Can we interpret the Einstein field equations to mean that stress-energy *is* the curvature of spacetime?

There is a pure geometric definition of the Einstein tensor $G_{\mu\nu}$ in terms of derivatives of the metric. Independent of any physics.

Likewise, given a field theory, you could in principle calculate the stress energy tensor.

GR is a physical theory which couples the geometry, through the Einstein tensor, to the matter content, via the stress energy tensor. There are other self-consistent theories which couple geometry to matter in different ways. In this sense, the equation $G=8\pi T$ is model dependent. Just like the ideal gas law $PV=NRT$, which only applies for ideal gases, not interacting gases.


In order to elucidate a potential answer to your question, it is helpful to consider the Einstein field equations in the context of field theory. The action,

$$S=\frac{1}{16\pi G}\int d^4 x \, \sqrt{|g|} \, R + \int d^4x \, \sqrt{|g|} \, \mathcal L_M$$

gives rise to the Einstein field equations through the principle of stationary action, with a right hand side corresponding to the usual symmetric definition of the stress-energy tensor of a field theory described by $\mathcal L_M$. In this context, we can think of,

$$R_{\mu\nu}-\frac12 g_{\mu\nu}R = 8\pi G \, T_{\mu\nu}$$

as being the equations of motion for the metric, and any fields in $\mathcal L_M$ coupled to gravity. So it's not so much that stress-energy is curvature but rather the fields (or other quantities) which contribute to the stress-energy are coupled to gravity.


Short answer:

  • If you like, you can say that the Einstein field equations define the active gravitational mass (or, more precisely, the active gravitational mass-energy-momentum-stress).

  • But active gravitational mass equals inertial and passive gravitational mass, so this makes the EFE more than a definition.

  • The EFE also contain all kinds of information that has nothing to do with sources. For example, they say that certain vacuum fields are not possible, and they predict the existence of gravitational waves.

  • There are some ambiguities that come into play in the case of dark energy.

Long answer:

The Einstein tensor $G$ is measurable. For example, when I drop a pencil and see how long it takes to hit the ground, I am finding out something about the Riemann tensor in a certain region of space. By doing enough measurements, I can measure the entire Riemann tensor and then determine $G$. This constitutes an operational definition of $G$. We could define $G$ in some other way, but we don't need to, it seems undesirable, and nobody does it.

The Einstein field equations relate the Einstein tensor to the stress-energy tensor. In the nonrelativistic limit, this is simply equivalent to the Newtonian equation $g=Gm_a/r^2$, which relates the active gravitational mass $m_a$ to the gravitational field. This can be taken as the definition of active gravitational mass in Newtonian physics, and there is no other way to define it. However, this does not make Newton's law of gravity a tautology or a definition, because in Newtonian gravity the active gravitational mass is strictly equal to both the passive gravitational mass and the inertial mass. Since we have other types of experiments that can measure inertial mass, there is no circularity involved. Furthermore, Newton's law of gravity specifies the distance dependence of the field, which is not a matter of definition. This $1/r^2$ form of the force law results, for example, in the prediction of elliptical orbits.

Similarly, in GR, the active gravitational mass (or, more precisely, active gravitational mass-energy-momentum-stress) is defined as $G/8\pi$, and there is no other way to define it. However, this does not make the Einstein field equations tautological or a matter of definition, for the same reasons as in the Newtonian theory.

Note that in GR, the equality of inertial, active, and passive gravitational masses is not just an optional feature as in Newtonian gravity. If any of these equalities fails, then GR is falsified and cannot be fixed by tinkering. (E.g., it's a theorem in GR that test particles follow geodesics.)

One place where I think it gets a little trickier to make proper operational definitions is in the case of dark energy. We have no way to measure the inertia or passive gravitational mass of dark energy. This is basically because our model of dark energy is a cosmological constant, and the Einstein field equations do not allow us to simply make solutions in which the cosmological constant varies from point to point. Such solutions always violate the field equations. Therefore the cosmological constant is ordinarily modeled as a constant -- it has no dynamics. (You can have a dynamical dark energy, but doing so requires something more elaborate than just letting $\Lambda$ vary.)

This lack of dynamics in $\Lambda$ prevents us from measuring dark energy's inertial or passive gravitational mass. For this reason, it's not uncommon to see different people making different choices about whether or not to include the dark energy piece as part of the stress-energy tensor.