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

Question

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 1

Just to show that the question makes little sense, let's take two of your equations:

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

Imagine a particle moving along the x axis. Now imagine a force accelerating along the x axis. The equation is not true.

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

Imagine a particle moving in a plane, let's with a 45 degrees angle with respect to the x and y axis (a diagonal). Not imagine a force (let's say gravity) accelerating it in the y-direction. The particle continues to be free in the x direction. Again the equation is wrong.

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Another example: consider the movement of a particle attached to an infinite wire (so a 1 dimensional problem). We define what we call the "velocity-phase-space" which is 2 dimensional. Imagine it as a plane where one axis is the position and the other is the speed of the particle.

Then the trajectory of the particle will be a curve in that place parametrized by a free parameter t: at each instant the particle has a given location and a given speed. So even if "speed" is a concept defined in every point of the plane, the speed of the particle takes only a set of values which is a 1-dimensional path defined in the velocity-phase-space.

When you take derivatives, you therefore have to follow the path $q(t), \dot{q}(t)$.

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A last comment: Two things can be dependent even if in some cases it appear they are not. In theory you can write $y=f(x)$ so that $y$ depends on $x$, then you consider the example $f(x) = a$ and you say "look $y$ does not depend on $x$", OK ... but only in that case.

Answer 2

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.

Answer 3

I'm still not sure quite what you're after, but the following statements might be helpful/relevant.

For a conservative vector field $\vec{F}$, the curl is always zero, i.e. $\nabla \times \vec{F} = \vec{0}$.

The divergence of the curl of any vector field is always zero, i.e. $\nabla \cdot \nabla \times \vec{F} = 0$.

Answer 4

I don't know whether the second sentence makes any sense. Let's omit it.

Electron in a constant homogeneous magnetic field moves on a circle with a constant value of velocity (in this meaning the velocity is independent of the location) but with a changing velocity vector. When we add an electric field, it will become only a bit more complicated.

The answer is, the velocity vector is constant in space (independent of the location) only for very specific settings, when a body moves along a straight line.

Answer 5

For a single particle motion $div(v) = [div(dr/dt)] = d/dt [div(r)]$. $div(r) = 3 = const.$ But the time-derivative of a constant is zero, so $div(v) = 0$. It is not an automatic assumption, it is a calculation result ;-).

EDIT: somebody voted down. OK, v = v(t) so any $\partial v_i/\partial x_k = 0$ by definition of v(t). What else do you need as explanation?