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The quadratic gravitational effective stress-energy tensor depends on the split between the fixed background and the metric perturbation. A residual gauge symmetry of linearized gravity changes $h$ without changing the first-order tidal field, but generally changes $G^{(2)}[h,h]$ and hence the local value of $t$. At second order the compensating change of $j$ restores the same physical geometry; assigning only the quadratic term to an energy density loses this compensation.

Moreover, a stress-energy superpotential can change the local energy expression while preserving its divergence and, with suitable boundary behavior, its integrated charges. Symmetry and flat conservation do not make $t$ a unique, gauge-independent local gravitational energy density. Averaging in an appropriate short-wavelength regime can lead to a useful physical gravitational-wave energy flux, but that is a qualified approximation, not a cure for the local definition requested here.

Expand the Einstein tensor about the Minkowski metric, keeping a second-order metric perturbation as well:Here $G^{(2)}[h,h]$ includes all quadratic terms from the Einstein tensor, including the inverse-metric corrections in its contractions. In vacuum, the first-order Linearized Einstein equations give $G^{(1)}[h]=0$, while the next order gives $G^{(1)}[j]=-G^{(2)}[h,h]$. Define the quadratic gravitational effective stress-energy tensor byIt is symmetric because the Einstein tensor is symmetric, and is quadratic in $h$ and its derivatives. The linearized contracted Bianchi identity is the off-shell identity ${\partial }^{\mu }{G}_{\mu \nu }^{(1)}[j]=0$. The second-order vacuum equation therefore impliesOne can obtain the same conclusion directly by expanding the full contracted Bianchi identity to second order: every correction to the flat divergence multiplies $G^{(1)}[h]$ and vanishes on the first-order vacuum solution. Thus conservation holds for vacuum solutions of linearized gravity, rather than for an arbitrary off-shell $h$. Relative to the fixed Minkowski background, $t$ is a tensor under background changes of coordinates, but this does not make it invariant under the perturbative gauge freedom.