The dot product is commutative:
$(\vec{v} \cdot \vec{w})f(x,y,z) = (\vec{w} \cdot \vec{v})f(x,y,z)$
But, what about the dot product of a gradient operator $\nabla$??? For example:
$(\vec{v} \cdot \nabla)f(x,y,z) = (\nabla \cdot \vec{v})f(x,y,z)$ <== is this valid?
Is it valid to commute a gradient operator in a dot product?
it seems to be the gradient is order sensitive:
$(\vec{v} \cdot \nabla) = (v_1 \hat{\text{i}} + v_2 \hat{\text{j}} + v_3 \hat{\text{k}}) \cdot (\frac{\partial}{\partial x}\hat{\text{i}} + \frac{\partial}{\partial y}\hat{\text{j}} + \frac{\partial}{\partial z}\hat{\text{k}})$
$(\vec{v} \cdot \nabla) = v_1\frac{\partial}{\partial x} + v_2\frac{\partial}{\partial y} + v_3\frac{\partial}{\partial y}$
verses:
$(\nabla \cdot \vec{v}) = (\frac{\partial}{\partial x}\hat{\text{i}} + \frac{\partial}{\partial y}\hat{\text{j}} + \frac{\partial}{\partial z}\hat{\text{k}}) \cdot (v_1 \hat{\text{i}} + v_2 \hat{\text{j}} + v_3 \hat{\text{k}})$
$(\nabla \cdot \vec{v}) = \frac{\partial}{\partial x}v_1 + \frac{\partial}{\partial y}v_2 + \frac{\partial}{\partial y}v_3$
it seems for $(\vec{v} \cdot \nabla)$ you don't need to take a partial derivative of the components of $\vec{v}$. but for $(\nabla \cdot \vec{v})$ you need to take the partial of the components of $\vec{v}$ and anything else $(\nabla \cdot \vec{v})$ gets left multiplied to....
Is this correct? I wasn't sure if i should pull the partial derivatives before the $\vec{v}$ components for the $(\vec{v} \cdot \nabla)$. because its an entirely different answer if i do that...
No, they do not commute. This has to do with the fact that the dot product defined on $\mathbb{R}^n$ is what we refer to as a bi-linear form. The moment we try to extend this to operators, we have to view the dot product, not as a bi-linear form, but more as a convenient short-hand notation. While the dot product between vectors can commute, operators will in general not commute with one another.