What's the difference between $( A \cdot \nabla ) B$ and $B ( \nabla \cdot A )$?

Question

In the identity: $\nabla\times(A\times B) = (B \cdot\nabla)A-(A \cdot\nabla)B+A(\nabla \cdot B)-B(\nabla \cdot A)$

Answer 1

Assuming we are in $\mathbb R^n$,

$$(A \cdot \nabla) = \sum_i A^i \frac{\partial}{\partial x^i}$$

for coordinates $\{x^i\}$ on $\mathbb R^n$. This denotes the directional derivative along $\vec A$ and moreover $A \cdot \nabla$ is simply a way of writing a vector field as a derivation. The Lie derivative of a scalar is precisely this:

$$\mathcal L_A \phi = \sum_i A^i \frac{\partial \phi}{\partial x^i}.$$

For the case of a vector, one has $\mathcal L_A B = [A,B]$ with $A$ and $B$ expressed as derivations as above. Now for $(\nabla \cdot A)$, this is simply a scalar and equal to the divergence of $\vec A$, namely,

$$\nabla \cdot A = \sum_i \frac{\partial A^i}{\partial x^i}.$$