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Main Question or Discussion Point
Does ∇⋅A = A ⋅∇? If not then, what does the latter actually equal?
So they are equal? Since it is multiplication, they are the same then?if [itex]\vec{A}=(A_x,A_y,A_z)[/itex] then [itex]\vec{A}\cdot\nabla=A_x\frac{\partial}{\partial x}+A_y\frac{\partial}{\partial y}+A_z\frac{\partial}{\partial z}[/itex] that is A⋅∇ is still an operator and can be applied to a scalar or a vector. For example applied to scalar φ, [itex](\vec{A}\cdot\nabla)\phi=A_x\frac{\partial\phi}{\partial x}+A_y\frac{\partial\phi}{\partial y}+A_z\frac{\partial\phi}{\partial z}[/itex]
Nice thank you! Are you familiar with vector potentials in physics? I am working on proving the following but I have been running into trouble. Trying to prove that ∇xA=B!No they are not equal, [itex]\nabla\cdot\vec{A}[/itex] is not an operator itself , its the result of the operator [itex]\nabla[/itex]applied to vector [itex]\vec{A}[/itex] via the dot product and it is a scalar quantity refered as the divergence of vector A. It is [itex]\nabla\cdot\vec{A}=\frac{\partial{A_x}}{\partial x}+\frac{\partial{A_y}}{\partial y}+\frac{\partial{A_z}}{\partial z}[/itex].