Vector Field is Expressible as Gradient of Scalar Field iff Conservative
Theorem
Let $\mathbf V$ be a vector field acting over $R$.
Then $\mathbf V$ can be expressed as the gradient of some scalar field $F$ if and only if $\mathbf V$ is a conservative vector field.
Proof
Let $\mathbf V_F$ be a vector field which is the gradient of some scalar field $F$:
- $\mathbf V_F = \grad F = \nabla F$
Let $A$ and $B$ be two points in $R$.
Let $\text {Path $1$}$ be an arbitrary path from $A$ to $B$ lying entirely in $R$.
At the point $P$, let $\d \mathbf l$ be a small element of $\text {Path $1$}$.
Let $\mathbf V_F$ make an angle $\theta$ with $\d \mathbf l$.
Then at $P$:
- $V_F \cos \theta \d l = \mathbf V_F \cdot \d \mathbf l$
where $V_F$ and $\d l$ are the magnitudes of $\mathbf V_F$ and $\d \mathbf l$ respectively.
Let $\mathbf r$ be the position vector of the point $P$ as it passes from $A$ to $B$.
Then $\d \mathbf l$ is the same as $\d \mathbf r$, and so we can write:
\(\ds \mathbf V_F \cdot \d \mathbf l\) | \(=\) | \(\ds \paren {\nabla F} \cdot \d \mathbf r\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds \paren {\grad F} \cdot \d \mathbf r\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \d F\) |
Hence the contour integral of $\mathbf V_F$ from $A$ to $B$ is:
\(\ds \int_A^B \mathbf V_F \cdot \d \mathbf l\) | \(=\) | \(\ds \int_A^B \paren {\grad F} \cdot \d \mathbf r\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_A^B \d F\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_B - F_A\) | where $F_A$ and $F_B$ are the values of $F$ at $A$ and $B$ |
Since only the end values feature in this expression, it follows that the actual route through $R$ taken by $\text {Path $1$}$ is immaterial.
That is, the value of $\ds \int_A^B \mathbf V_F \cdot \d \mathbf l$ is independent of the actual path from $A$ to $B$ along which the contour integral is taken.
$\Box$
Let $\text {Path $2$}$ now be an arbitrary path from $B$ back to $A$, so that $\text {Path $1$}$ and $\text {Path $2$}$ together make a closed loop.
Since the limits of integration are reversed for $\text {Path $2$}$, we have:
- $\ds \int_B^A \mathbf V_F \cdot \d \mathbf l = F_A - F_B$
Hence we have:
- $\ds \oint \paren {\grad F} \cdot \d \mathbf l = 0$
That is, $\mathbf V_F$ is a conservative vector field.
$\blacksquare$
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 $\nabla S$: $(4.5)$