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Could someone explain to me the physical meaning of vector Laplacian of Electric field intensity?

Where vector Laplacian means: $$\nabla^2 \mathbf{E} = \nabla(\nabla \cdot \mathbf{E}) - \nabla \times (\nabla \times \mathbf{E})$$ Thanks!

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Please provide context. the electric field intensity is a scalar, so "vector Laplacian" probably just means "Laplacian", so that you are looking at $\nabla^2 |E|^2 $, but you can't be sure without a reference. – Ron Maimon Dec 5 '11 at 0:40
The vector laplacian is this one: And Vents, you should put what you wrote in your answer as an edit to your question, since its not an answer. – Arnoques Dec 5 '11 at 16:01
Do you mean vector Laplacian of E, or of E intensity, these are two different things. Please write the formula for what you want, nobody needs the formula defining the laplacian. Plus you wrote it in an unnatural three-dimensional specific way. – Ron Maimon Apr 5 '12 at 3:33

1 Answer 1

As Ron Maimon said, a laplacian is applied to scalar fields. A vector laplacian is appliad to vector fields. You can't apply a vector laplacian to a scalar field.

I have no idea what $\nabla^2|\mathbf{E}|^2$ can represent physically, though it should be related to the energy of an electrostatic field.

On the other hand, the electromagnetic wave equation in vacuum for $\mathbf{E}$ is $$\frac{\partial^2 \mathbf{E}}{\partial t^2} - c_0^2 \cdot \nabla^2 \mathbf{E} = 0$$ so you could say that, if there are no sources, the vector laplacian of the electric field is proportional to the second time derivative of the field.

I hope that this is somewhat useful.

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$\nabla^2 |E|^2$ is the "scalar" Laplacian of the electric field intensity. It doesn't have another popular meaning. – Ron Maimon Dec 5 '11 at 18:41

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