# Difference between $\nabla$ and $\vec \nabla$

From what i have seen different places on the web $$\nabla$$ and $$\vec \nabla$$ is being used on many of the same things. Is there a difference?

I thought that $$\nabla f$$ was an operator on a function. And $$\vec\nabla f = \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix} f$$ was a vector on a function / scalar. In this example i would think i should get the same result. But if i look at $$\vec u = (u,v,w)= \begin{pmatrix}u\\ v\\ w\end{pmatrix}$$ i get:

$$\nabla \vec u = \begin{pmatrix}\frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial y}\\ \frac{\partial w}{\partial z}\end{pmatrix}$$ (My teacher said so). But i dont understand why this would not be $$\nabla \vec u = \begin{pmatrix}\frac{\partial \vec u}{\partial x}\\ \frac{\partial \vec u}{\partial y}\\ \frac{\partial \vec u}{\partial z}\end{pmatrix}$$. Wich is still a vector.

And if i take $$\vec\nabla \vec u = \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix} \begin{pmatrix}u\\ v\\ w\end{pmatrix}$$ i get $$\begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix} \begin{pmatrix}u\\ v\\ w\end{pmatrix} = \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}$$ Wich is not a vector. And this seems to be $$\nabla \cdot \vec u$$

So from what i see and think should $$\nabla \vec u = \begin{pmatrix}\frac{\partial \vec u}{\partial x}\\ \frac{\partial \vec u}{\partial y}\\ \frac{\partial \vec u}{\partial z}\end{pmatrix} = \begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}&\frac{\partial w}{\partial x}\\ \frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}&\frac{\partial w}{\partial y}\\ \frac{\partial u}{\partial z}&\frac{\partial v}{\partial z}&\frac{\partial w}{\partial z}\end{pmatrix} = \vec \nabla \vec u^{T} = \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix}\begin{pmatrix}u&v&w\end{pmatrix}$$. What is wrong with this or why is it correct?

• Arrows over operators (or ordinary vectors) are an author's preference, based on whether they think the reader needs them to understand. Omitting the arrow doesn't change the meaning; it just expects more of readers.
– J.G.
Feb 9, 2022 at 14:40
• I'm very skeptical about your “my teacher said so” statement. I've never seen $\nabla$ used in a way where element-wise application makes sense. Except if you do the dot product, where you sum the results as well, as you did later in your post.
– MvG
Feb 9, 2022 at 14:48

Your teacher is wrong! You are indeed correct that, at least in the standard basis, $$(\nabla \vec u)^i{}_j=\frac{\partial u^i}{\partial x^j}$$ E.g in your example with $$\vec{u}=(u~~v~~w)^\intercal~~$$ and coordinates $$x,y,z$$, $$\nabla \vec{u} =\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{bmatrix}$$
And the divergence is simply $$\operatorname{div} \vec u=\operatorname{tr}\nabla \vec u$$
It's the same. The arrow is just used to clarify the fact that $$\nabla$$ is a vector.