Covariant Derivative and Cartesian Coordiantes I have a somewhat weird question regarding some problems I have been encountering lately when dealing with transformations of the general covariant derivative to cartesian coordinates and back.
In the following, I will use Latin indices for cartesian components and greek indices for the general curved case.
In general, I have two sets of independent variables $v^i$ and $x^i$ or in the curvilinear case $v^\sigma$, $\xi^\sigma$, where $v$ are components of a sample space velocity and $\xi$ and $x$ are a parameterization of a position. The dependant variable is a probability distribution $f$.
We now consider for simplicity a term in the form of
$\nabla_\nu (v^\nu f)$
where $\nabla$ denotes the covariant derivative. When transforming this expression to cartesian coordiantes and the covariant derivative reduces to a partial derivative. I the have, since $x^i$ and $v^i$ are independent variables $\partial_i v^j=0$ and thus
$\nabla_\nu (v^\nu f) = \nabla_i (v^i f) = v^i \nabla_if$.
However, I do not see what stops me from transforming this back to the curvilinear basis and I thus obtain
$\nabla_\nu (v^\nu f) = v^\nu \nabla_\nu f$
which can only be true if $\nabla_\nu v^\mu=0$, but we $\xi^\nu$ and $v^\nu$ should be independent variables, which I have always assumed to mean $\partial_\mu v^\nu=0$ not $\nabla_\nu v^\mu=0$. ( I know that $\nabla_i v^j=0$ leads naturally to  $\nabla_\nu v^\mu=0$ since the tensor equation should hold in any coordinate frame. However, it still seems a bit weird since we are explicitly seeking a form where the sample space velocity components and the position are independent of one another )
Maybe my assumption / interpretation of coordinate independence is flawed. I would greatly appreciate any input.
 A: Functional independence $\frac{\partial}{\partial y^\alpha} V^\beta =0$ and zero covariant derivative $\nabla_\alpha V^\beta=0$ are disjoint facts, generally speaking.
In orthonormal Cartesian coordinates they are however  equivalent, but this is false when passing to other coordinate systems.
Take $\mathbb{R}^2$ equipped with the standard Levi-Civita connection and consider a vector field $V$ which has constant components in Cartesian coordinates. In other words the Cartesian components of $V$ are not functions of the Cartesian coordinates. This is equivalent to say that $\nabla_\alpha V^\beta=0$. This second requirement is intrinsic, it does not depend on the choice of coordinates. However it does not mean that, for instance, the components of $V$ in polar plane coordinates $\theta, r$ are independent from $\theta, r$. This is obvious without a computation just by thinking to the shape of curves $r=$ constant and $\theta=$ constant.
Analytically speaking the dependence arises form the transformation law of components. For instance,
$$V^r = \frac{\partial r}{\partial x}V^x +  \frac{\partial r}{\partial y}V^y
= \frac{x}{\sqrt{x^2+y^2}} V^x + \frac{y}{\sqrt{x^2+y^2}}  V^y\:.$$
You see that, even if $V^x$ and $V^y$ are constant -- and this is just the requirement $\nabla_\mu V^\mu=0$ -- the component $V^r$ is not constant and the reason is due to the appearance of the non-constant factors $\frac{\partial r}{\partial x}$ and $\frac{\partial r}{\partial y}$.
A: If I got your question right, you're computing the gradient of a vector field $f \mathbf{v}$, and you want to find the components of the gradient, defined as the covariant derivatives, using:

*

*2 sets of coordinates $q^i$, $Q^i$, with $Q_i$ Cartesian, to describe the space $\mathbf{r} = \mathbf{r}^q(q^i) = \mathbf{r}^Q(Q^i)$, using this notation to explicitly write the position in space $\mathbf{r}$ as two functions $\mathbf{r}^q$, $\mathbf{r}^Q$ with different sets of coordinates; these two representations can be related, knowing the transformation rule $q^i(Q^k)$ as
$\mathbf{r} = \mathbf{r}^q(q^i) = \mathbf{r}^q(q^i(Q^k)) = \mathbf{r}^Q(Q^k) $.


*the vectors of the natural bases $\mathbf{b}_i = \frac{\partial \mathbf{r}}{\partial
   q^i}$,  $\mathbf{B}_i = \frac{\partial \mathbf{r}}{\partial Q^i}$, induced by the coordinates, so that we can write the vector field
$f \mathbf{v} = f v^i \ \mathbf{b}_i =  f V^k \ \mathbf{B}_k$.
These vectors are related using differentiation of composite functions
$\mathbf{B}_k (Q^a) = \dfrac{\mathbf{\partial \mathbf{r}}}{\partial Q^k}(Q^a) = \dfrac{\mathbf{\partial \mathbf{r}}^Q}{\partial Q^k}(Q^a) =  \dfrac{\mathbf{\partial \mathbf{r}}^q}{\partial Q^k}(q^i(Q^a)) = \dfrac{\mathbf{\partial \mathbf{r}}^q}{\partial q^i}(q^i(Q^a)) \dfrac{\partial  q^i}{\partial Q^k}(Q^a) = \mathbf{b}_i (q^i(Q^a)) \dfrac{\partial  q^i}{\partial Q^k}(Q^a)$,
thus, hiding the functional dependence for brevity, we have found the law of transformation between the vectors of the two bases,
$\mathbf{B}_k = \mathbf{b}_i \dfrac{\partial  q^i}{\partial Q^k}$.
Now, the gradient of this vector field $\nabla ( f \mathbf{v} )$ can be written using the natural bases and their reciprocal bases $\{\mathbf{b}^i\}_i$, $\{\mathbf{B}^k\}_k$  as
$\nabla (f \mathbf{v}) = \mathbf{b}^i \otimes \dfrac{\partial}{\partial q^i} (f \mathbf{v}) = \mathbf{B}^a \otimes \dfrac{\partial}{\partial Q^a} (f \mathbf{v})$
Using coordinates $q^i$, we get
$\nabla (f \mathbf{v}) = \mathbf{b}^i \otimes \dfrac{\partial}{\partial q^i} (f \mathbf{v}) = \mathbf{b}^i \otimes \dfrac{\partial}{\partial q^i} (f v^k  \mathbf{b}_k) = \\ \qquad \quad = \mathbf{b}^i \otimes \left[ \dfrac{\partial}{\partial q^i} (f v^k ) \mathbf{b}_k + f v^k \dfrac{\partial \mathbf{b}_k}{\partial q^i}  \right] = \\ \qquad \quad = \mathbf{b}^i \otimes \left[ \dfrac{\partial}{\partial q^i} (f v^k ) \mathbf{b}_k + f v^k \dfrac{\partial \mathbf{b}_k}{\partial q^i}  \right] =  \\ \qquad \quad = \mathbf{b}^i \otimes \left[ \dfrac{\partial}{\partial q^i} (f v^k ) \mathbf{b}_k + f v^k \Gamma^{(q)\ell}_{ik} \mathbf{b}_{\ell}  \right] =  \mathbf{b}^i \otimes \mathbf{b}_k \left[ \dfrac{\partial}{\partial q^i} (f v^k ) + f v^{\ell} \Gamma^{(q)k}_{i \ell}  \right] =  \mathbf{b}^i \otimes \mathbf{b}_k \nabla^{(q)}_{/i} v^k$,
while using Cartesian coordinates $Q^a$, for which the vectors of the natural basis are uniform in space and thus their derivatives w.r.t. the coordinates $Q^a$, i.e. Christoffel symbols $\Gamma^{Q,a}_{bc}$ are identically zero, we get
$\nabla (f \mathbf{v}) = \mathbf{B}^a \otimes \mathbf{B}_b \dfrac{\partial V^b}{\partial Q^a} = \mathbf{B}^a \otimes \mathbf{B}_b \nabla^{(Q)}_{/a} V^b$.
It's possible to prove that the two expressions of the gradient are equivalent, being just the representation of the same tensor object using 2 different sets of coordinates and bases.
(writing, and checking symbols...)
