Does Vector Calculus Allow Operators to Function Like Vectors?

[henpen]

Reading the Feynman Lectures,

$$\nabla \times (\nabla T)=(\nabla \times \nabla) T$$, is achieved by analogy to the analogous case for $\mathbf{A} \times (\mathbf{A} T)=(\mathbf{A} \times \mathbf{A}) T$,where T is a scalar field in all cases.

While this is obvious if $\nabla$ were to be replaced by a vector, I'm a complete novice when it comes to operators- is it a general rule in vector calculus that operators behave pretty much like vectors (apart from needing 'feeding', of course)?

 

[lurflurf]

So aside from some nit picky stuff about continuity, convergence, and connected spaces the main issue is these derivatives by convention act to the right. A vector identity with have an analogous vector calculus identity when it does not require reversing order.

$$(\vec{a}\cdot \vec{\nabla})\vec{b}\ne \vec{b}(\vec{\nabla}\cdot \vec{a})$$

It is possible to define a new bidirectional operator to work these things out. Another one that surprises some people is

$$\vec{a}\cdot (\vec{\nabla}\times \vec{a})\ne \vec{\nabla}\cdot (\vec{a}\times \vec{a})=0$$

but

$$\vec{a}\cdot (\vec{b}\times \vec{a})= \vec{b}\cdot (\vec{a}\times \vec{a})=0$$

 

[henpen]

lurflurf, thanks for that. Sorry for not seeing it yet, but how do these properties apply in proving $(\nabla \times \nabla) \mathbf{F}= \nabla \times (\nabla \mathbf{F})$, though?

 

[lurflurf]

Since it does not require reversing the direction the same rule apply as in

$$\vec{a}\times (\vec{a}T)=(\vec{a}\times \vec{a})T$$

Also curl of gradient is zero so del cross del is not a useful operator.

So again given a vector rule the rule with del will hold as long as nothing changes which side of del it is on. There is something about that in The Feynman Lectures. There is a partial del

$$\vec{\nabla}_b (\vec{a}\cdot\vec{b})=\vec{a}\times(\vec{\nabla} \times \vec{b} )+(\vec{a}\cdot \vec{\nabla})\vec{b}$$

a can change sides because this del ignores it while the usual one does not.