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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.

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

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