The gravitational current keeps its coordinate tag

The action supplies Einstein’s canonical gravitational current and its exact off-shell divergence. An explicit flat example shows why internal frame invariance does not make that current a coordinate tensor.

8. Gravitational energy and coordinate dependence

The action determines a canonical coordinate-translation current. Because $q(E)$ is pointwise, the chain rule gives

$$\frac{\partial\mathcal H_Z}{\partial(\partial_\alpha E^a{}_{\lambda})} \partial_\sigma E^a{}_{\lambda} =H_{\partial_\alpha q^{\mu\nu}}\partial_\sigma q^{\mu\nu}. \tag{8.1}$$

The dependence of $\omega(E)$ is already included in (7.5) and its variation. With Einstein’s normalization define

$$-2\kappa t^\alpha{}_{\sigma} =\Gamma_{\!\mathrm E}^\alpha{}_{\mu\nu}\partial_\sigma q^{\mu\nu} -\delta^\alpha_\sigma H. \tag{8.2}$$

The fixed nonzero $\kappa$ is normalized later by the Newtonian comparison. From the product and chain rules,

$$\boxed{-2\kappa\partial_\alpha t^\alpha{}_{\sigma} =W_{\mu\nu}\partial_\sigma q^{\mu\nu}.} \tag{8.3}$$

Second derivatives of $q$ cancel in the canonical divergence; the remaining coefficient is its Euler residual. Thus vacuum gives Einstein (49).

Substituting inverse-metric compatibility into (8.2) yields

$$\boxed{\kappa t^\alpha{}_{\sigma} =\tfrac12\delta^\alpha_\sigma H -q^{\mu\nu}C^\alpha{}_{\mu\beta}C^\beta{}_{\nu\sigma}.} \tag{8.4}$$

The two differentiated metric terms are equal after index relabelling. Calling the final product $K^\alpha{}_{\sigma}$ gives $\operatorname{tr}K=H$ and $\kappa\operatorname{tr}t=H$, which establish Einstein (50) and the mixed identity

$$\boxed{\partial_\alpha(q^{\sigma\beta}\Gamma_{\!\mathrm E}^\alpha{}_{\mu\beta}) +\kappa(t^\sigma{}_{\mu}-\tfrac12\delta^\sigma_\mu\operatorname{tr}t) =q^{\sigma\beta}W_{\mu\beta}.} \tag{8.5}$$

To prove it, expand the derivative, replace $\partial\Gamma_{\!\mathrm E}$ with $W-A$, and use inverse compatibility to split $(\partial q)\Gamma_{\!\mathrm E}$ into $qA+K$. The $qA$ terms cancel, and (8.4) cancels $K$. On the unit-volume vacuum sector this is Einstein (51); conversely invertibility of $q$ recovers $W=0$. Four conservation equations alone are weaker: the witness (7.12) has $W\ne0$ but $W_{\mu\nu}\partial_\sigma q^{\mu\nu}=0$ since $\nu=1$.

For a fixed spatial coordinate region $V$, vacuum conservation gives

$$\frac{\mathrm{d}}{\mathrm{d}t}\int_Vt^4{}_{\sigma}\mathrm{d}^3x =-\int_{\partial V}t^i{}_{\sigma}n_i^{\rm out}\mathrm{d}S =\int_{\partial V}t^i{}_{\sigma}n_i^{\rm in}\mathrm{d}S, \tag{8.6}$$

recovering the inward-normal convention of Einstein (49a). Moving boundaries require an additional transport term.

A retained non-tensor readout

Einstein explicitly states that $t$ is not a tensor. A determinant-one nonlinear chart change makes this limitation concrete:

$$X=e^x,\quad Y=y,\quad Z=z,\quad T=te^{-x}. \tag{8.7}$$

Take $E$ to be its Jacobian, the pullback of a Cartesian calibration. Its coframe is exact, $\omega=0$, and $C=E^{-1}\mathrm{d}E$. Direct calculation gives

$$R=0,\quad \nu=1,\quad W=0,\quad H=-2e^{-2x},\qquad \kappa(t^\alpha{}_{\sigma})=\operatorname{diag}(0,-e^{-2x},-e^{-2x},0). \tag{8.8}$$

The array is zero in the original Cartesian coordinates. A tensor cannot change from zero to nonzero by a coordinate transformation. Its divergence nevertheless vanishes, as (8.3) requires. Internal frame invariance is not coordinate tensoriality; the chart remains an essential tag on this energy readout. No positive local energy density or globally invariant charge is inferred.

Finally, vacuum is not flatness. The independent calibration

$$E=\operatorname{diag}(r^{1/3},r^{1/3},r^{-1/6},r^{-1/2}),\qquad x^4=r>0 \tag{8.9}$$

has $\nu=1$, $\operatorname{Ric}=W=0$ but $R^1{}_{414}=1/(18r^2)$. Both curvature implementations and the current identities agree on this fixture. It verifies the distinction motivating §14 without introducing a different vacuum law.