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

[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.