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There are two possibilities for making this calculation: to use the relation ∆Ψ = div ( grad Ψ) or derive directly using covariant derivatives. Let us follow the second (a bit more economic) option, and also use the diagonal form of the metric tensor ∆Ψ = g ij ∇i ∇j Ψ = g ij ∂i ∂j − Γkij ∂k Ψ = g rr 2 2 ∂2Ψ ∂Ψ ϕϕ ∂ Ψ θθ ∂ Ψ + g + g − g ij Γkij k . 2 2 2 ∂r ∂ϕ ∂θ ∂x Since we already have all components of the Christoffel symbol, it is easy to obtain g ij Γrij = g ϕϕ Γrϕϕ + g θθ Γrθθ = − 2 r and g ij Γθij = Γθϕϕ g ϕϕ = − 1 cotan θ r2 ij while Γϕ ij g = 0.

Use the existence of global Cartesian coordinates in the flat space and the tensor form of the transformations from one coordinates to another. Observation. Of course, the same commutator is non-zero if we consider a more general geometry, which does not admit a global orthonormal basis. g. for the sphere. 5) Verify that for the scalar field Ψ the following relation takes place: ∆Ψ = div ( grad Ψ) . Hint. Use the fact that grad Ψ = ∇Ψ = ei ∂i Ψ is a covariant vector. Before taking div , which is defined as an operation over the contravariant vector, one has to rise the index: ∇i Ψ = g ij ∇j Ψ.

In order to define its covariant derivative, we perform the following steps: 1) Transform it to the Cartesian coordinates X a Wba = ∂X a ∂xj Wi . ∂xi ∂X b j 2) Take partial derivative with respect to the Cartesian coordinates X c ∂c Wba = ∂ ∂X c ∂X a ∂xj Wi . ∂xi ∂X b j 3) Transform this derivative back to the original coordinates xi using the tensor rule. l = ∇k W m ∂X c ∂xl ∂X b ∂c Wba . ∂xk ∂X a ∂xm It is easy to see that according to this procedure the conditions i) and ii) are satisfied automatically.