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The cosmological constant $\Lambda$ is the constant term proportional to the metric in the Einstein field equations. Interpreted as vacuum energy, it has $P_{\Lambda }=-{\rho }_{\Lambda }$, so its energy density is constant during expansion.

Transforming a metric tensor and all matter fields together by the pullback of a diffeomorphism preserves the Einstein field equations and covariant matter equations. Passive coordinate changes only change the description of the fields; active pullbacks produce another representative of the geometry. The treatment of transformations acting on a boundary depends on the boundary conditions and possible charges. A diffeomorphism is not necessarily an isometry of a fixed metric.

The Null Raychaudhuri equation and Einstein field equations give ${\partial }_U(f^{-1}{r}_{,U})=-r{R}_{UU}/(2f)=-4\pi r{T}_{UU}/f$, with an exchanged-coordinate identity in $V$. Under the null energy condition, each rescaled radial derivative is nonincreasing along its corresponding future null direction.

The Einstein-Hilbert action is the integral of the Ricci scalar against the metric volume form. Varying it with respect to the metric gives the vacuum Einstein field equations, up to a boundary term.

For a flat background dominated by one constant-$w$ barotropic equation of state, assume $\delta P=w\delta \rho$ and no scalar anisotropic stress. The Einstein field equations then reduce the common Newtonian-gauge potential to the stated Fourier transform equation, with $\mathcal{H}=2/[(1+3w)\tau ]$. For radiation the regular solution is $3{\Phi }_0(\sin x-x\cos x)/x^3$, $x=k\tau /\sqrt{3}$, and oscillates with a decaying envelope after sound-horizon entry. For pressureless matter the two solutions are a constant and ${\tau }^{-5}$. A non-adiabatic pressure perturbation would supply an additional source, so constant background $w$ alone does not justify the homogeneous equation.

The Hamiltonian constraint is the normal-normal projection of the Einstein field equations onto a spatial hypersurface. With the extrinsic-curvature convention used here,

With the Riemann curvature tensor convention $R^{\mu }{}_{\nu \rho \sigma }={\partial }_{\rho }{\Gamma }^{\mu }{}_{\nu \sigma }-{\partial }_{\sigma }{\Gamma }^{\mu }{}_{\nu \rho }+\cdots$, harmonic coordinates remove the mixed second derivatives from the Ricci tensor. The remainder $Q_{ab}$ contains at most first derivatives of the metric tensor. The vacuum Einstein field equations become a system of quasilinear partial differential equations with a Lorentzian wave equation principal part. Harmonic coordinate constraints must be maintained when using the reduced system.

Komar angular momentum is the Komar charge associated with an axial Killing vector field in four-dimensional general relativity. In geometrized units its normalization is $(16\pi {)}^{-1}{\int }_S\star dm$, where $m$ is the metric-dual differential one-form. With zero cosmological constant, vacuum Einstein field equations make $d\star dm=0$, so homologous surrounding surfaces in the vacuum region give the same charge.

With zero cosmological constant and the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$, the Einstein field equations and Killing equation give $d\star dm=-16\pi \star (T_{ab}{m}^b-\frac{1}{2}T{m}_a)$. If $m$ is tangent to a spacelike hypersurface, the trace term pulls back to zero. For $\partial \Sigma =S_{\infty }-{\bigcup }_i{S}_i$, Stokes theorem yieldsThe inner Komar charges cannot be dropped for a vacuum Kerr black hole, whose angular momentum is nonzero although its stress-energy tensor vanishes. A matter-only volume formula requires a regular filling with no inner boundaries.

A Killing vector field gives an antisymmetric tensor ${\nabla }^{\mu }{\xi }^{\nu }$ whose divergence is this current, up to overall surface-orientation convention. The four-dimensional Einstein field equations express it as $8\pi G(T^{\mu }{}_{\nu }-\frac{1}{2}T{\delta }^{\mu }{}_{\nu }){\xi }^{\nu }$ at zero cosmological constant. The contracted Bianchi identity and Killing invariance of scalar curvature prove its divergence is zero. Stress conservation alone does not cancel the trace derivative without Killing invariance of the trace.

The momentum constraint is the equation obtained by varying the shift vector in the Hamiltonian formulation of general relativity. It expresses the tangential-normal projection of the Einstein field equations and generates spatial coordinate transformations. Writing $p^{ij}={\pi }^{ij}/\sqrt{h}$ avoids ambiguities about covariant derivatives of tensor densities.

The null convergence condition requires $R_{ab}{k}^a{k}^b\ge 0$ for every null vector. Through the Einstein field equations, it follows from the null energy condition, since all terms proportional to the metric tensor vanish on a null vector.

In a planar warped spacetime, contraction of the Ricci tensor with a null vector in an orthonormal frame gives $R_{ab}{k}^a{k}^b=-2(\log A{)}^{\prime \prime }(k^3{)}^2$. The Einstein field equations identify this with $8\pi T_{ab}{k}^a{k}^b$; metric terms, including a cosmological constant, vanish in a null contraction. Thus the null energy condition is equivalent to concavity of $\log A$ in the proper transverse coordinate $z$. Null directions confined to a planar slice saturate the condition.