# Gravitational energy-momentum tensor

1. Jan 11, 2013

### Worldline

why general relativity can't define any tensorial expression for Gravitational energy momentum density ?

2. Jan 11, 2013

### kevinferreira

That's a good question, I think. Any tensorial quantity has a local definition only (it is defined on the tangent space at one point in spacetime), and the energy-momentum-stress tensor is locally conserved, so it should be comprised on it.

3. Jan 11, 2013

### Worldline

but other fundamental fields have well defined tensorial expression. I think it is because in GR , we consider gravitational effects as inertial effects, and inertial effects can't be expressed with tensors.....

4. Jan 11, 2013

### bcrowell

Staff Emeritus
Wald has a good discussion of this in section 11.2.

The Newtonian gravitational energy density goes like the square of the gravitational field. But the equivalence principle guarantees that locally, there are always coordinates such that the gravitational field, and therefore the Newtonian gravitational energy density, is zero. A tensor that's zero in one set of coordinates is zero in all other coordinates, so we can't have a gravitational energy density that's a tensor and that reduces to the Newtonian version in the appropriate limit, unless it vanishes identically.

Another way of putting this is that the Newtonian expression is basically the square of the gradient of the time-time component of the metric. To make this into a tensor, we'd have to do something involving a covariant derivative of the metric. But the covariant derivative is defined exactly so that it gives zero when it acts on the metric.

Yet another way of getting at this is that you have to think about definitional issues. The mass-energy density of, say, hydrogen gas is something that I can measure locally with laboratory instruments. But the energy density of a gravitational wave isn't something we can measure locally. It's not possible to measure it, even in principle, because to measure a form of energy you have to be able to exchange it for some other form of energy. But the zero-divergence property of the stress-energy tensor says that the things that we ordinarily consider to be forms of mass-energy, i.e., the mass-energy tied up in matter fields, *can't* be locally traded in for anything else.

All of these arguments fail when you try to apply them non-locally, and that's we we can observe mechanical energy being lost from the Hulse-Taylor pulsar into gravitational waves.

5. Jan 11, 2013

### pervect

Staff Emeritus
Hilbert was among the first one to ask that question. He noticed what he called "the failure of the energy theorem"..

Hilbert asked for some help from Emily Noether to clarify this problem, and the result was Noether's theorem, which relates energy conservation to time translation symmetry.

Viewed in this standpoint, the lack of a global energy in GR is due to the lack, in general, of a global time translation symmetry.

http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html

has a good historically summary

6. Jan 11, 2013

### bcrowell

Staff Emeritus
Or isn't the problem that it has too *much* symmetry, i.e., full coordinate invariance rather than a set of more limited symmetries such as time translation?

There are really two separate issues, local and global, addressed by my #4 and your #5, respectively. The OP's question was stated locally, but it could also be interesting to talk about the contrast and interplay between local and global. I don't see any simple way to connect them, or any clear conceptual common ground that seems to underlie both of them.

Another way of looking at the issue of global energy conservation is that in SR, energy is part of the energy-momentum four-vector p. We typically don't expect to be able to construct globally conserved vector quantities in GR, because parallel transport is path-dependent, so there is no unambiguously well defined way to add up a vector quantity that occurs scattered around in space.

7. Jan 12, 2013

### Worldline

Thanks for ur answers, But in Teleparallel Gravity we can easy define an energy-momentum tensor for gravity. So it seems that the impossibility of defining an energy-momentum tensor isn't a characteristic of gravity, but a property of geometrical picture of GR

8. Jan 12, 2013

### dextercioby

Teleparallel gravity 'lives' in a manifold without curvature, so it should be almost like special relativity, whose manifold is also flat, but has no torsion.

9. Jan 12, 2013

### bcrowell

Staff Emeritus
Sure. Every reason offered in #4, 5, and 6 was geometrical.

I don't know much about teleparallel gravity, but aren't there versions that are equivalent to GR ( TEGR, http://arxiv.org/abs/gr-qc/0011087 ) and others that are inequivalent?

In general, there are various theories that are either partially or completely equivalent to GR, e.g., TEGR and spin-2 gravity on a flat background (MTW pp. 179ff and 424ff). Although these formalisms may not look geometrical, their equivalence to GR means that they incorporate the equivalence principle, so they can be interpreted geometrically, implying the impossibility of defining or measuring the local density of gravitational energy.

If you're talking about a version of teleparallel gravity which, unlike TEGR, isn't equivalent to GR and doesn't obey the equivalence principle, then you might be able to evade this. But the problem you're going to run into is that the equivalence principle is well verified experimentally.

10. Jan 12, 2013