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Vector differential identities

  1. Dec 16, 2013 #1

    ShayanJ

    User Avatar
    Gold Member

    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:[itex] \vec{\nabla}\times\dot{\vec{p}}=\Large{\frac{\vec{r}}{r}\times\frac{ \partial \dot{\vec{p}}}{\partial r}} [/itex] where [itex] \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})[/itex].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:
    [itex]\vec{\nabla}\cdot\vec{F}(r)=\large{\frac{\vec{r}}{r}\cdot\frac{d\vec{F}}{dr}} [/itex]

    Is there any suggestion?
    Thanks
     
  2. jcsd
  3. Dec 17, 2013 #2

    jedishrfu

    Staff: Mentor

    Your second identity doesn't seem strange if there is no theta or phi component to F:
    [tex]
    \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}.
    [/tex]

    then it would reduce to:

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

    and then if it was a very large r value you'd be left with your identity.
     
  4. Dec 18, 2013 #3
    Both are simple chain rule consequences, i'll
    illustrate with the divergence since it's quicker
    [tex] \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}[/tex]
    now [itex] \frac{\partial r}{\partial x^i}=\frac{x^i}{r} [/itex], so
    [tex]
    \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}.[/tex]

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