Conservative Vector Field (Irrotational Vector Fields)
Definition1 2
A vector field $\mathbf{F}$ is said to be a conservative field if the integral of $\mathbf{F}$ is path-independent.
$$ \underset{\text{path I}}{\int_{\mathbf{a}}^{\mathbf{b}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \underset{\text{path II}}{\int_{\mathbf{a}}^{\mathbf{b}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} $$
Explanation
By the theorem below, a conservative field is also called a curl-free/irrotational field. If the units of the vector field $\mathbf{F}$ correspond to force, the line integral physically represents work. In other words, the amount of work required to move an object between two given points does not depend on the path taken. Such a force is called a 🔒(26/04/23)conservative force.
The theorem below is a useful result about equivalent conditions for a conservative field; its content can be summarized as follows.
$$ \begin{array}{ccc} \mathbf{F} \text{ is conservative} & \iff & \nabla \times \mathbf{F} = \mathbf{0} \\[1em] \Updownarrow & & \Updownarrow \\[1em] \displaystyle \oint_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = 0 & \iff & \text{There exists $V$ such that } \mathbf{F} = -\nabla V \end{array} $$
Theorem
Let the vector field $\mathbf{F} = (F_{x}, F_{y}, F_{z})$ and the derivatives $\partial_{j}F_{i}$ of its components be continuous on a simply connected domain. Then the following are all equivalent.
(a) $\nabla \times \mathbf{F} = \mathbf{0}$ holds at every point in the domain.
(b) For every simple closed curve $C$ in the domain, $\oint_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = 0$ holds.
(c) $\mathbf{F}$ is a conservative field. That is, for any two points $\mathbf{a}$, $\mathbf{b}$ in the domain, the integral $\int_{\mathbf{a}}^{\mathbf{b}} \mathbf{F} \cdot \mathrm{d}\mathbf{r}$ is path-independent.
(d) There exists a scalar field $V$ such that $\mathbf{F} = -\nabla V$.
Proof
(a) $\implies$ (b)
Assume (a) holds. Then by Stokes’ theorem the following holds.
$$ \oint_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \int_{S} (\nabla \times \mathbf{F}) \cdot \mathrm{d}\mathbf{a} = \int_{S} \mathbf{0} \cdot \mathrm{d}\mathbf{a} = 0 $$
Here $S$ is the surface bounded by the closed curve $C$.
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(b) $\implies$ (c)
Assume (b) holds. Then for any closed curve $C$ and any two points $\mathbf{a}$, $\mathbf{b}$ inside it the following holds.
$$ \oint_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \underset{\text{path I}}{\int_{\mathbf{a}}^{\mathbf{b}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \underset{\text{path II}}{\int_{\mathbf{b}}^{\mathbf{a}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = 0 $$
Here $\text{path I}$ is the path within the closed curve $C$ from $\mathbf{a}$ to $\mathbf{b}$, and $\text{path II}$ is the remaining path (their particular choice does not affect the proof (../2720)). Moving the second integral to the right-hand side and rearranging gives:
$$ \underset{\text{path I}}{\int_{\mathbf{a}}^{\mathbf{b}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = - \underset{\text{path II}}{\int_{\mathbf{b}}^{\mathbf{a}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \underset{\text{path II}}{\int_{\mathbf{a}}^{\mathbf{b}}} \mathbf{F} \cdot \mathrm{d}\mathbf{r} $$
Since this equation holds for any closed curve $C$, the integral from $\mathbf{a}$ to $\mathbf{b}$ is path-independent and always has the same value.
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(c) $\implies$ (d)
Assume (c) holds. Define the scalar function $V : \mathbb{R}^{3} \to \mathbb{R}$ as follows.
$$ V(\mathbf{r}) := - \int_{\mathbf{r}_{0}}^{\mathbf{r}} \mathbf{F}(\mathbf{s}) \cdot \mathrm{d}\mathbf{s}, \quad \mathbf{r} = (x,y,z) \tag{1} $$
The partial derivatives of $V$ are computed as follows.
$$ \begin{align*} \dfrac{\partial V}{\partial x} &= \lim\limits_{h \to 0} \dfrac{V(x+h, y, z) - V(x, y, z)}{h} \\ &= \lim\limits_{h \to 0} \dfrac{1}{h} \left( -\int_{\mathbf{r}_{0}}^{(x+h,y,z)} \mathbf{F}(\mathbf{s}) \cdot \mathrm{d}\mathbf{s} + \int_{\mathbf{r}_{0}}^{(x,y,z)} \mathbf{F}(\mathbf{s}) \cdot \mathrm{d}\mathbf{s} \right) \end{align*} $$
For the $y$ and $z$ components there is no change, so $\mathrm{d}\mathbf{s} = \mathrm{d}x\mathbf{i} + 0\mathbf{j} + 0z\mathbf{k} = \mathrm{d}x\mathbf{i}$ and $\mathbf{F} \cdot \mathrm{d}\mathbf{s} = F_{x} \mathrm{d}x$. Hence we obtain:
$$ \begin{align*} \dfrac{\partial V}{\partial x} &= \lim\limits_{h \to 0} \dfrac{1}{h} \left( -\int_{\mathbf{r}_{0}}^{(x+h,y,z)} F_{x} \mathrm{d}x + \int_{\mathbf{r}_{0}}^{(x,y,z)} F_{x} \mathrm{d}x \right) \\ &= - \lim\limits_{h \to 0} \dfrac{1}{h} \left( \int_{(x,y,z)}^{(x+h,y,z)} F_{x} \mathrm{d}x \right) \end{align*} $$
By the fundamental theorem of calculus we have:
$$ \begin{align*} \dfrac{\partial V}{\partial x} &= -F_{x}(x,y,z) \end{align*} $$
Similarly, for the $y$ and $z$ components we have:
$$ \begin{align*} \dfrac{\partial V}{\partial y} &= -F_{y}(x,y,z) \\ \dfrac{\partial V}{\partial z} &= -F_{z}(x,y,z) \end{align*} $$
Therefore a scalar field $V$ satisfying $\mathbf{F} = -\nabla V$ exists, given by $(1)$.
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(d) $\implies$ (a)
Since the curl of a gradient is $\mathbf{0}$, the implication follows.
$$ \nabla \times \mathbf{F} = \nabla \times (-\nabla V) = \mathbf{0} $$
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