Energy not defined in GR?

[selfAdjoint]

Steve Carlip, on sci.physics.research has asserted (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=c9tajo%24d6h%242%40woodrow.ucdavis.edu) 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.

Exploring alternatives, Carlip suggests various physical possibilities that would generate meaningful energy definitions - it boils down to considering cases where physics breaks covariance, and the flat universe is the best bet here.

The other alternative is that energy in GR could be NONLOCAL. If energy can only be defined within open sets, but not at points, the covariance argument fails. To me this suggests Bologiubov's definition of the quantum potential (which critically involved smearing over an open set in spacetime). Maybe some path like this could be explored?

[pmb_phy]

Steve Carlip, on sci.physics.research has asserted (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=c9tajo%24d6h%242%40woodrow.ucdavis.edu) that energy is not well defined.

Yup. Very true indeed.

Paraphrasing him, if you want to have a concept of energy, you need to include gravitational potential energy, right?

Yup. With ya so far.

But you can always switch to a freely falling coordinate system in which gravitational potential is zero.

Nope. I don't see that to be true in general. That is at best only true in uniform gravitational fields.

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.

Gravitational potential energy is not a tensor quantity so I don't see what that statement means. Seems as if that arguement assumes that energy is a sum of energies, i.e. total energy = rest energy + kinetic energy + potential energy. That, of course, is not true in GR.

I've never considered energy to be localized. That has always proven to be a poor way of thinking of energy.

Pete

[selfAdjoint]

:

But you can always switch to a freely falling coordinate system in which gravitational potential is zero.


Nope. I don't see that to be true in general. That is at best only true in uniform gravitational fields.

Locally you always can. That's the equivalence principle.

[pmb_phy]

Locally you always can. That's the equivalence principle.
Yes. One can always transform away the gravitational field at any point in spacetime. But that is only at a point. Gravitational potential energy is an energy of position. All you've said is that you can get rid of the potential energy at any point. But that's really an empty statement especially since differences in potential energy are the only things that are physically meaningful.

What did all that talk have to do with tensors? Gravitational potential energy has nothing to do with the non-vanishing of tensors. It has to do with the variability of the tensor components.

Perhaps you can rephrase your point. Or was that Carlip's point?

[kurious]

pmb _ phys

Yes. One can always transform away the gravitational field at any point in spacetime. But that is only at a point

The uncertainty principle says that exact points don't exist

SELF ADJOINT:

This would make energy in GR non-local.

[russ_watters]

But you can always switch to a freely falling coordinate system in which gravitational potential is zero. Sure. You can also declare yourself stationary and make velocity, distance traveled, and kinetic energy zero as well. This doesn't in any way make the definitions weak - just relative.

[selfAdjoint]

kurious: The uncertainty principle doesn't come in to GR. GR is a classical theory, and these are classical results. Until there is a unification, GR physics has to do without it. My suggestion was an idea for a unifying direction.

russ: If "relative" would do the job, why would Steve Carlip be so sure? Go read his post on s.p.r.

[kurious]

kurious: The uncertainty principle doesn't come in to GR. GR is a classical theory, and these are classical results. Until there is a unification, GR physics has to do without it. My suggestion was an idea for a unifying direction.


But what if the uncertainty in the potential energy of a particle in a gravitational field depends on how long it is measured to be at a position in the field ( E x t = h bar).This gives a measure of the distribution of energy and is an idea for a unifying direction because to unify QM and GR, QM ideas must come into GR.

[pmb_phy]

[QUOTE=kurious]pmb _ phys

Yes. One can always transform away the gravitational field at any point in spacetime. But that is only at a point[QUOTE]

Not neccesarily. That is only the case if the region of spacetime of interest is curved. If the region is flat then you can completely transform the field away in that region of space.

Pete

[robphy]

Here is a somewhat technical review article on Quasi-Local Energy
http://relativity.livingreviews.org/Articles/lrr-2004-4/index.html

[turin]

Why are people so adamant that gravity must have an energy associated with it? I do believe that gravity requires stress-energy in order to exist in the first place, but I do not at all follow the argument that gravitational fields must have energy, or that there must be some gravitational potential energy. In fact, whatever I read about GR seems to put the gravitational field as a purely metrical construct. I have seen approaches that introduce non-linearity by suggesting that the gravitational field has energy and therefore must, itself, gravitate. But that is by no means the only way, and it is not what I have found to be even general accepted as a sound approach.

Can someone explain to me what's the deal?

[DW]

Why are people so adamant that gravity must have an energy associated with it? I do believe that gravity requires stress-energy in order to exist in the first place, ...

Actually that is the problem. Gravity alone has no real stress-energy tensor. The stress-energy tensor for a reageon with no matter but gravitation is zero. Also, while it is true that pmb's Newtonian concept of the gravitational field as an acceleration field can be transformed away, the modern relativistic paradigm which refers to spacetime curvature can not. The field of spacetime curvature is expressed as the Riemann tensor and as with any tensor, when this is not zero according to any frame it is not zero according to all frames. It can not be transformed away. It is the affine connection expressed by the Christophel symbols of the second kind which are analogous to the Newtonian field concept and can be transformed away, not the modern relativistic field concept of Riemannian spacetime curvature.

[pmb_phy]

The above comment by DW is in error. The error being the association of gravity with spacetime curvature as well as the claim that gravitational acceleration is a Newtonian concept. The other error is the claim that only modern relativistic field concept is Riemman curvature. The Rieman tensor defines tidal fields, not gravitational fields.

It is highly inappropriate and illogical to refer to gravitational acceleration as belonging to a particular theory of gravity and therefore it is highly incorrect to claim that its Newtonian. Terminology doesn't, normally, belong to a theory. Only relations among quantities belong to a theory. Exceptions include things such as new terms such as 4-velocity, which do not belong to earlier theories. The term "normally" used here refers to the fact that there are times when a term is created to be used in another more accurate theory as in the 4-vector example. Another example: Suppose someone says "The body is moving at 300,000 km/s relative the inertial frame S". I do not need to ask "what theory are you speaking of?" In this case if one is speaking of speed then one is not speaking of QM. It someone says "The body is acclerating relative to the Earth." then one still does not need to know what theory one is speaking about to comprehend what the person is saying. At best one might ask if the acceleration is coordinate acceleration etc.

In general relativity the metric tensor is a said set of gravitational potentials which describe all aspects of the gravitational field including gravitational acceleration (Through its first derivatives in the Christoffel symbols) and gravitational tidal gradients, i.e. spacetime curvature (through its second derivatives in the Rieman tensor. The existance of a gravitational field (i.e. gravity) is dictated by the non-vanishing of the Christoffel symbols. The existance of tidal gradiants is dictated by the non-vanishing of the Rieman tensor. In fact Einstein was opposed to the notion of the non-vanishing of the Rieman tensor being associated with the non-vanishing of the Reimann tensor. In fact he stated that explicitly (back in 1954 as I recall). The post above also fails to note that there are tidal forces (non-vanishing Rieman tensor) in Newtonian theory as well.

Einstein's definitions are quite different than what one might be lead to believe as phrased above. Nobody has ever proved Einstein wrong to date and nobody has ever proved that defining the term gravity/gravitational field such that gravity=curvature is an inherently better view than Einstein's. In fact GRists in the relativity literature speak of gravitational fields in regions of spacetime even in those cases where they state that the Rieman tensor vanishes in those regions. Note: Einstein's definitions are not always adhered to in GR or SR. More so in GR since many, but not all, people do associate gravity with curvature. But it is not quite right to do so.

For details please see

"Eintein's gravitational field" - http://xxx.lanl.gov/abs/physics/0308039

See also the notes from the seminar given at MIT by John Stachel (GR expert/historian)
arcturus.mit.edu/8.224/Seminars/SemReptWk3.pdf

See also Einstein's landmark 1916 paper on GR.

Pete

[turin]

pmb_phy,
Why don't you reserve your unwarranted personal attacks for PM?DW (aka davy waite) makes his usual mistake here ...
...
... as well as making his usual mistake of ...
...
waite also makes his usual mistake here claiming that ...
...
(waite's favorite buzz word when he doesn't like something).
...
waite also fails to note that ...
...
Is this really productive. First of all, I don't think DW said anything too terribly wrong. Secondly, even if he did, why do you feel so compelled to emphasize it? Do you feel that the things you have to say are inadequate unless you compare them to someone else's mistakes.




... mistake ... of associating gravity with spacetime curvature ...
...
The Rieman tensor defines tidal fields, not gravitational fields.So, tidal fields are not gravitational?




...
Terminology doesn't, normally, belong to a theory.
...I beg to differ. But, I don't see this as the issue anyway.




The existance of a gravitational field (i.e. gravity) is dictated by the non-vanishing of the Christoffel symbols.Not in the Newtonian theory. The fact of the matter is that you have to establish in the context whether "gravitational field" is being used as the zeroth, first, or second order field (i.e. potential, acceleration, or tidal force in the Newtonian theory, or metric tensor, Christoffel symbol, or Riemann tensor in GR).




... Einstein was opposed to the notion of the non-vanishing of the Rieman tensor being associated with the non-vanishing of the Reimann tensor.Is this in the publication Relativity? I must have missed that.




waite also fails to note that there are tidal forces (non-vanishing Rieman tensor) in Newtonian theory as well.So? I failed to note that it's Thursday (until now). What does that have to do with it?




In fact GRists in the relativity literature speak of gravitational fields in regions of spacetime even in those cases where they state that the Rieman tensor vanishes in those regions.I suppose you mean the Riemann curvature scalar (or possibly the Ricci tensor)? Otherwise, I don't recall ever running into that notion.

[russ_watters]

russ: If "relative" would do the job, why would Steve Carlip be so sure? Go read his post on s.p.r. Dunno, maybe its just beyond my understanding (quite possible) but I still don't see why energy should have to be absolute in GR.

[turin]

Actually that is the problem. Gravity alone has no real stress-energy tensor. The stress-energy tensor for a reageon with no matter but gravitation is zero.Possibly you misunderstood me (or, just as likely, I am confused). If there is absolutely zero stress-energy, that is, if the stress-energy density tensor is defined to be zero at entirely every point in space-time, then can there still be curvature according to GR? I guess the cosmological constant is still an open issue, but my question is more of just a mathematical nature, so, assuming Λ = 0. (I don't count black holes, because they have singularities, so the stress-energy density would not be defined to be zero everywhere since it is not defined at all at the singularity [AFAIK].)

[Russell E. Rierson]

Here is an interesting FAQ discussing energy conservation in GR:

http://www.weburbia.demon.co.uk/physics/energy_gr.html

[pmb_phy]

pmb_phy,
Why don't you reserve your unwarranted personal attacks for PM? <snip>

DW chose not to recieve PM messages. He does that everytime in every new incantation of a new handle. That was a post to correct errors, not to simply criticise a person. Also, please distinguish the difference between a personal attack and explaining errors posted by a particular person. However, to preserve the peace I've rephrased the post.

That said - feel free to PM me with any more criticism. I'll be glad to dicsuss your objections there, unless there is a reason you choose to mention this stuff here? Were there points that you wanted others to know? I have chosen the abilit to recieve PMs.

I beg to differ. But, I don't see this as the issue anyway.

In the inertial fame S there is a particle moving whose speed is 3km/s. What theory does that refer to?

At best, in this particular case, there is a distinction between acceleration (aka 3-acceleration, spatial acceleration) and 4-acceleration. There is no reason to call acceleration "newtonian acceleration". It definitely gives the false impression that one is no longer speaking about GR which is quite obviously wrong.

Not in the Newtonian theory.

This thread is about general relativity. There is no need to distinguish what order you're speaking of in oder to speak of the gravitational field. One quantity dictates gravitational acceleration and thus the gravitational field and one dictates tidal acceleration (I assume you understand the difference). These are two different things. An analogy is a static electric field (magnetic field = 0). If the electric potential is Phi then the E field is E = - grad Phi. I can take another derivative but the resulting quantity is not refered to the E-field but the non-vanishing of this second derivative indicates gradients in the electric field. Same in gravity.

Consider the analogous example in Newtonian physics: In Newtonian physics the gravtational field is given by the gradient of the gravitional potential and is a 3-vector. It describes the acceleration of a free particle at that particular location. Tidal forces are given by the tidal force 3-tensor. It describes the relative acceleration of two nearby free particles. The 3-tensor can vanish without the vanishing on the 3-vector. One should never be confused with the other.

A tidal gravitational field is simply a gravitational field with tidal gradients present. The quantity which dictates the presence of the g-field is, however, the Christoffel symbols. Or as Einstein explained it in a letter to Max Von Laue

... what characterizes the existence of a gravitational field from the
empirical standpoint is the non-vanishing of the components of the affine
connection], not the non-vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.


Is this in the publication Relativity? I must have missed that.

Yes, it's in his publication. i.e. Einstein wrote

... the quantities gab are to be regarded from the physical standpoint as the quantities which describe the gravitational field in relation to the chosen system of reference.
...
If the <Christoffel symbols> vanish, then the point moves uniformly in a straight line. These quantities therefore condition the deviation of the motion from uniformity. They are the components of the gravitational field.


I suppose you mean the Riemann curvature scalar (or possibly the Ricci tensor)? Otherwise, I don't recall ever running into that notion.

Nosiree. Had I meant curvature scalar I would have said curvature scalar. I said Riemann tensor. Some simple examples from the literature are a uniform gravitational field, the field of a vacuum domain wall and the field of an straight cosmic string. The Riemann tensor vanishes everywhere outside the matter which generates those fields.

If you'd like an example, and you have access to the physics literature, then see

Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173

Relativistic solutions to the falling body in a uniform gravitational field, Carl G. Adler, Robert W. Brehme, Am. J. Phys. 59 (3), March 1991

Gravitational field of vacuum domain walls and strings, Alexander Vilenkin, Phys. Rev. D, Vol 23(4), (1981), page 852-857

Cosmic strings: Gravitation without local curvature, T. M. Helliwell, D. A. Konkowski, Am. J. Phys. 55(5), May 1987, page 401-407

Pete

[pmb_phy]

If there is absolutely zero stress-energy, that is, if the stress-energy density tensor is defined to be zero at entirely every point in space-time, then can there still be curvature according to GR?

Yep. A good example is a gravitational wave. The EM analogy is an EM wave with a vanishing 4-current. So long as you don't care where the wave comes from that the wave is a solution of Einstein's equation in empty spacetime.

[DW]

The above comment by DW is in error.

I am not in error and you should know that by now. I and other physicists have explained to you why you are wrong too many times already. Get over it.

[pmb_phy]

I am not in error and you should know that by now. I and other physicists have explained to you why you are wrong too many times already. Get over it.

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. The most well know of those errors is your claim that a uniform g-field has spacetime curvature. I even I explained your errors to you and you ignored it. Yet you still, to this very day, incorrectly think a uniform gravitational field has spacetime curvarture.

Regarding "others". I'm a physicist too. Being a physicist can never be considered proof that something you claim is correct or incorrect. Arguements such as "so and so agrees with me therefore I'm right" are totally illogical. Einstein was certainly aware of such illogic. Recall a letter Einstein wrote to Jost Winteler [8 July 1901]

There is no exageration in what you said about German Professors. I
have got to know another sad specimen of this kind -- one of the
foremost physicists of Germany. To two pertinent objections which I
raised against one of his theories and which demonstrate a direct
defect in his conclusions, he responds by pointing out that another
(infallible) colleague of his shares his opinion. I'll soon make it
hot for the man with a skillful publication. Authority gone to one's
head is the greatest enemy of truth.

Also, you've chosen a small group consisting of a few people who agree with you from internet forums and newsgroups and you ignore everyone who disagrees with you, even Einstein and well know physicists/GR experts such as John Stachel. I've even pointed you to the GR literature which confirms the definitions and views of what I told you and yet you ignore them.

On any point of calculation, such as the curvature of a uniform g-field, it makes zero difference of who agrees with you. 1 + 1 = 2 regardless of who agrees or disagrees with it (and don't employ your incorrect arguement regarding base and claim that 1+2 is not always 2. Numbers are always in base 10 when the subscript on a number is omitted it or it is explicitly stated).

If, on the other hand, I were to forget logic and reason and go blindly into GR and other areas of physics and simply parrot what I see and hear from others then I choose to pay attention to physicists with a good reputation and that are well known such as Einstein and the well known GR expert/historian John Stachel.

[Sammywu]

Hi, Pete,

How are you? Just one question.

You mentioned, "In fact Einstein was opposed to the notion of the non-vanishing of the Rieman tensor being associated with the non-vanishing of the Reimann tensor".

What is the difference between Rieman tensor and Reimann tensor?

Thanks

[turin]

In the inertial fame S there is a particle moving whose speed is 3km/s. What theory does that refer to?I would say (perhaps demonstrating my own ignorance) that this does not belong (very well) to QM. I don't know too many of the more advanced versions of QFT, QCD and so on. But, actually, my point was more along the lines that the terminology can mean one thing in one theory or formalism, and then something just different enough to cause great confusion in another. Inertial, to Newton and Galileo, (AFAIK) meant uniform motion wrt the "fixed stars." Does it mean the same thing in GR? (I'm not asking rhetorically; I sincerely would like an answer.)




At best, in this particular case, there is a distinction between acceleration (aka 3-acceleration, spatial acceleration) and 4-acceleration.
There is no reason to call acceleration "newtonian acceleration".I think there is reason, somewhat related to the difference between 3- vs. 4- acceleration. Newton considered acceleration wrt the "fixed stars," did he not? I was under the impression that part of GR is an abandonment of this picture of acceleration. In other words, we have this notion of inertial frame in SR. What mechanism determines such a notion in GR?




There is no need to distinguish what order you're speaking of in oder to speak of the gravitational field.
...
An analogy is a static electric field (magnetic field = 0).I don't think E&M is a good analogy. The Coulomb force is real, the gravitational force is inertial. Maxwell's equations derive from a 1st rank tensor source whereas Einstein's equation derives from a 2nd rank tensor source. My point is that, if you mean the acceleration when you talk about the gravitational field, then it can be transformed away completely, but the electric field cannot be transformed away completely (AFAIK). I suppose I should make sure: given an electrostatic field in on frame, is there any frame to which you could transform that would have absolutely no electric field?




In Newtonian physics the gravtational field is given by the gradient of the gravitional potential and is a 3-vector.
...
Tidal forces are given by the tidal force 3-tensor.
...
One should never be confused with the other.I agree that one should never be confused with the other. But I also think that not specifying which aspect of the gravitational field will lead to a confusion. I you personally want to use "tidal gravitational" to refer to the second order effect and simply "gravitational" to refer to the first order effect, then I am not trying to say you're wrong, but I am used to "gravitational field" meaning the second order field in GR and the first order field in Newtonian gravity.




Had I meant curvature scalar I would have said curvature scalar. I said Riemann tensor.Well, it has been pounded into my head that a scalar is a tensor, so I just wanted to clarify that you did mean the 4th rank tensor as opposed to the 0th rank tensor (because they are both called "Riemann ..."). In that case, I am apparently confused on some point of the Riemann tensor. I think this is most pronounced in my reading of
S. Chandrasekhar. On the 'Derivation' of Einstein's Field Equations. Am. J. Phys., 40, 224 (1972).




Some simple examples from the literature are a uniform gravitational field, the field of a vacuum domain wall and the field of an straight cosmic string. The Riemann tensor vanishes everywhere outside the matter which generates those fields.I don't know anything about the last two examples you gave, so I don't have a comment about them. The first example doesn't seem physical to me. Are you speaking of the approximation (of a frame that is small enough to ignore tidal effects)? I just thought that, even though it may seem slightly picky, it is an important distinction in this context between approximately vanishing and exactly vanishing. I suppose there is the possiblity of a rocket ship accelerating through space at a uniform acceleration, but I thought that even in this case, the acceleration must have some variation (it must be greater at the boosters than it is at the nose of the rocket). I think this is called Kruskal space? Is this model oversimplified?




If you'd like an example, and you have access to the physics literature, then see
...Well, I have read the first two sources in your list, but the other two I have not. Perhaps I should read them again; I think I have copies somewhere.

[pmb_phy]

I think there is reason, somewhat related to the difference between 3- vs. 4- acceleration. Newton considered acceleration wrt the "fixed stars," did he not? I was under the impression that part of GR is an abandonment of this picture of acceleration. In other words, we have this notion of inertial frame in SR. What mechanism determines such a notion in GR?

Newton considered inertial frames to be referenced with reference to the "fixed stars." Those were considered "special" frames. Acceleration was then with respect to those frames. Einstein argued that there are no "special" frames. There are either frames with no gravitational field or those with gravitational fields. To Einstein, an accelerating frame of reference is indistinguishable to a uniform gravitational field. The presence of such a field, at least according to Einstein, is seen through the inertial acceleration of particles (aka acceleration of particles subject to no other forces except gravity). That acceleration is spatial acceleration. Whenever you see Einstein speaking of gravitational acceleration that is what he's refering to.

I don't think E&M is a good analogy. The Coulomb force is real, the gravitational force is inertial. Maxwell's equations derive from a 1st rank tensor source whereas Einstein's equation derives from a 2nd rank tensor source. My point is that, if you mean the acceleration when you talk about the gravitational field, then it can be transformed away completely, but the electric field cannot be transformed away completely (AFAIK). I suppose I should make sure: given an electrostatic field in on frame, is there any frame to which you could transform that would have absolutely no electric field?

There are instances where the electric field can be transformed away, yes. A good example is that of a neutral current carrying wire. In the rest frame of the wire there is no E-field. If you move relative to the wire there is an electric field.

However that is an entirely different point than that I was making. I was speaking of that mathematical quantity which is called the "gravitational field" and that criteria which dictates the presence of a g-field.

I you personally want to use "tidal gravitational" to refer to the second order effect and simply "gravitational" to refer to the first order effect,..

Bravo! Me too. :-)

Well, it has been pounded into my head that a scalar is a tensor, ..

Yes. It is. you mentioned the curvature scalar. When people use that term they almost always mean the Rici scalar. But I was speaking of the Riemann tensor.

...so I just wanted to clarify that you did mean the 4th rank tensor as opposed to the 0th rank tensor (because they are both called "Riemann ...").

Yes.

In that case, I am apparently confused on some point of the Riemann tensor. I think this is most pronounced in my reading of
S. Chandrasekhar. On the 'Derivation' of Einstein's Field Equations. Am. J. Phys., 40, 224 (1972).

In that article Chandrasekhar is using the term "gravitational field" to refer to spacetime curvature. He uses a different criteria than Einstein. However I believe that he uses a poor arguement in his reasoning. It's been a while since I read that but I do recall being irritated at something he wrote which I thought was very wrong. I'll dig it out and get back to you on this point.

I don't know anything about the last two examples you gave, so I don't have a comment about them. The first example doesn't seem physical to me. Are you speaking of the approximation (of a frame that is small enough to ignore tidal effects)?

If you mean the uniform g-field then you're probably not familiar with the following situation
http://www.geocities.com/physics_world/gr/grav_cavity.htm

In the weak field limit (i.e. ignore pressure as source of gravity) the Riemann tensor vanishes. This is different then Newtonian gravity though. However there is an extremly small deviation of the g-field from from being perfectly uniform. Such a deviation can't be detected by modern instrumentation. It's akin to seeing a perfectly flat polished metalic floor and then looking at the surface with an electron microscope. There will be deviations. But we still call the floor flat.

By the way, varying acceleration of a rocket is unrelated to spacetime curvature. There is no possible way to transform from a flat spacetime to a curved spacetime. But you probably know that.

Pete

[pmb_phy]

Hi Sammy

How've you been?

What is the difference between Rieman tensor and Reimann tensor?

My poor spelling. :biggrin:

[turin]

Newton considered inertial frames to be referenced with reference to the "fixed stars." Those were considered "special" frames. Acceleration was then with respect to those frames. Einstein argued that there are no "special" frames. There are either frames with no gravitational field or those with gravitational fields. To Einstein, an accelerating frame of reference is indistinguishable to a uniform gravitational field.Please forgive my stubborn confusion. I would like to appreciate your point, but I am still not seeing it. I think that this is a rather important distinction between Newtonian and GR acceleration. If a rocket is travelling at 0.99c, and it accelerates at 1 g, then I believe Newton would drastically disagree with Einstein on what the speed of the rocket would be some time later. Though this is not directly/obviously related to the original issue, in my mind I extend this discrepancy to an object falling into a Neutron star. It seems to me that the 4-acceleration could be transformed away in geodesic coordinates, however, Newton's 3-acceleration would act rather strangely in the sense of such a transformation.




There are instances where the electric field can be transformed away, yes. A good example is that of a neutral current carrying wire. In the rest frame of the wire there is no E-field. If you move relative to the wire there is an electric field.
... :redface: Whoops. Yes, I agree....
However that is an entirely different point than that I was making.Yeah, me too. Let me try again:
If you have 1 C of charge pulled by -1 C of charge, then the 1 C of charge will, of course, accelerate towards the -1 C of charge. If you transform to a coordinate system that maintains the 1 C of charge at the origin, then the 1 C of charge will see a uniform inertial force field (I think), which it feels. This inertial force field will explain to the 1 C charge why the -1 C charge has a greater acceleration than just that due to the Coulomb force. The main point that I intended to bring up was that, in this case, the Coulomb force is not transfromed away, whereas, in the case of gravity, the gravitational force (first order) is transformed away in the analogous treatment.




In that article Chandrasekhar ... uses a poor arguement in his reasoning.
...
... I do recall being irritated at something he wrote which I thought was very wrong.Excellent! I thought it was just me. I would definitely like to hear what you have to say about it. It draws an analogy from Newton's Universal Law of Gravitation. More specifically:

U = 0 -> Rijkl = 0

This he calls the vanishing of the gravitational field. I don't like this.

He also says that the ability to identically transform the Christoffel symbols away has Rijkl = 0 as the necessary and sufficient condition. I am not comfortable enough with my math to agree with this rigorously, but this is one thing from his paper that I do find reasonable.




... there is an extremly small deviation of the g-field from from being perfectly uniform. Such a deviation can't be detected by modern instrumentation. It's akin to seeing a perfectly flat polished metalic floor and then looking at the surface with an electron microscope. There will be deviations. But we still call the floor flat.I didn't mean a deviation in the sense of a bunch of microscopic bumps; I meant deviation in the sense of a smooth, gross, macroscopic variation (in this case monotonic).




By the way, varying acceleration of a rocket is unrelated to spacetime curvature. There is no possible way to transform from a flat spacetime to a curved spacetime. But you probably know that.Whoops. Did I say the space would be curved? I guess I kind of did. Yeah, I'll go ahead and take that back, now. I knew I was missing something there.

[pmb_phy]

Please forgive my stubborn confusion. I would like to appreciate your point, but I am still not seeing it. I think that this is a rather important distinction between Newtonian and GR acceleration. If a rocket is travelling at 0.99c, and it accelerates at 1 g, then I believe Newton would drastically disagree with Einstein on what the speed of the rocket would be some time later.

What you've just said is that the rocket accelerates at 1 g. Einstein would say that's its impossible for the rocket to do that for long and Newton would have no problem.

It seems to me that the 4-acceleration could be transformed away in geodesic coordinates, however, Newton's 3-acceleration would act rather strangely in the sense of such a transformation.

The 4-acceleration of a rocket with its engines turned on cannot vanish in any coordinate system. The 4-acceleration of the rocket in free fall is zero in all coordinate systems.

If you transform to a coordinate system that maintains the 1 C of charge at the origin, then the 1 C of charge will see a uniform inertial force field (I think), which it feels. This inertial force field will explain to the 1 C charge why the -1 C charge has a greater acceleration than just that due to the Coulomb force.

Sure. Given a charged particle in an E field transform to the field in which the particle is at rest. Place another particle of identical charge next to it but of different mass then it will acclerate in this new frame.


Excellent! I thought it was just me. I would definitely like to hear what you have to say about it. It draws an analogy from Newton's Universal Law of Gravitation. More specifically:

U = 0 -> Rijkl = 0

No. That is incorrect. The correct analogy is del2 U -> Riemann

I did a write up on this. See
www.geocities.com/physics_world

Click on "On the concept of mass in relativity" and see the section on passive gravitational mass.

This he calls the vanishing of the gravitational field. I don't like this.

You're a smart person then. :-)

He also says that the ability to identically transform the Christoffel symbols away has Rijkl = 0 as the necessary and sufficient condition. I am not comfortable enough with my math to agree with this rigorously, but this is one thing from his paper that I do find reasonable.

He wasn't that clear here. What he's refering to is that fact that if you can transform to a frame in which the Christoffel symbols vanish identically throughout a in a finite region of spacetime then the Riemann tensor vanishes in that region. He calls it "transforming to a coordinate system" but he means throughout a finite region in spacetime, not simply a point in spacetime. A poor description in my opininion.

Pete

[turin]

What you've just said is that the rocket accelerates at 1 g. Einstein would say that's its impossible for the rocket to do that for long and Newton would have no problem.I was under the impression that Einstein would consider the g as the magnitude of the 4-acceleration, and so he would see no problem with the situation. I was under the impression that Newton would consider the g as the magnitude of the 3-acceleration, and, since (I don't think that) Newton considered c as a limiting velocity, he would see no problem with the situation either. So, it just seems to me that an indication of Newtonian vs. GR acceleration is a non-trivial issue.




The 4-acceleration of a rocket with its engines turned on cannot vanish in any coordinate system. The 4-acceleration of the rocket in free fall is zero in all coordinate systems.I'll have to think about that one.




No. That is incorrect. The correct analogy is del2 U -> RiemannI think I agree with this. I think Chandrasekhar generalizes del2 U -> Ricci instead, but I understand that Ricci and Riemann tensors are closely related.




What he's refering to is that fact that if you can transform to a frame in which the Christoffel symbols vanish identically throughout a in a finite region of spacetime then the Riemann tensor vanishes in that region. He calls it "transforming to a coordinate system" but he means throughout a finite region in spacetime, not simply a point in spacetime. A poor description in my opininion.I guess this is a point on which I don't see the problem whereas you do. I don't remember his exact wording, but I did understand him to mean a finite region. I understood this from the stipulation of "identically." At any rate, what I am still a little confused about is how the vanishing of the Riemann tensor necessarily and sufficiently dictates the ability to transform away the Christoffel symbols.

[selfAdjoint]

Ricci is the contraction of the Riemann on its contravariant with one of its covariant indices. R_{\mu\nu} = R^{\lambda}_{\lambda\mu\nu} . Einstein convention.

So if R_{\mu\nu} = 0 then R^0_{0 \mu\nu} - R^1_{1 \mu\nu} - R^2_{2 \mu\nu} - R^3_{3 \mu\nu} = 0 which could happen if all the R^{\lambda}_{\lambda\mu\nu} = 0 or if R^0_{0\mu\nu} = R^l_{l \mu\nu}, l\in {1,2,3} , in which case the "column" of the Riemann tensor is null, in the relativistic sense.

[pmb_phy]

I was under the impression that Einstein would consider the g as the magnitude of the 4-acceleration, and so he would see no problem with the situation.

Sorry. I didn't know that you were refering to the magnitude of the 4-acceleration. I thought you were speaking of the magnitude of the 3-acceleration.

I'll have to think about that one.

This is one of the most important points to understand in relativity.

I think I agree with this. I think Chandrasekhar generalizes del2 U -> Ricci instead, but I understand that Ricci and Riemann tensors are closely related.

I made an error there. The true analogy is given in terms of the Newtonian tidal force 3-tensor tij. That 3-tensor is defined here

http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm

As you can seem, the Laplacian of Phi is the contraction of the Newtonian tidal force tensor. A GR/Newtonian analogy is between the tij and Ruvab[/sup]. Another analogy is the Laplacian and Eintein's tensor, i.e. del2U <=> G[sub]uv

At any rate, what I am still a little confused about is how the vanishing of the Riemann tensor necessarily and sufficiently dictates the ability to transform away the Christoffel symbols.

That's one those math things, there is some theorem which says this, i.e. its something that has to be proven, i.e. its a theorem.

Pete

[DW]

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.

[pmb_phy]

Mr. Moderator

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

Thanks

Pete

[pmb_phy]

<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!

[DW]

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.

[DW]

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.

[DW]

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.

[Janus]

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.

[pmb_phy]

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.

[enigma]

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.

[turin]

This was regarding the statement about the coordinate independence of tensors: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.

[DW]

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.

[turin]

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.

[Phobos]

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.

[blue_sky]

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?

[pervect]

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"


C_{abcd} u^{b}u^{d}


"Electrogravitic"

R_{abcd} u^{b}u^{d}


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.

[pervect]

Steve Carlip, on sci.physics.research has asserted (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=c9tajo%24d6h%242%40woodrow.ucdavis.edu) 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).

[Haelfix]

"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 dependance 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.

[pervect]

"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 dependance 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.

[pmb_phy]

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 existance 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

[pervect]

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 existance 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.

[pmb_phy]

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, refering 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 refering 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

[selfAdjoint]

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.

[pmb_phy]

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

[marcus]

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%24d6h%242%40woodrow.ucdavis.edu

[marcus]

Pete, I have an idea for you

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

look at post #15

here is the direct link to it:
http://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!

[pervect]

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 refering 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.

[pmb_phy]

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

[pmb_phy]

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

[pmb_phy]

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 seperate 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}+m h_{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 seperate 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

[Haelfix]

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.

[DW]

Steve Carlip is 100% right in this arguement that the Newtonian potential concept you are arguing over breaks down and is not general relativistic at all. Also relativistic mass as I have already proven is not the same thing as inertial mass. In fact it has no place in modern relativity.

[pervect]

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


If you can find it in MTW or Wald, I'll conceed the point. It's certainly not in the index of either one of them (of course MTW's index **** ###). I really don't see the point of giving the metric coefficients Yet Another Name, though. It's easy enough for me to remember that when you say "relativistic mass" I think "energy", and when you say "gravitational force" I think "Christofel symbol", but I'm starting to build up quite a dictionary here :-(.


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.


The issue we are talking about, energy, is trivial in PPN, but much less so in the general theory. Unforutnately PPN theory doesn't illuminate much of the general problem.


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?


My view is that when we talk about the energy-momentum 4 vector of a particle, we are talking about the energy relative to some specific observer and some specific coordinate system. We need to define the coordinate system and the observer to measure the energy in GR, just as we need to define the "frame" we are talking about when we measure the energy of a body in Newtonian physics.

I also believe that given an energy-momentum 4-vector, we have an invariant mass^2 that's equal to g_{uv} E^u E^v -- of course we do to know the metric coefficients as well as the energy-momentum 4-vector to compute the mass

I'll leave to you to do the translation into your own terminology, I hope my meaning is clear.

The other point I want to make is that issues do arise when attempting to parallel transport the energy-momentum 4-vectors in order to do some sort of intergal to get "total energy". So having a definition of the energy-momentum 4 vector at a point isn't all that's needed to be able to define the energy of a system. For many applications, if the region is small enough, the parallel transport issue can be ignored, but this isn't always true.

[pmb_phy]

If you can find it in MTW or Wald, I'll conceed the point.
Okay, but that seems to be a bit of a narrow view. The GR literature uses the same terminology. I'm just letting you know what's out there. But I found where I saw it in MTW. Its on page 434 where they are quoting Hilbert. I've e-mailed someone to find it in Wald. I'll let you know the result.

To see where Einstein used it in the most notable place, its in his Nobel lecture. See - http://www.nobel.se/physics/laureates/1921/einstein-lecture.pdf

It's certainly not in the index of either one of them (of course MTW's index **** ###).

Indexes don't always list all terms used in a text

I really don't see the point of giving the metric coefficients Yet Another Name, though.

John Stachel wrote a paper which touched on this point in How Einstein Discovered General Relativity: A Historical Tale with some Contemporary Morals. Stachel is a famous GRist and GR historian. He points out

The main difficulty at this stage was to grasp the dual nature of the metric tensor: it is both the mathematical object which represents the space-time structure (chronogeometry) and the set of 'potentials,' from which represents the gravitational field (Christoffel symbols) and the tidal 'forces' (Riemann tensor) may be derived.

I'm doing something bad right now, i.e. I'm sitting and typing against docs orders so I can't get into all the details of that paper and the context. If you want more then I'll get to it at a later date.

But there is the question of "Why would I want to know that?" wherein I'd respond "To understand what you're reading when you encounter it." For example. See - http://arcturus.mit.edu/8.962/notes/gr6.pdf

This was written by Edmund Bertschinger, MIT cosmologist (GR prof). What do you think he means when he says

Although the gravitational potentials represent physical metric perturbations,...

What are these "gravitational potentials" that he refers to?


It's easy enough for me to remember that when you say "relativistic mass" I think "energy", and when you say "gravitational force" I think "Christofel symbol", but I'm starting to build up quite a dictionary here :-(.

That's a good thing. :smile:

If you have MTW then see what they say about the stress-energy pseudo-tensor of the gravitational field. It's in MTW page 465 Eq. (20.18).

My view is that when we talk about the energy-momentum 4 vector of a particle, we are talking about the energy relative to some specific observer and some specific coordinate system.

That is inaccurate. The 4-vector itself has a geometric signifigance which has nothing to do with coordinates. When you wish to choose a coordinate system then you've chosen a basis and a particular value of the energy. Once you do that, my question remains - Is the rest energy of the particle part of that energy?

The other point I want to make is that issues do arise when attempting to parallel transport the energy-momentum 4-vectors in order to do some sort of intergal to get "total energy".

I have no idea what you're talking about. What is it that you're adding? Total energy of what? I was refering to the energy of a single particle in a G-field.

So having a definition of the energy-momentum 4 vector at a point isn't all that's needed to be able to define the energy of a system.

Who was speaking of the energy of a system? I was speaking of the energy of a particle in a g-field. Those are two different topics.

Pete

[pervect]

When you wish to choose a coordinate system then you've chosen a basis and a particular value of the energy. Once you do that, my question remains - Is the rest energy of the particle part of that energy?


As far as "rest energy" goes, do you mean something other than the invariant associated with the 4-vector? We both agree that the energy-momentum 4-vector exists, and that it has an associated invariant. It seems to me that that should be sufficient.


Who was speaking of the energy of a system? I was speaking of the energy of a particle in a g-field. Those are two different topics.


Most of my remarks in this thread have been addressed to the issue of energy conservation in GR. This requires that one consider the notion of the nergy of a system as well as the energy of a particle.

[pmb_phy]

As far as "rest energy" goes, do you mean something other than the invariant associated with the 4-vector?

I don't mean something other. I'm refering to rest energy, i.e. the product of a particle's proper mass with c2.

We both agree that the energy-momentum 4-vector exists, and that it has an associated invariant. It seems to me that that should be sufficient.

For what? The topic here is energy in GR, correct? There are different subtopics on that topic. One subtopic is P0 = energy of a particle in a g-field. Another is the mass-energy that creates the g-field, i.e, T00, and then there is the self energy of the g-field which is represented be a pseudo-tensor (which I'm not all that familiar with).

Most of my remarks in this thread have been addressed to the issue of energy conservation in GR. This requires that one consider the notion of the nergy of a system as well as the energy of a particle.So do you think that energy is not conserved?

By the way, I just looked at the index to Wald. It does not contain the term "invariant." However I don't think that Wald considered it to be useless in GR.

Pete

[pmb_phy]

Pete, I have an idea for you

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

look at post #15

here is the direct link to it:
http://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!Okay. I've changed my mind. Since its a moderated forum I'll take a look. But Carlip and I have discussed something similar before. It turned into a discussion of semantics. But he obviously knows GR better than most of us so I recommend that people pay close attention to what he says.

Pete

[selfAdjoint]

Good show, Pete. We'll be paying attention for sure.

[pmb_phy]

Good show, Pete. We'll be paying attention for sure.
Well, I didn't want to appear closed minded. This is a topic that I'm highly interested in. One problem in addressing it is that energy itself has never been a well-defined quantity so when one says that its not well-defined in GR I'm not really surprised. But I do want to learn more about these problems and the meaning of the stress-energy-momemtum pseudo-tensor of the gravitational field.

Pete

[kurious]

If energy is not conserved in GR but it is conserved in QM this would make it difficult to see how QM and GR can be reconciled in one theory.Energy is conserved in every other branch of physics so why should GR be the exception? Can't the energy of microwave background photons for example be absorbed by something else in the universe? This has got to be more likely than a lack of energy conservation.
And if one looks at the entropy of the universe as a whole,are we to believe that
heat energy is lost from the universe? This would imply that the second law of thermodynamics can be violated or that the energy passes into another universe.
There is no proof for either of these.

[pervect]

For what? The topic here is energy in GR, correct? There are different subtopics on that topic. One subtopic is P0 = energy of a particle in a g-field. Another is the mass-energy that creates the g-field, i.e, T00, and then there is the self energy of the g-field which is represented be a pseudo-tensor (which I'm not all that familiar with).
So do you think that energy is not conserved?


OK, we've discussed most of the other subtopics, but we haven't discussed how the stress-energy tensor arises from the energy-momentum 4 vector in this thread.

I'm not sure if we need to talk about the stress energy tensor or not. I would describe it informally as the density per unit volume of the energy-momentum 4 vector. I could make this description more formal if there is a point to doing so, but I won't bother unless there is some interest and/or flames about my informal description. Well, I will mention that one can measure volume with both a three-form and with the dual (Hodges dual) of a three form, which is a formal way of saying that any volume element defines a unique perpendicular time-like vector. Hopefully that terse description may be useful in communicating my meaning.

The fact that the self energy of the field is represented by a pseudo-tensor and not a real tensor means that the notion of self-energy of a gravitational field isn't a geometric object. Different observers define it differently. So, when we require a pseudo tensor to measure system energy, "red flags" should be raised, because we are combining geometric objects, and something that's not geometric. Unfortuately, we can't simply ignore or drop the self-energy term and retain a conserved energy.

There's an amusing quote from MTW on this topic, I'll try and find it, after I resolve some technical difficulties with my computer - look for a short separate post, hopefully coming soon, unless I can't find it - did I mention that the index in MTW **** ###?.

In any event, defining energy with the pseudo-tensor approach is, at least in my opinion, a little bit like adding together all the physical energies, plus the well-known "Finnagle Factor" from engineering, where the "Finnagle Factor" takes on whatever value is required to make the answer come out right. It's not surprising that one can come up with a conserved quantity by adding the Finnagle number F to *any* general quantity, conserved or not, but it's not particularly physically significant to be able to do so.

Fortunately, there are a number of more physicaly motivated ways of defnining the total energy of a system in a manner which retains the desired conservation property. One way of doing this is to use the approach of Cartan moments, described in MTW. Another way of doing this, also mentioned in MTW, is to look at the behavior of an object orbiting the system "a long ways away" from the system, and to use agreement with Newtonian theory to define the energy of the system.

A more modern approach ito defining energy in GR s to focus on the behavior of the gravitational field at infinity, the behavior of the field at null infinity gives the Bondi energy, and the behavior of the field at spacelike infinity gives the ADM energy. Look at my previous posts in the thread for a reference to the pages in Wald where these are discussed. Wald discusses the relationship between the Bondi and ADM energy as well, according to Wald they eventually turn out to be closely related, though it takes a considerable amount of work to establish the relationship. I believe that the Bondi energy is also closely related to the first two definitions of energy I mentioned in the previous paragraph, but I'm not absolutely positive on this partciular point (I'm still working on undertanding Wald's discussion more fully, it's pretty technical).

The unifying thread to all of the methods for dealing with the energy problem that I"ve mentioned is the notion of the observer at infinity. Apparently this isn't quite the last word, though, based on the comments of another poster. It's a bit scary to think that there are probably more definitions of energy in GR than I've posted even in this very long post - but probably accurate.

[Garth]

In general energy is not conserved in GR - energy-momentum is. This is a result of the EEP and the carrying forward of the SR "non-preferred frame of reference" interpretation of physical observations into GR.
It is energy-momentum that is invariant over transformation between frames. Energy is frame independent, not just in the SR sense of the variation of kinetic energy between non-co-moving frames, but also because of curvature effects.

Is energy described by P^0 or P_0 ? Is it naturally a contravariant or covariant quantity?
Weinberg uses P^0 e.g. G&C, pg 182 eq. 8.2.16 as do MTW, Gravitation, e.g. pg 158 and yet it is P_0 that is the conserved quantity when the gravitational potentials are not time dependent. The covariant form P_0 is often defined as energy because of that. Yet this usage is confusing. It only works because in the asymptotic limit of flat space-time P^0 = P_0 , and it is conserved with the condition specified.
In fact energy is neither covariant nor contravariant, its value is the scalar time component of the norm of the energy-momentum vector, E^2 = P^0.P_0 , (see MTW pg 463 eq. 20.10) and this is not conserved except under special conditions i.e. flat space-time.
Might it be that only when the non-conservation of energy in GR, as against its conservation in quantum theory, is fully recognised that the problems at their interface in quantum gravity may be resolved?

Energy may be defined in GR but it is not general conserved; if it is demanded that energy should be conserved then it can not be localised, because the energy density has to be integrated over the entire field out to the flat space-time asymptotic limit.

[pmb_phy]

Is energy described by P^0 or P_0 , is it naturally a covariant or contravariant quantity?

The energy of a particle which is moving in a gravitational field is proportional to P0 while the inertial mass is proportional to P0.

Weinberg uses P^0 e.g. G&C pg 182 eq. 8.2.16 and MTW Gravitation e.g. pg 158 and yet it is P_0 that is the conserved quantity when the gravitational potentials are not time dependent.

Weinberg uses P0 there to represent the mass/energy of the body which generates the gravitational field. That M is the proportional to the time component of the 4-momentum of the gravitating body. So you can think of that as a particle. He shouldn't have called it that but such is life. I don't see what you're refering to on that page in MTW since there is no mention of what you said there.

But, as I recall, MTW do explain this elsewhere. They explain that it is P0 which is the conserved quantity when the particle is in a time independant g-field (i.e. guv do not exlicitly depend on time).

The contravariant form P_0 is often defined as energy because of that. Yet this usage is confusing. It only works because in the asymptotic limit of flat space-time P^0 = P_0 , and it is conserved with the condition specified.

That is incorrect. It works under all conditions where one would expect the energy of a particle to be conserved, i.e. in a time independant g-field.
For proof see -
http://www.geocities.com/physics_world/gr/conserved_quantities.htm

In fact energy is neither covariant nor contravariant, its value is the scalar time component of the norm of the enegy-momentum vector, E^2 = P^0.P_0 , (see MTW pg 463 eq. 20.10) and this is not conserved except under special conditions i.e. flat space-time. That is incorrect. What you've just described is not energy. Energy is not a scalar since a scalar is a tensor of rank zero and therefore is independant of the system of coordinates. You've just described proper energy, E0. Energy is P0.

Also, what you've quoted is not from MTW. That equation in MTW you just quoted is really

M = (-P^{\mu}P_{\mu})^{1/2}

where they use M to mean the proper mass of the object. You've made the mistake of confusing proper mass with energy. The correct relationship between the two is given by E0 = Mc2.

You should never expect the energy of a particle to be constant when the field is time dependant. That doesn't even hold in classical mechanics or electrodynamics.

Pete

[DW]

The energy of a particle which is moving in a gravitational field is proportional to P0 while the inertial mass is proportional to P0.

Weinberg uses P0 there to represent the mass/energy of the body which generates the gravitational field. That M is the proportional to the time component of the 4-momentum of the gravitating body. So you can think of that as a particle. He shouldn't have called it that but such is life. I don't see what you're refering to on that page in MTW since there is no mention of what you said there.

But, as I recall, MTW do explain this elsewhere. They explain that it is P0 which is the conserved quantity when the particle is in a time independant g-field (i.e. guv do not exlicitly depend on time).

That is incorrect. It works under all conditions where one would expect the energy of a particle to be conserved, i.e. in a time independant g-field.
For proof see -
http://www.geocities.com/physics_world/gr/conserved_quantities.htm
That is incorrect. What you've just described is not energy. Energy is not a scalar since a scalar is a tensor of rank zero and therefore is independant of the system of coordinates. You've just described proper energy, E0. Energy is P0.

Also, what you've quoted is not from MTW. That equation in MTW you just quoted is really

M = (-P^{\mu}P_{\mu})^{1/2}

where they use M to mean the proper mass of the object. You've made the mistake of confusing proper mass with energy. The correct relationship between the two is given by E0 = Mc2.

You should never expect the energy of a particle to be constant when the field is time dependant. That doesn't even hold in classical mechanics or electrodynamics.

Pete
Actually no, energy is most commonly defined as the time element of the contravariant momentum four-vector, but neither the time element of the contravariant or covariant expression of the vector is what is conserved in general. In general it is energy and momentum parameters that are conserved instead and only in special cases is the energy parameter equal to the energy or the time element of the covariant momentum four-vector. It is also wrong to call the mass m proper mass because it is the length of p^{\mu } according to any frame, not just the proper frame. It is also not equal to the relativistic mass M which shouldn't even be used anymore.

[Garth]

Pete: The energy of a particle which is moving in a gravitational field is proportional to P0 while the inertial mass is proportional to P0.
E0 = Mc2.
So P0 = P0c2 then? I think not; actually when the metric is symmetric
P0 =g00P0c2 and therefore E0 = Mc2 is only true in Minkowski space-time in the absence of curvature.

I don't see what you're refering to on that page in MTW since there is no mention of what you said there.
Perhaps page 462 equations 20.6, 20.7, & 20.9 would have been a better example of MTW using P0 as energy.

That is incorrect. It works under all conditions where one would expect the energy of a particle to be conserved, i.e. in a time independant g-field.
For proof see -
http://www.geocities.com/physics_world/gr/conserved_quantities.htm
Your notes are useful but (more simply) they have shown that a free falling particle's covariant four velocity Ua is invariant if the metric components are not dependent on xa, where
Ua = gabUb. Multiplying by mass gives you Pawhich is also invariant and which therefore you define as energy. However this is a very special case and, as I have said, GR is an improper energy theorem. To introduce the concept of energy conservation is very attractive but it is not consistent with the general theory, it is 'bolting on' a classical (i.e. pre-GR) principle.

I was using "scalar" in its general and not specific meaning, i.e. as a mathematical 'object' having only magnitude and not as a "scalar invariant". I agree it is important to be specific about language and I should have been more accurate in terminology.

[pmb_phy]

Pete:
So P0 = P0c2 then?

Nothing I've ever posted ever hinted at that being true. You were using the term M to mean proper mass and you then claimed that the magnitude of 4-momentum was energy. That is not the case. I said that correct relationship between energy and M is given by E0 = Mc2.

I think not; actually when the metric is symmetric
P0 =g00P0c2 and therefore E0 = Mc2 is only true in Minkowski space-time in the absence of curvature.

That is incorrect. In the first place the metric is always symmetric. P0 is related to a particle's 4-momentum through the relation

P_0 = g_{0\mu}P^{\mu}

Perhaps page 462 equations 20.6, 20.7, & 20.9 would have been a better example of MTW using P0 as energy.

Yeah, I don't know why they do that. Seems strange and doesn't make sense.

Your notes are useful but (more simply) they have shown that a free falling particle's covariant four velocity Ua is invariant if the metric components are not dependent on xa, where
Ua = gabUb. Multiplying by mass gives you Pawhich is also invariant and which therefore you define as energy. However this is a very special case and, as I have said, GR is an improper energy theorem.

I know that's what you think. You're repeating yourself since we've discussed this point several times in the past. Do you want to rehash that whole discussion we had earlier on this one point?

Note: It appears that you're using the term invariant to mean "does not change with time" when, in relativity, it is always used to mean "remains unchanged by a change in coordinates".

Pete

[pervect]

Here's a definition from Wald for the total mass in a stationary, asymptotically flat space with a vacuum at infinity. It expresses the mass in terms of the time translation symmetry, represented by the killing vector \xi , and a volume intergal.


M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV


Here Tab is the stress-energy tensor, gab is the metric tensor, na is the unit future normal to \Sigma, and \xi^b is the Killing vector representing the time translation symmetry of the static system.

This is about as simple of a definition of the concept of mass as I've seen in GR, which is why I picked it.

One can also write the intergal in terms of the Riemann tensor


M = \frac{1}{4 \pi}\int_{\Sigma} R_{ab} n^a \xi^b dV

[Garth]

In the first place the metric is always symmetric.
Pete
Of course, you are quite correct, I meant "diagonal",

Garth

[hellfire]

According to Noether's theorem, there is a procedure to determine the stress-energy tensor (as the Noether current associated to spacetime translations) of a given action. What happens if one applies this procedure to the Einstein-Hilbert action? Does the resulting energy tensor fulfill the requirements imposed by general relativity? and how far does it have a physical meaning as the energy associated to the gravitational field? I have not seen this aspect discussed in this thread, isn't it relevant? Could anyone elaborate on this?

Thanks.

[pervect]

According to Noether's theorem, there is a procedure to determine the stress-energy tensor (as the Noether current associated to spacetime translations) of a given action. What happens if one applies this procedure to the Einstein-Hilbert action? Does the resulting energy tensor fulfill the requirements imposed by general relativity? and how far does it have a physical meaning as the energy associated to the gravitational field? I have not seen this aspect discussed in this thread, isn't it relevant? Could anyone elaborate on this?

Thanks.

I may be way off base here, but if I'm interpreting my textbook properly, this tensor is known as the canonical energy-momentum tensor (Wald, pg 457). The term stress-energy tensor is reserved for Tab, your tensor gets a different name.

Apparently, when you add terms to the Einstein-Hilbert action to generate the gravity associated with electromagnetic fields, you find that the cannonical energy momentum tensor depends on the Maxwell gauge, and is not symmetric. There are apparently some other problems with it as well.

[Chronos]

There is that little covariance issue to deal with, as well.

[pmb_phy]

I may be way off base here, but if I'm interpreting my textbook properly, this tensor is known as the canonical energy-momentum tensor (Wald, pg 457). The term stress-energy tensor is reserved for Tab, your tensor gets a different name.There is a procedure to make them identical. Its not a good idea to think of them as seperate things. It is important to know the relationship though. This tensor is obtained from the Lagrangian of the field. It is a good way to obtain the stress-energy-momentum tensor when one only has the Lagrangian. This is how the stress-tensor if obtained for a cosmic string and a vacuum domain wall.

Pete

[pervect]

I was going to ask about the procedure for finding T_ab from the cannonical energy momentum tensor, but I ran across

http://bolvan.ph.utexas.edu/~vadim/Classes/2004f.homeworks/hw01.pdf

which seems to cover most of it. It's not clear how one finds the correct divergence to add to the Lagrangian to get a symmetric and gauge invaraint stress energy tensor, though, unless you happen to know the right answer already.

[gptejms]

I know too little of GR to make serious comments in such a discussion,but here's what I wish to say(tell me if I'm wrong):-In a non-inertial frame,there are pseudo forces ; these pseudo forces would do work on particles/objects present in such a frame and energy is bound not to be conserved in such a frame.But if I watch these obects from outside i.e. from an inertial frame,energy would come out to be nicely conserved.Now in Einstein's model,gravitation is equivalent to a set of non-inertial frames(EEP)--so it should be of no surprise that energy is not conserved.I read in one of the posts above that an observer in asymptotically flat regions sees energy to be conserved---well this is your observer in the inertial frame.Comments please.

Jagmeet Singh

[pmb_phy]

I know too little of GR to make serious comments in such a discussion,but here's what I wish to say(tell me if I'm wrong):-In a non-inertial frame,there are pseudo forces ; these pseudo forces would do work on particles/objects present in such a frame and energy is bound not to be conserved in such a frame.

That is incorrect. Work can be done on a particle and still leave the energy conserved. What changes is the potential energy and the potential energy. Its the sum that is constant. Whenever the components of the metric are not functions of time the energy of a particle in free-fall will remain constant.

I read in one of the posts above that an observer in asymptotically flat regions sees energy to be conserved---well this is your observer in the inertial frame.Comments please.

Jagmeet Singh
Any observer who is in a frame of reference such that the metric is not an explicit function of time will reckon the energy of a particle in free-fall to be constant, regardless of whether the is an asymptotically flat region. If we're speaking of a g-field like the Earth then no observer will be an an inertial frame unless that inertial frame is a local one. The spacetime curvature of such a field prevents one from finding an extended inertial frame of arbitrary size.

Pete

[pervect]

Asymptotic flatness is still important for defining energy in the special case Pete is talking about, the case of a particle with a time-like killing vector, i.e. a small particle "in free fall" following a geodesic in a space-time where the metric coefficients are all constant.

The covariant energy of this partical will indeed be constant, so one has a conserved quantity independent of asymptotic flatness. This is not surprising, because one has a timelike symmetry, therfore one expects a conserved energy.

The problem is that the magnitude of this conserved energy isn't well defined. Multiplying a conserved quantity by a constant gives another conserved quantity. So one winds up with a conserved quantity, but no way of scaling it.

The usual way of scaling it is to say that the metric coefficients go to +/-1 at infinity, i.e. the metric becomes Minkowskian at infinity. But without asymptotic flatness, we can't scale the metric in this way. The value of the component E0 will depend directly on the scale factor of the metric. As one sets the overall "scale factor" for the metric, (one can think of this process as picking a particular point where g_00 = 1, a condition that one can set arbitrarily), the numerical value of the conserved "energy" will change, depending on how this choice is made.

Thus, without asymptotic flatness, there is no convenient way to set the scale factor of the metric, and hence the scale factor of the conserved conjugate quantity, the energy.

So yes, one can have a conserved energy in the special case of a timelike killing vector, even without asymptotic flatness. But in general one won't have any good way of setting the scale factor for this conserved quantity unless one also has asymptotic flatness.

If one does have asymptotic flatness, it is sufficient on its own to define energy, though the details are rather technical.

[pmb_phy]

The covariant energy ..

Please define your term covariant energy.

The problem is that the magnitude of this conserved energy isn't well defined.
Since when??
Multiplying a conserved quantity by a constant gives another conserved quantity. So one winds up with a conserved quantity, but no way of scaling it.That is not quite correct. One must demand that their choice of "scaling" leads one to obtain the Newtonian values in the Newtonian limit. But what you say is the same in non-relativistic mechanice,. Or do you know a reason why I can't call W = 13*mv2 + 26*U the "energy" of a particle in non-relativistic mechanics? Its still a constant isn't it?

The usual way of scaling it is to say that the metric coefficients go to +/-1 at infinity, i.e. the metric becomes Minkowskian at infinity. But without asymptotic flatness, we can't scale the metric in this way.
One does not need asymptotic flatness since one can always go to the weak field limit by choosing coordinates such that hab << 1 at any event in spacetime.

The value of the component E0 will depend directly on the scale factor of the metric. As one sets the overall "scale factor" for the metric, (one can think of this process as picking a particular point where g_00 = 1, a condition that one can set arbitrarily), the numerical value of the conserved "energy" will change, depending on how this choice is made.
Why are you so hot and bothered about this scale factor? Do you think its a problem in Newtonian mechanics? The magnitude of energy is of no real use. All that one needs is the fact that it is constant for it to be useful. One can even add a constant since only changes are meaningful.

So yes, one can have a conserved energy in the special case of a timelike killing vector, even without asymptotic flatness. But in general one won't have any good way of setting the scale factor for this conserved quantity unless one also has asymptotic flatness.
This is clearly not the case. One does not need asymptotic flatness to use the Newtonian limit.

Pete

[gptejms]

That is incorrect. Work can be done on a particle and still leave the energy conserved. What changes is the potential energy and the potential energy. Its the sum that is constant. Whenever the components of the metric are not functions of time the energy of a particle in free-fall will remain constant.

Pete

What's potential energy in a non-inertial frame?Potential energy in a gravitational field is fine,but we are talking of non-inertial frames here.Inspite of the equivalence principle,I don't see the concept of potential energy carrying over.

[Garth]

The problem is we find the classical treatment of energy so useful and persuasive that it is hard to let it go. We do work and then expect the energy we have expended to show somewhere in the 'energy accounts', either as kinetic energy, heat (same thing on a microscopic scale) or potential energy. (Yes I know this isn't an exhaustive list) The classical treatment (and SR) balances these 'energy accounts' by conserving energy. Furthermore, SR gives mass an energy equivalent, enabling, in some cases, energy conversion to/from mass.

The problem in GR is that GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which, when space-time curvature is introduced, is different. GR is an example of one of Noether's improper energy theorems.

The desire to conserve energy in GR is where confusion may arise.

It is true in a static field, where the metric components do not depend on time, that the covariant time element of a free-falling particle's four-velocity is conserved. Therefore it is tempting to multiply it by the mass (rest-mass to some) of a particle and obtain a conserved covariant time element of four-momentum. Because it is conserved it may then be used as a definition of energy.

However, as I posted above, you find several authorities, such as Weinberg and MTW, define energy, at times, as the contra-variant time element of four-momentum. As Pete said above
Yeah, I don't know why they do that. Seems strange and doesn't make sense.
Four-momentum is naturally described by a contra-variant vector (one form), hence if energy is its (frame dependent) time element then that too is more naturally described as a contra-variant element. However as such it is not conserved.
Quite.
“GR does not conserve energy. It doesn't intend to. It conserves energy-momentum”

The problem resolves itself in asymptotic flatness, because there the covariant and contra-variant elements of a four-vector converge. Hence system energy is said to be defined only at the null infinity of asymptotic flatness.
[But note this is a little artificial as the Schwarzschild solution is embedded in Minkowski, flat, space-time, whereas real gravitational systems are embedded in a cosmological background in which asymptotic flatness does not exist]

If energy is the time element of four-momentum and if the scalar value of four-momentum is given by its norm then an alternative definition of energy (frame-dependent) may be more consistently given by:

E = (-P^{0}P_{0})^{1/2} ,

because 3-momentum may be defined by

p = (-P^{i}P_{i})^{1/2} (i = 1,2,3),

and

|P|^{2} = p^{2} + E^{2} = -g_{\nu}_{\mu}P^{\nu}P^{\mu} .

However this E will not be conserved in GR, but then it shouldn't be; “GR does not conserve energy. It doesn't intend to. It conserves energy-momentum”
But note it is in SCC.
Garth

[meteor]

My knowledge of GR is quite meagre, but if it can helps, i was reading today the TASI lectures of cosmology, and equation 25 is referred as the "Energy conservation equation"
http://arxiv.org/abs/astro-ph/0401547

[Garth]

My knowledge of GR is quite meagre, but if it can helps, i was reading today the TASI lectures of cosmology, and equation 25 is referred as the "Energy conservation equation"
http://arxiv.org/abs/astro-ph/0401547

Yes meteor, but that equation is the cosmological case energy conservation equation in the special co-moving R-W cosmological frame, in which it is really a conservation of energy-momentum. (In the cosmological case they are equal - there are no local motions, just Hubble expansion) The above discussion is in the more general case of local gravitational fields and local definitions of energy.
Garth

[juju]

Hi all,

I know very little of the math specific concepts of GR, but is it possible that the missing energies are bound up in the space/time curvature itself.

juju

[Garth]

Hi all,

I know very little of the math specific concepts of GR, but is it possible that the missing energies are bound up in the space/time curvature itself.

juju

Yes that is the normal way it is explained, the energy is absorbed by the gravitational field, but it then ceases to appear in the "accounts", it has disappeared, and will re-appear again when the process is reversed. The question is whether this is an appropriate way to deal with energy.
Garth

[juju]

The question is whether this is an appropriate way to deal with energy.
Garth

I would think this is an OK way to deal with energy.

It would be just like a battery or capacitor. You store the energy in a potential field and it is not manifest until drawn upon.

Thanx Garth.

juju

[Garth]

It would be just like a battery or capacitor. You store the energy in a potential field and it is not manifest until drawn upon.
juju
That is a good example of what I mean. As you charge a capacitor the charge on the plates increases and the electric potential between the plates. The energy does show in the electric 'potential' "accounts".
However with the gravitational field in GR this is not so clear.
The Schwarzschild solution describes a spherically symmetric and static gravitational field around a central mass at any radius from that centre which depends on that central mass alone. So long as there is spherical symmetry it does not matter how the mass is distributed. However you can redistribute the mass, keeping spherical symmetry, using up/releasing energy in the process and the exterior field will not change at all. There will not even be gravitational waves generated by such a process. The Sun could turn into a black hole and the Earth's orbit would not be affected.
So, using a 'skyhook', lift a shell around a spherical mass and prop it up high off the ground. According to GR the total energy of the system, rest mass and gravitational energy is equal to P^{0}, which is equal to M before and after the process. The energy you used has simply "disappeared"! It has been absorbed into the field somehow. The standard explanation is to say that energy cannot be localised, it is a mistake to try to locate it. However this is just an example of the fact GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which is different (generally).
Garth

[juju]

So long as there is spherical symmetry it does not matter how the mass is distributed. However you can redistribute the mass, keeping spherical symmetry, using up/releasing energy in the process and the exterior field will not change at all.

But don't the internal gravitational potentials change as a result of this redistribution. This would change the potenial energy structure internal to the mass,

I think I have read of experiments that have been done that say they have proven that this internal structure affects the effective gravitational mass of the object. I believe I read this in Scientific American but I am not sure.

juju

[Garth]

You find in the Schwarzschild solution that internally the mass M that appears in the gravitational potential is redefined and normalised so that its value is as determined by Kepler at 'infinity', i.e. determined to agree with Kepler's period of an orbit in a region approaching flatness.
Garth

[juju]

Hi Garth,

But then doesn't this changing and renormalizing of the mass M, actually prevent apriori any possible energy accounting. And isn't this a bad way of going about this.

juju

[Garth]

The key issue is that energy is not conserved in GR - energy-momentum is, which is different. There is no other way of defining the M you put in the gravitational potentials, is there?
Garth

[juju]

Hi Garth,

I think I understand what you are saying. Since the mass and the symmetry don't change, the external fields do not change. I get that. However, the internal gravitational binding energy of the particles making up the mass does change.

juju

[pervect]

However you can redistribute the mass, keeping spherical symmetry, using up/releasing energy in the process and the exterior field will not change at all. There will not even be gravitational waves generated by such a process. The Sun could turn into a black hole and the Earth's orbit would not be affected.
So, using a 'skyhook', lift a shell around a spherical mass and prop it up high off the ground. According to GR the total energy of the system, rest mass and gravitational energy is equal to P^{0}, which is equal to M before and after the process. The energy you used has simply "disappeared"! It has been absorbed into the field somehow. The standard explanation is to say that energy cannot be localised, it is a mistake to try to locate it. However this is just an example of the fact
Garth

The specific example here is flawed, I think. Neither P^{0} or P_{0} represents the total energy of a system as I have been trying to explain for some time. Neither one of them is even coordinate independent! It's true that P0 is an invariant of motion for a particle following a geodesic in a static space-time. But this doesn't mean that summing up the P0 makes a good measure of system energy.

Here's one of the definitions from Wald for the total mass in a stationary, asymptotically flat space with a vacuum at infinity, which I've located from earlier in the thread and cut & pasted. It's expressed as a volume intergal of the stress-energy tensor.


M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV


Here Tab is the stress-energy tensor, gab is the metric tensor, na is the unit future normal to the volume element, and \xi^b is the Killing vector representing the time translation symmetry of the static system, sutiably normalized so that |\xi^a \xi_a| equals unity at infinity.

Note that

T_{ab} n^a dV

is going to give the energy-momentum 4-vector P, because the stress-energy tensor is the density of 4-momentum per a vector-valued unit volume, and na dV is just that vector-valued unit of volume.

In a static space time, \xi^b is going to be a unit vector. This means that

T_{ab} n^a \xi^b dv

is going to pick out P_{0} in a coordinate system that's flat at asymptotic infinity.

However, note that

M = \int_{\Sigma} T_{ab} n^a \xi^b dV


is not the system energy! Because of the second term in the intergal, the total system energy does change when you "lift up" mass from a region with a different metric curvature g00.

To put this all in words, you have to integrate the following quantity:

twice the energy momentum P0 minus the trace of T_{ab} multiplied by g00.

to get the system energy.

You can, I think, find a similar formula in MTW, if you have MTW and not Wald I can _probably_ find it for you (but I'm not going to bother to look through that book's terrible index system unless you both have the book and are interested enough that finding it would be worthwhile).

[pervect]

Please define your term covariant energy.


That's the covariant energy component of the energy-momentum 4-vector, i.e. P0


Since when??
That is not quite correct. One must demand that their choice of "scaling" leads one to obtain the Newtonian values in the Newtonian limit.


The Newtonian limit exists when you have an asymptotically flat space-time- then you can look at the "far field". But you're trying to get energy without asymptotic flatness, so it's not clear that the Newtonian limit exists.


But what you say is the same in non-relativistic mechanice,. Or do you know a reason why I can't call W = 13*mv2 + 26*U the "energy" of a particle in non-relativistic mechanics? Its still a constant isn't it?
One does not need asymptotic flatness since one can always go to the weak field limit by choosing coordinates such that hab << 1 at any event in spacetime.


Unfortunately, the value of energy you get is going to depend on which point in space-time you choose to make this happen at, i.e. at what point in space_time is g00 going to be nearly equal to 1.

To put it another way that may (or may not) be simpler, energy is conjugate to time, but the clocks are all running at different rates. Usually we pick the clock at infinity - without asymptotic flatness, we don't have this ability.

[gptejms]

The problem in GR is that GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which, when space-time curvature is introduced, is different. GR is an example of one of Noether's improper energy theorems.

The desire to conserve energy in GR is where confusion may arise.

Garth

Thanks all of you for your inputs.I searched on the internet and found an article "Energy not conserved in GR"(by Baez,I guess) which explains that the stress-energy tensor doesen't satisfy Gauss's law if there is spacetime curvature.Is this not in contradiction with your statement above "GR conserves energy-momentum"?

Another question:- One of you(I guess pmb_phy) has said that energy is conserved in static fields. Why so?----a static field would also have accompanying spacetime curvature and Gauss's law wouldn't be satisfied.

[pmb_phy]

What's potential energy in a non-inertial frame?Potential energy in a gravitational field is fine,but we are talking of non-inertial frames here.

Recall the equivalence principle: A uniformly accelerating frame of reference is equivalent to a uniform gravitational field.

There are other versions but this one gives you a clearer idea of what I was referikng to. In such a field (1) There here is somethikng which you can clearly call a "potential" and the total energy is a constant. In the weak field limit you can express the total energy as the sum of kinetic, potential and rest energy. I've never calculated the exact value so I'm not sure what it is in the strong field case. I'll get back to that on Friday when I'm near a computer again.

In the weak field limit of a uniformly accelerating frame of reference the gravitational potential is V(z) = mgz where g = gravitational acceleration as measured at z = 0.

But in all of relativity "energy" is a frame dependant concept and as such the value changes when you change frames.

Inspite of the equivalence principle,I don't see the concept of potential energy carrying over.Why?

The problem in GR is that GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which, when space-time curvature is introduced, is different.Wrong. GR does not say this. This is simply Garth's opinion and it can be shown to be false be mere calculation. Conservation of energy-momentum means Tab;b = 0. This equation contains both the conservation of energy and the conservation of momentum. In fact in order to prove that Tab;b = 0 one starts with the conservation of energy and the conservation of momentum and one [b]derives[/sub] this expression. It simply puts two conservation theorems in one tiny package. But each, i.e. energy and momentum, are conserved independantly of each other.

That's the covariant energy component of the energy-momentum 4-vector, i.e. P0
Yipes! That is an odd way to phrase that. Energy is energy and there is little reason to qualify it with "covariant". Energy is the time component of the energy-momentum 1-form.

The Newtonian limit exists when you have an asymptotically flat space-time- then you can look at the "far field".
Sufficient but not neccesary. Consider an infinitely long straight string of uniform mass density. The potential does not go to zero at infinity, its logarithmic. One defines the zero reference at a finite distanc, r0, from the string. If you stay close to 0 then you can choose r close to 0 such that hab is as small as you like and therefore the weak field limit will apply - or even the Newtonian limit.

But you're trying to get energy without asymptotic flatness, so it's not clear that the Newtonian limit exists.Energy is always there in a static g-field. There is no requirement for the g-field to go to zero for one to be able to use the weak field limit (where, similar to the Newtonian limit, one can define a potential energy term). I can always change my reference point to any location that I like so that if I stay near it the potential is small. The weak field limit only demands that hab is small. It doesn't demand asymptotic flatness. The string is a great example of this.

Unfortunately, the value of energy you get is going to depend on which point in space-time you choose to make this happen at, i.e. at what point in space_time is g00 going to be nearly equal to 1.
Why is that unfortunate? Energy has always been a relative quantity and a quantity whose magnitude has never been of great importance. Only differences in magnitude are of any importance. And energy is always frame dependant in relativity.

To put it another way that may (or may not) be simpler, energy is conjugate to time, but the clocks are all running at different rates. Usually we pick the clock at infinity - without asymptotic flatness, we don't have this ability.I'm sorry but I don't know what your point is here.

Thanks all of you for your inputs.I searched on the internet and found an article "Energy not conserved in GR"(by Baez,I guess) which explains that the stress-energy tensor doesen't satisfy Gauss's law if there is spacetime curvature.Is this not in contradiction with your statement above "GR conserves energy-momentum"?

You're speaking about something else. That is the total energy of the closed system consisting of the source of the g-field. We're talking anout the energy of a particle moving in a g-field. These are different topics.

Another question:- One of you(I guess pmb_phy) has said that energy is conserved in static fields. Why so?----a static field would also have accompanying spacetime curvature and Gauss's law wouldn't be satisfied.First off - The presence of a gravitational field does not imply the existace of spacetime curvature. In the second place you're refering to the energy of the source of gravity and not the energy of a particle moving in a static gravitational field. In all cases where there is a static gravitational field P0 is a constant. This constant is called the "energy" of the particle.

As to why this is a constant see the derivation at
http://www.geocities.com/physics_world/gr/conserved_quantities.htm

Pete

[Garth]

The nature of the signature of the metric in SR/GR means that the energy time-component (squared) and the momentum space-component (squared) are subtracted from each other to obtain the norm (squared) of the energy-momentum.

In the particular case when energy is conserved and momentum is conserved then energy-momentum is conserved.

However in the general case both energy and momentum individually may not be conserved and yet their difference of squares, the norm squared of energy-momentum, may still be conserved.

Therefore the conservation of energy-momentum does not require the conservation of both energy and momentum. In fact in most cases of freely falling frames of reference through space-time with curvature there are no Killing vectors and individually they are not conserved.

The desire to conserve energy is a natural one but it does not sit lightly with the principles of GR.

Garth

[pmb_phy]

The nature of the signature of the metric in SR/GR means that the energy time-component (squared) and the momentum space-component (squared) are subtracted from each other to obtain the norm (squared) of the energy-momentum.
That is only true in SR. It is quite false in GR.

In the particular case when energy is conserved and momentum is conserved then energy-momentum is conserved.
Yep.

Your comments are confusing since you switch topics in a heartbeat with no mention that you're doing so. On the one hand there is the energy-momentum of the source of gravity. That is described by the energy-momentum tensor. Then there is the energy-momentum of a particle which is moving in the G-field. That is something quite different and is described by the 4-momentum.


However in the general case both energy and momentum individually may not be conserved and yet their difference of squares, the norm squared of energy-momentum, may still be conserved.


So? That only means that the proper energy of a particle is constant, e.g. not a function of proper time. That is something quite different than refering to conservation of energy. The term conservation of energy refers only to the total energy of a particle and not part part of the total energy of which proper energy is. You're refering to proper energy which is part of total energy. And even then you're speaking of SR only (inertial frame) and then seeming to claim that it somehow applies to GR. The magnitude of 4-momentum is proportional to proper energy. The square of proper energy = E^2 - (pc)^2 only when you're in an inertial frame of reference, i.e. when you're using Lorentz coordinates.

Therefore the conservation of energy-momentum does not require the conservation of both energy and momentum.

That is totally wrong. The very phrase "conservation of energy-momentum" quite literally means "conservation of inertial energy and conservation of 3-momentum".

In fact in most cases of freely falling frames of reference through space-time with curvature there are no Killing vectors and individually they are not conserved.
So what? What does that have to do with the meaning and definition of the terms we're discussing?

Pete

[gptejms]

You're speaking about something else. That is the total energy of the closed system consisting of the source of the g-field. We're talking anout the energy of a particle moving in a g-field. These are different topics.


Ok.


First off - The presence of a gravitational field does not imply the existace of spacetime curvature.


Why?



In the second place you're refering to the energy of the source of gravity and not the energy of a particle moving in a static gravitational field. In all cases where there is a static gravitational field P0 is a constant. This constant is called the "energy" of the particle.

Pete

Ok.

[pervect]

Thanks all of you for your inputs.I searched on the internet and found an article "Energy not conserved in GR"(by Baez,I guess) which explains that the stress-energy tensor doesen't satisfy Gauss's law if there is spacetime curvature.Is this not in contradiction with your statement above "GR conserves energy-momentum"?

Another question:- One of you(I guess pmb_phy) has said that energy is conserved in static fields. Why so?----a static field would also have accompanying spacetime curvature and Gauss's law wouldn't be satisfied.

I would assume that you found the sci.physics.faq about energy in GR

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

a good source of information.

What you'll need to appreciate the argument here is in the FAQ, look for the line:


In certain special cases, energy conservation works out with fewer caveats. The two main examples are static spacetimes and asymptotically flat spacetimes.


Pete's been talking abut static space-times. There is indeed a conserved quantity that can be defined for a static space-time, even one that is highly curved, one that is not asymptotically flat. The only slightly odd thing you run into is that it's hard to set the right "scale factor" if you have a static space-time without asymptotic flatness. (AFAIK it's impossible to find any generally preferred scale factor for this conserved quantity without asymptotic flatness).

The simplest cases, BTW, like the Schwarzschild black hole, are both static AND asymptotically flat. That's the simple case I've been presenting the formulas for.

Of the two concepts, asymptotic flatness is, in my opinion, by far the most useful, and you'll find that that's how Wald, for instance, approaches the problem of energy in GR. MTW takes a different route to the same path - they discuss energy in terms of pseudo-tensors (also mentioned in the FAQ) - an approach that winds up working if and only if you have asymptotically flat space-times.

The two concepts do not conflict - when they both apply, they yield the same answer.

So Garth is right when he says that energy is not always conserved in GR - you do need some additional conditions, you can't define energy for an arbitrary space-time. But the specific example he gave to illustrate this was flawed - rather than going through the details, saying that we know that energy is conserved in asymptotically flat space-times should be enough to illustrate why the example is flawed.

Pete is also right when he says that energy is conserved if you have a static space-time (though I'm unclear if he's ever acknowledged that energy can also be conserved if you have asymptotic flatness in a non-static space-time.) Regardless of whether Pete acknowledges it or not, energy is conserved in this case.

[Garth]

pervect - Agreed. But about your post #100 in this thread my understanding of the total system energy was taken from Weinberg, eq. 8.2.16 pg. 182.

Pete - From your post #103Originally Posted by pervect
That's the covariant energy component of the energy-momentum 4-vector, i.e. P_{0}



Yipes! That is an odd way to phrase that. Energy is energy and there is little reason to qualify it with "covariant". Energy is the time component of the energy-momentum 1-form.

Elsewhere you define energy as P_{0} as it is this time element of four-momentum that is conserved if the metric time component is static. Yet a 1-form is the contravariant form of a vector.

Garth

[gptejms]

Pete is also right when he says that energy is conserved if you have a static space-time (though I'm unclear if he's ever acknowledged that energy can also be conserved if you have asymptotic flatness in a non-static space-time.) Regardless of whether Pete acknowledges it or not, energy is conserved in this case.

Thanks pervect for your answer.So,energy is conserved in the case of a star collapsing to a black hole(non-static spacetime with asymptotic flatness).(right?).
What about RW spacetimes?(I've picked up these terms reading the discussions here!)

[pmb_phy]

Why?Why would it? Miknd you - I'm going by Einstein's definition of "gravitational field" and nobody elses. Spacetime curvature, aka tidal force (i.e. Non-vanishing Riemman tensor) only means that there is a gravitational field present which can't be transformed away in a finite region of spacetime. The presence of a gravitational field, aka gravitational acceleration, is dictated by the non-vanishing of the Christoffel symbols (when the spatial coordinates are expressed in Cartesian coordinates). This does not require spacetime curvature. Thus a gravitational field can be "produced" by a change in spacetime coordinatges. But this is not true when the spacetime is curved, i.e. I can't produce spacetime curvature by changing coordinates.
..though I'm unclear if he's ever acknowledged that energy can also be conserved if you have asymptotic flatness in a non-static space-time.In the first place I think that you're refering to the conservation of the mass of source of the gravitational field and not the conservation of the energy of a particle moving in the gravitational field. I was speaking of the later. In the second place I have not acknowledged what you speak of since I have not proved it to myself nor have I seen a proof ... and if its in Wald then I doubt I'd be able to follow it since that is an extremely hard text to follow. Someday. :smile:

Pete - From your post #103 ... Elsewhere you define energy as P_{0} as it is this time element of four-momentum that is conserved if the metric time component is static.I understand that and yes, that how its defined. By the way - why do you say that its how "I" defined it? Don't assume that it was I who presented this definition to the world of GR. :biggrin:

I simply don't think its a good idea to call it "covariant" energy or something like that - its simply "energy" and calling it something else like "covariant energy" gives the impression that its something different than "energy".

Yet a 1-form is the contravariant form of a vector.That is incorrect. A 1-form is a very different animal than a vector so its not a good idea to think of it as being a component of a different "form" of a vector. What you said here is like saying that a ket is a different form of a bra and that is certainly a bad way to look at bras and kets.

Pete

[pervect]

Thanks pervect for your answer.So,energy is conserved in the case of a star collapsing to a black hole(non-static spacetime with asymptotic flatness).(right?).
What about RW spacetimes?(I've picked up these terms reading the discussions here!)

Basically, yes. This works in the real world because our spacetime appears to be reasonably flat (on a global scale, it's obviously curved near massive bodies). So if you have a black hole that's reasonably isolated, energy is conserved as it collapses, when you add together the energy of the BH itself, any ejecta (looking at the Crab nebula as an example, there will be a fair amount of this in an actual collapse), and the energy of the radiation it emits (both electromagnetic and gravitational). All of this is measured from the viewpoint of a far-away observer, one where space-time is reasonably flat. A non-rotating black-hole wouldn't emit gravitational radiation as it collapses, but it rotating black hole is expected to emit gravitational radiation as it collapses.

BY RW space-time, you mean "Robertson-Walker"?
[edit]
I'm deleting my previous response here, I actually have to *think* about this one more before I answer!
[end edit]

[pmb_phy]

I'm *think* that the spatially flat Robertson-Wlaker universe is also asymptotically flat, but I'm not 100% positive.The spacetime curvature for the RW spacetime has the same value everywhere. Since the spacetime is not flat anywhere it can't be asymptotically flat.

Pete

ps - I say "has the same value everywhere" but I'm not sure what that means per se. If I change frames from one inertial frame to another I will change the value of the spacetime curvature. But due to the homogeneity of the spacetime I think the curvature will still be the same everywhere. Yes? No? Maybe? To much to drink last night? :biggrin:

[pervect]

I wasn't quite fast enough on the "edit" button, I see.....

I am currently tending to think that energy is not conserved in Robertson-Walker space-times. The reason I think it's not conserved is cosmological redshift. If you emit some light, wait a bit, and measure it's energy in isotropic coordinates, you find that the light has lost energy.

If you don't use isotropic coordinates, you find that masses are accelerating away from each other which is also no good for energy conservation.

However, I can't quite pin down the reason why a spatially flat Robertson-Walker space-time wouldn't be asymptotically flat in the sense needed for energy conservation. It's very clear that there is a conformal isometry to flat space-time, because the metric is just -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2).

However, there are 5 additional conditions that the conformal isometry must satisfy for a metric to be considered asymptotically flat. I suspect one of them is violated, but I can't pin down which one.

[juju]

Hi,

It may just be that energy/momentum conservation is the basic conservation rule everywhere, rather than two seperate rules one for energy and one for momentum.

Since in most cases energy and momentum are both conserved seperately in most cases, this might be an appropriate idea.

In this way the lack of energy conservation on its own is no big deal.

This seems to suggest to me that the concept of forces arises from one continuous energy/momentum field, whose mode of energy/momentum transfer is a function of the basic structure of space/time in the neighborhood where such forces arise.

juju

juju

[Haelfix]

"However, I can't quite pin down the reason why a spatially flat Robertson-Walker space-time wouldn't be asymptotically flat in the sense needed for energy conservation."

Bingo, you came around to my original posts back on page 3 of this thread. Look up what k asymptotic freedom means, things will become clearer.

But yes in general there is no ADM solution to FRW solutions. Let me say that again..

Global energy (defined canonically from the hamiltonian for the pedants) does not make sense in *general* in GR! Only in very specific formulations of asymptotic flatness where we can far field it, or in stationary metrics.

Worse.. gravitational energy due to matter is not relative in the strict sense we are used too classically. Why? B/c matter particles fix their definition of energy at infinity relative to the ground state vacuum (otherwise you get infinity for say the harmonic oscillator for a scalar particle). But in curved spacetime, there is no good way to even define a vacuum at some fixed point in space. Quantum mechanics bites us in the *** again.

[gptejms]

All of this is measured from the viewpoint of a far-away observer, one where space-time is reasonably flat.


So I was right after all when I said that energy was conserved from the viewpoint of an observer in an inertial frame(corresponding to the asymptotically flat region)looking at the non-inertial frames from outside--the reason being that there are no pseudo forces in the inertial frame.

I also said that energy is not conserved for observers in the non-inertial frames(regions where spacetime is not flat) because there are pseudo forces in such frames which do work.However I'm not sure of this latter statement since pmb_phy and pervect didn't seem to agree.

[gptejms]

The presence of a gravitational field, aka gravitational acceleration, is dictated by the non-vanishing of the Christoffel symbols (when the spatial coordinates are expressed in Cartesian coordinates). This does not require spacetime curvature.

Pete

You mean non-vanishing Christoffel symbols associated with a gravitational field do not necessarily imply spacetime curvature-----this is news to me.Do others agree?

[pervect]

You mean non-vanishing Christoffel symbols associated with a gravitational field do not necessarily imply spacetime curvature-----this is news to me.Do others agree?

If you take "space-time curvature" to mean a non-vanishing Riemann tensor, the statement is definitely true - an example is an accelerating rocket. The metric for an accelerating rocket is the Rindler metric

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

where g is a constant, the acceleration of the rocket. (This is in geometric units where c=1).

if you calculate the Riemann tensor and the Christoffel symbols for this metric, you'll find the Riemann is zero and the Christoffel symbols are not. (FYI some texts vary on exactly what they call the Rindler metric, but it doesn't matter to the point I"m making which uses the above metric as an example regardless of what name you choose to give it).

Cheat and use your favorite tensor program to calculate this sort of calculation, it's a pain to do by hand.

However, sometimes the term "space-time curvature" is loosely used to mean that the metric coefficients vary as a function of the coordinates. The metric above is curved in that lose sense, but not in the strict sense of having a non-zero Riemann curvature tensor.

[pervect]

Bingo, you came around to my original posts back on page 3 of this thread. Look up what k asymptotic freedom means, things will become clearer.



What text would I look this up in? I don't think it's in Wald :-(, his treatment of ADM mass is fairly superficial (his treatment of the Bondi mass is better).

I hope k asymptotic freedom is simpler than conformal infinity - the basic idea of making infinity a "place" is pretty clear, but the mathematical fine print is rather intricate.

[Garth]

So I was right after all when I said that energy was conserved from the viewpoint of an observer in an inertial frame(corresponding to the asymptotically flat region)looking at the non-inertial frames from outside--the reason being that there are no pseudo forces in the inertial frame.

I also said that energy is not conserved for observers in the non-inertial frames(regions where spacetime is not flat) because there are pseudo forces in such frames which do work.However I'm not sure of this latter statement since pmb_phy and pervect didn't seem to agree.

An inertial frame is a freely falling coordinate system. In such a frame of reference particles do not suffer accelerations unless there are specific non-gravitational forces acting on them. Such a frame can only be defined for a sufficiently small region around its origin, otherwise tidal forces will be experienced.

However, measured in such a frame, the energy of other particles outside this sufficiently small region will not be conserved in general. There is a confusion even in the definition of the energy of a freely falling particle in GR; as to whether it is P_{0} or P^{0}, because although P^{0} is the natural definition it is not conserved, even though no work is being done on the particle. Although P_{0} is conserved in the frame of reference of the centre of mass of a static system, in general it is not so and this identification with energy is restricted and rather artificial.

The overall insight is that GR does not in general conserve energy, it is an improper energy theorem, it conserves energy-momentum instead.
The principle of the conservation of energy-momentum is not a concatenation of the principle of the conservation of energy and principle of the conservation of momentum; energy-momentum is a geometric concept in its own right, invariant under Lorentz transformations.

Energy and momentum are frame dependent concepts; therefore it is necessary to define a frame of reference, a preferred frame. in order to restore the principle of the conservation of energy.

Garth

[pervect]

An inertial frame is a freely falling coordinate system. In such a frame of reference particles do not suffer accelerations unless there are specific non-gravitational forces acting on them. Such a frame can only be defined for a sufficiently small region around its origin, otherwise tidal forces will be experienced.


The tidal forces aren't a problem, if they approach the Newtonian tidal forces in the limit as you go to infinity. The standard definition of energy and energy conservation in GR can deal with tidal forces that approach Newtonian tidal forces as one goes to infinity.

It does appear to me that the expansion of the universe is not something that can be dealt with in this (standard) manner, however. This problem can only be dealt with by dealing with sections of the universe small enough that the cosmological expansion isn't important over the timescale studied.


The overall insight is that GR does not in general conserve energy, it is an improper energy theorem, it conserves energy-momentum instead.
The principle of the conservation of energy-momentum is not a concatenation of the principle of the conservation of energy and principle of the conservation of momentum; energy-momentum is a geometric concept in its own right, invariant under Lorentz transformations.

Energy and momentum are frame dependent concepts; therefore it is necessary to define a frame of reference, a preferred frame. in order to restore the principle of the conservation of energy.

Garth

The notion of a preferred frame of course requires a rather fundamental re-write of GR - one which a certain author just happens to have done :-).

We'll see how this new theory works out when the Gravity probe B results get back.

Meanwhile, I have to say that it is quite possible that the universe is screwy enough that the standard GR notion of energy conservation is the correct one - something that works over human time and distance scales, but something that doesn't work over cosmological time and distance scales.