ismaili said:
hmm... so you mean mathworld writes
\partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu
But mathworld's metric is lower-indexed, which is NOT upper-indexed, this is contradiction.
Moreoever, I still don't get why their eq(61), i.e. D_\theta A_\theta has the partial derivative term like this
\frac{1}{r}\frac{\partial A_\theta}{\partial \theta}
If the metric is involved, since g_{\theta\theta} = 1/r^2,
I don't understand how they could get the scale factor 1/r.
Thanks for your discussion!
You seem to have gotten it all confused!
They use an standard metric for a space associated with spherical coordinates-based metric. For example, to compute
A_{r;r}
one must know that if there are one-forms and basis vectors involved in the equation, then it is also mandatory to know how to deal with their transformations. Here obviously the second term in the covariant derivative vanishes and the first gives
A_{r;r}=\frac{\partial}{\partial x^1} A_r
where the coordinates are taken to be (x^1,x^2,x^3). Of course the reason why I have written \frac{\partial}{\partial x^1} independently of the r-component of the tensor A_a is that it is the r-component of a basis vector, to wit,
\frac{\partial}{\partial x^1}=e_1=e_r,
is a component of the basis vector e_a.
Therefore to transform it into its dual vector (or one-form) \omega^a we can use the fact that
\omega^b e_a=\delta^b_a,
and that g_{ab}=e_ae_b
to obtain
g_{ab} \omega^b=e_a.
Since the metric chosen for our coordinates is diagonal, this turns out to be
g_{bb} \omega^b=e_b,
and finally
g_{rr} \omega^r=e_r \Rightarrow A_{r;r}= g_{rr} \frac{\partial}{\partial x_1}.
But don't be tricked into laying your finger at this because Mathworld uses the matrix multiplication and so, in MW's sense, g_{rr} is equivalent to g^{-1}_{rr}. Now if you got this right, everything would be well set!
AB