Div of curl identity

The Divergence of any twice-differentiable vector field’s Curl is identically zero:

$\nabla \cdot (\nabla \times \mathbf{F})=0.$

Holds for every smooth $\mathbf{F}$. Curl fields are always divergence-free.

Why it’s true

In Cartesian, $\nabla \times \mathbf{F}$ has components

$({\partial }_y{F}_z-{\partial }_z{F}_y,{\partial }_z{F}_x-{\partial }_x{F}_z,{\partial }_x{F}_y-{\partial }_y{F}_x).$

Taking the divergence:

${\partial }_x({\partial }_y{F}_z-{\partial }_z{F}_y)+{\partial }_y({\partial }_z{F}_x-{\partial }_x{F}_z)+{\partial }_z({\partial }_x{F}_y-{\partial }_y{F}_x).$

Expanding, every term appears twice with opposite signs (using equality of mixed partials, Clairaut’s theorem): ${\partial }_x{\partial }_y{F}_z-{\partial }_y{\partial }_x{F}_z=0$, and similarly for the other pairs. All six terms cancel.

The identity is one face of $\nabla \cdot (\nabla \times )=0$ (treating $\nabla$ as a vector, $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{w})=0$ from the Scalar triple product).

Physical content: solenoidal fields

A vector field with zero divergence is called solenoidal (or “divergence-free”): no point sources or sinks, field lines are closed loops or escape to infinity.

So every curl field is solenoidal. The converse holds on simply-connected domains: if $\nabla \cdot \mathbf{G}=0$, then $\mathbf{G}=\nabla \times \mathbf{F}$ for some vector field $\mathbf{F}$ (the vector potential for $\mathbf{G}$). This is the Helmholtz decomposition-style result that every divergence-free field comes from a curl.

In magnetostatics

Gauss’s law for magnetism says $\nabla \cdot \mathbf{B}=0$: magnetic fields are solenoidal because there are no magnetic monopoles. By the converse of the div-of-curl identity, $\mathbf{B}$ can be written as the curl of a vector field:

$\mathbf{B}=\nabla \times \mathbf{A},$

with $\mathbf{A}$ the Vector magnetic potential. The identity guarantees:

• Defining $\mathbf{B}$ as a curl automatically makes $\nabla \cdot \mathbf{B}=0$, so Gauss for magnetism comes for free.
• $\mathbf{B}$ being divergence-free guarantees a vector potential exists.

In the continuity equation

Take the divergence of Ampère’s law (with displacement current):

$\nabla \cdot (\nabla \times \mathbf{H})=0=\nabla \cdot \mathbf{J}+\nabla \cdot \frac{\partial \mathbf{D}}{\partial t}.$

The left side is identically zero by this identity. Rearranging gives the Charge continuity equation:

$\nabla \cdot \mathbf{J}=-\frac{\partial {\rho }_v}{\partial t}.$

So charge conservation is forced by the structure of Maxwell’s equations, specifically by the div-of-curl identity applied to Ampère with displacement current. Without the displacement-current term, charge wouldn’t be automatically conserved. You don’t postulate charge conservation, you derive it from Maxwell.

In the magnetic flux argument

The identity gives the surface-independence of magnetic flux:

${\oint }_S\mathbf{B}\cdot d\mathbf{s}={\oint }_S(\nabla \times \mathbf{A})\cdot d\mathbf{s}=0$

for any closed $S$, since the flux of a curl out of a closed surface vanishes by Divergence theorem + this identity. Equivalently, the flux through any open surface depends only on the boundary loop, not on which surface spans the loop.

Companion identities

Three identities involving second-order $\nabla$ operations:

• $\nabla \cdot (\nabla \times \mathbf{F})=0$, divergence of curl.
• [[Curl of gradient identity|$\nabla \times (\nabla \varphi )=0$]], curl of gradient.
• $\nabla \times (\nabla \times \mathbf{F})=\nabla (\nabla \cdot \mathbf{F})-{\nabla }^2\mathbf{F}$, curl of curl.

The first two are “identically zero” identities; the third reduces double-curl to gradient-of-divergence minus Laplacian. Between them they cover every second-order combination of $\nabla$ on scalar or vector fields.