In general relativity without matter, the equation of motion of the metric field is described by the Hilbert's action or the Einstein tensor $G$. It's natural to lead to this conclusion once one realizes how natural the Riemannian scalar curvature is. In short,

scalar curvature ==>     Einstein tensor    $G$

Enter the matter field. By spacetime symmetries we have the energy-momentum tensor $T$, following Noether's theorem. This is also natural to me. In short,

      symmetry   ==> Energy-Momentum tensor $T$

What is not natural to me yet is: why we should couple metric and matter as Einstein did?

$$ G = T \mbox{ (Einstein equation)}$$

Of course, at the end we care about if the theory matches our observation, and as well-known it turned to be pretty good. But I'm still curious if there's some explanation that justifies, if any, why this method is canonical/natural/unique in any sense.


Einstein derived his equation by the requirement that it is covariant, reduces to the Poisson equation $\Delta \phi = 4 \pi G \rho$ in the weak-field, Newtonian limit, and that matter energy is conserved. The correspondence between $T^{\mu\nu}$ and matter density is established through field theory in flat space-time as $\rho \sim T^{00}$, and between the Newtonian potential and the metric as $g_{00} \sim 1-2\phi$. The matter-energy conservation is expressed as the requirement that the stress-energy tensor is divergence-less, $T^{\mu\nu}_{\;\;\;;\nu} = 0$. These requirements do not specify the equations for gravity uniquely, but they do limit the options considerably with Einstein equations being the simplest choice where $T^{\mu\nu}_{\;\;\;;\nu} = 0$ is not only consistent with the equations, but also automatically implied by the equations. For more details see this answer.

However, if you know nothing about the Poisson equation, you can still understand the gravity-matter coupling as "natural" or at least "simple" (in fact, the choice of the matter-energy coupling in the Einstein equations is often called minimal). This is because it comes automatically when adding matter fields to the Einstein-Hilbert action $$S_{\rm E-H} = \int \kappa R \sqrt{-g}\,d^4 x$$ As you probably know, the variation of this is proportional to the Einstein tensor. What you probably do not know is that any action for matter on a curved background $$S_{\rm m} = \int \mathcal{L}(g_{\mu\nu}, \psi_{,\mu},\psi,...)\sqrt{-g} \,d^4 x$$ the variation with respect to $g_{\mu\nu}$ (showing up in $\mathcal{L}$ for instance in contractions of kinetic terms $\sim g^{\mu\nu} \psi_{,\mu} \psi_{,\nu}$) is also proportional to the stress-energy tensor of the matter fields $\psi$. When one then takes the sum of the actions $S_{\rm E-H} + S_{\rm m}$ and requires vanishing variation with respect to $g_{\mu\nu}$, the resulting equations are automatically the Einstein equations (at least if your actions have physical normalizations).

  • $\begingroup$ Why do we take the sum of the two actions, namely $S_{EH} + S_m$? $\endgroup$ – Student Jul 9 '20 at 17:57
  • $\begingroup$ In principle you could use a different "addition" and get different physics. This is what happens with quantum corrections in the path integral formulation where the "addition" of actions is in an exponential $e^{iS/\hbar}$ (the classical limit is still addition, though). All classical interacting field theories are constructed by adding the two actions for the free (self-interacting) fields and an interaction term. The GR procedure is simpler since there is no need to come up with an interaction term and the coupling comes naturally. $\endgroup$ – Void Jul 10 '20 at 8:29
  • $\begingroup$ Thanks for your explanation! But why there's no need to add in interaction term in GR? Is it simply that the easier version (that without interaction term) has worked? $\endgroup$ – Student Jul 10 '20 at 11:38
  • $\begingroup$ @Student Of course, there can be other interaction terms, it is just that the GR version is simplest and yields the Newtonian limit (and other already observed GR corrections in the solar system) without any hassle. Additional interaction terms such as contractions of fields and the curvature tensor will lead to some sort of modified gravity which often has new degrees of freedom (the equations of motion have more than two derivatives of the metric in them) and a new coupling constant that often has to be fine-tuned so that the new theory fits our existing observations. $\endgroup$ – Void Jul 10 '20 at 12:31

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