4
$\begingroup$

Einstein's field equations (EFEs) describe the pointwise relation between the geometry of the spacetime and possible sources described by an energy-momentum tensor $T^{\mu\nu}$. As well known, such equations can be derived from a variational principle applied to the following action: \begin{equation} S= \int \left( L_{EH} + L_M \right)\sqrt{-g}\, d^4x\, , \end{equation} where $L_{EH}=\frac{1}{2\kappa} R$ is the Einstein-Hilbert Lagrangian and $L_M$ is the Lagrangian describing the matter/field sources. Thus EFEs are obtained by requiring the stationarity of $S$ with respect to variations of the metric, and they are given by \begin{equation} R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \kappa\, T_{\mu\nu} \end{equation} Such equations are automatically compatible with energy-momentum conservation due to the validity of the doubly-contracted Bianchi identites, which imply essentially the vanishing of the covariant divergence of the left-hand side of EFEs.

However, EFEs take into account only the gravitational degrees of freedom that couple to matter: the "free part" of the gravitational field is said to be encoded in the Weyl tensor $C_{abcd}$ and its evolution/propagation is given by the trace-free part of the once-contracted (second) Bianchi identities, which are given by \begin{equation} \nabla^d C_{abcd} = \nabla_{[a}\left( -R_{b]c} +\frac{1}{6}R\, g_{b]c} \right)\, . \end{equation} In a 1+3 covariant splitting of the spacetime, such equations take a form which is strikingly similar to Maxwell's equations for electromagnetism (see here for more details on such analogy).

Hence the question: would it be possible to obtain the equations of motion for the free gravitational field (the Bianchi identities above) from a variational principle? This, of course, implies the more detailed question: what would be the action and with respect to which "field" one has to vary such action in order to obtain the aforementioned equations of motion?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.