Cartesian Definition of Gradient defines Gradient Operator

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Theorem

Let $R$ be a region of Cartesian $3$-space $\R^3$.

Let $\map F {x, y, z}$ be a scalar field acting over $R$.

Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.

Let $\grad F$ be defined according to the Cartesian $3$-space definition of the gradient of $F$:

\(\ds \grad F\) \(:=\) \(\ds \nabla F\)
\(\ds \) \(=\) \(\ds \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } F\) Definition of Del Operator
\(\ds \) \(=\) \(\ds \dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k\)


Then $\grad F$ is a gradient operator as defined by the geometrical representation:

$\grad F = \dfrac {\partial F} {\partial n} \mathbf {\hat n}$

where:

$\mathbf {\hat n}$ denotes the unit normal to the equal surface $S$ of $F$ at $A$
$n$ is the magnitude of the normal vector to $S$ at $A$.


Proof

The vector rates of increase of $F$ in the directions of the $3$ axes are:

$\dfrac {\partial F} {\partial x} \mathbf i$, $\dfrac {\partial F} {\partial y} \mathbf j$, $\dfrac {\partial F} {\partial z} \mathbf k$

Their sum will be a vector with the magnitude and direction of the most rapid rate of increase of $F$.

It remains to show that this expression:

$\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k$

is equivalent to:

$\dfrac {\partial F} {\partial n} \mathbf {\hat n}$

Let us take the dot product of both sides of the gradient equation with the position vector $\d \mathbf r$ of an arbitrary point $A$ on an equal surface $S$ of $F$.

Thus:

\(\ds \paren {\grad F} \cdot \d \mathbf r\) \(=\) \(\ds \dfrac {\partial F} {\partial n} \mathbf {\hat n} \cdot \d \mathbf r\)
\(\ds \) \(=\) \(\ds \dfrac {\partial F} {\partial n} \d r \cos \theta\) where $\d r = \norm {\d \mathbf r}$: Definition of Dot Product
\(\ds \) \(=\) \(\ds \dfrac {\partial F} {\partial n} \d n\)
\(\ds \) \(=\) \(\ds \d F\) as $\dfrac {\partial F} {\partial n}$ is the total rate of change of $F$ with respect to $n$


In Cartesian coordinates:

\(\ds \d F\) \(=\) \(\ds \dfrac {\partial F} {\partial x} \d x + \dfrac {\partial F} {\partial y} \d y + \dfrac {\partial F} {\partial z} \d z\)
\(\ds \leadsto \ \ \) \(\ds \paren {\grad F} \cdot \d \mathbf r\) \(=\) \(\ds \dfrac {\partial F} {\partial x} \d x + \dfrac {\partial F} {\partial y} \d y + \dfrac {\partial F} {\partial z} \d z\)
\(\ds \) \(=\) \(\ds \paren {\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k} \cdot \paren {\d x \mathbf i + \d y \mathbf j + \d z \mathbf k}\)
\(\ds \) \(=\) \(\ds \paren {\nabla F} \cdot \d \mathbf r\)
\(\ds \leadsto \ \ \) \(\ds \grad F = \nabla F\) \(=\) \(\ds \dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k\)

Thus the operations $\grad$ and $\nabla$ applied to a point in a scalar field are identical.

$\blacksquare$


Sources

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