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Equality of line integrals on two selected paths is a necessary test for a conservative vector field, but it is not sufficient. The paths may both miss the component responsible for nonzero curl.

For an explicit example, take $W(x,y,z)=(z-y,x-z,y-x{)}^T$. Its curl is the nonzero constant $2(1,1,1{)}^T$, so it is not a gradient field on ${\mathbb{R}}^3$. Nevertheless, on the plane $x=y$, the field equals $(z-x,x-z,0{)}^T$. Every tangent vector to that plane has equal first and second components, giving zero dot product with $W$. Thus every line integral of $W$ along a path confined to that plane is zero, regardless of endpoints or the path's shape.

In particular, a broken straight path and a genuinely curved path in that plane can have matching line integrals while revealing nothing about global path independence. To establish a conservative vector field, construct a vector-field potential, or verify the appropriate global criterion such as zero curl on a simply connected domain.

The Helmholtz decomposition separates the peculiar velocity into a gradient field and a transverse vector field:With decaying or periodic spatial boundary conditions this is equivalently the Fourier transform projection, for $k\ne 0$,The transverse projection operator annihilates a gradient, because $({\delta }_{ij}-k_i{k}_j/k^2)k_j=0$. Both forces in the linear cosmological Euler equation are gradients, with spatially homogeneous coefficients. Applying the transverse projection operator therefore givesSince $H=\dot{a}/a$, this is $d(a{\mathbf{v}}_{\perp })/dt=0$, and henceAt this order the scalar pressure and gravitational forces cannot generate vorticity; an initially vanishing transverse vector field remains zero. Its nonzero initial value is damped by cosmic expansion.

The position vector field is a gradient field:Along any piecewise smooth curve, $\mathbf{r}\cdot d\mathbf{r}=d(|\mathbf{r}{|}^2/2)$. Therefore a closed line integral has identical endpoint values andFor a loop bounding an admissible oriented surface, Stokes theorem gives the same result because $\nabla \times \mathbf{r}=0$. The gradient field proof also applies directly to arbitrary piecewise smooth closed loops without choosing a spanning surface.

Letbe the vertical absolute vorticity. Taking the vertical curl of the momentum equation removes both the tidal potential and the isothermal pressure force, because each is a gradient field. The two-dimensional vorticity equation isThe surface-density equation isThe quotient rule now givesSince $\zeta =W/\Sigma$ is the vortensity,

The $y$ component integrates to $\varphi =xy{e}^{z-x}+g(x,z)$. Matching the $x$ component gives $g_x=2x$, and matching the $z$ component then gives the remaining derivative zero. ConsequentlyDirect differentiation recovers all three components, proving this is a gradient field globally on ${\mathbb{R}}^3$.

A differentiable gradient field must have equal mixed derivatives. Here $\partial F_y/\partial z=3z^2$ but $\partial F_z/\partial y=-x$, which are not identically equal on any open region. Equivalently,Thus no scalar potential exists for this field on an open region. Vanishing of the curl on a special lower-dimensional set would not supply a three-dimensional potential.

No scalar potential exists on the full three-dimensional domain. A sufficiently smooth gradient field has zero curl, since mixed partial derivatives commute. The calculated curl is not identically zero. The fact that it vanishes at some individual points does not make the whole field a gradient field.

Direct calculation of the curl givesSince $\mathbf{a}\ne 0$, the first vector field cannot be a gradient field: the curl of a gradient is zero. The second is a smooth polynomial field on all of the simply connected space ${\mathbb{R}}^3$, so its vanishing curl makes it a conservative vector field. A potential is also constructed explicitly in the next part. Thus ${\mathbf{v}}_2$ is a conservative vector field and ${\mathbf{v}}_1$ is not. A vanishing divergence alone, as for ${\mathbf{v}}_1$, is a different property and does not establish conservativity.

The curl of $\mathbf{G}=\mathbf{a}\times \mathbf{r}$ is $2\mathbf{a}$. Thus for $\mathbf{a}\ne 0$ it cannot be a gradient field, while for $\mathbf{a}=0$ it is zero and all its vector-field potentials are constants. This zero-vector case is allowed by the question.

For the radial vector field $\mathbf{H}$, write $\rho =|\mathbf{r}|$. Choose a radial primitiveFor $\rho >0$, the chain rule givesHence all vector-field potentials on a connected domain where the field is defined areThe assumed existence of the needed derivatives and integrals justifies this construction. If the domain includes the origin and the field is regular there, choose ${\rho }_0=0$ and extend the primitive there. Any two vector-field potentials differ by a constant on each connected component.

For $\mathbf{G}$, the product rule for the curl givesSince $\mathbf{a}\ne 0$, this is not identically zero on ${\mathbb{R}}^3$. A gradient field has zero curl, so no scalar potential exists for $\mathbf{G}$ on the whole space.

For the radial vector field $\mathbf{K}$, observe thatThus all vector-field potentials areThese are smooth everywhere, including the origin. Any two differ by a constant because their gradient difference vanishes on the connected whole space.