Need to prove $\operatorname{div}(\nabla u)=\nabla ^2 u$ where $u=g(x,y,z)$
The RHS is the Lapacian which we were told is a vector. But $\nabla u=(g_x,g_y,g_z)$ and the divergence of that is $g_{xx}+g_{yy}+g_{zz}$ which is not a vector. I don't get how it can equate then...
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$\begingroup$The "Laplacian" is an operator that can operate on both scalar fields and vector fields. The operator on a scalar can be written,
$$\nabla^2 \{\} = \nabla \cdot (\nabla \{\})$$
which will produce another scalar field.
The operator on a vector can be expressed as
$$\nabla^2 \{\} = \nabla (\nabla \cdot \{\})\,\,-\nabla \times (\nabla \times \{\})$$
which will produce another vector field.
In Cartesian coordinates, both operators can be written
$$\nabla^2 \{\} = \frac{\partial^2 \{\}}{\partial x^2}+\frac{\partial^2 \{\}}{\partial y^2}+\frac{\partial^2 \{\}}{\partial z^2}$$
where it is evident that operation on a scalar (vector) field transforms into a scalar (vector) field.
$\endgroup$ 5 $\begingroup$Lapacian of an Nth Rank Tensor is another Nth Rank Tensor. This means Lapacian of a Scalar Field is another Scalar Field. Laplacian of Vector Field is another Vector Field and so on.
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