# Stress-energy tensors in GR

samalkhaiat
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

vanhees71
samalkhaiat
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

samalkhaiat
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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vanhees71 and ShayanJ
Gold Member
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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atyy
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 $$\frac{ 1 }{ \sqrt{ - g } } \partial_{ \sigma } ( \sqrt{ - g } \ g^{ \sigma \mu } \ \delta^{ \rho }_{ \mu } ) = \frac{ 1 }{ \sqrt{ - g } } \partial_{ \sigma } \left( ( \sqrt{ - g } g^{ \sigma \mu } \ \partial_{ \mu } ) x^{ \rho } \right) = 0 .$$
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 possess (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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martinbn
(2) In http://arxiv.org/abs/0910.2975 Deser says "Another non-uniqueness pseudo-problem is that free gauge fields of spin > 1 cannot possess (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.

pervect
Staff Emeritus
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?

WannabeNewton
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.

2) Is this then further equivalent to the Bondi Mass...
Yes.

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.

WannabeNewton
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.

Jonathan Scott
Gold Member
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.

pervect
Staff Emeritus
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.

WannabeNewton
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

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.

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.

pervect
Staff Emeritus
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.

samalkhaiat
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

ShayanJ
samalkhaiat
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 possess (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

samalkhaiat
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?

Jonathan Scott
Gold Member
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.

Gold Member
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?

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!!!

samalkhaiat