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Gradient equation with retarded time

  1. May 24, 2005 #1
    For a radiation problem,
    i am desperate about the expansion of the following equation:
    [tex]\nabla ( \hat{r} /r^2 \cdot \vec{p}(t_o))[/tex]

    where [tex]t_o[/tex] is the retarded time at the center
    [tex]t_o=t-r/c[/tex]

    and [tex]\vec{p}(t_o)[/tex] is the electric dipole moment at [tex]t_o[/tex]

    actually, it expands to 4 main parts and i am unable to figure out the last one, namely:
    [tex]\hat{r} /r^2 \times (\nabla \times \vec{p}(t_o))[/tex]

    it would be magnifique if anyone can figure out the expansion of that term
     
    Last edited: May 24, 2005
  2. jcsd
  3. May 24, 2005 #2

    dextercioby

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    [tex]\nabla\left[\frac{\vec{r}}{r^{3}}\cdot\vec{p}\left(t_{o}\right)\right] [/tex]

    [tex] =\left(\frac{\vec{r}}{r^{3}}\cdot\nabla\right)\vec{p}\left(t_{o}\right)+\left[\vec{p}\left(t_{o}\right)\cdot\nabla\right]\frac{\vec{r}}{r^{3}}\ +\frac{\vec{r}}{r^{3}}\times\left[\nabla\times\vec{p}\left(t_{o}\right)\right][/tex]

    [tex] +\vec{p}\left(t_{o}\right)\times\left(\nabla\times\frac{\vec{r}}{r^{3}}\right) [/tex]

    All 4 of them,okay?

    Daniel.
     
    Last edited: May 24, 2005
  4. May 24, 2005 #3
    ok, it's actually [tex]\nabla\left[\frac{\vec{r}}{r^{2}}\cdot\vec{p}\left(t_{o}\right )\right] [/tex], but really fine, thanks a lot.

    so may i ask which one of these terms give out [tex](\hat{r}\cdot\dot{p}(t_o))\hat{r}[/tex] and how?
     
  5. May 24, 2005 #4

    dextercioby

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    Nope,u said

    [tex] \frac{\hat{r}}{r^{2}}\equiv\frac{\vec{r}}{r^{3}} [/tex]

    The first.

    Daniel.
     
  6. May 24, 2005 #5
    yes, :)
    you caught me, i've misread yours

    but what about [tex](\hat{r}\cdot\dot{p}(t_o))\hat{r}[/tex] ? do you happen to figure out which of four gives it and how?
     
  7. May 24, 2005 #6

    dextercioby

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    I've told you,use the chain rule for the first of the 4.

    Daniel.
     
  8. May 24, 2005 #7
    i find this out of the first one
    [tex](p(t_o)-3(\hat{r}\cdot p(t_o))\hat{r})/r^3[/tex]

    approved?
     
  9. May 24, 2005 #8
    no, okay this's from the second one i fnd
     
  10. May 24, 2005 #9
    i can't expand [tex] =\left(\frac{\vec{r}}{r^{3}}\cdot\nabla\right)\vec {p}\left(t_{o}\right)[/tex] as it should be. mine doesn't satisfy the given answer

    can anyone help???

    i am going NUTS here
     
  11. May 24, 2005 #10

    dextercioby

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    [tex] I=\left(\frac{\vec{r}}{r^{3}}\cdot\nabla\right)\vec{p}\left(t_{o}\right) [/tex] (1)

    Use cartesian tensors

    [tex] I=\frac{x_{i}}{r^{3}}\partial_{i}p_{j}\left(t_{o}\right)\vec{e}_{j} [/tex] (2)

    [tex] \partial_{i}p_{j}=-\frac{1}{c}\frac{dp_{j}}{dt_{o}}\frac{\partial r}{\partial x_{j}}=-\frac{\dot{p}_{j}\left(t_{o}\right)}{c} \frac{x_{i}}{r} [/tex] (3)

    [tex] I=-\frac{\dot{p}_{j}\left(t_{o}\right)}{c} \frac{x_{i}}{r^{3}}\frac{x_{i}}{r} \vec{e}_{j} =-\frac{\vec{p}\left(t_{o}\right)}{c r^{2}} [/tex]

    Daniel.
     
    Last edited: May 24, 2005
  12. May 25, 2005 #11
    thanx a lot, that's exactly what i found too

    then my mistake is certainly in the expansion of [tex]\left[\vec{p}\left(t_{o}\right)\cdot\nabla\right]\frac{\vec{r}}{r^{3}}[/tex] term
     
  13. May 25, 2005 #12

    dextercioby

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    That cannot give a derivative of the moment of dipole vector.It's the one with a curl acting on the vector.

    The one u mentioned is quite easy to compute,just use the Leibniz rule carefully.So what is

    [tex]\partial_{i}\left(\frac{x_{j}}{r^{3}}\right) [/tex]

    equal to...?


    Daniel.
     
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