Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 68 / 3

The Brans-Dicke theory couples a scalar field $\Phi$ to the Ricci scalar. Assume $\Phi \ne 0$, constant $\omega$, matter independent of $\Phi$, and variations of compact support so boundary terms may be discarded. Use the paper's curvature sign convention and signature $(+---)$. Set $X=g^{ab}{\partial }_a\Phi {\partial }_b\Phi$ and $\square \Phi ={\nabla }_a{\nabla }^a\Phi$. Define the stress-energy tensor consistently byThis convention fixes the sign of the matter term.

Vary with respect to $g^{ab}$. The variation of metric volume density contributes $-g_{ab}/2$ times each scalar Lagrangian, while $\delta R=R_{ab}\delta g^{ab}+g^{ab}\delta R_{ab}$. Applying the Palatini identity followed by two rounds of integration by parts gives the metric variation of a scalar-curvature coupling,The differentiated $\Phi$ terms remain because the curvature multiplier is not constant. Varying the kinetic term at fixed $\Phi$ givesCombining these terms with the matter variation, the Euler-Lagrange equation isRaising both free indices reproduces the contravariant form, because ${\nabla }_a{g}_{bc}=0$.

The independent scalar variation givesor $R+\omega {\Phi }^{-2}X-2\omega {\Phi }^{-1}\square \Phi =0$. The trace of the metric tensor equation in four dimensions isMultiplying the scalar equation by $\Phi$ eliminates the same combination $\Phi R+\omega {\Phi }^{-1}X$. Thus the scalar equation of Brans-Dicke theory isAt the degenerate coupling of Brans-Dicke theory, $\omega =-3/2$, the combined equations impose $T=0$ instead; division by $3+2\omega$ is unavailable. If matter depended directly on $\Phi$, its scalar variation would add a source and the displayed scalar equation would change.

The following lettered sections justify the supplied variation identities; they are supporting identities in the PDF, rather than three independent field-equation problems.

Differentiate $g^{ac}{g}_{cb}={\delta }_b^a$ while holding the coordinate system fixed. The variation of inverse metric obeysMultiplication by the inverse metric tensor givesThis is the matrix inverse derivative written with tensor indices. A variation of the metric tensor is symmetric, so the right-hand side is symmetric in $c,d$.

Write $g=\det (g_{ab})$. The Jacobi formula gives $\delta g=g{g}^{ab}\delta g_{ab}$. Since a Lorentzian metric of signature $(+---)$ has negative determinant,The variation of metric volume density therefore has the opposite sign when expressed using the inverse metric tensor: $\delta \sqrt{-g}=-\sqrt{-g}{g}_{ab}\delta g^{ab}/2$. This follows immediately from the variation of inverse metric and explains the volume term in the field-equation derivation.

Although connection coefficients are not tensor components, their variation $C^a{}_{bc}=\delta {\Gamma }^a{}_{bc}$ is tensorial: the difference of affine connections is a tensor. Varying the Levi-Civita connection at fixed coordinates givesIn the paper's curvature sign convention, varying the derivative and quadratic-connection terms and grouping them into covariant derivatives gives the Palatini identityOne may verify the grouping in normal coordinates at a point, where the affine connection vanishes and the expression is just the difference of two partial derivatives; both sides are tensors, so it holds in every coordinate frame. Reversing the definition of the Riemann curvature tensor reverses this identity's right-hand side.

For completeness, put $q_{ab}=\delta g_{ab}$ and $q=g^{ab}{q}_{ab}$. The contracted variation uses $C^a{}_{ba}={\nabla }_{b}q/2$ and $g^{bc}{C}^a{}_{bc}={\nabla }_b{q}^{ab}-{\nabla }^{a}q/2$. After two integrations by parts,up to the discarded boundary term. Substituting $q^{ab}=-\delta g^{ab}$ gives the differentiated-scalar terms in the metric variation of a scalar-curvature coupling.