After spending some time looking at tensors, we can now expose the problem of how to differentiate a tensor.

##### Covariant differentiation for a contravariant vector

Consider a vector V = V^{α}e_{α} (ie the tensor has contravariant components V^{α} and coordinate basis vectors e_{α}). Using the product rule of derivation, the rate of change of the components Vα (of the vector V) with respect to x^{β}.

But we recall from our article Christoffel Symbol or Connection coefficient that the connection coefficients are defined by:

Substituing this expression in the above equation gives

The right hand term has two dummy indices (ie indices to be summed over) α and γ. We can improve the formula by changing *α* to* γ* and* γ* to *α* to give:

and factoring out e_{α} gives

This expression indicates the rate of change of V^{α} in each of the directions β of the coordinate system x^{β}, and is known as the **covariant derivative** of the contravariant vector V. The * nabla* symbol is used to denote the covariant derivative

In words: the covariant derivative is the usual derivative along the coordinates with correction terms which tell how the coordinates change.

The intesting property about the covariant derivative is that, as opposed to the usual directional derivative, **this quantity transforms like a tensor, **i.e. it **is independant of the manner in which it is expressed in a coordinate system**.

**Remark 1**: As we have seen in our articles Local Flatness or Local Inertial Frames and SpaceTime curvature and Local Inertial Frame (LIF), in a inertial frame of reference, the vanishing of the partial derivatives of the metric tensor at any point of M **is equivalent to the vanishing of Christoffel symbols**, and then we can write this fundamental equality in the context of any inertial or local inertial frame:

**Remark 2**: the fact that the Christoffel symbol by itself does NOT transform as a tensor can be easily deduced from the fact that we can always find an (local) inertial frame in which its value equals zero, which should not be possible for a tensor.

**Remark 3**: we can also find these equivalent notations for the covariant differentiation. In particular, common notation for the covariant derivative is to use a semi-colon (;) in front of the index with respect to which the covariant derivative is being taken (β in this case)

##### Covariant differentiation for a covariant vector

Let's take the scalar product A^{μ}Bμ of two arbitrary vectors, one covariant A and the other contravariant B. We have then, applying the derivation rules:

But as the value of a scalar in a point in spacetime does not depend on the basis vectors, then the covariant derivative of a scalar equals to its ordinary derivative:

Comparing these two last equations gives by renaming some of the mute indices:

As this equation should hold true for each arbitrary A vector, the quantity in brackets should be necessary null. So we have shown that the expression of the covariant derivative of the covariant components of a vector B is as below:

Note that the term involving Christoffel symbols is substracted in this case.

In the same manner as with the contravariant vectors, the second term vanishes in a context of inertial frame of reference. We have then:

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