Are Vector Differential Identities in Electromagnetic Theory Consistent?

[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

 

[jedishrfu]

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.

 

[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 $\phi$.