Difference between the $\nabla\cdot a$ and $a\cdot\nabla$

Question

Hello I am new to vector calculus and I have a basic question . The del operator which is defined as $\nabla = \Bigl(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\Bigr)$ which is basically a vector.

Let $a = (a_1,a_2,a_3)$ be a vector(or a vector function) . Then $a\cdot\nabla = a_1\frac{\partial }{\partial x}+a_2\frac{\partial }{\partial y}+a_3\frac{\partial }{\partial z}$ and $\nabla\cdot a =\frac{\partial a_1}{\partial x}+\frac{\partial a_2}{\partial y}+\frac{\partial a_3}{\partial z} $ obviously both are not equal i.e $\nabla\cdot a \neq a \cdot\nabla$ . But both $a,\nabla$ are vectors and for vectors dot product is commutative i.e $\nabla\cdot a = a \cdot\nabla$ should be true. I find this confusing , please explain.

Answer 1

Yes as you noticed they are not equal.

It is true as you say that: $$a\cdot \nabla = a_1\cdot \frac{\partial}{\partial x}+a_2\cdot \frac{\partial}{\partial y}+a_3\cdot \frac{\partial}{\partial z}$$

And that $$\nabla \cdot a = \frac{\partial a_1}{\partial x}+\frac{\partial a_2}{\partial y}+\frac{\partial a_3}{\partial z}$$

$\nabla$ is an operator, not a vector. It works on functions, not vectors or scalars. It is still practical to use this notation because differentiation is a linear operation which means that almost always we can represent it with linear algebras by a matrix somehow.

The linearity how differentiation behaves on functions is somehow "compatible" with the language of linear algebra.

Answer 2

The dot product is commutative if its arguments commute. But $\nabla$ is an operator and $a$ is a vector function; they do not commute.

This is however only one part of the explanation. The other part is that $\nabla \cdot a$ is defined by $\nabla$ only working on $a$. Had it been defined as an operator acting on a scalar function by $$(\nabla \cdot a) \phi := \sum_k \frac{\partial}{\partial x_k}(a_k \phi) \quad \quad\text{(NB: not the correct definition!)}$$ then the non-commutativity would have had more significance.