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Let the symmetric metric perturbation $h^{\alpha \beta }$ be varied with compact support. Integrating the mixed derivative product by parts makes its integral equal to that of $A_{\sigma }{A}^{\sigma }$. Thus the supplied massless Fierz-Pauli action has, up to a boundary term, densityFor example, the equivalence of the mixed term follows from commuting flat-space partial derivatives after moving one derivative off $h^{\rho \sigma }$. Here $\delta h={\eta }_{\alpha \beta }\delta h^{\alpha \beta }$ and $\delta A^{\nu }={\partial }_{\mu }\delta h^{\mu \nu }$.

Integrating each variation by parts, the four terms contribute respectivelyThe symmetrization in the second term is required because the varied field is symmetric. Comparing this coefficient with the Einstein tensor perturbation above givesThus arbitrary compactly supported variations give $\delta G_{\alpha \beta }=0$, exactly the vacuum Linearized Einstein equations. The minus sign and overall normalization of the action do not change those equations; boundary conditions justify the discarded total derivatives.

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

The contracted Bianchi identity gives ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$. Combining the proposed equation with stress-energy conservation therefore imposes ${\nabla }_{b}R=0$. Taking its trace gives $R=\kappa T$, with $T=g^{ab}{T}_{ab}$, so for nonzero $\kappa$ the matter trace must also be constant. Ordinary matter need not satisfy this extra condition. For example, if asymptotic flatness forces the constant scalar curvature to vanish, the proposal would require $T=0$ everywhere.

The problem is an unjustified restriction on the matter trace, rather than the absence of every possible solution. The divergence-free Einstein tensor avoids it:The Einstein field equations $G_{ab}+\Lambda g_{ab}=\kappa T_{ab}$, with constant cosmological constant $\Lambda$, are consequently compatible with stress-energy conservation.

Define the stress-energy tensor by metric variation with the scalar held fixed:Use compactly supported variations to remove boundary terms. The first covariant derivative of a scalar field is an ordinary derivative, so it has no connection variation. Varying the kinetic term and the volume form gives

For the curvature term set $f={\Phi }^2$. Using part (a), its variation isApplying integration by parts twice moves each pair of derivatives from the metric variation onto $f$. The result iswhere $G_{ab}=R_{ab}-R{g}_{ab}/2$ is the Einstein tensor and $\square ={\nabla }^c{\nabla }_c$ is the d'Alembert operator. Comparing with the defining variation gives the stress-energy tensor of a nonminimally coupled scalar field:The factor $2\xi$ follows from the action's coupling $-\xi R{\Phi }^2$; a convention using $-\xi R{\Phi }^2/2$ would instead give $\xi$ in that bracket. At $\xi =0$ this reduces to the massless Klein-Gordon scalar stress-energy tensor.

The Einstein tensor is $G_{\alpha \beta }=R_{\alpha \beta }-\frac{1}{2}{g}_{\alpha \beta }R$. Taking the metric trace of the Einstein field equations in four dimensions givesand substitution back gives the Trace-reversed Einstein field equations in D dimensions specialized to $D=4$:For linearized gravity, the given equation and $G_{\mu \nu }=8\pi T_{\mu \nu }$ implySince $\bar{h}=-h$, its trace is $\square h=16\pi T$. Using ${\bar{h}}_{\mu \nu }=h_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }h$ then yields

Contracting the Ricci tensor gives the Ricci scalarHence the time-time component of the Einstein tensor is $G_{00}=R_{00}-\frac{1}{2}{g}_{00}R=3H^2$. A comoving perfect fluid in general relativity has $T_{00}=\rho$, so the time-time Einstein field equations give the flat Friedmann equation

At any point choose orthonormal coordinates. The antisymmetries and pair symmetry of the Riemann curvature tensor leave only $R_{1212}$ independent. In these coordinates,while $R=2R_{1212}$. Since this is a tensor identity, it follows in every coordinate system that the Ricci tensor in two dimensions satisfiesThe Einstein tensor consequently vanishes identically. Hence the two-dimensional vacuum field equation contains no local gravitational dynamics, while coupling the unmodified equation to matter would require $T_{\alpha \beta }=0$. This is the basic degeneracy of two-dimensional general relativity.

Contract the Riemann tensor to obtainand define the Einstein tensorContracting the differential Bianchi identity twice, using the curvature symmetries and metric compatibility, gives the contracted Bianchi identityIt follows immediately thatThe Einstein equations with cosmological constant areSince $\nabla g=0$, the left side is divergence-free, so consistency requires and implies the local conservation law ${\nabla }^{\nu }{T}_{\mu \nu }=0$.

Expand the Einstein tensor about the Minkowski metric. The quadratic term in the metric perturbation defines $t_{\mu \nu }=-G_{\mu \nu }^{(2)}[h,h]/(8\pi )$, a symmetric tensor relative to the background. For a first-order vacuum solution, the expanded contracted Bianchi identity gives ${\partial }^{\mu }{t}_{\mu \nu }=0$. Equivalently, a second-order correction $j$ obeys $G^{(1)}[j]=8\pi t$, whose flat divergence vanishes identically. This conservation is an on-shell statement. The local $t$ depends on perturbative gauge and the background split; conservation alone does not make it a gauge-independent local gravitational stress-energy tensor.

The algebraic symmetries of the Riemann curvature tensor leave only one independent curvature component in two dimensions and forceConsequently the Einstein tensor vanishes identically.