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I have seen here a question asking for the physical interpretation of the Laplace operator for a scalar field. However, there is also a vectorial version of this operator, the vector laplace operator, which is defined as follows:

$$ \nabla^{2} \mathbf{A}=\nabla(\nabla \cdot \mathbf{A})-\nabla \times(\nabla \times \mathbf{A}) $$

being both $\mathbf {A}$ and $\nabla^{2} \mathbf{A}$ vector fields. In particular, in Cartesian coordinates it would take this form: $$ \nabla^{2} \mathbf{A}=\left(\nabla^{2} A_{x}, \nabla^{2} A_{y}, \nabla^{2} A_{z}\right) $$

Since the Laplacian $\nabla^2 f$ of a scalar field $f$ at a point $p$ measures by how much the average value of $f$ over small balls centered at $p$ deviates from $f(p)$, what would be the physical or intuitive meaning of the vector Laplacian?

Invenietis
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1 Answers1

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The meaning is exactly the same. The Laplacian of a vector field at a point $p$ measures the amount by which the average of the vector over small balls centered at $p$ differs from the vector at $p$. In fact, since scalars and vectors are tensors of rank $(0,0)$ and $(1,0)$ respectively, the Laplacian can be applied to tensors of any rank.

For the sake of completeness, the Laplacian in tensor notation (curved space without non-metricity) is:

$$\nabla^i \nabla_i = g^{ij} \nabla_i \nabla_j$$

Vincent Thacker
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