# 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.