Find an expression for the second covariant derivative, written in semicolon notation as vi;jk≡(vi;j);k, of a vector vi. By interchanging the order of differentiation and then subtracting the two expressions, we define the components Rijk^l of the Riemann tensor as vi;jk−vi;kj≡Rijk^l vl. Show that

Question

Find an expression for the second covariant derivative, written in semicolon notation as [latex]{v}_{i; jk} ≡ ({v}_{i; j} )_{; k}[/latex], of a vector [latex]{v}_{i}[/latex]. By interchanging the order of differentiation and then subtracting the two expressions, we define the components [latex]{ R }_{ ijk }^{ l }[/latex] of the Riemann tensor as

[latex]{v}_{i; jk} − {v}_{i; kj} ≡{ R }_{ ijk }^{ l }{v}_{l}[/latex].

Show that in a general coordinate system [latex]{u}^{i}[/latex] these components are given by

[latex]R_{i j k}^l=\frac{\partial \Gamma_{i k}^l}{\partial u^j}-\frac{\partial \Gamma_{i j}^l}{\partial u^k}+\Gamma_{i k}^m\Gamma_{m j}^l-\Gamma^m{ }_{i j} \Gamma_{m k}^l.[/latex]

By first considering Cartesian coordinates, show that all the components [latex]{ R }_{ ijk }^{ l }≡ 0[/latex] for any coordinate system in three-dimensional Euclidean space.

In such a space, therefore, we may change the order of the covariant derivatives without changing the resulting expression.

Accepted Answer

For the covariant derivative of the covariant components of a vector, we have

[latex]v_{i ; j}=\frac{\partial v_i}{\partial u^j}-\Gamma_{i j}^k v_k,[/latex]

where [latex]\Gamma_{i j}^k[/latex] is a Christoffel symbol of the second kind. Hence,

[latex]\begin{aligned}v_{i ; j k} & \equiv\left(v_{i ; j} ight)_{; k} \& =\left(\frac{\partial v_i}{\partial u^j}-\Gamma_{i j}^l v_l ight)_{; k} \& =\frac{\partial}{\partial u^k}\left(\frac{\partial v_i}{\partial u^j}-\Gamma_{i j}^l v_l ight)-\Gamma_{i k}^m\left(\frac{\partial v_m}{\partial u^j}-\Gamma_{m j}^l v_l ight) \& =\frac{\partial^2 v_i}{\partial u^k \partial u^j}-\Gamma_{i j}^l \frac{\partial v_l}{\partial u^k}-v_l \frac{\partial \Gamma_{i j}^l}{\partial u^k}-\Gamma_{i k}^m \frac{\partial v_m}{\partial u^j}+\Gamma_{i k}^m \Gamma_{m j}^l v_l .\end{aligned}[/latex]

Interchanging subscripts j and k,

[latex]v_{i ; k j}=\frac{\partial^2 v_i}{\partial u^j \partial u^k}-\Gamma_{i k}^l \frac{\partial v_l}{\partial u^j}-v_l \frac{\partial \Gamma_{i k}^l}{\partial u^j}-\Gamma_{i j}^m \frac{\partial v_m}{\partial u^k}+\Gamma_{i j}^m \Gamma_{m k}^l v_l .[/latex]

When these two expressions are subtracted to define the Riemann tensor, the first, second and fourth terms (the second of one with the fourth of the other and vice versa) on the two RHSs cancel in pairs to yield

[latex]R_{i j k}^l v_l \equiv v_{i ; j k}-v_{i ; k j}=\left(\frac{\partial \Gamma_{i k}^l}{\partial u^j}-\frac{\partial \Gamma_{i j}^l}{\partial u^k}+\Gamma_{i k}^m \Gamma_{m j}^l-\Gamma_{i j}^m \Gamma_{m k}^l ight) v_l .[/latex]

Now, in three-dimensional Euclidean space, one possible coordinate system is the Cartesian one. In this system g = 1 and all of its derivatives are zero. Thus all Christoffel symbols and their derivatives are zero, as are all components of the Riemann tensor. As all the components vanish in this Cartesian coordinate system, they must do so in any coordinate system in this space.

Question 26.27: Find an expression for the second covariant derivative, writ...

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