# Homework Help: Gradient equation with retarded time

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.