Is $\nabla$ a vector?

Question

The following passage has been extracted from the book "Mathematical methods for Physicists":

A key idea of the present chapter is that a quantity that is properly called a vector must have the transformation properties that preserve its essential features under coordinate transformation; there exist quantities with direction and magnitude that do not transform appropriately and hence are not vectors.

Cross product: $\nabla \times (Vector)=Vector$

From the above equation of cross product we can say that $\nabla$ is a vector (specifically vector operator). However, a vector generally has magnitude and an associated direction. While in case of $\nabla$, it might satisfy essential features under transformation to be a vector, but I don't see whether it has magnitude or not? Does it has magnitude? If so, what is it? Or otherwise is it that a vector need not have magnitude?

I did it before posting. When I did I found no reliable results. Did you find any? — Sensebe, Nov 19 '14 at 16:55
As in every branch of mathematics, one should define $X$ before asking "is $Y$ an example of $X$?". If you find yourself unwilling to do this, then you are instead asking the question "Should we consider $Y$ an example of $X$, and thus formulate our definition of $X$ accordingly?". I think you meant to ask the latter but accidentally phrased it as the former. — RghtHndSd, Nov 19 '14 at 17:48
Read The Feynman Lectures on Physics second volume, second chapter, sections 4 and 5. — Tony Piccolo, Nov 20 '14 at 08:01
@TonyPiccolo: To be frank, I read it before posting. He has made the analysis, but he didn't reach where I expected him to reach. — Sensebe, Nov 20 '14 at 14:42
In general, vectors need not have "magnitude and direction". The mathematical concept of a vector is just anything that you can add with similar things and multiply by a constant. — Ben Grossmann, Nov 20 '14 at 15:47
@Omnomnomnom: vectors need not have "magnitude and direction" What is the source for your statement? — Sensebe, Nov 20 '14 at 16:06
I'm not sure any one source will do here, but the gist is this: "direction" requires a notion of angle between two vectors, which requires an inner product. Most normed vector spaces (vector spaces with "magnitude") are not inner product spaces (vector spaces with "direction"), and only some normed vector spaces can be assigned an inner product that makes sense. — Ben Grossmann, Nov 20 '14 at 16:17
In fact, not all vector spaces have a notion of "magnitude" (i.e. a norm) that makes sense. So, we can indeed have vectors where no concept of either magnitude or direction "fits" appropriately. — Ben Grossmann, Nov 20 '14 at 16:21
The bright side of this development is that it may yet be possible to conceptualize $\nabla$ as some sort of "vector". However, it is clear that in such a structure, the usual notion of "length" falls apart, since (as you seem to have inferred) no value of $|\nabla|$ really makes sense here. — Ben Grossmann, Nov 20 '14 at 16:24
Did you see this question? : http://math.stackexchange.com/questions/1032382/what-is-the-product-of-magnitudes-frac-partial-partial-x-and-x — Sensebe, Nov 21 '14 at 16:35

score 11 · Answer 1 · answered Nov 19 '14 at 18:16

Let $f=f(x,y,z)$ be a scalar function and $\mathbf F=\langle F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)\rangle$ be a vector field in $\mathbb{R}^3$. Then we can think of $f$ or $\mathbf F$ (as appropriate) as the inputs to the operators grad, div, curl, and even laplacian with the resulting outputs indicated:

\begin{align} f\longrightarrow &\ \color{blue}{{\LARGE\boxed{\nabla}}} \longrightarrow \langle f_x,f_y,f_z\rangle\\ \mathbf F\longrightarrow &\ \color{blue}{{\LARGE\boxed{\nabla\cdot}}} \longrightarrow {\partial F_1\over \partial x}+{\partial F_2\over \partial y}+{\partial F_3\over \partial z}\\ \mathbf F\longrightarrow &\ \color{blue}{{\LARGE\boxed{\nabla\times}}} \longrightarrow \left\langle {\partial F_3\over \partial y}-{\partial F_2\over \partial z},{\partial F_1\over \partial z}-{\partial F_3\over \partial x},{\partial F_2\over \partial x}-{\partial F_1\over \partial y}\right\rangle\\ f\longrightarrow &\ \color{blue}{{\LARGE\boxed{\nabla\cdot\nabla}}} \longrightarrow f_{xx}+f_{yy}+f_{zz}.\\ \end{align}

Thus $\nabla$ is not a vector, but rather indicates an operator whose action on the input $f$ results in the output $\langle f_x,f_y,f_z\rangle$. Similarly for the others.

If you find the del notation counterproductive, just abandon that notation/nomenclature for this:

\begin{align} f\longrightarrow &\ \color{blue}{{\LARGE\boxed{\text{grad}}}} \longrightarrow \langle f_x,f_y,f_z\rangle\\ \mathbf F\longrightarrow &\ \color{blue}{{\LARGE\boxed{\text{div}}}} \longrightarrow {\partial F_1\over \partial x}+{\partial F_2\over \partial y}+{\partial F_3\over \partial z}\\ \mathbf F\longrightarrow &\ \color{blue}{{\LARGE\boxed{\text{curl}}}} \longrightarrow \left\langle {\partial F_3\over \partial y}-{\partial F_2\over \partial z},{\partial F_1\over \partial z}-{\partial F_3\over \partial x},{\partial F_2\over \partial x}-{\partial F_1\over \partial y}\right\rangle\\ f\longrightarrow &\ \color{blue}{{\LARGE\boxed{\text{lap}}}} \longrightarrow f_{xx}+f_{yy}+f_{zz}.\\ \end{align}

mfl · Answer 2 · 2014-11-19T17:44:38.300

6

$\nabla=(\partial_x, \partial_y,\partial_z)$ is not a vector. It is an operator that maps a differentiable function $f$ at a point $p$ to a vector: $$\nabla f(p)=((\partial_xf)(p), (\partial_yf)(p),(\partial_zf)(p)).$$

If $V=(u,v,w)$ is a vector function (that is, $u,v,w$ are functions of $(x,y,z))$ then we can consider $\nabla \cdot V,$ defined as $\partial_xu+\partial_yv+\partial_zw.$ It is not the dot product of two vectors. If we consider the product formally we such a result, which is, $\mathrm{div}(V).$

In a similar way, we can consider $\nabla \times V,$ defined formally as $$\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \partial_x & \partial_y & \partial_z \\ u & v & w\end{array}\right|.$$ Thus $\nabla \times$ is an operator ($\mathrm{curl})$ that maps a vector function to a vector function, but not a vector.

edited Nov 19 '14 at 17:44

answered Nov 19 '14 at 16:56

mfl

29,399

If $\nabla$ is not a vector, why to make use of it in cross product? – Sensebe Nov 19 '14 at 16:58
It is a notation for the operator given in the answer. $\nabla \times v$ doesn't mean the cross product of a vector $v$ with a vector $\nabla.$ You only have a vector, $v,$ and an operator that acts on it. – mfl Nov 19 '14 at 17:00
As I know $\nabla$ is a vector operator, and not just operator. – Sensebe Nov 19 '14 at 17:03
Do you agree with my previous statement? – Sensebe Nov 19 '14 at 17:05
2

@Godparticle: $\nabla$ is an operator. Thinking of $\nabla$ as the "vector" of differential operators $\langle \partial/\partial x, \partial/\partial y, \partial/\partial z\rangle$ is just a useful notational device. This enables us to make sense of the package of symbols $\nabla\times\mathbf F$. $\nabla$ alone is analogous to writing ${d\over dx}$ or $\int$ alone in that the symbolize actions to be done to an appropriate input object. – JohnD Nov 19 '14 at 17:09
@JohnD: Just because it is useful, we can't define things in the way we want. Particularly, in cross products we can have only vector quantities as possible candidates as I know. And these vectors have magnitude and direction. If $nabla$ is not a vector, we can't use it in cross product. If it is a vector, what is it's magnitude? – Sensebe Nov 19 '14 at 17:25
1

You really just need to grasp the concept of what an operator is. The notational formalism will follow described in this nice answer above will then follow. – JohnD Nov 19 '14 at 17:31
@JhonD: Thank you for the comment. You didn't answer the questions in my last comment. Please answer those to have clear picture. – Sensebe Nov 19 '14 at 18:04

score 3 · Accepted Answer · answered Nov 28 '14 at 00:19

The $\times$ is a symbol in "$\nabla \times$". Only that. It is used because it helps to remember the formulas, but it is only a symbol. The definition of the symbol "$\nabla \times f$" is $$\displaystyle \left\langle {\partial F_3\over \partial y}-{\partial F_2\over \partial z},{\partial F_1\over \partial z}-{\partial F_3\over \partial x},{\partial F_2\over \partial x}-{\partial F_1\over \partial y}\right\rangle$$

You could call it anything. For example, "$curl f$" or even "$\nabla + f$", or "$MATH ~ f$". But the standard notation is convenient.

score 2 · Answer 4 · answered Mar 17 '17 at 16:00

2

It is nowhere near a vector, it is an operator. Eg:differential operator operates on a function and gives another function.Now it makes no sense to call differential operator a function. Same thing here function replaced by vector and operator as it is written above. And here it not like a vector cross product it makes the process of finding the curl of a vector which very much can be made easier by the cross product rule which is assumed to be already known and more

answered Mar 17 '17 at 16:00

Vasanth Kumar B

21

1

Welcome to Math Stack Exchange. I see you posted two answers to this question, while you can also [edit] it to provide more information. Also, your answers are a bit hard to follow and I'm not sure they add any value to the site. – Glorfindel Mar 17 '17 at 16:12

score 1 · Answer 5 · 2014-11-19T17:49:41.113

1

With $f$ a scalar function of the coordinates, $\nabla f$ is a vector called the gradient of $f$.

With $f$ a vector function of the coordinates, $\nabla.f$ is a scalar called the divergence of $f$.

With $f$ a vector function of the coordinates, $\nabla\times f$ is a vector called the curl of $f$.

These three symbols ($\nabla,\nabla.,\nabla\times$) are differential operators and represent no quantity by themselves.

If you really want to see $\nabla$ as a vector, then it is $$\nabla=i\frac\partial{\partial x}+j\frac\partial{\partial y}+k\frac\partial{\partial z}$$

edited Nov 19 '14 at 17:49

answered Nov 19 '14 at 17:05

1

Thank you for the answer. I think you didn't understand me or I didn't understand you. To make sure we understand each other, I want to know whether $\nabla$ is a vector or not. – Sensebe Nov 19 '14 at 17:07
4

No, $\nabla$ alone is meaningless. – Nov 19 '14 at 17:12
Sorry, particularly I want to know whether $\nabla$ is a vector or not. – Sensebe Nov 19 '14 at 17:18
1

I just said no. – Nov 19 '14 at 17:19
Then why to use non-vector in a cross product? – Sensebe Nov 19 '14 at 17:20
There is no cross product. $\nabla\times$ denotes the curl differential operator. – Nov 19 '14 at 17:22
Let us continue this discussion in chat. – Sensebe Nov 19 '14 at 17:25

Is $\nabla$ a vector?

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