1
$\begingroup$

I've been trying to derive the covariant divergence of a vector field and I've ran into 2 problems.

Basically I have $\nabla_aA^a=\partial _{a}A^a+\Gamma^{a}{}_{ab}A^b$, and then I found $\Gamma^{\mu}{}_{\mu\sigma}=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^\sigma}\sqrt{g}$. So, putting these together I am hoping to have $A^\sigma_{\:\:;\sigma}=A^\sigma_{\:\:,\sigma}+\frac{1}{\sqrt{g}}(\sqrt{g}\:A^\sigma)_{,\sigma}$. However, the answer does not contain the "ordinary derivative" term, which is why I'm confused. Does this mean $\partial _{a}A^a$ vanishes once I have chosen a coordinate basis to work with?

Looking back into the chapter of the book I found the book I am using I found the definition of the "ordinary (coordinate-dependent) derivative", which is, for example,

$$\partial_aT^b{}_{c}:=(dx^\mu)_a\left ( \frac{\partial}{\partial x^\nu} \right ) ^b(dx^\sigma)_c\:\partial _\mu T^\nu{}_{\sigma},$$ where $a$, $b$, and $c$ are abstract indices. I then stumbled upon a statement that $\partial_a\left ( \frac{\partial }{\partial x^\nu} \right ) ^b=0$, or that the ordinary derivatives of any coordinate basis vector vanish, and I'm thinking that this somehow relates to my confusion about the covariant divergence. However I also don't understand why this statement holds. For example, in 3D Euclidean space, the partial derivative with respect to $\theta$ of $e_\theta$ is not vanishing. I'm hoping someone could help me out.

$\endgroup$

1 Answer 1

1
$\begingroup$

Not sure how you put those two formulas together, because your bracketing is wrong. We have \begin{align} \text{div}(A)=(\nabla_{\mu}A)^{\mu}&\equiv\nabla_{\mu}A^{\mu}\\ &=\frac{\partial A^\mu}{\partial x^\mu}+\frac{1}{\sqrt{|g|}}\frac{\partial \sqrt{|g|}}{\partial x^\mu}A^{\mu}\\ &=\frac{1}{\sqrt{|g|}}\frac{\partial(\sqrt{|g|}A^{\mu})}{\partial x^{\mu}}, \end{align} where the last step is simply by the product rule (write it out explicitly). In the special case of $\Bbb{R}^n$, and using Cartesian coordinates, we have $\sqrt{|g|}=1$, so the divergence formula reduces to what you expect.

I will never understand why one would want to mix abstract indices with “actual” ones when writing equations when using only one or the other is so much clearer, but this seems to be part of your confusion. The only thing they’re saying at the end is that the partial derivatives of the constant functions $\delta^{\mu}_{\nu}$ vanishes (in your question, we have explicitly that $e_{\theta}=0\cdot e_r+1\cdot e_{\theta}+0\cdot e_{\phi}$, so you’re just calculating the derivative of the function $1$ (or $0$) with respect to $\theta$, and this is obviously just $0$). They’re not saying anything about derivatives of the vectors themselves.

$\endgroup$
2
  • $\begingroup$ indices are always a hot topic, so take my opinion for whatever you think it’s worth: but seriously don’t write equations with both types in them. Stick to one which is more convenient, and use that! $\endgroup$
    – peek-a-boo
    Commented Jan 6, 2023 at 1:43
  • $\begingroup$ Thanks for answering both questions! Now that you've said it I guess I have really confused myself with what I was supposed to be looking at (the components) when I even have the definition of the ordinary derivative written down $\endgroup$ Commented Jan 6, 2023 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.