Vector differential identities

1. Dec 16, 2013

ShayanJ

Vector differential identities!!!

In chapter 20 of "Foundations of Electromagnetic theory" by Reitz,Milford and Christy,there is calculation which seems to make use of:$\vec{\nabla}\times\dot{\vec{p}}=\Large{\frac{\vec{r}}{r}\times\frac{ \partial \dot{\vec{p}}}{\partial r}}$ where $\dot{\vec{p}}=\large{\frac{d}{d \tau}} \int_V \vec{r}'\rho(\vec{r}',t-\frac{r}{c})dv' \ (\tau=t-\frac{r}{c})$.But I can't prove it and worse is that it seems to be inconsistent with the formula for curl in spherical coordinates.
There is also another identity mentioned in the problems of chapter 1 which seems as strange:
$\vec{\nabla}\cdot\vec{F}(r)=\large{\frac{\vec{r}}{r}\cdot\frac{d\vec{F}}{dr}}$

Is there any suggestion?
Thanks

2. Dec 17, 2013

Staff: Mentor

Your second identity doesn't seem strange if there is no theta or phi component to F:
$$\operatorname{div}\, \mathbf F = \nabla\cdot\mathbf F = \frac1{r^2} \frac{\partial}{\partial r}(r^2 F_r) + \frac1{r\sin\theta} \frac{\partial}{\partial \theta} (\sin\theta\, F_\theta) + \frac1{r\sin\theta} \frac{\partial F_\phi}{\partial \phi}.$$

then it would reduce to:

$$\operatorname{div}\, \mathbf F = \nabla\cdot\mathbf F = \frac1{r^2} \frac{\partial}{\partial r}(r^2 F_r)$$

and then if it was a very large r value you'd be left with your identity.

3. Dec 18, 2013

qbert

Both are simple chain rule consequences, i'll
illustrate with the divergence since it's quicker
$$\nabla \cdot {\bf F}(r) = \sum_i \frac{\partial F_i(r)}{\partial x^i} = \sum_i \frac{\partial F_i(r)}{\partial r} \frac{\partial r}{\partial x^i}$$
now $\frac{\partial r}{\partial x^i}=\frac{x^i}{r}$, so
$$\nabla \cdot {\bf F}(r) = \sum_i \frac{\partial F_i(r)}{\partial r} \frac{x^i}{r} = \frac{{\bf r}}{r} \cdot \frac{\partial {\bf F}(r)}{\partial r}.$$

As an aside to get the result from the spherical formula you have to keep all three terms.
even though ${\bf F}$ only depends on $r$, when you break it into spherical
components, for example $F_\theta = {\widehat \theta} \cdot {\bf F}$, depends on
$r, \theta$ and $\phi$.