Partial Derivatives of the Position and Velocity Vectors of a Particle

[Zag]

Hello guys!

Lately I've been studying some topics in Physics which require an extensive use vector calculus identities and, therefore, the manipulation of partial redivatives of vectors - in particular of the position and velocity vectors. However, I am not sure if my understanding of partial derivatives of positions and velocity vectors is correct. Below I present my reasoning and some doubts I have. I would really appreciate comments on what I've written so i can either confirm this is correct, or learn the right way of visualizing these derivatives.

To illustrate the situation, I will start by writing a small part of a calculation I had to make related to the angular momentum carried be the eletromagnetic fields.

$\frac{\partial}{\partial t}\vec{r}\times(\vec{E}\times\vec{B}) = \frac{\partial\vec{r}}{\partial t}\times(\vec{E}\times\vec{B}) + \vec{r}\times\frac{\partial\vec{E}\times\vec{B}}{\partial t} = \vec{r}\times\frac{\partial\vec{E}\times\vec{B}}{\partial t}$

So, to get to the result above, the following relationship had to be used: $\frac{\partial\vec{r}}{\partial t} = \vec{0}$

This is quite obvious if you think about the position vector as $\vec{r} = (x, y, z)$ and its partial derivative with respect to time (and, consequently, with coordinates x, y and z fixed) as being given by the limit

$\frac{\partial\vec{r}}{\partial t} = lim_{t → t_{o}}\frac{(x, y, z) - (x, y, z)}{t - t_{o}} = \vec{0}$

So, in other words, the partial derivative of the position vector of a particle, $\frac{\partial\vec{r}}{\partial t}$, has to be zero because $\vec{r}$ is the very embodiment of the coordinates x, y and z, and since differentiating partialy with respect to time fixes these very coordinates, the vector suffers no change at all. (Note that this is different from the case of a total derivative in which $\frac{d\vec{r}}{dt} = \vec{v}$)

Also, it seems that it would be easy to expand this reasoning to the partial derivatives $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial z}$, as these also lock the particle (or the position vector) in two coordinates of space and also in time, inevitably fixing the position vector of the particle.

Now, I am having considerably more problems trying to understand what happens with the partial derivatives of the velocity vector of a single particle. Following the same reasoning that was presented above, it seems that the partial derivatives of the velocity vector single particle should also be zero, since these always derivatives lock three members of the ordered quadruple $(x, y, z, t)$. Would this be correct?

Thank you everyone!

 

[mikeph]

Without any explanation of what r is, I beg to differ that it's obvious that dr/dt = 0.

r = r(x,y,z,t), and these coordinates are orthogonal. You can't use mere terminology such as "r is the embodiment of x, y and x" to deduce that dr/dt = 0. What stops me from permuting the arguments in your line of reasoning to deduce that dr/dx = dr/dy = dr/dz = 0 and that the particle is fixed forever more?

Treat time and space as independent coordinates.

Sidenote: your first principles expression for dr/dt is wrong. The numerator should read r(x,y,z,t + t0) - r(x,y,z,t); this is where your differential with respect to one particular coordinate arises from.

 

[Zag]

MikeyW, $\vec{r}$ is the position vector of a particle. I am sorry if I did not make it clear.

Before I start exlaining myself better, please notice that I am talking about partial derivatives. I say that because on your reply you wrote all the derivatives with a regular "d", which makes me think you may be talking about total derivatives. Note that I included an extra note on the body of my text to avoid any confusion.

Also, I've been trying to understand why these things are "true" exactly because I see it being used, specially at graduate level class on Eletromagnetism. So, please, I would very much like an explanation for this thing which is really annoying me!

About my reasoning, I actually I did not use any terminology to prove my point, I just described what $\vec{r}$ really is. When I write $(x, y, z)$, I am talking about the components of the vector $\vec{r}$, not the functional dependence as you put it. I am talking of $\vec{r}$ in the sense that $\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$.

And I agree with your point that you could say that the particle would be stuck forever in the same place, since the total derivative with respect to time would be

$\frac{d\vec{r}}{dt} = \frac{\partial\vec{r}}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial\vec{r}}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial\vec{r}}{\partial z}\frac{\partial z}{\partial t} + \frac{\partial\vec{r}}{\partial t} = \vec{0}$

in case the partial derivatives with respect to the space coordinates were zero too. But, if this is not the case, why do people use these properties? I repeat, I see it being used, so I am looking for an explanation for this.

 

[mikeph]

Ah I see.

r(t) = [x(t), y(t), z(t)], in vector notation.

Substituting in your expression, r = ix + jy + kz, to the expression for the total derivative:

dr/dt = (dr/dx)(dx/dt) + (dr/dy)(dy/dt) + (dr/dz)(dz/dt) = i(dx/dt) + j(dy/dt) + k(dz/dt)

so dr/dt = [dx/dt, dy/dt, dz/dt];

There is no reason that any of these three velocity components should be zero based on what you've told me. If r is the position of a particle, then why can't that particle move? There must have been a step in the derivation which made the assumption of constant conditions, or maybe you're in a coordinate system in which the particle happens to be at rest.

 

[mikeph]

I wrote a long explanation about partials, I don't use them here because it's a pain without the specific key on a keyboard, and I'm used to it being assumed that the partial is meant by a normal d.

 

[Zag]

Thank you for you reply MikeyW!

I've been reading some references on this issue and re-reading my class notes as well, and I think I have finally understood what happens with these derivatives.

We can describe a particle's trajectory its the position vector

$\vec{r} = x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}$.

By taking the partial derivative with respect to time of this vector, we consider the variables $x$, $y$, $z$ and $t$ to be independent. Therefore, there is no explicit dependence on the variable $t$, and one can indeed write: $\frac{\partial\vec{r}}{\partial t} = \vec{0}$

However, as you have pointed out, MikeyW, this reasoning would not be applicable to the partial derivatives $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial z}$, since this would make the total time derivative with respect to time equal to zero:

$\frac{d\vec{r}}{dt} = \frac{\partial\vec{r}}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial\vec{r}}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial\vec{r}}{\partial z}\frac{\partial z}{\partial t} + \frac{\partial\vec{r}}{\partial t} = \vec{0}$

This would then imply that any particle would be stuck forever in the same position, which we simply know is not true. Indeed, the partial derivatives with respect to the spatial coordinates will not be zero, and will actually be given by

$\frac{\partial\vec{r}}{\partial x} = \vec{x}$, $\frac{\partial\vec{r}}{\partial y} = \vec{y}$ and $\frac{\partial\vec{r}}{\partial z} = \vec{z}$

Or, in a more compact notation

$\frac{\partial r_{i}}{\partial x_{j}} = δ_{ij}$

Having said that, what was really happening on my class notes was that the terms involving the partial derivatives with respect to the spatial coordinates were miraculously cancelling out other contributions on the equation. The fact that the entire calculation was so abstruse made it more complicated to see that, so I apologize for the confusion. I am glad that this doubt seems to have been finally eliminated though. :)

Thanks again MikeyW!