Conservative Fields and Potential Functions

A conservative vector field is the gradient of a scalar potential, and line integrals in such fields depend only on endpoints. This concept unifies path independence, exactness, and energy conservation.

Characterizations

Definition

A vector field $\mathbf{F}$ on a domain $D$ is conservative if there exists a scalar function $f : D \to \mathbb{R}$ (the potential function or scalar potential) such that $\mathbf{F} = \nabla f$ . The potential $f$ is unique up to an additive constant.

Theorem10.1Characterizations of Conservative Fields

For a $C^1$ vector field $\mathbf{F}$ on an open, simply connected domain $D \subseteq \mathbb{R}^3$ , the following are equivalent:

$\mathbf{F}$ is conservative: $\mathbf{F} = \nabla f$ for some $f$
$\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ for every closed curve $C$ in $D$
$\int_C \mathbf{F} \cdot d\mathbf{r}$ is path-independent (depends only on endpoints)
$\nabla \times \mathbf{F} = \mathbf{0}$ throughout $D$ (the field is irrotational)

Finding Potential Functions

ExampleFinding a potential function

For $\mathbf{F} = (2xy + z^2, x^2, 2xz)$ : check $\nabla \times \mathbf{F} = \mathbf{0}$ (verified). Then:

$f_x = 2xy + z^2 \implies f = x^2 y + xz^2 + g(y,z)$
$f_y = x^2 + g_y = x^2 \implies g_y = 0 \implies g = h(z)$
$f_z = 2xz + h'(z) = 2xz \implies h'(z) = 0 \implies h = C$

So $f(x,y,z) = x^2 y + xz^2 + C$ .

Path Independence

ExampleWork done by a conservative force

The work done by $\mathbf{F} = \nabla f$ along any path from $\mathbf{a}$ to $\mathbf{b}$ is $W = \int_C \mathbf{F} \cdot d\mathbf{r} = f(\mathbf{b}) - f(\mathbf{a})$ This is the Fundamental Theorem of Line Integrals. For gravity near Earth's surface, $\mathbf{F} = (0, 0, -mg)$ and $f = -mgz$ , so work equals $-mg(z_b - z_a)$ regardless of the path taken.

RemarkSimply connected hypothesis

The condition that $D$ be simply connected is essential for the equivalence $\nabla \times \mathbf{F} = \mathbf{0} \iff \mathbf{F}$ is conservative. On domains with "holes," irrotational fields may fail to be conservative. The classic example is $\mathbf{F} = \frac{(-y, x)}{x^2+y^2}$ on $\mathbb{R}^2 \setminus \{0\}$ : $\nabla \times \mathbf{F} = 0$ but $\oint \mathbf{F} \cdot d\mathbf{r} = 2\pi$ around the origin. This connects vector calculus to the topology of the domain (de Rham cohomology).