# Physical meaning of the vector Laplace operator

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?

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