Vector Operator Identities

^head #TODO

Examples §

1: Proving identity 11

Prove that if $f$ is a $C^2$ smooth 3D scalar function, then:

$$\nabla \times (\nabla f)=\vec{0}$$

Solution

We know that:

$$\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$$

Then the curl returns:

$$\nabla \times \nabla f=\det \left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{\partial f}{\partial x}& \frac{\partial f}{\partial y}& \frac{\partial f}{\partial z}\end{array}\right]=\left(\frac{\partial f}{\partial y\partial z}-\frac{\partial f}{\partial z\partial y},\frac{\partial f}{\partial x\partial z}-\frac{\partial f}{\partial z\partial x},\frac{\partial f}{\partial x\partial y}-\frac{\partial f}{\partial y\partial x}\right)$$

We can use Fubini’s theorem to swap the partial derivatives, since we assumed that $f$ is $C^2$, it must be continuous.

$$=\left(\frac{\partial f}{\partial y\partial z}-\frac{\partial f}{\partial y\partial z},\frac{\partial f}{\partial x\partial z}-\frac{\partial f}{\partial x\partial z},\frac{\partial f}{\partial x\partial y}-\frac{\partial f}{\partial x\partial y}\right)=\vec{0}$$