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As far as I know, in General Relativity, an expression of the kind $\nabla_{\mu} X = 0$ states that, associated to $X$, there exist a charge which is conserved. The first example that comes to mind is the conservation of energy-momentum $\nabla_{\mu} T^{\mu\nu} = 0$, as a consequence of the contracted Bianchi identities and assuming Einstein's field equations.

My question regards the meaning of the following expression:

$\nabla_{\mu} \left( R^{\mu\nu} K_{\nu} \right) = 0$

where $R^{\mu\nu}$ is the covariant Ricci tensor and $K_{\nu}$ is a generic vector field.

Does this equation mean something as it stands? Or do I have to impose a metric (and hence a form of the Ricci tensor) in order to find more informations?

The fact is that I don't understand what kind of conserved quantity could be related to the vector $R^{\mu\nu} K_{\nu}$.

The equation comes from a rather peculiar way to obtain a modified theory of gravity. The reference is here, but the original source is there

I admit that the procedure is not entirely clear to me, but basically it seems to appeal to a "gauge theory"-like approach in order to obtain modified gravitational field equations. It's kind of lengthy, but ultimately they consider commutators of covariant derivatives of successive orders to obtain relevant quantities, such as the Riemann tensor (first-order commutator) and matter current (second-order commutator). For the latter, the analogy with gauge theories is necessary. Finally they get the modified gravitational field equations, eq.(16) in the first reference.

The very nature of the approach makes it necessary to use a vector field $K_{\nu}$ in order to carry on the calculations, but the character of such a vector field is never specified (there's no need to, in fact). Only after eq.(16) they comment that Einstein's case is recovered if $\nabla_{\mu}K_{\nu}=0$.

Going back to the equation that I posted... in order to have some hints about the meaning of it all, starting from eq.(16) in the paper, I just impose the vacuum condition $T^{\mu\nu}=0$ and see what happens:

$\left(\nabla_{\mu}R^{\mu\nu}\right) K_{\nu} + R^{\mu\nu} \left( \nabla_{\mu} K_{\nu} \right) = 0$

and that's it.

I'm not at all convinced about the comments they make about the vector field (they call it "substratum") but the overall procedure doesn't seem entirely bad.

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Just one idea, but $\nabla_\mu R^\mu{}_\nu = \frac{1}{2} \nabla_\nu R$ (since the Einstein tensor has zero divergence). –  Vibert May 15 '13 at 20:32
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Where did you find your expression? It only holds for a generic $K_\nu$ if $R^{\mu\nu}=0$! –  Michael Brown May 15 '13 at 20:43
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1 Answer 1

There's a great discussion of this sort of thing in the first few pages of a paper by Penrose. Basically, to get an integral conservation law, you need the divergence of a vector to be zero. The energy-momentum tensor satisfies a differential conservation law, of course. But there's no associated quantity that you can generally integrate over a volume on a time-slice and say: that's a conserved charge of some kind.

As Michael Brown pointed out, $\nabla_\mu(R^{\mu\nu}K_\nu)$ being true for all (generic) vector fields $K_\nu$ pretty much means $R^{\mu\nu}=0$ (and presumably the same for the energy-momentum tensor). So maybe that's not the most interesting interpretation.

On the other hand, if $K$ is a Killing vector field, you might have something. In that case, your equation is equivalent to $\nabla_\mu R^{\mu\nu} = 0$. And, as Vibert pointed out, this means that the Ricci scalar is constant.

An interesting related note is something Penrose discussed. Again assuming that $K$ is a Killing vector field, then the divergence law you cited above ($\nabla_\mu T^{\mu\nu}=0$) implies that \begin{equation} \nabla_\mu(T^{\mu\nu}K_\nu) = 0~. \end{equation} Penrose points out that you then have various conserved quantities in $T^{\mu\nu}K_\nu$, like energy-momentum and angular momentum, depending on the particular Killing field. Moreover, this equation implies your equation ($\nabla_\mu(R^{\mu\nu}K_\nu) = 0$) if you assume metric compatibility and Einstein's equations (with no cosmological constant).

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Thanks, Mike. I'm editing my question in order to explain better the situation. –  Dude May 15 '13 at 22:42
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