What would it mean for a dipole moment not to be constant? It is fairly easy to show that, for a neutral the electrostatic force on a neutral charge distribution with a dipole moment $\bf{p}$ is given by
$\bf{F}$$ = $$(\bf{p}$$\cdot \nabla)$$\bf{E}(r)$ if we can take the applied field $\bf{E}(r')$ as a two term Taylor expansion about a reference point $\bf{r}$ somewhere in the charge distribution (i.e. if the applied field varies slowly over the distribution).
In my textbook (Zangwill, Chapter 4.2.3), it is then noted that "When $\bf{p}$ is a constant vector..." we may use a standard vector identity (and the fact that we are in an electrostatic situation) to obtain $\bf{F}$$ = \nabla(\bf{p}\cdot\bf{E}(r))$.
I'm wondering when the dipole moment would NOT be constant (in terms of derivatives with respect to the reference point $\bf{r}$ which are taken to be zero in using the vector identity noted above)? Isn't the dipole vector always constant?
 A: In the macroscopic Maxwell's equation the macroscopic displacement vector is defined by the dipole moment density $\mathbf P$ as $\mathbf D = \epsilon_0 \mathbf E + \mathbf P$ material. This means that locally we have an infinitesimal dipole moment $d \mathbf p= \mathbf P dV$ where both $\mathbf P$ and $d\mathbf p$ are functions of the location variable. The infinitesimal force acting on this infinitesimal dipole moment is $d\mathbf F = (d\mathbf p \cdot \nabla) \mathbf E$ reflecting the local variation of the dipole moment density and the force acting on it.
An obvious practical example of this would be an inhomogeneous dielectric whose permittivity is varying with location.
A: Many atoms/molecules do not have an inherent dipole moment.  Instead, their dipole moment is itself proportional to the external electric field experienced by the molecule:  $\mathbf{p} = \alpha \mathbf{E}$, where $\alpha$ is the atomic polarizability.  Such a molecule moving in a non-uniform electric field would not have a constant dipole moment.  The resulting force on the molecule would be $$\mathbf{F} = \alpha (\mathbf{E} \cdot \nabla) \mathbf{E} = \frac{1}{2} \alpha \nabla (E^2)$$ while we would have $$\nabla (\mathbf{p} \cdot \mathbf{E}) = \alpha \nabla (E^2)$$
which is obviously not the same.
