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With the stated curvature convention, the Ricci tensor is $R_{bd}=R^a{}_{bad}$ and the scalar curvature is $R=g^{bd}{R}_{bd}$. Contract the differential Bianchi identity once to give the divergence of the Riemann tensor, then again with the inverse metric. The curvature antisymmetries yield ${\nabla }^a{R}_{ab}-{\nabla }_{b}R+{\nabla }^a{R}_{ab}=0$, so ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$.

The Einstein field equations are $G_{ab}+\Lambda g_{ab}=(8\pi G/c^4)T_{ab}$, with $G_{ab}=R_{ab}-\frac{1}{2}R{g}_{ab}$. The contracted Bianchi identity makes the left side divergence-free, enforcing the local stress-energy conservation law ${\nabla }^a{T}_{ab}=0$ and ensuring consistency of the matter equations with the gravitational constraints.

For a scalar, second covariant derivatives commute. Commuting a further derivative past the resulting covector introduces Ricci curvature, giving ${\nabla }^2{\nabla }_a\varphi ={\nabla }_a{\nabla }^2\varphi +R_{ab}{\nabla }^b\varphi$.

Now $R_{ab}={\nabla }_a{\nabla }_b\varphi$ implies $R={\nabla }^2\varphi$. Taking its divergence and using that commutation formula gives $\frac{1}{2}{\nabla }_{b}R={\nabla }_{b}R+R_{ba}{\nabla }^a\varphi$, hence ${\nabla }_{b}R=-2R_{ba}{\nabla }^a\varphi$. Meanwhile ${\nabla }_b({\nabla }_a\varphi {\nabla }^a\varphi )=2R_{ba}{\nabla }^a\varphi$. Adding provesIn Riemannian signature this is the steady gradient Ricci soliton scalar identity, with soliton potential $f=-\varphi$.

For a homogeneous scalar field, every spatial derivative vanishes. The stress-energy tensor consequently has no momentum density or directional stress:The only spatial tensor left is ${\delta }^{ij}$, so the stress is isotropic. This is the perfect fluid form in its comoving frame. Now expand the time component of stress-energy conservation using the covariant derivative:The first term is $\dot{\rho }$. In the second, only $\lambda =0$ contributes, and ${\sum }_i{\Gamma }^i{}_{i0}=3H$, giving $3H\rho$. The third isHere the quoted Christoffel symbols include their lower-index symmetry ${\Gamma }^i{}_{j0}={\Gamma }^i{}_{0j}$. Therefore the cosmological continuity equation isAs a check using the scalar field equation, $\dot{\rho }=\dot{\varphi }(\ddot{\varphi }+V_{,\varphi })=-3H{\dot{\varphi }}^2=-3H(\rho +P)$, with exactly the same sign.

Use metric signature $(+---)$ throughout this solution. Stress-energy conservation for the perfect fluid is ${\nabla }_b{T}^{ab}=0$. Expanding it and using metric compatibility givesThe normalization of the four-velocity implies $U_a{U}^b{\nabla }_b{U}^a=0$. Contracting with $U_a$ therefore gives the relativistic perfect-fluid energy equation,while subtracting the component parallel to $U$ gives the relativistic Euler equation,Equivalently, apply the orthogonal projection ${\delta }_a{}^c-U_a{U}^c$ to the conservation equation. This form makes clear that a pressure gradient orthogonal to the four-velocity supplies the fluid's four-acceleration.

Take $z$ positive outward and define the separation as upper minus lower position. The Newtonian potential of a point mass gives the acceleration $a(z)=-{\partial }_z\varphi =-GM/z^2$. Expanding its difference across a separation $\zeta \ll z$ gives the tidal force equationThe upper particle falls slightly less rapidly, so the separation increases. If the upper particle carries the electric charge and the electric field points outward, the extra relative acceleration is $qE/m$. Reversing the field or placing the charge on the lower particle changes its sign. Write this signed acceleration as $a_e$ and put $k=2GM/z^3$. With initial relative velocity zero and nearly constant $k$ and $a_e$, the ordinary differential equation has solutionFor $k{t}^2\ll 1$, the leading changes due to the tidal force and electric field are respectivelywith the stated sign convention. Thus, for a nonzero fixed electric acceleration,A laboratory of size $L$ has $|{\zeta }_0|\le L$, so this ratio tends to zero as $L\to 0$. This is local tidal acceleration scaling: gravitational relative acceleration is proportional to separation, whereas the electric difference persists even at coincident positions.

The duration also matters. At fixed time, $\Delta {\zeta }_g\to 0$ in absolute size, but $\Delta {\zeta }_g/{\zeta }_0\simeq (GM/z^3)t^2$ need not tend to zero. To keep the electrically displaced particles in the shrinking laboratory, take $|a_e|t^2\lesssim L$; then $|\Delta {\zeta }_g|/L\lesssim GML/(z^3|a_e|)\to 0$. Alternatively a stationary terrestrial laboratory has a free-fall time of order $\sqrt{L/g}$, giving $|\Delta {\zeta }_g|/|{\zeta }_0|=O(L/z)$. Both make precise the small spacetime-region limit. An extended, long-running experiment can still measure arbitrarily small tidal forces.

The weak equivalence principle says that freely falling test particles with the same initial data have composition-independent motion. In particular, a freely falling frame removes the common gravitational acceleration of both particles. The electric accelerations depend on charge and cannot be removed for charged and uncharged particles simultaneously by the same change of frame.

The Einstein equivalence principle extends the freely falling laboratory statement to all local nongravitational experiments: their outcomes obey special relativity, with local Lorentz invariance and local position invariance. The strong equivalence principle also includes local gravitational experiments and bodies with significant gravitational binding energy. It asserts universality of their free fall and independence of local experimental outcomes from the laboratory's location and velocity. The two-particle experiment illustrates the local distinction between gravity and a nongravitational force; by itself it does not establish this stronger claim about self-gravitating bodies.

In a metric theory, the common freely falling trajectories are timelike geodesics of a Lorentzian metric. Normal coordinates make $g_{ab}={\eta }_{ab}$ and the Christoffel symbols zero at one point. Fermi normal coordinates do the same along a reference timelike geodesic. The residual effects are controlled by the Riemann curvature tensor through geodesic deviation, and enter the local metric at quadratic order in distance. Curvature is a physical tidal field, so no coordinate choice removes it over a finite region.

A relativistic theory should consequently couple matter universally to the metric tensor, recover nongravitational laws in local inertial frames and recover the Newtonian limit. General relativity implements this with the Einstein field equations; the contracted Bianchi identity is consistent with stress-energy conservation. The equivalence principle motivates this geometric description but does not alone uniquely determine the field equations. Extra gravitational fields may satisfy the Einstein equivalence principle for ordinary matter while failing the strong equivalence principle because local gravitational measurements or self-gravitating trajectories depend on those fields.

Let $I^{ij}(t)=\int T^{00}(t,\mathbf{x})x^i{x}^j{d}^{3}x$, using spatial indices $i,j$. Stress-energy conservation, symmetry of the stress-energy tensor and integration by parts giveand a second derivative givesAll boundary terms vanish by compact support. This proves the stress-energy quadrupole identity,The support is understood to remain bounded over the time interval used for differentiating these integrals.

For the radiation calculation set $c=1$, write $g_{ab}={\eta }_{ab}+h_{ab}$ and use the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ in Lorenz gauge in linearized gravity. With the curvature and field-equation signs in this paper, the linear equation isThe retarded solution with no incoming radiation is consequentlyLet $r=|\mathbf{x}|$ be the distance from the source's centre of mass, and let $d$ be its size. In the radiation zone, $r\gg d$. The leading multipole expansion additionally assumes slow internal motion, or $d$ small compared with the radiation wavelength. These assumptions allow $|\mathbf{x}-{\mathbf{x}}^{\prime }|\simeq r$ in the denominator and the common retarded time $u=t-r$ at leading quadrupole order. ThusDefine the trace-free mass quadrupole moment byTo make the requested normalization explicit, write the trace-free spatial metric coefficient as $h_{ij}^{\mathrm{TF}}=-2E_{ij}$. Trace reversal only adds a spatial trace, so the preceding equations give the far-zone quadrupole strain coefficientThe supplied scan does not contain the referenced perturbation handout. The definition above fixes the normalization of $E$ directly. If $E$ denotes the physical transverse-traceless tensor in that handout, the right-hand side is also to be transversely projected for the observation direction. Trace-free alone does not imply transverse. With $\mathbf{n}=\mathbf{x}/r$, putThe transverse-traceless projector gives $E_{ij}^{\mathrm{TT}}={\Lambda }_{ij,kl}{E}_{kl}$ and the physical spatial strain is $H_{ij}^{\mathrm{TT}}=2E_{ij}^{\mathrm{TT}}$. This also states the result without relying on omitted handout notation. The retarded construction and direction-dependent transverse projection are described in Carroll's general relativity lectures, with the appropriate changes of metric-sign convention.

For the equal-mass binary star, its centre of mass is at the origin and the separation is $2R$. The gravitational force on either star is $G{m}^2/(2R{)}^2$, while its centripetal force is $mR{\omega }^2$. HenceAt leading Newtonian order, $T^{00}$ is the sum of the two point-mass densities. Its moments areso the trace-free mass quadrupole moment isThe Newtonian bound system is treated in the usual leading weak-field quadrupole approximation; an exactly conserved effective source also includes the interaction stresses. Accelerated point-mass dust alone would not satisfy the conservation hypothesis.

Differentiating twice and evaluating at $u=t-r$ givesFor an observer on the $z$ axis this is already transverse and traceless. Its alternating plus polarization and cross polarization have a phase difference of $\pi /2$, giving circularly polarized gravitational radiation at twice the orbital frequency. The amplitude decays as $1/r$, as expected for outgoing radiation in three spatial dimensions.

More generally take the line of sight in the $xz$ plane at inclination $\iota$ to the orbital angular-momentum axis. In the transverse basis ${\mathbf{e}}_{\theta }=(\cos \iota ,0,-\sin \iota )$, ${\mathbf{e}}_{\varphi }=(0,1,0)$, define $A=4Gm{R}^2{\omega }^2/r$. Projection gives the physical gravitational wave polarizationsThus a generic view is elliptically polarized, the face-on view is circularly polarized and the edge-on view is linearly polarized. This is circular-binary quadrupole radiation; an overall sign or transverse-axis rotation only changes the polarization convention.

Write $\rho =\bar{\rho }(1+\delta )$, take a constant barotropic equation-of-state parameter $w$, and keep first-order terms in $\delta$, $v^i$ and $h_{ij}$. In synchronous gauge, the inverse metric is $g^{00}=-a^{-2}$, $g^{0i}=0$, $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$. The normalization $u^{\mu }{u}_{\mu }=-1$ gives $u^0=a^{-1}+O(v^2)$ and $u^i=a^{-1}{v}^i$. Substitution in the perfect fluid stress tensor gives the linear perfect-fluid stress tensor in synchronous gauge:In the first component the pressure contributions cancel; in the last, the velocity product is second order and the inverse-metric perturbation supplies the minus sign.

Use projections of stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, to retain all connection terms efficiently. Let $\mathcal{H}=a^{\prime }/a$ be the conformal Hubble parameter and $h={\delta }^{ij}{h}_{ij}$. Since $\sqrt{-g}=a^4(1+h/2)$, the fluid expansion isThe energy projection is $u^{\mu }{\partial }_{\mu }\rho +(\rho +P)\vartheta =0$. Its background is ${\bar{\rho }}^{\prime }+3\mathcal{H}(1+w)\bar{\rho }=0$. Expanding and subtracting that background leaves the synchronous perfect-fluid density equationFor momentum, the spatial projection is $(\rho +P)u^{\nu }{\nabla }_{\nu }{u}_i+({\delta }_i{}^{\nu }+u_i{u}^{\nu }){\partial }_{\nu }P=0$. The given connections yield $u^{\nu }{\nabla }_{\nu }{u}_i=v_i^{\prime }+\mathcal{H}{v}_i$ at first order. The pressure projection contains both ${\partial }_i\delta P$ and $v_i{\bar{P}}^{\prime }$; retaining the latter is essential. With $\delta P=w\bar{\rho }\delta$ and ${\bar{P}}^{\prime }=-3w\mathcal{H}(1+w)\bar{\rho }$, division by $\bar{\rho }(1+w)$ gives the Cosmological Euler equation in synchronous gauge,For Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}}$, these results areThe velocity equation assumes nonzero enthalpy $\bar{\rho }+\bar{P}$, so a cosmological constant with $w=-1$ is not a propagating fluid-velocity case. A time-dependent $w$ would also introduce additional terms and is not the constant-barotropic system used here.

The strong equivalence principle asserts universality of free fall even for bodies with gravitational self-binding, together with independence of local experimental outcomes, including gravitational experiments, from the location and velocity of a freely falling laboratory. “Local” means that external tidal variation across the apparatus can be neglected. It does not mean that all curvature, or a laboratory's own gravitational field, can be removed. The weak principle concerns freely falling test bodies without appreciable self-gravity; laboratory composition tests alone therefore do not establish the strong version.

Lunar laser ranging supplies evidence specifically sensitive to the strong version. The Earth and Moon have different fractions of gravitational self-energy. If those fractions affected gravitational and inertial mass differently, their accelerations toward the Sun would differ and the lunar orbit would acquire a periodically varying polarization, the Nordtvedt effect. Combining lunar ranging with laboratory bounds on composition-dependent differential acceleration separates this self-energy effect. A determination available before this examination found the self-energy contribution to the Earth–Moon difference in $m_G/m_I$ to be $(-2.0\pm 2.0)\times 10^{-13}$, consistent with zero; see the 2004 experimental analysis. Universality-of-free-fall experiments, gravitational redshift measurements and tests of local Lorentz invariance support the weaker constituent principles, but should not be confused with a direct self-gravity test.

In general relativity, all matter couples to one metric tensor, and freely falling test particles follow its geodesics. At any event, Riemann normal coordinates give $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$. Nongravitational local equations reduce to their special-relativistic forms in that frame. Gravitational binding is governed by the same metric field equations rather than a separate composition-dependent force, and the local gravitational coupling is universal. These features implement the strong equivalence principle. Curvature still governs geodesic deviation, so this construction removes uniform acceleration, not tidal gravity.

To geometrize Newtonian mechanics, use a four-dimensional manifold with coordinates $(t,x^1,x^2,x^3)$, absolute time ${\tau }_a=(dt{)}_a$ and a degenerate spatial inverse metric $h^{ij}={\delta }^{ij}$, $h^{0a}=0$. For a gravitational potential $\Phi (t,x)$ define the torsion-free Newton–Cartan gravitational connection byIt satisfies ${\nabla }_a{\tau }_b=0$ and ${\nabla }_a{h}^{bc}=0$. An affine geodesic has $d^{2}t/d{\lambda }^2=0$; choosing $\lambda =t$ makes its spatial equationsThis is Newton's free-fall law. The geometry is Newton–Cartan theory, with an affine connection and degenerate metric data, rather than a nondegenerate four-dimensional Lorentzian metric. The spatial slices themselves remain Euclidean.

Use the curvature conventionIn these coordinates the quadratic connection terms vanish, and the only nonzero curvature components, apart from antisymmetry in $c,d$, areIn particular $R_{00}={\nabla }^2\Phi$ and all other components of the Ricci tensor vanish. This Newtonian tidal curvature gives relative acceleration $\delta {\ddot{x}}^i=-{\partial }_i{\partial }_j\Phi \delta x^j$. Spatially constant gravitational acceleration can be removed by an accelerating frame because its Hessian vanishes. The Poisson equation for Newtonian gravity becomes $R_{ab}=4\pi G\rho {\tau }_a{\tau }_b$. Fluid mechanics is compatible with the same connection: the covariant continuity equation is the usual mass-conservation equation, and the spatial Euler equation is $u^b{\nabla }_b{u}^i=-{\rho }^{-1}{\partial }_{i}p$, which includes the gravitational force through the connection.

Now replace this absolute-time geometry by the Lorentzian metric of general relativity, with signature $(-,+,+,+)$. A perfect fluid haswhere $\epsilon$ is rest-frame energy density. In units $c=1$, $\epsilon \simeq \rho$ for a nonrelativistic fluid. The Newtonian equation suggests a covariant curvature equation sourced by $T_{ab}$. However $R_{ab}=\kappa T_{ab}$ alone conflicts with generic stress-energy conservation, because the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$. Among expressions linear in Ricci curvature, conservation therefore selectsA constant multiple of the metric is also divergence-free, giving the possible cosmological constant term. This is a heuristic selection using locality, universal metric coupling and second-order metric equations; Newtonian matching by itself does not prove uniqueness among every possible gravity theory.

To fix the coefficient, first set the cosmological constant to zero. The trace-reversed equation would read $R_{ab}=\kappa (T_{ab}-\frac{1}{2}T{g}_{ab})$. In the fluid rest frame, $T=-\epsilon +3p$, so $R_{00}=\frac{\kappa }{2}(\epsilon +3p)$. In the weak static Newtonian limit, $g_{00}=-(1+2\Phi )$ and $R_{00}\simeq {\nabla }^2\Phi$. For negligible pressure the Poisson equation for Newtonian gravity then requires $\kappa /2=4\pi G$. Restoring $c$ uses $x^0=ct$, $R_{00}\simeq {\nabla }^2\Phi /c^2$ and $T_{00}\simeq \rho c^2$, giving $\kappa =8\pi G/c^4$. ThusPressure contributes to relativistic active gravity, while the low-pressure weak-field limit recovers Newton's Poisson equation. The comparison fixes the Newtonian normalization but does not fix $\Lambda$, which is an independent parameter.

Use metric compatibility and a torsion-free connection. The coordinate formula for the Lie derivative of the metric tensor isSubstituting ${\partial }_a{g}_{bc}={\Gamma }^d{}_{ab}{g}_{dc}+{\Gamma }^d{}_{ac}{g}_{bd}$ and collecting connection terms gives $({\mathcal{L}}_{\xi }g{)}_{ab}={\nabla }_a{\xi }_b+{\nabla }_b{\xi }_a$. Thus a Killing vector obeys ${\nabla }_{(a}{\xi }_{b)}=0$, and conversely this Killing equation makes its flow preserve the metric. Its contraction gives ${\nabla }_a{\xi }^a=0$.

With the curvature convention fixed in Q1, contract the vector Ricci identity to obtainThe second term vanishes because the Killing divergence is zero. Metric compatibility converts the first term to ${\nabla }^b{\nabla }_a{\xi }_b$, and the Killing equation changes this to $-{\nabla }^b{\nabla }_b{\xi }_a$. Therefore the second covariant derivative of a Killing vector satisfiesThis derivation fixes the sign by the stated Ricci identity rather than by an unspecified convention for curvature.

For the axial Killing vector set $F^{ab}={\nabla }^b{\xi }^a$. It is antisymmetric and its divergence is ${\nabla }_b{F}^{ab}=-R^a{}_b{\xi }^b$. Work in geometrized units $G=c=1$ and with zero cosmological constant, as required by the asymptotically flat setting. If a spatial shell $\Omega$ has oriented boundary $S_2-S_1$, the given form of Stokes theorem applied to the Komar angular momentum givesThe vacuum Einstein field equations give $R_{ab}=0$ everywhere in the shell outside the matter source. The shell can still contain gravitational waves, since those have nonzero Weyl curvature rather than a nonzero Ricci tensor. HenceThis is angular momentum conservation in an axisymmetric vacuum region. Moving the outer surface through the radiative region adds no angular momentum. In an exactly axisymmetric spacetime there is no outward transport of the axial angular momentum by the waves. Axisymmetry does not imply stationarity, so this conclusion does not exclude radiation of energy.

For completeness the Einstein equations express the bulk current asOn an axisymmetric spacelike slice, $\xi$ is tangent to the slice and its contraction with $d^3{\Sigma }_a$ vanishes; there the trace contribution to the angular-momentum volume integral drops out. One must retain any inner surface charges if the integration region has inner boundaries. The vacuum shell argument itself does not require a matter-only representation of the total charge.

The current $Y$ is conserved on arbitrary slices, not just those to which $\xi$ is tangent. Indeed symmetry of the Ricci tensor and antisymmetry of ${\nabla }_a{\xi }_b$ giveThe last equality follows because a metric-preserving Killing flow also preserves the scalar curvature. Equivalently, the stress-energy conservation law and invariance of the stress trace give the same conserved current.

Let two spacelike hypersurfaces terminate on large enclosing spheres $S_1,S_2$ at the same asymptotic end. Join the spheres by a timelike tube lying wholly in the vacuum exterior. On this tube ${\nabla }_b{F}^{ab}=0$, so Stokes theorem gives zero difference of their Komar surface integrals, independently of the tube's shape. Alternatively, use the given Gauss theorem on a spacetime volume between the slices: ${\nabla }_a{Y}^a=0$, and the lateral flux is zero because $Y=0$ in the exterior vacuum tube. Taking the enclosing spheres to infinity with the usual asymptotic falloff therefore provesAll surface orientations in these equations use the same convention as the stated Komar integral. The argument concerns enclosing surfaces at the same asymptotic end, so it does not discard additional charges from other ends or inner boundaries.

For a fixed coordinate cell, the spatial volume element isThus its perturbation relative to the background expansion, or its perturbed comoving volume, is $\Delta V_c/V_c=h/2$. In the cold-matter rest frame each such cell contains a fixed amount of matter. Since density is mass divided by physical volume,Equivalently, stress-energy conservation gives ${\delta }_c^{\prime }=-h^{\prime }/2$; the stated relation fixes the spatially dependent integration constant by the initial mass-coordinate convention. A contraction of comoving volume gives a positive density contrast.

In the matter era $a\propto {\tau }^2$, so $\mathcal{H}=2/\tau$ and the matter density fraction is approximately one. The evolution equation reduces toFor ${\delta }_c\propto {\tau }^p$, the indicial equation is $p(p-1)+2p-6=(p-2)(p+3)=0$. ThereforeThe growing mode is proportional to $a$, and the decaying mode to $a^{-3/2}$. These are the density solutions after the comoving coordinate choice; the residual time mode found in part (i) explains why a similar power in an unfixed density variable cannot be interpreted in isolation.

There is a normalization omission in the printed density parameter if ${\bar{\rho }}_c$ denotes physical density. Its usual definition in conformal time isThe printed expression lacks $a^2$, unless its density is already rescaled by that factor. The matter-era statement ${\Omega }_c\simeq 1$ and the growth equation use the physical density fraction above.

On superhorizon scales during radiation domination, the regular adiabatic cosmological perturbation has approximately constant primordial curvature. The comoving density perturbation is gradient-suppressed, of order $(k/\mathcal{H}{)}^2$ times that curvature. Since $\mathcal{H}\propto {\tau }^{-1}$, it grows as ${\delta }_c\propto k^2{\tau }^2$ for fixed mode amplitude. This is the regular superhorizon solution in the specified slicing, not causal collapse driven by communication across the entire wavelength.

After radiation-era horizon entry, the dominant radiation develops pressure-supported acoustic oscillations. Its potential no longer provides a steady growing gravitational source, while cold matter is too small a fraction of the total density to produce rapid self-gravitating growth. The result is the Mészáros effect: growth is strongly suppressed until equality. The supplied stagnation approximation replaces that slow evolution by a constant. More accurately, ignoring the rapidly oscillating source at leading order gives ${\delta }_c^{\prime \prime }+{\delta }_c^{\prime }/\tau \simeq 0$, with a constant and a logarithmic solution. Thus radiation-era subhorizon growth is at most slow logarithmic growth, rather than the matter-era power law; treating it as constant is the explicit approximation used for the simple transfer estimate below.

The gravitational variable and the fields held fixed in a variation must be specified. For bosonic matter a second-order metric tensor formulation is sufficient. For spinorial matter a vierbein and a spin connection are needed to define the local Lorentz representation and its covariant derivative. “Second order” refers to a connection already expressed through derivatives of the vierbein; “first order” treats that connection as an independent variable.

For a metric tensor, a scalar field and an Abelian gauge field, consider the Einstein-Hilbert action with matter,Here $e=\sqrt{-g}$ and the connection in $R$ is Levi-Civita. The metric tensor variation uses $\delta e=-e{g}_{\mu \nu }\delta g^{\mu \nu }/2$ andMetric compatibility makes the last two terms a divergence. ThusStationarity for arbitrary compactly supported variations givesExplicitly,Scalar field and gauge-potential variations give ${\nabla }^2\varphi -V^{\prime }=0$ and ${\nabla }_{\mu }{F}^{\mu \nu }=0$. Their equations imply stress-energy conservation, consistently with the contracted Bianchi identity. If the boundary metric tensor is fixed instead of using compactly supported variations, add the Gibbons–Hawking–York boundary term, ${\kappa }^{-2}{\int }_{\partial M}{\epsilon }_{\partial M}\sqrt{|h|}K$, with the outward-normal orientation. Fixing the metric tensor alone does not otherwise remove the normal derivatives of its variation.

For spinorial matter, write $g_{\mu \nu }=e_{\mu }{}^a{e}_{\nu }{}^b{\eta }_{ab}$ and ${\nabla }_{\mu }\chi ={\partial }_{\mu }\chi +{\omega }_{\mu ab}{\gamma }^{ab}\chi /4$. In the second-order theory set $\omega =\omega (e)$ before varying. A symmetrized Dirac action, in the same no-$i$, mostly-plus convention as the earlier solutions, isVarying $\bar{\chi }$ and integrating by parts gives $(/\nabla +m)\chi =0$; varying $\chi$ gives the adjoint equation. The spinor components held fixed in a gravitational variation are components in the Lorentz frame. Both ${\gamma }^{\mu }=e_a{}^{\mu }{\gamma }^a$ and $\omega (e)$ vary with the vierbein. The former gives the canonical stress, and integrating the spin-connection variation supplies its spin-current improvement. On the Dirac equations the resulting symmetric stress tensor isThe off-shell expression also contains $g_{\mu \nu }{\mathcal{L}}_{\chi }/e$; this vanishes on the matter equations for the displayed symmetrized action. Treating the coordinate gamma matrices as fixed while varying the metric tensor would miss the displayed stress. Local Lorentz invariance removes the independent antisymmetric vierbein equation, leaving the symmetric Einstein field equations with this matter source.

In a first-order vierbein formulation, $e^a$ and an antisymmetric Lorentz connection ${\omega }^{ab}$ are independent. Defineand use $R(e,\omega )=e_a{}^{\mu }{e}_b{}^{\nu }{R}_{\mu \nu }{}^{ab}$. The curvature form of a connection variation is $\delta R^{ab}=D_{\omega }\delta {\omega }^{ab}$. For the Einstein-Hilbert term, integration by parts in form notation gives a connection equation proportional towhen matter has no connection dependence. For an invertible vierbein this forces $T^a=0$, so $\omega =\omega (e)$. The vierbein equation then becomes the second-order Einstein field equations. This applies to the scalar and Maxwell examples: use $F=dA$, which is independent of $\omega$ and remains gauge invariant even if torsion is present.

A Dirac field changes the connection equation. Directly varying its two covariant derivatives givesThus its spin density sources totally antisymmetric torsion. This is Einstein-Cartan theory, and the connection equation is algebraic rather than a propagation equation for an extra field. To see both its solution and its effect, writeFor the totally antisymmetric component sourced by this field, the Ricci scalar is $R(e,\omega )=R(e)-K_{abc}{K}^{abc}$ up to a divergence. The other irreducible contorsion components have no spinor source and their connection equations set them to zero. The auxiliary part of the Lagrangian is thereforeIts algebraic variation gives the algebraic torsion of a Dirac field,Here the first index of $K_{abc}$ labels the derivative, and the torsion convention is $T_{abc}=K_{bac}-K_{cab}$; specifying this convention fixes the sign. Eliminating $K$ leaves the second-order effective action with contact interactionIt is a local axial-current four-fermion interaction. Hence first-order gravity with Dirac spin density is equivalent to torsion-free second-order gravity with this additional interaction. It is not equivalent to simply declaring the spin connection torsion-free and dropping the contact term. In the effective formulation the Einstein field equations have the stress tensor of the full effective matter action. Before eliminating the connection, the antisymmetric stress and its spin current are related by the local Lorentz Noether identity.

Finally, the 1.5-order formalism is a convenient way to vary this effective action. Let $\hat{\omega }(e,\chi )$ solve the algebraic connection equation and let $S_2(e,\chi )=S_1(e,\hat{\omega }(e,\chi ),\chi )$. The chain rule givesThus vary as in first order, then insert the solved connection. One does not need to calculate the complicated induced connection variation. This procedure is justified by its connection equation, not by discarding that variation arbitrarily. In supergravity the gravitino sources a contorsion quadratic in fermions; substituting it generates the four-fermion terms, while the 1.5-order method simplifies the local supersymmetry cancellation through the lower orders considered above.