In the first chapter of Griffiths' Electrodynamics, he introduces some math that will be used.

There is a section on second derivatives.

One type of second derivative is $\nabla \cdot(\nabla T)$, where $T$ is a scalar function.

$$\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right )\cdot \left ( \frac{\partial T}{\partial x}\hat{i}+\frac{\partial T}{\partial y}\hat{j}+\frac{\partial T}{\partial z}\hat{k}\right )$$

$$= \frac{\partial T^2}{\partial^2 x}\hat{i}+\frac{\partial T^2}{\partial^2 y}\hat{j}+\frac{\partial T^2}{\partial^2 z}\hat{k}\tag{1}$$

which we call the Laplacian of $T$.

Then he says

occasionally, we shall speak of the Laplacian of a vector $\nabla^2\vec{v}$. By this we mean a vector quantity whose x-component is the Laplacian of $v_x$, and so on.

$$\nabla^2 \vec{v} \equiv \nabla^2v_x\hat{i} + \nabla^2 v_y \hat{j} + \nabla^2 v_z \hat{k}\tag{2}$$

$(1)$ is obtained via definitions of dot product and gradient (del) operator. $(2)$ on the other hand seems to be a definition. After all, in $\nabla \cdot (\nabla \vec{v})$, the term $\nabla \vec{v}$ doesn't seem to make sense. We can't use simple multiplication between two vectors. Is $(2)$ derivable as $(1)$ is from previous definitions?

  • $\begingroup$ That's because $\nabla \vec{v}$ is an entirely new object that isn't a scalar or a vector, it is an object know as a tensor. $\endgroup$
    – Triatticus
    Mar 20, 2022 at 19:33

1 Answer 1


For vectors (and other higher order tensors), the laplacian is defined as $$\nabla^2\mathbf{T} :=(\nabla\cdot\nabla)\mathbf{T}$$ For scalars $(\nabla\cdot\nabla)T = \nabla\cdot(\nabla T)$, but for vectors this is only true if the dot product is interpreted to be a left dot product. This can be easily seen in index notation where $(\nabla \vec{v})_{ij}=\partial_i v_j $ and you can take the dot product either with the left or the right index. The left dot product is the laplacian, $\partial_i\partial_i\,v_j=[(\nabla\cdot\nabla)\vec{v}]_j$ but the right one is not, $\partial_j\partial_i\,v_j=\partial_i\partial_j v_j=[\nabla(\nabla\cdot \vec{v})]_i$.


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