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A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

The first equation expresses horizontal incompressibility. The other two express steady balance of the Coriolis force, a geopotential gradient, linear drag and a prescribed zonal acceleration $F_u(y)$; nonlinear advection has been neglected. The parameter $r>0$ is an inverse drag time, so the frictional acceleration is $-r(u,v)$. On the equatorial beta plane, $f=\beta y$.

Use the streamfunction conventionTake ${\partial }_x$ of the meridional momentum equation minus ${\partial }_y$ of the zonal one. The mixed geopotential derivatives cancel, andHorizontal incompressibility removes the first term, leavingThe right side is the vertical curl of the imposed acceleration, with units ${\mathrm{s}}^{-2}$. The left side balances planetary vorticity advection and frictional removal of relative vorticity. The streamfunction has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, since it describes horizontal velocity, rather than depth-integrated volume flux.

For a consistently oriented surface and boundary, Ampère's law isStokes theorem gives the left side as ${\int }_S(\nabla \times \mathbf{B})\cdot \mathbf{n}dS$, so the given curl equation proves the law.

On a circle of radius $r$ around the axis, the assumed azimuthal field is tangent and constant in magnitude. Thus the circulation is $2\pi r{B}_{\theta }(r)$. For $r\le a$, the enclosed electric current is $I{r}^2/a^2$; for $r\ge a$, it is $I$. ThereforeThe field is continuous at $r=a$ and vanishes regularly at the axis. Its direction follows the right-hand rule for positive electric current along ${\mathbf{e}}_z$. The radial component of the curl equation also gives ${\partial }_z{B}_{\theta }=0$, so the expression is valid at every height; no unmentioned axial dependence remains.

Take the curl of the Euler equations for an inviscid fluid. The gradient on the right has zero curl. Use the stated vector identity with $\mathbf{F}=\mathbf{\omega }$, $\mathbf{G}=\mathbf{u}$:incompressible flow gives $\nabla \cdot \mathbf{u}=0$, while a curl always has zero divergence, so $\nabla \cdot \mathbf{\omega }=0$. The remaining terms giveFor a two-dimensional velocity independent of $z$, $\mathbf{\omega }=(0,0,v_x-u_y)$ and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}={\omega }_z{\partial }_z\mathbf{u}=0$. Thus the scalar vorticity is conserved following each fluid particle. It can still vary from particle to particle; this is advection, not a claim of spatially uniform vorticity.

Use International System of Units, with $c^{-2}={\mu }_0{\epsilon }_0$. The Maxwell equations with charge density $\rho$ and current density $\mathbf{J}$ areLocally, or in a suitable contractible region, the zero divergence of $\mathbf{B}$ allows a magnetic vector potential $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law then makes $\mathbf{E}+{\partial }_t\mathbf{A}$ curl-free, so a electric potential can be chosen withThese formulas automatically satisfy the two homogeneous Maxwell equations. The local topology assumption rules out obstructions to expressing a closed field as a gradient or curl.

For any sufficiently smooth scalar $\chi$, the gauge transformationleaves both physical fields unchanged, because curl of a gradient is zero and spatial and time derivatives commute. In such a region this accounts for the freedom between potentials for the same fields.

The Lorenz gauge is $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\Phi =0$. Substitution into the sourced Maxwell equations, using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$, gives the decoupled wave equationsFor example, Gauss's law gives $-{\nabla }^2\Phi -{\partial }_t\nabla \cdot \mathbf{A}=\rho /{\epsilon }_0$; the gauge condition turns this into the scalar equation. The remaining gauge freedom preserving the condition has $({\nabla }^2-c^{-2}{\partial }_t^2)\chi =0$.

Put $\theta =\omega t-Kz$ for the given wave. Direct differentiation gives $\nabla \cdot \mathbf{E}=0$ and $\nabla \times \mathbf{E}=K\mathbf{E}$. Integrating Faraday's law in time yields the wave's magnetic fieldOne could superpose a time-independent divergence-free and curl-free magnetic field; the expression displayed chooses no such extra background. It has zero divergence and ${\partial }_t\mathbf{B}=-K\mathbf{E}$, so the magnetic Gauss and Faraday equations hold. MoreoverSince the time derivative is nonzero, the vacuum Ampere-Maxwell law holds precisely whenThis verifies all four source-free equations, including both divergence constraints. The sign of $\omega /K$ selects the direction of propagation.

A wave has circular polarization when, at a fixed spatial point, its transverse electric vector has constant nonzero magnitude and rotates uniformly through a circle. Here $E_x^2+E_y^2=E^2$ and the angle changes at rate $\omega$, so the wave is circularly polarized. Its magnetic vector is perpendicular to the electric vector, has magnitude $|E|/c$ under the dispersion relation, and rotates with it. A handedness label additionally requires a convention for the observer's viewing direction.

The Lorenz-gauge potentials for a circularly polarized plane wave give a convenient choice. Take $\Phi =0$ and $\mathbf{A}={\omega }^{-2}{\partial }_t\mathbf{E}$. Because ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$,Also $\nabla \cdot \mathbf{A}={\omega }^{-2}{\partial }_t\nabla \cdot \mathbf{E}=0$. Together with ${\partial }_t\Phi =0$, this proves these potentials satisfy the Lorenz gauge. They satisfy the source-free potential wave equations because every component has the same dispersion relation just obtained.

Only $A_z$ depends on position. Taking the curl,The field reverses across $x=0$. Its divergence is zero, as required for a magnetic field; the vanishing field on the midplane is important for the crossing and separatrix trajectories below.

For a twice continuously differentiable scalar potential, mixed second partial derivatives commute. Taking the spatial curl of the equation from part (i) givesFor example, its first component is $-{\partial }_2{\partial }_3\xi +{\partial }_3{\partial }_2\xi =0$. The curl acts on the spatial representation of the material acceleration, not directly on its expression in reference coordinates.

This is a necessary local condition. It is sufficient locally, and sufficient globally on a simply connected domain under the usual smoothness assumptions. On a domain with holes, a global scalar potential additionally requires every closed-loop line integral of the material acceleration to vanish. The homogeneous motion in part (c) admits an explicit global scalar potential, so this topological qualification causes no obstruction there.

Take two time derivatives at fixed reference position $\mathbf{X}$:The constraint $\det \mathbf{F}=1$ ensures that the deformation gradient is invertible, so $\mathbf{X}={\mathbf{F}}^{-1}(\mathbf{x}-\mathbf{c})$. Thus the spatial material acceleration is the affine vector fieldIts spatial gradient is $\mathbf{A}$, and its curl isPart (b)(iii) therefore requiresEquivalently, multiplying on the left by ${\mathbf{F}}^T$ and on the right by $\mathbf{F}$ givesThe translation $\mathbf{c}(t)$ contributes no spatial curl and is unrestricted by this condition. A useful equivalent first integral isthe antisymmetric matrix in parentheses must be constant, and need not vanish. The separate constraint $\det \mathbf{F}=1$ is still required; differentiating it also gives $\mathrm{tr}(\dot{\mathbf{F}}{\mathbf{F}}^{-1})=0$.

The condition is understood in the setting of part (b), where a static family already exists under the prescribed body force. For a homogeneous elastic deformation gradient, the elastic stress $\mathbf{S}(\mathbf{F})$ is spatially constant. Static balance of linear momentum consequently gives $\nabla p_0=\rho \mathbf{b}$, ensuring that $\mathbf{b}$ has a scalar potential. If the static-family assumption were dropped and an arbitrary time-independent body force allowed, the general necessary condition would instead be $\nabla \times \ddot{\mathbf{x}}=\nabla \times \mathbf{b}$.

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.

For a material volume, mass conservation is $d{\int }_{V(t)}\rho dV/dt=0$. The transport theorem and arbitrariness of that volume give $\boxed{{\rho }_t+\nabla \cdot (\rho u)=0}$. Balance of linear momentum includes the pressure traction and body force:Apply transport, the divergence theorem and the mass equation to obtainFor small perturbations of the uniform resting barotropic state, write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=p_0+c_0^2{\rho }^{\prime }+\cdots$. Linearized mass and momentum areTaking the curl shows that initially vanishing vorticity remains zero, so the acoustic velocity has a potential $u=\nabla \varphi$. Integrating momentum, using the additive freedom of a velocity potential to remove the spatially constant integration term, gives ${\rho }^{\prime }=-({\rho }_0/c_0^2){\varphi }_t$. Continuity now yields $\boxed{{\varphi }_{tt}=c_0^2{\nabla }^2\varphi }$.

For spherical symmetry, set $\varphi =e^{-i\omega t}F(r)$ and $k=\omega /c_0$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }+k^2{r}^{2}F=0$, or $(rF{)}^{\prime \prime }+k^2(rF)=0$. HenceRigid boundaries require radial velocity $F^{\prime }=0$ at both radii. A form automatically satisfying the condition at $a$ isAt $b$, the condition $b(rF{)}^{\prime }-(rF)=0$ becomesTherefore the nonzero oscillatory eigenfrequencies obeyThe undivided sine-cosine form avoids artificial issues from dividing by a cosine. The formal $k=0$ root is the spatially constant zero-frequency potential additive freedom of a velocity potential, not a propagating acoustic oscillation. This is the radial acoustic spectrum between rigid concentric spheres.

For a vector $a$ described in axes rotating at constant angular velocity $\Omega$, the rotating-axes theorem is $(da/dt{)}_{\mathrm{inertial}}=(da/dt{)}_{\mathrm{rotating}}+\Omega \times a$. Inertial particle velocity is $u+\Omega \times x$. Applying the theorem once more gives inertial acceleration $Du/Dt+2\Omega \times u+\Omega \times (\Omega \times x)$. Thus the rotating momentum equation isIf $g=-\nabla \Phi$ and density is constant, the centrifugal term can be absorbed into the reduced pressure in a rotating fluid:For small disturbances with $\Omega =\Omega e_z$, linearize to $u_t-2\Omega v=-P_x/\rho$, $v_t+2\Omega u=-P_y/\rho$, $w_t=-P_z/\rho$, together with incompressibility. Put $\zeta =v_x-u_y$. Taking the vertical curl gives ${\zeta }_t=2\Omega w_z$. Taking divergence gives ${\nabla }^{2}P=2\rho \Omega \zeta$. Applying the Laplacian to the vertical momentum equation therefore gives ${\partial }_t{\nabla }^{2}w=-2\Omega {\zeta }_z$. Another time derivative yieldsAll three velocity components are retained, even for fields independent of $y$; rotation produces the transverse component.

Set $m=\pi /h$ and seek $w=a{e}^{-i\omega t}W(x)\sin mz$. The rigid-wall conditions are then automatic and the scalar equation givesFor $0<|\omega |<2|\Omega |$, define $q=m\sqrt{4{\Omega }^2/{\omega }^2-1}$. The two possible oscillatory factors are distinguished by energy propagation. With horizontal wavenumber $k$, the dispersion relation is ${\omega }^2=4{\Omega }^2{m}^2/(k^2+m^2)$ and $c_{g,x}=-\omega k/(k^2+m^2)$. An outgoing wave in $x>0$ must have $\omega k<0$, givingFor $|\omega |>2|\Omega |$, define $\kappa =m\sqrt{1-4{\Omega }^2/{\omega }^2}$. Reject the exponentially growing branch, leaving the evanescent responseThe real part is the physical velocity. The radiation or bounded-decay condition is needed in addition to the wavemaker boundary value to pick a unique half-channel response. Steady forcing ($\omega =0$) is a singular limit: with nonzero rotation a steady geostrophic balance and impermeable top/bottom require $w=0$, incompatible with a nonzero imposed sine profile. At the cutoff equality, the bounded outgoing polarization also degenerates. These exceptional cases cannot be obtained by blindly substituting into the two strict-inequality formulas. The propagating sign is the radiation sign for an inertial wave.

Linearize about rest, with the equilibrium forces already absorbed in the reduced pressure. The quadratic convective acceleration drops out. Since $\mathbf{k}\times (u\mathbf{i}+v\mathbf{j}+w\mathbf{k})=-v\mathbf{i}+u\mathbf{j}$, the component equations areThe scan prints $v_w$ in the continuity equation; the derivative required by incompressibility is $v_y$.

To eliminate pressure, put $\zeta =v_x-u_y$ and ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$. Taking the vertical component of the curl of momentum givesTaking horizontal divergence of momentum and substituting continuity givesDifferentiate this in $z$ and use $P_z/\rho =-w_t$:One more time derivative, followed by the vorticity relation, yields the Poincaré equation for inertial wavesThis derivation retains all three velocity components, including the transverse component coupled by rotation.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

Linear incompressible motion about rest in a frame rotating with $\mathbf{\Omega }=\Omega \mathbf{k}$ obeys ${\mathbf{u}}_t+2\mathbf{\Omega }\times \mathbf{u}=-\nabla P/\rho$. If $\zeta =v_x-u_y$ is its vertical vorticity, taking curl gives ${\zeta }_t=2\Omega w_z$, while eliminating pressure using continuity gives ${\partial }_t{\nabla }^{2}w+2\Omega {\zeta }_z=0$. Differentiating in time yields the displayed scalar equation. All three velocity components must be retained, even when the fields depend on only two spatial coordinates. The Coriolis acceleration generally generates a transverse velocity that is absent in a strictly two-component description.

A spherical poloidal-toroidal decomposition represents a solenoidal vector field with zero flux through each sphere by scalar potentials $S,T$. The second term is tangent to concentric spheres; the first carries the radial component. Requiring zero angular means for the potentials fixes their otherwise irrelevant radial additions. In a plane layer, the analogous fluctuating components are $\nabla \times \nabla \times (\varphi \hat{\mathbf{z}})$ and $\nabla \times (\psi \hat{\mathbf{z}})$, with any horizontally uniform components handled separately. These constructions enforce the divergence-free condition because the divergence of a curl vanishes.

// Target: fluid-mechanics.bigb

Let ${\nabla }_{\tan }F=\nabla F-{\mathbf{e}}_r{\partial }_{r}F$ be the tangential surface gradient on a sphere of radius $r>0$. For smooth scalar fields, set $\mathbf{f}=R{\mathbf{e}}_r+{\nabla }_{\tan }G+{\mathbf{e}}_r\times {\nabla }_{\tan }F$. Its radial curl detects only $F$:To prove this, use the radial curl formula $(r\sin \theta {)}^{-1}[{\partial }_{\theta }(\sin \theta f_{\varphi })-{\partial }_{\varphi }{f}_{\theta }]$. The radial component $R$ never enters it. The $G$ terms cancel as $G_{\varphi \theta }-G_{\theta \varphi }=0$ by symmetry of second derivatives, while the $F$ terms give the unit-sphere Laplace-Beltrami operator. Taking the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$ gives the displayed $r^{-1}$ version. This tangential gradient decomposition does not assert that the whole vector field is solenoidal.

For a smooth divergence-free field on ${\mathbb{R}}^3$, the differential form $F={\iota }_{\mathbf{B}}(d{x}^1\wedge d{x}^2\wedge d{x}^3)$ is closed. The radial homotopy formula gives a global magnetic vector potential $\mathbf{A}(\mathbf{x})={\int }_0^{1}s\mathbf{B}(s\mathbf{x})\times \mathbf{x}ds$ with curl $\mathbf{B}$. For constant $\mathbf{B}$, this reduces to $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

A vector field directed along the radius from a fixed origin, with magnitude depending only on radius. In three-dimensional Euclidean space and away from the origin, $\mathbf{F}=f(r)\mathbf{r}=rf(r)\hat{\mathbf{r}}$. The product rule gives $\nabla \cdot \mathbf{F}=3f+r{f}^{\prime }$, and the curl vanishes because the derivative matrix is symmetric. These identities alone do not establish regularity at the origin.

When the only nonpotential force points radially, its curl has no radial component. The radial absolute vorticity therefore obeys a local flux conservation law, although vortex stretching can change it along a fluid trajectory.