# Strees-energy-momentum tensor

1. Dec 30, 2012

### kevinferreira

Hello everyone,

I have some questions concerning the stress-energy-momentum tensor. I know it is a far more general object than the one used in GR, but I guess no one will disagree it plays a far more important role here.

So, firstly, what is the proof of its existence and its tensor properties (I guess I have the same problem with the electromagnetic tensor). Everytime I read something on this it seems that the tensor just falls from the sky to accomplish our wishes. Secondly, the usual way of writing this tensor, e.g. the 00 component being the relativistic mass density, etc., is it purely conventional or not?
Concerning GR, given that this tensor has different forms for different observers (but that's the whole point of the principle of general covariance), they will see differently the effects of the presence of massive bodies (for example an observer in uniform motion will measure a different momentum density). How does this difference is understood in GR?

Last edited: Dec 30, 2012
2. Dec 30, 2012

### friend

Those are good questions. One of the questions I have is whether the stress-energy-momentum tensor can always be put in the form of particles and interactions like photons, leptons and quarks and the energy associated with their speed. I assume that the stress tensor never involves gravitational energy, right?

3. Dec 30, 2012

### Altabeh

No it does not fall from sky. It is basically a tensor quantity that has conserved components following the fact that almost every physical theory must be invariant under a global set of transformations assigned to the Poincare group (translations in space and time and rotations and boosts). The conservation follows Noether theorem by which one can claim that any global symmetry in the theory in consideration has to associate the theory with a conserved quantity called "Noether charge". Note that Poincare group is global (not local).

It is sort of conventional. In some sense, you can define it in a way that the component 33 can indicate the energy density. But there are some subtleties involved. Sometimes it happens that your energy stress tensor is not symmetric (does not generally mean that it is not torsion free). Then the description of continuity equation, $∇_{\mu}T^{\mu\nu}=0$ becomes flawed as switching $\mu$ with ${\nu}$ is not allowed. (For example, there is a need for such symmetry to let the conservation of angular momentum hold.) Also one is not allowed to make the component 01 define encode density since it has to have a counterpart (10 component) which means the energy is now a vector quantity embedded in this tensor, again invalid. So for our best practical purposes, let it be defined the way it is.

Following principle of general covariance, all components in this tensor are only conserved locally in GR. So to all observers they look the same. Remember that energy-momentum tensor in GR does not always give you conserved charges as there may be interactions between gravitational field and matter which requires transfer of momentum and energy thus destroying their invariance under translations even locally. Instead, one uses some other object called "Landau–Lifgarbagez pseudotensor" to include all conservation laws at least locally to make sure that the theory has a valid description of conservation laws in it. Note also that in GR conservation laws only hold in asymptotically flat space-times which is by itself a big disadvantage for GR as a theory of gravity in generic curved spacetimes.

P

4. Dec 30, 2012

### Altabeh

You have stress tensor for pure Maxwell theory, remember?

5. Dec 30, 2012

### bcrowell

Staff Emeritus
I think this question is the easy part, and the second parenthetical in your quote shows that you already understand it pretty well.

This is a foundational question, and foundational questions are always tricky because the answer depends completely on what you take as your initial assumptions, i.e., your definitions and axioms. For instance, if I ask why the Pythagorean theorem is true, you could say it's true because it can be proved from Euclid's five postulates. But you could just as well take Cartesian geometry to define your set of starting assumptions, in which case the Pythagorean theorem would just be an axiom. Here's a relativistic example along these lines: https://www.physicsforums.com/showthread.php?t=534862 [Broken] .

In the case of the stress-energy tensor T, I could start by *defining* it as $T_{ab}=(1/8\pi)G_{ab}$. Then its existence and tensor properties are automatically established, since those have already been established for the Einstein tensor G. This would be exactly analogous to what we do in Newtonian mechanics when we define the active gravitational mass of a particle as $m_a=gr^2/G$, where g is the gravitational field it produces at a distance r. I can't think of any other way of defining ma -- can you? Note that both the Newtonian definition and the relativistic one are nontrivial, falsifiable statements. The Newtonian one could be falsified, for example, by any experiment that showed a deviation from the 1/r2 behavior of gravitational fields. The relativistic one could be falsified, for example, by any experiment that showed a nonvanishing value of G in a region of vacuum. (Actually, it *has* been falsified in this sense, because we now know we need to add a cosmological constant term.)

In this approach, what remains is to establish the familiar interpretation of T:

(1) $T^{ab}$ equals the flux of four-momentum pa across a surface of constant xb.

(2) In the Newtonian limit, let x0 be the universal time coordinate that everyone agrees on. Then $T^{00}$ is the mass density.

Interpretation 2 follows by taking the weak-field limit of the Schwarzschild metric and comparing it with the Newtonian definition of active gravitational mass. In other words, it follows from the correspondence principle, so any experiment that falsified it would be in conflict with the centuries' worth of observations that had already established the validity of Newtonian gravity within its own domain of applicability.

I would consider 1 to be a theoretical conjecture by Einstein, which was verified by experiment. That is, in the approach I've suggested, where the Einstein field equations are taken to be true by definition, the only way to falsify GR (or actually GR with $\Lambda=0$) is to falsify statement 1.

Statement 1 is highly theoretically plausible. That is, if experiments falsified 1, then I think it would be hard to salvage GR as a theory of four-dimensional spacetime in which matter determines curvature and curvature determines the geodesics along which material particles move. For example, if you believe in these basic ideas of GR, then energy-momentum has to be a four-vector, i.e., you can't have a scalar energy that acts as a source of gravity. Given these transformation properties of energy-momentum, you can, for example, transform dust from its rest frame to some other frame, and verify statement 1: http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html [Broken] (example 1). This verifies 1 in the case where the matter is dust, up to first order in the velocity relative to the dust's rest frame.

To justify 1 more generally, I think you have to start by observing that the Einstein field equation makes T divergenceless as a matter of geometrical definition. This is what allows you to interpret T as the flux of some quantity that is locally conserved.

As an alternative, Carroll gives a treatment in which statement 1 is taken as a definition, in which case the Einstein field equations are not simply true as a matter of definition: http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll1.html

Last edited by a moderator: May 6, 2017
6. Dec 30, 2012

### pervect

Staff Emeritus
One way I ran across recently for deriving the stress energy tensor formally for a swarm of particles is that it's the tensor product of the number-flux vector of the swarm and the energy-momentum 4-vector of each particle.

http://web.mit.edu/edbert/GR/gr2b.pdf has a little bit in the same vein, though the treatment I really liked was in google books (but I'm not sure where).

You might have to look up "number-flux 4 vector",or ask more about it if it's unfamiliar, but I'm going to wait for questions rather than try to volunteer an explanation.

Swarms of particles are not the most general model, but serve IMO as reasonable motivation for constructing the more general stress-energy tensor. The swarm of particle model doesn't handle fields, but there isn't any problem including them in the stress-energy tensor.

7. Dec 30, 2012

### friend

Well, if I'm right, the stress-energy-momentum tensor does not involve potentials or fields of the gravitational force itself. That would lead to an infinite series of corrections as the gravitational field contributed to the tensor which contributes to the curvature, which contributes to the tensor again and again. Therefore, I assume that that stress tensor is only due to the other 3 forces of the strong, weak, and electromagnetic forces. But these 3 forces and all their effects can be accounted for in quantum field theory. So my question is what is the stress-energy tensor in terms of the quantum field theory of these 3 forces and fields? Is that just a higher dimensional version of the energy in the fields? Then how is that become a tensor?

8. Dec 30, 2012

### jfy4

A side note not totally related to this question; infinite series of corrections are not necessarily a bad thing, if the series is convergent. If it isn't, as long as one pays attention to the radius of convergence it is still useful.

9. Dec 30, 2012

### kevinferreira

I understand this lack of conservation laws on a curved space, present on GR. And about this Landau-Lifgarbagez pseudo-tensor, does it express something only locally? If not, how does it relate to the curvature and the presence of momentum/mass/energy densities?

10. Dec 30, 2012

### kevinferreira

But I'm still pretty confused with something: if different observers express the energy-mom tensor differently, and Einstein's equations hold on each of them equally, they will also perceive differently the spacetime structure. It seems odd that this doesn't lead to contradictions, but on the other hand we're used to this kind of things in special rerlativity too... Also, can we include on this set of observers non-inertial observers?

I can see from the definition you gave above in terms of the Einstein tensor (that is simply Einstein's eqns, just as you did with Newton's) that T becomes divergenceless. But I don't understand that this "allows' us to interpret T as the flux of some locally conserved quantity, I think about this as being true only the other way around, that is, if you have some locally conserved quantity then you can build a divergenceless object. But then, from your explanation, I kinda see the 'intuition' behind this construction and what you said.
Also, I don't see why this divergenceless property is useful. You obtain 4 covariantly constant quantities (for each $\nu$ in $\nabla_{\mu}T^{\mu\nu}=0$), which means that they are Lorentz scalars even in a curved spacetime... So what?

11. Dec 31, 2012

### Staff: Mentor

Yes, you can include non-inertial observers and coordinate systems.

The key to understanding this, at least for me, was to recognize the difference between coordinate values and geometric (covariant or invariant) objects. Although different observers using different coordinate systems will express the geometric objects using different coordinate values, they are in fact talking about the same geometric objects. Because they are talking about the same geometric objects they never get contradictions, although their descriptions vary.

The same thing happens with ordinary vectors in flat spacetime. Alice can use Cartesian coordinates, Bob can use rotated Cartesian coordinates, and Chris can use spherical coordinates to describe a set vectors and their sums and products. As long as they all do their math correctly they will not get any contradictions, although they will use different numbers to describe the same things.

The only difficult thing is to realize that energy, by itself, is not one of these geometric objects. Similarly with momentum, etc.

12. Dec 31, 2012

### kevinferreira

Aah, now I see, I didn't think about that, but you're absolutely right, of course, this is the whole point of general covariance, of course.
Thanks!

13. Dec 31, 2012

### bcrowell

Staff Emeritus
Let's say you are able to make an accurate census each day of how many U.S. pennies exist in Canada and the U.S., and say the pennies always stay within those two countries. If, on a given day, the number of pennies in the U.S. increases by 738, you should be able to verify that the number in Canada changes by -738. The two numbers have to add up to zero, because pennies are conserved. You can infer that that is the net number of pennies that crossed the border during that 24-hour period. That is, you can interpret the 738 as a flux of pennies.

If you try to do it with people instead of pennies, it doesn't work. That's because people aren't conserved. They're born, and they die. The populations of the U.S. and Canada can change both because of births and deaths and because people are crossing the border. The change in population from day to day cannot be interpreted as a flux across the border.

The divergenceless property of T says that energy-momentum is (locally) conserved. This is closely analogous to the continuity equation for electromagnetism, which says that charge is conserved: $\nabla\cdot J = -\partial\rho/\partial t$, or $\nabla_a J^a=0$.

14. Dec 31, 2012

### atyy

It does fall from the sky if you have an action, and the action falls from the sky in the sense that you can take it as a definition of the theory. https://www.physicsforums.com/showpost.php?p=4213595&postcount=1

In specifying an action for general relativity, we of course put in the Einstein-Hilbert term. However, without specifying what matter there is - ie. the form of the enery-momentum tensor and the equations of motion for that matter - the Einstein equation is meaningless or "tautological" - like F=ma without saying what F is. One way of specifying these additional quantities needed to define GR is by specifying the matter action - eg. the action of the electrons, quarks etc (or if you don't need such detailed modelling, you can use less detailed matter such as a perfect fluid). By the principle of equivalence that SR is an excellent approximation locally, the matter action should couple to the metric and not its derivatives ("minimal coupling"), and the GR form should be the general covariant form of the SR form.

In GR, in arbitrary coordinates, the components don't have any special meaning. However, since in GR we can locally make coordinates to look like SR, then in those coordinates the local components have their SR meaning.

Last edited: Dec 31, 2012
15. Dec 31, 2012

### pervect

Staff Emeritus
GR, like Newtonian mechanics, can be derived as a Lagrangian theory.

Hence, knowing the Lagrangian density (or the action) of the matter fields is all that one needs to get the stress energy tensor.

I will assume people know what a Lagrangian density is (at least well enough) and not explain it further unless asked. (I'm sorry, but I don't know people's background here, I have to guess.)

Wald has a brief discussion of the process.

To devlop GR as a Lagrangian theory, first you start out without matter fields, and you find that the Einstein-Hilbert action is just sqrt(R), R being the Ricci scalar.

There are other, more complex, actions that people use, but this simple action is the basis of GR, anticipating the question of "why this action".

Applying variational principles one wishes to find the metric $g_{ab}$ that makes the action stationary. One gets the usual Euler-lagrange equatios from this, which basically say that G_ab = 0, where G is the Einstein Tensor.

To add matter into the theory, one writes $L_{tot} = L_{space} + \alpha L_{matter}$, where I've used L for the Lagrangian density.

Then applying variational principle as before, one gets $G_{ab}$ = (some constant) $T_{ab}$ where T_{ab} is naturally defined as the variation of the Lagrangian density with respect to the metric coefficeints, $g_{ab}$. As mentioned in some other thread, this is just

$T_{ab} = \frac{\delta L}{\delta g_{ab}}$

Last edited: Dec 31, 2012
16. Dec 31, 2012

### Naty1

Kevin: I'm not sure the discussion so far answers your real question: pervect and Crowell addressed the SET admirably.....I continue learn a lot from their posts.

If they don't exclaim 'idiot' or some such about this post and this is in fact of interest, I can provide some other details.

But I wonder if this interests you:

The answer to the discrepancy between the previous discussion and these statements is here:

From Misner,Thorne and Wheeler:

[That 'the amount of gravity produced by an object' is frame invariant is easy to prove:
Right now I am in some frame approaching the speed of light. Yet in my frame all is normal. Another way to say this: no matter how fast an object moves it will not become a black hole. ]

17. Dec 31, 2012

### pervect

Staff Emeritus
I would say the stress-energy tensor as a whole is frame invariant, this comes from its tensor nature - it's a geometric object.

Wiki seems to offer the same definition (though their discussion is nontelativistic, so the details don't strictly apply here). But they also say a quantity is frame invariant if it transforms correctly.

http://en.wikipedia.org/w/index.php?title=Objectivity_(frame_invariance)&oldid=524091607

The individual components of the stress-energy tensor are not frame-invariant, however.

This is similar to the way that the energy-momentum 4 vector is frame-invariant, but energy is not.

I don't know how one would define "the amount of gravity" produced by an object. If one looks at the residual velocities present after a high speed flyby, though, GR predicts that the residual velocities observed after a flyby will increase as the flyby velocity increases for an object of the same rest mass.

If measuring the velocites induced by a flyby is at all comparable to your idea of "amount of gravity", then the "amount of gravity" does depend on your velocity.

18. Dec 31, 2012

### kevinferreira

Yes, I see that, all objects used in Einstein equations (and the associated mathematical treatment) have an 'independet reality', just like a vector in euclidean space have a unique defining direction and magnitude no matter how we wish to represent it via our coordinates. So, I just have to think that the Riemann curvature tensor, the Ricci tensor, the metric tensor, etc. have a mathematically objective reality on spacetime to get to the conclusion with the help of Einstein's to see that the energy-momentum tensor shares the same properties.

But....
If I'm not mistaken, all these objects are defined locally, i.e. through their action on elements of the tangent space to a point (I take here the definition of tensor as a function acting on any number of cartesian products between the tangent and the cotangent spaces).
Therefore, even if we may define a tensorfield in every point of our spacetime as the collection of these definitions at each tangent space for each point, how can GR tell us something about large scale phenomena if all objects in our treatment are defined on tangent spaces, which are nothing else but 'nice' and local aproximations of our manifold?
Also, another thing that bothers me is that since locally the laws of special relativity hold, and we can 'transform away' gravity by just considering an accelerated frame (just as Einstein presented it in his 1916 paper), how is it that this locality and the locality used to define our tensors may in the end tell us something about the manifold itself?
This bothers me alot, we use the word 'locally' everywhere, but in the end we can actually tell something about spacetime itself.
Is it implicit in our definitions and mathematical treatment of the theory that what we actually do is the same as for example the way we would study the curvature of a graph, just observe the tangent lines at each point and then compare them in order to know how 'non-flat' is our graph? I think this might be somehow true, as the Riemann and Ricci tensors' components are calculated in some particular coordinates with the derivatives of the metric components...
And I also remember having done an exercise where we show that locally the metric is the Minkowski flat metric up to a second order term, related with the curvature tensors. But again, this 'locally' is not of the same kind as the previous ones, or is it?

I'm sorry I'm getting out of the topic, but I think all this has to do with the energy-momentum tensor, it's the other side of the equation!

19. Dec 31, 2012

### atyy

The first "locally" just means that the objects in the theory are tensor fields - they exist at every point on the manifold, and act on tangent vectors. (Actually, I've never seen this called "locally" until now.)

The second "locally the metric is the Minkowski flat metric up to a second order term" shows that the EP is true only at a point, and only up to first derivatives of the metric. It's the second derivatives that are crucial in distinguishing curved and flat spacetime. This is why in writing the matter action that Pervect and I talked about in posts 14 and 15, the matter fields must be coupled to the metric and not its derivatives. This prescription from the equivalence principle is also called "minimal coupling".

When you differentiate the full GR action LEH+Lmatter wrt the various fields, you then get the Einstein field equation with the correct definition of the stress tensor, and the equations of motion for the various matter fields. Because the equations came from an action in which matter was minimally coupled to the metric, the stress tensor derived in this way obeys the EP in the sense that in coordinates in which the metric (but not its second derivatives) have the same form as a Lorentz inertial frame, the components of the stress-energy tensor then have their SR meaning.

20. Dec 31, 2012

### Altabeh

Good question. Well, the general approach is based on a covariant formalism, providing one with global conservation laws in asymptotically flat spacetimes. For example, it can be used to give the mass of an isolated black-hole (which is asymptotically flat). A more recent and general approach can be found in arXiv:1202.1905.

LL pseudo-potential is related to curvature via Einstein field equations with some modifications. More technically, it is related with Ricci tensor density and scalar curvature density through an identity derived according to Noether's prescription for finding the conserved quantities of a Lagrangian which is a scalar density.

I hope this is enough. If you need more details, we can discuss it.

P