Follow exchange and boundary flux

Conservation follows through an explicit double-divergence chain, with its volume defect retained. Matter and gravitational contributions exchange equal and opposite amounts on the selected sector.

10. Total conservation, exchange and boundary flux

The original double-divergence chain

To follow Einstein’s actual §17 argument, define

$$P_\mu^{\alpha\sigma}=q^{\sigma\beta}\Gamma_{\!\mathrm E}^\alpha{}_{\mu\beta} -\tfrac12\delta^\sigma_\mu V^\alpha, \qquad V^\alpha=q^{\lambda\beta}\Gamma_{\!\mathrm E}^\alpha{}_{\lambda\beta}. \tag{10.1}$$

Here $V$ is the indicated contraction of $\Gamma_{\!\mathrm E}$, distinct from the boundary vector in (7.15). Trace reversal of (9.8) gives

$$\partial_\alpha P_\mu^{\alpha\sigma}+\kappa\mathcal J^\sigma{}_{\mu} =q^{\sigma\beta}\mathsf J_g(U)_{\mu\beta}. \tag{10.2}$$

On $U=0,\nu=1$ this is Einstein (52a).

Theorem 10.1 (Double divergence with its volume defect)

On the compatible metric-calculus domain in arbitrary coordinates,

$$\begin{aligned} \partial_\alpha\partial_\sigma (q^{\sigma\beta}\Gamma_{\!\mathrm E}^\alpha{}_{\mu\beta}) &=\tfrac12\partial_\alpha\partial_\beta\partial_\mu q^{\alpha\beta},&&\text{(10.3)}\\ \partial_\alpha\partial_\sigma P_\mu^{\alpha\sigma} &=-\tfrac12\partial_\mu\partial_\alpha(q^{\alpha\beta}c_\beta). &&\text{(10.4)}\end{aligned}$$

Proof.

Expansion of the recovered Christoffel expression gives

$$q^{\sigma\beta}\Gamma_{\!\mathrm E}^\alpha{}_{\mu\beta} =\tfrac12\partial_\mu q^{\sigma\alpha}+L_\mu^{\alpha\sigma},\qquad L_\mu^{\alpha\sigma}=\tfrac12q^{\alpha\lambda}q^{\sigma\beta} (\partial_\lambda g_{\mu\beta}-\partial_\beta g_{\mu\lambda}). \tag{10.5}$$

Exchange $\alpha,\sigma$ and relabel $\lambda,\beta$ to see that $L$ is antisymmetric in its upper pair. Its double divergence vanishes, proving (10.3). Inverse compatibility gives $V^\alpha=\partial_\beta q^{\alpha\beta}+q^{\alpha\beta}c_\beta$. Insert this in (10.1); the triple derivatives cancel and the displayed defect remains. No field equation has been used.

$\square$

The first identity is Einstein (54); the second reduces to Einstein (55) at $\nu=1$. Combining with (10.2) yields the full off-shell relation

$$\kappa\partial_\sigma\mathcal J^\sigma{}_{\mu} =\partial_\sigma\!\left(q^{\sigma\beta}\mathsf J_g(U)_{\mu\beta}\right) +\tfrac12\partial_\mu\partial_\alpha(q^{\alpha\beta}c_\beta). \tag{10.6}$$

Consequently the selected original sector obeys

$$\boxed{\partial_\sigma(t^\sigma{}_{\mu}+T^\sigma{}_{\mu})=0,} \tag{10.7}$$

recovering Einstein (56). Outside the volume specialization the extra term is real: $E=\operatorname{diag}(1,1,1,e^{x^3})$, where $x$ is the first spatial coordinate, gives $\partial_\alpha\partial_\sigma P_\mu^{\alpha\sigma}=(3,0,0,0)_\mu$.

An antisymmetric representation can replace $P$ for this operation, but only after its difference has been accounted for. Define

$$\mathcal A_\mu^{\alpha\sigma}=L_\mu^{\alpha\sigma} +\tfrac12\delta^\alpha_\mu\partial_\beta q^{\sigma\beta} -\tfrac12\delta^\sigma_\mu\partial_\beta q^{\alpha\beta}. \tag{10.8}$$

It is antisymmetric, and direct differentiation gives

$$\partial_\alpha P_\mu^{\alpha\sigma} =\partial_\alpha\mathcal A_\mu^{\alpha\sigma} -\tfrac12\delta^\sigma_\mu\partial_\alpha(q^{\alpha\beta}c_\beta). \tag{10.9}$$

At $\nu=1$ their divergences agree. Pointwise equality of the potentials is not claimed. This is an instance of operation-specific information removal in Lemma 2.1.

Opposite matter and gravitational exchange

Define

$$e_\mu=\tfrac12T_{\rho\sigma}\partial_\mu q^{\rho\sigma}. \tag{10.10}$$

Substitute $W=U-\kappa\mathsf J_g(T)$ into (8.3) and use $g_{\rho\sigma}\partial_\mu q^{\rho\sigma}=-2c_\mu$. The complete identity is

$$\partial_\alpha t^\alpha{}_{\mu} =e_\mu+\tfrac12\tau c_\mu -\frac1{2\kappa}U_{\rho\sigma}\partial_\mu q^{\rho\sigma}. \tag{10.11}$$

On the coupled unit-volume sector this becomes

$$\boxed{\partial_\alpha t^\alpha{}_{\mu}=e_\mu, \qquad \partial_\alpha T^\alpha{}_{\mu}=-e_\mu.} \tag{10.12}$$

The second equality follows from total conservation and is numbered Einstein (57). For symmetric stress, $C^\rho{}_{\alpha\mu}T^\alpha{}_{\rho}=-e_\mu$ by (5.15), so

$$\partial_\alpha T^\alpha{}_{\mu} =-\Gamma_{\!\mathrm E}^\alpha{}_{\mu\beta}T^\beta{}_{\alpha}, \tag{10.13}$$

recovering Einstein (57a). The inconsistent unnumbered gravitational line before Einstein (57) has the opposite sign in the supplied copy; its nonzero test is included below and in Appendix A.

At the source level put $\Sigma^a{}_b=\eta^{ac}\Sigma_{cb}$ and

$$\mathscr C_b=D_a\Sigma^a{}_b =\sum_{a,\mu}(E^{-1})^\mu{}_a \left(\partial_\mu\Sigma^a{}_b+ \omega^a{}_{c\mu}\Sigma^c{}_b-\omega^c{}_{b\mu}\Sigma^a{}_c\right). \tag{10.14}$$

The tensor-derivative intertwiner and contraction give

$$\boxed{E^b{}_{\mu}\mathscr C_b =\nabla_\alpha T^\alpha{}_{\mu} =\nu^{-1}\partial_\alpha(\nu T^\alpha{}_{\mu})+e_\mu.} \tag{10.15}$$

Thus the matter conservation statement is $D_a\Sigma^a{}_b=0$, equivalently $\nabla_\alpha T^\alpha{}_{\mu}=0$. In a general chart its density expression is $\partial_\alpha(\nu T^\alpha{}_{\mu})=-\nu e_\mu$, not the formula with the volume factor omitted.

A separate curvature-identity proof

The source comparison also supplies an independent algebraic route, without replacing the original argument above. Expansion gives

$$D\mathcal F=\mathrm{d}\mathcal F+\omega\wedge\mathcal F-\mathcal F\wedge\omega=0, \qquad \mathcal F\wedge\theta=D\Theta=0. \tag{10.16}$$

The first follows by the product rule, $\mathrm{d}^2=0$ and associativity; the second uses zero torsion. Their coordinate readouts give the differential and algebraic Bianchi identities. To make the contraction explicit, metric compatibility and zero torsion give the lowered curvature pair symmetries. They can be verified at a point where $C=0$: there

$$R_{\rho\sigma\mu\nu}=\tfrac12( g_{\rho\nu,\mu\sigma}-g_{\sigma\nu,\mu\rho} -g_{\rho\mu,\nu\sigma}+g_{\sigma\mu,\nu\rho}),$$

which is antisymmetric in each pair and invariant under pair exchange. Tensor character extends the identity to every chart. Contracting $\nabla_\lambda R^\rho{}_{\sigma\mu\nu} +\nabla_\mu R^\rho{}_{\sigma\nu\lambda} +\nabla_\nu R^\rho{}_{\sigma\lambda\mu}=0$ first in $\lambda,\rho$ and then in $\sigma,\mu$ yields

$$2\nabla_\rho\operatorname{Ric}^\rho{}_{\nu}=\partial_\nu\mathcal R, \qquad \nabla_\alpha G^\alpha{}_{\mu}=0. \tag{10.17}$$

Consequently the complete coupled residual satisfies

$$\nabla_\alpha\mathcal K^\alpha{}_{\mu} =-\kappa\nabla_\alpha T^\alpha{}_{\mu},\qquad D_a(\mathscr G-\kappa\Sigma)^a{}_b=-\kappa\mathscr C_b. \tag{10.18}$$

This proves covariant matter conservation on every coupled configuration, in every chart, while preserving the distinct chart-tagged total-current construction.

Finite regions and explicit coupled checks

For a fixed bounded spatial coordinate region $V$ in a unit-volume chart,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int_V\mathcal J^4{}_{\mu}\mathrm{d}^3x &=-\int_{\partial V}\mathcal J^i{}_{\mu}n_i^{\rm out}\mathrm{d}S,&&\text{(10.19)}\\ \frac{\mathrm{d}}{\mathrm{d}t}\int_VT^4{}_{\mu}\mathrm{d}^3x &=-\int_{\partial V}T^i{}_{\mu}n_i^{\rm out}\mathrm{d}S-\int_Ve_\mu\mathrm{d}^3x,&&\text{(10.20)}\\ \frac{\mathrm{d}}{\mathrm{d}t}\int_Vt^4{}_{\mu}\mathrm{d}^3x &=-\int_{\partial V}t^i{}_{\mu}n_i^{\rm out}\mathrm{d}S+\int_Ve_\mu\mathrm{d}^3x. &&\text{(10.21)}\end{aligned}$$

These follow by differentiation under the finite integral and the divergence theorem. No boundary field is required to vanish for the balance identities; constancy of a regional total requires zero net flux. The exchange cancellation requires the same realization in both summands.

A same-realization nonzero-exchange witness is

$$E_d=\operatorname{diag}(r^{2/9},r^{2/9},r^{2/9},r^{-2/3}),\quad \epsilon_d=\frac4{27\kappa}r^{-2/3},\quad \Sigma_d=\operatorname{diag}(0,0,0,\epsilon_d),\qquad x^4=r>0. \tag{10.22}$$

The stress is specified before computing curvature. The calculation gives $\nu=1$, $\mathscr G=\kappa\Sigma_d$, $T^\alpha{}_{\mu}=\operatorname{diag}(0,0,0,\epsilon_d)$ and $t^\alpha{}_{\mu}=\operatorname{diag}(\epsilon_d,\epsilon_d,\epsilon_d,-\epsilon_d)$. In particular

$$e_4=\frac8{81\kappa}r^{-5/3},\qquad \partial_\alpha t^\alpha{}_4=e_4, \qquad \partial_\alpha T^\alpha{}_4=-e_4. \tag{10.23}$$

This distinguishes the exchange signs in a way a zero-exchange fixture cannot.

For a finite-flux test apply the determinant-one affine transformation $X=x+t$, $R=x+2t$, $Y=y$, $Z=z$ to (10.22). Recompute from $E_d(R)J$ and $\Sigma_d(R)$. The total current has $\mathcal J^4{}_4=-\epsilon_d(x+2t)$ and $\mathcal J^1{}_4=2\epsilon_d(x+2t)$. On the unit spatial cube,

$$Q_4=-\frac4{9\kappa}\big[(1+2t)^{1/3}-(2t)^{1/3}\big], \quad \Phi_4^{\rm out}=2[\epsilon_d(1+2t)-\epsilon_d(2t)], \quad Q_4'+\Phi_4^{\rm out}=0 \tag{10.24}$$

for $t>0$. The primitive and endpoints verify the integral rather than assuming flux vanishes.

Conservation is not the full field equation. Constant nonzero stress on a flat calibration is conserved but has $G-\kappa T\ne0$. Doubling the stress in (10.22) preserves its covariant conservation and breaks the gravitational equation. Conversely, ordinary matter divergence can vanish for a special varying metric and stress: (7.12) with $\Sigma=\eta/(3\kappa)$ has $T^\alpha{}_{\mu}=\delta^\alpha_\mu/(3\kappa)$ and $e=0$. A constant metric is sufficient to eliminate all such exchange terms, not necessary for every individual stress cancellation. Finally, four divergence equations do not determine ten arbitrary stress functions. Constitutive variables and their laws are addressed next rather than inferred from an equation count alone.