Covariant vs partial derivative

Question

Heads up: this is a duplicate from physics S.E. where it is being treated as off topic. I hope its not a problem that I repost here ?

Question

$\partial_\mu \vec{V} = 0$, does this imply that $\nabla_\mu V^\kappa = 0$ ?

My argument

For any vector $\vec{V}$ we can write that $V=V^\kappa \vec{g_\kappa}$ with $\vec{g_\kappa} = \partial \vec{r}/\partial r^\kappa$ such that:

$$0= \partial_\mu \vec{V} = \partial_\mu(V^\kappa \vec{g}_\kappa) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \partial_\mu(\vec{g_\kappa}) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \Gamma_{\mu\kappa}^\rho \vec{g_\rho} = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\rho \Gamma_{\mu\rho}^\kappa \vec{g_\kappa} = \nabla_\mu (V^\kappa) \vec{g_\kappa}$$

projecting onto the different basis vectors than gives the result that $\nabla_\mu V^\kappa = 0$

Additional question

If this is true, and we consider a translation in space $\vec{x} \rightarrow \vec{x} + \vec{a}$ with $\vec{a}$ constant. Do we than find that $\nabla_\mu a_\nu = 0$?

Answer 1

Not quite, and for two reasons:

1) The quantity $\partial_\mu V $ is not well-defined for $V$ a vector, that is, it transforms in a non-tensorial way under a change of coordinates. Since $\partial _\mu V= (\partial _\mu V^\nu) \partial _\nu$ it is enough to check that $\partial_\mu V^\nu$ does not transform tensorially. To see this consider the following example. Let $V$ be the vector field in $\mathbb{R}^2$, $V=\partial _x$ where $(x,y)$ are the usual Cartesian coordinates. With respect to the coordinate basis $\{\partial_x,\partial_y\}$ we have $V^x=1,V^y=0$ hence all the terms $\partial_\mu V^\nu$, $\mu,\nu\in\{x,y\}$ vanish. Consider however the same vector field in polar coordinates, \begin{equation} \partial_x= \cos\theta\ \partial_r -(\sin\theta/r)\ \partial_\theta. \end{equation} With respect to the coordinate basis $\{\partial_r,\partial_\theta\}$ e.g. $\partial_\theta V^r=-\sin\theta \neq 0$ hence it cannot hold $\partial_\theta V^r = J_\theta ^\mu J^r _\nu\partial _\mu V^\nu$, where $J^\mu_\theta =\partial_\theta V^\mu$, $J^r_\nu = \partial_\nu V^r$, $\mu,\nu\in\{x,y\}$ are (entries of) the Jacobian of the coordinate transformation and its inverse. You can still define and use $\partial_\mu V$ in a fixed coordinate system but that is not very useful. In fact this is the reason for introducing the covariant derivative.

Instead, the partial derivative of a function transforms tensorially as we have, by the chain rule (or the transformation law for 1-forms if you prefer) \begin{equation*} \partial_a f = J^\mu_a \partial_\mu f, \end{equation*} so if it vanishes in a coordinate system it vanishes in all. If it is non-zero it transforms in a controlled way, namely by the Jacobian of the change of bases. This is why we want to work with quantities which transform tensorially.

2) The covariant derivative of a vector field is well-defined, that is independent of the coordinate system. Applied to a function (as opposed to a tensor) however, it is simply the ordinary partial derivative, so \begin{equation*} \nabla_\mu V^\nu=\partial_\mu V^\nu \neq \nabla_\mu V=\nabla_\mu (V ^\nu \partial_\nu) =(\partial_\mu V^\nu)\partial_\nu +V^\nu \nabla_\mu \partial_\nu =\Big( \partial_\mu V^\rho + V^\nu \Gamma^\rho_{\mu\nu}\Big) \partial_\rho \end{equation*} In fact, the extra term in the covariant derivative \begin{equation} \Gamma^\rho_{\mu\nu}= \mathrm{d}x^\rho \Big(\nabla_\mu (\partial_\nu)\Big) \end{equation} is the correction required to obtain an expression which transforms in a tensorial way under changes of coordinates.