Conservative vector field

A vector field $\mathbf{F}$ on a connected open region $\Omega$ is conservative if it has a potential function on $\Omega$ — a scalar $\varphi$ such that $\mathbf{F}=\nabla \varphi$ throughout $\Omega$. Equivalently (these are all the same condition):

1. $\mathbf{F}$ is conservative on $\Omega$ ($\mathbf{F}=\nabla \varphi$).
2. Line integrals of $\mathbf{F}$ are path-independent in $\Omega$.
3. The circulation of $\mathbf{F}$ around every closed curve in $\Omega$ is zero.

How the three characterizations are equivalent

(1) ⇒ (2) by the Fundamental theorem of line integrals: ${\int }_C\nabla \varphi \cdot d\mathbf{r}=\varphi (B)-\varphi (A)$ depends only on the endpoints.

(2) ⇒ (3): a closed loop is a path from a point to itself; by path-independence, it equals the trivial path of zero length, giving zero.

(3) ⇒ (2): given two paths $C_1,C_2$ from $A$ to $B$, the composite $C_1-C_2$ is a closed loop. Zero circulation means ${\int }_{C_1}-{\int }_{C_2}=0$, so the two integrals agree.

(2) ⇒ (1) (the trickiest direction): fix a base point $P_0$ and define $\varphi (P)={\int }_{P_0}^P\mathbf{F}\cdot d\mathbf{r}$. Path-independence makes this well-defined; a calculation shows $\nabla \varphi =\mathbf{F}$.

The cross-partial test

For $\mathbf{F}=⟨P,Q,R⟩$ with continuous first partials: if $\mathbf{F}$ is conservative, then by equality of mixed partials (Clairaut’s theorem) applied to $\varphi$,

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x},\frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y},\frac{\partial R}{\partial x}=\frac{\partial P}{\partial z}.$

In 2D, only the first condition. Compactly: $\nabla \times \mathbf{F}=0$.

This is necessary but not always sufficient. On a simply connected region with continuous first partials, the cross-partial test (zero curl) is sufficient: $\nabla \times \mathbf{F}=0$ on a simply connected domain implies $\mathbf{F}$ conservative.

On non-simply-connected domains the converse can fail dramatically; see the 2D rotational field example below.

Finding a potential function — 2D recipe

Given $\mathbf{F}=⟨P,Q⟩$ with $P_y=Q_x$:

1. Integrate ${\varphi }_x=P$ with respect to $x$, treating $y$ as constant: $\varphi (x,y)=\int Pdx+g(y)$, with unknown $g$.
2. Differentiate with respect to $y$ and set equal to $Q$: $\partial \varphi /\partial y=(\text{from step 1})+g^{\prime }(y)=Q$. Solve for $g^{\prime }(y)$.
3. Integrate $g^{\prime }(y)$, with the constant of integration absorbed.

Worked example. $\mathbf{F}=⟨2x{y}^3,3x^2{y}^2⟩$ on ${\mathbb{R}}^2$.

Cross-partials: $P_y=6x{y}^2$, $Q_x=6x{y}^2$. Equal. Domain simply connected. Conservative.

Integrate ${\varphi }_x=2x{y}^3$: $\varphi =x^2{y}^3+g(y)$. Differentiate w.r.t. $y$: ${\varphi }_y=3x^2{y}^2+g^{\prime }(y)=3x^2{y}^2$. So $g^{\prime }(y)=0$, $g=0$. Potential: $\varphi (x,y)=x^2{y}^3$.

3D recipe

Given $\mathbf{F}=⟨P,Q,R⟩$ passing all three cross-partial checks:

1. Integrate $P$ with respect to $x$: $\varphi =\int Pdx+g(y,z)$.
2. Differentiate w.r.t. $y$ and equate to $Q$, solve for $g$ up to a function $h(z)$.
3. Differentiate w.r.t. $z$ and equate to $R$, solve for $h$.

Worked example. $\mathbf{F}=⟨yz+2x,xz,xy+2z⟩$ on ${\mathbb{R}}^3$.

All cross-partials match. Integrate $P=yz+2x$: $\varphi =xyz+x^2+g(y,z)$. Differentiate w.r.t. $y$: $xz+g_y=xz$, so $g_y=0$, $g=h(z)$. Differentiate w.r.t. $z$: $xy+h^{\prime }(z)=xy+2z$, so $h^{\prime }(z)=2z$, $h=z^2$. Potential: $\varphi =xyz+x^2+z^2$.

The cautionary 2D rotational field

$\mathbf{F}=⟨\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}⟩,\Omega ={\mathbb{R}}^2\setminus \{(0,0)\}.$

Cross-partials: $P_y=(y^2-x^2)/(x^2+y^2{)}^2=Q_x$. Satisfied everywhere on $\Omega$.

But circulation around the unit circle: parameterize $\mathbf{r}(t)=⟨\cos t,\sin t⟩$, ${\mathbf{r}}^{\prime }(t)=⟨-\sin t,\cos t⟩$. On the circle, $x^2+y^2=1$, so $\mathbf{F}=⟨-\sin t,\cos t⟩$. Dot: ${\sin }^{2}t+{\cos }^{2}t=1$. Integral: ${\int }_0^{2\pi }1dt=2\pi \ne 0$.

Nonzero closed-loop circulation. $\mathbf{F}$ is not conservative, despite the cross-partials.

What went wrong: cross-partials are local; conservativeness is global. The field “winds around” the puncture at the origin, and the circulation detects the winding. The domain isn’t simply connected — a loop around the origin can’t be contracted to a point without crossing the removed singularity.

Restricted to a simply-connected sub-domain (say, $\{x>0\}$), $\mathbf{F}$ is conservative there, with potential $\varphi =\arctan (y/x)$. But $\arctan (y/x)$ is multi-valued globally — going around the origin changes it by $2\pi$. The circulation $2\pi$ is exactly the winding number.

This is the prototype of all topological obstructions in vector calculus. The complex-analytic version — $1/(z-z_0)$ with a contour integral that depends on whether the path encircles $z_0$ — is the same phenomenon in different notation, and gives the Residue theorem.

Practical use

When you see a vector field, the first questions to ask are:

1. Is it conservative? Apply the cross-partial test.
2. If yes, find the potential, and use $\varphi (B)-\varphi (A)$ for any line integral.
3. If no, parameterize and integrate the hard way, or apply Green’s theorem / Stokes’ theorem if it helps.

For physical fields: gravitational and electrostatic fields are conservative (in regions with no time-varying magnetic flux); magnetic fields are not (they have non-zero curl wherever current flows).