The force on a dipole (with moment $\vec{p}_1$) in an electric field is given as $\vec{F} = (\vec{p}_1\cdot\vec{\nabla})\vec{E}$
Now lets say we have another dipole with moment $\vec{p}_2$ at the origin, exerting a force on the first dipole. Jackson (3rd edition), equation (4.13) has the electric field due to a dipole (at the origin) being
$$\vec{E}(\vec{r}) = \frac{\hat{n}(\vec{p}_2\cdot\hat{n}) – \vec{p}_2}{4\pi\epsilon_0|\vec{r}|^3}$$
After substituting this into the force equation, we get
$$\vec{F}(\vec{r}) = (\vec{p}_1\cdot\vec{\nabla})\Bigg(\frac{\hat{n}(\vec{p}_2\cdot\hat{n}) – \vec{p}_2}{4\pi\epsilon_0|\vec{r}|^3}\Bigg)$$
After which, I am a bit afraid to proceed. Now, the $(\vec{p}_1\cdot\vec{\nabla})$ term needs to be taken through the electric field term, although I am not sure how the gradient here interacts with everything inside the electric field term. I know $\hat{n}$ can be written as $\hat{n} = \vec{r}/|\vec{r}|$, so does the gradient affect $\hat{n}$? How do we deal with that fact that $|\vec{r}|^3$ is in an absolute value when we take the derivative? $\vec{p}_2$ should be unaffected because it doesn't depend on $\vec{r}$, right? Also, having $(\vec{p}_1\cdot\vec{\nabla})$ being a non-vector makes me nervous.
These sort of vector calculus calculations have always made me very uncomfortable. I need to iron this out. Can anyone help out?
Best Answer
First of all, no reason to be nervous about $(\vec{p}_1\cdot\vec{\nabla})$; just like $\vec\nabla$, it is an operator. $(\vec{p}_1\cdot\vec{\nabla})\vec{A}$ is a directional derivative and can be thought of as the rate of change of $\vec{A}$ as you move along $\vec{p}_1$ (with "velocity" $\vec{p}_1$).
Using vector calculus identities can simplify such computations considerably. You will need the identities $$ (\vec v\cdot\vec\nabla)(\phi\vec{A}) = (\vec v\cdot\vec\nabla\phi)\vec{A} + \phi(\vec v\cdot\vec\nabla)\vec{A} \tag{1}$$ $$ \vec\nabla f(\phi) = f'(\phi)\nabla\phi \tag{2}\ \ \ \text{(chain rule)} $$ $$ \vec\nabla(\vec A\cdot\vec B) = (\vec A\cdot\vec\nabla)\vec B + (\vec B\cdot\vec\nabla)\vec A + \vec A\times(\vec\nabla\times\vec B) + \vec B\times(\vec\nabla\times\vec A) \tag{3} $$ and the useful relation $\vec\nabla r=\hat{n}$ (prove this for yourself), where I have used the notation $r=|\vec r|$ for simplicity.
In particular, identity 2 is useful for gradients such as $$\vec\nabla (r^m)=mr^{m-1}\vec\nabla r = mr^{m-1}\hat n $$ Finally, to make your life easier, consider rewriting $\hat{n}$ as $\vec{r}/r$, since $\vec{r}$ is often easier to work with when derivative operators are involved.