# What is $\nabla$ $\vec{F}$?

Everywhere I look online I can see $\nabla$ applied to a scalar. I'm wondering what happens when del ($\nabla$) is applied to a vector, $\vec{F}$. I can't find it in any book or online for some reason. I'm sure it's something simple I'm missing.

$\nabla$ $f$ (where scalar is $f$): $df/dx + df/dy + df/dz$

What is $\nabla$ $\vec{F}$ (where vector is $\vec{F}$ = $(x,y,z)$ )

is it: $d$ $\vec{F}$ $/dx$ + $d$ $\vec{F}$ $/dy$ + $d$ $\vec{F}$ $/dz$? And is that equal to $(dx/dx + dy/dx + dz/dx) + (dx/dy + dy/dy + dz/dy) + (dx/dz + dy/dz + dz/dz)$ which is equal to (1+0+0) + (0+1+0) + (0+0+1) = 3?

I don't think this is the case as it doesn't coincide with my lecture notes so it would be helpful to know what I'm misunderstanding.

I'm really wanting to know the difference between $\vec{A} ( \nabla. \vec{F})$ and $(\vec{A} . \nabla) \vec{F}$, where $\vec{A}$ is constant. So if you could explain that as well that would be helpful, thanks.

Edit: The answer is $3 \vec{A}$ - $\vec{A}$ = $2 \vec{A}$ I don't see it so if someone would explain that, that would be helpful, thanks.

• What is "del"? Proper formatting is expected; for information about writing math at this site see e.g. here, here, here and here.
– mlc
Apr 9, 2017 at 19:10
• Hi and welcome to the site. In the long run you are encouraged to learn LaTeX (MathJax) typesetting and also to show what you have tried if a question looks like it could be homework. Those two things will go a long way to get positive responses. Apr 9, 2017 at 19:10
• I have formatted the question now Apr 9, 2017 at 19:29

"Nabla" $\nabla$ is a vector. You can do inner or outer products with vectors.

• Divergence is dot product (inner product) with a vector.
• Gradient is outer product with a scalar or a vector.

Outer products "blow up" the space, adding an index:

• If we start with a scalar (0 indexes) we get a vector (1 index) - often called a gradient.
• If we start with a vector (1 indexes) we get a matrix (2 indexes) - often called a jacobian et.c.
• If we start with a scalar (0 indexes) and do outer product with gradient twice the index count increase twice and we end up with a matrix (2 indexes) which is called a hessian.

So the simplest way is to treat it like a "special vector" which elements do partial differentiating on whatever they happen to hit in the matrix multiplication.

• Thanks @mathreadler for your comment. I don't feel this has answered my question though. If I use bullet point two, I would get 3 - 0 = 3 which isn't the answer. Maybe I am misunderstanding you though. Apr 9, 2017 at 19:39

When $\vec{F}$ $= (x,y,z)$

$\nabla$ $\vec{F}$ =

dx/dx + dx/dy + dx/dz
dy/dx + dy/dy + dy/dz
dz/dx + dz/dy + dz/dz


= (1,1,1)

so $\vec{A} . \nabla$ $\vec{F}$ where $\vec{A}$ is constant and is $(a1,a2,a3)$ is:

a1*dx/dx + a2*dx/dy + a3*dx/dz
a1*dy/dx + a2*dy/dy + a3*dy/dz
a1*dz/dx + a2*dz/dy + a3*dz/dz


= (A1,A2,A3) = $\vec{A}$

• I just worked it out because I realised I understand the NSEs. How silly. Hopefully this will help someone else though. Apr 9, 2017 at 20:39

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}\newcommand{\Del}{\nabla}$tl; dr The $\Del$ operator (i.e., the gradient operator on functions and the curl and divergence operators on vector fields in $\Reals^{3}$) turns out to be an artifact of the exterior derivative operator acting on differential forms.

At the same level of generality (first-order differential operators depending only on the smooth structure of a manifold, not on a Riemannian metric), there is no operator corresponding to $\Del F$ for a vector field $F$.

In more detail, if $f$ is a smooth, real-valued function on some non-empty open subset $U \subset \Reals^{3}$, define $$df = \frac{\dd f}{\dd x}\, dx + \frac{\dd f}{\dd y}\, dy + \frac{\dd f}{\dd z}\, dz. \tag{1}$$

If $\alpha = \alpha_{1}\, dx + \alpha_{2}\, dy + \alpha_{3}\, dz$ is a smooth $1$-form in $U$, define \begin{align*} d\alpha &= d\alpha_{1} \wedge dx + d\alpha_{2} \wedge dy + d\alpha_{3} \wedge dz \\ &= \left[\frac{\dd \alpha_{3}}{\dd y} - \frac{\dd \alpha_{2}}{\dd z}\right] dy \wedge dz + \left[\frac{\dd \alpha_{1}}{\dd z} - \frac{\dd \alpha_{3}}{\dd x}\right] dz \wedge dx + \left[\frac{\dd \alpha_{2}}{\dd x} - \frac{\dd \alpha_{1}}{\dd y}\right] dx \wedge dy. \tag{2} \end{align*}

If $\omega = \omega_{1}\, dy \wedge dz + \omega_{2}\, dz \wedge dx + \omega_{3}\, dx \wedge dy$ is a smooth $2$-form in $U$, define \begin{align*} d\omega &= d\omega_{1} \wedge dy \wedge dz + d\omega_{2} \wedge dz \wedge dx + d\omega_{3} \wedge dx \wedge dy \\ &= \left[\frac{\dd \omega_{1}}{\dd x} + \frac{\dd \omega_{2}}{\dd y} + \frac{\dd \omega_{3}}{\dd z}\right] dx \wedge dy \wedge dz. \tag{3} \end{align*}

(Exercise: Use (1) and anti-symmetry of the wedge product to deduce (2) and (3).) In a sense that should be formally apparent, (1) corresponds to the gradient $\Del f$ of $f$, (2) corresponds to the curl $\Del \times F$ of the vector field $F = (\alpha_{1}, \alpha_{2}, \alpha_{3})$, and (3) corresponds to the divergence $\Del \cdot F$ of the vector field $F = (\omega_{1}, \omega_{2}, \omega_{3})$.

It turns out that these operators on differential forms are natural, while the classical vector operators are...not, really (and secretly they rely on extra structure, the Euclidean metric, in their definitions).

From the perspective pf physics or multivariable calculus, the $\Del$ operator should be viewed as a mnemonic, a formal way of remembering how to compute grad, curl, and div.

Finally, assuming I understand your intent, you're expecting something like $$\Del F = (\Del F_{1}, \Del F_{2}, \Del F_{3}),$$ which would formally be a vector field whose components are vector fields, or a matrix-valued function. In this sense, for the example $F = (x, y, z)$ you give, you'd have $$\Del F = \left[ \begin{array}{@{}ccc@{}} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right].$$