Thread: Covariant derivative View Single Post
 P: 111 Well, lots of texts show the covariant derivative of a rank 1 tensor (vector) first to establish kind of the idea for what we're going for before developing the concept for higher rank tensors, so here is a very simple case for what the covariant derivative is for vectors from an article I'm writing (this might be too simple for you I'm not really sure what you're looking for, and it's too simple to show all of the essential features of the covariant derivative clearly, it's just a very simple specific case): Consider the derivative of a vector in general curvilinear coordinates such that the derivative of the basis vectors may change: $$\frac{{\partial}{\vec{A}}}{{\partial}{x^{\alpha}}} = \frac{{\partial}{(A^{i} \vec{e}_{i})}}{{\partial}{x^{\alpha}}} = \frac{{\partial}{A^{i}}}{{\partial}{x^{\alpha}}} \vec{e}_{i} + A^{i} \frac{{\partial}{\vec{e}_{i}}}{{\partial}{x^\alpha}}$$ Now, the next step is going to be in anticipation that we can find a way to factor the basis vector '$\vec{e}_{i}$' out of the equation such that our derivative is an operator on this basis vector. Thus we need to do something about the derivative on the far right. We know that the result of this derivative is going to be some weighted combination of the basis vectors, thus we instead represent this derivative as some sum of each unit vector times some weighing factor: $$\frac{{\partial}{\vec{e}_{i}}}{{\partial}{x^\alpha}} = \sum_{j}{\Gamma^{j}_{{i}{\alpha}} \vec{e}_{j}}$$ These coefficients aren't weird looking on accident, they appear frequently and are called "Christoffel symbols of the $2^{nd}$ kind". There are 3 indices for these symbols, the first bottom index (here 'i') indicates the basis vector being differentiated, the second bottom index (here '$\alpha$') indicates the direction in which the basis vector is being differentiated in, and finally the top index (here 'j') indicates the direction in which this derivative points. Thus we can write: $$= \frac{{\partial}{A^{i}}}{{\partial}{x^{\alpha}}} \vec{e}_{i} + A^{i} (\Gamma^{j}_{{i}{\alpha}} \vec{e}_{j})$$ Now comes a trick that allows us to factor out the basis vector '$\vec{e}_{i}$', the result of our Christoffel symbol notation. We simply notice that 'i' and 'j' are simply dummy summation variables, and thus we can swap them without changing a thing: $$= \frac{{\partial}{A^{i}}}{{\partial}{x^{\alpha}}} \vec{e}_{i} + A^{j} (\Gamma^{i}_{{j}{\alpha}} \vec{e}_{i})$$ Thus we can now factor out our basis vector '$\vec{e}_{i}$' and define an operation that we call the "covariant derivative": $$= ( \frac{{\partial}{A^{i}}}{{\partial}{x^{\alpha}}} + A^{j} \Gamma^{i}_{{j}{\alpha}} ) \vec{e}_{i}$$ The symbol we use to denote this derivative is the semi-colon before the variable being differentiated: $$A^{i}_{{;}{\alpha}} = \frac{{\partial}{A^{i}}}{{\partial}{x^{\alpha}}} + \Gamma^{i}_{{j}{\alpha}} A^{j}$$ $$\frac{{\partial}{\vec{A}}}{{\partial}{x^{\alpha}}} = A^{i}_{{;}{\alpha}} \vec{e}_{i}$$ And thus the covariant derivative becomes something like the Christoffel symbols, weighing factors that link the derivative to each of the basis vectors representing the space.