In this section we shall consider the vector space ${\displaystyle \mathbb {R} ^{3}}$ over reals with the basis ${\displaystyle {\hat {x}},{\hat {y}},{\hat {z}}}$.

We now wish to deal with some of the introductory concepts of vector calculus.

Vector and Scalar Fields

The Gradient

Let ${\displaystyle C}$ be a scalar field. We define the gradient as an "operator" ${\displaystyle \nabla }$ mapping the field ${\displaystyle C}$ to a vector in ${\displaystyle \mathbb {R} ^{3}}$ such that

${\displaystyle \nabla C=\left({\frac {\partial C}{\partial x}},{\frac {\partial C}{\partial y}},{\frac {\partial C}{\partial z}}\right)}$, or as is commonly denoted ${\displaystyle \nabla C={\frac {\partial C}{\partial x}}{\hat {x}}+{\frac {\partial C}{\partial y}}{\hat {y}}+{\frac {\partial C}{\partial z}}{\hat {z}}}$

We shall encounter the physicist's notion of "operator" before defining it formally in the chapter Hilbert Spaces. It can be loosely thought of as "a function of functions"

Gradient and the total derivative

Recall from multivariable calculus that the total derivative of a function ${\displaystyle f:\mathbb {R} ^{3}\to \mathbb {R} }$ at ${\displaystyle \mathbf {a} \in \mathbb {R} ^{3}}$ is defined as the linear transformation ${\displaystyle A}$ that satisfies

${\displaystyle \lim _{|\mathbf {h} |\to 0}{\frac {f(\mathbf {a} +\mathbf {h} )-f(\mathbf {a} )-A\mathbf {h} }{|\mathbf {h} |}}=0}$

In the usual basis, we can express as the row matrix ${\displaystyle f'(\mathbf {a} )=A=\displaystyle {\begin{pmatrix}{\tfrac {\partial f}{\partial x}}&{\tfrac {\partial f}{\partial y}}&{\tfrac {\partial f}{\partial z}}\\\end{pmatrix}}}$

It is customary to denote vectors as column matrices. Thus we may write ${\displaystyle \nabla f=\displaystyle {\begin{pmatrix}{\tfrac {\partial f}{\partial x}}\\{\tfrac {\partial f}{\partial y}}\\{\tfrac {\partial f}{\partial z}}\\\end{pmatrix}}}$

The transpose of a matrix given by constituents ${\displaystyle a_{ij}}$ is the matrix with constituents ${\displaystyle a_{ij}^{T}=a_{ji}}$

Thus, the gradient is the transpose of the total derivative.

Divergence

Let ${\displaystyle \mathbf {F} :\mathbb {R} ^{3}\to \mathbb {R} ^{3}}$ be a vector field and let ${\displaystyle \mathbf {F} }$ be differentiable.

We define the divergence as the operator ${\displaystyle (\nabla \cdot )}$ mapping ${\displaystyle \mathbf {F} }$ to a scalar such that

${\displaystyle (\nabla \cdot \mathbf {F} )={\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}}$

Curl

Let ${\displaystyle \mathbf {F} :\mathbb {R} ^{3}\to \mathbb {R} ^{3}}$ be a vector field and let ${\displaystyle \mathbf {F} }$ be differentiable.

We define the curl as the operator ${\displaystyle (\nabla \times )}$ mapping ${\displaystyle \mathbf {F} }$ to a linear transformation from ${\displaystyle \mathbb {R} ^{3}}$ onto itself such that the linear transformation can be expressed as the matrix

${\displaystyle (\nabla \times \mathbf {F} )_{ij}={\frac {\partial F_{j}}{\partial x_{i}}}-{\frac {\partial F_{i}}{\partial x_{j}}}}$ written in short as ${\displaystyle (\nabla \times \mathbf {F} )_{ij}=\partial _{i}F_{j}-\partial _{j}F_{i}}$. Here, ${\displaystyle x_{1},x_{2},x_{3}}$ denote ${\displaystyle x,y,z}$ and so on.

the curl can be explicitly given by the matrix: ${\displaystyle \nabla \times \mathbf {F} ={\begin{pmatrix}0&\partial _{1}F_{2}-\partial _{2}F_{1}&\partial _{1}F_{3}-\partial _{3}F_{1}\\\partial _{2}F_{1}-\partial _{1}F_{2}&0&\partial _{2}F_{3}-\partial _{3}F_{2}\\\partial _{2}F_{1}-\partial _{1}F_{3}&\partial _{3}F_{2}-\partial _{2}F_{3}&0\\\end{pmatrix}}}$

this notation is also sometimes used to denote the vector exterior or cross product, ${\displaystyle \nabla \times \mathbf {F} =(\partial _{2}F_{3}-\partial _{3}F_{2}){\hat {x}}+(\partial _{1}F_{3}-\partial _{3}F_{1}){\hat {y}}+(\partial _{1}F_{2}-\partial _{2}F_{1}){\hat {z}}}$