Prove $ \nabla \cdot (f \nabla \psi ) = \nabla f \cdot \nabla \psi + f \nabla ^2 \psi $ in general curvilinear coordinates.

Question

Prove $ \nabla \cdot (f \nabla \psi ) = \nabla f \cdot \nabla \psi + f \nabla ^2 \psi $ in general curvilinear coordinates.

I have been attempting to do this using general curvilinear dot products and other associated formulas. I am getting quite confused when it comes to scale factors, do they differ between $f$ and $\psi$?

If someone could show me how to do this without getting overwhelmed with the number of vectors and scalars and scale factors and partial differentials that would be fantastic.

Answer 1

Just use the product rule and let $a\cdot b=a_ib_i$.

$\nabla\cdot(f\nabla\psi)=\nabla_i(f\nabla\psi)_i=\nabla_i(f(\nabla\psi)_i)=(\nabla_if)(\nabla\psi)_i+f(\nabla_i(\nabla\psi)_i)=(\nabla f)_i(\nabla\psi)_i+f(\nabla\cdot(\nabla\psi))=(\nabla f)\cdot(\nabla\psi)+f(\nabla^2\psi)$

The above is written in Einstein notation, that is $\sum_{i}a_i\equiv a_i$.