Does the gradient of the gradient make sense?

Question

We know that for a vector-valued function $A$ we can define

$$\operatorname{curl}(\operatorname{curl}(A))$$

where the curl operator is $\operatorname{curl}(A) = \nabla \times A$.

But given an scalar function $f$ what does

$$\operatorname{grad}(\operatorname{grad}(f)$$ mean?

Answer 1

Here's one interpretation: In geometric calculus we have $$\nabla = \sum_k\mathbf e_k\partial_k$$ where $\{\mathbf e_1, \dots, \mathbf e_n\}$ is an orthonormal basis (we get the same result (as we should) if our basis is curvilinear, but it's easier just to assume orthonormality).

So $$\begin{align}\nabla f(\mathbf x) &= \sum_k\mathbf e_k\partial_kf(\mathbf x) \\ \implies \nabla\nabla f(\mathbf x) &= \sum_j\mathbf e_j\partial_j\sum_k\mathbf e_k \partial_kf(\mathbf x) \\ &= \sum_{j,k} \mathbf e_j\mathbf e_k\partial_j\partial_k f(\mathbf x) \\ &= \sum_{j}{\partial_j}^2f(\mathbf x) + \sum_{j\lt k} \mathbf e_j\wedge \mathbf e_k (\partial_j\partial_k-\partial_k\partial_j)f(\mathbf x) \\ &= \sum_{j}{\partial_j}^2 f(\mathbf x)\end{align}$$ where that last step is due to the equality of mixed partials.

Thus in GC, $\nabla\nabla f(\mathbf x) = \nabla \cdot \nabla f(\mathbf x) = \Delta f(\mathbf x)$.

Answer 2

The gradient is an operator with input a scalar field and output a vector field. Therefore the gradient of a gradient is undefined.