# Stress-energy tensors in GR

Gold Member
It is often stated that when one tries to find a stress-energy tensor of gravitational field in GR, the resulting quantity is zero because we can always make the metric zero at a point by a coordinate transformation. So there is no local measure of energy-momentum for gravitational fields. But I can't find a text which, in addition to just stating it, actually shows this. I mean, it may need some mathematical calculations Or maybe its not that much mathematical and only an application of equivalence principle will do. Whether the former or the latter, I wanna actually see the conclusion coming out from some detailed reasoning. Can anyone suggest a text?

Another question is, because the stress-energy tensor present in the EFEs is only for the matter(I mean, you know, sources!) present in the region, and because there will be some energy-momentum exchange between the matter and space-time, we should have $\partial_\mu T^{\mu \nu} \neq 0$. Is this right?
But then I see that actually $\nabla_\mu T^{\mu \nu}=0$, which seems strange to me.
If we say this is a conservation law for the stress-energy tensor, then this should be in contradiction with the above paragraph and also we should ask why making the derivative covariant makes a non-conserved quantity conserved, which seems non-sense to me! And it can't be true because making the derivative covariant still doesn't account for the gravitational energy!
But if we say this isn't a conservation law for stress-energy tensor, we should ask what's its meaning?(Actually I see it being called a conservation law!)
I'm really confused about it and can't find a way out. I need urgent help!!! Thanks

PAllen
I assume you are aware that EM radiation, and EM fields in general, appear in the stress energy tensor? Every text on GR I've seen has a section on how EM fields contribute to the stress energy tensor.

As for a proof about your first question, it's more of a mathematical construction. The rules for what goes T include all matter and fields, while gravitational waves are propagating changes in the Weyl curvature, which is the part of curvature after removing the Ricci curvature; the latter is what is directly related to T. If you tried to add Weyl curvature to T somehow, you would have to change the field equations, because it couldn't be represented in the Einstein tensor built from Ricci curvature.

Jonathan Scott
Gold Member
The concept of "gravitational energy" is extremely tricky! There are many ways of describing the "effective" gravitational energy, usually in terms of a "pseudotensor" in a specific frame, including for example Einstein's original one or the Landau-Liftshitz pseudotensor. In another thread a few weeks ago, I mentioned that I recently spotted a MNRAS paper "Gravitational field energy density for spheres and black holes" from 1985 by D Lynden-Bell and J Katz which takes the GR Schwarzschild solution (for the static spherical case) and concludes that the effective energy density of the field is ##g^2/8 \pi G## (positive) and the effective energy of the central mass is reduced by the time dilation (which as they point out reduces it by twice the potential energy).

As far as I know, the covariant divergence gives a conservation law for the flow of energy and momentum taking into account the shape of space-time as well as internal forces, which is essentially equivalent to including the effect of gravitational forces in the conservation laws.

martinbn
You cannot make the metric zero! You can make the connection coefficients zero at a point, equivalently you can make the metric that of Minkowski space-time. This isn't hard and you can find it in most books, but it is what the equivalence principle say.

• dextercioby
Gold Member
I assume you are aware that EM radiation, and EM fields in general, appear in the stress energy tensor? Every text on GR I've seen has a section on how EM fields contribute to the stress energy tensor.

As for a proof about your first question, it's more of a mathematical construction. The rules for what goes T include all matter and fields, while gravitational waves are propagating changes in the Weyl curvature, which is the part of curvature after removing the Ricci curvature; the latter is what is directly related to T. If you tried to add Weyl curvature to T somehow, you would have to change the field equations, because it couldn't be represented in the Einstein tensor built from Ricci curvature.
I don't mean that. I meant finding some $t^{\mu \nu}$ associated with the gravitational field so that we have $\partial_{\mu}(T^{\mu \nu}+t^{\mu \nu})=0$ and then showing that such a thing is troublesome as mentioned! In fact I wanna see those problems in work, as equations!

The concept of "gravitational energy" is extremely tricky! There are many ways of describing the "effective" gravitational energy, usually in terms of a "pseudotensor" in a specific frame, including for example Einstein's original one or the Landau-Liftshitz pseudotensor. In another thread a few weeks ago, I mentioned that I recently spotted a MNRAS paper "Gravitational field energy density for spheres and black holes" from 1985 by D Lynden-Bell and J Katz which takes the GR Schwarzschild solution (for the static spherical case) and concludes that the effective energy density of the field is ##g^2/8 \pi G## (positive) and the effective energy of the central mass is reduced by the time dilation (which as they point out reduces it by twice the potential energy).

As far as I know, the covariant divergence gives a conservation law for the flow of energy and momentum taking into account the shape of space-time as well as internal forces, which is essentially equivalent to including the effect of gravitational forces in the conservation laws.
I suddenly remembered a blog post by Sean Carroll! What I understand from his explanations is that because we don't have time translation symmetry in GR, then energy doesn't have to be conserved and the equation $\nabla_\mu T^{\mu \nu}=0$ is telling us how the energy is changing! Is that what you mean too? So technically this equation isn't a conservation law, right?
Otherwise this is really strange because we can't find a local measure for gravitational energy but suddenly, like a miracle, the energy becomes conserved locally! You know, its hard for me to see how covariant derivative can work here. And...I can't explain!!! God, maybe I should wait and learn more before tackling such problems!!!
You cannot make the metric zero! You can make the connection coefficients zero at a point, equivalently you can make the metric that of Minkowski space-time. This isn't hard and you can find it in most books, but it is what the equivalence principle say.
Yeah, sorry. But I actually wanted to see the kind of dependence on metric tensor that the proposed stress-energy tensors for gravitational field have and how it makes things troublesome!

haushofer
A quick and easy reply: gravity and strings by Ortin has some nice discussion on this :)

• dextercioby and ShayanJ
Gold Member
And useful!;)
Thanks all!

pervect
Staff Emeritus
It is often stated that when one tries to find a stress-energy tensor of gravitational field in GR, the resulting quantity is zero because we can always make the metric zero at a point by a coordinate transformation. So there is no local measure of energy-momentum for gravitational fields. But I can't find a text which, in addition to just stating it, actually shows this. I mean, it may need some mathematical calculations Or maybe its not that much mathematical and only an application of equivalence principle will do. Whether the former or the latter, I wanna actually see the conclusion coming out from some detailed reasoning. Can anyone suggest a text?

martinbn has provided such a good answer here that I will refer you to it, and only add that if you need a text, MTW's "Gravitation" discusses this issue. If you have acces to this text I'll try to track down a more specific page or chapter reference, the size of the book may make it difficult, but you can probably find a discussion in many texts if you look

Another question is, because the stress-energy tensor present in the EFEs is only for the matter(I mean, you know, sources!) present in the region, and because there will be some energy-momentum exchange between the matter and space-time, we should have $\partial_\mu T^{\mu \nu} \neq 0$. Is this right?

Nope, it's not right. Einstein's field equations give you ##\nabla_a T^{ab}=0##, because ##T^{ab}## is proportional to the Einstein tensor ##G^{ab}## and ##\nabla_a G^{ab}=0##.

This is closely related to the first point - when you find a local set of coordinates in which the Christoffel symbols aka the connection are zero, you conclude that the in those coordinates, there is no energy storage in "space-time". One argument (not very formal) is that the effects of curvature can be neglected in a sufficiently small region.. And there isn't any sensible way to regard the flat tangent space as having some sort of energy density other than zero.

The end result, which you will find in many texts, using both the above arguments and more sophisticated ones based on what sort of tensors one can construct and looking at all the candidates, is that there is no tensor expression for the idea of the "energy of space-time".

Some people have introduced non-tensor expressions, such as psuedotensors, to try to get around this. Other texts point out that a better interpretation of the lack of a tensor expression for the "energy of space time" is that the concept itself is suspect.

Given that the concept of the energy density of space-tie is suspect, we look at how it arises. It was introduced to try to find a globally conserved energy. So the conclusion is that the existence of a globally conserved energy is suspect.

I doubt I can dot all my t's and cross all my i's - ooops, I mean dot my I's an cross my t's - accurately enough to treat t his complex subject formally and accurately. So you'll have to go to the texts for that. But I hope this general overview helps you interpret what's going on.

Jonathan Scott
Gold Member
Otherwise this is really strange because we can't find a local measure for gravitational energy but suddenly, like a miracle, from somewhere we don't expect, covariant derivative comes and makes the energy conserved locally!

The covariant divergence doesn't explain where the gravitational forces come from, so it doesn't conserve total energy in that sense. It says that if you have some local energy and momentum then the flow and change in that energy and momentum depends on the internal forces and the shape of space-time. The internal stuff follows the usual conservation laws but the term due to the shape of space-time effectively describes external "forces" acting without any relationship to a supply of momentum or energy.

At least that's how I think it works.

Gold Member
Nope, it's not right. Einstein's field equations give you ∇aTab=0\nabla_a T^{ab}=0, because TabT^{ab} is proportional to the Einstein tensor GabG^{ab} and ∇aGab=0\nabla_a G^{ab}=0.
You're talking about the contraction of covariant derivative of the stress-energy tensor(Its covariant divergence). The one I said not equal to zero, was the divergence!

The end result, which you will find in many texts, using both the above arguments and more sophisticated ones based on what sort of tensors one can construct and looking at all the candidates, is that there is no tensor expression for the idea of the "energy of space-time".

Some people have introduced non-tensor expressions, such as psuedotensors, to try to get around this. Other texts point out that a better interpretation of the lack of a tensor expression for the "energy of space time" is that the concept itself is suspect.
That's it. I wanted to see those "more sophisticated ones". and all tensor candidates and their fall and the rise of pseudotensors and all that. The book mentioned by haushofer seems to have a complete account of this issue.

The covariant divergence doesn't explain where the gravitational forces come from, so it doesn't conserve total energy in that sense. It says that if you have some local energy and momentum then the flow and change in that energy and momentum depends on the internal forces and the shape of space-time. The internal stuff follows the usual conservation laws but the term due to the shape of space-time effectively describes external "forces" acting without any relationship to a supply of momentum or energy.

At least that's how I think it works.
Yeah, that makes sense!

Thanks people!

pervect
Staff Emeritus
I haven't seen the other treatment mentioned. The treatment of energy in GR I'm familiar with is in Wald, "General Relativity", section 11.2 pg 285 in my text. It has a quick overview of the difficulties, then focuses on the positive.

However, despite the absence of energy density of the gravitational field, there does exist a useful and meaningful definition of the total energy of an isolated system, i.e., more precisely, the total energy-momentum 4-vector present in an asymptotically flat space-time.

This would be more directly applicable to cosmology if the FLRW metric was asymptotically flat. But it isn't.

PeterDonis
Mentor
we should have ##\partial_\mu T^{\mu \nu} \neq 0## . Is this right?

In general, yes.

actually ##\nabla_\mu T^{\mu \nu}=0## , which seems strange to me.

Why? ##\nabla## is a different operator than ##\partial## . The ##\partial## operator only has a reasonable physical meaning in particular coordinates, and in particular spacetimes with symmetries that match those coordinates. The ##\nabla## operator has a coordinate-independent physical meaning. It's perfectly reasonable that ##\nabla## would obey a conservation law that ##\partial## does not.

PeterDonis
Mentor
the covariant divergence gives a conservation law for the flow of energy and momentum taking into account the shape of space-time as well as internal forces

This is not wrong, but I would put it differently. I would say that the covariant divergence gives a conservation law that, since it is independent of coordinates, can be given a direct physical meaning, because we can always express it in a local inertial frame in which the "shape of spacetime", locally, is just the ordinary flat spacetime of SR, so the conservation law has the same direct physical interpretation as it would in SR: no "stuff" is being created or destroyed in that infinitesimal patch of spacetime.

Another way of putting the above would be to note that, in a local inertial frame, the ##\nabla## operator is equivalent to the ##\partial## operator; but, unlike the ##\partial## operator, the ##\nabla## operator can be used to write equations that remain valid when we change coordinate charts.

• ShayanJ
PeterDonis
Mentor
because we don't have time translation symmetry in GR, then energy doesn't have to be conserved and the equation ##\nabla_\mu T^{\mu \nu}=0## is telling us how the energy is changing!

Yes. (More precisely, it's telling us that in spacetimes that don't have time translation symmetry. Some spacetimes do--for example, the spacetime around a gravitating body like the Earth which can be modeled as sufficiently isolated from other bodies. But other spacetimes don't--for example, the FRW spacetimes that we use to model the universe as a whole.)

So technically this equation isn't a conservation law, right?

It is, but what's conserved isn't energy; it's stress-energy, a more general concept that includes energy, momentum, pressure, and other stresses. The covariant divergence of the stress-energy tensor says that stress-energy can't be created or destroyed; it can only change form, so it could be energy at one point, momentum at another, pressure or stress at a third, and the changes in form have to happen in a particular, well-defined way.

PeterDonis
Mentor
the term due to the shape of space-time effectively describes external "forces" acting without any relationship to a supply of momentum or energy

I would put this differently as well. I would say that the term due to the "shape of spacetime" is there to ensure that we are counting the stress-energy properly. The components of the stress-energy tensor are "densities" of one sort or another--energy, momentum, etc. But a "density" of anything, to be counted correctly, has to be counted per unit of proper volume, not per unit of coordinate volume. The "shape of spacetime" term ensures that we do in fact count that way. Again, an easy way to see that is to look at the equation in a local inertial frame, where the "shape of spacetime" term disappears, and we can interpret coordinate volume as proper volume the way we do in SR.

• martinbn
Jonathan Scott
Gold Member
It is, but what's conserved isn't energy; it's stress-energy, a more general concept that includes energy, momentum, pressure, and other stresses. The covariant divergence of the stress-energy tensor says that stress-energy can't be created or destroyed; it can only change form, so it could be energy at one point, momentum at another, pressure or stress at a third, and the changes in form have to happen in a particular, well-defined way.

I don't know where you got that idea (I've seen it several times before) but it's not right.

The stress-energy tensor describes the density and flow (per area per time) of the energy and components of momentum.

The four components of the ordinary divergence correspond to energy, x-momentum, y-momentum and z-momentum, and express a separate equation of continuity for each of those four quantities, which are separately conserved locally.

As far as I know, the covariant divergence simply adds in the effect of gravitational acceleration on each of the four components.

stevendaryl
Staff Emeritus
The four components of the ordinary divergence correspond to energy, x-momentum, y-momentum and z-momentum, and express a separate equation of continuity for each of those four quantities, which are separately conserved locally.

As far as I know, the covariant divergence simply adds in the effect of gravitational acceleration on each of the four components.

I don't like putting it that way, because even in the absence of gravity, there's a difference between the covariant derivative and the partial derivative, if you are using non-rectangular coordinates.

For example, in 2D consider the constant vector field $\vec{A} = A \hat{x}$. Obviously, the divergence should be zero, so $\partial_\mu A^\mu = 0$. But now, if we switch to polar coordinates $r, \theta$, then this same vector field looks like this: $\vec{A} = A cos(\theta) \hat{r} + A sin(\theta) \hat{\theta}$. In polar coordinates, it will certainly not be the case that $\partial_\mu A^\mu = 0$.

The use of a covariant derivative (whether there is gravity or not) insures that $\nabla_\mu A^\mu$ is independent of which coordinate system you are using.

stevendaryl
Staff Emeritus
I don't like putting it that way, because even in the absence of gravity, there's a difference between the covariant derivative and the partial derivative, if you are using non-rectangular coordinates.

I guess you could say that the covariant derivative takes into account acceleration due to gravitational and inertial forces.

Jonathan Scott
Gold Member
I don't like putting it that way, because even in the absence of gravity, there's a difference between the covariant derivative and the partial derivative, if you are using non-rectangular coordinates.

OK, I accept that's true in the general case. But one way or another we are talking about conservation of the energy and momentum quantities whose density and flow are described by the tensor, not conservation of "stress-energy".

stevendaryl
Staff Emeritus
OK, I accept that's true in the general case. But one way or another we are talking about conservation of the energy and momentum quantities whose density and flow are described by the tensor, not conservation of "stress-energy".

I guess I'm not certain about the distinction, but I do agree that $T^{\mu \nu}$ is the rate of flow of $\mu$-momentum density in the $\nu$-direction. I don't know what "stress-energy" is, since I've never seen that phrase outside the phrase "stress-energy tensor".

PeterDonis
Mentor
The stress-energy tensor describes the density and flow (per area per time) of the energy and components of momentum.

Yes, but "area" and "time" are coordinate-dependent, so if you want to count density and flow correctly (i.e., invariantly), you have to compensate for that. As stevendaryl pointed out, this is true even in flat spacetime if you use non-rectilinear coordinates.

The four components of the ordinary divergence correspond to energy, x-momentum, y-momentum and z-momentum, and express a separate equation of continuity for each of those four quantities, which are separately conserved locally.

If by "locally" you mean "at a given event, as seen in a local inertial frame", then yes. Otherwise, no, because the ordinary divergence in non-Minkowski coordinates does not express any physical conservation law, because of the coordinate dependence.

As far as I know, the covariant divergence simply adds in the effect of gravitational acceleration on each of the four components.

That's one way of looking at it, but the very phrase "gravitational acceleration" should be a red flag, because in GR gravity is not a force and does not cause any acceleration.

The other way of looking at it is that the extra terms in the covariant divergence compensate for the coordinate dependence of the ordinary divergence, so that we are counting actual physical densities and flows, per unit of physical area and proper time instead of per unit of coordinate area and coordinate time.

You may not prefer the geometric viewpoint, but that doesn't mean it's not right.

Jonathan Scott
Gold Member
You may not prefer the geometric viewpoint, but that doesn't mean it's not right.

I have no problem with the geometric viewpoint. My specific point in the context of this thread was only to point out that the covariant divergence being zero means simply that locally energy and momentum components are each conserved separately, including the local effect of gravity. This was in response to the following statement which I consider to be incorrect:

It is, but what's conserved isn't energy; it's stress-energy, a more general concept that includes energy, momentum, pressure, and other stresses.

Also, when I first referred to "ordinary" divergence in this thread, I meant as for typical coordinates in Minkowski space, and it was intended to be an illustration rather than a precise physical statement. I guess I need to make such things clear, but I was really only trying to address the original post, not write a text book.

PeterDonis
Mentor
the covariant divergence being zero means simply that locally energy and momentum components are each conserved separately, including the local effect of gravity.

But that's not what the covariant divergence actually says, except in spacetimes that have particular symmetries. For example, if we take the covariant divergence of the stress-energy tensor used in cosmology (a perfect fluid in an expanding universe), the "energy" component does not say that energy is conserved; it says that energy changes in a particular way. That's what the blog post by Sean Carroll, that Shyan linked to earlier, was getting at. Similar remarks apply to the momentum components. (It's also worth remembering that "energy" and "momentum" are coordinate-dependent, so the "energy" that Carroll says is changing in a particular way is energy relative to the standard FRW coordinates; in a local inertial frame, the divergence being zero means energy, relative to that frame, is conserved, at least within that small patch of spacetime--but that's because that small patch of spacetime has approximate time translation symmetry, since we're modeling it as a small patch of Minkowski spacetime.)

This was in response to the following statement which I consider to be incorrect:

It probably wasn't the best way to put it, true (at least in the sense that it doesn't seem to have helped the OP's understanding). It's always hard to translate precise mathematical statements into vague ordinary language. My point was simply that "energy conservation" in the usual sense doesn't just depend on the covariant divergence being zero; it also depends on the spacetime in question having a particular symmetry (time translation symmetry). Similar remarks apply to momentum conservation (and angular momentum conservation, and other conservation laws). So if you want to interpret the covariant divergence being zero as "conservation" of something, it has to be something more general than "energy" or "momentum".

pervect
Staff Emeritus
If you consider the specialized case of a swarm of point particles that can interact when they touch, you can interpret the divergencelesness of the stress energy tensor as a continuity condition of the flow of momentum and energy carried by those particles.'

Thus the rate at which energy flows into a region of infinitesimal volume must equal the rate at which energy flows out of the infinitesimal volume. The same is true for the various components of the momentum. You get the stress-energy tensor by taking the tensor product of the particle flux 4-vector, and the energy momentum 4-vector. Setting the divergence of the particle-flux 4-vector to zero gives the continuity equation for the creation of particles - it ensures there are no particles created or destroyed in an infiintesimal region. Doing the same for the stress energy tensor gives the continuity equations for the momentum and energy carried by those particles.

However, as the sci.physics.faq points out, while the lack of divergence means you can say there is no energy created in an infinitesimal volume, you can't say the same thing about a finite volume. The divergence of the stress-energy tensor being zero isn't sufficient to guarantee that energy is conserved in a region of finite volume, it only guarantees that it's conserved in an infinitesimal volume.

• ShayanJ
Gold Member
I now get it guys and I really can't keep myself from saying it here(Also I feel I'm going to say it better than you all:p).
At first we should understand that when we're in curved spacetime(or using non-Cartesian coordinate systems), $\partial_\mu T^{\mu \nu}$ is meaningless and we shouldn't care about its value and it should be replaced by $\nabla_\mu T^{\mu \nu}$. But what's the meaning of this quantity being zero? Now this is the nice part! The little smart flea living in a small patch of spacetime, should conceive the spacetime in his\her vicinity as Minkowskian and so s\he shouldn't be able to find any evidence of the gravitational field s\he is in. But if s\he is able to measure a violation of stress-energy conservation in matter, and s\he is much smarter than other fleas, s\he will find out that s\he is located in a gravitational field but that violates equivalence principle and so we should have $\nabla_\mu T^{\mu \nu}=0$.
This also helps us to understand why there is no (and there shouldn't be any) stress-energy tensor associated to gravitational field. Because that would mean that gravitational energy is playing a part in the local things and that leads again to a violation of equivalence principle. So the point is not that people searched for a stress-energy tensor and didn't find any and they said so it doesn't exist. The point is that we know a priori, from equivalence principle, that such a thing should not exist and even if one claims s\he has found it, we shouldn't listen because we love equivalence principle much more than any stress-energy tensor of gravitational fields!
Thanks all of you people. This was really interesting.

samalkhaiat
It is often ......
But if we say this isn't a conservation law for stress-energy tensor, we should ask what's its meaning?(Actually I see it being called a conservation law!)
I'm really confused about it and can't find a way out. I need urgent help!!! Thanks

You can rewrite the equation $\nabla_{ \mu } T^{ \mu \nu } = 0$ in the form
$$\partial_{ \mu } \left( \sqrt{ - g } \ T^{ \mu }{}_{ \nu } \right) = \sqrt{ - g } \Gamma^{ \rho }_{ \nu \mu } \ T^{ \mu }{}_{ \rho } . \ \ \ \ (1)$$
Using the field equation, you can replace $T^{ \mu }{}_{ \rho }$ in the RHS by $( - 1 / k \ G^{ \mu }{}_{ \rho } )$. With some algebra, you can transform Eq(1) into a genuine conservation law
$$\partial_{ \mu } \left( \sqrt{ - g } \ ( T^{ \mu \nu } + \frac{ 1 }{ 2 k } t^{ \mu \nu } ) \right) = 0 , \ \ \ (2)$$ where $t^{ \mu \nu }$ is defined by $$\partial_{ \rho } ( \sqrt{ - g } \ t^{ \rho }{}_{ \sigma } ) \equiv \partial_{ \rho } \left( \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ \rho } g^{ \mu \nu } ) } \partial_{ \sigma } g^{ \mu \nu } - \delta^{ \rho }_{ \sigma } \mathcal{ L } \right) = - \sqrt{ - g } \ G_{ \mu \nu } \ \partial_{ \sigma } g^{ \mu \nu } ,$$ and $\mathcal{ L } ( g , \partial g )$ is the non-covariant part of the E-H scalar density $( \sqrt{ - g } R )$. Explicitly, it is given by $$\mathcal{ L } = \sqrt{ - g } \ g^{ \mu \nu } \ ( \Gamma^{ \sigma }_{ \rho \sigma } \ \Gamma^{ \rho }_{ \mu \nu } - \Gamma^{ \sigma }_{ \mu \rho } \ \Gamma^{ \rho }_{ \nu \sigma } ) .$$
Notice that Eq(2) holds in all reference frames, i.e., it is a generally covariant statement, even though $t^{ \mu \nu }$ is not a tensor. It only transforms as a tensor under linear transformations.
So, we can make the following observations: In a closed system of matter and gravitational field, the matter energy-momentum tensor is not conserved. On the other hand, while the object
$$\tau^{ \mu \nu } \equiv \sqrt{ - g } \ ( T^{ \mu \nu } + \frac{ 1 }{ 2 k } t^{ \mu \nu } ) ,$$ is conserved, it is not a tensor density.
Exactly the same thing occurs in Yang-Mills theories: the matter field current $j^{ a }_{ \mu }$ is gauge-invariant, but it is not conserved. On the other hand, the total current (matter plus gauge fields)
$$J^{ a }_{ \mu } = j^{ a }_{ \mu } + C^{ a }{}_{ b c } \ F^{ b }_{ \mu \nu } \ A^{ \mu c } ,$$ is conserved but not gauge-invariant.

Ok, let us go back to GR and integrate Eq(2) with appropriate boundary conditions and obtain the following time-independent object
$$\mathcal{ P }_{ \mu } = \int d^{ 3 } x \sqrt{ - g } \ ( T^{ 0 }{}_{ \mu } + \frac{ 1 }{ 2 k } t^{ 0 }{}_{ \mu } ) = \mbox{ const. } \ \ \ (3)$$
Inspired by Eq(2) and Eq(3), Einstein called $t_{ \mu \nu }$ the energy-momentum components of the gravitational field and $\mathcal{ P }_{ \mu }$ the total energy and momentum of the closed system. Back then this interpretation was problematic at first sight. In the final analysis, all difficulties stem from the fact that $t_{ \mu \nu }$ does not transform as a tensor. Since $t_{ \mu \nu }$ contains only first derivatives of the metric tensor $g_{ \mu \nu }$, it can be made to vanish at an arbitrary point by a suitable choice of the coordinates. In other words, it is always possible to find a class of frames of reference relative to which the gravitational field vanishes locally and, therefore, the $t_{ \mu \nu }$ vanishes locally too. On the other hand, in a perfectly flat spacetime it is possible to find a frame of reference relative to which we detect “inertial forces”. However, these inertial forces are locally equivalent to a gravitational field (principle of equivalence). Therefore, the components of $t_{ \mu \nu }$ do not vanish in a non-inertial coordinate system. Also, since $t_{ \mu \nu }$ is not symmetric, difficulties arise in defining conserved angular momentum. However, this deficiency is less disturbing. Indeed, dy adding appropriate “super-potential” an alternative symmetric pseudo-tensor can be defined (Landau & Lifshitz).
In spite of all these difficulties, it was hard to abandon the idea that an analogue to energy and momentum should exist. A final resolution was eventually made by Einstein (and subsequently completed by F. Klein). Einstein proved that the total 4-momentum $\mathcal{ P }_{ \mu }$ of a closed system (matter plus field) is, to large extent, independent of the choice of coordinate system, although (in general) the localization of energy will be different for different coordinate systems.
The spirit of Einstein’s proof is, again, similar to the Yang-Mills case: While the total current, $J^{ a }_{ \mu } = j^{ a }_{ \mu } + C^{ a }{}_{ b c } F^{ b }_{ \mu \nu } A^{ \nu c }$, has no simple gauge transformation properties, the integrated time-independent charge,
$$Q^{ a } = \int d^{ 3 } x \ J^{ a }_{ 0 } ( x ) ,$$ is gauge covariant with respect to all gauge transformations which tend to a definite (angle-independent) limit at spatial infinity. Similarly, in GR, the quantity $\mathcal{ P }_{ \mu }$ behaves as a 4-vector with respect to all coordinate transformations which approach Lorentz transformations at infinity.
Finally, I can summarize our current understanding as follow:
“It is impossible to localize energy and momentum in a gravitational field in a generally covariant and physically meaningful way, i.e., $t_{ \mu \nu }$ has no physical meaning. However, the integral expression, Eq(3), for the total energy-momentum 4-vector, $\mathcal{ P }_{ \mu }$, has a definite physical meaning”.

Sam

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• haushofer and ShayanJ
samalkhaiat
The covariant divergence doesn't explain where the gravitational forces come from,
At least that's how I think it works.
It does in the following sense. The expression $\nabla_{ \mu \nu } T^{ \mu \nu } = 0$ alone (i.e. with no extra assumptions) is sufficient to derive the gravitational force law.

Gold Member
Very nice, I was beginning to miss you and your nice mathematical treatments sam ! Thanks for posting.;)

stevendaryl
Staff Emeritus
I now get it guys and I really can't keep myself from saying it here(Also I feel I'm going to say it better than you all:p).
At first we should understand that when we're in curved spacetime(or using non-Cartesian coordinate systems), $\partial_\mu T^{\mu \nu}$ is meaningless and we shouldn't care about its value and it should be replaced by $\nabla_\mu T^{\mu \nu}$.

Yes, there is a sense in which $\partial_\mu T^{\mu \nu} = 0$ is not that physically meaningful. If it's true in one coordinate system, then it's not necessarily true in another coordinate system. However, it's sort of nice if it is true in one coordinate system, because in that coordinate system, you can define $P^\nu = \int T^{0 \nu} dV$ (where $dV$ is a volume integral over a spatial slice of spacetime), and the 4 numbers $P^\nu$ will be time-indendent in that coordinate system. The $P^\nu$ won't actually be a 4-vector, because it's only meaningful in that one coordinate system.

Gold Member
You can rewrite the equation $\nabla_{ \mu } T^{ \mu \nu } = 0$ in the form
$$\partial_{ \mu } \left( \sqrt{ - g } \ T^{ \mu }{}_{ \nu } \right) = \sqrt{ - g } \Gamma^{ \rho }_{ \nu \mu } \ T^{ \mu }{}_{ \rho } . \ \ \ \ (1)$$
Using the field equation, you can replace $T^{ \mu }{}_{ \rho }$ in the RHS by $( - 1 / k \ G^{ \mu }{}_{ \rho } )$. With some algebra, you can transform Eq(1) into a genuine conservation law
$$\partial_{ \mu } \left( \sqrt{ - g } \ ( T^{ \mu \nu } + \frac{ 1 }{ 2 k } t^{ \mu \nu } ) \right) = 0 , \ \ \ (2)$$ where $t^{ \mu \nu }$ is defined by $$\partial_{ \rho } ( \sqrt{ - g } \ t^{ \rho }{}_{ \sigma } ) \equiv \partial_{ \rho } \left( \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ \rho } g^{ \mu \nu } ) } \partial_{ \sigma } g^{ \mu \nu } - \delta^{ \rho }_{ \sigma } \mathcal{ L } \right) = - \sqrt{ - g } \ G_{ \mu \nu } \ \partial_{ \sigma } g^{ \mu \nu } ,$$ and $\mathcal{ L } ( g , \partial g )$ is the non-covariant part of the E-H scalar density $( \sqrt{ - g } R )$. Explicitly, it is given by $$\mathcal{ L } = \sqrt{ - g } \ g^{ \mu \nu } \ ( \Gamma^{ \sigma }_{ \rho \sigma } \ \Gamma^{ \rho }_{ \mu \nu } - \Gamma^{ \sigma }_{ \mu \rho } \ \Gamma^{ \rho }_{ \nu \sigma } ) .$$
Notice that Eq(2) holds in all reference frames, i.e., it is a generally covariant statement, even though $t^{ \mu \nu }$ is not a tensor. It only transforms as a tensor under linear transformations.
So, we can make the following observations: In a closed system of matter and gravitational field, the matter energy-momentum tensor is not conserved. On the other hand, while the object
$$\tau^{ \mu \nu } \equiv \sqrt{ - g } \ ( T^{ \mu \nu } + \frac{ 1 }{ 2 k } t^{ \mu \nu } ) ,$$ is conserved, it is not a tensor density.
Exactly the same thing occurs in Yang-Mills theories: the matter field current $j^{ a }_{ \mu }$ is gauge-invariant, but it is not conserved. On the other hand, the total current (matter plus gauge fields)
$$J^{ a }_{ \mu } = j^{ a }_{ \mu } + C^{ a }{}_{ b c } \ F^{ b }_{ \mu \nu } \ A^{ \mu c } ,$$ is conserved but not gauge-invariant.

Ok, let us go back to GR and integrate Eq(2) with appropriate boundary conditions and obtain the following time-independent object
$$\mathcal{ P }_{ \mu } = \int d^{ 3 } x \sqrt{ - g } \ ( T^{ 0 }{}_{ \mu } + \frac{ 1 }{ 2 k } t^{ 0 }{}_{ \mu } ) = \mbox{ const. } \ \ \ (3)$$
Inspired by Eq(2) and Eq(3), Einstein called $t_{ \mu \nu }$ the energy-momentum components of the gravitational field and $\mathcal{ P }_{ \mu }$ the total energy and momentum of the closed system. Back then this interpretation was problematic at first sight. In the final analysis, all difficulties stem from the fact that $t_{ \mu \nu }$ does not transform as a tensor. Since $t_{ \mu \nu }$ contains only first derivatives of the metric tensor $g_{ \mu \nu }$, it can be made to vanish at an arbitrary point by a suitable choice of the coordinates. In other words, it is always possible to find a class of frames of reference relative to which the gravitational field vanishes locally and, therefore, the $t_{ \mu \nu }$ vanishes locally too. On the other hand, in a perfectly flat spacetime it is possible to find a frame of reference relative to which we detect “inertial forces”. However, these inertial forces are locally equivalent to a gravitational field (principle of equivalence). Therefore, the components of $t_{ \mu \nu }$ do not vanish in a non-inertial coordinate system. Also, since $t_{ \mu \nu }$ is not symmetric, difficulties arise in defining conserved angular momentum. However, this deficiency is less disturbing. Indeed, dy adding appropriate “super-potential” an alternative symmetric pseudo-tensor can be defined (Landau & Lifshitz).
In spite of all these difficulties, it was hard to abandon the idea that an analogue to energy and momentum should exist. A final resolution was eventually made by Einstein (and subsequently completed by F. Klein). Einstein proved that the total 4-momentum $\mathcal{ P }_{ \mu }$ of a closed system (matter plus field) is, to large extent, independent of the choice of coordinate system, although (in general) the localization of energy will be different for different coordinate systems.
The spirit of Einstein’s proof is, again, similar to the Yang-Mills case: While the total current, $J^{ a }_{ \mu } = j^{ a }_{ \mu } + C^{ a }{}_{ b c } F^{ b }_{ \mu \nu } A^{ \nu c }$, has no simple gauge transformation properties, the integrated time-independent charge,
$$Q^{ a } = \int d^{ 3 } x \ J^{ a }_{ 0 } ( x ) ,$$ is gauge covariant with respect to all gauge transformations which tend to a definite (angle-independent) limit at spatial infinity. Similarly, in GR, the quantity $\mathcal{ P }_{ \mu }$ behaves as a 4-vector with respect to all coordinate transformations which approach Lorentz transformations at infinity.
Finally, I can summarize our current understanding as follow:
“It is impossible to localize energy and momentum in a gravitational field in a generally covariant and physically meaningful way, i.e., $t_{ \mu \nu }$ has no physical meaning. However, the integral expression, Eq(3), for the total energy-momentum 4-vector, $\mathcal{ P }_{ \mu }$, has a definite physical meaning”.

Sam

As mentioned by Sean Caroll in this blog post, the fact that we don't have time translation in some spacetimes is involved here too. Can you say how your calculations and reasonings change if we assume that the spacetime we're talking about is time translation invariant? Or more general how can you enter time translation symmetry here?
Thanks