Covariant derivatives and Christoffel symbols

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In Minkowski spacetime with Minkowski coordinates (ct,x,y,z) the derivative of a vector is just

since the basis vectors do not vary. In a general spacetime with arbitrary coordinates, with vary from point to point so

Since is itself a vector for a given it can be written as a linear combination of the bases vectors:

The 's are called Christoffel symbols [ or the metric connection  ]. Thus we have:

so

Thus we can write

where

Let us now prove that are the components of a 1/1 tensor. Remember in section 3.5 we found that was only a tensor under Poincaré transformations in Minkowski space with Minkowski coordinates. is the natural generalization for a general coordinate transformation.

Writing , we have:

Now

therefore

so we obtain:

Now using , and we obtain:

so

We have shown that are indeed the components of a 1/1 tensor. We write this tensor as

It is called the covariant derivative  of . Using a Cartesian basis, the components are just , but this is not true in general; however for a scalar we have:

since scalars do not depend on basis vectors.

Writing , we can find the transformation law for the components of the Christoffel symbols .

This is just

We can calculate the covariant derivative of a one- form  by using the fact that is a scalar for any vector :

We have

Since and are tensors, the term in the parenthesis is a tensor with components:

We can extend this argument to show that

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