Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 17; Box 17.2.

We’ve seen how to define the absolute gradient or **covariant derivative** of a contravariant vector, giving the formula:

The covariant derivative of this vector is a tensor, unlike the ordinary derivative. Here we see how to generalize this to get the absolute gradient of tensors of any rank.

First, let’s find the covariant derivative of a covariant vector . The starting is to consider . The quantity is a scalar, and to proceed we *require* two conditions:

- The covariant derivative of a scalar is the same as the ordinary derivative.
- The covariant derivative obeys the product rule.

These two conditions aren’t derived; they are just required as part of the definition of the covariant derivative.

Using these rule 2, we have

We now apply rule 1 to the LHS:

Equating 3 and 5 we get

In 8 we’ve relabelled the dummy index to in the second and third terms so we could factor out in the last line. Since the vector is arbitrary, the factors multiplying it on each side must be equal, so we get

from which we can get the covariant derivative of :

To extend this argument to a tensor of higher rank with mixed indices, we generalize this argument. First, we contract all the indices of the tensor with covariant or contravariant vectors, as appropriate, and then apply the product rule. So for example

We now substitute for the covariant derivatives of the vectors in 12 and set the result equal to 13, and cancel terms. We then use the fact that , and are all arbitrary so the factors on each side of the equation multiplying them must be equal. The result is

In general, the rule is that for each contravariant (upper) index in the tensor, there is a positive term with a Christoffel symbol, and for each covariant (lower) index, there is a negative term.

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