On the Euclidean plane take the scalar function

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Tensors and differential geometry · Tensors and differential geometry

Invariance, Ch The metric, the connection and geodesics

Problem

On the Euclidean plane take the scalar function f(x,y)=xf(x,y) = x. Its gradient, the covector ωi=∂if\omega_i = \partial_i f, has Cartesian components ωx=1\omega_x = 1, ωy=0\omega_y = 0, so the second covariant derivative ∂i∂jf\partial_i\partial_j f is 00 in Cartesian coordinates, where Γ=0\Gamma = 0. The covariant derivative ∂μων−Γλμνωλ\partial_\mu\omega_\nu - \Gamma^\lambda{}_{\mu\nu}\omega_\lambda is a (0,2)(0,2) tensor, as shown in the chapter. If its components vanish in one coordinate system, they vanish in every one. In plane polar coordinates x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, f=rcos⁡θf = r\cos\theta, so ωr=cos⁡θ\omega_r = \cos\theta and ωθ=−rsin⁡θ\omega_\theta = -r\sin\theta. Using the Christoffel symbols Γrθθ=−r\Gamma^r{}_{\theta\theta} = -r, Γθrθ=Γθθr=1/r\Gamma^\theta{}_{r\theta} = \Gamma^\theta{}_{\theta r} = 1/r (all others zero), compute the four components ∇μων=∂μων−Γλμνωλ\nabla_\mu\omega_\nu = \partial_\mu\omega_\nu - \Gamma^\lambda{}_{\mu\nu}\omega_\lambda for μ,ν∈{r,θ}\mu,\nu \in \{r,\theta\} and verify each one is zero, as the tensor transformation law requires.

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