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.
[itex]\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}[/itex]
So, to get to the result above, the following relationship had to be used: [itex]\frac{\partial\vec{r}}{\partial t} = \vec{0}[/itex]
This is quite obvious if you think about the position vector as [itex]\vec{r} = (x, y, z)[/itex] and its partial derivative with respect to time (and, consequently, with coordinates x, y and z fixed) as being given by the limit
[itex]\frac{\partial\vec{r}}{\partial t} = lim_{t → t_{o}}\frac{(x, y, z) - (x, y, z)}{t - t_{o}} = \vec{0}[/itex]
So, in other words, the partial derivative of the position vector of a particle, [itex]\frac{\partial\vec{r}}{\partial t}[/itex], has to be zero because [itex]\vec{r}[/itex] 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 [itex]\frac{d\vec{r}}{dt} = \vec{v}[/itex])
Also, it seems that it would be easy to expand this reasoning to the partial derivatives [itex]\frac{\partial}{\partial x}[/itex], [itex]\frac{\partial}{\partial y}[/itex] and [itex]\frac{\partial}{\partial z}[/itex], 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 [itex](x, y, z, t)[/itex]. Would this be correct?
Thank you everyone!
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.
[itex]\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}[/itex]
So, to get to the result above, the following relationship had to be used: [itex]\frac{\partial\vec{r}}{\partial t} = \vec{0}[/itex]
This is quite obvious if you think about the position vector as [itex]\vec{r} = (x, y, z)[/itex] and its partial derivative with respect to time (and, consequently, with coordinates x, y and z fixed) as being given by the limit
[itex]\frac{\partial\vec{r}}{\partial t} = lim_{t → t_{o}}\frac{(x, y, z) - (x, y, z)}{t - t_{o}} = \vec{0}[/itex]
So, in other words, the partial derivative of the position vector of a particle, [itex]\frac{\partial\vec{r}}{\partial t}[/itex], has to be zero because [itex]\vec{r}[/itex] 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 [itex]\frac{d\vec{r}}{dt} = \vec{v}[/itex])
Also, it seems that it would be easy to expand this reasoning to the partial derivatives [itex]\frac{\partial}{\partial x}[/itex], [itex]\frac{\partial}{\partial y}[/itex] and [itex]\frac{\partial}{\partial z}[/itex], 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 [itex](x, y, z, t)[/itex]. Would this be correct?
Thank you everyone!