Past exam of the general relativity course of the University of Oxford / 2011 / 1

The paper uses $T_{\mu \nu }{}^{\lambda }={\Gamma }_{\nu \mu }{}^{\lambda }-{\Gamma }_{\mu \nu }{}^{\lambda }$, the negative of a common torsion tensor convention. Keep that sign throughout. Also, ${\nabla }_{\mu }{\nabla }_{\nu }\phi$ denotes the covariant Hessian, so the second derivative acts on the lower index $\nu$ of the covector ${\nabla }_{\nu }\phi$; it is not just a composition of two directional derivatives on a scalar.

Since the covariant derivative of a scalar field is its ordinary derivative,Commuting the ordinary partial derivatives therefore givesFor a vector field, write $D_{\nu }{X}^{\lambda }={\partial }_{\nu }{X}^{\lambda }+{\Gamma }_{\nu \rho }{}^{\lambda }{X}^{\rho }$. This is a tensor with one covariant and one contravariant index. HenceSubtracting the expression with $\mu ,\nu$ exchanged, the first two terms combine toThe coefficients of first derivatives of $X$ in those two terms cancel. The last terms give $({\Gamma }_{\nu \mu }{}^{\sigma }-{\Gamma }_{\mu \nu }{}^{\sigma })D_{\sigma }{X}^{\lambda }$. Identifying the curvature coefficient yieldsHere the bracket denotes antisymmetrization of the tensorial second covariant derivative, consistently with the preceding calculation.

To check tensoriality directly, antisymmetrize the supplied transformation law of the connection coefficients in the two lower indices. The inhomogeneous term vanishes because mixed coordinate second derivatives are symmetric. The remaining transformation law iswhich is precisely that of a $(1,2)$ tensor. Intrinsically, with the paper's sign,The derivative terms involving the components of $X,Y$ cancel against their Lie bracket, leaving $({\Gamma }_{\nu \mu }{}^{\lambda }-{\Gamma }_{\mu \nu }{}^{\lambda })X^{\mu }{Y}^{\nu }$. This also shows that the expression is linear over smooth functions in each argument.

Finally take the torsion-free connection and the commuting parameter fields $Y,Z$ along the geodesic family. Vanishing torsion gives ${\nabla }_{Y}Z={\nabla }_{Z}Y$, while affine geodesic parametrization gives ${\nabla }_{Y}Y=0$. The curvature operator in the specified convention isThereforeIn components this is the geodesic deviation equationThe connecting field represents infinitesimal separation within the family, so this is a linear equation for relative free-fall acceleration, rather than an exact coordinate difference for widely separated worldlines.