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