time-varying metric, and energy is not conserved.

 

[A.T.]

Why conservation of energy is relevant here is because

• In the case of a book sitting on a table in an accelerating rocket in flat spacetime, the table is doing work on the book (as viewed in free-fall coordinates) and we can account for where the energy comes from: The energy comes from burning rocket fuel.
  
• In the case of a book sitting on a table on a planet in curved spacetime, the table is doing work on the book (as viewed in free-fall coordinates) but we can't account for where the energy comes from: No fuel is being used up. But in this case, there is a time-varying metric, and energy is not conserved.

What if you replace the the planet with an infinite wall?
http://www.mathpages.com/home/kmath530/kmath530.htm

Will an observer in free fall towards the wall observe a time-varying metric too?

 

[TrickyDicky]

[*]In the case of a book sitting on a table on a planet in curved spacetime, the table is doing work on the book (as viewed in free-fall coordinates) but we can't account for where the energy comes from: No fuel is being used up. But in this case, there is a time-varying metric, and energy is not conserved.
[/LIST]

Energy is conserved in the Schwarzschild spacetime by definition. The fact that time-varying coordinates can be chosen doesn't mean energy is not conserved as long as there is some coordinate system that is time invariant. We've been thru this already. A possible cause of misunderstanding is in the use of the term "metric". This link might be useful: http://badphysics.wordpress.com/2009/10/21/metric/

 

[TrickyDicky]

we can't account for where the energy comes from

Sure we can, gravitational energy, which happens to be well defined in the Schwarzschild solution (although not generally in GR).

 

[stevendaryl]

Sure we can, gravitational energy, which happens to be well defined in the Schwarzschild solution (although not generally in GR).

No, in the free-falling coordinates, there is no gravitational potential energy.

 

[TrickyDicky]

No, in the free-falling coordinates, there is no gravitational potential energy.

First, once more the choice of coordinates doesn't affect the physics, second, you must be referring to the free falling frame, it is simply the fact that we are using as observer one whose motion is geodesic in Schwarzschild geometry.
Gravitational energy is simply that associated to a gravitational field and as I said in this particular case is well defined thanks to the timelike KV.

 

[stevendaryl]

Energy is conserved in the Schwarzschild spacetime by definition.

There are two very different meanings to "energy is conserved". One is true for any spacetime, in any coordinates, another is only true for special spacetimes for special coordinates. The differential form of conservation of energy is true in any coordinate system, in any coordinates: [itex]\nabla_{\mu} T^{\mu \nu} = 0[/itex]. In this form of conservation of energy, there is no "gravitational potential". It's a local statement.

A second form of conservation of energy is in terms of conserved quantities; quantities whose values do not change with time. This notion of conserved quantity is very coordinate-dependent. If the metric is time-varying, then there is no quantity, in general, that is conserved and independent of time.

The fact that time-varying coordinates can be chosen doesn't mean energy is not conserved as long as there is some coordinate system that is time invariant.

You need to be clear about what you mean by "energy" and by "is conserved". The meaning of energy in the stress-energy tensor does not include "gravitational energy".

We've been through this already.

Yes, I know.

A possible cause of misunderstanding is in the use of the term "metric".

Nope.

 

[stevendaryl]

First, once more the choice of coordinates doesn't affect the physics

That's true, but so was what I said. "Gravitational potential energy" is a coordinate-dependent quantity. In free-fall coordinates, it's zero.

second, you must be referring to the free falling frame, it is simply the fact that we are using as observer one whose motion is geodesic in Schwarzschild geometry.
Gravitational energy is simply that associated to a gravitational field and as I said in this particular case is well defined thanks to the timelike KV.

GR has no notion of "gravitational potential energy", in general. You're right, that there is a coordinate-independent quantity associated with a timelike Killing Vector field, but it's extremely misleading to call it "gravitational potential energy". It plays that role in Schwarzschild coordinates, but in other coordinates, it's not meaningful to call it a potential energy.

 

[stevendaryl]

Sure we can, gravitational energy, which happens to be well defined in the Schwarzschild solution (although not generally in GR).

You're getting way off track. The particular scenario that we are talking about is that of a table sitting on the surface of the Earth. In freefalling coordinates, the table does work on the book. But in those coordinates, there is no "source" for the energy imparted to the book by the table. There is no rocket fuel being burned.

For you to say that the source of the energy is gravitational energy just makes no sense. Or, at least, I don't think it makes any sense. What exactly do you mean by that? How is the table transferring gravitational energy to the book?

You bring up the existence of a Killing Vector field in Schwarzschild geometry, but what does that have to do with the table imparting energy (doing work) on the book?

I don't think your explanation makes any sense.

 

[Austin0]

In flat spacetime with an accelerating rocket, we can look at things from two different points of view: from the point of view of an inertial observer, the only energy is kinetic energy, and work must be done in order to keep the rocket accelerating.

From the point of view of an accelerating observer (Rindler coordinates), the rocket is stationary, and no work is being done on the rocket. So where does the energy used up by the rocket go? It goes into throwing exhaust gases backwards. Those exhaust gases are NOT stationary, and so work is done to get them moving backwards.

Hi I have a couple of questions.
Within the accelerating system is the total energy from the (assuming) chemical reaction
exactly equivalent to the imparted momentum to the ejected mass??
This scenario as presented seems to contradict the 3rd law of motion , no?

Also it would seem that the observers would be aware of the system acceleration both through their accelerometer and through observed increased velocities of inertial objects.

While I am here: my knowledge of electrodynamics is vague but you might know.
An electromagnet picks up (accelerates ) a ferrous mass off the ground through the air.
Work is done increasing it's PE. Once in contact the mass is still being held static against the downward acceleration of G but no work is being done because of constant PE and KE=0
My question is does this condition require additional electric draw compared to the magnet's base electric draw without the mass?

Thanks

 

[pervect]

Well, if we define "does work" as "changes energy", which seems to me to be a reasonable reading of the intent of the question, the first thing we have to answer is "which energy are we talking about?", given that GR has several different notions of energy.

It seems pretty clear that Tricky Dicky , at least, is talkig about the conserved Komar energy, also sometimes known as "the energy at infinity" ala MTW , which is the sort of energy defined in GR whenever you have a static or stationary space-time, such as the Schwarzschild metric. It may have other names, too, but Komar energy and "energy at infinity" are the two names I've heard this sort of energy called most often.

It should also be clear from using this definition that if we have a particle (book) with r = theta = phi = constant, that this Komar energy , aka the energy at infinity, is constant.

I suspect that some of the other readers in this thread are not familiar with the basic definitions of the different sorts of energy in GR. This makes it rather difficult to have a meaningful discussion, alas - the thread just wanders around in circles, which is what I think is happening here.

 

[Dale]

the choice of coordinates doesn't affect the physics

Then local energy isn't physics since the choice of coordinates does change energy. Not that there is anything wrong with taking that position.

 

[stevendaryl]

Hi I have a couple of questions.
Within the accelerating system is the total energy from the (assuming) chemical reaction
exactly equivalent to the imparted momentum to the ejected mass??
This scenario as presented seems to contradict the 3rd law of motion , no?

Yes, you're exactly right. That could be taken to be a distinguishing difference between "inertial forces" and "real forces": inertial forces don't obey the 3rd law.

Also it would seem that the observers would be aware of the system acceleration both through their accelerometer and through observed increased velocities of inertial objects.

Yes, only in inertial, Cartesian coordinates is it the case that geodesics (the paths of freefalling objects) are "straight lines" (when you plot position as a function of time).

While I am here: my knowledge of electrodynamics is vague but you might know.
An electromagnet picks up (accelerates ) a ferrous mass off the ground through the air.
Work is done increasing it's PE. Once in contact the mass is still being held static against the downward acceleration of G but no work is being done because of constant PE and KE=0
My question is does this condition require additional electric draw compared to the magnet's base electric draw without the mass?

Thanks

That's a good question. I don't know the answer off the top of my head.

 

[stevendaryl]

It seems pretty clear that Tricky Dicky , at least, is talkig about the conserved Komar energy, also sometimes known as "the energy at infinity" ala MTW , which is the sort of energy defined in GR whenever you have a static or stationary space-time, such as the Schwarzschild metric. It may have other names, too, but Komar energy and "energy at infinity" are the two names I've heard this sort of energy called most often.

It should also be clear from using this definition that if we have a particle (book) with r = theta = phi = constant, that this Komar energy , aka the energy at infinity, is constant.

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.

 

[PeterDonis]

inertial forces don't obey the 3rd law.

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.

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.

Yes, only in inertial, Cartesian coordinates is it the case that geodesics (the paths of freefalling objects) are "straight lines" (when you plot position as a function of time).

This is only true in a local coordinate patch. You can't set up inertial coordinates covering the entire Earth such that the paths of all freely falling objects are straight lines.

That's a good question. I don't know the answer off the top of my head.

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.

 

[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 [itex]\partial / \partial t[/itex] 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 ([itex]\partial / \partial t[/itex] 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.