# Stress-energy tensors in GR

1. Dec 12, 2014

### ShayanJ

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

2. Dec 12, 2014

### 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.

3. Dec 12, 2014

### Jonathan Scott

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.

4. Dec 12, 2014

### 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.

5. Dec 12, 2014

### ShayanJ

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!

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!!!
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!

6. Dec 12, 2014

### haushofer

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

7. Dec 12, 2014

### ShayanJ

And useful!;)
Thanks all!

8. Dec 12, 2014

### pervect

Staff Emeritus
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

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.

9. Dec 12, 2014

### Jonathan Scott

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.

10. Dec 12, 2014

### ShayanJ

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!

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.

Yeah, that makes sense!

Thanks people!

11. Dec 12, 2014

### 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.

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

12. Dec 12, 2014

### Staff: Mentor

In general, yes.

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.

13. Dec 12, 2014

### Staff: Mentor

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.

14. Dec 12, 2014

### Staff: Mentor

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.)

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.

15. Dec 12, 2014

### Staff: Mentor

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.

16. Dec 13, 2014

### Jonathan Scott

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.

17. Dec 13, 2014

### 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.

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.

18. Dec 13, 2014

### stevendaryl

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

19. Dec 13, 2014

### Jonathan Scott

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".

20. Dec 13, 2014

### stevendaryl

Staff Emeritus
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".