Why do we generally write $(\vec{v} \cdot \nabla )\vec{u}$ as opposed to $\vec{v} \cdot (\nabla \vec{u})$?

Question

It seems a bit strange that we would write it this way, given that $(\vec{v} \cdot \nabla )$ doesn't really have any inherent meaning, whereas $(\nabla \vec{u})$ does.

Answer 1

$\vec v \cdot \nabla$ has an inherent meaning in the same way that $\nabla$ has an inherent meaning: both denote operators on differentiable functions.

We can think of $\nabla$ as the operator $$ f \xrightarrow{\mspace{4mu}\fbox{$\nabla$}\mspace{4mu}} \nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) $$ Likewise, for $\vec u = (u_1,u_2,u_3)$, we can think of $\vec u \cdot \nabla$ as the operator $$ f \xrightarrow{\mspace{4mu}\fbox{$\vec u \cdot \nabla$}\mspace{4mu}} \vec u \cdot \nabla f = \vec u \cdot \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = u_1 \frac{\partial f}{\partial x} + u_2\frac{\partial f}{\partial y} + u_3 \frac{\partial f}{\partial z} $$ Note that the meaning of $\vec u \cdot \nabla$ is quite distinct from that of $\nabla \vec u$ or $\nabla \cdot \vec u$.