Where Can I Find a Detailed Explanation of Stress-Energy Tensors in GR?

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  • #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.
 
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  • #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 \partial_\rho(\sqrt{-g} \ t^\rho_{\ \sigma}) and I got:
<br /> \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<br />

\nabla_{ \mu } ( \sqrt{ - g } \ T^{ \mu }{}_{ \nu } ) = 0, implies \partial_{ \mu } ( \sqrt{ - g } \ T^{ \mu }{}_{ \nu } ) = \frac{ 1 }{ 2 } \sqrt{ - g } \ T^{ \rho \sigma } \ \partial_{ \nu } g_{ \rho \sigma } . \ \ \ \ (1)
Hence \partial_{ \mu } ( \sqrt{ - g } \ T^{ \mu }{}_{ 0 } ) = \frac{ 1 }{ 2 } \sqrt{ - g } \ T^{ \rho \sigma } \ \partial_{ 0 } g_{ \rho \sigma } .
Therefore, in a closed interacting system of matter and gravitational field, the energy of the matter component is conserved if \partial_{ 0 } g_{ \mu \nu } = 0 holds everywhere. Otherwise, the conservation of the total energy (matter plus gravitational fields) follows from,
\partial_{ \mu } \left( ( - g ) \ ( T^{ \mu \nu } + t^{ \mu \nu } ) \right) = 0 . \ \ \ \ ( 2 ) 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): \partial_{ \mu } \Theta^{ \mu }{}_{ \nu } = F^{ \mu }{}_{ \nu } \ j_{ \mu }. However, since GR is a nonlinear theory, it is not correct to associate T_{ \mu \nu} solely with matter and t_{ \mu \nu } with the gravitational field. This is because, the T is constructed out of the metric tensor g_{ \mu \nu } as well as the matter field, and the t 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 \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) for \nu = 0, 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 P^{ \nu } = \int_{ V } d^{ 3 } x \ ( - g ) \ ( T^{ 0 \nu } + t^{ 0 \nu } ) = \mbox{ const. } \ \ \ (4) Here is what Einstein said about this equation, “… the quantities P^{ \mu }, 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, P^{ \mu } 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 GL(4), 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 \frac{ 1 }{ 16 \pi G } \ \partial_{ \rho } h^{ \mu \nu \rho } = ( - g ) \ ( T^{ \mu \nu } + t^{ \mu \nu } ) , \ \ \ \ (5) where h^{ \mu \nu \rho } = - h^{ \mu \rho \nu }. Then, we may add/choose an appropriate super-potential to h^{ \mu \nu \rho } 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 h^{ \mu \nu \rho } = \partial_{ \sigma } \left( ( - g ) \ ( g^{ \mu \nu } \ g^{ \rho \sigma } - g^{ \mu \rho } \ g^{ \nu \sigma } ) \right) . \ \ \ \ (6) Inserting (5) in (4), we get 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 } , where h^{ \nu 0 0 } = 0 has been used in the last step. So, by the divergence theorem, we find P^{ \nu } = \frac{ 1 }{16 \pi G } \oint d S_{ i } \ h^{ \nu 0 i } . \ \ \ \ (7) 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 m = P^{ 0 } = \frac{ 1 }{16 \pi G } \oint d S_{ i } \ h^{ 0 0 i } . \ \ \ \ (8)Now, we will use (8) to solve an age-old mystery in physics. Using (6) in (8), we find
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) So, the inertial mass of a static, spherically symmetric object can be calculated using the Schwarzschild metric which has the following asymptotic behaviour g^{ 0 0 } \sim 1 + \frac{ 2 M G }{ r } ; \ \ \ g^{ i j } \sim - \delta^{ i j } ( 1 + \frac{ 2 M G }{ r } ) . Where, r^{ 2 } = x^{ 2 } + y^{ 2 } + z^{ 2 } and M is the GRAVITATIONAL MASS of the object. It is also called the heavy mass. Also note that
g^{ 0 0 } \ g^{ i j } = - \delta^{ i j } ( 1 + \frac{ 4 M G }{ r } ) , and \partial_{ j } ( g^{ 0 0 } \ g^{ i j } ) = 4 M G \frac{ x^{ i } }{ r^{ 3 } } . \ \ \ \ (10) Substituting (10) in (9), we find 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 . 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 r \to 0 about any singularities, \oint d S_{ i } h^{ 0 0 i } 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 \rho 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 \rho 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 \rho 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 \rho 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.
 
  • #71
  • #72
haushofer said:
So how well-defined is the energy density of a gravitational field in Newtonian gravity ??

It is only well-defined when we have a preferred Galilean frame (which is essentially what you said regarding the preferred space-like foliation).

If one formulates Newtonian gravity in such a way that no preferred Galilean frames exist, thus making ##\vec{\nabla}\varphi## gauge dependent, then one has the same problem as in GR of an ill-defined gravitational energy density.
 
  • #73
Also c.f. section 96 of "Classical Theory of Fields"-Landau Lifshitz for an extremely detailed calculation of the LL pseudotensor and the conserved currents of the conservation law ##\partial_{\mu}(T^{\mu\nu} + t^{\mu\nu}) = 0## in arbitrary spacetimes, which of course reduce to ADM currents in asymptotically flat spacetimes.
 
  • #74
WannabeNewton said:
Also c.f. section 96 of "Classical Theory of Fields"-Landau Lifshitz for an extremely detailed calculation of the LL pseudotensor ...
I originally bought that book specifically to try to understand that pseudotensor. It has never seemed quite right to me as it is clearly incompatible with the Newtonian approximation. I was very interested to see in the Lynden-Bell / Katz paper the following comment on it, which appears to support my suspicions:

D Lynden-Bell and J Katz said:
Landau and Lifshitz's expression in isotropic coordinates does not give the correct result. Even in a stationary metric, their expression denies the possibility of evaluating their "matter energy" before coordinates are chosen. This is because their expression has the wrong weight in ##\sqrt{-g}##.
 
  • #75
PAllen said:
Let me point again at:http://ptp.oxfordjournals.org/content/75/6/1351.full.pdf

Which argues that while coordinates don't affect observables, using De Donder gauge gives a consistent local interpretation of a pseudo-tensor as local stress/energy of the gravitational field.
I sometimes feel physics likes to play tricks on us!
 
  • #76
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 \rho 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?

1) There is no such thing as “mass of gravity”. Gravitational field is massless, spin-2 field.
2) The mass of an object at rest is determined entirely by the number of its constituent Nucleons. And the mass of each Nucleon is due (almost entirely) to QCD and QED effects. Since we do not yet have a consistent theory of QG, we cannot determined (if any) the gravitational contribution to the Nucleon mass. In principle, however, if the mass of the Nucleon is defined by, m = P^{ 0 } = \int d^{ 3 } x \ \Theta^{ 0 }{}_{ 0 } ( x ) , then it is reasonable that ALL of the 4 fundamental interactions contribute to the TOTAL SYMMETRIC energy-momentum TENSOR \Theta_{ \mu \nu }.
3) The “gravitational mass” (also called heavy mass) which I talked about, was an historical concept. You see, Newton had no reasons to equate the inertial mass of an object (the one which enters in F = m_{ I } a) with its gravitational mass (the one which appears in F = G \frac{ m_{ G } \ M_{ G } }{ r^{ 2 } }). We now know that the equality m_{ I } = m_{ G } is consistent only with LOCAL (CAUSAL) field theory based on the principle of equivalence (recall that Newton’s theory is an action-at-a-distance theory).

Sam
 
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  • #77
PAllen said:
Let me point again at:

http://ptp.oxfordjournals.org/content/75/6/1351.full.pdf
, using De Donder gauge gives a consistent local interpretation of a pseudo-tensor as local stress/energy of the gravitational field.
This is true for two reasons: (1) Nakanishi is very able and trusted expert in the field. (2) If you transform the Schwarzschild metric so that it satisfies the de Donder condition and then redo my calculation in post #62, you will arrive at his conclusion.

Sam
 
  • #78
Shyan said:
I sometimes feel physics likes to play tricks on us!

Physics tries to uncover the rules of the game in nature. GR is a theory of constraint system. Because of the contracted Bianchi identity \nabla_{ \nu } G^{ \mu \nu } = 0, only six out of 10 (field) equations are dynamical ones. The remaining four are constraints. In fact, the Einstein equations involve \ddot{ g }_{ i j } but not \ddot{ g }_{ 0 \rho }. Indeed, the invariance of the action under a general coordinate transformations shows that an action principle and with it the field equations cannot determine the metric tensor unless a coordinate system is specified in some non-covariant way.
So, in order to determine g_{ \mu \nu }, it is necessary to impose a coordinate condition, explicitly violating general coordinate invariance. This procedure is often called “gauge fixing”. However, this is an abuse of language: one must keep in mind that gauge fixing has no observable effects, while the choice of a coordinate system has physical meaning (detectable inertial forces depend on it). In fact, the properties of the solution of Einstein equation depend on the choice of a coordinate.
So, in order to make physically acceptable predictions, one must supplement the Einstein equation with a coordinate condition, itself regarded as a field equation. On theoretical grounds, the most natural coordinate condition seems to be the de Donder condition g^{ \mu \nu } \ \Gamma^{ \rho }_{ \mu \nu } = 0 , \ \ \ \ \ \ (1) which is also called harmonic coordinate condition: For any rank-2 tensor T^{ \mu \nu }, it is easy to show the following \nabla_{ \mu } T^{ \mu \nu } = \frac{ 1 }{ \sqrt{ - g } } \partial_{ \mu } ( \sqrt{ - g } \ T^{ \mu \nu } ) + \Gamma^{ \nu }_{ \rho \sigma } \ T^{ \rho \sigma } . So, taking T^{ \mu \nu } = g^{ \mu \nu }, and using \nabla_{ \mu } g^{ \mu \nu } = 0, we find g^{ \mu \nu } \ \Gamma^{ \rho }_{ \mu \nu } = - \frac{ 1 }{ \sqrt{ - g } } \ \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \rho } ) , and the de Donder condition, Eq(1), now reads \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \rho } ) = 0 . \ \ \ \ \ \ (2) This can be rewritten as \frac{ 1 }{ \sqrt{ - g } } \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \mu } \ \delta^{ \rho }_{ \mu } ) = \frac{ 1 }{ \sqrt{ - g } } \partial_{ \sigma } \left( ( \sqrt{ - g } g^{ \sigma \mu } \ \partial_{ \mu } ) x^{ \rho } \right) = 0 . Thus, the term “harmonic coordinate condition” is understood, because the operator ( - g )^{ - 1 / 2 } \ \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \mu } \ \partial_{ \mu } ) is nothing but the Riemannian form of the flat space d’Alembertian operator \eta^{ \rho \sigma } \partial_{ \rho } \partial_{ \sigma }.
The privileged nature of harmonic (de Donder) coordinate condition is explained nicely in Fock’s textbook:

(i) Note that the condition (1) is covariant under the transformation group GL(4). Indeed, in an arbitrary coordinate system \{ \bar{ x }^{ \mu } \}, one can show that \bar{ g }^{ \mu \nu } \ \bar{ \Gamma }^{ \sigma }_{ \mu \nu } = \frac{ \partial \bar{ x }^{ \sigma } }{ \partial x^{ \rho } } ( g^{ \mu \nu } \ \Gamma^{ \rho }_{ \mu \nu }) - g^{ \mu \nu } \frac{ \partial^{ 2 } \bar{ x }^{ \sigma } }{ \partial x^{ \mu } \partial x^{ \nu } }. \ \ \ (3) Under GL(4), the second term vanishes implying the covariant nature of the condition (1), i.e. the de Donder condition holds in all coordinate frames reached by the action of the group GL(4).

(ii) It is always possible to impose the de Donder condition. Indeed, suppose that g^{ \mu \nu } \Gamma^{ \rho }_{ \mu \nu } \neq 0. According to (3), we can always find coordinate system \bar{ x } such that \bar{ g }^{ \mu \nu } \bar{ \Gamma }^{ \sigma }_{ \mu \nu } = 0. This is done by solving the following 4 uncoupled second order PDE’s g^{ \mu \nu } \frac{ \partial^{ 2 } \bar{ x }^{ \sigma } }{ \partial x^{ \mu } \partial x^{ \nu } } = \frac{ \partial \bar{ x }^{ \sigma } }{ \partial x^{ \rho } } g^{ \mu \nu } \ \Gamma^{ \rho }_{ \mu \nu } . \ \ \ \ (4)

(iii) If the de Donder condition holds in a frame x, it also holds in all frames \bar{ x } satisfying g^{ \mu \nu } \frac{ \partial^{ 2 } \bar{ x }^{ \sigma } }{ \partial x^{ \mu } \partial x^{ \nu } } = 0 . \ \ \ \ \ (5) Indeed, from (3) and (4) we can show that (5) is a necessary and sufficient condition for the de Donder.

Finally, repeat the above argument replacing g^{ \mu \nu } by the energy-momentum tensor T^{ \mu \nu } or (equivalently from Einsten equation) by the Einstein tensor G^{ \mu \nu }. Such class of coordinate frames is called “non-rotating” frames. This class can be viewed as the INERTIAL FRAMES of curved spacetime. Recall that unlike the case in special relativity, in curved spacetime an inertial frame relative to an observer, is not necessarily inertial relative to another observer (however, it is non-rotating). You can easily see that in such frames, the matter energy-momentum tensor density satisfies GLOBALLY an ORDINARY conservation law \partial_{ \mu } ( \sqrt{ - g } T^{ \mu }{}_{ \nu } ) = 0. It follows from this equation that, for physical system confined to a spatial volume V, i.e. T_{ \mu \nu } = 0 outside V, the integrals P_{ \nu } = \int_{ V } d^{ 3 } x \ \sqrt{ - g } \ T^{ 0 }{}_{ \nu } , define 4 time-independent physical quantities and transform as a vector under the group of linear coordinate transformations.

Sam
 
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  • #79
So...coordinate systems are more than just mathematical tools? What is in physics that makes them more than just mathematical tools? I mean, they're only our language! Things should change when we speak in a different language. How can it be that when we just change the coordinates, phenomena change too?
I remember reading the wikipedia page about the hole argument. Einstein worried about the fact that a point has two different field values in two different coordinate systems. But the solution came along when he realized that such changes don't affect physical phenomena because events of spacetime have no meaning by themselves and only when e.g. there are particles moving in spacetime, the points acquire meaning by the positions of the particles. Its hard for me to think there is a difference here. Maybe using the de Donder gauge makes somethings easier but how in the earth, a different language predicts different things? The only escape route I see is attaching a physical meaning to g^{\mu \nu}\Gamma^\rho_{\mu\nu}=0,i.e. interpreting it as a special kind of reference frame. As I understood from Sam's post, for G^{\mu\nu}\Gamma^\rho_{\mu\nu}=0, the interpretation is that its a non-rotating frame(w.r.t. what? Because we can go to a coordinate system that rotates but cancels the inertial forces by gravitational forces!or not? and how? is there a demonstration somewhere?). But what about g^{\mu \nu}\Gamma^\rho_{\mu\nu}=0?
Please feel free to tell me my knowledge is not enough for this discussion. Actually I never finished a book on GR!(Which is my first priority in my to-do-list right now!)
 
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  • #80
samalkhaiat said:
On theoretical grounds, the most natural coordinate condition seems to be the de Donder condition g^{ \mu \nu } \ \Gamma^{ \rho }_{ \mu \nu } = 0 , \ \ \ \ \ \ (1) which is also called harmonic coordinate condition: For any rank-2 tensor T^{ \mu \nu }, it is easy to show the following \nabla_{ \mu } T^{ \mu \nu } = \frac{ 1 }{ \sqrt{ - g } } \partial_{ \mu } ( \sqrt{ - g } \ T^{ \mu \nu } ) + \Gamma^{ \nu }_{ \rho \sigma } \ T^{ \rho \sigma } . So, taking T^{ \mu \nu } = g^{ \mu \nu }, and using \nabla_{ \mu } g^{ \mu \nu } = 0, we find g^{ \mu \nu } \ \Gamma^{ \rho }_{ \mu \nu } = - \frac{ 1 }{ \sqrt{ - g } } \ \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \rho } ) , and the de Donder condition, Eq(1), now reads \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \rho } ) = 0 . \ \ \ \ \ \ (2) This can be rewritten as \frac{ 1 }{ \sqrt{ - g } } \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \mu } \ \delta^{ \rho }_{ \mu } ) = \frac{ 1 }{ \sqrt{ - g } } \partial_{ \sigma } \left( ( \sqrt{ - g } g^{ \sigma \mu } \ \partial_{ \mu } ) x^{ \rho } \right) = 0 .

samalkhaiat said:
This class can be viewed as the INERTIAL FRAMES of curved spacetime. Recall that unlike the case in special relativity, in curved spacetime an inertial frame relative to an observer, is not necessarily inertial relative to another observer (however, it is non-rotating). You can easily see that in such frames, the matter energy-momentum tensor density satisfies GLOBALLY an ORDINARY conservation law \partial_{ \mu } ( \sqrt{ - g } T^{ \mu }{}_{ \nu } ) = 0. It follows from this equation that, for physical system confined to a spatial volume V, i.e. T_{ \mu \nu } = 0 outside V, the integrals P_{ \nu } = \int_{ V } d^{ 3 } x \ \sqrt{ - g } \ T^{ 0 }{}_{ \nu } , define 4 time-independent physical quantities and transform as a vector under the group of linear coordinate transformations.

Since I have an expert to ask, I have two questions:

(1) Is de Donder gauge always applicable globally, eg. in cosmological solutions with cosmological constant?

(2) In http://arxiv.org/abs/0910.2975 Deser says "Another non-uniqueness pseudo-problem is that free gauge fields of spin > 1 cannot possesses (abelian) gauge-invariant stress tensors; this truism actually turns out to be a plus: only full GR recaptures the initial invariance, but now in non-abelian form, at the (satisfactory!) price of forfeiting any physical significance for its own stress-tensors, a fact also known as the equivalence principle." Is Deser saying that in GR there is a non-Abelian stress energy tensor? If so, what is a non-Abelian stress energy tensor?
 
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  • #81
atyy said:
(2) In http://arxiv.org/abs/0910.2975 Deser says "Another non-uniqueness pseudo-problem is that free gauge fields of spin > 1 cannot possesses (abelian) gauge-invariant stress tensors; this truism actually turns out to be a plus: only full GR recaptures the initial invariance, but now in non-abelian form, at the (satisfactory!) price of forfeiting any physical significance for its own stress-tensors, a fact also known as the equivalence principle." Is Deser saying that in GR there is a non-Abelian stress energy tensor? If so, what is a non-Abelian stress energy tensor?

Just from the way it is written it seems to me that he is not talking about a non-abelian tensor. He is talking about a gauge-invariant tensor, where the gauge is non-abelian. About the first question, I think it is in general only local.
 
  • #82
I've only skimmed this thread, but I'd like to see if I'm interpreting a few things correctly. And a question.

1) Is the use of the DeDonder gauge to define energy equivalent to the Landau (Lifshitz) psuedotensor approach?
2) Is this then further equivalent to the Bondi Mass, given the results in http://link.springer.com/article/10.1007/BF00762133#page-1 "A covariant formulation of the Landau-Lifschitz complex" Persedies and Papadopoulos 1979, assuming the space-time is such that both can be computed?
3) Are there circumstances where the Bondi mass can be computed and the LL pseudotensor mass cannot, or the Bondi mass cannot be computed by the LL pseudotensor mass can?
 
  • #83
pervect said:
1) Is the use of the DeDonder gauge to define energy equivalent to the Landau (Lifshitz) psuedotensor approach?

The LL formulation of GR is valid for any gauge; the DeDonder gauge (harmonic gauge) ##\partial_{\mu}\mathfrak{g}^{\mu\nu} = 0 \Leftrightarrow \partial_{\mu}t^{\mu\nu}_{\text{LL}} = 0## is but one such choice when working with the LL pseudotensor.

pervect said:
2) Is this then further equivalent to the Bondi Mass...

Yes.

pervect said:
3) Are there circumstances where the Bondi mass can be computed and the LL pseudotensor mass cannot, or the Bondi mass cannot be computed by the LL pseudotensor mass can?

The mass ##M(V) = \int_V d^3 x(-g)(T^{00} + t^{00}_{\text{LL}})## can always be computed in any space-time. The Bondi mass however requires asymptotic flatness.
 
  • #84
Jonathan Scott said:
It has never seemed quite right to me as it is clearly incompatible with the Newtonian approximation.

In what sense is it incompatible with the Newtonian approximation? The LL formulation is heavily used in Post-Newtonian theory.
 
  • #85
WannabeNewton said:
In what sense is it incompatible with the Newtonian approximation? The LL formulation is heavily used in Post-Newtonian theory.
Sorry, I have to admit I don't recall the specific details, but a few years ago when I added up the total LL field energy with the "matter energy" of the source, the total wasn't equal to the rest mass minus the potential energy. I thought perhaps I'd made a mistake in calculating the LL energy density and asked a friend of mine (a professor of physics at Southampton University) to check it; he agreed with my conclusion and found it puzzling, but didn't have time to investigate any further. In contrast, the density given by Lynden-Bell matches up exactly with the semi-Newtonian model.
 
  • #86
WannabeNewton said:
The LL formulation of GR is valid for any gauge; the DeDonder gauge (harmonic gauge) ##\partial_{\mu}\mathfrak{g}^{\mu\nu} = 0 \Leftrightarrow \partial_{\mu}t^{\mu\nu}_{\text{LL}} = 0## is but one such choice when working with the LL pseudotensor.

Hold up a bit - refreshing my memory from Wald (p 85), if you change the gauge the value of the LL pseudotensor changes - otherwise it'd be a real tensor, right?

Wald said:
Furthermore ##t_{ab}## isn't even gauge invariant, i.e. if we replace ##\gamma_{\alpha\beta}## by ..., then ##t_{ab}## does _not_ remain unchanged.

But under certain conditions, which seem to be related to some notion of asymptotic flatness, (I'm not sure if it's the same notion that is used for the Bondi mass), the total energy defined by integrating the LL pseudotensor is gauge invariant even though the pseudotensor itself is not.

If I am understanding the DeDonder approach correctly (and I'm not sure I am) , rather than demand gauge invariance, we fix the gauge instead.
 
  • #87
pervect said:
Hold up a bit - refreshing my memory from Wald (p 85), if you change the gauge the value of the LL pseudotensor changes - otherwise it'd be a real tensor, right?

Right but that doesn't mean the LL pseudotensor can only be used in the harmonic gauge. One can choose any gauge when working in the LL formulation. As long as we are only interested in the global charges obtained from the conserved currents of ##\partial_{\mu}(T^{\mu\nu} + t^{\mu\nu}_{\text{LL}}) = 0## the lack of gauge invariance is inconsequential, the choice of gauge is simply a matter of convenience

pervect said:
But under certain conditions, which seem to be related to some notion of asymptotic flatness, (I'm not sure if it's the same notion that is used for the Bondi mass), the total energy defined by integrating the LL pseudotensor is gauge invariant even though the pseudotensor itself is not.

Yes this is true. C.f. Wald exercise 4.7.

pervect said:
If I am understanding the DeDonder approach correctly (and I'm not sure I am) , rather than demand gauge invariance, we fix the gauge instead.

Yes because the harmonic gauge yields the relaxed Einstein equation, which is a wave equation, and makes order reduction in Post-Minkowskian theory much simpler to compute.
 
  • #88
WannabeNewton said:
Yes because the harmonic gauge yields the relaxed Einstein equation, which is a wave equation, and makes order reduction in Post-Minkowskian theory much simpler to compute.

Interesting - it looks to me like the relaxed wave equation, being a wave equation, has a sort of time translation symmetry, and it is this time translation symmetry that is associated (by Noether's theorem) with the notion of energy calculated by this sort of gauge fixing.
 
  • #89
Shyan said:
So...coordinate systems are more than just mathematical tools? What is in physics that makes them more than just mathematical tools? I mean, they're only our language! Things should change when we speak in a different language. How can it be that when we just change the coordinates, phenomena change too?
We would like the general formalism to be coordinates (language) independent, but we end up paying a price for doing that. Coordinates independence introduces, almost always, unphysical degrees of freedom which can be gauged away from the physical solutions. In ordinary gauge theories, choosing certain coordinate system (choosing a gauge) has no observable effects. However in GR, the quantities g_{ \mu \nu } have two meanings, being both the variables of the “gauge” field and the metric tensor. As a result of this physical and geometrical duality, a coordinate choice may have observable effects.
As I understood from Sam's post, for G^{\mu\nu}\Gamma^\rho_{\mu\nu}=0, the interpretation is that its a non-rotating frame(w.r.t. what? Because we can go to a coordinate system that rotates but cancels the inertial forces by gravitational forces!or not? and how? is there a demonstration somewhere?). But what about g^{\mu \nu}\Gamma^\rho_{\mu\nu}=0?
The name can be justified, but that requires both tedious and complicated mathematics. It simply means the following: any frame R rotating with respect to any member of the class K, which is defined by G^{ \mu \nu } \partial_{ \mu } \partial_{ \nu } \bar{ x }^{ \rho } = 0 , or equivalently T^{ \mu \nu } \partial_{ \mu } \partial_{ \nu } \bar{ x }^{ \rho } = 0 , \ \ \ \ \ (1) does not belong to the class K. As an example, consider a cloud of non-interacting dust particles. The energy-momentum tensor is given by T^{ \mu \nu } = \rho \ u^{ \mu } \ u^{ \nu } . \ \ \ \ \ \ (2) Using \nabla_{ \mu } T^{ \mu \nu } = 0, we can show that the dust particles follow the geodesic lines u^{ \nu } \nabla_{ \nu } u^{ \mu } = 0 . \ \ \ \ \ \ (3) Using (2), we rewrite (3) as u^{ \nu } \ \partial_{ \nu } u^{ \mu } + \frac{ 1 }{ \rho } \ T^{ \alpha \beta } \Gamma^{ \mu }_{ \alpha \beta } = 0 . Thus, it follows that a coordinate system is non-rotating if and only if u^{ \nu } \ \partial_{ \nu } u^{ \mu } = 0 . Clearly, a co-moving frame defined by u^{ \mu } = \delta^{ \mu }_{ 0 } , \ \ \ \ \ \ (4) is non-rotating. From this one particular frame, an arbitrary member of the non-rotating class K can be obtained by solving (1). Using (4) and (2), the defining equation (1) of the class K becomes \frac{ \partial^{ 2 } \bar{ x }^{ \mu } }{ \partial t^{ 2 } } = 0 . The most general solution of this equation is given by \bar{ x }^{ \mu } = x^{ 0 } \ A^{ \mu } ( x^{ j } ) + B^{ \mu } ( x^{ j } ) , where A^{ \mu } and B^{ \mu } are arbitrary functions of the spatial coordinates. Notice that the general linear group GL(4) is contained in the class K as a subgroup with A^{ \mu }’s are constants and B^{ \mu } are linear functions in x^{ j }.
Please feel free to tell me my knowledge is not enough for this discussion. Actually I never finished a book on GR!(Which is my first priority in my to-do-list right now!)
In this case, I suggest you postpone stepping into this treacherous and controversial territory until you finished one of the good textbooks on GR.
Sam
 
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  • #90
atyy said:
Since I have an expert to ask,
No, you don't. I don't regard myself as "expert". I just know few things.
I have two questions:
(1) Is de Donder gauge always applicable globally, eg. in cosmological solutions with cosmological constant?
This is one instance where ordinary language does not make sense. As stated, here the adverb “globally” is devoid of any precise mathematical meaning. What does “gauge fixing is applicable globally” actually mean? Is it a question about whether or not the de Donder condition is valid in a finite (as appose to infinitesimal) region in spacetime? Or the question is about whether or not all admissible spacetimes admit the de Donder condition? Well, in both cases the answer is yes. The condition can be imposed in a finite region of any conceivable spacetime, and the presence of cosmological constant has little to no effect on the answer.
(2) In http://arxiv.org/abs/0910.2975 Deser says "Another non-uniqueness pseudo-problem is that free gauge fields of spin > 1 cannot possesses (abelian) gauge-invariant stress tensors; this truism actually turns out to be a plus: only full GR recaptures the initial invariance, but now in non-abelian form, at the (satisfactory!) price of forfeiting any physical significance for its own stress-tensors, a fact also known as the equivalence principle."
I respect Deser and admire his contribution to our theoretical knowledge. But on this program (which is not connected to the issue of this thread) I have to disagree with his efforts. I believe the program (at least in ordinary gauge theories) has no merit for it raises more questions (about its consistency) than it can answer. I have never put any efforts to follow the details of the program in GR because similar program in ordinary gauge theories fails (except for su(2)) to reproduce the structure constants of the relevant gauge group. This means that a theory based on copies of U(1) gauge fields cannot be equivalent to self-interacting non-abelian gauge theory.
Is Deser saying that in GR there is a non-Abelian stress energy tensor? If so, what is a non-Abelian stress energy tensor.

He means that the energy-momentum tensor becomes a source for self-interacting rank-2 world tensor field. The word “world” here refers to the non-abelian nature of the group of arbitrary coordinate transformations.

Sam
 
  • #91
Jonathan Scott said:
Sorry, I have to admit I don't recall the specific details, but a few years ago when I added up the total LL field energy with the "matter energy" of the source, the total wasn't equal to the rest mass minus the potential energy. I thought perhaps I'd made a mistake in calculating the LL energy density and asked a friend of mine (a professor of physics at Southampton University) to check it; he agreed with my conclusion and found it puzzling, but didn't have time to investigate any further. In contrast, the density given by Lynden-Bell matches up exactly with the semi-Newtonian model.
You certainly have made a mistake. Can you reproduce your calculations in here?
 
  • #92
samalkhaiat said:
You certainly have made a mistake. Can you reproduce your calculations in here?
It's quite likely that I made a mistake, but that was some time ago and it's unlikely I kept notes for something that didn't work; I certainly don't have them in my file of interesting notes, although I may be able to find some of my correspondence on the subject. I seem to remember the effective field energy density being 7/2 times the square of the Newtonian field instead of 1/2, but that may have been when using a non-equivalent coordinate system.
 
  • #93
samalkhaiat said:
In this case, I suggest you postpone stepping into this treacherous and controversial territory until you finished one of the good textbooks on GR.

Actually I have some doubts here. Which book should I read? Ryder? Zee? Weinberg? Straumann? Carroll? MTW(just kidding!:D)?
I have the problem that since I know things about GR, I become bored on some sections. Also I want a book that covers advanced and exciting topics in a mathematically serious way. So I need a book that, in addition to being good, should be a bit advanced too. Can you suggest one?

samalkhaiat said:
No, you don't. I don't regard myself as "expert". I just know few things.
Its good to be in this forum and see people like you saying such a sentence. Because if I were to only look at the physics students around myself, I would do a really bad mistake in overestimating my level of knowledge!
 
  • #94
Shyan said:
Actually I have some doubts here. Which book should I read? Ryder? Zee? Weinberg? Straumann? Carroll? MTW(just kidding!:D)?
I have the problem that since I know things about GR, I become bored on some sections. Also I want a book that covers advanced and exciting topics in a mathematically serious way. So I need a book that, in addition to being good, should be a bit advanced too. Can you suggest one?
I can only suggest what I believe the golden rule in learning: Read the book that you understand and think it is nice. A book that one finds "good and nice" might not be as "good and nice" for others.
Good luck
 
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