Show by explicit calculation that is a tensor: the second-derivative term in the…

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Relativity & Gravitation · Differential geometry of curved spacetime

The Geometry of Gravity, Ch The covariant derivative, parallel transport and geodesics

Problem

Show by explicit calculation that ∇μVν=∂μVν+ΓνμλVλ\nabla_\mu V^\nu=\partial_\mu V^\nu+\Gamma^\nu{}_{\mu\lambda}V^\lambda is a tensor: the second-derivative term in the transformation of ∂μVν\partial_\mu V^\nu and the one in the transformation of ΓνμλVλ\Gamma^\nu{}_{\mu\lambda}V^\lambda cancel. (a) Starting from V′ν′=(∂x′ν′/∂xν)VνV'^{\nu'}=(\partial x'^{\nu'}/\partial x^\nu)V^\nu, use the product rule and the chain rule to write ∂μ′V′ν′\partial_{\mu'}V'^{\nu'} as the tensor law for ∂μVν\partial_\mu V^\nu plus one extra term built from a second derivative of the coordinate change and VνV^\nu itself. (b) Using the transformation law of Γ\Gamma given earlier in the chapter, Γ′λ′μ′ν′=∂x′λ′∂xρ∂xσ∂x′μ′∂xτ∂x′ν′Γρστ+∂x′λ′∂xρ∂2xρ∂x′μ′∂x′ν′,\Gamma'^{\lambda'}{}_{\mu'\nu'}=\frac{\partial x'^{\lambda'}}{\partial x^\rho}\frac{\partial x^\sigma}{\partial x'^{\mu'}}\frac{\partial x^\tau}{\partial x'^{\nu'}}\Gamma^\rho{}_{\sigma\tau}+\frac{\partial x'^{\lambda'}}{\partial x^\rho}\frac{\partial^2x^\rho}{\partial x'^{\mu'}\partial x'^{\nu'}}, and the vector law again, write Γ′ν′μ′λ′V′λ′\Gamma'^{\nu'}{}_{\mu'\lambda'}V'^{\lambda'} as the tensor law for ΓνμλVλ\Gamma^\nu{}_{\mu\lambda}V^\lambda plus one extra term, also built from a second derivative and VV. (c) Differentiate the chain-rule identity (∂x′ν′/∂xμ)(∂xμ/∂x′μ′)=δμ′ν′(\partial x'^{\nu'}/\partial x^\mu)(\partial x^\mu/\partial x'^{\mu'})=\delta^{\nu'}_{\mu'} with respect to xνx^\nu to obtain a relation between the two extra terms, and use it to show they cancel exactly, so that ∇μ′V′ν′=(∂xμ/∂x′μ′)(∂x′ν′/∂xν)∇μVν\nabla_{\mu'}V'^{\nu'}=(\partial x^\mu/\partial x'^{\mu'})(\partial x'^{\nu'}/\partial x^\nu)\nabla_\mu V^\nu: a genuine type (1,1)(1,1) tensor.

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