physical meaning of the vector $(A \cdot \nabla) A$

Question

If we consider the vector $\left ( A \cdot \nabla \right) \: B$, we have in Cartesian coordinates

$$\left ( A \cdot \nabla \right) \: B = \left ( A \cdot \nabla B_x \right ) e_x + \left ( A \cdot \nabla B_y \right ) e_y + \left ( A \cdot \nabla B_z \right ) e_z,$$ which gives in full writing:

$$\left ( A \cdot \nabla \right) \: B = \left (A_x \frac{\partial \: B_x}{\partial \: x} + A_y \frac{\partial \: B_x}{\partial \: y} + A_z \frac{\partial \: B_x}{\partial \: z} \right )e_x + \left (A_x \frac{\partial \: B_y}{\partial \: x} + A_y \frac{\partial \: B_y}{\partial \: y} + A_z \frac{\partial \: B_y}{\partial \: z} \right )e_y + \left (A_x \frac{\partial \: B_z}{\partial \: x} + A_y \frac{\partial \: B_z}{\partial \: y} + A_z \frac{\partial \: B_z}{\partial \: z} \right )e_z$$

Now, if $B=A$ and $A=\left (0, \: 0, \: A_z \right )$ (i.e. $A_x=A_y=0$), we have from the above definition, only the following term along the unit vector $e_z$:

$$\left ( A \cdot \nabla \right) \: A = A_z \frac{\partial \: A_z}{\partial \: z}$$

But this is equal to zero since $\frac{\partial A_z}{\partial \: z}=0$, despite $A_z \neq 0$.

So we get $$\left ( A \cdot \nabla \right) \: A =0$$

I would like to know the physical interpretation of this result. What is the significance of this vector being zero in terms of field lines?

Thanks a lot...

Answer 1

$A\cdot\nabla$ is the derivative in the direction of $A$ (times the length of $A$), so $(A\cdot\nabla) A=0$ means that $A$ itself does not change when you move along $A$. Which in turn implies that the field lines are straight lines.

More generally, the field lines are straight lines if and only if $(A\cdot\nabla)A$ is a (not necessarily constant) multiple of $A$.