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For the positive-expansion convention for extrinsic curvature, write ${\nabla }_{\mu }{n}_{\nu }=K_{\mu \nu }-n_{\mu }{a}_{\nu }$ with normal acceleration $a_{\nu }=n^{\rho }{\nabla }_{\rho }{n}_{\nu }$. The Ricci commutator applied to $n$ gives the difference of its two second-derivative contractions. Projecting tangentially, the acceleration term in that commutator cancels exactly the acceleration term converting a spacetime divergence of $K$ into its fully projected spatial covariant derivative. Einstein's equations then identify the displayed equation as the momentum constraint.

Let $a_{\mu }=n^{\nu }{\nabla }_{\nu }{n}_{\mu }$ be the normal acceleration. It is tangential by normalization. Resolving the derivative's first index into normal and tangential parts givesThe extrinsic curvature is symmetric because the normal is hypersurface orthogonal: locally $n_{\mu }=-N{\partial }_{\mu }t$, and the antisymmetric derivative of $n$ vanishes after projecting both indices tangentially.

Apply the supplied Ricci commutator to $n^{\nu }$ and project its free index. With the positive-expansion convention it givesTo identify the first two terms, expand the fully projected divergence:Tangency gives $n_{\lambda }{K}^{\lambda }{}_{\mu }=0$. Differentiating it shows that the last term is $-P^{\mu }{}_{\alpha }{K}^{\lambda }{}_{\mu }{n}^{\rho }{\nabla }_{\rho }{n}_{\lambda }=-K_{\alpha \lambda }{a}^{\lambda }$. Substituting into the commutator proves the contracted Codazzi equation with positive expansion:No assumption of geodesic normals was needed; the acceleration terms cancel rather than being discarded.

The metric-trace term in the Einstein field equations drops out of a tangential-normal projection, since $P^{\mu }{}_{\alpha }{g}_{\mu \nu }{n}^{\nu }=0$. Define the normal-frame momentum density $j_{\alpha }=-P^{\mu }{}_{\alpha }{T}_{\mu \nu }{n}^{\nu }$. The equation becomesThus Codazzi supplies the momentum constraint on each spatial slice. In flat homogeneous FRW, $K^i{}_j=H{\delta }^i{}_j$ and $H$ has no spatial gradient, so its left side vanishes. A comoving homogeneous perfect fluid likewise has $j_i=0$. The background satisfies the constraint automatically; the Friedmann expansion law is instead obtained from the normal-normal constraint.

Suitable scalar variables are lapse and shift potentials $A,B$, spatial curvature and scalar shear potentials $\psi ,E$, and a matter momentum potential $J$ with $j_i={\partial }_{i}J$. For example, chooseA density contrast and pressure perturbation can be added for the remaining scalar matter equations. With this positive-shift convention, the linear extrinsic-curvature perturbation isIts scalar shear cancels out of the momentum-constraint combination, leavingThese are the scalar momentum constraint with positive extrinsic curvature variables; a specified gauge can then remove two of the four scalar metric potentials.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.

The definition of the extrinsic curvature of a spatial hypersurface and ${\perp }^{\rho }{}_{\mu }={\delta }^{\rho }{}_{\mu }+n^{\rho }{n}_{\mu }$ givewhere $a_{\nu }=n^{\rho }{\nabla }_{\rho }{n}_{\nu }$ is the normal acceleration. HenceDifferentiating $n^{\nu }{n}_{\nu }=-1$ shows $n^{\nu }{a}_{\nu }=0$, so the acceleration is spatial as required.

First take the spacetime trace of the generalized field equation. Since $n_{\mu }{Z}^{\mu }=-\Theta$ and spacetime has dimension four, this givesNext contract twice with the unit normal. The result isAdding the trace equation to twice this normal projection cancels $T$. The Scalar Gauss equation then converts the curvature terms to

For the derivative terms, part (iii) givesDifferentiating $n^{\nu }{Z}_{\nu }=-\Theta$ along $n^{\mu }$ yieldsCombining these identities produces

Because $\Theta$ is a scalar field and $n^{\mu }={\alpha }^{-1}(1,-{\beta }^i)$,The normal acceleration is spatial, so $Z_{\mu }{a}^{\mu }={\Theta }_{\mu }{a}^{\mu }$. Part (iv) also gives $D^{\mu }{Z}_{\mu }=D^{\mu }{\Theta }_{\mu }-K\Theta$. Solving the preceding constraint for ${\partial }_t\Theta$ givesThus the constants in this Z4 formulation evolution equation are