Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 309 / 4

Put $h_{ab}=\delta g_{ab}$, $h=g^{ab}{h}_{ab}$, and $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$. Throughout this part use the Riemann curvature tensor convention printed in the paper. Varying $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives the inverse-metric variation. The determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ gives the volume form variation. Together these areIn Lorentzian coordinates the latter reads $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}h$.

For the connection, vary metric compatibility ${\nabla }_a{g}_{bc}=0$:Add the equations with derivatives $a,b$, subtract that with derivative $c$, and use symmetry of the lower indices of the Levi-Civita connection. This givesThe difference of two connections is a tensor, so this formula is covariant.

Varying the coordinate curvature formula yields the Palatini identityThe contractions of the connection variation areFinally, $\delta R=(\delta g^{ab})R_{ab}+g^{ab}\delta R_{ab}$ gives the metric variation of scalar curvature:The double divergence may equivalently be written ${\nabla }_a{\nabla }_b{h}^{ab}$: relabeling the contracted dummy indices and using symmetry of $h^{ab}$ gives the same expression. The derivative terms are a covariant divergence, useful when varying a curvature-dependent action.

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 Euler-Lagrange field equation obtained by varying the nonminimally coupled scalar field isOne can check stress-energy conservation directly. The divergence of the kinetic contribution is $(\square \Phi ){\nabla }_b\Phi$, because second derivatives of a scalar commute. With $f={\Phi }^2$, the contracted Bianchi identity gives ${\nabla }^a{G}_{ab}=0$, while the Ricci identity givesConsequentlySince ${\nabla }_{b}f=2\Phi {\nabla }_b\Phi$, the full stress-energy tensor satisfies

The general reason is the diffeomorphism Noether identity for a scalar field. Under the Lie derivative along a compactly supported vector field $X$, $\delta g_{ab}=2{\nabla }_{(a}{X}_{b)}$ and $\delta \Phi =X^a{\nabla }_a\Phi$. Invariance of the action under diffeomorphisms gives, after integration by parts,Arbitrariness of $X$ gives the same divergence identity. Conservation requires the scalar equation of motion, with no need to impose the Einstein field equations for the background metric.