 Solution

The qualification concerns gravitational energy. Local matter energy exists: an observer $u$ measures $T_{\mu \nu }{u}^{\mu }{u}^{\nu }$. What is absent is a generally covariant local gravitational stress-energy density with the required universal conservation properties. The Equivalence principle allows the connection to vanish at a point in freely falling coordinates; standard gravitational energy expressions depend on coordinates. Curvature does remain, but supplies no unique gravitational energy tensor of the required kind. A nonstationary geometry also lacks a preferred time-translation Killing vector field. For a trial timelike $t$, stress-energy conservation gives ${\nabla }_{\mu }(T^{\mu \nu }{t}_{\nu })=T^{\mu \nu }{\nabla }_{(\mu }{t}_{\nu )}$, not generally zero.

An internal charge instead has a current ${\nabla }_{\mu }{j}^{\mu }=0$ independently of spacetime time translations. Its hypersurface flux is conserved with suitable boundary conditions. For electric charge, the Gauss law expresses it as a surface flux. Observer-dependent charge density does not prevent a well-defined total charge.

For an asymptotically flat spacetime, take asymptotically Cartesian slice coordinates with ${\gamma }_{ij}={\delta }_{ij}+h_{ij}$ and appropriate falloff. In $c=1$ units, the ADM energy isIt is an asymptotic gravitational energy. The dominant energy condition requires $-T^{\mu }{}_{\nu }{u}^{\nu }$ to be future-directed nonspacelike or zero for every future timelike $u$, in particular $T_{\mu \nu }{u}^{\mu }{u}^{\nu }\ge 0$. Together with the constraint equations, completeness and appropriate asymptotic/boundary hypotheses, it yields $E_{\mathrm{ADM}}\ge |{\mathbf{P}}_{\mathrm{ADM}}|\ge 0$. The energy condition alone is not the complete positive-energy theorem.