classical mechanics - Why do we automatically assume that the velocity vector $vec{v}$ and location vector $vec{r}$ are independent?

I'm not sure if it's relevant, but I'm talking about a situation where a particle is moving in an electro-magnetic field.

As I understand, if we see the term $\nabla \cdot \vec{v}$ or $\nabla \times \vec{v}$ we can say it is equal to 0, because the velocity vector is independent of the location vector.

My question is this: Is this always true, or only in specific settings (particle in an electro-magnetic field for example)?

If it is always true: What about a state where an electron's velocity changes with respect to x? Then they can't be independent, can they?

In other words, are the following always true:

$\partial v_{x}/\partial x = 0$

$\partial v_{y}/\partial x = 0$

$\partial v_{z}/\partial x = 0$

$\partial v_{x}/\partial y = 0$

$\partial v_{y}/\partial y = 0$

$\partial v_{z}/\partial y = 0$

$\partial v_{x}/\partial z = 0$

$\partial v_{y}/\partial z = 0$

$\partial v_{z}/\partial z = 0$

Answer

Despite your notation, I'm guessing that you're asking about partial derivatives. To tex a partial derivative use "\partial x" as in "$\partial x$".

This sort of confusion arises in Lagrangian calculations and I'll bet that's where you came upon it. The effect arises from our mathematics. It appears with only a single dimension x so I'll illustrate the effect that way.

If we left everything in terms of position (with velocity defined as $dx/dt$, then the gradient of the velocity wouldn't be zero. That's because the velocity does, in fact, depend on position.

Instead, when we do classical mechanics with Lagrange's equation, we think of the Lagrangian as being a function of position, velocity, and perhaps time. So we write it as $L(x,\dot{x},t)$. The effect of this way of looking at the problem is that we double the number of variables (from $x$ to $x,\dot{x}$), but we eliminate the need for second derivatives. This makes the problem actually easier to solve, but to be consistent, we have to make our partial derivatives (i.e. partials with respect to position or velocity) apply only to the position or velocity.

You can quickly verify that the Lagrangian
$L(x,y,\dot{x},\dot{y}) = 0.5m\dot{x}^2 +0.5m\dot{y}^2 - mgy$

leads to the equations of motion that you expect for a body of mass m moving in a gravitational field:
$m\ddot{x} = 0,\;\;m\ddot{y} = -g$,
and that this happens only if you follow the rules you've been given for calculating partial derivatives. Maybe this will ease your mind. Also see:
Why does Calculus of Variations work?
Why does calculus of variations work?

By the way, the place where partial derivatives used to bother me the most was in thermodynamics.

So in short, we don't automatically assume it. It happens when we use math in certain ways.