Are Vector Differential Identities in Electromagnetic Theory Consistent?

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SUMMARY

The discussion centers on the consistency of vector differential identities in electromagnetic theory, specifically referencing chapter 20 of "Foundations of Electromagnetic Theory" by Reitz, Milford, and Christy. The identity involving the curl of the time derivative of momentum, \(\vec{\nabla}\times\dot{\vec{p}}=\frac{\vec{r}}{r}\times\frac{ \partial \dot{\vec{p}}}{\partial r}\), raises questions regarding its validity in spherical coordinates. Additionally, the divergence identity \(\vec{\nabla}\cdot\vec{F}(r)=\frac{\vec{r}}{r}\cdot\frac{d\vec{F}}{dr}\) is clarified as consistent under certain conditions, particularly when there are no angular components in \(\vec{F}\).

PREREQUISITES
  • Understanding of vector calculus, particularly curl and divergence operations.
  • Familiarity with spherical coordinate systems in vector fields.
  • Knowledge of electromagnetic theory as presented in "Foundations of Electromagnetic Theory".
  • Proficiency in applying the chain rule in multivariable calculus.
NEXT STEPS
  • Study the derivation of vector identities in spherical coordinates.
  • Explore the implications of the curl and divergence in electromagnetic applications.
  • Review the relevant sections of "Foundations of Electromagnetic Theory" for deeper insights.
  • Learn about the chain rule in the context of vector fields and its applications in physics.
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Students and professionals in physics, particularly those focusing on electromagnetic theory, vector calculus, and mathematical physics. This discussion is beneficial for anyone seeking to understand the nuances of vector differential identities.

ShayanJ
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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
 
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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.
 

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