Working in 3D. I know that the gradient is a vector operator defined as $\nabla = [\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}]$.
The gradient of a scalar scalar-valued function $f(\vec{x})\in\mathbb{R}$ is $\nabla f(\vec{x}) = [\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}]f(\vec{x}) = [\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}]$. This makes sense to me.
But if we take the gradient of a vector field, say $\vec{f} = [f_1,f_2,f_3]$, I know that this is
$\displaystyle
\nabla \vec{f} =
\begin{bmatrix}
\frac{\partial f_1}{\partial x} &
\frac{\partial f_1}{\partial y} &
\frac{\partial f_1}{\partial z} \\
\frac{\partial f_2}{\partial x} &
\frac{\partial f_2}{\partial y} &
\frac{\partial f_2}{\partial z} \\
\frac{\partial f_3}{\partial x} &
\frac{\partial f_3}{\partial y} &
\frac{\partial f_3}{\partial z} \\
\end{bmatrix}
$.
But how did we get to this? Since both $\nabla$ and $\vec{f}$ are vectors, this seems a bit like an outer product, but writing $\nabla \otimes\vec{f}$ turns out to be the transpose of what I want i.e.
$\displaystyle
\nabla \otimes\vec{f}=\nabla\vec{f}^T
=
\begin{bmatrix}
\frac{\partial}{\partial x}\\
\frac{\partial}{\partial y}\\
\frac{\partial}{\partial z}
\end{bmatrix}
\begin{bmatrix}
f_1 & f_2 & f_3
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial f_1}{\partial x} &
\frac{\partial f_2}{\partial x} &
\frac{\partial f_3}{\partial x} \\
\frac{\partial f_1}{\partial y} &
\frac{\partial f_2}{\partial y} &
\frac{\partial f_3}{\partial y} \\
\frac{\partial f_1}{\partial z} &
\frac{\partial f_2}{\partial z} &
\frac{\partial f_3}{\partial z} \\
\end{bmatrix}
$
Am I not understanding something correctly? What am I doing wrong?
Best Answer
From my limited understanding you want the dyadic tensor product, not the outer product.
In more complex examples this is not always the case, but for ${\bf a},{\bf b}\in$ Euclidean space, the dyadic product is related to the outer product by $$ {\bf a}\otimes{\bf b}\equiv{\bf a}{\bf b}^T $$
Which agrees with your example.
When placed next to a scalar-valued function or vector field alike $\nabla$ is considered an operator. Subsequently, for the vector field case you want the dyadic product of $\nabla$ and the vector field it is acting on. Check out the section marked, three dimensional space on the wikipedia page for dyadics https://en.wikipedia.org/wiki/Dyadics