Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 52 / 2 / b

Contract the Christoffel symbol formula for the Levi-Civita connection. The first and third derivative terms cancel after relabelling their summed indices:The Jacobi determinant derivative formula gives ${\partial }_{\mu }g=g{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. In a Lorentzian chart $g<0$, and hence ${\partial }_{\mu }\sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. Thus the trace is the logarithmic derivative of the volume density:In particular, the covariant divergence formula for a vector field is ${\nabla }_{\mu }{A}^{\mu }=(-g{)}^{-1/2}{\partial }_{\mu }(\sqrt{-g}{A}^{\mu })$.

Vary the inverse metric tensor while holding the matter fields fixed. Differentiating $g_{a\mu }{g}^{\mu b}={\delta }_a{}^b$ gives $\delta g_{ab}=-g_{a\mu }{g}_{b\nu }\delta g^{\mu \nu }$. The Jacobi determinant derivative formula consequently implies the inverse metric volume variationSince the matter Lagrangian in this question contains no derivatives of the metric tensor, its metric variation is algebraic. The product rule givesMultiplying by $-2/\sqrt{-g}$ proves the equivalence of the stress-energy definitions:The variations of the symmetric metric tensor are understood symmetrically. If the matter action contained metric derivatives, the definition would instead require a functional derivative and the corresponding integration by parts.

For the minimally coupled scalar field, keep the metric tensor fixed and vary $\varphi$. A scalar's first covariant derivative equals its ordinary derivative, soApply integration by parts for tensor fields. The boundary contribution is proportional to $\delta \varphi$ and vanishes under the given boundary condition, leavingThe interior variation is arbitrary. Thus the scalar equation of motion isHere the d'Alembert operator contains the Levi-Civita connection in the second covariant derivative, even though the first derivative of the scalar field is ordinary.

The metric dependence of the minimally coupled scalar field Lagrangian is $\partial L/\partial g^{\alpha \beta }=-\frac{1}{2}{\nabla }_{\alpha }\varphi {\nabla }_{\beta }\varphi$. Part (ii) therefore gives the stress-energy tensorUse metric compatibility and the symmetric Hessian of a scalar field to computeThis is the scalar stress-energy divergence identity, an instance of the diffeomorphism Noether identity for a scalar field. On the scalar equation of motion it gives stress-energy conservation, $\boxed{{\nabla }_{\mu }{T}^{\mu }{}_{\alpha }=0}$.