How do we know that the stress-energy is a tensor?

1. Mar 15, 2009

jdstokes

Einstein used the requirement of general covariance to motivate the field equations of general relativity.

Suppose I define $T^{\mu\nu}$ as the component of $\mu$-momentum across a surface of constant $\nu$, relative to some coordinate system $x^\mu$. If we change coordinates to $x'^\mu = x'^\mu(x)$, how do we know that the components of stress energy will look like

$T'^{\mu\nu} = \frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}T^{\alpha\beta}$.

I ask this because I've been thinking about how the stress-energy would look in a coordinate system other than cartesian (e.g., spherical coordinates) and it occurred to me that it is not obvious that it will transform in the desired way.

2. Mar 15, 2009

HallsofIvy

??? Any good book on general relativity will show that.

3. Mar 15, 2009

jdstokes

I have seen a lot of textbooks on GR, and I have never seen such a derivation. Can you name one which contains it?

I'm not talking about some abstract statement like the energy-momentum tensor is a multilinear map and is therefore coordinate invariant'. I'm looking for something along the lines of what Einstein was thinking.

4. Mar 15, 2009

atyy

The components of the stress-energy tensor only have their nice interpretation in a Cartesian (Lorentz) frame. However nabla.T=0 is a frame independent equation.

5. Mar 15, 2009

jdstokes

The stress-energy certainly transforms in the correct way under Lorentz transformations which you can see by examining the components of e.g., the electromagnetic stress-energy.

Are you saying that the tensor nature of the stress-energy follows from the requirement that it be covariantly conserved (i.e., that $\nabla_\mu T^{\mu\nu}$?

It seems like there should be a way of showing that it is a tensor, similar to the way that one would show that $A_\mu \equiv \partial A/\partial x^\mu$ is a tensor (where A is a scalar).

6. Mar 15, 2009

atyy

My understanding is that in electromagnetism the stress-energy tensor is defined using the vector potential. So the tensor property should follow from there.

Einstein's field equation alone doesn't give the stress-energy tensor meaning, except tautologically. There has to be a separate equation of state that defines the matter content of the universe (eg. Maxwell's equations, perfect fluid, etc.), and also a definition of the stress-energy tensor in terms of the state variables. The stress-energy tensor appropriate for each sort of matter has to be constructed so that it is symmetric and covariantly conserved. A Lagrangian is a way of getting an equation of state which automatically has an appropriate stress-energy tensor to go into Einstein's field equation.

Maybe look at section 3 of Andersson and Comer's http://relativity.livingreviews.org/Articles/lrr-2007-1/ [Broken].

Last edited by a moderator: May 4, 2017
7. Mar 15, 2009

jdstokes

Hi atyy.

In em, the stress-energy tensor is manifestly Lorentz covariant. It is not clear, however, whether or not it is generally coordinate invariant.

It is an interesting point you raise that the stress-energy loses the interpretation in terms of fluxes if we depart from a cartesian frame. In some sense it is meaningless to ask how the stress-energy transforms from cartesian to curvilinear coordinates, since we only know how to define it relative to a cartesian frame.

My feeling now is that general coordinate transformation properties of the stress-energy cannot be derived but must be postulated as an upgraded form of Lorentz covariance. Unless, of course, we define the stress energy in terms of tensors to begin with (as is done in modern treatments) in which case general covariance follows as a corollary.

8. Mar 16, 2009

Haelfix

Weinberg discusses this at length in his GR book and is probably the most nuts and bolts explanation you will find out there.

"My feeling now is that general coordinate transformation properties of the stress-energy cannot be derived but must be postulated as an upgraded form of Lorentz covariance."

Mmm, it depends a little on how you go about it, there are many possible and distinct definitions and ways to get at the object (which is a *pseudo*tensor the second you turn gravity on, so *not* generally covariant). You can sometimes single it out uniquely, as Weinberg shows when you use it as a source or if you get it from an action principle.

Last edited: Mar 16, 2009
9. Mar 16, 2009

jdstokes

Hi Haelfix,

Can you give a particular page/chapter reference? I have learned some GR from Weinberg but don't recall any discussion about the stress-energy tensor being a pseudo-tensor nor on the general covariance of it.

Why is the stress-energy not a tensor? Surely being the right-hand side of Einstein's equations has the same transformation properties as the Einstein tensor and is therefore generally covariant?

10. Mar 16, 2009

Haelfix

The matter/fluid/EM stress energy tensor by itself *is* usually a tensor (except for one possible sublety related to chiral ambiguities and signature issues). Eg the thing appearing on the right hand side of Einsteins eqns. But thats not the only way the object arises, and you have to be careful. In fact, even with the geometric interpretation of the field equation, coordinates appear in the components of the stress energy tensor. Its quite a nuissance to show everything is self consistent.

The problem is that while that of course transforms in the way you want, its not necessarily whats conserved either (say for energy conservation) or what may appear in generalized versions of Noethers law. If you want to include gravity as a source (say gravitational waves and their interactions) you either have to deal with the full nonlinear Einstein's equation directly or construct an object that also includes gravities self contribution. Thats where pseudo tensors/tensor densities often arise.

http://en.wikipedia.org/wiki/Stress-energy_tensor#Variant_definitions_of_stress-energy

or alternatively an advanced GR text like Weinberg or even Landau/Lifgarbagez.

Last edited: Mar 16, 2009
11. Mar 16, 2009

dx

It depends on what kind of source you're talking about. For example, if you have a gas of particles (treated as a continuum), you can deduce the transformation properties of the energy-momentum the fact that the particles are relativistic. Or if you have a classical relativistic field, then the energy-momentum tensor is defined in terms of the Lagrangian for the field and you can deduce its transformation properties from the definition. But the point is that whatever kind of source you have, if its a properly relativistic theory, the energy-momentum tensor will transform in the correct way. The theory of the material/matter/field source is not given by general relativity of course, so you can't prove anything about the energy-momentum tensor of the EM field for example from GR. This is essentially the same as the fact that Newton's theory of gravity can't prove that mass is a scalar.

Last edited: Mar 16, 2009
12. Mar 16, 2009

samalkhaiat

Yes there is a way. The functional derivative of a SCALAR functional with respect to a rank-2 TENSOR is another rank-2 TENSOR. In any field theory, the symmetrized canonical energy-momentum tensor coincides with the functional derivative of the scalar action with respect to the metric tensor,evaluated in flat space, i.e., it can be defined by the covariant statement;

$$T_{ab} \equiv 2 \frac{\delta S[\phi]}{\delta g^{ab}}$$

regards

sam

13. Mar 16, 2009

dx

Yes, but the fact that energy-momentum is the functional derivative of the scalar action with respect to the metric tensor in flat space is not a fact of general relativity, it is a fact of classical field theory. Also, the classical relativistic fields, although a very general set of things, do not encompass all possible sources.

Last edited: Mar 16, 2009
14. Mar 16, 2009

samalkhaiat

I think you misunderstood meaning of "evaluated in flat space". When you evaluate the above generally covariant expression in flat space, it gives you the energy-momentum tensor of the MATTER (including the electromagnetic) fields in flat space.

In GR the action is the sum of the actions of the gravitational field S[g] and matter field $S[\phi]$. To derive the equations of the gravitational field, we use the action principle

$$\delta \left( S[g] + S[\phi] \right) = 0$$

where

$$\delta S[g] = - \frac{1}{16 \pi k} \int d^{4}x G_{ab} \delta g^{ab} (-g)^{1/2}$$

and

$$\delta S[\phi] = \frac{1}{2} \int d^{4}x T_{ab} \delta g^{ab} (-g)^{1/2}$$

or

$$(-g)^{1/2}T_{ab} = 2 \frac{\delta S[\phi]}{\delta g^{ab}}$$

So, it is a "fact of general relativity". Indeed, the above covariant definition of the energy-momentum tensor was not even known before GR.

regards

sam

Last edited: Mar 16, 2009
15. Mar 23, 2009

jdstokes

Hi samalkhaiat,

Thanks for your reply. I sort of independently arrived at the same answer but it's good to have reassurance.

I knew it had to be something simple like this.

Don't you find it surprising that something as fundamental as this is not discussed in any of the textbooks? E.g., there is no mention of this in Landau, D'inverno, Weinberg, Carrol, MTW... They seem to prefer to define the stress-energy as a linear function which maps basis vectors to fluxes of 4-momentum.

Although I can sympathise with their definition, I do think that it obscures Einstein's original motivation for the field equations.

16. Mar 23, 2009

jdstokes

I doubt that this is true. If you allow the field to form delta functions then you should be able to construct any source you want.

17. Mar 23, 2009

George Jones

Staff Emeritus
Again, I recommend chapter 4, Lagrangian and Hamiltonian formulations of general relativity, from the book A Relativist's Toolkit: The Mathematics of Black-Hole Mechanics by Eric Poisson. Samalkhaiat's treatment is given on page 125.

18. Mar 28, 2009

samalkhaiat

19. Mar 28, 2009

jdstokes

What do you mean we cannot prove it'? I know that this identity definitely holds for the electromagnetic field since I have checked it. I suspect very strongly that the proof generalizes to arbitrary (nonabelian) vector fields, and probably also for scalars and fermions.

20. Mar 30, 2009

samalkhaiat

By "arbitrary field theory" I mean ANY field theory:

$$S[\phi] = \int d^{4}x \sqrt{-g} \ \mathcal{L}(\phi , \partial \phi )$$

where $\phi$ is a field of GENERAL (unspecified) nature, i.e., the explicit functional form of $\mathcal{L}$ is not given. For such ( arbitrary or model-independent) action integral, we cannot show that

$$2 \ \frac{\delta S}{\delta g_{ab}} = \sqrt{-g} T^{ab} \ \ \ (1)$$

follows from some physical principle! I find this fact very annoying.

Consider, for example, the invariance of S under the diffeomorphism $x \rightarrow x + \epsilon$;

$$\delta S = \int \ \delta (d^{4}x \sqrt{-g}) \ \mathcal{L} \ + \int d^{4}x \sqrt{-g} \ \delta \mathcal{L} = \int d^{4}x \sqrt{-g} \ \delta \mathcal{L}$$

(the first integral vanishes because ($\sqrt{-g}d^{4}x$) is a scalar). Now with some algebra on ($\delta \mathcal{L}$), we can show, after neglecting surface terms, that

$$\delta S[\phi] = \int d^{4}x \left( \frac{\delta S}{\delta g_{ab}} - \frac{1}{2} \sqrt{-g} T^{ab}\right) \ \delta g_{ab}$$

But the ten quantities $\delta g_{ab}$ are not all independent, because they were brought about by only four diffeomorphisms. Therefore, the invariance of the action under diffeomorphism DOES NOT leads to Eq(1). It does though justify regarding Eq(1) as an alternative "definition" for the symmetric energy momentum tensor.

regards

sam

Last edited: Mar 30, 2009
21. Mar 30, 2009

jdstokes

Interesting.

Personally, I don't find it annoying that there exist field theories which don't satisfy condition (1). I would regard such theories as pathological and use this condition as a criterion for selecting physically meaningful theories.