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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:

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

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

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

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

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

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

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

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

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

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    $\begingroup$ I don't understand you question. In any case the speed v is the time derivative of x. $\endgroup$
    – Cedric H.
    Commented Nov 9, 2010 at 18:11
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    $\begingroup$ Gradient of a velocity vector? (Do you expect a rank-2 tensor?) $\endgroup$
    – kennytm
    Commented Nov 9, 2010 at 18:27
  • $\begingroup$ I don't get how $v$ would be a vector field in this scenario. You just have one particle and hence just one $\vec{v}$ $\endgroup$
    – Mark Eichenlaub
    Commented Nov 9, 2010 at 18:42
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    $\begingroup$ This question really makes very little sense I'm afraid, for the reasons Cedric and Kenny pointed out. I for one have never taken the gradient of a vector in all the physics I've done! $\endgroup$
    – Noldorin
    Commented Nov 9, 2010 at 19:21
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    $\begingroup$ well, $\partial\vec{v}/\partial t + \vec{v}\cdot\nabla\vec{v}$ is a useful physics quantity. It's the acceleration of a bit of mass in a flowing fluid. That's referencing vector field, though. $\endgroup$
    – Mark Eichenlaub
    Commented Nov 9, 2010 at 20:41

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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.

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  • $\begingroup$ Indeed this question arose from Lagrangian mechanics. Let me see if I understood (it sounds right): You mean that because we chose x and x' as new variables, it is also assumed they are independent, so partial derivatives give 0, and this is merely a formal-mathematical choice despite it contradicting the meaning of a derivative in the usual sense. Is that right? $\endgroup$
    – Uri
    Commented Feb 24, 2011 at 16:14
  • $\begingroup$ Yep. By the way, this is one of the reasons I would prefer that the foundations of physics be built on the sturdy surface of "simple equations of motion" rather than the theoretical surface of "simple Lagrangians", which actually hasn't resulted in a simple Lagrangian, just the simplest 30,000 person-years of effort could find. See: nuclear.ucdavis.edu/~tgutierr/files/stmL1.html for the Standard Model Lagrangian, written out. $\endgroup$
    – Carl Brannen
    Commented Feb 25, 2011 at 1:14
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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.

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 $q(t), \dot{q}(t)$.

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.

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  • $\begingroup$ So you're saying that sometimes it can happen and sometimes not, and that in the specific case of an electron in the electro-magnetic field the speed is independent of location? $\endgroup$
    – Uri
    Commented Nov 10, 2010 at 16:52
  • $\begingroup$ Just forget about all these things about the link between v and x. I think there is a big confusion about all this in your question. Start from the basics and try to understand the second part of my answer. $\endgroup$
    – Cedric H.
    Commented Nov 10, 2010 at 16:55
  • $\begingroup$ And think about real cases as in the first part of my answer. You can then easily deduce if v is independent of x or not. $\endgroup$
    – Cedric H.
    Commented Nov 10, 2010 at 16:56
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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. $\nabla \cdot \nabla \times \vec{F} = 0$.

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    $\begingroup$ Maybe that's the problem: It's a particle's velocity vector, usually a function of t, not a field vector (there are many places in the plane where the particle can't reach, so it is useless to talk about its speed there). Div and curl are used here as operators on a vector And I'm asking whether it's possible, physically speaking, for this vector to be a function of the location of a particle, and not just the time. In that case deriving the x component of this vector v will give us a non zero value, so the divergence will not be 0. $\endgroup$
    – Uri
    Commented Nov 10, 2010 at 1:02
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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.

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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?

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