Differential Forms 04: Line Integrals and Conservative Fields

Differential forms / 04

Line integrals and conservative fields

A vector-field line integral is the integral of a 1-form; exactness is the condition that the integral only sees endpoints.

Calculus root: antiderivatives and path integrals

In one variable, if \(F'=f\), then

\[\int_a^b f(x)\,dx=F(b)-F(a).\]

Line integrals ask when the same endpoint principle survives after the input is allowed to move through a plane or a manifold. A 1-form \(\eta\) is exact when \(\eta=df\), and then every path integral satisfies

\[\int_\gamma \eta=f(\gamma(b))-f(\gamma(a)).\]

Example: ordinary integral as a line integral

On the real line, the 1-form \(x^2\,dx\) has primitive \(x^3/3\). Along the interval path \(\gamma(t)=t\) from \(1\) to \(3\),

\[\int_\gamma x^2\,dx=\int_1^3 x^2\,dx={26\over3}.\]

This is the fundamental theorem for line integrals in dimension one.

Example: path dependence is the new phenomenon

For \(\eta=-y\,dx+x\,dy\), the path from \((0,0)\) to \((1,1)\) matters. Along the diagonal \(r(t)=(t,t)\) the integral is \(0\). Along the broken path \((0,0)\to(1,0)\to(1,1)\), the first segment contributes \(0\) and the second contributes \(\int_0^1 1\,dy=1\). The failure of endpoint-only behavior is exactly the failure of having a global potential.

Section 4.2 of the PDF rewrites planar vector calculus in form language. A vector field \(F=(P,Q)\) on a domain \(D\subset\mathbb R^2\) corresponds to the 1-form

\[\eta=P\,dx+Q\,dy.\]

If \(C\) is parametrized by \(r(t)=(x(t),y(t))\), \(a\le t\le b\), then

\[\int_C \eta =\int_a^b \left(P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)\right)\,dt.\]

This is exactly \(\int_C F\cdot dr\).

Conservative fields

The field \(F\) is conservative when \(F=\nabla f\). In form language,

\[\eta=P\,dx+Q\,dy=df.\]

Then

\[\int_C df=f(r(b))-f(r(a)).\]

This is the fundamental theorem for line integrals. The integral over every closed curve is zero.

Example 1: potential found by integration

Let

\[\eta=(3x^2+6xy)\,dx+(3x^2+6y)\,dy.\]

Since \(P_y=6x=Q_x\), the form is closed. On \(\mathbb R^2\) this closedness is enough for exactness. Integrating \(f_x=P\) gives

\[f=x^3+3x^2y+h(y).\]

Then \(f_y=3x^2+h'(y)=3x^2+6y\), so \(h(y)=3y^2\). Thus \(\eta=df\) with

\(f=x^3+3x^2y+3y^2.\)

Example 2: path dependence

For \(\eta=y\,dx\), integrate from \((0,0)\) to \((1,1)\). Along the line \(r(t)=(t,t)\),

\[\int_C y\,dx=\int_0^1 t\,dt={1\over 2}.\]

Along the broken path \((0,0)\to(1,0)\to(1,1)\), the first segment has \(y=0\) and the second has \(dx=0\), so the integral is \(0\). The form is not exact; indeed \(d\eta=-dx\wedge dy\ne0\).

Closed need not mean exact

On a simply connected planar domain, a closed 1-form is exact. On a domain with a hole, closedness still gives local potentials, but it need not give one global potential.

The standard form on the punctured plane is

\[\eta={-y\,dx+x\,dy\over x^2+y^2}.\]

Compute

\[d\eta=0\qquad\text{on }\mathbb R^2\setminus\{0\}.\]

Yet on the circle \(C_R(t)=(R\cos t,R\sin t)\),

\[dx=-R\sin t\,dt,\qquad dy=R\cos t\,dt,\]

so

\[\eta=dt,\qquad \int_{C_R}\eta=2\pi.\]

If \(\eta=df\) globally, every closed-loop integral would vanish. Hence \(\eta\) is closed but not exact.

Topology detected by periods

For closed cycles homologous inside the domain, the number \(\int_\gamma \eta\) depends only on the homology class of \(\gamma\) when \(\eta\) is closed. For the punctured plane, the homology class of a loop is measured by its winding number around the missing point.

Displaced circles

The same closed form gives zero around loops that do not wind around the origin. For the circle \(C(t)=(2+\cos t,\sin t)\),

\[\int_C \eta=\int_0^{2\pi}{1+2\cos t\over 5+4\cos t}\,dt=0.\]

This calculation is not a contradiction. The form is closed; the integral is determined by winding, and this circle has winding number zero about \(0\).

Practical test

For \(\eta=P\,dx+Q\,dy\):

1. Compute \(d\eta=(Q_x-P_y)\,dx\wedge dy\).
2. If \(d\eta\ne0\), the form is not exact.
3. If \(d\eta=0\), check the domain. On a star-shaped domain, or on a simply connected planar domain in this two-dimensional setting, find a potential. On a punctured domain, test periods around holes.