# Stress-energy tensor in curved spacetime

1. Oct 20, 2006

### notknowing

In any textbook on relativity, one finds the classical expression for the stress-energy tensor of a perfect fluid. In generalizing this tensor to curved spacetime, one just replaced the flat-spacetime metric tensor by the metric tensor of curved spacetime. It seems logical to do so, but in my view there is no certainty that this is indeed the correct generalisation. Maybe the generalisation could be defined differently. So, how can we know for sure that we use the correct form of the stress-energy tensor in Einsteins Field equations ? Does there exist experimental evidence that we have indeed the correct tensor ? Also for the stress-energy tensor of the electromagnetic field, one could pose the same questions.

2. Oct 20, 2006

### pmb_phy

Why would you need to 'generalize' it? A tensor is a geometrical object which exists independantly of the presence of spacetime curvature. The definition of the tensor doesn't change when going from flat spacetime to curved spacetime. Can you please clarify this "generalizing" that you're refering to? Thanks.

Pete

3. Oct 20, 2006

### notknowing

I'm afraid it is not so simple as you indicated.
In the book Gravity by J. Hartle, one can read on page 480 : "There is some ambiguity but little difficulty in generalizing stress-energy tensors to curved spacetime". The way to generalize seems to be described by "the principal of minimal coupling" which is briefly described in wikipedia : "Also, under the principle of minimal coupling, the physical equations of special relativity can be turned into their general relativity counterparts by replacing the Minkowski metric (ηab) with the relevant metric of spacetime (gab) and by replacing any partial derivatives with covariant derivatives. "
Now, I'm not sure whether this "principal of minimal coupling" is to be understood as a recepy (which works most of the time), or as a strict mathematical principle. I found some other internet references where one mentions that there exist other approaches too, such as "conformal coupling", though I must admit that I don't understand this. In any case, there seems to be an ambiguity in the exact form of a stress-energy tensor in GR. If this is the case, why are certain stress-energy tensors (like the one of a perfect fluid) given as facts in most textbooks ? Could it just be wrong ?

4. Oct 20, 2006

### pervect

Staff Emeritus
I'm not sure what Hartle is getting at either.

I've seen the "minimal substitution rule" described as follows (Wald:70)

$$\eta_{ab} \rightarrow g_{ab} \hspace{.5 in} \partial_a \rightarrow \nabla_a$$

I.e. one replaces the Minkowski metric with the space-time metric, and ordinary partial derivatives with the covariant derviatve.

This process usually works, but not always, because covariant derivatives don't in general commute. The above process doesn't guarantee the correct ordering. I could dig up an example of where it fails if there is further interest, but I'm not sure I want to go that deep yet.

Unfortunately, if covariant differentiation has anything to do with the stress-energy tensor, I'm not aware of the connection.

It would be worthwhile looking at Hartle to see if his notion of "minimal substitution" is the same as Wald's and to see what he has to say about it to gain some illumination on what he's trying to say.

Onto the stress-energy tensor. I don't know what Hartle has in mind, but I'll talk about some generalities anyway, in the next post.

5. Oct 20, 2006

### pervect

Staff Emeritus
The stress-energy tensor

Any curved manifold can be seen as "locally flat". In GR, in a local region, we can always replace the metric gab with a Minkowskian metric $$\eta_{ab}$$. This ties in with the principle of "minimal substitution" as mentioned earlier, so hopefully it's the sense that Hartle hand in mind.

We can use the local Minkowskian metric to define the amount of energy and momentum (the energy-momentum 4-vector) contained in a small volume element at any given time by the rules of special relativity in this locally flat section of the manifold.

I'm not aware of any ambiguities here. The volume elements will be very small, so little "gotchas" regarding the relativity of simultaneity in a volume element shouldn't be an issue. The metric will naturally define a local coordinate system which can be used to give the specific components of the energy and momentum. This will in general not be an orthonormal coordinate system of course.

The remaining issues boil down to - "how do we naturally specify a volume element on a 4-d manifold"?

The standard way of doing this is to represent a volume in a 4-d manifold as a vector - a vector orthogonal to the volume element. Perhaps this is ambiguous, but I'm not aware of how. It can be justified a bit more formally in terms of geometric algebra.

Note: the following is optional and a bit advanaced, but I think it's interesting. If you are already familiar with vectors and their duals, one-forms, it can be quite illuminating. But the whole purpose of this section of the post is only to discuss in more detail how we can represent a volume element by the vector orthogonal to it.

In geometric algebra, one can represent a volume element as the interior region of a parallelpiped formed by three vectors. One can also think of it as the interior region of three one-forms. If you have MTW, you should have run across the visual image of one-forms as "stacks of plates'.

The idea of a volume element can be expressed as the geometric product, also known as the wedge product

http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html

of three one-forms. It's actually an oriented volume. This means that volumes can be represented as three-forms, because the wedge product of three one-forms is a three-form.

Rather than work with the three-forms, we work with their duals (this is the Hodges dual). There is a long complicated discussion in MTW that's not very intuitive, the geometric picture of a dual as described in

http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node8.html#SECTION00026000000000000000

is a lot easier. Every three-form has a unique "hodges dual', which is an ordinary one form. Thus the "natural" way to represent volumes is with a one form (or equivalently, a vector). This is the Hodges dual of the three form.

Note that there are two types of duality here - there is a duality between one forms and vectors, this is a different sort of duality than the Hodges duality, which is the duality between one-forms and three-forms.

So, in conclusion - the stress-energy tensor simply takes a vector-valued volume element and converets it into the local energy-momentum 4-vector. It can therefore be regarded as the "density" of energy and momentum.

6. Oct 20, 2006

### pmb_phy

Do you have a scanner? If so then please scan in that page and any other page which requires to be included for context. I find it hard to believe that Hartle is saying what you think he is. Thanks. You can either post the pages here or e-mail them to me.
That applies to tensor equations, not to the tensors themselves.
No ambiguity exists. A tensor can be defined as a multi-linear function of vectors and 1-forms to real numbers. This function does not depend on the components of the vectors/1-forms nor does it depend on the components of the tensor itself. You can evaluate the tensor in any system of coordinates to your liking. The locally Minkowski coordinates are usually the easiest since it has the same form as in flat time in globally Minkowski coordinates. As such the components of the tensor are well-defined making the tensor well-defined.

Best wishes

Pete

ps - I just e-mailed the author for clarification. Most authors are often willing to respond and clarify. Is this a text which is in depth (i.e. tensors, differential geometry, etc.)? I think I may pick up a copy.

Last edited: Oct 20, 2006
7. Oct 20, 2006

### pmb_phy

This is the response I got from the author
Now if someone can explain to me what he means when he said that .... :rofl:

Do you like this text? I'm thinking about picking up a new text on GR.

Best wishes

Pete

Last edited: Oct 20, 2006
8. Oct 20, 2006

### Stingray

It is useful to think of the decomposition of a stress-energy tensor into density and pressure terms as an eigenvalue problem. In particular, look at the solutions of
$$T^{a}{}_{b} v^{b} = \lambda_{v} v^{a}$$

There will be 4 mutually orthogonal eigenvectors. For reasonable matter, one will be timelike, and the others spacelike. Now the eigenvalue associated with the timelike eigenvector is just the (negative of) the density $\rho$. The eigenvectors associated with the three spacelike eigenvectors are called the principle pressures (or stresses). A perfect fluid is defined to be an object where all principle pressures are identical. Calling this $p$, it's easy to show that the only way of writing a stress-energy tensor with these properties is
$$T^{a}{}_{b} = \rho u^{a} u_{b} + p (\delta^{a}_{b} + u^{a} u_{b} )$$
Incidentally, the four-velocity $u^{a}$ is the timelike eigenvector.

Last edited: Oct 20, 2006
9. Oct 21, 2006

### notknowing

Thanks for contacting the author. I think it is very clear what he means (how could he explain it better ?). I'm curious about the opinion of the other experts (Pervect, ..)

10. Oct 21, 2006

### pervect

Staff Emeritus
I'm wondering about Hartle's approach. Is he perhaps deriving the stress-energy tensor from a Lagrangian density formalism? If he is, this might give some insight into his comments. There's nothing in particular in his resonse to indicate if he is doing this - but that would be one logical way to define energy and momentum.

The stress-energy tensor one gets via this procedure isn't unique, one can add a divergence term to it. I believe that this divergence term is usually used to make the tensor symmetric and gauge-invariant, which doesn't always happen automatically.

11. Oct 21, 2006

### pmb_phy

Yeah. I see what he means now.

Pete

12. Oct 21, 2006

Staff Emeritus
This is clear enough. In flat spacetime there are obviously no terms proportional to the local curvature. In any given coordinate system you will have a bunch of zeros there. But in curved spacetime - curved spacetime whise curvature is a function of this very tensor! - those zeros will become non-zero amounts that depend on the curvature. And you can't tell from the flat version of the tensor which zeros are going to do what.

But look, the tensor multiplication with the metric tensor is what goes through, just replacing Minkowski's diagonal metric with the fiull tensor of pseudo-Riemannian geometry. That will automatically fix the energy-momentum tensor.

13. Oct 21, 2006

### pmb_phy

We must have posted at the same time because I had already posted that I understood what the author meant. Thanks anyway selfadjoint.
Aplication of the metric tensor to the stress-energy-momentum tensor will not change the tensor itself. Only the components in a given coordinate system. It can change the type of tensor though and perhaps this was what you were thinking. But normally we assign the same name to the tensor A^ub as we do to A^u_v.

Best wishes

Pete

Last edited: Oct 21, 2006
14. Oct 21, 2006

### notknowing

I also guess that Hartle is working with the Lagrangian density formalism. After reading the various comments, I'm not sure what to think now. Is it the Lagrangian formalism only which leads to the ambiguity, while other approaches lead to a unique stress-energy tensor ?

15. Oct 21, 2006

### Stingray

Even if you just take the point of view I outlined (with eigenvalues/vectors), there can still be some ambiguity. I showed how to define the 4-velocity, density, principle pressures, etc. That's pretty much fixed.

But that may not completely specify the stress-energy tensor. For example, you might want to find the equations of motion of the system from $\nabla_{a} T^{ab}=0$. The perfect-fluid expression isn't quite sufficient for that. There are more variables that you need to solve for than there are equations. So you need to bring in an "equation of state" relating the pressure to the density. More complicated materials naturally need much more involved constitutive relations, and it may be natural to write these in terms of (say) the Riemann tensor, some potentials, etc.

You also might not want to decompose the stress-energy tensor into a density and pressures. For example, it would look a little odd to describe long-range fields in that way (despite that it's possible). The "natural variables" which characterize the system may be related to the stress-energy tensor in a complicated way.

In general, most people prefer to define the behavior of an object through a Lagrangian formalism. You get a different material for each action you write down, but there isn't necessarily a unique way of going from a flat-spacetime Lagrangian to a curved-spacetime one.

16. Oct 21, 2006

### samalkhaiat

Last edited: Oct 21, 2006
17. Oct 21, 2006

### Stingray

Minimal coupling is not the same as the equivalence principle. It says that the form of an equation that people like to write down in SR is essentially the same as the form of that equation in GR. In other words, the equations involve the curvature as little as possible. This is just a rule-of-thumb.

The equivalence principle is also a rule-of-thumb (there is no precise nontrivial definition which is known to be correct). But it has so much historical baggage associated with it that it means different things to almost everyone. Some of those are similar to minimal coupling, but many are not. It's better to use a term which everyone agrees on.

"Every non-inertial frame is locally equivalent to some gravitational field" is only true if you can ignore all derivatives of the metric of order 2 or higher. That's often possible, but it certainly isn't universal.

18. Oct 21, 2006

### pervect

Staff Emeritus
What motivates this decomposition? To me this equation is requiring that the energy-momentum 4-vector is a multiple of the 4-velocity? I suspect I'm missing the point here :-(.

19. Oct 21, 2006

### Stingray

$v^{a}$ is any eigenvector. So there are actually four of them in any decomposition. If you normalize their magnitudes, you might get a unique set. But for perfect fluids, there will always be a 2-parameter ambiguity representing spatial rotations.

Anyway, the point of the decomposition is to extract intuitive information from the stress-energy tensor. As you mentioned, in the case where $v^{a}=u^{a}$, the eigenvalue equation is basically stating that the momentum is a multiple of the velocity. The proportionality "constant" is usually called $\rho$. But if you don't know $u^{a}$ or $\rho$ beforehand, you can extract both of them uniquely from $T^{ab}$.

The spacelike eigenvectors represent the directions of principal stress. These are a carryover from the Newtonian treatment of (3x3) stress tensors. There, one often discusses the frame which diagonalizes the stress tensor. In other words, what frame removes all shear stresses (at a point)? The associated eigenvalues are the magnitudes of the stress in each of these directions.

In the end you get 4 unique and easily interpretable scalars. You also get 4 vectors, at least one of which is unique. Also, people sometimes state energy conditions in terms of the eigenvalues (e.g. Wald p. 219).

Last edited: Oct 21, 2006
20. Oct 26, 2006

### armandowww

In general relativity the spacetime is curved, so the Poincare' invariance fails because different points of spacetime are not equivalent.:surprised
In flat spacetime the energy tensor T_(ab) is defined as the conserved charges of the translation invariance, that's not valid in curved spacetime.
The general definiton of T_(ab) is simply the derivative of the matter part of the action S with respect to the metric g_(ab).

These words describe only a part of the question so...to be continued.

21. Oct 29, 2006

### samalkhaiat

Last edited: Oct 29, 2006
22. Oct 30, 2006

### Stingray

A classic textbook tells you that. And even there, it says that there are caveats. MTW is a very wishy-washy textbook. More serious ones devote much less (if any) space to the EP. Wald's mentions it mainly in historical context, for example. It isn't even in the index of Hawking and Ellis. Synge's book probably has my favorite quote on the subject, which is as follows:
The EP was of fundamental importance to Einstein, as it guided his thoughts to the right direction. But we have much better ways of thinking about things nowadays. It is mainly of historical interest at this point, although it can still be helpful as a rule-of-thumb. I don't know of any field of science besides relativity where people still try to cling to the methods of its founders as gospel. Einstein was certainly brilliant, but progress has been made.

Anyway, I'll agree that MTW's description of the EP is identical to minimal coupling. But it is not the only one floating around.

Hardly. You can write down theories of gravity on curved spacetimes which violate the EP in (say) the Newtonian limit. Just because the EP was the historical motivation for curved spacetime doesn't mean that it can't be useful if it were false.

Pauli's version of the principle is wrong, or at best a tautology. Consider a rotating particle. Even in the test-mass limit, such an object falling freely will have a worldline with a nonzero 4-acceleration. That doesn't happen in flat spacetime, so there is already a difference. Measure the stresses in a small solid material (via changes in proper volume, optical properties, etc). Put some weights and viscous fluid in a jar, and have a gravitational wave pass through. Measure the temperature change in the fluid due to those weights moving around. Let these experiments fall freely from different places and see if identical results are obtained. According to GR, that won't happen. But GR does satisfy the EP, doesn't it?

Huh? The metric can be made Minkowski at a point, and its first derivatives can be made to vanish at the same time. But that's it. You still have second and higher derivatives of the metric no matter what coordinates you choose. Those do directly couple to some physical processes. They also comprise what most relativists call the "gravitational field" (the Riemann tensor or occasionally the Weyl tensor). If the ability to transform the metric into a nice form is the "EP," that's really just a simple mathematical result on pseudo-Riemannian manifolds. Why not state it as such.

I never mentioned geodesics... But geodesic motion isn't universal in GR anyway. Of course it's an excellent approximation in the cases you mentioned (and many others).

Last edited: Oct 30, 2006
23. Nov 19, 2006

### Chris Hillman

Sounds like you already indirectly received a response from Hartle, so that you know that the short answer is: "the former".

Well, be careful of InterNet sources, including this one. Textbooks are more reliable. This might be the exception which proves the rule, since you obtained an answer to your question from Hartle! Anyway, I just wanted to point out that the principle of minimal coupling is discussed in section 16.3 of MTW. In particular, the connection with "factor ordering" arises because interchaning covariant derivatives introduces a term involving the Riemann tensor (in fact, this is taken as the definining property of the Riemann tensor in many textbooks). In Box 16.1, example1 should answer your original question, and example2 discusses the problem of determining the correct contribution from the EM field to the stress-energy tensor.

One place where curvature coupling can arise is in theories in which one postulates a field equation for a scalar field. In Hawking and Ellis you can find a discussion of the Lagrangian, the field equation, and the stress-energy contribution for a massless or massive scalar field. Although they don't bother to mention this, their example is minimally coupled. You can modify the Lagrangian for the massless scalar field to obtain a theory which is conformally invariant (like Maxwell's theory of EM) and this is called "conformal coupling" of the scalar field to the gravitational field. For example, this changes the field equation from the standard (curved spacetime) wave equation $$\Box h = 0$$ to $$\Box h = \frac{h}{6} \, R$$

Oh dear, I think you might have misunderstood MTW. In their customary helical approach, they say something naive in section 16.2 and immediately provide an important caveat in section 16.3. Their point, I think, was actually that in some situations, the EP should be supplemented by NON-minimal coupling--- but that this possibility is unlikely to arise except in Track 3! :-/

Chris Hillman

24. Nov 21, 2006

### notknowing

Hi Chris, thanks for all your comments. I noticed now that you commented on some previous topics as well (I still have to check out some ..)