In Newtonian theory, gravity is described by the gravitational potential $\Phi$. It can be shown that tidal acceleration in the $\uu$ direction is just the difference in the acceleration due to gravity at nearby points, which can be expressed as a directional derivative of the gravitational field, namely \[ -(\uu\cdot\grad)(\grad\Phi) \] More precisely, this expression describes the vector change in the displacement between two freely falling objects, which are (originally) separated in the $\uu$ direction; this is geodesic deviation. In relativity, geodesic deviation is described by an expression of the form \[ -R^i{}_{mjn} v^m u^j v^n \] This suggests that there is a correspondence between second derivatives of $\Phi$ (in the $i$ and $j$ directions) and $R^i{}_{mjn} v^m v^n$.

As we have seen, energy density in relativity is described by \begin{equation} \rho = g(\Tvec,\vv\cdot d\rr) \cdot\vv = T^i{}_j \,v_i \,v^j \end{equation} whereas in Newtonian theory matter enters through Poisson's equation \begin{equation} \Delta \Phi = 4\pi\rho \end{equation} where $\rho$ is the matter density (energy density). The Laplacian can be thought of as the trace of a matrix of second derivatives. Thus, if second derivatives of $\Phi$ correspond to the Riemann tensor, then the Laplacian corresponds to the trace of the Riemann tensor, which is known as the Ricci tensor, whose components are given by \begin{equation} R_{ij} = R^m{}_{imj} \end{equation} Thus, a reasonable guess for Einstein's equation, the relationship between curvature and matter, would be \begin{equation} R_{ij} = 4\pi T_{ij} = \frac{4\pi G}{c^2} T_{ij} \label{guess} \end{equation}

This guess turns out to be correct only in the absence of matter, when $\rho=0$; Einstein's equation in vacuum is indeed (equivalent to) \begin{equation} R_{ij} = 0 \end{equation} However, ($\ref{guess}$) turns out to be awkward mathematically, and incorrect physically. The mathematical objection is that the Ricci tensor is naturally a (rank 2, covariant) tensor, while the energy-momentum tensor is naturally a vector-valued 1-form. We are accustomed to raising and lowering indices by now, so this objection is not necessarily serious, but it turns out that we can do better.

The physical objection is far more serious: Conservation of matter (discussed in § Conservation Laws) together with the contracted Bianchi identity (§ Bianchi Identities) would force the Ricci scalar \begin{equation} R = R^m{}_m \end{equation} to be constant. It is unreasonable to expect that highly nonsymmetric configurations of matter would correspond to spacetime geometries with constant (scalar) curvature.