The dot product is not associative, so, no, they are not necessarily equivalent.Does
\(\vec{\nabla} \cdot \vec{E} = 0\)
imply
\(\vec{\nabla}^2 \cdot \vec{E} = 0\)
?
Is this true:
\(\vec{\nabla}^2 \cdot \vec{E} = \vec{\nabla}(\vec{\nabla} \cdot \vec{E})\)
Wow. I'd never even heard of this. It sounds very interesting.Knot theory.
The bookKnot theory
by Aaron Carman
by Duane Benson
by Jake Hertz
by Aaron Carman