Line Integrals and Conservative Fields

1Introduction

This lecture defines line integrals as accumulated quantities along curves and explains when the work integral of a vector field depends only on the endpoints. The distinction between local differential conditions and the global geometry of the domain is essential.

2Definitions

Let $U\subset R^n$U\subset\mathbb R^n be open and let $\mathit{r}:[a,b]\to U$\boldsymbol r:[a,b]\to U be an oriented piecewise $C^1$C^1 curve. For a continuous scalar field $g:U\to \mathbb{R}$g:U\to\mathbb R, define the scalar line integral by

For a continuous vector field $\mathit{F}:U\to R^n$\boldsymbol F:U\to\mathbb R^n, define the work line integral by

This lecture primarily concerns the second integral. A vector field is conservative if there is a function $\varphi \in C^1\left(U\right)$\phi\in C^1(U) such that $\mathit{F}=\nabla \varphi$\boldsymbol F=\nabla\phi. The function $\varphi$\phi is called a potential.

3Geometric Interpretation

In a work line integral, only the component of the vector field tangent to the direction of motion contributes. This explains the dot product $\mathit{F}\left(\mathit{r}\right(t\left)\right)·r^{\prime }\left(t\right)$\boldsymbol F(\boldsymbol r(t))\cdot\boldsymbol r'(t). For a conservative field, the work depends on the initial and terminal points rather than on the path.

4Reparameterization and Orientation

Let $\psi :[\alpha ,\beta ]\to [a,b]$\psi:[\alpha,\beta]\to[a,b] be a continuous piecewise $C^1$C^1 bijection and set $\mathit{\rho }=\mathit{r}\circ \psi$\boldsymbol\rho=\boldsymbol r\circ\psi. If $\psi$\psi is increasing, substitution shows that the work integral is unchanged. If $\psi$\psi is decreasing, the curve orientation reverses and the work integral changes sign. A scalar line integral uses $\parallel r^{\prime }\parallel$\|\boldsymbol r'\|, so reversing orientation does not change its sign.

5Fundamental Theorem for Line Integrals

Suppose $\mathit{F}=\nabla \varphi$\boldsymbol F=\nabla\phi. On each $C^1$C^1 segment of the curve, the chain rule gives

Summing over all segments and applying the one-variable fundamental theorem yields

Thus the integral depends only on the endpoints.

In this lecture, a domain $U$U is path-connected if every two points of $U$U can be joined by a piecewise $C^1$C^1 curve contained in $U$U. For open sets, this condition is equivalent to the usual definition using continuous curves. More generally, let $U$U be open and path-connected and let $\mathit{F}$\boldsymbol F be continuous. The following conditions are equivalent:

1. There is a $\varphi \in C^1\left(U\right)$\phi\in C^1(U) such that $\mathit{F}=\nabla \varphi$\boldsymbol F=\nabla\phi.
2. The work integral between any two points of $U$U is independent of the piecewise $C^1$C^1 path.
3. The work integral around every piecewise $C^1$C^1 closed curve in $U$U is zero.

Condition 1 implies Condition 2 by the chain-rule calculation above. To derive Condition 1 from Condition 2, fix a base point $p_0$\boldsymbol p_0 and define

Path independence makes this definition unambiguous. Because $U$U is open, for each $\mathit{p}\in U$\boldsymbol p\in U and sufficiently small $h$h, the line segment from $\mathit{p}$\boldsymbol p to $\mathit{p}+h{e}_i$\boldsymbol p+h\boldsymbol e_i lies in $U$U. Hence

by continuity of $\mathit{F}$\boldsymbol F. Thus ${\mathit{\partial }}_i\varphi =F_i$\partial_i\phi=F_i for every $i$i, so $\varphi \in C^1\left(U\right)$\phi\in C^1(U) and $\nabla \varphi =\mathit{F}$\nabla\phi=\boldsymbol F. Condition 2 implies Condition 3 by comparing any closed curve with the constant curve having the same initial and terminal point. Conversely, under Condition 3, two paths with the same endpoints form a closed curve when one is traversed in reverse, and the zero integral around that closed curve proves that their integrals are equal.

6Domain Conditions and a Counterexample

Let $U\subset R^2$U\subset\mathbb R^2 or $U\subset R^3$U\subset\mathbb R^3 be open and simply connected, and let $\mathit{F}\in C^1\left(U\right)$\boldsymbol F\in C^1(U). A domain is simply connected if every closed curve in it can be continuously contracted to a point without leaving the domain. In two dimensions, ${\mathrm{curl}}_{2D}\mathit{F}=0$\operatorname{curl}_{2D}\boldsymbol F=0 throughout $U$U implies that $\mathit{F}$\boldsymbol F is conservative. In three dimensions, the corresponding sufficient condition is $\nabla \times \mathit{F}=0$\nabla\times\boldsymbol F=\boldsymbol0 throughout $U$U.

The assumption on the domain cannot be omitted. On $U=R^2\setminus \left\{0\right\}$U=\mathbb R^2\setminus\{\boldsymbol0\}, consider

The field is smooth and has ${\mathrm{curl}}_{2D}\mathit{F}=0$\operatorname{curl}_{2D}\boldsymbol F=0 at every point of $U$U. For the counterclockwise unit circle $\mathit{r}\left(t\right)=(\cos t,\sin t)$\boldsymbol r(t)=(\cos t,\sin t), $0\text{[PARSE ERROR: Undefined("Command("le")")]}t\text{[PARSE ERROR: Undefined("Command("le")")]}2\pi$0\le t\le2\pi, however, $\mathit{F}\left(\mathit{r}\right(t\left)\right)=r^{\prime }\left(t\right)$\boldsymbol F(\boldsymbol r(t))=\boldsymbol r'(t), so

Therefore the field is not conservative. Vanishing curl is a local condition and does not detect the hole in the domain. Simple connectivity is a standard sufficient condition for the implication from zero curl to conservativity; it is not necessary for every individual conservative field.

7Physical Interpretation

If a force field is conservative, its work is determined by the initial and terminal points. Under the physical convention that $V$V is potential energy, $\mathit{F}=-\nabla V$\boldsymbol F=-\nabla V, so the mathematical potential in this lecture is $\varphi =-V$\phi=-V. A conservative force performs zero work around every closed curve.

8Diagnostic Procedure

• If every piecewise $C^1$C^1 closed curve in the domain has zero work integral, then the field is path-independent and conservative.
• If a $C^1$C^1 field has zero curl on an open, simply connected domain, then it is conservative.
• On a domain with holes, zero curl does not by itself guarantee path independence.

9Example 1: A Conservative Field

Let $\varphi (x,y)=x^{2}y$\phi(x,y)=x^2y. Then $\nabla \varphi =(2xy,x^2)$\nabla\phi=(2xy,x^2). For every piecewise $C^1$C^1 curve $C$C from $(0,0)$(0,0) to $(1,1)$(1,1),

The value is the same for a line segment, a polygonal path, or any other admissible path.

10Example 2: A Nonconservative Field

Let $\mathit{F}=(-y,x)$\boldsymbol F=(-y,x) and integrate around the counterclockwise unit circle $\mathit{r}\left(t\right)=(\cos t,\sin t)$\boldsymbol r(t)=(\cos t,\sin t), $0\text{[PARSE ERROR: Undefined("Command("le")")]}t\text{[PARSE ERROR: Undefined("Command("le")")]}2\pi$0\le t\le2\pi. Since

we obtain

The nonzero integral around a closed curve proves that the field is not conservative.

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