$\nabla^2\mathbf{V}$ versus $\nabla (\nabla \cdot \mathbf{V})$. They seem pretty much the same to me. Both give back a vector.
$\endgroup$ 22 Answers
$\begingroup$Let me break this out in components. I let $\partial_i~=~\frac{\partial}{\partial x_i}$. Clearly the divergence of a vector ${\bf V}~=~{\bf i}V_x~+~{\bf j}V_y~+~{\bf k}V_z$ and the gradient operator $\nabla~=~{\bf i}\partial_x~+~{\bf j}\partial_y~+~{\bf k}\partial_z$ is $$ \nabla\cdot{\bf V}~=~({\bf i}\partial_x~+~{\bf j}\partial_y~+~{\bf k}\partial_z)\cdot({\bf i}V_x~+~{\bf j}V_y~+~{\bf k}V_z) $$ $$ =~\partial_xV_x~+~\partial_yV_y~+~\partial_zV_z~=~\sum_i\partial_iV_i. $$ So far so good. Now let us take the divergence of this $$ \nabla\nabla\cdot{\bf V}~=~\sum_j{\bf e}_j\partial_j\sum_i\partial_iV_i $$ $$ =~\sum_{i,j}{\bf e}_j\partial_j\partial_iV_i~=~\sum_i{\bf e}_i\partial_i\partial_iV_i~+~\sum_{i\ne j}{\bf e}_j\partial_j\partial_iV_i. $$ The first term on the right on the equal sign is $\nabla^2\bf V$, but the second term has mixed partials.
If instead you take the gradient of a scalar $\nabla\phi$ this is $$ \nabla\phi~=~{\bf i}\partial_x\phi~+~{\bf j}\partial_y\phi~+~{\bf k}\partial_z\phi $$ The divergence of this will produce $\nabla^2\phi$
$\endgroup$ 1 $\begingroup$The Laplacian is a scalar function and returns a scalar value. The gradient of a function returns a vector value.
$\endgroup$ 2
