Stress-energy tensors in GR

In summary, the stress-energy tensor of the gravitational field in GR is often considered to be zero due to the ability to make the metric zero at a point through coordinate transformations. However, there is a lack of texts that actually prove this statement and it may require mathematical calculations or an application of the equivalence principle. The stress-energy tensor present in the Einstein field equations (EFEs) only takes into account the matter present in the region and does not account for the exchange of energy and momentum between matter and space-time. This is shown by the equation \partial_\mu T^{\mu \nu} \neq 0 , but surprisingly, the covariant divergence of the stress-energy tensor is zero, leading to confusion about its meaning.
  • #36
PeterDonis said:
No, this isn't correct. What time translation symmetry does is give you a "preferred" slicing of spacetime into space and time (the one that is compatible with the symmetry). That preferred slicing in turn gives you a preferred way of defining the pseudo-tensor, and therefore a preferred way of writing the conservation law. But that way of writing it is still not covariant--it's still only valid for that particular slicing of spacetime into space and time.
But how are you defining time translation symmetry? I am defining it in the sense the Carroll seemed to mean: mass/energy don't exchange energy with spacetime. To me, that implies stationary spacetime. If you define it differently, then it seems you can't possibly have conservation without accounting from gravitational energy.
 
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  • #37
PAllen said:
But how are you defining time translation symmetry? I am defining it in the sense the Carroll seemed to mean: mass/energy don't exchange energy with spacetime. To me, that implies stationary spacetime. If you define it differently, then it seems you can't possibly have conservation without accounting from gravitational energy.

I'm defining time translation symmetry as stationary spacetime. So is Carroll, as far as I can tell. I don't think that means no energy exchange between matter and spacetime; a body free-falling radially towards a stationary gravitating body gains kinetic energy which comes from spacetime (at least that's how you would interpret it using the general scheme under discussion). So you still need to include gravitational energy for conservation. The key fact about stationary spacetime is that there is a "preferred" definition of "gravitational energy" because of the time translation symmetry.
 
  • #38
PeterDonis said:
I'm defining time translation symmetry as stationary spacetime. So is Carroll, as far as I can tell. I don't think that means no energy exchange between matter and spacetime; a body free-falling radially towards a stationary gravitating body gains kinetic energy which comes from spacetime (at least that's how you would interpret it using the general scheme under discussion). So you still need to include gravitational energy for conservation. The key fact about stationary spacetime is that there is a "preferred" definition of "gravitational energy" because of the time translation symmetry.
How can a body falling be represented in stationary solution? There is no timelike killing vector, and GW are emitted.
 
  • #39
PAllen said:
How can a body falling be represented in stationary solution? There is no timelike killing vector, and GW are emitted.

I was talking about a test body falling. If we insist on including the gravitational effects of all bodies, then yes, the only possible stationary solutions are for a single isolated body with nothing else present.

However, I don't think Carroll was being that restrictive. I think the distinction he was drawing between stationary and not was referring to the "background" spacetime in which test bodies move.
 
  • #40
PeterDonis said:
I was talking about a test body falling. If we insist on including the gravitational effects of all bodies, then yes, the only possible stationary solutions are for a single isolated body with nothing else present.

However, I don't think Carroll was being that restrictive. I think the distinction he was drawing between stationary and not was referring to the "background" spacetime in which test bodies move.
But then you need to add something to the stress energy tensor to represent (at least) gravitational potential energy. If you add the pseudo-tensor, then you don't need to place any limits on the spacetime. I think I see where Shyan might be struggling: either you say there is no conservation at all except in totally trivial cases in GR, or you have it quite generally at the expense of non-localizability (either via a pseudo-tensor or using ADM formulation for AF spacetime).

[edit: you do need to place limits on the spacetime. For the integral including pseudo-tensor to produce a coordinate independent total energy-momentum of a closed system, you need to be able to have the coordinate transforms approach Lorentz at infinity. I think this basically means asymptotic flatness. Then, the real meaning of time symmetry becomes 'time symmetric at infinity' in some sense, which is not true in cosmology. There are more baroque ways to get a conserved energy even for cosmolgy - Gibbs is a proponent of these.]
 
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  • #41
PAllen said:
But then you need to add something to the stress energy tensor to represent (at least) gravitational potential energy.

Not if all you want to do is calculate answers; the standard EFE, with the standard stress-energy tensor on the RHS, works just fine for that. You only need to define a pseudo-tensor for "gravitational energy" if you insist on rearranging the EFE to support a particular interpretation of "energy" (the pseudo-tensor then becomes the particular piece of the Einstein tensor that you move from the LHS to the RHS of the EFE).

PAllen said:
If you add the pseudo-tensor, then you don't need to place any limits on the spacetime.

Not just to define the pseudo-tensor, no. But as I understand it, if you don't have a stationary spacetime, and you don't choose coordinates compatible with that symmetry, the ordinary divergence of SET plus pseudo-tensor won't be zero. However, it's possible that I'm thinking of asymptotic flatness here, not stationarity; in your edit to your post, you say:

PAllen said:
For the integral including pseudo-tensor to produce a coordinate independent total energy-momentum of a closed system, you need to be able to have the coordinate transforms approach Lorentz at infinity. I think this basically means asymptotic flatness.

I'll have to think about this some more. A good test case to consider would be a star that runs out of nuclear fuel and collapses to a white dwarf. The overall spacetime is clearly not stationary (though it is approximately so in the far past and far future of the collapse), but it is asymptotically flat.
 
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  • #42
PeterDonis said:
Not if all you want to do is calculate answers; the standard EFE, with the standard stress-energy tensor on the RHS, works just fine for that. You only need to define a pseudo-tensor for "gravitational energy" if you insist on rearranging the EFE to support a particular interpretation of "energy" (the pseudo-tensor then becomes the particular piece of the Einstein tensor that you move from the LHS to the RHS of the EFE).
The issue is conservation of energy- momentum. The quantities as defined by stress-energy tensor are not conserved in any but the most totally trivial case without including gravitational energy. If you want to claim a broad class of (non cosmological) solutions have conservation of energy, you need a modified version of stress energy tensor.
PeterDonis said:
Not just to define the pseudo-tensor, no. But as I understand it, if you don't have a stationary spacetime, and you don't choose coordinates compatible with that symmetry, the ordinary divergence of SET plus pseudo-tensor won't be zero.

No, that's not true. The ordinary divergence of this sum is zero in all coordinates, in all spacetimes. What is more restricted is a (near) coordinate independent notion of total energy-momentum of a closed system. So ordinary divergence zero has no restrictions at all. The integral law does not require anything like stationary, but does require an asymptotic condition at infinity, similar to ADM formulation.

[edit: here is reference consistent with my understanding:

http://www3.nd.edu/~kbrading/Research/Brading%20HGR.pdf

I also note that Samalkhaits presentation is absolutely identical to that in P.G.Bergmann's 1942 book except to modernize the notation.]
 
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  • #43
PAllen said:
The quantities as defined by stress-energy tensor are not conserved in any but the most totally trivial case without including gravitational energy.

You're talking about global conservation, right?

I'll take a look at the reference you linked to.
 
  • #44
PeterDonis said:
You're talking about global conservation, right?

I'll take a look at the reference you linked to.
No, I mean the the energy density, and momentum flow density you put in the stress energy tensor are not conserved at all without accounting for gravity (even in the Newtonian potential sense). Yet gravitational potential does not figure in the covariant divergence law at all. An ordinary divergence is not conserved (i.e. zero). Maybe you are noting that strictly locally, in a free fall cartesion coordinates, the covariant divergence is the same as the ordinary? But that doesn't capture conservation even in the sense of an apple falling to the ground, in the ground frame.
 
  • #45
PAllen said:
the the energy density, and momentum flow density you put in the stress energy tensor are not conserved at all without accounting for gravity (even in the Newtonian potential sense). Yet gravitational potential does not figure in the covariant divergence law at all.

If gravitational potential does not figure in the covariant divergence law, yet that law holds, what do you mean by the first sentence in the above quote? I don't understand what "the energy density and momentum flow density you put in the stress-energy tensor" means if it doesn't mean the SET components that obey the covariant divergence law.

PAllen said:
Maybe you are noting that strictly locally, in a free fall cartesion coordinates, the covariant divergence is the same as the ordinary?

I mentioned that earlier, but that wasn't the distinction I was trying to get at with the question you quoted.

PAllen said:
that doesn't capture conservation even in the sense of an apple falling to the ground, in the ground frame.

Of course not, because a local inertial frame doesn't cover enough of the apple's trajectory. In other words, in order to assess the kind of conservation you appear to be talking about, you have to look at a global coordinate chart--or at least a chart that covers enough of the apple's trajectory. But of course the apple isn't the only object falling in the Earth's field, and we also want to be able to assess cases like the star collapsing to a white dwarf that I mentioned earlier.

So in the end, if we're looking for something beyond the standard local conservation law (covariant divergence of SET = 0), we end up looking for something that is valid globally, in the sense of being some sort of integral over a spacelike hypersurface which in some sense covers "all of space". (For spacetimes that are globally hyperbolic, which covers all of the important standard ones, this means the hypersurface should be a Cauchy surface, i.e., a surface that every inextendible timelike worldline intersects exactly once.) Then the question becomes, what integrals of that sort are conserved, i.e., do not change from one spacelike hypersurface to another, and under what conditions? That seems to me to be what all the fuss is about.
 
  • #46
PeterDonis said:
If gravitational potential does not figure in the covariant divergence law, yet that law holds, what do you mean by the first sentence in the above quote? I don't understand what "the energy density and momentum flow density you put in the stress-energy tensor" means if it doesn't mean the SET components that obey the covariant divergence law.
Sean Carroll argues that you should consider energy as not conserved in GR, basically at all, because integral of (e.g.) energy density (as defined in stress energy tensor) is not conserved. This is true even for a lab scale observation of an apple falling to Earth (using any reasonable ground coordinates). To get a conserved energy, even for this simple case, you have to add something the stress-energy energy density term. There is nothing in the stress energy tensor to capture gravitational potential energy that gets converted to KE.

I have alway felt his position in this blog is way overblown, and the energy is fully conserved in AF spacetimes and for the approximation of treating an isolated system as if embedded in AF spacetime. His approach of integrating stress energy tensor components and claiming non-conservation leads to the idea that GR cannot handle conservation of energy for a falling apple.
PeterDonis said:
Of course not, because a local inertial frame doesn't cover enough of the apple's trajectory. In other words, in order to assess the kind of conservation you appear to be talking about, you have to look at a global coordinate chart--or at least a chart that covers enough of the apple's trajectory. But of course the apple isn't the only object falling in the Earth's field, and we also want to be able to assess cases like the star collapsing to a white dwarf that I mentioned earlier.

So in the end, if we're looking for something beyond the standard local conservation law (covariant divergence of SET = 0), we end up looking for something that is valid globally, in the sense of being some sort of integral over a spacelike hypersurface which in some sense covers "all of space". (For spacetimes that are globally hyperbolic, which covers all of the important standard ones, this means the hypersurface should be a Cauchy surface, i.e., a surface that every inextendible timelike worldline intersects exactly once.) Then the question becomes, what integrals of that sort are conserved, i.e., do not change from one spacelike hypersurface to another, and under what conditions? That seems to me to be what all the fuss is about.

On this, I pretty much agree. But you were the one who brought up an apple falling and claimed you didn't need anything fancy to express conservation in that case. I claim that already is a problem in Sean Carroll's sense, and that to solve it, you already need something beyond stress energy tensor to describe it.

In case I am missing something, can you explain how, using only the stress energy tensor, you represent that apple's KE grows due to decrease in gravitational potential energy?
 
  • #47
PAllen said:
There is nothing in the stress energy tensor to capture gravitational potential energy that gets converted to KE.

That's because, in a local inertial frame, the apple's KE doesn't change. (I'm assuming that the apple starts from a low enough height that a single local inertial frame can cover its entire fall; I failed to include this possibility when I said in a previous post that a local inertial frame can't cover the apple's fall. I was thinking then of falls that are long enough that the "acceleration due to gravity" changes from start to finish. If the height change is small enough we can treat the "acceleration due to gravity" as constant and the experiment fits in a single local inertial frame.) The KE of the ground changes in the local inertial frame because it's being subjected to a force and feels nonzero proper acceleration; a full accounting of the SET of the ground includes the energy responsible for that force (although you will have to impose some boundary conditions since a local inertial frame obviously can't include the entire Earth, and ultimately it's the entire Earth that's pushing the ground upward). There is no gravitational potential energy anywhere.

If you want to talk about "conversion of gravitational potential energy to KE", then you have to adopt a global coordinate chart from the outset, because the concept of "gravitational potential energy" only makes sense in terms of the global time translation symmetry of the spacetime. There's no way to tell, within a local inertial frame, that the apple, at rest in that frame, is not following an integral curve of the global time translation symmetry.

PAllen said:
you were the one who brought up an apple falling and claimed you didn't need anything fancy to express conservation in that case.

You don't, because the concept of "conservation" I was talking about is local. See above.
 
  • #48
PeterDonis said:
You don't, because the concept of "conservation" I was talking about is local. See above.

Here is what I started responding to:

"I'm defining time translation symmetry as stationary spacetime. So is Carroll, as far as I can tell. I don't think that means no energy exchange between matter and spacetime; a body free-falling radially towards a stationary gravitating body gains kinetic energy which comes from spacetime (at least that's how you would interpret it using the general scheme under discussion). So you still need to include gravitational energy for conservation."

I claim there is no way to do this using the covariant divergence of T. You already need to introduce something else to describe this. If you do, in a reasonable way, you get can at least cover any AF spacetime with a global, coordinate independent, conservation law. Further, locally [in ground frame] in a non-AF spacetime, you still cover this case with a pseudo-tensor that still represents such conversion. In non-AF spacetime, you run into a problem (as I see it) only trying to get total energy of the universe in a coordinate independent way. The problem is not so much non-conservation, as inability to define total energy before you can ask about conservation.
 
  • #49
PAllen said:
I claim there is no way to do this using the covariant divergence of T.

Yes, because the covariant divergence of T is local, and the concept of energy conservation referred to in the quote you gave, and in all the other discussion in this thread that talks about integrals and pseudotensors and so on, is global. I've talked about both of these concepts in different posts in this thread; perhaps I wasn't clear enough about when I was switching back and forth.

Carroll, btw, talks about integrating the standard SET (without any pseudotensors); I don't think he ever makes any claim that the covariant divergence of T is anything but local. His main point, at least in the article I've read about energy not being conserved in GR, is that, in a stationary spacetime, you can view "space" as not changing with time (by picking a slicing of spacetime into space and time that is compatible with the time translation symmetry), and you can view the conservation of energy that arises from the time translation symmetry as being a consequence of that (i.e., a consequence of the fact that "space", the background against which things happen, is unchanging). In a non-stationary spacetime, however, you are forced to view "space" as changing with time (because there is no slicing of spacetime into space and time that makes "space" unchanging), and the failure of energy conservation is a consequence of that. It seems to me that this distinction corresponds to the fact that gravitational potential energy can only be defined in a stationary spacetime, which is why I mentioned that earlier. (I'm still not sure how this fits with the pseudotensors.)

Another thing to keep in mind is that, as I said many posts ago, all of this talk about global energy conservation is unnecessary if all you want to do is calculate observables. You can do that without ever talking about global energy conservation, or defining pseudotensors, or any of that. The reason we talk about such things is our intuitions: we can't help trying to rearrange the equations so that something appears that intuitively seems like "energy" and obeys some kind of intuitively appealing conservation law.
 
  • #50
PeterDonis said:
Carroll, btw, talks about integrating the standard SET (without any pseudotensors); I don't think he ever makes any claim that the covariant divergence of T is anything but local. His main point, at least in the article I've read about energy not being conserved in GR, is that, in a stationary spacetime, you can view "space" as not changing with time (by picking a slicing of spacetime into space and time that is compatible with the time translation symmetry), and you can view the conservation of energy that arises from the time translation symmetry as being a consequence of that (i.e., a consequence of the fact that "space", the background against which things happen, is unchanging). In a non-stationary spacetime, however, you are forced to view "space" as changing with time (because there is no slicing of spacetime into space and time that makes "space" unchanging), and the failure of energy conservation is a consequence of that. It seems to me that this distinction corresponds to the fact that gravitational potential energy can only be defined in a stationary spacetime, which is why I mentioned that earlier. (I'm still not sure how this fits with the pseudotensors.)
What I still don't see is how integrating just the SET with any representation of a falling test body in a stationary spacetime shows conservation of energy. The Komar mass integral (for example) could not, to my knowledge, include the falling body. This case can be handled either by integrating a pseudo-tensor, or using ADM energy (with limitations on the asymptotic behavior of the spacetime).
 
  • #51
PAllen said:
What I still don't see is how integrating just the SET with any representation of a falling test body in a stationary spacetime shows conservation of energy.

I don't think it does, because the SET doesn't include the test body, by definition. I'm not sure Carroll was claiming it does either; from what I remember, the main reason he brought up integrating the SET at all was to say what it doesn't show, not what it does show.

PAllen said:
The Komar mass integral (for example) could not, to my knowledge, include the falling body.

No, because, once again, the SET doesn't include the test body, and the Komar mass integral is based on the SET (but with the appropriate factor included in the integral to correct for spacetime curvature).

PAllen said:
This case can be handled either by integrating a pseudo-tensor, or using ADM energy (with limitations on the asymptotic behavior of the spacetime).

Can you elaborate? AFAIK neither the pseudo-tensor integral nor the ADM energy includes the test body either. They are integrals over the metric and its derivatives.
 
  • #52
PeterDonis said:
Can you elaborate? AFAIK neither the pseudo-tensor integral nor the ADM energy includes the test body either. They are integrals over the metric and its derivatives.

The test body could be put it as a tiny metric perturbation following a radial geodesic, for example.
 
  • #53
PAllen said:
The test body could be put it as a tiny metric perturbation following a radial geodesic, for example.

Then it isn't a test body. Anyway, has this been done? I'd be interested to see such an analysis if anyone has done one.
 
  • #54
PeterDonis said:
Then it isn't a test body. Anyway, has this been done? I'd be interested to see such an analysis if anyone has done one.
I don't know if it has been done as such, but general theorems establish that the result would show conservation.
 
  • #55
PAllen said:
general theorems establish that the result would show conservation

Are these general theorems covered in the references already given in this thread?
 
  • #56
PeterDonis said:
Are these general theorems covered in the references already given in this thread?
The results on ADM mass are covered in many sources, including starting from MTW. For the pseudo-tensor integral, the presentation by Samalkhait matches that in P.G. Bergmann's 1942 book. He can probably provide more references.

This also makes reference to integral theorems for pseutotensors, as described by Samaklhait:

http://cwp.library.ucla.edu/articles/noether.asg/noether.html
(see esp. discussion and references cited after eqn. 13).

[edit: This paper argues an intriguing result that a pseutotensor is locally physically meaningful in De Donder gauge:

http://ptp.oxfordjournals.org/content/75/6/1351.full.pdf
]
 
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  • #57
The paragraph below is written in the Wikipedia page about Emmy Noether.
Noether was brought to Göttingen in 1915 by David Hilbert and Felix Klein, who wanted her expertise in invariant theory to help them in understanding general relativity, a geometrical theory of gravitation developed mainly by Albert Einstein. Hilbert had observed that the conservation of energy seemed to be violated in general relativity, due to the fact that gravitational energy could itself gravitate. Noether provided the resolution of this paradox, and a fundamental tool of modern theoretical physics, with Noether's first theorem, which she proved in 1915, but did not publish until 1918.[102] She not only solved the problem for general relativity, but also determined the conserved quantities for every system of physical laws that possesses some continuous symmetry.

This seems strange to me. Here we were talking about the fact that generally in GR we don't have time translation symmetry and so energy isn't conserved but here things are attributed to Noether's theorems!
 
  • #58
Shyan said:
The paragraph below is written in the Wikipedia page about Emmy Noether.This seems strange to me. Here we were talking about the fact that generally in GR we don't have time translation symmetry and so energy isn't conserved but here things are attributed to Noether's theorems!
It seems strange to me too. The "History" section of:

http://en.wikipedia.org/wiki/Mass_in_general_relativity#History

actually seems like good summary of consensus knowledge, consistent with discussion here.
 
  • #59
Shyan said:
The paragraph below is written in the Wikipedia page about Emmy Noether.


This seems strange to me. Here we were talking about the fact that generally in GR we don't have time translation symmetry and so energy isn't conserved but here things are attributed to Noether's theorems!

It doesn't seem strange to me. In fact it seems quite accurate.
 
  • #60
Shyan said:
The paragraph below is written in the Wikipedia page about Emmy Noether.This seems strange to me. Here we were talking about the fact that generally in GR we don't have time translation symmetry and so energy isn't conserved but here things are attributed to Noether's theorems!

Well, there are two different forms of a conservation law:
  1. As a statement about the time-independence of some global quantity.
  2. As a statement about the flow of some local quantity.
These two forms are equivalent in flat spacetime, but not in general. Noether's theorem (or at least the one I've seen) is about local flows, and only implies global conservation in flat spacetime.

Actually, now that I think about it, you don't need flat spacetime, you just need a global coordinate system, so that you can relate quantities at different locations and so that you can integrate over all space.
 
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  • #61
Shyan said:
As mentioned by Sean Caroll in this blog post, the...
Thanks

I’m sorry my friend, I don’t bother myself with non-mathematical blogs. In theoretical physics, ordinary language is confusing and often misleading. I don’t “understand it”, I don’t like it and I don’t have time for it. This is why my posts, in here, are filled with equations. Because “everybody” understands mathematical treatment better, it shuts people up and, hence, reduces the number of posts in a thread. Even for trivial and often meaningless question, a thread can drag on and on with countless number of conflicting, confusing and self-contradicting posts.

Sam
 
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  • #62
Shyan said:
Things don't seem to be that simple. I started from Sam's equation 2 and the definition of [itex] \partial_\rho(\sqrt{-g} \ t^\rho_{\ \sigma}) [/itex] and I got:
[itex]
\partial_\mu(\sqrt{-g}T^{\mu\lambda})+\frac{\sqrt{-g}}{2k}\left(t^\rho_{ \ \sigma} \partial_\rho g^{\lambda \sigma}-G_{\mu\nu} g^{\lambda \rho} \partial_\rho g^{\mu \nu} \right)=0
[/itex]

[itex]\nabla_{ \mu } ( \sqrt{ - g } \ T^{ \mu }{}_{ \nu } ) = 0[/itex], implies [tex]\partial_{ \mu } ( \sqrt{ - g } \ T^{ \mu }{}_{ \nu } ) = \frac{ 1 }{ 2 } \sqrt{ - g } \ T^{ \rho \sigma } \ \partial_{ \nu } g_{ \rho \sigma } . \ \ \ \ (1)[/tex]
Hence [tex]\partial_{ \mu } ( \sqrt{ - g } \ T^{ \mu }{}_{ 0 } ) = \frac{ 1 }{ 2 } \sqrt{ - g } \ T^{ \rho \sigma } \ \partial_{ 0 } g_{ \rho \sigma } .[/tex]
Therefore, in a closed interacting system of matter and gravitational field, the energy of the matter component is conserved if [itex]\partial_{ 0 } g_{ \mu \nu } = 0[/itex] holds everywhere. Otherwise, the conservation of the total energy (matter plus gravitational fields) follows from,
[tex]\partial_{ \mu } \left( ( - g ) \ ( T^{ \mu \nu } + t^{ \mu \nu } ) \right) = 0 . \ \ \ \ ( 2 )[/tex] It is, therefore, clear that neither matter nor gravitational field obey separate conservation laws. This is not surprising, it happens even in ordinary electrodynamics: the divergence of the electromagnetic stress tensor does not vanish in presence of charges. Indeed, it obeys equation very similar to Eq(1): [tex]\partial_{ \mu } \Theta^{ \mu }{}_{ \nu } = F^{ \mu }{}_{ \nu } \ j_{ \mu }.[/tex] However, since GR is a nonlinear theory, it is not correct to associate [itex]T_{ \mu \nu}[/itex] solely with matter and [itex]t_{ \mu \nu }[/itex] with the gravitational field. This is because, the [itex]T[/itex] is constructed out of the metric tensor [itex]g_{ \mu \nu }[/itex] as well as the matter field, and the [itex]t[/itex] depends on the matter distribution through the metric tensor.
Integrating Eq(2) over 3-volume and assuming the absence of material fluxes through the boundary, we obtain [tex]\frac{ d }{ d t } \int_{ V } d^{ 3 } x \ ( - g ) \ ( T^{ 0 \nu } + t^{ 0 \nu } ) = - \oint_{ \partial V } \ d S_{ j } \ ( - g ) \ t^{ j \nu } . \ \ \ (3)[/tex] for [itex]\nu = 0[/itex], Einstein said that the RHS “… is surely the energy lost by the material system”.
The RHS of (3) vanishes when the gravitational energy-momentum fluxes do not leak out of the boundary. In this case, one finds an expression for the total, time-independent, 4-momentum of the system [tex]P^{ \nu } = \int_{ V } d^{ 3 } x \ ( - g ) \ ( T^{ 0 \nu } + t^{ 0 \nu } ) = \mbox{ const. } \ \ \ (4)[/tex] Here is what Einstein said about this equation, “… the quantities [itex]P^{ \mu }[/itex], a four-momentum of field and matter, have a definite meaning, i.e. they appear to be independent of the particular choice of coordinate system to the extent necessary from a physical stand point”. Indeed, [itex]P^{ \mu }[/itex] behaves like a vector with respect to all general coordinate transformations which approach Lorentz transformations at infinity. Otherwise, it is only a vector with respect to [itex]GL(4)[/itex], hence the last part of Einstein’s statement “to the extent necessary ….”.
So, what does Eq(4) buy us? And how can we use it? In order to make Eq(4) ready to use and useful, we use Noether’s second theorem to rewrite Einstein equation in the following (Maxwell-like) form [tex]\frac{ 1 }{ 16 \pi G } \ \partial_{ \rho } h^{ \mu \nu \rho } = ( - g ) \ ( T^{ \mu \nu } + t^{ \mu \nu } ) , \ \ \ \ (5)[/tex] where [itex]h^{ \mu \nu \rho } = - h^{ \mu \rho \nu }[/itex]. Then, we may add/choose an appropriate super-potential to [itex]h^{ \mu \nu \rho }[/itex] so that, for example, the resulting pseudo-tensor is symmetric. Of course, there are more super-potential than are physicists. But for our purpose, the following (Landau) super-potential is sufficient [tex]h^{ \mu \nu \rho } = \partial_{ \sigma } \left( ( - g ) \ ( g^{ \mu \nu } \ g^{ \rho \sigma } - g^{ \mu \rho } \ g^{ \nu \sigma } ) \right) . \ \ \ \ (6)[/tex] Inserting (5) in (4), we get [tex]P^{ \nu } = \frac{ 1 }{16 \pi G } \int d^{ 3 } x \ \partial_{ \rho } h^{ \nu 0 \rho } = \frac{ 1 }{16 \pi G } \int d^{ 3 } x \ \partial_{ i } h^{ \nu 0 i } ,[/tex] where [itex]h^{ \nu 0 0 } = 0[/itex] has been used in the last step. So, by the divergence theorem, we find [tex]P^{ \nu } = \frac{ 1 }{16 \pi G } \oint d S_{ i } \ h^{ \nu 0 i } . \ \ \ \ (7)[/tex] The equations (4) and (7) should make you jump and scream! Indeed, because of these equations, the notion of an INERTIAL MASS shows up unexpectedly in a theory of GRAVITY through [tex]m = P^{ 0 } = \frac{ 1 }{16 \pi G } \oint d S_{ i } \ h^{ 0 0 i } . \ \ \ \ (8)[/tex]Now, we will use (8) to solve an age-old mystery in physics. Using (6) in (8), we find
[tex]m = \frac{ 1 }{16 \pi G } \oint d S_{ i } \partial_{ j } \left( ( - g ) \ ( g^{ 0 0 } \ g^{ i j } - g^{ 0 i } \ g^{ 0 j } ) \right) . \ \ \ (9)[/tex] So, the inertial mass of a static, spherically symmetric object can be calculated using the Schwarzschild metric which has the following asymptotic behaviour [tex]g^{ 0 0 } \sim 1 + \frac{ 2 M G }{ r } ; \ \ \ g^{ i j } \sim - \delta^{ i j } ( 1 + \frac{ 2 M G }{ r } ) .[/tex] Where, [itex]r^{ 2 } = x^{ 2 } + y^{ 2 } + z^{ 2 }[/itex] and [itex]M[/itex] is the GRAVITATIONAL MASS of the object. It is also called the heavy mass. Also note that
[tex]g^{ 0 0 } \ g^{ i j } = - \delta^{ i j } ( 1 + \frac{ 4 M G }{ r } ) ,[/tex] and [tex]\partial_{ j } ( g^{ 0 0 } \ g^{ i j } ) = 4 M G \frac{ x^{ i } }{ r^{ 3 } } . \ \ \ \ (10)[/tex] Substituting (10) in (9), we find [tex]m = \frac{ 1 }{16 \pi G } \oint \left( d \Omega \ r^{ 2 } \frac{ x^{ i } }{ r } \right) \left( 4 M G \frac{ x^{ i } }{ r^{ 3 } } \right) = M .[/tex] This establishes the (forever was mysterious) equality between inertial and gravitational masses. It is assumed that either a non-singular interior solution exists or else, as [itex]r \to 0[/itex] about any singularities, [itex]\oint d S_{ i } h^{ 0 0 i }[/itex] goes to zero.

Sam
 
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  • #63
PAllen said:
The results on ADM mass are covered in many sources, including starting from MTW. For the pseudo-tensor integral, the presentation by Samalkhait matches that in P.G. Bergmann's 1942 book. He can probably provide more references.

This also makes reference to integral theorems for pseutotensors, as described by Samaklhait:

http://cwp.library.ucla.edu/articles/noether.asg/noether.html
(see esp. discussion and references cited after eqn. 13).

[edit: This paper argues an intriguing result that a pseutotensor is locally physically meaningful in De Donder gauge:

http://ptp.oxfordjournals.org/content/75/6/1351.full.pdf
]

I must have good memory, for I studied that book some twenty years ago.
As for references: Between 1949 and 1960, Bergmann and his co-workers (at Syracuse University) produced a wealth of paper on “conservation laws in GR”, “integrating strong conservation laws”, “the second Noether theorem” and other relevant issues.
Here are few of them:
J. L. Anderson & P. G. Bergmann, Phys. Rev. 83, 1018(1951).
P. G. Bergmann & R. Schiller, Phys. Rev. 89, 4(1953).
J. N. Goldberg, Phys. Rev. 89, 263 (1953).
P. G. Bergmann & R. Thomson, Phys. Rev. 89, 400(1953). Bergmann, Goldberg, Janis & Newman ,Phy. Rev. 103, 807(1956).
Of course, the-must-read classic works on the subject are the following
P. G. Bergmann, Phys. Rev., 112, 287(1958).
C. Moller, Ann. Phys., 4, 347(1958).
A. Komar, Phys. Rev., 113, 934(1959).
J. G. Fletcher, Rev. Mod. Phys., 32, 65(1960).
Sam
 
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  • #64
Still one question remains. If there is no unique general way of defining mass in GR (I mean gravitational mass, because as Sam demonstrated, inertial mass can be defined and turns out to be equal to gravitational mass), what is that thing people use when they construct the stress-energy tensor of a matter distribution(example here, where a mass-energy density [itex] \rho [/itex] appears in the definition of stress-energy tensor of a fluid)? If we need to know the spacetime geometry in order to define what we mean by mass, how can we have mass before actually finding out what the geometry is and then use it to find geometry?
 
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  • #65
Shyan said:
Still one question remains. If there is no unique general way of defining mass in GR (I mean gravitational mass, because as Sam demonstrated, inertial mass can be defined and turns out to be equal to gravitational mass), what is that thing people use when they construct the stress-energy tensor of a matter distribution(example here, where a mass-energy density [itex] \rho [/itex] appears in the definition of stress-energy tensor of a fluid)? If we need to know the spacetime geometry in order to define what we mean by mass, how can we have mass before actually finding out what the geometry is and then use it to find geometry?
The difficulty with mass in GR is the curvature that occurs within extended objects which means there isn't an agreed unique way to integrate density over a volume. However density is a local concept; it's defined via a small volume tending to zero, so you can replace GR by SR in a small enough volume. Locally you can assign the stress-energy tensor in locally-inertial coordinates (using the SR definitions of mass, energy, momentum, etc), then transform to your chosen GR coordinates.
 
  • #66
Shyan said:
Still one question remains. If there is no unique general way of defining mass in GR (I mean gravitational mass, because as Sam demonstrated, inertial mass can be defined and turns out to be equal to gravitational mass), what is that thing people use when they construct the stress-energy tensor of a matter distribution(example here, where a mass-energy density [itex] \rho [/itex] appears in the definition of stress-energy tensor of a fluid)? If we need to know the spacetime geometry in order to define what we mean by mass, how can we have mass before actually finding out what the geometry is and then use it to find geometry?

One way to think about is that we define an action for gravity (Hilbert action) and matter (say the electromagnetic Lagrangian). So the primary quantitites in the theory are the metric field (up to the gauge redundancy) and the electromagnetic field (also up to the gauge redundancy if one is using potentials). Then the correct equations of motion for the matter fields and the metric field are given by minimization of the action. This minimization automatically produces the Einstein equation of motion, and a stress-energy tensor that is defined in terms of the more fundamental quantities such as the metric field and the electromagnetic field, eg. http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf (Eq 1.48). In the case of the perfect fluid, one can take ##\rho## to be defined in terms of the particle number density, eg. http://arxiv.org/abs/gr-qc/0605010 (discussion following Eq 82).

So if by "mass" one means "stress-energy tensor", there is a general way of defining it in general relativity, so GR is a well-defined theory. The stress-energy tensor formulates the concept of "exactly localizable energy" for the matter fields. The stress-energy tensor does not include the concept of the "(quasi)localizable mass of gravity", which would seem to be a useful concept to have, and various possibilities for their definition are discussed in http://relativity.livingreviews.org/Articles/lrr-2009-4/ and http://arxiv.org/abs/1001.5429.
 
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  • #67
Shyan said:
Still one question remains. If there is no unique general way of defining mass in GR (I mean gravitational mass, because as Sam demonstrated, inertial mass can be defined and turns out to be equal to gravitational mass), what is that thing people use when they construct the stress-energy tensor of a matter distribution(example here, where a mass-energy density [itex] \rho [/itex] appears in the definition of stress-energy tensor of a fluid)? If we need to know the spacetime geometry in order to define what we mean by mass, how can we have mass before actually finding out what the geometry is and then use it to find geometry?

I'd suggest that the way to avoid circularity in the definition of the stress energy tensor is to regard the stress energy tensor as describing the distribution of energy and momentum in space-time, to be more specific as the amount of energy and momentum in a unit volume.

Momentum and energy are really a unified concept in relativity, very similar to the way that space and time are unified into space-time. The "slicing" of momentum-energy into a momentum part and an energy part is observer dependent, much as is the "slicing" of space-time into space and time. A few textbook authors have created the term "momenergy" to stress this unification, but the terminology never really caught on.

The non-tensor concepts needed to describe the stress-energy tensor are the observer-dependent concepts of energy, momentum, and volume. To understand the tensor nature of the stress energy tensor, you need to be familiar with tensors in general, and some particular specific tensors, the energy-momentum 4-vector, and 4-velocity tensors. It is also very helpful to be familiar with the charge-current tensor and it's particle equivalent , called the number-flux 4-vector, which assigns each particle a dimensionless unit number for particle number, rather than a dimensionfull scalar for charge as does the charge-current 4-vector.

Using a "swarm of particles" model, you can describe the stress energy tensor as the tensor product of the number-flux 4-vector and the energy-momentum 4-vector, calculated for each particle in the swarm, then the tensor products of these two 4-vectors are calculated per particle and summed together to give the overall stress-energy tensor. While you can assign each particle in the swarm an invariant mass (the magnitude of it's energy-momentum 4-vector), there is no direct need to do this to create the stress-energy tensor.

Given the above, you arrive with the textbook result that the amount of energy and momentum contained in a unit volume is given by the stress energy tensor, multipled by the 4-velocity of the observer. The least documented part in deriving the above relationship is the concept of the unit volume of an observer, in particular the tensor representation of such - it does not appear to be a terribly hard concept, but for some reason the textbooks don't go into the details. I'm not sure of the reason for the omission, if it's a case of there being some hidden trickiness, or if they just didn't get around to it.
 
  • #68
Another question: in textbooks you often see the statement that the energy density of a gravitational field is ill-defined due to the equivalence principle, as was treated by Sam here before. But in good old Newtonian gravity (without even going to Newton-cartan formulation) we also have the EP; we can always "gauge away" the Newton potential by going to an accelerated observer. So how well-defined is the energy density of a gravitational field in Newtonian gravity ? Do we circumvent problems because the field is static or/and there is a preferred spacetime foliatio ?

Just wondering about this ("how would I answer this if a student would ask me this" :p )
 
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  • #69
haushofer said:
So how well-defined is the energy density of a gravitational field in Newtonian gravity ?
In a near-static situation, there is a conserved total effective energy in a Newtonian system, and the Lynden-Bell / Katz paper suggests that this carries over to GR as well, although they were only able to demonstrate this for spherically symmetrical situations.

At some level, we divide up the system into "masses", where the energy of each "mass" takes into account any internal binding energy and kinetic energy. The effective energy of each mass is reduced by the local potential due to all other masses. The field has positive effective energy density given by ##g^2/8\pi G##. The energy of each mass is effectively reduced by its potential energy relative to all other masses, so each contribution to potential energy is counted twice, but the energy of the field adds back in the potential energy again (as can be seen by integrating by parts), so the total energy is equal to the energy of the original masses (including any kinetic energy) minus the potential energy of the configuration. Changes in the field locally preserve continuity.

I've always been puzzled about the assertions that if we adopt a free-fall frame of reference, no energy is being transferred, so there cannot be any energy in the field. In Newtonian mechanics, we have perfectly adequate descriptions of energy in accelerated systems, for example as applied for example to a rotating space station. Even though we can transform our view to the point of view of an observer on the station, we do not seem to have problems with that, mainly because we consider any non-inertial frame (from the Newtonian point of view) to be effectively a "simulation" of gravity rather than the real thing.

It is clear that transforming to or from an accelerated frame of reference shows that gravitational energy does not transform locally like energy. However, it is possible to obtain various pictures of how the "effective" gravitational energy is distributed using various pseudotensor representations, and although this doesn't represent some "stuff" I think it is still a useful picture.

One thing that causes confusion about such models is the question about whether gravitational energy is a source of gravity itself. In one sense it must be, in that if you consider the gravitational effect of some system which includes gravitational binding energy, then clearly we expect the internal potential energy to affect the overall total energy of the system as a gravitational source. However, if you take a system of masses and change their configuration, the total energy of the configuration does not change unless you transfer energy into or out of the system, so for example you cannot change to a more tightly bound static system without either extracting energy or converting it to some other internal form. In that sense, gravitational energy is merely an accounting label for energy converted via the mechanism of gravity.

There is also an alternative semi-Newtonian model which gives the correct total energy without any energy in the field, which is that the effective energy of every mass is actually reduced by only half of its potential energy relative to every other mass. This then allows conventional energy to be treated as the only source of gravity while at the same time preserving the correct total energy, although not with a locally conserved flow.
 
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  • #70
The two models of Newtonian gravitational energy which I mentioned above have a simple mathematical connection. I don't know whether it has any physical significance, but I think it provides a nice illustration of looking at the same situation in different ways.

If we let ##\Phi## represent the multiplicative Newtonian potential, approximately equal to ##1 - \sum Gm/rc^2## for all sources, then the source density is as usual given by ##\nabla^2 \Phi##, which is proportional to the local rest mass density in Newtonian theory, without any correction for potential energy.

One way to reduce the effect of each source by half of the potential energy is to replace the operator ##\nabla## by ##\Phi^{1/2} \nabla##. This then gives the following for the effective energy density from this new point of view:
$$\Phi^{1/2} \nabla (\Phi^{1/2} \nabla \Phi) = \Phi^{1/2} (\frac{1}{2} \Phi^{-1/2} (\nabla \Phi)^2 + \Phi^{1/2} \nabla^2 \Phi) = \frac{1}{2} (\nabla \Phi)^2 + \Phi \nabla^2 \Phi$$
The result is the other way of looking at the energy, consisting of 1/2 the square of the field plus the rest mass reduced by the full potential.
 

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