# Why does no physical energy-momentum tensor exist for the gravitational field?

+ 14 like - 0 dislike
166 views

Starting with the Einstein-Hilbert Lagrangian

$$L_{EH} = -\frac{1}{2}(R + 2\Lambda)$$

one can formally calculate a gravitational energy-momentum tensor

$$T_{EH}^{\mu\nu} = -2 \frac{\delta L_{EH}}{\delta g_{\mu\nu}}$$

$$T_{EH}^{\mu\nu} = -G_{\mu\nu} + \Lambda g_{\mu\nu} = -(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) + \Lambda g_{\mu\nu}$$

But then, in the paragraph below Eq(228) on page 62 of this paper, it is said that this quantity is not a physical quantity and that it is well known that for the gravitational field no (physical) energy-momentum tensor exists.

To me personally, this fact is rather surprising than well known. So can somebody explain to me (mathematically and/or "intuitively") why there is no energy-momentum tensor for the gravitational field?

A conserved energy-momentum tensor means $\partial_\mu( \sqrt{-g} T^{\mu\nu}) = 0$, but this is not a covariant equation. And the covariant equation $\nabla _\mu( T^{\mu\nu}) = 0$, does not correspond to a conserved energy-momentum tensor.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Trimok
See stress-energy momentum pseudotensor (en.wikipedia.org/wiki/…)

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Trimok
thinking out loud what if energy-momentum isn't really a tensor quantity. What if it is actually something else, that doesn't transform independently on each point, but needs to transform as a whole in an open set at the same time? Imagine such geometrical things, and take the limit of open sets to individual points, then you recover normal tensors.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user lurscher
related: physics.stackexchange.com/q/68382

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell

+ 7 like - 0 dislike

The canonical energy-momentum tensor is exactly zero, due to the Einstein equation. The same holds for any diffeomorphism invariant theory.

By saying ''it doesnt exist'' one just means that it doesn't contain any useful information.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Arnold Neumaier
answered Oct 25, 2012 by (13,660 points)
This answer is fine, +1, but seems to me to mystify an idea that should be straightforward. For fields like the electric and magnetic fields -- or the gravitational field in Newtonian mechanics -- we have an energy density that goes like the square of the field. In relativity this clearly isn't going to work: by the equivalence principle, we can always say that the gravitational field is zero at a given point, simply by adopting a free-falling frame of reference.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell
I would guess you can define a non-zero stress-energy tensor for gravity if you consider perturbations about a fixed background spacetime. (Almost certainly if these perturbations are small.) Can anyone with expertise confirm this?

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Jess Riedel
@JessRiedel: You can do that, but the object you end up with isn't a tensor. See ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll6.html , at "It is natural at this point to talk about the energy emitted via gravitational radiation."

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell
+ 6 like - 0 dislike

The energy-momentum tensor is defined locally, and it's a tensor. In electromagnetism, or in Newtonian gravity, the way we form a local energy density is basically by squaring the field.

The problem with applying this to GR is that the gravitational field $\mathbf{g}$ is zero, locally, in an inertial (i.e., free-falling) frame of reference, so any energy density we form by squaring it is going to be something that can be made to be zero at any given point, simply by a choice of coordinates. But a tensor that's zero for one choice of coordinates is zero for any choice of coordinates, so the whole idea doesn't work for GR.

The other kind of thing you could try would be taking derivatives of the field and using them as ingredients in such a locally defined tensor. This doesn't help, though. There's a discussion of this in Wald, section 11.2. The basic problem is that if you want the result to be a tensor, the derivatives have to be covariant derivatives operating on a tensor. But the only tensor we have available is the metric, and the defining characteristic of the covariant derivative is that it gives zero when you differentiate the metric. (There's a loophole in Wald's argument, however, that bothers me. When forming a tensorial quantity by differentiation, it's sufficient but not necessary that the derivative be a covariant derivative. When we form a curvature tensor from the metric, we do it by taking non-covariant derivatives on the metric in order to form the Christoffel symbols, and then doing further operations involving non-covariant derivatives to get the Riemann curvature tensor -- which is a valid tensor.)

None of this prevents the definition of nonlocal measures of the energy carried by gravitational fields in a certain region. That's why, for example, we can talk about the energy carried by a gravitational wave, but we have to talk about a region that's big compared to a wavelength. However, that won't allow us to define something that can go into the Einstein field equations, which are local because they're a differential equation.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Ben Crowell
answered Jun 16, 2013 by (1,070 points)
+ 3 like - 0 dislike