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?

Answer 1

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.

Answer 2

$\nabla=(\partial_x, \partial_y,\partial_z)$ 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.

Answer 3

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 $$\nabla=i\frac\partial{\partial x}+j\frac\partial{\partial y}+k\frac\partial{\partial z}$$

Answer 4

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}

Answer 5

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