Are Vector Differential Identities in Electromagnetic Theory Consistent?

ShayanJ
Science Advisor
Insights Author
Messages
2,801
Reaction score
606
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
 
Physics news on Phys.org
Your second identity doesn't seem strange if there is no theta or phi component to F:
<br /> \operatorname{div}\, \mathbf F <br /> = \nabla\cdot\mathbf F <br /> = \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}.<br />

then it would reduce to:

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

and then if it was a very large r value you'd be left with your identity.
 
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}<br /> = \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
<br /> \nabla \cdot {\bf F}(r) <br /> = \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.
 
Back
Top