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.


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 $\sum_{i}a_i\equiv a_i$.

  • $\begingroup$ A very clever way maintain generality. Thank you. $\endgroup$ – MDOG Sep 23 '14 at 4:42

Let us use Einstein notation. The vector $f\nabla\psi$ has coordinates $(f\nabla\psi)_i = f\psi_{,i}$. Taking the divergence we have \begin{aligned} \nabla\cdot (f\nabla\psi) &= (f\psi_{,i})_{,i} \\ &= f_{,i}\psi_{,i} + f\psi_{,ii} = \nabla f\cdot \nabla\psi + f\nabla^2 \psi \, .\\ \end{aligned}


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