Vector Field is Expressible as Gradient of Scalar Field iff Conservative

Theorem
Let $R$ be a region of space.

Let $\mathbf V$ be a vector field acting over $R$.

Then $\mathbf V$ can be expressed as the gradient of some scalar field $F$ $\mathbf V$ is a conservative vector field.

Proof
Let $\mathbf V_F$ be a vector field which is the gradient of some scalar field $F$:


 * $\mathbf V_F = \grad F = \nabla F$


 * Line-Integrals-in-Lamellar-Field.png

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:

Hence the contour integral of $\mathbf V_F$ from $A$ to $B$ is:

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.

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.