Conservative vector field

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.

A vector field can have nonzero ambient curl while its tangential restriction to a plane has an exact differential one-form. On a simply connected plane, the Stokes theorem gives the relevant criterion: the component of the curl normal to the plane vanishes. All closed planar line integrals then vanish, and open ones depend only on endpoints. For example, $(y-z,z-x,x-y)$ restricts on $x-z=1$ to $F\cdot dr=d(x-y)$, despite ambient curl $-2(1,1,1)$.

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 $-\pi <\theta <\pi$.

If $\psi (A_1,A_2)$ is a conservative vector field, equality of the mixed derivatives of its potential gives ${\partial }_y(\psi A_1)={\partial }_x(\psi A_2)$. Expanding proves the displayed necessary equation. For continuously differentiable data on the whole plane, the vanishing planar curl is also sufficient for a potential. A useful integrating factor is nonzero on the region under consideration.

If a periodic vector field has a periodic potential of a conservative vector field, the line integral between points separated by a period vector must vanish, since it is the potential difference. A periodic gradient field need not satisfy this: the constant field $(1,2)$ on the plane has potential $x+2y$, but its integrals over the two unit translations are $1$ and $2$. On the periodic quotient these translation integrals are nonzero periods, obstructing a single-valued potential there.

A potential for a conservative vector field $X$ is a function with scalar values $\Phi$ whose gradient is $X$. Potentials differ by a constant on each connected component. The fundamental theorem for line integrals gives ${\int }_{\gamma }X\cdot d\mathbf{x}=\Phi (\gamma (1))-\Phi (\gamma (0))$. In mechanics a conservative force is conventionally written $-\nabla V$, so its potential energy is $V=-\Phi$. This spatial potential differs from the scalar potential $V(\varphi )$ that denotes the derivative-free energy density of a scalar field.

For a continuously differentiable curl-free vector field $F$ on a domain star-shaped about the origin, define $\Phi$ by the displayed integral. Differentiation and ${\partial }_i{F}_j={\partial }_j{F}_i$ give ${\partial }_i\Phi ={\int }_0^1[F_i(tx)+t{x}_j{\partial }_j{F}_i(tx)]dt={\int }_0^1{\partial }_t[t{F}_i(tx)]dt=F_i(x)$. Thus $F=\nabla \Phi$. To adopt the sign convention $F=-\nabla \varphi$, take $\varphi =-\Phi$. The star-shaped-domain hypothesis ensures every segment used in the construction lies in the domain; curl-free fields need not have global potentials on arbitrary domains.