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For complete asymptotically flat three-dimensional Riemannian initial data with nonnegative scalar curvature and an outermost compact minimal surface boundary of area $A$, the Riemannian Penrose inequality gives the displayed bound. Equality is realized by the spatial exterior of Schwarzschild spacetime. Time-symmetric gravitational initial data have this form when the Einstein field equations and energy hypothesis give the required nonnegative curvature. This theorem does not justify substituting arbitrary non-time-symmetric apparent horizon area into the same formula.

The stress-energy tensor describes local energy density, momentum density, and stress. Compatibility with the Einstein field equations requires covariant conservation ${\nabla }^{\mu }{T}_{\mu \nu }=0$.

The strong energy condition requires $(T_{ab}-T{g}_{ab}/2)u^a{u}^b\ge 0$ for every timelike vector $u$ in four spacetime dimensions. Through the Einstein field equations, it implies the timelike convergence condition $R_{ab}{u}^a{u}^b\ge 0$ when the cosmological constant vanishes.

In the leading radiation domination limit, neglect $k^2{\delta }_r$ and the matter contribution to gravity. The radiation equation gives $S^{\prime \prime }=0$. Substitution into ${\delta }_c^{\prime \prime }+{\delta }_c^{\prime }/\tau =3{\delta }_r/{\tau }^2$ givesThus adiabatic initial conditions select both $S_0=0$ and $S_1=0$, leaving the growing and decaying power-law modes. The four constants describe the leading supplied coupled system; regularity and additional initial Einstein field equations constraints can further restrict their physical interpretation.

In four spacetime dimensions, the trace reversal of a stress-energy tensor has the displayed form, where $T=g^{\mu \nu }{T}_{\mu \nu }$. Its trace is $-T$, so applying trace reversal twice returns the original tensor. Taking the trace of the Einstein field equations without a cosmological constant gives $R=-\kappa T$; substitution yields $R_{\mu \nu }=\kappa {\overline{T}}_{\mu \nu }$. In a general dimension $d>2$, solving for the Ricci tensor instead uses $T_{\mu \nu }-g_{\mu \nu }T/(d-2)$.

With no cosmological term, the two-dimensional Einstein field equations reduce identically on the geometric side to $G_{\alpha \beta }=0$. They impose $T_{\alpha \beta }=0$ and supply no local metric dynamics; the Einstein-Hilbert action is topological.

In a region with vanishing stress-energy tensor, the Einstein field equations reduce to $R_{\mu \nu }=0$ when the cosmological constant vanishes.