Help on the proof related to the vectorial wave equation

AI Thread Summary
The discussion revolves around proving that if \(\vec{M} = \nabla \times \vec{c} \psi\) satisfies the equation \(-\nabla \times \nabla \times \vec{M} + k^2 \vec{M} = 0\), then it follows that \(\nabla^2 \psi + k^2 \psi = 0\). A user suggests using vector identities, specifically \(\nabla \times c \psi = c \nabla \times \psi + \nabla c \times \psi\) and \(\nabla \times \nabla \times A = \nabla (\nabla \cdot A) - \nabla^2 A\), to facilitate the proof. The conversation touches on the appropriateness of the question for a homework section, but ultimately, the user resolves the problem with the provided guidance. The thread highlights the application of vector calculus in solving wave equation proofs.
elgen
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Dear forum users,

I need some help on the following proof that appears in a book (pp. 84 in Bohren' Absorption and Scattering of light by Small Particles). This is no a home work problem.

The problem statement:

\vec{M} = \nabla\times\vec{c}\psi, where \vec{c} is some constant vector and \psi is a scalar function, then if -\nabla\times\nabla\times\vec{M}+k^2\vec{M}=0 where k is the wave number, then prove that \nabla^2 \psi + k^2\psi=0.

I could prove for the cases by using a curve-linear coordinate system, etc. rectangular, cylindrical, etc. I am seeking a general proof. I suspect that there is some vectorial identity applicable here.

Thank you for the attention.Elgen
 
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What happens if you use ## \nabla \times c \psi = c \nabla \times \psi +\nabla c \times \psi ## and ## \nabla \times \nabla \times A = \nabla (\nabla \cdot A) - \nabla^2 A##?
 
The (Cartesian) Ricci Calculus is your friend here. But isn't this more homework like? So I'd say, this thread should be moved to the homework section of these forums!
 
Worked it out. Thank you for the pointer.
 
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