Is Energy Conserved in General Relativity?

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Energy is not well defined in general relativity (GR), as asserted by Steve Carlip, due to the inclusion of gravitational potential energy, which can be transformed away in freely falling coordinate systems. This leads to the conclusion that either energy is zero everywhere or lacks a consistent definition across different frames. Alternatives for defining energy in GR include considering nonlocal definitions or scenarios where covariance breaks down, with flat spacetime being a potential candidate. The discussion also touches on the distinction between gravitational fields and spacetime curvature, emphasizing that gravitational potential energy is not a tensor quantity. Overall, the debate highlights the complexities of defining energy within the framework of general relativity.
  • #31
pmb_phy said:
And I'm the king of the universe. Hmmmm. You're right! It's very easy to make claims.

You've chosen a poor, albeit popular, view of gravity. One that has led you to make errors in the past. ...
Regarding "others". I'm a physicist too. ...

No I mean real physicists, not someone who has done secretarial work for a physicist. As for the popular and correct view, no the correct view has not led me to errors, as I and other physicists have told you in the past the modern relativistic paradigm is not the Newtonian paradigm. Others may see examples at
http://tinyurl.com/2aq3z
And why do you use 9 different email addresses in google anyway? Are you so dermined not to be killfiled?
The true analogy is given in terms of the Newtonian tidal force 3-tensor...
Truth, lol. As I was saying the Newtonian paradigm is not the modern relativistic paradigm. And no Newtonian quantities are rank 3 tensors. You obviously mean a rank 2 pseudo-tensor whose indeces run 1 through 3. That is not a tensor at all in modern relativity.
 
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  • #32
Mr. Moderator

Can you do something about DW? Perhaps explain to him that flaming is unwelcome here?

Thanks

Pete
 
  • #33
DW said:
<snipped flames>
If you can't resist posting lies and flames then please leave. You are not wanted here when you post lies and flames like you're doing now.
And no Newtonian quantities are rank 3 tensors. You obviously mean a rank 2 pseudo-tensor whose indeces run 1 through 3.
You're confusing the term 3-tensor with the term tensor of rank 3. I've already explained the difference to you many times in the past.

The term 3-tensor refers to a tensor in R3. It does not mean a 3rd rank tensor. Tell you what waite. Turn to page 160 in MTW . At the top you'll see MTW explain the following
This equation suggests that one call mjk the "inertial mass per unit volume" of a stressed medium at rest. In general mjk is a symetric 3-tensor.
When you learn what that statement means then you'll have learned the meaning of the term "3-tensor". Take a hint from the term "4-vector". That's a tensor of rank 1 in a 4-d space. A 3-vector is a tensor of rank 1 in a 3-d space.

In general, the term n-tensor of rank (l,m) is a tensor of rank (l,m) in an n-dimensional space. Haven't you ever heard of a 3-vector? That's a tensor of rank 1 in R3.

Also, you're incorrectly in calling it a pseudo-tensor. You're abusing the terminology. To learn the definition of the term pseudo-tensor please see -- http://mathworld.wolfram.com/Pseudotensor.html

Unless, that is, you consider "pseudo-tensor" and "pseudotensor" to be two different terms with two different meanings. If so then the former is undefined. If you like to define terms of your own then please state the definition before you use it.
That is not a tensor at all in modern relativity.
In the first place that is totally irrelevant since I said Consider the analogous example in Newtonian physics:... so whether that had to do with relativity is neither here nor there.

In the second place that is incorrect. That's like saying that the 3-velocity 3-vector is not part of relativity. Both claims are incorrect. The tidal force 3-tensor is part of the Riemann tensor just as the 3-velocity 3-vector dr/dt is part of the 4-velocity, the 3-force 3-vector is part of 4-force, and the current density vector 3-vector is part of 4-current.

Tell me waite - why is it that all you've ever been able to do is to post lies and accusations and claims of proof without ever actually posting a proof of any shape or kind? It's like that time you flamed Tom and I and Tom had to ban you from here. You did the same thing then as you're doing not and you've always done. All you did was claim he and I were wrong and then when you couldn't force your misconceptions on us you reverted to flaming.

Stop stalking me for cripes sake! The only time you post on the internet is to cause trouble at the places I'm posting. Grow up dude!
 
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  • #34
pmb_phy said:
Mr. Moderator

Can you do something about DW? Perhaps explain to him that flaming is unwelcome here?

Thanks

Pete

What lies? What flames? You said that you were king of the universe denoting sarcasm and in that context said that you could just as well claim to be a physicist. I then clarified and said that I meant real physicists have explained it to you. And I know that you have done secretarial work for Taylor&Wheeler in contributing to a glossary of terms so I acknowledged that. In no way did I flame you.
 
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  • #35
pmb_phy said:
In general, the term n-tensor of rank (l,m) is a tensor of rank (l,m) in an n-dimensional space.
I already corrected you. AGAIN, see the "modern" relativistic version of that problem as problem 3.1.4
at
http://www.geocities.com/zcphysicsms/chap3.htm#BM25
And believe it or not the dimensions in Newtonian mechanics are also 4. If that is where you picked up the bad terminology then complain to them, not me.
 
  • #36
pmb_phy said:
Unless, that is, you consider "pseudo-tensor" and "pseudotensor" to be two different terms with two different meanings.

I can't resist anymore, you coersed me into telling everyone else this joke.
Q: What is the definition of a pseudointellectual?
A: One who pretends to know what pseudo means.
Look it up.
 
  • #37
TO pmb_phy & DW:

You know what? I've just about had it up to here with the both of you. If you can't leave your personal differences at the door and stick to the subject, maybe neither of you should be here.

Consider this a warning.
 
  • #38
Janus said:
TO pmb_phy & DW:

You know what? I've just about had it up to here with the both of you. If you can't leave your personal differences at the door and stick to the subject, maybe neither of you should be here.

Consider this a warning.

You're kidding me. dw flames me that you're warning me?

Sorry. I can't think of a reason to want to post at a place where a moderator warns people who are being flamed.

Good luck with dw. You'll certainly need it.
 
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  • #39
pmb_phy said:
You're kidding me. dw flames me that you're warning me?

That's funny. He's saying the exact same thing about you.

Both of you have got your fangs out and your talons sharpened.

Both of you read back what you posted, including the earlier post before you edited it, pmb_phy.

Can you honestly say that you'd talk to someone in the same room as you the same way you've been talking to each other on this thread?

The little flamewar which you're both participating in started with a little jab and then quickly escalated, as it appears to me. That being said, I'll echo Janus and say:

Knock it off, the both of you.
 
  • #40
This was regarding the statement about the coordinate independence of tensors:
pmb_phy said:
This is one of the most important points to understand in relativity.
I have heard this (concerning tensors), and I think I agree, but I don't think I quite have the idea behind 4-acceleration (as a tensor). If it is a tensor (invariant, so vanishing is vanishing in any coordinate system), then I guess I'm just having a little trouble understanding how acceleration can be a tensor in the first place. I can sit on the patio and watch something fall out of the sky. Then I would say, "that thing is accelerating." Then, I could be sitting right on top of the thing and fall with it and say, "this thing is not accelerating."

I guess this has something to do with the proper time dilating more as the speed increases. I can kind of get the idea mathematically, by my intuitive picture seems to be lacking.
 
  • #41
turin said:
This was regarding the statement about the coordinate independence of tensors:I have heard this (concerning tensors), and I think I agree, but I don't think I quite have the idea behind 4-acceleration (as a tensor). If it is a tensor (invariant, so vanishing is vanishing in any coordinate system), then I guess I'm just having a little trouble understanding how acceleration can be a tensor in the first place. I can sit on the patio and watch something fall out of the sky. Then I would say, "that thing is accelerating." Then, I could be sitting right on top of the thing and fall with it and say, "this thing is not accelerating."

I guess this has something to do with the proper time dilating more as the speed increases. I can kind of get the idea mathematically, by my intuitive picture seems to be lacking.

The "length" of a tensor is frame invariant. So is a tensor equation. The tensor itself is frame covariant. If a tensor is not zero according to one frame it is not zero according to all and this goes for the acceleration four-vector as well. The length of the velocity four-vector is c which is also invariant. This means that in a relativistic interperetation of the motion anything always moves at the speed c in four dimensional spacetime. It is just the direction in four-dimensional spacetime that can be deflected. Naturally an object travels along a geodesic. When a real force is exerted it is expressed as a four-vector force on the object. This never changes the length of the velocity four-vector or the four dimensional speed which is c. What it does is change the deflect the motion or rotate it in spacetime from the geodesic. The four-vector acceleration is a description of this change in motion. When your motion is changed from that of a geodesic all observers will agree on that, even you.
If you must think of the forced and falling object as changing speed then consider its motion with respect to a local free fall frame observer. The length of the acceleration four-vector according to any frame is equal to the magnitude of the coordinate acceleration for a local free fall frame according to which the object is instantaneously at rest. If you are in free fall along with a reference object and then experience a real force, the relative velocity between the two of you will become nonzero and you will feel the force that pushed you so you will be able to determine that you were accelerated just as anyone else will say.
 
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  • #42
DW said:
When your motion is changed from that of a geodesic all observers will agree on that, even you.
Yes, of course! Thank you, DW. It is starting to become a bit more organized. A geodesic is a geodesic, regardless of how one describes it.

I still think that this is drastically different than the pre-relativity (Newton's) picture.
 
  • #43
pmb & DW - - We've been through this before. I'm sorry to see that the cease fire has ended. Further flaming will be locked/deleted.
 
  • #44
Energy concept

May I propose a simple problem that could help in understanding the energy concept?
Let consider an astronaut out of his spaceship and connected to it by a wire. He as a tool that can put the wire under tension and let him move towards the spaceship.
The tool can be, for exemplum, a spring loaded so with that tool the astronaut has a definite amount of energy (E*) available.
The astronaut use that energy E* to accelerate towards the spaceship.
If E* is known what will be the relative velocity between the astronaut and the spaceship in Newtonian physics?
 
  • #45
I'd like to comment on the tidal tensor business. There is a very useful 3x3 tensor in GR that's sometimes called the "Electric part of the Weyl Tensor" (E[v] in GRTensorII software package). Other times a closely related tensor has been called the "electrogravitic tensor"

http://www.lns.cornell.edu/spr/2002-03/msg0039921.html

You basically compute

"Electric part of Weyl"

<br /> C_{abcd} u^{b}u^{d}<br />

"Electrogravitic"
<br /> R_{abcd} u^{b}u^{d}<br />

Where u is the 4-velocity of an observer.

If the matter density is zero where the measurement is made, both of these will be equal.

If you eliminate the index for time, you get a 3x3 tensor that describes the tidal force near the object. If you feed the 3x3 tensor a direction, the result is an acceleration vector - or a force/unit mass. The diagonal terms of this tensor represent "stretching" forces, the off diagonal terms represent torques.
 
  • #46
selfAdjoint said:
Steve Carlip, on sci.physics.research has asserted that energy is not well defined. Paraphrasing him, if you want to have a concept of energy, you need to include gravitational potential energy, right? But you can always switch to a freely falling coordinate system in which gravitational potential is zero. And a tensor, if zero in any coordinate system is zero in every coordinate system. So since this is true at every point of spacetime, either energy is identically zero everywhere, or else it is not well defined, because only tensors are well defined (covariant) in GR.

Going back to the original question/point

I would say that the energy of an isolated system in asymptotically flat space-time is is well defined in GR, but the location of the energy is not.

There's a whole chapter in MTW's "Gravitation"

"Why the energy of the gravitatioanl field cannot be localized" on p 466.

I'd like to add, though, that though I'm fairly confident that the energy and momentum of an isolated but moving system are well defined in GR, I'm currently having a heck of the time with the details for the case in which the system is not stationary (i.e. the gravitational field of a moving black hole).
 
  • #47
"I'm currently having a heck of the time with the details for the case in which the system is not stationary (i.e. the gravitational field of a moving black hole)."

Good luck finding a sensible definition of energy for a nonstationary system. You don't have timelike isometries, and you will run into problems over and over again far away from the linearized theory where the time dependence of the real metric becomes important.

Its really quite simple if you think about it. A metric has ten independant components, and any transformation allows you to introduce 4 extra degrees of freedom.. Only in very particular types of metrics, will it allow you to make your components time independant.

You can sometimes define things in such a way as to give you a conserved quantity that can be interpreted as energy, but its never a general solution, and is case by case for metrics.
 
  • #48
Haelfix said:
"I'm currently having a heck of the time with the details for the case in which the system is not stationary (i.e. the gravitational field of a moving black hole)."

Good luck finding a sensible definition of energy for a nonstationary system. You don't have timelike isometries, and you will run into problems over and over again far away from the linearized theory where the time dependence of the real metric becomes important.

Its really quite simple if you think about it. A metric has ten independant components, and any transformation allows you to introduce 4 extra degrees of freedom.. Only in very particular types of metrics, will it allow you to make your components time independant.

You can sometimes define things in such a way as to give you a conserved quantity that can be interpreted as energy, but its never a general solution, and is case by case for metrics.

Well, as near as I can make out, there are supposed to be not just one, but two notions of the energy of a system, that apply in general to asymptotically flat space-times. This is all rather confusingly discussed in one of my textbooks, Wald's "General Relativity" around p 291.

One is called the Bondi energy, the concept originated with Bondi before the reformulation in terms of asymptotic flatness, and is associated with null infinity. The other is the ADM energy, and is associated with spatial infinity.

This is in addition to some special-case things one can do, which are supposed to give one a Poincare subgroup of the more general infinite-dimensional groups that arise at the appropriate infinity(ies). These special case things do require some conditions on the metric, though.

The Bondi energy actually isn't strictly speaking an invariant - because it's at null infinity the system can lose energy due to gravitational radiation. I'm describing the known loss of energy at null infinity due to gravitational radiation as "energy being conserved". The ADM energy, being at spatial infinity, doesn't have this feature (or problem, if you prefer), so it's a true invariant. The Bondi and ADM energies can even be compared, with sensible results, they turn out to be the same if there is no gravitational radiation.

Anyway, that's supposed to be the theory. So far I haven't actually managed to calculate anything.
 
  • #49
pervect said:
Steve Carlip, on sci.physics.research has asserted that energy is not well defined. Paraphrasing him, if you want to have a concept of energy, you need to include gravitational potential energy, right? But you can always switch to a freely falling coordinate system in which gravitational potential is zero. And a tensor, if zero in any coordinate system is zero in every coordinate system. So since this is true at every point of spacetime, either energy is identically zero everywhere, or else it is not well defined, because only tensors are well defined (covariant) in GR.
That is not a reasonable statement. Transforming to a freefall frame and thus transforming to a frame in which the gravitational field vanishes does not mean that the potential energy of a particle vanishes. For example: If you're in a uniform gravitational field and you transform to a freefall frame then the gravitational potentials, gjk become constants. Thus the metric does not vanish. The existence of a gravitational field does not depend on the vanishing of the metric tensor but on the spacetime variation of the gravitational potentials (when the spatial coordinates are Cartesian). Of course this was all explained/described by Einstein in his 1916 GR paper.

Pete
 
  • #50
pmb_phy said:
That is not a reasonable statement. Transforming to a freefall frame and thus transforming to a frame in which the gravitational field vanishes does not mean that the potential energy of a particle vanishes. For example: If you're in a uniform gravitational field and you transform to a freefall frame then the gravitational potentials, gjk become constants. Thus the metric does not vanish. The existence of a gravitational field does not depend on the vanishing of the metric tensor but on the spacetime variation of the gravitational potentials (when the spatial coordinates are Cartesian). Of course this was all explained/described by Einstein in his 1916 GR paper.
Pete

I believe Self Adjoint was the one quoting Steve Carlip, not I. This may seem a minor point, but it does mean that I haven't even seen Mr Carlip's full argument personally.

Regardless of whether the argument was spelled out in enough detail to be believed on its own, I do agree with the conculsion, which is that there is no way to describe potential energy in GR with a tensor quantity. I thought you had agreed with this too, now I'm rather unclear as to your position on the matter.
 
  • #51
pervect said:
I believe Self Adjoint was the one quoting Steve Carlip, not I.
Sorry. I was simply addressing his comments. My mistake if I got things a bit mixed up.
Regardless of whether the argument was spelled out in enough detail to be believed on its own, I do agree with the conculsion, which is that there is no way to describe potential energy in GR with a tensor quantity. I thought you had agreed with this too, now I'm rather unclear as to your position on the matter.
There is a subtle thing that many people miss here. Notice exactly the comment I was addressing. It was, exactly, this
But you can always switch to a freely falling coordinate system in which gravitational potential is zero.
There is a difference between the gravitational potential and, referring to a particle in a gravitational field, the gravitational potential energy of a particle. There is also gravitational self energy which is the energy related to the gravitational field itself.

The gravitational potential is related to the gravitational force. This means that the gravitational force, in general relativity, is a combination of the derivatives of the gravitational potentials, i.e. guv (aka components of the metric tensor) and the velocity of the particle. See Eq. (8a) in

http://www.geocities.com/physics_world/gr/grav_force.htm

The Christoffel symbols (capital gammas) are functions of the gravitational potentials, guv. The gravitational potentials are well defined quantities in GR.

The gravitational potential energy of a particle, at least to me, is the energy of the particle by virtue of its position in a gravitational field. The energy of a particle as a function of position is also well defined in GR. Just because the energy is not a linear sum of rest, kinetic and potential energy, it doesn't mean that they are not well defined or meaningless.

I believe that what Carlip was referring to was the fact that if you tell me the position and velocity of a particle in a strong gravitational field that I will not be able to give you a specific value for something and meaningfully call it "potential energy". However if the field is weak I can do this and do it in general relativity.

Pete
 
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  • #52
Pete, why don't you post on s.p.s in that Carlip thread and let's get the issues cleared up. With all respect, I trust his claims more than yours, but I think it's probably a semantic difference. But it should be resolved, and I don't see how I can improve the conversation as a middleman.
 
  • #53
selfAdjoint said:
Pete, why don't you post on s.p.s in that Carlip thread and let's get the issues cleared up.
What is "s.p.s"? If its a newsgroup then I don't post in newsgroups anymore. Carlip and I have discussed this before anyway.
With all respect, I trust his claims more than yours, but I think it's probably a semantic difference. But it should be resolved, and I don't see how I can improve the conversation as a middleman.
I don't wish to post anywhere outside this forum again. If someone wants to discuss this within a forum here then I'll be glad to chime in and clarify/backup anything I say/post.

Pete
 
  • #54
pmb_phy said:
What is "s.p.s"? If its a newsgroup then I don't post in newsgroups anymore. Carlip and I have discussed this before anyway.

Pete I think it was SPR (sci.physics.research) that was meant and that by mistake SPS got typed----that would stand for sci.physics.strings
but that would not make sense in this context.

SelfAdjoint originally gave a link to something on SPR by Carlip.
June 6, 2004
Re: Tired Light
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=c9tajo$d6h$2@woodrow.ucdavis.edu
 
  • #55
Pete, I have an idea for you

Go to this thread here at PhysicsForums
https://www.physicsforums.com/showthread.php?t=26625

look at post #15

here is the direct link to it:
https://www.physicsforums.com/showthread.php?p=227804&posted=1#post227804

PF has "eaten" your old Alma Mater newsgroup sci.physics.research

The look and feel is nicer in the PF version

Now you can enjoy corresponding with Steve Carlip in the comfort of PF accomodations! Have fun!
 
  • #56
pmb_phy said:
The gravitational potential is related to the gravitational force. This means that the gravitational force, in general relativity, is a combination of the derivatives of the gravitational potentials, i.e. guv (aka components of the metric tensor) and the velocity of the particle. See Eq. (8a) in

http://www.geocities.com/physics_world/gr/grav_force.htm

The Christoffel symbols (capital gammas) are functions of the gravitational potentials, guv. The gravitational potentials are well defined quantities in GR.

Now that I know that you are calling the metric coefficients guv "gravitational potentials", some of your remarks make more sense. But I should note that this usage is not standard in any of my textbooks, and I would imagine it would confuse most readers as much as it confused me. Since we have a perfectly good name for the metric coefficients already ("metric coefficients"), and since this usage causes *no* confusion, I would like to suggest that we continue to call metric coefficients metric coefficients.

The gravitational potential energy of a particle, at least to me, is the energy of the particle by virtue of its position in a gravitational field. The energy of a particle as a function of position is also well defined in GR. Just because the energy is not a linear sum of rest, kinetic and potential energy, it doesn't mean that they are not well defined or meaningless.

Since the energy of a particle will in general depend on the path it takes, I don't see how to define "the energy of the particle by virtue of its position".

I believe that what Carlip was referring to was the fact that if you tell me the position and velocity of a particle in a strong gravitational field that I will not be able to give you a specific value for something and meaningfully call it "potential energy". However if the field is weak I can do this and do it in general relativity.

Oh, weak fields. Sure, one can define potential energy for weak fields, the first thought that comes to mind is to use PPN. But I thought we were talking about GR, not the weak-field version therof.
 
  • #57
pervect said:
Going back to the original question/point

I would say that the energy of an isolated system in asymptotically flat space-time is is well defined in GR, but the location of the energy is not.
You mentioned the chapter in MTW on this point. Note that they say that its the gravitational energy that is not localizable, not the energy (i.e. other forms such as mass-energy, EM energy etc.).

That's a nice section in MTW by the way.

Pete
 
  • #58
marcus said:
Now you can enjoy corresponding with Steve Carlip in the comfort of PF accomodations! Have fun!
Thanks but no thanks. I have no desire to post to any of those newsgroups for my own personal reasons. Thanks anyway.

Pete
 
  • #59
pervect said:
Now that I know that you are calling the metric coefficients guv "gravitational potentials", some of your remarks make more sense. But I should note that this usage is not standard in any of my textbooks, and I would imagine it would confuse most readers as much as it confused me.
You're kidding me! Almost all of my GR texts call them that as did Einstein. For example: The following texts refer to guv as "gravitational potentials" -

Gravitation and Spacetime, Ohanian & Ruffini, WW Norton n& Co., (1994)

Introducing Einstein’s Relativity, D’Inverno, Oxford Univ. Press, (1992)

Basic Relativity, Mould, Springer Verlag, (1994)

I also recall seeing it in MTW and in Wald but I can't locate it at the moment

Since we have a perfectly good name for the metric coefficients already ("metric coefficients"), and since this usage causes *no* confusion, I would like to suggest that we continue to call metric coefficients metric coefficients.
I've never heard them called that. They are called the components of the metric. But you're free to call them what you like. But when you start to discuss gravitational potentials in GR then you're talking about guv whether you want to call them that or not.
Since the energy of a particle will in general depend on the path it takes, I don't see how to define "the energy of the particle by virtue of its position".
The energy of a particle is a function of velocity, position and rest mass. The functionality of position is what I mean by potential energy. I did not say that you can separate these energies into individual pieces. Let me quote Ohanian, page 157
P_0 = \simeq \frac{m}{\sqrt{1-v^2}} - \frac{1}{2}\frac{m}{(1-v^2)^{3/2}} h_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}+mh_{0\alpha}u^{\alpha}

The first term on the right side is of the form of the usual rest-mass and kinetic energy; the other terms represent gravitational interaction (potential energy).

Oh, weak fields. Sure, one can define potential energy for weak fields, the first thought that comes to mind is to use PPN. But I thought we were talking about GR, not the weak-field version therof.
You understand that by weak I do not mean Newtonian, right? Why do you think that the weak field approximation is not part of general relativity. I don't see any need to make such a distinction myself. I understand that you think that they are different but I don't see them that way.

In any case there are cases where, even in strong gravitational fields, the gravitational force is given in terms of the gravitational potential as

\bold G = -m\nabla \Phi

where m = inertial mass (aka relativistic mass). See derivation (and meaning of phi = gravitational potential) at
http://www.geocities.com/physics_world/gr/grav_force.htm

Its my turn to ask you something - A particle in a gravitational field has energy P0 where P is the 4-momentum of the particle. Do you think that a particle in a gravitational field has rest energy? If so then do you think that the rest energy is part of the energy P0? Its a mixture of these energies which one cannot separate into nifty pieces. Notice that Carlip said that potential energy is not well defined. He did not say it does not exist or that it is totally meaningless.

By the way, I posted a web page quoting that part of D'Inverno on this metric = potentials in case you don't have that text. See
http://www.geocities.com/physics_world/gr_potential.htm

As you can see, the term "potential" is used here in a similar way to its used in EM where the magnetic field is the curl (which also involves derivatives) of something called the magnetic vector potential. The force on a charged particle can therefore be written in terms of the derivatives of potentials, i.e. the Coulomb potential and the magnetic vector potential. That is the reason for calling the components of the metric "potentials".

Pete
 
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  • #60
Pervect.. Ok.

The ADM paradigm is a little bit tricky, but it does reproduce somewhat sensible results.

The problem is essentially trying to reproduce covariant hamiltonians for a suitable definition of asymptotic flatness called k-asymptotic flatness (believe it or not, there are several definitions of this, the weakest being the usual flat hypersurface at infinity). As this will lead to a method of constructing an asymptotic noether current, both for energy-density and angular momentum, etc.

AFAIR, the weakest form does indeed reproduce a good quantity for energy-density for a large subclass of metrics k > 1/2. It had some problems, I think, b/c it wasn't in agreement with a few other methods, but again AFAIR it ended up being ok.

Now, the non tensorial nature of this quantity had people perplexed for awhile, so they felt the whole thing was contrived and useless for more general metrics.

Others disagreed, as it was important in several fields including quantum gravity... Particuraly to find sensible definitions of observables in GR (the socalled quasi-locality theorems that Witten and others worked on). So believe it or not, ADM works in some situations even with nonstationary metrics with no asymptotic flatness assumptions.
 

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