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

[TrickyDicky]

This discussion started as a side clarification of something in this thread https://www.physicsforums.com/showthread.php?p=3971013#post3971013 and although tangentially related it probably deserves a thread of its own so anyone can participate without reservations.
Please keep in mind I am not referring to the question in classical mechanics with the Earth's frame of reference and gravity acting as a real force in which it is obvious there is no work done on a book placed on a table.

Also to simplify things we are using the Schwarzschild exterior solution of GR, so that we can consider the geodesic free-falling frame of reference as static.

The question is, given this setting, with gravity considered as fictitious force in GR does the table do work on a book sitting on it?
And if the book falls off the table and while is falling, does gravity fictitious force do work on the book?
I would like to take a more neutral stance in this thread,and just explain why I would consider good arguments to say that work is done in the particular situation I described.

First by analogy to other fictitious forces like centrifugal force. In my GR scenario we are considering the free-falling frame the "inertial" frame (remember the terms inertial and non-inertial are used differently in Newtonian mechanics versus GR), and the book is kept in the non-inertial frame of the surface of the table on the earth.
In the centrifugal force example this force is the apparent outward force that draws a rotating body away from the center of rotation and is caused by the inertia of the body.


By analogy I consider the gravitational (fictitious) force as the apparent downward force that draws the book to the centre of the Earth and is caused by the mass of the body.

Now what allows to make the analogy between this two fictitious forces is the equivalence principle that equals the inertial mass to the gravitational mass.


In the above linked thread stevendaryl argued that the equivalence failed because he could find coordinates in the schwarzschild spacetime that made the metric appear time varying.
Being obvious that coordinates can introduce artifacts that don't affect the physics of the situation I still don't know what his point was wrt the physics of the problem at hand.
I believe he was trying to make a point about being able to choose coordinate systems making the metric tensor appear time varying but that is such an obvious thing and so disconnected with the physics of the problem I chose not to follow that distracting path.

Following the analogy with the centrifugal force, I maintain that the table is doing work on the book against the gravitational fictitious force so that as soon as the book falls and as long as no object is in its way the gravitational fictitious force does work on the book as it tries catching up the free-falling frame.
Any objections?

 

[A.T.]

Work is frame dependent.

I maintain that the table is doing work on the book against the gravitational fictitious force

In the inertial free falling frame, there is no gravitational fictitious force on the book. Fictitious forces exist only in non-inertial frames. Here the only force on the book is the table force. And yes, it is doing work on the upwards accelerating book, as soon as it moves in that frame. Note that inertial frames exist only locally in curved space time.

In the non-inertial rest frame of the table & book there is no work done on the book, because there no displacement.

as soon as the book falls and as long as no object is in its way the gravitational fictitious force does work on the book

This true in the non-inertial frame rest frame of the table.

In the inertial rest frame frame of the book there are no forces acting on the book, so there is no work done on the book.

 

[TrickyDicky]

Work is frame dependent.

Exactly and I clearly referred to what is observed in the non-inertial frame of the book.

In the inertial free falling frame, there is no gravitational fictitious force on the book. Fictitious forces exist only in non-inertial frames.

In the free-falling (geodesic) frame there are no forces. That's correct. We must remember we are always referring to apparent forces in the given scenario.

Here the only force on the book is the table force. And yes, it is doing work on the upwards accelerating book, as soon as it moves in that frame.

Right. This work is felt by the non-inertial observer.

In the non-inertial rest frame of the table & book there is no work done on the book, because there no displacement.

This seems to contradict what you just wrote. The non-inertial observer is not at rest wrt to the inertial frame.

This true in the non-inertial frame rest frame of the table.

I'm not sure what you mean by this.

In the inertial rest frame frame of the book there are no forces acting on the book, so there is no work done on the book.

Exactly.

 

[A.T.]

Let's deconfuse the two scenarios and two frames.

Book rests on the table:

- Rest frame of table: no work is done on the book
- Free falling frame: table force is doing work on the book

Book falls from the table:

- Rest frame of table: inertial force of gravity is doing work on the book
- Free falling rest frame of the book: no work is done on the book

 

[PeterDonis]

Also to simplify things we are using the Schwarzschild exterior solution of GR, so that we can consider the geodesic free-falling frame of reference as static.

I'm not sure I understand what this means. I understand what it means for a spacetime to be static, and the Schwarzschild spacetime outside the horizon is; but what does it mean for a free-falling frame of reference to be static?

 

[TrickyDicky]

I'm not sure I understand what this means. I understand what it means for a spacetime to be static, and the Schwarzschild spacetime outside the horizon is; but what does it mean for a free-falling frame of reference to be static?

I meant the inertial rest frame, and to give a well defined rest notion it is simpler to choose a static solution, instead of say an expanding one where it is trickier.

 

[TrickyDicky]

Let's deconfuse the two scenarios and two frames.

Book rests on the table:

- Rest frame of table: no work is done on the book
- Free falling frame: table force is doing work on the book

Book falls from the table:

- Rest frame of table: inertial force of gravity is doing work on the book
- Free falling rest frame of the book: no work is done on the book

Looks like a good summary.

 

[PeterDonis]

I meant the inertial rest frame, and to give a well defined rest notion it is simpler to choose a static solution, instead of say an expanding one where it is trickier.

But an inertial frame can only be at rest relative to a static observer (one who stays at a constant radius, or, to put it in invariant terms, one who follows an orbit of the timelike Killing vector field) for an instant. So it seems strange to call the frame itself "static".

Also, there is a difference between "a local inertial frame at some particular event on a free-falling observer's worldline" and "a frame in which the free-falling observer is at rest for his entire fall". The former acts just like an inertial frame in SR (within its local range); the latter does not. I'm still not sure which of the two you mean by "the inertial rest frame".

 

[TrickyDicky]

But an inertial frame can only be at rest relative to a static observer (one who stays at a constant radius, or, to put it in invariant terms, one who follows an orbit of the timelike Killing vector field) for an instant. So it seems strange to call the frame itself "static".

It seems strange because as I explained it was just a slip, I meant a rest frame in a static spacetime rather than a static frame (I don't think that is even a conventional form to refer to a frame of reference)

Also, there is a difference between "a local inertial frame at some particular event on a free-falling observer's worldline" and "a frame in which the free-falling observer is at rest for his entire fall". The former acts just like an inertial frame in SR (within its local range); the latter does not. I'm still not sure which of the two you mean by "the inertial rest frame".

The latter. The geodesic path in a exterior Schwarzschlld geometry.


What is your answer to the OP question?

 

[PeterDonis]

I meant a rest frame in a static spacetime

...

The latter. The geodesic path in a exterior Schwarzschlld geometry.

Ok, got it.

What is your answer to the OP question?

I agree with A.T.'s summary.

 

[Dale]

Let's deconfuse the two scenarios and two frames.

Book rests on the table:

- Rest frame of table: no work is done on the book
- Free falling frame: table force is doing work on the book

Book falls from the table:

- Rest frame of table: inertial force of gravity is doing work on the book
- Free falling rest frame of the book: no work is done on the book

I also agree with this summary, but would add a little detail as follows:

Book rests on the table:

- Rest frame of table: two forces on book (upwards real, downwards fictitious), no motion, no work is done on the book, KE constant, PE constant.
- Free falling frame: one force (upwards real), book accelerates upward, table force is doing work on the book, KE increases, PE does not exist.

Book falls from the table:

- Rest frame of table: one force (downwards fictitious), acceleration downwards, inertial force of gravity is doing work on the book, KE increases, PE decreases
- Free falling rest frame of the book: no forces on book, no motion, no work is done on the book, KE constant, PE does not exist

Hope I got those right.

 

[TrickyDicky]

I also agree with this summary, but would add a little detail as follows:

Book rests on the table:
- Free falling frame: one force (upwards real), book accelerates upward, table force is doing work on the book, KE increases, PE does not exist.

This was the only point come to think of it that you (and stevendaryl) were disagreeing with me in Q-reeus thread, I can't really understand why if you admitted that in the falling book case gravity force was doing work on the book, even less I uderstand how could you not admit you were wrong, I can only in your case attribute it to something personal.
By the way what do you mean potential energy doesn't exist? It does if you want to conserve energy.

 

[Dale]

even less I uderstand how could you not admit you were wrong, I can only in your case attribute it to something personal.

Please point out exactly where you think that anything I said there conflicts in any way with anything I said here.

 

[TrickyDicky]

Please point out exactly where you think that anything I said there conflicts in any way with anything I said here.

Just a few:

This is an incorrect understanding of energy. Once the small deformation is done there is no motion, so no work is being done.

None of that energy is going into the book whose KE and PE are remaining constant.

No, I am not considering the table an inertial object, if the table were inertial then the force on the book would be 0.

So, the book, sitting on the table at rest, has no change in KE. It also has no change in PE. No work is being done on it.

However here:

Book rests on the table:
- Free falling frame: one force (upwards real), book accelerates upward, table force is doing work on the book, KE increases, PE does not exist.

But none of this quoting was really necessary, I started this thread because you were saying in the other thread that my claim that-in a different frame than the usually considered- work was done on the book was nonsense. And now you are saying that "you hope you got those right".

 

[A.T.]

Just a few:

I checked the whole posts by DS you quoted, and it is quite obvious that he talks about the non-inertial rest frame of the table. So there is no contrdiction to this:

Book rests on the table:
- Free falling frame: one force (upwards real), book accelerates upward, table force is doing work on the book, KE increases, PE does not exist.

 

[TrickyDicky]

I checked the whole posts by DS you quoted, and it is quite obvious that he talks about the non-inertial rest frame of the table. So there is no contrdiction to this:

So then why was he saying I was wrong when I claimed that in the free-falling frame work was done on the book. That's pretty absurd. I kept saying I was not referring to the Earth's non-inertial frame. And even made a heuristic formula for work in the geodesic (free-falling) frame that he said was wrong.

 

[A.T.]

So then why was he saying I was wrong when I claimed that in the free-falling frame work was done on the book.

Where?

 

[TrickyDicky]

Where?

Where what? where was he saying I was wrong? In every post he answered mine.
where was I saying I was considering the inertial frame, implicitly in most of my posts. Explicitly in 317# and the posterior discussion with stevendaryl made it even clearer. Besides we were only considering two frames the free-falling and the non-inertial, what else could I mean if I kept saying that in the Earth's frame it was clear work was not done.

 

[TrickyDicky]

In any case if he actually knew it and kept saying I was wrong that can only mean he was not acting on good faith.

 

[A.T.]

where was I saying I was considering the inertial frame, implicitly in most of my posts. Explicitly in 317#

Where in post 317# do you explicitly state that you consider the free falling frame?

 

[TrickyDicky]

Where in post 317# do you explicitly state that you consider the free falling frame?

"In which the table would be preventing the book from following its geodesic path. And in order to do that Work in the form of a quantity proportional to the EM force in the table material times distance from the theoretical geodesic path of the book, must be done." My bold.

In this case geodesic path and free-falling path are accepted synonims. If here it is not obvious I'm considering the geodesic frame, in the context of my insistence on not considering the Earth's non-inertial...

This is a not very productive exchange, you may think what you like. I can see there is certain subjective bias on your part to favor what is obvious DS thinks.

 

[A.T.]

Where in post 317# do you explicitly state that you consider the free falling frame?

"In which the table would be preventing the book from following its geodesic path. And in order to do that Work in the form of a quantity proportional to the EM force in the table material times distance from the theoretical geodesic path of the book, must be done." My bold.

In this case geodesic path and free-falling path are accepted synonims.

That says nothing about the reference frame you are considering. Not even implicitly. The table is preventing the book from flowing a geodesic path in any frame.

You seem to confuse:
- objects being inertial / non-inertial
- reference frame being inertial / non-inertial

The non-zero proper acceleration of the book on the table is frame invariant. So the book on the table is non-inertial in any frame. So stating that the book is non-inertial (or doesn't follow a geodesic path) doesn't imply anything about the reference frame you are considering.

This seems to be the root of the misunderstanding between you and DS about which frame is being considered.

 

[TrickyDicky]

That says nothing about the reference frame you are considering. Not even implicitly. The table is preventing the book from flowing a geodesic path in any frame.

Are you serious? I was referring to the next sentence(we have different notions of what is obvious it seems): when considering the usual formula W=F.d you need to reference the distance d to some reference, in the case of the book if the frame is the table distance is zero and no work is done, if the reference is the free falling frame the distance is not zero and that is why there is work.

 

[A.T.]

we have different notions of what is obvious it seems

We definitely have different notions of what "explicitly" means. Anyway, the whole misunderstanding between you two could have been cleared up quickly by stating the frames explicitly, like I did in post #4.

 

[Dale]

Just a few:

However here:

As A.T. mentioned, all of those quotes in the other thread referred to the table frame since that was always the topic of that thread. (Q-reeus refuses to consider any coordinates other than Schwarzschild and is interested in coordinate effects, so all of his threads deal only with the non-inertial frames). So they cannot possibly be in conflict with the quote above regarding physics in the free-falling frame.

 

[Dale]

This is a not very productive exchange, you may think what you like. I can see there is certain subjective bias on your part to favor what is obvious DS thinks.

It is amazing how quickly you go to a personal attack. Nobody who disagrees with you can possibly have any legitimate reason to do so, therefore it must be nefarious personality defects and conspiracies. What nonsense! A.T. and I disagree vehemently with each other when we think the other is wrong, we agree with each other when we think the other is right. There is no bias on either part.

You should stop assuming that every correction is a personality flaw and actually learn something.

 

[Dale]

So then why was he saying I was wrong when I claimed that in the free-falling frame work was done on the book.

Because you didn't say that. Q-reeus and I were discussing the non-inertial frame. As a result Austin0 and I had a discussion about the non-inertial frame. You jumped in with:

This would be correct only if the table is considered an inertial object

which was wrong, I was not considering the table an inertial object. And

the table is non-inertial and in continuous motion so there is work done.

which is wrong, simply because the table is non-inertial neither implies that it is in motion nor that there is work done. Both of those depend on the reference frame and are not true in the table's frame. If you were considering an inertial frame, you certainly didn't specify.

Besides misunderstanding that GR uses fictitious forces, the next thing you got wrong was:

And in order to do that Work in the form of a quantity proportional to the EM force in the table material times distance from the theoretical geodesic path of the book, must be done.

which is a nonsense definition of work. I challenged you to find a referece, which of course you couldn't.

Then this one:

No work is done on the book in the Earth's frame and considering gravity a real force. But you are mixing the concept of gravity as real force and fictitious force according to your interest.

Here you are explicitly talking about the non-inertial rest frame of the table where gravity is a fictitious force, but somehow you think I was considering gravity as a real force, which is wrong, I was always considering gravity as a fictitious force.

I believe that covers every place where I said you were wrong. Everything else was either a misunderstanding of what I wrote, which I corrected, or a misunderstanding of the Wikipedia article which I identified but didn't particularly correct.

 

[TrickyDicky]

Oh, I see, you never misunderstood anything and it never ocurred to you I was talking about a different frame.
How about you being open to do some learning for a change?

 

[Dale]

Oh, I see, you never misunderstood anything and it never ocurred to you I was talking about a different frame.
How about you being open to do some learning for a change?

I certainly may have misunderstood your intention, it happens often, but that communication failure goes two ways. I can only respond to what you wrote, which was wrong multiple times.

When I realized that you had misunderstood me about the GR vs Newtonian POV I corrected the misunderstanding factually without any personal vitriol and with references.

If you were thinking about a different frame than the rest of the conversation then you should have said so, which you never did. Furthermore, you demanded that I admit I was wrong when I wasn't, and tried to turn my inability to read your mind into a personality defect.

While I am always open to learning, but learning to develop psychic powers for the sole purpose of telepathically reading your mind seems like a useless effort all around.

 

[TrickyDicky]

I can only respond to what you wrote, which was wrong multiple times. If you were thinking about a different frame than the rest of the conversation then you should have said so, which you never did.

While I am always open to learning, but learning to develop psychic powers for the sole purpose of telepathically reading your mind seems like a useless effort all around.

Well you at least are trying hard to be funny. But I get the drift , you need to have everything written down explicitly and in the most conventional textbook way, no matter the context.
Well, I just disagree with that philosophy. We seem to have very different perspectives about science, I disagree with most of what you write. That is not bad in itself as long as finally the physics part comes out right. And that has come out right, since we agree there is work done on the sitting book in the free-falling frame. I simply was perplexed by your attitude but I understand PF is not the place to tackle that.
Time to move on. Can you?
Edit: written before you edited your post.

 

[Dale]

you need to have everything written down explicitly and in the most conventional textbook way, no matter the context.
Well, I just disagree with that philosophy.

And I disagree with your philosophy of making personal attacks and demanding that people, who have done nothing but make correct statements in a conventional textbook manner, should nevertheless admit they were wrong. But yes, I can move on.

 

[TrickyDicky]

And I disagree with your philosophy of making personal attacks

That's not my philosophy as you well know. But it's ironic that you say you disagree with that philosophy by making a personal judgement.

But yes, I can move on.

Really? Wish you could.

 

[stevendaryl]

In the above linked thread stevendaryl argued that the equivalence failed because he could find coordinates in the schwarzschild spacetime that made the metric appear time varying. Being obvious that coordinates can introduce artifacts that don't affect the physics of the situation I still don't know what his point was wrt the physics of the problem at hand.

I was only pointing out a disanalogy between the case of (1) flat spacetime and (2) Schwarzschild spacetime. In the first case, there is a coordinate system in which an inertial observer is at rest and the metric is time-independent. In the second case, there is no such coordinate system. In Schwarzschild spacetime, any coordinate system in which a freefalling observer is at rest will have a time-varying metric.

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

 

[A.T.]

In Schwarzschild spacetime, any coordinate system in which a freefalling observer is at rest will have a time-varying metric.

What about an observer at rest in the center of the spherical mass? Isn't the metric in his rest frame time independent?

 

[stevendaryl]

What about an observer at rest in the center of the spherical mass? Isn't the metric in his rest frame time independent?

You're right. that's an interesting case, since it's the only case in which the notion of "inertial observer" and "observer at rest at constant radius" coincide.

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.

 

[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: \nabla_{\mu} T^{\mu \nu} = 0. 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.