# Energy not defined in GR?

1. Jun 9, 2004

Staff Emeritus
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.

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?

2. Jun 9, 2004

### pmb_phy

Yup. Very true indeed.
Yup. With ya so far.
Nope. I don't see that to be true in general. That is at best only true in uniform gravitational fields.
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

3. Jun 9, 2004

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

4. Jun 9, 2004

### pmb_phy

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?

5. Jun 9, 2004

### 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

This would make energy in GR non-local.

6. Jun 9, 2004

### Staff: Mentor

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.

Last edited: Jun 9, 2004
7. Jun 9, 2004

Staff Emeritus
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.

8. Jun 9, 2004

### 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.

Last edited: Jun 9, 2004
9. Jun 9, 2004

### pmb_phy

10. Jun 9, 2004

### robphy

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

Last edited by a moderator: May 1, 2017
11. Jun 9, 2004

### 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?

12. Jun 10, 2004

### DW

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.

13. Jun 10, 2004

### 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.

"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

Pete

Last edited: Jun 11, 2004
14. Jun 10, 2004

### turin

pmb_phy,
Why don't you reserve your unwarranted personal attacks for PM?
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.

So, tidal fields are not gravitational?

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

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).

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

So? I failed to note that it's Thursday (until now). What does that have to do with it?

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

15. Jun 10, 2004

### Staff: Mentor

Dunno, maybe its just beyond my understanding (quite possible) but I still don't see why energy should have to be absolute in GR.

16. Jun 10, 2004

### turin

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 &Lambda; = 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].)

17. Jun 11, 2004

### Russell E. Rierson

Here is an interesting FAQ discussing energy conservation in GR:

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

Last edited by a moderator: May 1, 2017
18. Jun 11, 2004

### pmb_phy

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

Last edited: Jun 11, 2004
19. Jun 11, 2004

### pmb_phy

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.

20. Jun 11, 2004

### DW

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.

21. Jun 11, 2004

### pmb_phy

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]
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.

Last edited: Jun 11, 2004
22. Jun 11, 2004

### 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

23. Jun 11, 2004

### turin

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.)

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?

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?

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.

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).

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?

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.

24. Jun 11, 2004

### pmb_phy

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.
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.
Bravo! Me too. :-)
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.
Yes.
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.
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

25. Jun 11, 2004

### pmb_phy

Hi Sammy

How've you been?
My poor spelling.