1. May 24, 2005

ucclarke

i am desperate about the expansion of the following equation:
$$\nabla ( \hat{r} /r^2 \cdot \vec{p}(t_o))$$

where $$t_o$$ is the retarded time at the center
$$t_o=t-r/c$$

and $$\vec{p}(t_o)$$ is the electric dipole moment at $$t_o$$

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

it would be magnifique if anyone can figure out the expansion of that term

Last edited: May 24, 2005
2. May 24, 2005

dextercioby

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

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

$$+\vec{p}\left(t_{o}\right)\times\left(\nabla\times\frac{\vec{r}}{r^{3}}\right)$$

All 4 of them,okay?

Daniel.

Last edited: May 24, 2005
3. May 24, 2005

ucclarke

ok, it's actually $$\nabla\left[\frac{\vec{r}}{r^{2}}\cdot\vec{p}\left(t_{o}\right )\right]$$, but really fine, thanks a lot.

so may i ask which one of these terms give out $$(\hat{r}\cdot\dot{p}(t_o))\hat{r}$$ and how?

4. May 24, 2005

dextercioby

Nope,u said

$$\frac{\hat{r}}{r^{2}}\equiv\frac{\vec{r}}{r^{3}}$$

The first.

Daniel.

5. May 24, 2005

ucclarke

yes, :)
you caught me, i've misread yours

but what about $$(\hat{r}\cdot\dot{p}(t_o))\hat{r}$$ ? do you happen to figure out which of four gives it and how?

6. May 24, 2005

dextercioby

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

Daniel.

7. May 24, 2005

ucclarke

i find this out of the first one
$$(p(t_o)-3(\hat{r}\cdot p(t_o))\hat{r})/r^3$$

approved?

8. May 24, 2005

ucclarke

no, okay this's from the second one i fnd

9. May 24, 2005

ucclarke

i can't expand $$=\left(\frac{\vec{r}}{r^{3}}\cdot\nabla\right)\vec {p}\left(t_{o}\right)$$ as it should be. mine doesn't satisfy the given answer

can anyone help???

i am going NUTS here

10. May 24, 2005

dextercioby

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

Use cartesian tensors

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

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

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

Daniel.

Last edited: May 24, 2005
11. May 25, 2005

ucclarke

thanx a lot, that's exactly what i found too

then my mistake is certainly in the expansion of $$\left[\vec{p}\left(t_{o}\right)\cdot\nabla\right]\frac{\vec{r}}{r^{3}}$$ term

12. May 25, 2005

dextercioby

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

$$\partial_{i}\left(\frac{x_{j}}{r^{3}}\right)$$

equal to...?

Daniel.