Is Energy Conserved in General Relativity?

[selfAdjoint]

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?

 

[pmb_phy]

Steve Carlip, on sci.physics.research has asserted 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 you 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 argument 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 into 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 into 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]

pmb _ phys

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

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 existence of a gravitational field (i.e. gravity) is dictated by the non-vanishing of the Christoffel symbols. The existence 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 existence 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 receive PM messages. He does that everytime in every new incantation of a new handle. That was a post to correct errors, not to simply criticize 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 receive 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 referred 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 gravitational 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 g_ab 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 argument 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 gravitational 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 referring 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 argument 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 extremely 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.

 

[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 traveling 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.
...

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 argument 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 -> Rⁱ_jkl = 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 Rⁱ_jkl = 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 extremely 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 traveling 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 -> Rⁱ_jkl = 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 Rⁱ_jkl = 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 referring 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 -> Riemann

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.

What he's referring 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 referring 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 t_ij. 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 t_ij and R^uvab[/sup]. Another analogy is the Laplacian and Eintein's tensor, i.e. del²U <=> G_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 m^jk the "inertial mass per unit volume" of a stressed medium at rest. In general m^jk 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.