Cartesian Definition of Gradient defines Gradient Operator
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 $n$ |
\(\ds \d F\) | \(=\) | \(\ds \dfrac {\partial F} {\partial x} \d x + \dfrac {\partial F} {\partial y} \d y + \dfrac {\partial F} {\partial z} \d z\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\grad F} \cdot \d \mathbf r\) | \(=\) | \(\ds \dfrac {\partial F} {\partial x} \d x + \dfrac {\partial F} {\partial y} \d y + \dfrac {\partial F} {\partial z} \d z\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k} \cdot \paren {\d x \mathbf i + \d y \mathbf j + \d z \mathbf k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\nabla F} \cdot \d \mathbf r\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \grad F = \nabla F\) | \(=\) | \(\ds \dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k\) |
Thus the operations $\grad$ and $\nabla$ applied to a point in a scalar field are identical.
$\blacksquare$
Sources
- 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $2 a$. The Operation $\nabla S$: $(4.3)$