Vector field

A vector field assigns a vector to every point of its domain.

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.

A vector field on an open subset of Euclidean space whose component functions have continuous first partial derivatives. Its Jacobian matrix is continuous; this is a standard regularity assumption for the divergence theorem and Stokes theorem.

The local flow of a smooth vector field is the family of solutions of its ordinary differential equation, with ${\Phi }_0(p)=p$. Uniqueness gives ${\Phi }_{s+t}={\Phi }_s\circ {\Phi }_t$ wherever defined. A compactly supported smooth vector field on a manifold without boundary is complete, so its flow exists for all real times.

A smooth family $Y_t$ assigns a tangent vector at each space-time point. Its flow maps solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0$ the identity. For a varying differential form, differentiation gives ${\partial }_t({\rho }_t^{\ast }{\alpha }_t)={\rho }_t^{\ast }({\dot{\alpha }}_t+{\mathcal{L}}_{Y_t}{\alpha }_t)$.

A smooth time-dependent vector field on a compact manifold without boundary has flow maps throughout each compact time interval on which the field is defined. Finite-time escape is impossible on the compact state space. Uniqueness and the backward-time equation supply smooth inverses, so the flow is a smooth isotopy.

Differentiate the displayed homogeneity relation in $\lambda$ at $1$. On every regular point of $H=0$, the radial vector field $x$ is therefore perpendicular to the normal gradient and tangent to the level surface. Its flow $x(t)=e^{t}x(0)$ preserves that zero set. At a singular vertex this flow formulation remains meaningful even if there is no tangent plane.

For $r=|x|>0$, the gradient satisfies ${\partial }_{i}r=x_i/r$. The product rule and Einstein notation then give the displayed divergence formula in dimension $n$. In dimension three, its derivative matrix is symmetric, so the curl vanishes. Zero divergence forces $f(r)=C{r}^{-n}$ on each positive radial interval. This formula does not assert regularity at the origin.

Let $V=\{ϵ<r<R\}$ with $ϵ>0$ and $\mathbf{F}=\hat{\mathbf{r}}/r^{\gamma }$. Direct differentiation gives $\nabla \cdot \mathbf{F}=(2-\gamma )r^{-\gamma -1}$. For $\gamma \ne 2$, integration against $4\pi r^{2}dr$ gives the displayed expression. Independently, the two boundary spheres contribute $4\pi R^{2-\gamma }$ and $-4\pi {ϵ}^{2-\gamma }$, since the outward normal of the shell at its inner boundary points toward the origin. For $\gamma =2$, the divergence is zero in the shell and the surface contributions cancel exactly. The excluded origin must not be silently added to this regular-domain calculation.

The vector field $\mathbf{G}=(-y,x,0)/(x^2+y^2)$ has zero curl away from its excluded axis, but has circulation $2\pi$ around a positively oriented circle enclosing that axis. Stokes theorem gives zero total circulation for a smooth annular surface avoiding the axis, whose inner and outer boundary circles have opposite orientations. It cannot be applied directly to a disk meeting the axis, because the field is undefined there.

An integral curve follows a vector field by solving the displayed ordinary differential equation. For a smooth vector field, integral curves through initial points assemble into its local flow.

For a continuously differentiable nonzero vector field $F$, the curvature of a space curve following its direction isTake the unit tangent vector $t=\pm F/|F|$ and differentiate with respect to arc length. The derivative of $|F{|}^{-1}$ is parallel to $F$ and disappears from $t\times t^{\prime }$. Since $t\cdot t^{\prime }=0$, we have $|t\times t^{\prime }|=|t^{\prime }|=\kappa$, proving the formula. It remains valid with zero curvature, even where the usual Frenet frame is undefined.

An axisymmetric vector field is invariant under rotations about a chosen axis. In cylindrical coordinates, its components are independent of the azimuthal angle.

A solenoidal vector field has zero divergence. Magnetic fields are solenoidal because $\nabla \cdot \mathbf{B}=0$.

For a solenoidal vector field $J$ with compact support and suitable integrability, apply the divergence theorem to $x_i{x}_{j}J$. It gives zero boundary term and hence the displayed antisymmetry. Equivalently, for any constant vector $C$, the vector triple product identity yields $\int (C\cdot r)JdV=\frac{1}{2}C\times \int (J\times r)dV$. A piecewise smooth field must have no normal jump at its support boundary; otherwise its weak divergence contains a surface contribution and the argument fails.

Suppose two oriented surfaces have the same oriented boundary and $S_1\cup (-S_2)$ bounds a volume inside a domain where the vector field $\mathbf{F}$ is smooth and solenoidal. The divergence theorem gives zero flux through their closed union, proving the displayed equality. If a global vector potential $\mathbf{F}=\nabla \times \mathbf{G}$ exists on the relevant surfaces, the Stokes theorem proves the equality directly from their common boundary. The domain condition matters: a singularity inside the joining volume can obstruct the divergence argument.

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

For an isolated solenoidal magnetic field, zero flux through every sphere removes the degree-zero spherical harmonic of $P=\mathbf{B}\cdot \mathbf{x}$. In a poloidal-toroidal decomposition, let ${\mathcal{L}}^2=-{\Delta }_{S^2}$. The spherical poloidal component obeysIf $P={\sum }_{l\ge 1,m}{p}_{lm}(r)Y_{lm}$, put $D_{lm}={\int }_0^{\infty }[r^2|p_{lm}^{\prime }{|}^2+l(l+1)|p_{lm}{|}^2]dr$. The two sides of the bound are respectively $\sum D_{lm}/[l(l+1)]$ and $\frac{1}{2}\sum D_{lm}$. Since $l(l+1)\ge 2$, the bound follows, with equality for degree-one radial scalars. Together with the radial-flow dynamo energy bound it gives $(q/\eta {)}^2\ge 2E_P/E_{\mathrm{total}}$ for a marginal or growing dynamo field.

// Target: fluid-mechanics.bigb