- #1
henpen
- 50
- 0
Reading the Feynman Lectures,
[tex]\nabla \times (\nabla T)=(\nabla \times \nabla) T[/tex], is achieved by analogy to the analogous case for [itex]\mathbf{A} \times (\mathbf{A} T)=(\mathbf{A} \times \mathbf{A}) T[/itex],where T is a scalar field in all cases.
While this is obvious if [itex] \nabla [/itex] 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)?
[tex]\nabla \times (\nabla T)=(\nabla \times \nabla) T[/tex], is achieved by analogy to the analogous case for [itex]\mathbf{A} \times (\mathbf{A} T)=(\mathbf{A} \times \mathbf{A}) T[/itex],where T is a scalar field in all cases.
While this is obvious if [itex] \nabla [/itex] 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)?