Why does gravity have to be non-linear?

[Jonathan Scott]

I know that Newtonian gravity as a theory is linear and GR is not, but I'm trying to get a feel for why textbooks often insist relativistic gravity must be non-linear, in order to account for the gravitational effect of "gravitational energy" itself (which is however not part of the source term in GR).

What specific example could they have in mind of a case where "gravitational energy" must have an existence separate from the existing mass-energy of the objects involved?

I'm asking mainly because it seems to me that apparent non-linearity very similar to that in GR can be introduced very easily into a Newtonian-style gravitational theory as follows:

Let [itex]\phi = - Gm/r[/itex], the Newtonian potential.

Let [itex]\Phi = (1 + \phi/c^2)[/itex], the time dilation potential factor.

Suppose that what we assume to be the effective field for our relativistic equation of motion is not actually identified with the Newtonian [itex]\nabla \Phi[/itex] but rather with [itex]\Phi \, \nabla \Phi[/itex], which is of course very similar, as [itex]\Phi \approx 1[/itex].

The divergence of this effective field (giving the effective source density) is then as follows:
[tex]
\nabla (\Phi \, \nabla \Phi) = \Phi \, \nabla^2 \Phi + (\nabla \Phi)^2
[/tex]
This differs from the original linear Newtonian picture in two ways:

1. The Newtonian source mass density [itex]\nabla^2 \Phi[/itex] is now modified by the local time dilation factor, so the effective energy of the source is reduced by that factor. This implies a non-linear effect, as the gravitational source is causing a potential which is reducing its own effective strength as a gravitational source.
2. There is an extra term proportional to the square of the field which describes a source density within the field.
   This implies a non-linear effect, as the field is acting as an additional gravitational source.

However, the only actual difference was in (perhaps incorrectly) identifying the field with a slightly different function of the Newtonian potential rather than just the gradient.

We can turn this round and check what effective time dilation potential [itex]\Phi_E[/itex] in terms of the Newtonian potential would cause the effective field to be given by [itex]\nabla \Phi_E = \Phi \, \nabla \Phi[/itex]. The result is as follows:
[tex]
\Phi_E = \frac{(1 + \Phi^2)}{2} = \left (1 + \frac{\phi}{c^2} + \frac{1}{2}\left ( \frac{\phi}{c^2} \right )^2 \right )
[/tex]
The additional term in the effective potential time dilation factor is exactly equal to the corresponding term in the isotropic form of the Schwarzschild solution, corresponding to the PPN [itex]\beta[/itex] non-linearity parameter being equal to 1, as in GR.

I think that as for electromagnetic energy, one can transform this integral into a boundary one and show that the integral over a large enough region is the same as the total amount of source mass (as seen locally, not reduced by time dilation) within the region, so despite the different identification of the field, the total energy still seems to come out right anyway.

So this seems to illustrate that a linear theory could give an illusion of non-linearity closely matching GR.

Does anyone know of a simple way to prove that gravity really must be a non-linear theory, without assuming that it's non-linear because GR says it is?

 

[bcrowell]

Let [itex]\phi = - Gm/r[/itex], the Newtonian potential.

Let [itex]\Phi = (1 + \phi/c^2)[/itex], the time dilation potential factor.

One thing to watch out for here is that [itex]\Phi = (1 + \phi/c^2)[/itex] is only an approximation to first order in the potential difference. In general, the time dilation factor is [itex]e^\phi[/itex] (in units with c=1). Since you're trying to show later on that you get agreement with GR to second order in [itex]\phi[/itex], I don't think it's necessarily valid to ignore this difference. Also, [itex]\Phi[/itex] is only well defined in a static spacetime, and r isn't well defined for a general spacetime in GR (there is no uniquely defined notion of the spatial distance between points).

It seems like you're saying that [itex]\Phi[/itex] is the fundamental field in your theory. What is the wave equation that it satisfies? How does it relate to the metric?

In general, I think there is a fundamental problem with trying to construct a relativistic metric theory of gravity that is based on a scalar field. If it's a scalar field, then waves have no polarization properties. But it's pretty straightforward to show that gravitational waves must be transverse waves. (In terms of their effect on the metric, they can't be longitudinal because a longitudinal wave can be converted into Minkowski spacetime by a change of coordinates. They have to be transverse because they have to conserve volume, so you have to have expansion along one transverse axis and compression along the other.)

Re the fundamental reason that GR has to be nonlinear, it's that mass and energy are equivalent, so if mass gravitates, gravitational potential energy has to gravitate as well.

 

[atyy]

http://arxiv.org/abs/gr-qc/0611100

In Eq 3,4,8 it looks like a linear theory is proposed. But the proposal is corrected for universal coupling in Eq 16. It looks like the correction does not involve the gravitational field equation which is still Eq 4, and linear at first sight.

Eq 17 seems close to something Jonathan Scott wrote in the OP.

 

[Jonathan Scott]

One thing to watch out for here is that [itex]\Phi = (1 + \phi/c^2)[/itex] is only an approximation to first order in the potential difference. In general, the time dilation factor is [itex]e^\phi[/itex] (in units with c=1). Since you're trying to show later on that you get agreement with GR to second order in [itex]\phi[/itex], I don't think it's necessarily valid to ignore this difference.

I'm showing that if you start from an assumption of a relativistic theory which matches Newtonian linear theory to first order, but you identify the field in this theory with something that isn't quite the same as the gradient of the Newtonian potential, then you get non-linear effects that look just like a better approximation to GR, including matching [itex]\Phi_E = e^\phi[/itex] to the term in [itex]\phi^2[/itex].

Also, [itex]\Phi[/itex] is only well defined in a static spacetime, and r isn't well defined for a general spacetime in GR (there is no uniquely defined notion of the spatial distance between points).

For purposes of the comparison, I'm assuming GR to be represented by the Schwarzschild solution in isotropic coordinates, as that's the closest relativistic equivalent we have to Newtonian gravity.

It seems like you're saying that [itex]\Phi[/itex] is the fundamental field in your theory. What is the wave equation that it satisfies? How does it relate to the metric?

In general, I think there is a fundamental problem with trying to construct a relativistic metric theory of gravity that is based on a scalar field. If it's a scalar field, then waves have no polarization properties. But it's pretty straightforward to show that gravitational waves must be transverse waves. (In terms of their effect on the metric, they can't be longitudinal because a longitudinal wave can be converted into Minkowski spacetime by a change of coordinates. They have to be transverse because they have to conserve volume, so you have to have expansion along one transverse axis and compression along the other.)

I'm not trying to develop a new theory of gravity here, and I don't see any reason to assume that the timelike scale factor in a static situation wouldn't become a tensor in a fully dynamic one. For the moment, I'm specifically interested in linearity.

Re the fundamental reason that GR has to be nonlinear, it's that mass and energy are equivalent, so if mass gravitates, gravitational potential energy has to gravitate as well.

Sorry, but that's right back to where I started - the sort of vague assertion I find in many places, as mentioned in the first sentence of my original post, and which I'd like turned into a more concrete example.

If you call some part of the total energy of a system "gravitational potential energy" then of course that part has to contribute to the gravitational source, as Einstein pointed out from the start.

However, it's not at all clear that "gravitational potential energy" has an independent existence. For example, if I lower a weight, I can extract potential energy from it, but from a GR point of view the energy which I extract is equal to the difference in its energy between the higher and lower positions, which arises because of time dilation.

 

[SteamKing]

Newtonian gravity is decidedly a non-linear phenomenon. Three-body problem anyone? Bueller?

 

[Jonathan Scott]

Newtonian gravity is decidedly a non-linear phenomenon. Three-body problem anyone? Bueller?

By "non-linear", I mean the idea that the gravitational effect is not exactly linearly proportional to the strength of the source. GR is certainly non-linear in that respect, and Newtonian gravity is linear.

 

[TrickyDicky]

Does anyone know of a simple way to prove that gravity really must be a non-linear theory, without assuming that it's non-linear because GR says it is?

This is purely heuristic, and it doesn't prove anything, it is really simple though ;)
In GR gravity and acceleration are linked by the Equivalence principle thru General covariance, and the potentials are replaced by the metric; unlike in Newton theory the potentials (thru the metric) in GR are dependent on time as well as on spatial parameters.

So if a uniform acceleration locally mimics a gravitational field then the coordinate transformation for an acceleration along the x-axis (x'=x+at^2) is non-linear and since GR demands the generalization of the coordinate transformations from the linear ones involving uniform velocity of SR to the non-linear of non-uniform motion that require tensor equations I guess a geometric theory of gravitation must be non-linear.
Maybe the fac that GR must include non-linear transformations doesn't really explain GR non-linearity but it works for me.

The usual reference to gravitational energy or to the "gravity of gravity" are unrelated to GR non-linearity IMHO.

 

[TrickyDicky]

By "non-linear", I mean the idea that the gravitational effect is not exactly linearly proportional to the strength of the source. GR is certainly non-linear in that respect, and Newtonian gravity is linear.

I think you're thinking of gravity in terms of Newtonian force, but in a geometric theory that describes gravity as curvature, it seems logical that it has to be non-linear, maybe this interpretation is too naive?

 

[Q-reeus]

...Re the fundamental reason that GR has to be nonlinear, it's that mass and energy are equivalent, so if mass gravitates, gravitational potential energy has to gravitate as well.

Not true in GR - from http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html :
"One other complaint about the pseudo-tensors deserves mention. Einstein argued that all energy has mass, and all mass acts gravitationally. Does "gravitational energy" itself act as a source of gravity? Now, the Einstein field equations are

Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below. Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy". So one can argue that "gravitational energy" does NOT act as a source of gravity. On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory). So one can argue that "gravitational energy" IS a source of gravity."

Ignoring the GW issue, 'gravity does NOT gravitate' wrt the field of a massive object in GR. Which is an interesting example of 'an elephant in the living room' imo - it implies a failure of the equivalence principle, but is quietly ignored.

 

[atyy]

Ignoring the GW issue, 'gravity does NOT gravitate' wrt the field of a massive object in GR. Which is an interesting example of 'an elephant in the living room' imo - it implies a failure of the equivalence principle, but is quietly ignored.

Not to defend the EP as a principle, but how about the Nordtvedt effect experiments?

 

[atyy]

While it looks like scalar gravity could have a linear field equation if I'm reading Giulini's paper in post #3 correctly, the argument for metric gravity being nonlinear is given by http://arxiv.org/abs/gr-qc/0411023 .

 

[Q-reeus]

Not to defend the EP as a principle, but how about the Nordtvedt effect experiments?

Yes I agree there is no experimental evidence that 'gravitational field energy/mass' of strongly self-gravitating bodies behaves any differently to 'ordinary matter' wrt orbital dynamics eg http://arxiv.org/abs/astro-ph/0404270. So what's the answer? I don't know!
EDIT: Could be that negative Nordtvedt effect findings only requires m_p = m_i holds for 'gravity', while m_a 'goes missing'? Need to study that one.

 

[PAllen]

Not true in GR - from http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html :
"One other complaint about the pseudo-tensors deserves mention. Einstein argued that all energy has mass, and all mass acts gravitationally. Does "gravitational energy" itself act as a source of gravity? Now, the Einstein field equations are

Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below. Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy". So one can argue that "gravitational energy" does NOT act as a source of gravity. On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory). So one can argue that "gravitational energy" IS a source of gravity."

Ignoring the GW issue, 'gravity does NOT gravitate' wrt the field of a massive object in GR. Which is an interesting example of 'an elephant in the living room' imo - it implies a failure of the equivalence principle, but is quietly ignored.

There are two answers to this:

1) Almost any realistic stress energy tensor includes dependencies on g (for example, unit volumes). Thus g is on both sides, even though there is not an explicit term for gravitational energy.

2) The stress energy tensor and Einstein tensor are equated (bar noise factors). Thus, looking at an initial value problem, T encodes G, so the evolution of G depends on initial G, non-linearly.

The upshot is that, for reasonable definitions of mass (which is, of course, a complex topic in GR), the mass of a bunch of particles far apart will be greater than the mass of the same particles close together. In disguised form, gravitational energy does gravitate.

 

[Q-reeus]

...The upshot is that, for reasonable definitions of mass (which is, of course, a complex topic in GR), the mass of a bunch of particles far apart will be greater than the mass of the same particles close together. In disguised form, gravitational energy does gravitate.

I'm no GR buff, but just on that last point, is this any different in essence to Newtonian gravity? Assuming we are talking about the final 'condensed' system being in static equilibrium, is it not so that energy has been given off as heat etc during collapse, so simply invoking coe requires the remainder to have reduced mass. But this is obviously nothing peculiar to GR.

 

[Jonathan Scott]

I'm no GR buff, but just on that last point, is this any different in essence to Newtonian gravity? Assuming we are talking about the final 'condensed' system being in static equilibrium, is it not so that energy has been given off as heat etc during collapse, so simply invoking coe requires the remainder to have reduced mass. But this is obviously nothing peculiar to GR.

I agree; a condensed configuration of masses has less potential energy, but can only get from the expanded configuration to the condensed configuration if the potential energy goes somewhere. If the energy is removed from the configuration, it is that process which reduces the mass, so there is no role for "gravitational energy" except as a label for accounting purposes.

 

[PAllen]

I'm no GR buff, but just on that last point, is this any different in essence to Newtonian gravity? Assuming we are talking about the final 'condensed' system being in static equilibrium, is it not so that energy has been given off as heat etc during collapse, so simply invoking coe requires the remainder to have reduced mass. But this is obviously nothing peculiar to GR.

Newtonian mechanics has no mass/energy equivalence, nor is energy a source term for gravity. So, no, this is true in GR and related theories, not Newtonian gravity (unless you stick it in by adjusting masses based on binding energy).

The point was simply to show that lack of an explicit gravitational energy contribution to stress-energy tensor does not imply that gravitational energy does not gravitate.

 

[Jonathan Scott]

While it looks like scalar gravity could have a linear field equation if I'm reading Giulini's paper in post #3 correctly, the argument for metric gravity being nonlinear is given by http://arxiv.org/abs/gr-qc/0411023 .

Interesting paper, but not something I can immediately latch on to!

 

[Jonathan Scott]

Newtonian mechanics has no mass/energy equivalence, nor is energy a source term for gravity. So, no, this is true in GR and related theories, not Newtonian gravity (unless you stick it in by adjusting masses based on binding energy).

The point was simply to show that lack of an explicit gravitational energy contribution to stress-energy tensor does not imply that gravitational energy does not gravitate.

Yes, even though the source term for GR doesn't include gravitational energy (as I mentioned in the first paragraph of this thread) the PPN [itex]\beta[/itex] parameter is normally described loosely as the strength of the non-linearity in relativistic gravity theory, and is equal to 1 for GR.

 

[atyy]

Interesting paper, but not something I can immediately latch on to!

Actually, I've never read that. I learned about Deser's work in part from Straumann's http://arxiv.org/abs/astro-ph/0006423 , which describes both scalar and metric approaches starting from linear equations and then seeing how universal coupling to matter would work or not.

 

[Q-reeus]

Newtonian mechanics has no mass/energy equivalence, nor is energy a source term for gravity. So, no, this is true in GR and related theories, not Newtonian gravity (unless you stick it in by adjusting masses based on binding energy).

Sure in 'pure' Newtonian theory there is only mass as source term, but typically binding energy is taken into consideration - eg http://www.engineering.uco.edu/wwilson/Talks/Grav Field Energy.ppt

The point was simply to show that lack of an explicit gravitational energy contribution to stress-energy tensor does not imply that gravitational energy does not gravitate.

This issue is a perennial mindfield it seems, this past thread shows agreement even on the sign of gravitational field energy is still in dispute:
https://www.physicsforums.com/showthread.php?t=142298 :zzz:

 

[PAllen]

Sure in 'pure' Newtonian theory there is only mass as source term, but typically binding energy is taken into consideration - eg http://www.engineering.uco.edu/wwilson/Talks/Grav Field Energy.ppt

Of course, that's why I said you can 'put it in' by hand. However, it doesn't follow from Newon's law, and can't be incorporated without substantially changine Newton's law. Thus, the linearity of Newton's law really is because it doesn't encapsulate (in *any* way) the fact that gravitational energy is a source of gravity.

This issue is a perennial mindfield it seems, this past thread shows agreement even on the sign of gravitational field energy is still in dispute:
https://www.physicsforums.com/showthread.php?t=142298 :zzz:

I don't see that. The only knowledgeable commenters were Pervect and Chris Hillman. They both agreed. If you think they disagreed, you are mis-understanding.

 

[TrickyDicky]

If one wants a mathematical reason for "gravity energy" not acting as a source of curvature just have to look at the Einstein equation as the quote from Q-reeus does. I just don't know why people would want to be blind to the very equations of GR and the fact that the gravitational field can be coordinate-transformed away.
If one is more inclined to word explanations the famous Wheeler words are very clear:
It is "Matter tells spacetime how to curve", not "curvature tells spacetime how to curve, repeat this ad infinitum" in an eternal loop, otherwise spacetime wouldn't be able to ever "tell matter how to move".

 

[TrickyDicky]

And yet it is obvious by the existence of GW that gravitational energy gravitates, isn't it a nice puzzle guys?!

 

[PAllen]

If one wants a mathematical reason for "gravity energy" not acting as a source of curvature just have to look at the Einstein equation as the quote from Q-reeus does. I just don't know why people would want to be blind to the very equations of GR and the fact that the gravitational field can be coordinate-transformed away.
If one is more inclined to word explanations the famous Wheeler words are very clear:
It is "Matter tells spacetime how to curve", not "curvature tells spacetime how to curve, repeat this ad infinitum" in an eternal loop, otherwise spacetime wouldn't be able to ever "tell matter how to move".

The graviational field, g, appears on both sides of the equation for most any reasonable T. You have to look 'inside' the equations.

As for poetry, the complete poetry was "matter tells spacetime how to curve, spacetime tells matter how to move". This clearly leads to non-linearity and gravitational self-action: spacetime tells matter how to move, then matter tells spacetime how to curve, which tells matter how to move, which tells space time how to curve, ad infinitum. This the poetic formulation of saying the *equality* between stress energy and curvature enforces non-linear mutual dependence.

 

[PAllen]

And yet it is obvious by the existence of GW that gravitational energy gravitates, isn't it a nice puzzle guys?!

It is bit of a puzzle, but one that has been resolved thoroughly and long ago, in many sources.

 

[George Jones]

he only knowledgeable commenters were Pervect and Chris Hillman.

Actually, except for the original poster, all posters in the thread are extremely knowledgeable (real experts).

 

[PAllen]

Actually, except for the original poster, all posters in the thread are extremely knowledgeable (real experts).

Sorry, I just noticed those two.

 

[Q-reeus]

Originally Posted by Q-reeus View Post

This issue is a perennial mindfield it seems, this past thread shows agreement even on the sign of gravitational field energy is still in dispute:
https://www.physicsforums.com/showthread.php?t=142298

I don't see that. The only knowledgeable commenters were Pervect and Chris Hillman. They both agreed. If you think they disagreed, you are mis-understanding.

In the faint hope this could be a golden opportunity to clear up some confusion on this subject in general, let's take this point first. Reading through all the entries, there is afaik tellingly no definitive statement from Hillman et al re sign of gravitational field energy density itself. In fact pervect suggests avoiding the issue. Rather they discuss the 'positive energy theorem' (ie - source + field en toto always adds to a net positive energy, yes?), and the sign of gravitational binding energy (negative) - which leaves the question of the split between matter and field unresolved. So on that issue, consider one cubic meter of one Earth g 'space' right in front of you while sitting at your PC reading this. A simple situation with static observer and static gravitational source. Presumably we agree there is an energy density within this volume of 'vacuum' - but what is it's sign?

 

[TrickyDicky]

The graviational field, g, appears on both sides of the equation for most any reasonable T. You have to look 'inside' the equations.

As for poetry, the complete poetry was "matter tells spacetime how to curve, spacetime tells matter how to move". This clearly leads to non-linearity and gravitational self-action: spacetime tells matter how to move, then matter tells spacetime how to curve, which tells matter how to move, which tells space time how to curve, ad infinitum. This the poetic formulation of saying the *equality* between stress energy and curvature enforces non-linear mutual dependence.

I was only referring to the "gravitational self-action", and it clearly doesn't lead to it. Of course non-linearity is implicit here, but as I said you only need curved spacetime to get non-linearity.

It is bit of a puzzle, but one that has been resolved thoroughly and long ago, in many sources.

You must have looked very superficilally at the sources and left out many authors to get that impression, otherwise why would we be discussing it if it's been completely solved for years? At least for the OP is not completely solved it seems.

 

[PAllen]

In the faint hope this could be a golden opportunity to clear up some confusion on this subject in general, let's take this point first. Reading through all the entries, there is afaik tellingly no definitive statement from Hillman et al re sign of gravitational field energy density itself. In fact pervect suggests avoiding the issue. Rather they discuss the 'positive energy theorem' (ie - source + field en toto always adds to a net positive energy, yes?), and the sign of gravitational binding energy (negative) - which leaves the question of the split between matter and field unresolved. So on that issue, consider one cubic meter of one Earth g 'space' right in front of you while sitting at your PC reading this. A simple situation with static observer and static gravitational source. Presumably we agree there is an energy density within this volume of 'vacuum' - but what is it's sign?

The way I interpret that discussion and other sources is that fully resolved issues are:

1) gravitational binding energy is negative
2) Several positive energy theorems are proven (slightly different assumptions and generality)

It is unsolved how (or even whether it is meaningfully possible) to make statements about the energy content of a specific region of spacetime with no matter or E/M fields.

 

[Q-reeus]

The way I interpret that discussion and other sources is that fully resolved issues are:

1) gravitational binding energy is negative
2) Several positive energy theorems are proven (slightly different assumptions and generality)

It is unsolved how (or even whether it is meaningfully possible) to make statements about the energy content of a specific region of spacetime with no matter or E/M fields.

OK well thanks for that frank response. Evidently GR is not a simple topic!

 

[Jonathan Scott]

From the point of view of semi-Newtonian static approximations, there is a simple self-consistent model for the location of conserved gravitational energy as follows:

1. Placing a test object in a potential decreases its energy by the time dilation factor. It also decreases the energy of the source of the potential by exactly the same amount because of the potential due to the test object acting on the source.

2. There is an additional positive energy density of [itex](1/{8\pi G}) \, \mathbf{g}^2[/itex] where [itex]\mathbf{g}[/itex] is the field, like the Coulomb energy density in an electric field, and with the same sign. Integrated over all space, this gives a positive term equal to the potential energy in part 1, so the overall potential energy of the system comes out correct.

These two effects seem closely related to the "non-linear" illusion in my first post, except for irritating factors of 2.

 

[Q-reeus]

From the point of view of semi-Newtonian static approximations, there is a simple self-consistent model for the location of conserved gravitational energy as follows:

1. Placing a test object in a potential decreases its energy by the time dilation factor. It also decreases the energy of the source of the potential by exactly the same amount because of the potential due to the test object acting on the source.

2. There is an additional positive energy density of [itex](1/{8\pi G}) \, \mathbf{g}^2[/itex] where [itex]\mathbf{g}[/itex] is the field, like the Coulomb energy density in an electric field, and with the same sign. Integrated over all space, this gives a positive term equal to the potential energy in part 1, so the overall potential energy of the system comes out correct.

These two effects seem closely related to the "non-linear" illusion in my first post, except for irritating factors of 2.

This makes sense - the net field energy is positive and precisely half the net decrease in matter potential energy. Hence the system binding energy is equal and opposite to the total field energy - correct? Not sure if this is 'widely accepted' but then amazingly from a recent thread even whether a uniformly accelerated charge radiates is controversial after more than a century of debate! Anyway so the sign of field energy density is positive in both the static field and GW cases - the latter seemingly confirmed by match between Hulse-Taylor binary neutron star data and Einsteins quadrupole formula. A nice consistency here imo.
One more piece of this gravitational energy triangle is still problematic for me at least. It seems most cosmologists subscribe to the view the universe has zero total energy: http://en.wikipedia.org/wiki/Zero-energy_universe. Given the enormous positive energy tied up in matter+radiation+maybe dark energy, the assumption is 'gravity acts as a kind of negative energy' (prevalent and annoying use of vague language), exactly cancelling it all out to zero. So this ADFW (or similar acronym) universe is clearly non-static, but is that the 'magical ingredient' that allows a negative sign for gravitational energy in this case?

 

[Jonathan Scott]

This makes sense - the net field energy is positive and precisely half the net decrease in matter potential energy. Hence the system binding energy is equal and opposite to the total field energy - correct?

Yes, that's right.

The fact that this model works doesn't necessarily mean that the energy is "really" in the field in that way.

If anyone knows what the Landau-Lifgarbagez pseudotensor gives as the effective energy density in this simple case, I'd be interested to know. I don't have any symbolic maths software. I tried using pencil and paper on a couple of occasions but was confused by the conventions in the sources, and the answer typically came out 8 or 16 times more than I had been hoping. I know more about tensors now, but I have less patience!