Category: General Relativity
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In the precedent article Covariant differentiation exercise 1: calculation in cylindrical coordinates, we have deduced the expression of the covariant derivative of a tensor of rank 1, i.e of a contravariant vector - type (1,0) or of a covariant vector - type (0,1).

It can be shown that the covariant derivatives of higher rank tensors are constructed from the following building blocks:

 Following the three rules given above, we obtain for tensors of rank 2, respectively of type (1,1) Tμν , (0,2) Tμν and (2,0) Tμν:

We recall from our article Minkowski's Four-Dimensional Space-Time the Euclidean metric tensor's expression  for Cartesian coordinates

 And substituting gij in the second above equality dedicated to type (0,2) tensor gives

 As all the terms gij are constant, the first term δgij/δxβ is null. And because there is no curvature in the Euclidean space, all the Christoffel symbols vanish, so that the conplete right-hand side of the equation equals zero, and therefore

But this is the magic and clever thing about tensors, if this equation holds true in a particular coordinates system, here the Cartesian coordinates, it must be true for the Euclidean metric in ALL coordinate systems.

Covariant derivation of the Euclidean metric in spherical coordinates

Let's try to verify this by calculating one component of the covariant differentiation in the spherical coordinates.

We recall from our article that in spherical coordinates, the metric's expression is

If we were to calculate the component gΦΦ;θ, we should then write

But gαφ !=0 only if α = φ, based on the above expression, so we can simplify this equation

 

Or

 And from previous article Christoffel symbol exercise: calculation in polar coordinates part II, we know the expression of the Christoffel symbols ΓΦθΦ = ΓΦΦθ = cosθ/sinθ

 so that

Finally, we confirm that this component of the covariant derivative with respect to θ equals also zero in a polar coordinates system, as expected