Curl of gradient identity

The Curl of any twice-differentiable scalar field’s Gradient is identically zero:

$\nabla \times (\nabla \varphi )=0.$

Holds for every smooth scalar $\varphi$. A constraint on which vector fields can arise as gradients.

Why it’s true

Writing out the curl in Cartesian coordinates:

$\nabla \times (\nabla \varphi )=\left|\begin{array}{ccc}\hat{\mathbf{x}}& \hat{\mathbf{y}}& \hat{\mathbf{z}}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ {\partial }_x\varphi & {\partial }_y\varphi & {\partial }_z\varphi \end{array}\right|.$

The $\hat{\mathbf{x}}$ component is ${\partial }_y({\partial }_z\varphi )-{\partial }_z({\partial }_y\varphi )$. By equality of mixed partials (assuming $\varphi \in C^2$), ${\partial }_y{\partial }_z\varphi ={\partial }_z{\partial }_y\varphi$, so the $\hat{\mathbf{x}}$ component vanishes. Same for the $\hat{\mathbf{y}}$ and $\hat{\mathbf{z}}$ components. Every term cancels.

The identity is one face of the algebraic fact that $\nabla \times \nabla =0$ (treating $\nabla$ as a vector and using $\mathbf{v}\times \mathbf{v}=0$).

Physical content: conservative fields

A vector field $\mathbf{F}$ that is the gradient of a scalar, $\mathbf{F}=\nabla \varphi$, is a conservative field (or a Gradient field). The identity says every conservative field has zero curl.

The converse holds on simply-connected domains: if $\nabla \times \mathbf{F}=0$ and the domain has no holes, then $\mathbf{F}=\nabla \varphi$ for some scalar $\varphi$. On non-simply-connected domains (a torus, a region with a missing cylinder), zero-curl fields exist that aren’t gradients. The prototypical example is $\mathbf{F}=(-y/r^2,x/r^2)$ around a removed origin.

In electrostatics

In the static case Faraday’s law reads $\nabla \times \mathbf{E}=0$, so $\mathbf{E}$ is curl-free, hence (on simply-connected regions) the gradient of a scalar, the Electric potential:

$\mathbf{E}=-\nabla V.$

The identity makes this work both ways:

• Defining $\mathbf{E}=-\nabla V$ automatically gives a curl-free $\mathbf{E}$, no need to check Faraday separately.
• Static $\mathbf{E}$ comes from some potential, given by $V=-\int \mathbf{E}\cdot d\mathbf{l}$ along any path (path-independence is exactly equivalent to $\nabla \times \mathbf{E}=0$).

In dynamics ($\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}\ne 0$), $\mathbf{E}$ is no longer a pure gradient. It picks up the $-{\partial }_t\mathbf{A}$ term (see Electromagnetic potentials).

In magnetostatics

In a current-free region, $\nabla \times \mathbf{H}=0$, so $\mathbf{H}$ is also expressible as a gradient: $\mathbf{H}=-\nabla V_m$, with $V_m$ the magnetic scalar potential. Useful for permanent-magnet problems.

In the four-potential formalism

The identity supports gauge freedom of the Electromagnetic potentials. Under the transformation $\mathbf{A}\to \mathbf{A}+\nabla \psi$:

$\mathbf{B}=\nabla \times \mathbf{A}\to \nabla \times (\mathbf{A}+\nabla \psi )=\nabla \times \mathbf{A}+\nabla \times \nabla \psi =\nabla \times \mathbf{A}.$

The added $\nabla \psi$ produces no change in $\mathbf{B}$ because $\nabla \times \nabla \psi =0$. This is what makes the gauge transformation physically invisible.

Companion identities

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

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

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