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Use the negative curvature convention defined in curvature-sign convention in Killing derivative identities, and $C_{ab}=C^c{}_{acb}$. Covariant derivatives of a vector obey $[{\nabla }_b,{\nabla }^a]A^b=-C^a{}_b{A}^b$. Hence the flat-space equivalent expressions $\square A^a-{\partial }_b{\partial }^a{A}^b$ and $\square A^a-{\partial }^a{\partial }_b{A}^b$ acquire a difference $+C^a{}_b{A}^b$ under the comma-to-semicolon prescription. Covariantizing $F_{ab}=2{\nabla }_{[a}{A}_{b]}$ and its Maxwell equations gives $\square A^a-{\nabla }^a({\nabla }_b{A}^b)+C^a{}_b{A}^b={\mu }_0{J}^a$. This form respects gauge invariance. Dropping the Ricci term is generally not gauge invariant. The two expressions coincide in a Ricci-flat vacuum spacetime; vanishing matter alone with nonzero cosmological constant does not imply Ricci flatness.

For a Killing vector field $K$, its metric-dual differential one-form produces the electromagnetic field $F_{ab}=2{\nabla }_{[a}{K}_{b]}=2{\nabla }_a{K}_b$. The exterior derivative gives $dF=0$, while the second covariant derivative of a Killing vector gives ${\nabla }^a{F}_{ab}=-2R_{bd}{K}^d$. Thus this construction solves the source-free Maxwell equations in a vacuum spacetime with zero cosmological constant. The field is a test field: adding its stress-energy tensor would generally change the original metric tensor.

The strong equivalence principle asserts universality of free fall, including gravitational self-energy, and local independence of nongravitational and gravitational experiments from a freely falling laboratory's location and velocity. In a sufficiently small local inertial frame, uniform gravitational acceleration is removed and local physics takes its special-relativistic form. The Riemann curvature tensor remains measurable through tidal effects; a finite laboratory is not made exactly flat. Freely falling test bodies follow geodesics, clocks measure proper time, and minimally coupled light follows null geodesics in geometrical optics. Local nongravitational laws retain their special-relativistic form even when different laboratories observe gravitational redshift.

The comma-to-semicolon prescription replaces flat partial derivatives by the Levi-Civita connection, and the Minkowski metric by the curved metric tensor. It implements this local correspondence for minimally coupled matter. However, an identity obtained by commuting partial derivatives need not survive unchanged when they become covariant derivatives.

Raising the indices of the field-strength definition in flat spacetime gives $F^{ab}={\partial }^a{A}^b-{\partial }^b{A}^a$. Substitution into the sourced Maxwell equations yieldsso the first potential equation isPartial derivatives commute, giving the second formNo Lorenz gauge assumption was used.

Apply the comma-to-semicolon prescription before commuting derivatives. The two candidate left-hand sides becomeUse the cover curvature $C$ from Question 1, with $C_{ab}=C^c{}_{acb}$. The vector commutator is $[{\nabla }_c,{\nabla }_d]A^a=-C^a{}_{bcd}{A}^b$. HenceThis is the derivative-order ambiguity in covariant Maxwell potential equations. At a point in Riemann normal coordinates, the connection itself vanishes but its derivatives need not; the strong equivalence principle does not dispose of this curvature term by a coordinate choice.

Covariantizing the antisymmetric field strength and its Maxwell equations directly gives ${\mathcal{P}}_1^a={\mu }_0{J}^a$, orThis formulation has gauge invariance under $A_a\mapsto A_a+{\nabla }_a\chi$, since $F_{ab}$ is unchanged. In contrast, the uncorrected ${\mathcal{P}}_2$ acquires $-C^a{}_b{\nabla }^b\chi$ under that transformation. Thus preserving the original electromagnetic gauge symmetry singles out the directly covariantized Maxwell equations, rather than treating the two potential equations as interchangeable in matter. The equivalence principle alone is not a unique recipe for every possible higher-derivative or curvature coupling.

In a Ricci-flat vacuum spacetime, $C_{ab}=0$, so ${\mathcal{P}}_1={\mathcal{P}}_2$ and this particular ambiguity disappears even though the full Riemann curvature tensor may be nonzero. This uses vacuum with zero cosmological constant; an Einstein vacuum with nonzero cosmological constant is not Ricci flat.

For the terrestrial estimate, the relevant effect is Ricci curvature inside matter, not merely the tidal field outside the Earth. If a typical material density is $\rho \sim 5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, the Einstein field equations give a characteristic scaleFor a field varying over scale $L$, the derivative terms are $O(A/L^2)$ whereas the disputed term is $O(CA)$. Their relative size is $O(C{L}^2)$: about $10^{-22}$ for a metre-scale experiment. Even substituting the Earth's radius gives onlyA curvature length comparable to $C^{-1/2}\sim 10^{13}\mathrm{cm}$ would be needed for an order-one effect. In principle non-vacuum measurements could distinguish different curvature couplings, but an ordinary terrestrial laboratory has an exceedingly small signal, overwhelmed by its ordinary electromagnetic response to matter. This is a practical scale estimate, not a theorem that an arbitrarily precise experiment is impossible.

The Levi-Civita connection has symmetric Christoffel symbols, so antisymmetrization removes its connection terms. Hence the Papapetrou electromagnetic field iswhere the last equality uses the Killing equation. As a differential two-form, $F=d{V}^{♭}$, so the identity $d^2=0$ for the exterior derivative proves ${\nabla }_{[a}{F}_{bc]}=0$. Contracting the preceding second covariant derivative of a Killing vector givesA vacuum spacetime here has $R_{ab}=0$, with no cosmological constant. Therefore both source-free Maxwell equations hold:This is a test electromagnetic field on the given vacuum spacetime; its own stress-energy tensor is not included in that background's Einstein field equations.

Use $G=c=1$, the Einstein field equations with zero cosmological constant, and metric signature $(-,+,+,+)$. Write $m=m_{a}d{x}^a$ for the differential one-form dual to the axial Killing vector field. To fix the sign of the volume formula, use the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$ and $(\star F{)}_{ab}=\frac{1}{2}{ϵ}_{abcd}{F}^{cd}$. The Killing equation gives $(dm{)}_{ab}=2{\nabla }_a{m}_b$, and its contracted curvature identity givesTo see the curvature step, tracing the Killing equation gives ${\nabla }_b{m}^b=0$. The second covariant derivative of a Killing vector then gives ${\nabla }^b{\nabla }_b{m}_a=-R_{ab}{m}^b$, while commuting the covariant derivatives gives ${\nabla }^b{\nabla }_a{m}_b=R_{ab}{m}^b$. Subtracting these two terms is precisely ${\nabla }^b(dm{)}_{ab}$ above. The component Hodge star operator converts this divergence to $d\star dm$ with the displayed minus sign.

In a vacuum spacetime region $R_{ab}=0$, so $d\star dm=0$. If $S_1$ and $S_2$ are enclosing spacelike submanifolds in the same homology class bounding a vacuum three-dimensional region $W$, Stokes theorem yieldsThus the Komar angular momentum is independent of the enclosing vacuum spacelike two-manifold, provided the surfaces have the same orientation and enclose the same sources and inner boundaries. The vacuum region need not be stationary: an axial Killing vector field suffices.

For a regular filling hypersurface $\Sigma$ with $\partial \Sigma =S$, the Einstein field equations giveThe pullback of $\star m$ to $\Sigma$ vanishes because $m^a$ is tangent to $\Sigma$: the dual three-form measures the normal component, which is zero. ThereforeThis is the stress-energy current from a Killing vector integrated over the slice, with the signs fixed by the stated Hodge star operator convention.

There is a necessary boundary qualification omitted from the printed formulation. If the slice has inner boundaries $S_i$, orient them so $\partial \Sigma =S-{\bigcup }_i{S}_i$. The actual Komar angular momentum with inner boundaries identity isFor a vacuum Kerr black hole, $T_{ab}=0$ on the exterior slice but $J\ne 0$; its horizon supplies precisely the inner boundary contribution. Thus the matter-only formula is false for arbitrary exterior slices. It is valid when a nonsingular filling with no inner boundaries exists, or when all omitted inner charges vanish. The Komar angular momentum independence likewise concerns homologous surfaces in the same vacuum region, not an unrestricted comparison of differently enclosed objects.

A vacuum spacetime has vanishing matter stress-energy tensor. In $D>2$ dimensions, tracing the Vacuum Einstein equations gives $R=2D\Lambda /(D-2)$ and hence $R_{ab}=2\Lambda g_{ab}/(D-2)$; in four dimensions this reduces to $R_{ab}=\Lambda g_{ab}$. With zero cosmological constant it is Ricci flat, although its full Riemann curvature tensor can be nonzero: a vacuum Kerr black hole and vacuum gravitational waves illustrate that distinction.