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Let $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. In the metric variation specified here, the covariant metric tensor $g_{\mu \nu }$ is varied while the scalar field is held fixed. Differentiating the inverse and the determinant givesThus $\delta \mathcal{L}=\frac{1}{2}{\partial }^{\mu }\varphi {\partial }^{\nu }\varphi \delta g_{\mu \nu }$, and the action varies asThe functional derivative therefore gives $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$. Lowering both indices yields the stress-energy tensor of a canonical scalar field:The determinant variation has a positive sign because the variable is $g_{\mu \nu }$; varying the inverse metric tensor instead would reverse that intermediate sign.

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.

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.

Write $h_{ab}=\delta g_{ab}$ and $h=g^{ab}{h}_{ab}$. Standard metric variation givesand variation of metric compatibility gives

Vary $g^{ac}{g}_{cb}={\delta }_b^a$. The product rule givesand multiplication by $g^{bd}$ yields the metric variation

Put $h_{ab}=\delta g_{ab}$. Varying $g^{ac}{g}_{cb}={\delta }_b^a$ givesThe determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ givesFinally, varying the formula for the Christoffel symbols and rewriting partial derivatives covariantly gives the tensorThese are the basic metric variation identities.

For action $S=\int \sqrt{-g}[-\frac{1}{2}(\nabla \Phi {)}^2-\xi R{\Phi }^2]d^{4}x$, metric variation defines $\delta S=\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}{d}^{4}x$ and gives the displayed tensor. The contracted Bianchi identity and the scalar derivative commutator yield ${\nabla }^a{T}_{ab}=(\square \Phi -2\xi R\Phi ){\nabla }_b\Phi$, so the stress-energy tensor is conserved on the field equation.