Identities: div(curl) and curl(grad)

⚡ Key Takeaways

One-line statement: The divergence of any curl is zero, ∇⋅(∇×F)=0\nabla\cdot(\nabla\times\mathbf{F})=0, and the curl of any gradient is zero, ∇×(∇ϕ)=0\nabla\times(\nabla\phi)=\mathbf{0}, for sufficiently smooth fields.

Key result in words: Curls are always divergence-free (solenoidal); gradients are always curl-free (irrotational). These two identities are exact algebraic consequences of the equality of mixed partial derivatives.

Where it applies: Vector calculus on R3\mathbb{R}^3 (and beyond via differential forms), electromagnetism, fluid dynamics, the Helmholtz decomposition, and the Poincaré lemma.

Who/when: Implicit in the work of Gauss, Green, and Stokes (early–mid 19th century); crystallized in Maxwell's electromagnetic theory (1860s) and the vector-analysis notation of Gibbs and Heaviside (1880s).

💡 The Core Insight

Both identities are nothing more than the statement

∂2∂xi ∂xj=∂2∂xj ∂xi\frac{\partial^2}{\partial x_i\,\partial x_j} = \frac{\partial^2}{\partial x_j\,\partial x_i}

dressed in vector notation. The curl and divergence operators combine partial derivatives with antisymmetric coefficients (the cross product, the alternating symbol εijk\varepsilon_{ijk}). When you compose two such operators, you contract a quantity that is symmetric in two indices (the mixed second derivative) against one that is antisymmetric in those same indices — and any symmetric-times-antisymmetric contraction is identically zero. That single observation proves both identities at once.

Formal Statement

DefinitionThe three first-order operators

For a scalar field ϕ:R3→R\phi:\mathbb{R}^3\to\mathbb{R} and a vector field F=(F1,F2,F3)\mathbf{F}=(F_1,F_2,F_3) on R3\mathbb{R}^3, with ∇=(∂x,∂y,∂z)\nabla=(\partial_x,\partial_y,\partial_z):

1
∇ϕ=(∂ϕ∂x, ∂ϕ∂y, ∂ϕ∂z)\quad \nabla\phi=\left(\frac{\partial\phi}{\partial x},\ \frac{\partial\phi}{\partial y},\ \frac{\partial\phi}{\partial z}\right)

Derivation hint: the gradient stacks the three first partials of a scalar into a vector pointing in the direction of steepest ascent.

2
∇⋅F=∂F1∂x+∂F2∂y+∂F3∂z\quad \nabla\cdot\mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}

Derivation hint: the divergence is the trace of the Jacobian ∂Fi/∂xi\partial F_i/\partial x_i, summed over the diagonal.

3
∇×F=(∂F3∂y−∂F2∂z, ∂F1∂z−∂F3∂x, ∂F2∂x−∂F1∂y)\quad \nabla\times\mathbf{F}=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z},\ \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x},\ \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)

Derivation hint: expand the symbolic determinant det⁡[e^i, ∂j, Fk]\det[\hat e_i,\ \partial_j,\ F_k]; each component is an antisymmetric difference of cross-partials. Here ϕ\phi is assumed C2C^2 and each FiF_i is assumed C2C^2 so that all second partials are continuous.

TheoremThe two second-order identities

Let ϕ∈C2(U)\phi\in C^2(U) and F∈C2(U;R3)\mathbf{F}\in C^2(U;\mathbb{R}^3) on an open set U⊆R3U\subseteq\mathbb{R}^3. Then:

4
∇×(∇ϕ)=0(curl of a gradient vanishes)\quad \nabla\times(\nabla\phi)=\mathbf{0}\qquad\text{(curl of a gradient vanishes)}

Derivation hint: the kk-th component is εkij ∂i∂jϕ\varepsilon_{kij}\,\partial_i\partial_j\phi; the symmetric Hessian contracts to zero against ε\varepsilon.

5
∇⋅(∇×F)=0(divergence of a curl vanishes)\quad \nabla\cdot(\nabla\times\mathbf{F})=0\qquad\text{(divergence of a curl vanishes)}

Derivation hint: write it as εijk ∂i∂jFk\varepsilon_{ijk}\,\partial_i\partial_j F_k; antisymmetry in i,ji,j kills the symmetric ∂i∂j\partial_i\partial_j. In index notation with the Einstein summation convention and the Levi-Civita symbol εijk\varepsilon_{ijk}:

6
(∇×∇ϕ)k=εkij ∂i∂jϕ=0,∇⋅(∇×F)=εijk ∂i∂jFk=0\quad (\nabla\times\nabla\phi)_k=\varepsilon_{kij}\,\partial_i\partial_j\phi=0,\qquad \nabla\cdot(\nabla\times\mathbf{F})=\varepsilon_{ijk}\,\partial_i\partial_j F_k=0

Derivation hint: ε\varepsilon is antisymmetric under i↔ji\leftrightarrow j while ∂i∂j\partial_i\partial_j is symmetric; swapping dummy indices shows each sum equals its own negative.

Why It Holds / Derivation

:::proof[Identity [4]: curl(grad) = 0] Step 1. Compute the first component of ∇×(∇ϕ)\nabla\times(\nabla\phi) directly from [3], substituting Fi=∂iϕF_i=\partial_i\phi:

7
(∇×∇ϕ)1=∂∂y ⁣(∂ϕ∂z)−∂∂z ⁣(∂ϕ∂y)=∂2ϕ∂y ∂z−∂2ϕ∂z ∂y.\quad \big(\nabla\times\nabla\phi\big)_1=\frac{\partial}{\partial y}\!\left(\frac{\partial\phi}{\partial z}\right)-\frac{\partial}{\partial z}\!\left(\frac{\partial\phi}{\partial y}\right)=\frac{\partial^2\phi}{\partial y\,\partial z}-\frac{\partial^2\phi}{\partial z\,\partial y}.

Derivation hint: replace F3→∂zϕF_3\to\partial_z\phi and F2→∂yϕF_2\to\partial_y\phi in the first curl component.

Step 2. Invoke Clairaut–Schwarz: since ϕ∈C2\phi\in C^2, the mixed partials are equal, ∂y∂zϕ=∂z∂yϕ\partial_y\partial_z\phi=\partial_z\partial_y\phi. Hence the first component is 00.

Step 3. The second and third components,

8
(∇×∇ϕ)2=∂z∂xϕ−∂x∂zϕ,(∇×∇ϕ)3=∂x∂yϕ−∂y∂xϕ,\quad \big(\nabla\times\nabla\phi\big)_2=\partial_z\partial_x\phi-\partial_x\partial_z\phi,\qquad \big(\nabla\times\nabla\phi\big)_3=\partial_x\partial_y\phi-\partial_y\partial_x\phi,

vanish identically by the same theorem. Derivation hint: each component is a difference of two orderings of the same mixed partial.

Step 4 (index form). All three components are summarized by (∇×∇ϕ)k=εkij∂i∂jϕ(\nabla\times\nabla\phi)_k=\varepsilon_{kij}\partial_i\partial_j\phi. Rename dummies i↔ji\leftrightarrow j: εkij∂i∂jϕ=εkji∂j∂iϕ=−εkij∂i∂jϕ\varepsilon_{kij}\partial_i\partial_j\phi=\varepsilon_{kji}\partial_j\partial_i\phi=-\varepsilon_{kij}\partial_i\partial_j\phi. A quantity equal to its own negative is zero. ■\blacksquare

:::proof[Identity [5]: div(curl) = 0] Step 1. Write the curl from [3] and apply the divergence [2]:

9
∇⋅(∇×F)=∂∂x ⁣(∂F3∂y−∂F2∂z)+∂∂y ⁣(∂F1∂z−∂F3∂x)+∂∂z ⁣(∂F2∂x−∂F1∂y).\quad \nabla\cdot(\nabla\times\mathbf{F})=\frac{\partial}{\partial x}\!\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)+\frac{\partial}{\partial y}\!\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)+\frac{\partial}{\partial z}\!\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right).

Derivation hint: take ∂x\partial_x of component 1, ∂y\partial_y of component 2, ∂z\partial_z of component 3, then add.

Step 2. Expand and group by field component:

10
(∂x∂yF3−∂y∂xF3)+(∂y∂zF1−∂z∂yF1)+(∂z∂xF2−∂x∂zF2).\quad \left(\partial_x\partial_y F_3-\partial_y\partial_x F_3\right)+\left(\partial_y\partial_z F_1-\partial_z\partial_y F_1\right)+\left(\partial_z\partial_x F_2-\partial_x\partial_z F_2\right).

Derivation hint: collect the two terms in F1F_1, the two in F2F_2, and the two in F3F_3.

Step 3. Each parenthesized pair is a difference of equal mixed partials (Clairaut–Schwarz, since Fi∈C2F_i\in C^2), so each pair vanishes; the total is 00.

Step 4 (index form). ∇⋅(∇×F)=∂i εijk∂jFk=εijk ∂i∂jFk\nabla\cdot(\nabla\times\mathbf{F})=\partial_i\,\varepsilon_{ijk}\partial_j F_k=\varepsilon_{ijk}\,\partial_i\partial_j F_k. Swapping dummies i↔ji\leftrightarrow j gives εjik∂j∂iFk=−εijk∂i∂jFk\varepsilon_{jik}\partial_j\partial_i F_k=-\varepsilon_{ijk}\partial_i\partial_j F_k, so the sum equals its negative and is therefore 00. ■\blacksquare

⚠️ The hypothesis is essential

Both identities require continuity of the second partials (class C2C^2). If ϕ\phi or F\mathbf{F} has discontinuous or non-equal mixed partials at a point, the identities can fail at that point. The classic example ϕ(x,y)=xy(x2−y2)x2+y2\phi(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2} (with ϕ(0,0)=0\phi(0,0)=0) has ∂x∂yϕ(0,0)=−1≠1=∂y∂xϕ(0,0)\partial_x\partial_y\phi(0,0)=-1\neq 1=\partial_y\partial_x\phi(0,0), so ∇×∇ϕ≠0\nabla\times\nabla\phi\neq\mathbf 0 at the origin. The identities are theorems about smooth fields, not unconditional truths.

JC
James Clerk Maxwell
1831–1879

Maxwell did not "discover" these identities in isolation, but he wielded them as structural pillars of electromagnetic theory. Working partly in quaternion notation inherited from Hamilton, he recognized that the magnetic field's source-free nature (∇⋅B=0\nabla\cdot\mathbf{B}=0) is guaranteed if B=∇×A\mathbf{B}=\nabla\times\mathbf{A} for a vector potential A\mathbf{A} — precisely identity [5]. Likewise, that a static electric field is curl-free (∇×E=0\nabla\times\mathbf{E}=\mathbf 0) permits the scalar potential E=−∇ϕ\mathbf{E}=-\nabla\phi — the converse of identity [4]. His Treatise on Electricity and Magnetism (1873) embedded these relations into physics permanently, and Gibbs and Heaviside later recast them into the clean ∇\nabla-operator notation students use today.

Building Intuition

🧠 Intuition

Think of the three operators as a chain of "differentiation flavors": gradient turns a scalar into a vector, curl turns a vector into a vector, divergence turns a vector into a scalar. The identities say two consecutive links in a particular order produce nothing.

  • curl(grad)=0: A gradient field always flows "downhill" along a single potential surface stack. Such a field can't circulate — there is no loop along which the potential keeps increasing and returns higher than it started. No circulation means no curl.
  • div(curl)=0: A curl field is built from rotation. Rotation neither creates nor destroys material — what swirls in one side of any tiny box swirls out the other. No net outflow means no divergence; a pure swirl has no sources or sinks.

Geometrically, both reduce to the same picture: the boundary of a boundary is empty. The "boundary" of a closed surface is nothing; the "boundary" of a closed loop is nothing. Curl integrates over loops bounding surfaces (Stokes), divergence over surfaces bounding volumes (Gauss), and applying one after the other tries to integrate over an empty boundary.

Figure
The operator chain: each composition of two consecutive arrows collapses to zero
scalar φ vector ∇φ vector F scalar / vector ∇ (grad) ∇× → 0 ∇× (curl) ∇· → 0 Gold paths: any two consecutive operators in these orders annihilate the field.

Worked Examples

ExampleVerifying curl(grad)=0 for a concrete potential

Given: ϕ(x,y,z)=x2y+yz3+5x\phi(x,y,z)=x^2 y + y z^3 + 5x.

Step 1 — gradient:

11
∇ϕ=(2xy+5,  x2+z3,  3yz2).\quad \nabla\phi=\big(2xy+5,\ \ x^2+z^3,\ \ 3yz^2\big).

Derivation hint: differentiate ϕ\phi separately with respect to xx, yy, zz.

Step 2 — curl, component 1: ∂y(3yz2)−∂z(x2+z3)=3z2−3z2=0.\partial_y(3yz^2)-\partial_z(x^2+z^3)=3z^2-3z^2=0.

Step 3 — component 2: ∂z(2xy+5)−∂x(3yz2)=0−0=0.\partial_z(2xy+5)-\partial_x(3yz^2)=0-0=0.

Step 4 — component 3: ∂x(x2+z3)−∂y(2xy+5)=2x−2x=0.\partial_x(x^2+z^3)-\partial_y(2xy+5)=2x-2x=0.

12
∇×(∇ϕ)=(0, 0, 0)\quad \boxed{\nabla\times(\nabla\phi)=(0,\,0,\,0)}

Derivation hint: each component cancels because the relevant cross-partials of ϕ\phi are equal.

Units check: if ϕ\phi has units of (say) electric potential V, then ∇ϕ\nabla\phi has V/m and its curl has V/m² — and that quantity is identically the zero vector. ✓

ExampleVerifying div(curl)=0 for a concrete vector field

Given: F=(xyz,  x2−y2,  z ex)\mathbf{F}=(xyz,\ \ x^2-y^2,\ \ z\,e^{x}).

Step 1 — curl:

13
∇×F=( ∂y(zex)−∂z(x2−y2),  ∂z(xyz)−∂x(zex),  ∂x(x2−y2)−∂y(xyz) ).\quad \nabla\times\mathbf{F}=\big(\,\partial_y(ze^x)-\partial_z(x^2-y^2),\ \ \partial_z(xyz)-\partial_x(ze^x),\ \ \partial_x(x^2-y^2)-\partial_y(xyz)\,\big).

Derivation hint: apply the determinant formula [3] to the three components of F\mathbf F. Evaluating: ∇×F=(0−0,  xy−zex,  2x−xz)=(0, xy−zex, 2x−xz).\nabla\times\mathbf F=\big(0-0,\ \ xy-ze^x,\ \ 2x-xz\big)=(0,\ xy-ze^x,\ 2x-xz).

Step 2 — divergence of that curl:

14
∇⋅(∇×F)=∂x(0)+∂y(xy−zex)+∂z(2x−xz).\quad \nabla\cdot(\nabla\times\mathbf F)=\partial_x(0)+\partial_y(xy-ze^x)+\partial_z(2x-xz).

Derivation hint: differentiate each curl component by its matching coordinate.

Step 3 — evaluate: ∂x(0)=0\partial_x(0)=0,  ∂y(xy−zex)=x\ \partial_y(xy-ze^x)=x,  ∂z(2x−xz)=−x\ \partial_z(2x-xz)=-x. Sum =0+x−x=0+x-x.

15
∇⋅(∇×F)=0\quad \boxed{\nabla\cdot(\nabla\times\mathbf F)=0}

Derivation hint: the +x+x and −x-x terms cancel exactly — a manifestation of the symmetry of mixed partials of F\mathbf F.

Units check: if F\mathbf F is a velocity (m/s), its curl is in s⁻¹ and the divergence of that is m⁻¹·s⁻¹ — and equals zero everywhere. ✓

Common Misconceptions

⚠️ Six common errors

1. "The converse is automatic." ∇×F=0\nabla\times\mathbf F=\mathbf 0 does not by itself imply F=∇ϕ\mathbf F=\nabla\phi, and ∇⋅G=0\nabla\cdot\mathbf G=0 does not by itself imply G=∇×A\mathbf G=\nabla\times\mathbf A. The converses need the domain to be simply connected / have trivial topology (Poincaré lemma). 2. "Identities hold for any field." They require C2C^2 smoothness. Drop continuity of mixed partials and they can fail pointwise. 3. Confusing the two zeros. ∇×(∇ϕ)\nabla\times(\nabla\phi) is the zero vector 0\mathbf 0; ∇⋅(∇×F)\nabla\cdot(\nabla\times\mathbf F) is the scalar 00. Different objects. 4. Wrong composition order. ∇⋅(∇ϕ)=∇2ϕ\nabla\cdot(\nabla\phi)=\nabla^2\phi (the Laplacian) is generally nonzero; only div-of-curl and curl-of-grad vanish. 5. Assuming ∇×(∇×F)=0\nabla\times(\nabla\times\mathbf F)=\mathbf 0. It is not; instead ∇×(∇×F)=∇(∇⋅F)−∇2F\nabla\times(\nabla\times\mathbf F)=\nabla(\nabla\cdot\mathbf F)-\nabla^2\mathbf F, generally nonzero. 6. Forgetting dimension. Curl as a vector is special to 3D; in nn dimensions the right framework is the exterior derivative dd, where the identities become d2=0d^2=0.

CounterexampleA "potential" whose curl is nonzero at a point

Let ϕ(x,y)=xy(x2−y2)x2+y2\phi(x,y)=\dfrac{xy(x^2-y^2)}{x^2+y^2} for (x,y)≠(0,0)(x,y)\neq(0,0) and ϕ(0,0)=0\phi(0,0)=0. The first partials exist everywhere, but

16
∂2ϕ∂x ∂y(0,0)=1,∂2ϕ∂y ∂x(0,0)=−1.\quad \frac{\partial^2\phi}{\partial x\,\partial y}(0,0)=1,\qquad \frac{\partial^2\phi}{\partial y\,\partial x}(0,0)=-1.

Derivation hint: compute ∂yϕ\partial_y\phi then differentiate in xx along the axis, and vice versa; the two limits differ because the second mixed partials are discontinuous at the origin. The third component of ∇×(∇ϕ)\nabla\times(\nabla\phi) at the origin is ∂x∂yϕ−∂y∂xϕ=1−(−1)=2≠0\partial_x\partial_y\phi-\partial_y\partial_x\phi = 1-(-1)=2\neq 0. The identity [4] fails precisely because ϕ\phi is not C2C^2 at the origin — confirming the hypothesis cannot be dropped.

Connections

connection

These identities sit at the crossroads of analysis, geometry, and physics.

  • <span class="wiki-link-missing">Helmholtz Decomposition</span> — Splits a field into a curl-free (gradient) part plus a divergence-free (curl) part; the two identities guarantee each piece behaves as advertised.
  • <span class="wiki-link-missing">Poincaré Lemma</span> — Provides the converses: on a contractible domain, curl-free ⇒\Rightarrow gradient and divergence-free ⇒\Rightarrow curl. The identities are the "easy direction."
  • <span class="wiki-link-missing">Exterior Derivative and Differential Forms</span> — Both identities are the single statement d2=0d^2=0: d(dω)=0d(d\omega)=0 for any form ω\omega. Gradient is dd on 00-forms, curl is dd on 11-forms, divergence is dd on 22-forms (up to Hodge stars).
  • <a class="wiki-link" href="/wiki/ndli-iit-kgp-ma10002-mathematics-ii-stokes-theorem">Stokes&#39; Theorem</a> and <span class="wiki-link-missing">Divergence Theorem</span> — Integral counterparts; "boundary of a boundary is empty" (∂2=∅\partial^2=\varnothing) is the integral shadow of d2=0d^2=0.
  • <a class="wiki-link" href="/wiki/maxwells-equations">Maxwell&#39;s Equations</a> — ∇⋅B=0\nabla\cdot\mathbf B=0 permits B=∇×A\mathbf B=\nabla\times\mathbf A; ∇×E=0\nabla\times\mathbf E=\mathbf 0 (statics) permits E=−∇ϕ\mathbf E=-\nabla\phi.
  • <span class="wiki-link-missing">Vector Potential</span> and <span class="wiki-link-missing">Scalar Potential</span> — Their very existence rests on these identities.

Real-World Applications

🔬 Where this appears

Historical Development

⏳ History
- 1746 / 1740s — Clairaut & Euler: the equality of mixed partial derivatives (Clairaut's theorem) emerges in studies of differentials and exact equations — the analytic seed of both identities.
- 1813 — Gauss: formulates the divergence (flux) theorem in work on attraction of ellipsoids, foundational for the divergence operator.
- 1828 — George Green: An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism introduces potential functions, implicitly using irrotationality of gradients.
- 1831–1839 — Carl Friedrich Gauss / Mikhail Ostrogradsky: rigorous statements of the divergence theorem (Ostrogradsky 1826/1831).
- 1849 — George Stokes: poses the curl–circulation theorem (the "Stokes theorem") in a Cambridge exam, set by Kelvin (1850); ties curl to loop integrals.
- 1860s–1873 — James Clerk Maxwell: systematizes the operators in his electromagnetic work and Treatise; uses B=∇×A\mathbf B=\nabla\times\mathbf A and E=−∇ϕ\mathbf E=-\nabla\phi, embedding both identities into physics.
- 1880s — J. Willard Gibbs & Oliver Heaviside: independently develop modern vector analysis and the ∇\nabla notation; the identities take the form taught today.
- 1899 — Henri Poincaré: establishes the Poincaré lemma, giving the topological converses and revealing the identities as one half of a deep cohomological duality.
- 1930s–1945 — Élie Cartan / Georges de Rham: the exterior calculus recasts everything as d2=0d^2=0, and de Rham cohomology measures exactly when the converses fail.

Subtleties & Limits

open-problem · Where the easy identities meet hard mathematics

The identities ∇×∇=0\nabla\times\nabla=0 and ∇⋅(∇×)=0\nabla\cdot(\nabla\times)=0 are elementary, but their failure of converse opens onto genuinely deep terrain. On a domain with holes, a curl-free field need not be a gradient (e.g., the angle-field F=(−y,x,0)/(x2+y2)\mathbf F=(-y,x,0)/(x^2+y^2) is curl-free off the zz-axis yet has nonzero circulation), and a divergence-free field need not be a global curl. Measuring this gap is the subject of de Rham cohomology: the dimensions of the cohomology groups count the independent obstructions. Open and active questions live in the regularity-limited setting (when F\mathbf F is only weakly differentiable, as in PDE theory and weak solutions of Maxwell's and Navier–Stokes equations), in discrete settings (constructing finite-element complexes that preserve d2=0d^2=0 for high-order and on general meshes — "finite element exterior calculus"), and in infinite-dimensional or singular geometries where the smooth Poincaré lemma must be replaced by sheaf-theoretic or distributional machinery.

See Also

Gradient, Divergence, and Curl · Clairaut's Theorem on Mixed Partials · Helmholtz Decomposition · Poincaré Lemma · Exterior Derivative and Differential Forms · Stokes' Theorem · Divergence Theorem · Maxwell's Equations · Vector Potential · Scalar Potential · Levi-Civita Symbol and Index Notation