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Einstein's vacuum equations, that is without matter, allows the possibility of curvature without matter. For instance, we may consider gravitational waves.

The question is: Is there some link between the Riemann curvature tensor, and/or the Weyl tensor, and some gravitational "physical" quantities (as stress-energy tensor or total energy)?

Of course, at first glance, there is no covariant gravitational stress-energy tensor, so it seems there is no relation, but maybe things are more subtle?

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Are you asking why the equations relate only the Ricci tensor and the stress-energy, but not the Weyl tensor? But why would expect something like that? For example only two of the classic Maxwell equations couple with the sources. This may not be a good analogy but I still think you could say something about your motivation. –  MBN Sep 6 '13 at 16:14
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@MBN : The motivation is that, without matter, there is still gravitational energy in some way (even if it is non localizable), so I was asking about any interesting relation between Riemann/Weyl tensor, or operations on them, and gravitational energy quantities. –  Trimok Sep 6 '13 at 16:18
    
In vacuum, the stress-energy tensor is $0$ by definition. Hence, for cosmological constant $0$, the Riemann tensor equals the Weyl tensor, $R_{abcd}=C_{abcd}$. Some ideas of getting matter from vacuum were explored by Wheeler (search for "geons"). You can also take a look, with a grain of salt, at fqxi.org/data/essay-contest-files/… –  Cristi Stoica Sep 6 '13 at 18:29
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@brightmagus One of two things. Firstly, you could be thinking about a vacuum region where there is no matter, but other oscillating mass far removed from the region under consideration is the "source". This would be wholly analogous to studying the electromagnetic field in empty space. Sometimes, we think of plane waves or other wave solutions that take up all space disembodied from any source: this is a valid solution of the Maxwell equations. You can make perfectly valid conclusions about a light beam from such models. Physicists do analogous things with the EFE. –  WetSavannaAnimal aka Rod Vance Nov 16 at 22:57
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You can think of anything you want, guys. I like, however, things I think of in physics to be physical. Trimok's does not obviously refer to a certain region only, because there would bo sense to mention such a trivial case. And as to the BB without matter ... no comment ... –  bright magus Nov 17 at 6:40

2 Answers 2

Is there some link between the Riemann curvature tensor [...] and some gravitational "physical" quantities*

Maybe you could clarify what you want that would qualify as "physical." Curvature is observable, and IMO is physical. Projects like LIGO are designed to detect gravitational waves. Gravity Probe B was a project that accomplished its purpose of essentially verifying GR's predictions of spacetime curvature in the neighborhood of a gravitating, spinning body. In the simplest terms, curvature can be measured by transporting a gyroscope around a closed path. This is essentially what GPB did.

Of course, at first glance, there is no covariant gravitational stress-energy tensor

But that's only a prohibition on defining a local measure of gravitational-wave energy. For example, in an asymptotically flat spacetime, the ADM energy includes energy being radiated away to null infinity by gravitational waves. If LIGO-like projects succeed, they will measure the energy of gravitational waves.

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I am interested in any detailed example of a relation between gravitational physical quantites like energy, radiated energy, stress-energy tensor, and Riemann/Weyl tensor, with Einstein vacuum equations. –  Trimok Sep 6 '13 at 16:05
    
OK, then I think the second half of my question should provide an example of such a relation. However, I would object to this definition of "physical" as excluding curvature, for the reasons given in the first half of my answer. –  Ben Crowell Sep 6 '13 at 16:07

The first point to consider is that the Riemann tensor can be expressed in terms of the Weyl tensor and the Ricci Tensor: $$R_{abcd}=C_{abcd}-g_{a[d}R_{c]b}-g_{b[c}R_{d]a}-\frac{1}{3}Rg_{a[c}g_{d]b}$$

The Ricci tensor is given by Einstein's equation:

$$R_{ab}-\frac{1}{2}g_{ab}R+\Lambda g_{ab}=8\pi T_{ab}$$

Now the Weyl tensor is not specified by the EFE. However, it can not be arbitary as the Riemann tensor must satisfy the Bianchi identities: $$R_{ab[cd;e]}=0$$

Applying this last condition to the first equation we obtain that the Weyl tensor must satisfy: $$C^{abcd}_{;d}=J^{abc} $$ where $J^{abc}=R^{c[a;b]}+\frac{1}{6}g^{c[b}R^{;a]}$.

You can find a proof here.

You can now make the interpretation of this field equations as determining that part of the curvature at a point that depends on the matter distribution at other points.

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