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Let $g$ be a continuous $2\pi$-periodic function and $\mathbf{F}=g(\theta ){\mathbf{e}}_{\theta }/r$ on $r>0$. On a circle about the axis, $d\mathbf{r}=r{\mathbf{e}}_{\theta }d\theta$, giving the displayed line integral. A single-valued potential of a conservative vector field on the full punctured space exists if and only if this integral is zero. Necessity is the fundamental theorem for line integrals. For sufficiency, the primitive $P(\theta )={\int }_0^{\theta }g(s)ds$ is $2\pi$-periodic when the mean is zero, and the gradient in cylindrical coordinates gives $\nabla P=\mathbf{F}$.

On a domain with one angular cut removed, the same primitive gives a potential of a conservative vector field even when the mean is nonzero, because the angle has a single-valued continuous branch there. For example $g(\theta )=\sqrt{1+\cos \theta }$ has integral $4\sqrt{2}$ around the full circle but primitive $2\sqrt{2}\sin (\theta /2)$ on $-\pi <\theta <\pi$.

A vector field $X$ is conservative when $X=\nabla \Phi$ for a potential of a conservative vector field $\Phi$. On a simply connected open subset of ${\mathbb{R}}^3$, every continuously differentiable curl-free vector field is conservative. Locally this follows from the Poincaré lemma applied to the differential one-form $X_{i}d{x}^i$; globally, simple connectivity makes its line integral independent of the path. The fundamental theorem for line integrals expresses that path independence as the difference of the potential at the endpoints.

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

On the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$, consider the vector fieldWith $r^2=x^2+y^2$, its partial derivatives satisfyThus the field has zero scalar curl everywhere it is defined. Nevertheless, on any counterclockwise circle ${\gamma }_R(t)=(R\cos t,R\sin t)$ with $R>0$,The radius cancels exactly.

Applying Green theorem to the enclosed disk would be invalid: $G$ is undefined at the origin, inside that disk. Zero curl on the path alone is insufficient.^[ref]

Now take the annulus $A=\{(x,y):1\le r\le 2\}$. Its positive boundary orientation is counterclockwise on the outer circle and clockwise on the inner circle. The complete boundary line integral is thereforeHere Green theorem applies: an open set containing the closed annulus can avoid the origin. The inner contribution remains $-2\pi$ even for an arbitrarily small positive inner radius. Decreasing the hole's area does not decrease this line integral.

The cancellation has a useful proof picture. Cut the annulus along two radial segments into two half-annuli. Each piece has a single simple closed curve as its boundary; along each cut, the two induced boundary orientations oppose, so those line integrals cancel when the two applications of Green theorem are added. Only the outer and inner circles remain.^[ref]

Locally, a choice of angle $\theta$ in polar coordinates has gradient $\nabla \theta =G$. Globally, following a circle once increases that angle by $2\pi$, so it cannot be a single-valued potential of a conservative vector field on the punctured plane. The nonzero closed-path line integral also rules out such a potential by the fundamental theorem for line integrals.

On a simply connected open region, a continuously differentiable planar vector field with zero scalar curl does have a global potential. The hole is what separates the local derivative condition from the global conclusion.^[ref]

Write the planar velocity field as $\mathbf{u}=(u,v,0)$, with $\mathbf{k}=(0,0,1)$. Expanding the curl givesConsequently, for a piecewise smooth path $\gamma$ from the origin to $\mathbf{x}=(x,y)$,The arbitrary additive constant $\psi (0)$ may be set to zero. The integrand is exactly $\nabla \psi \cdot d\mathbf{r}$, so the fundamental theorem for line integrals proves path independence immediately: the difference between two paths is the line integral of a gradient around a closed curve, which is zero.

One can also see the local integrability condition from Green's theorem:because the given representation enforces incompressibility. This second proof applies whenever the region between the paths lies in the regular fluid domain. The question already assumes a globally defined streamfunction; without that assumption, incompressibility alone would guarantee a global streamfunction on a simply connected domain, but a hole carrying nonzero volume flux could obstruct it.

Integrate the first component of $\mathbf{F}=\nabla \varphi$ with respect to $x$:Matching the $y$ component gives $g_y=0$, and matching the $z$ component gives $g_z=0$. HenceThe fundamental theorem for line integrals now yieldsThe potential is a single-valued polynomial everywhere, so any piecewise smooth path with these endpoints has the same integral. The plus-gradient convention here should not be confused with the minus gradient of a mechanical potential energy.

Both paths start at $(0,0,0)$ and end at $(1,2,1)$. The potential values are $\varphi (0,0,0)=C$ and $\varphi (1,2,1)=4+C$, soThis is the fundamental theorem for line integrals: an exact one-form integrates to the difference of its single-valued potential at the endpoints. The agreement is guaranteed for every piecewise smooth connecting path, rather than being an accident of these two examples. In particular the integral around a closed path assembled from one route and the reverse of the other is zero.

A one-form $A(x,y)dx+B(x,y)dy$ is an exact differential if a single-valued differentiable potential of a conservative vector field $f$ satisfies $A=f_x$ and $B=f_y$ on the domain. Then it equals $df$, and its integral along a path is the endpoint difference of $f$.

Integrate $f_x=10x^4-2x{y}^3$ with respect to $x$:Its $y$ derivative is $-3x^2{y}^2+g^{\prime }(y)$. Matching the required second component forces $g^{\prime }(y)=0$, so the additive constant is arbitrary andThe fundamental theorem for line integrals now givesindependently of the convenient segmented path used before. Merely checking equality of mixed partial derivatives on an unspecified domain would not by itself construct this globally defined potential.

Construct a global vector-field potential rather than relying only on vanishing curl:Thus $\mathbf{F}$ is a conservative vector field. The endpoints of the given parametric curve are $\mathbf{r}(0)=(3,0,0)$ and $\mathbf{r}(\pi /2)=(0,2,1)$. The fundamental theorem for line integrals givesThe intervening turns of the curve have no effect because this line integral is path independent.

The field is the gradient of a globally defined polynomial:It is therefore a conservative vector field on all of ${\mathbb{R}}^3$. The fundamental theorem for line integrals givesfor every closed piecewise smooth loop. Constructing the global vector-field potential avoids relying on a zero-curl test without checking the topology of the domain.

The field is the gradient of $\Phi =x^2-3y^2+4z^2$ on all of ${\mathbb{R}}^3$. It is therefore a conservative vector field, and the fundamental theorem for line integrals gives ${\int }_{C}F\cdot dr=\Phi (r_{\mathrm{end}})-\Phi (r_{\mathrm{start}})$. The endpoints are $(3,0,0)$ and $(0,2,-1)$, soThe complicated intermediate path does not affect the work. Directly, $\nabla \times F=0$, consistent with the globally defined potential function.

An exact differential is an expression $Pdx+Qdy$ that equals $df=f_{x}dx+f_{y}dy$ for a single-valued differentiable function $f$ on the domain. Its line integral is an endpoint difference by the fundamental theorem for line integrals.

Integrating $f_x=P$ with respect to $x$ gives $f=2x^5-x^2{y}^3+g(y)$. Matching $f_y=-3x^2{y}^2+g^{\prime }(y)$ to $Q$ requires $g^{\prime }(y)=0$. HenceThe arbitrary additive constant cancels. This constructs the vector-field potential globally and provides a second proof of path independence.

An exact differential is a one-form equal to $d\Phi$ for a single-valued scalar function $\Phi$ on the domain. In Euclidean coordinates this means its vector coefficients form $\nabla \Phi$. Integrating the first component of $F$ with respect to $x$, then matching the other components, givesDifferentiating checks all three components. Since the domain ${\mathbb{R}}^3$ is connected, two such potentials differ only by a constant, so this is the most general potential.

At both specified endpoints the nonconstant terms vanish. The fundamental theorem for line integrals gives zero for the line integral along every path joining the two points.

For the planar field, choose a continuous lift $\theta (t)$ of the polar angle along a curve avoiding the origin. Differentiating the angle givesOn the closed curve the angle increases by $2\pi$ times its winding number. ThusThe angle differential is locally exact but not globally exact on the punctured plane: the angle lift need not return to its initial value. This is the nonexact angular one-form, whose nonzero closed-path integral obstructs a global potential.

When $\alpha =2\beta$, all three values reduce to $-\beta /3$. To prove path independence for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ givesIndeed ${\Phi }_x=2\beta (x^2+xy)=\alpha (x^2+xy)$ and ${\Phi }_y=\beta (x^2+y^2)$. The integrand is the exact differential $d\Phi$ throughout ${\mathbb{R}}^2$. The fundamental theorem for line integrals therefore gives, for every such path from $A$ to $B$,

Parametrize the smooth path by $\gamma :[0,1]\to G$. The chain rule turns the line integral of the gradient into a one-variable derivative:This is the fundamental theorem for line integrals; in particular the integral has path independence.

For a twice continuously differentiable potential of a conservative vector field, the curl of its gradient vanishes because mixed partial derivatives commute:A smooth gradient field is necessarily curl-free.

A sufficient condition is that $G$ be an open simply connected domain, with $\mathbf{F}$ continuously differentiable. Under this condition a curl-free vector field has a global potential of a conservative vector field and hence path independence by the fundamental theorem for line integrals. One way to see the global step is to deform the closed loop formed by two paths into a point inside $G$: the integral is unchanged during the deformation because the curl is zero, as expressed by Stokes theorem. Therefore simple connectivity is a sufficient domain hypothesis. It is not a claim that every particular curl-free field needs such a domain; a given field can have a global potential on a domain with holes.

Differentiating the assumed primitive identity and using Cartan's magic formula givesApply the pullback of a differential form by ${\rho }_t^{-1}$ and rearrange:This smooth family of Hamiltonian functions proves that ${\rho }_t$ is a Hamiltonian isotopy in the convention of part (b).

To connect this with the preceding assertion that every ${\rho }_t^{\ast }\lambda -\lambda$ is an exact differential form, the smooth choice of primitives causes no additional obstruction. Fix $p_0\in M$ and normalize $F_t(p_0)=0$. Define $F_t(p)$ by the line integral of ${\rho }_t^{\ast }\lambda -\lambda$ from $p_0$ to $p$. The fundamental theorem for line integrals makes this independent of the path. Near any $p$, use one fixed path to a nearby chart center followed by coordinate line segments; this expression is smooth jointly in $t,p$. Thus these normalized primitives form the required smooth family.

On the simply connected domain ${\mathbb{R}}^3$, a continuously differentiable differential one-form is exact if its mixed partial derivatives agree, equivalently if its associated vector field has zero curl. Here one can verify exactness directly by observing thatsatisfiesThus $udx+vdy+wdz=dF$. By the fundamental theorem for line integrals, its integral is path independent and equalsConsequently, for every path with the stated endpoints,

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

The field is $\nabla f$ for $f=\rho \sin \varphi e^{-{\rho }^2-z^2}$. Hence $I=0$, either directly or by the fundamental theorem for line integrals.