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Curl and Divergence · Tutorial 908 of 999

Divergence of Curl and Curl of Curl

Compute divergence of curl and curl of curl in Cartesian coordinates, justify the identities, and apply them to simplify vector-field expressions.

Advanced 9 min read

What You'll Learn

  • State the smoothness needed for the Cartesian identities
  • Compute divergence of a curl from its components
  • Expand curl of a curl and relate it to gradient, divergence, and the componentwise Laplacian
  • Verify the identities by direct calculation in polynomial examples
  • Use the identities to simplify expressions involving a vector field and its derivatives

Why These Identities Matter

A vector field can be differentiated in more than one way. Divergence measures a scalar change associated with a field, while curl produces another vector field. Combining these operations gives expressions such as the divergence of a curl and the curl of a curl.

These combinations are useful because their component calculations simplify in systematic ways. Under suitable smoothness assumptions, the divergence of a curl vanishes, and the curl of a curl can be rewritten using the gradient of the divergence and the Laplacian. The assumptions matter: the calculations involve second partial derivatives, and rearranging mixed partials requires those derivatives to be well behaved.

Central identities: For a twice continuously differentiable vector field \(\mathbf{F}\) on an open subset of \(\mathbb{R}^3\), \(\nabla\cdot(\nabla\times\mathbf{F})=0\), and \(\nabla\times(\nabla\times\mathbf{F})=\nabla(\nabla\cdot\mathbf{F})-\nabla^2\mathbf{F}\). Here the vector Laplacian \(\nabla^2\mathbf{F}\) applies the scalar Laplacian to each component.

Definitions and Smoothness

Write a Cartesian vector field as \(\mathbf{F}=(P,Q,R)\), where \(P,Q,R\) are functions of \(x,y,z\). Its divergence and curl are

$$ \nabla\cdot\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}, \qquad \nabla\times\mathbf{F} = \left( \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right). $$

The scalar Laplacian of a twice differentiable scalar function \(g\) is \(\nabla^2g=g_{xx}+g_{yy}+g_{zz}\). For a vector field, the vector Laplacian is defined componentwise:

$$ \nabla^2\mathbf{F} = \left( \nabla^2P,\nabla^2Q,\nabla^2R \right). $$

For the identities in this tutorial, assume that \(P,Q,R\) have continuous second partial derivatives on an open set. This \(C^2\) condition lets us use equality of mixed partial derivatives, such as \(P_{xy}=P_{yx}\). The open-set condition ensures the partial derivatives are defined in a neighborhood of each point where the identities are asserted.

Worked Example: Divergence of a Curl

Worked Example: Divergence of a Curl

Let \(\mathbf{F}=(x^2y,y^2z,z^2x)\). Find \(\nabla\cdot(\nabla\times\mathbf{F})\) by first computing the curl.

Set \(P=x^2y\), \(Q=y^2z\), and \(R=z^2x\). The first component of the curl is \(R_y-Q_z=0-y^2=-y^2\). The second is \(P_z-R_x=0-z^2=-z^2\). The third is \(Q_x-P_y=0-x^2=-x^2\). Thus

$$ \nabla\times\mathbf{F}=(-y^2,-z^2,-x^2). $$

Now differentiate each component with respect to its corresponding coordinate:

$$ \nabla\cdot(\nabla\times\mathbf{F}) = \frac{\partial(-y^2)}{\partial x} + \frac{\partial(-z^2)}{\partial y} + \frac{\partial(-x^2)}{\partial z} = 0+0+0 = 0. $$

This verifies the identity for this field. In particular, each component of the curl depends only on a coordinate different from the one used to take its divergence derivative.

Why the Divergence of a Curl Is Zero

The identity follows by expanding the definitions and pairing the mixed partial derivatives. Let \(\mathbf{F}=(P,Q,R)\) be \(C^2\). First write the curl, then take its divergence:

$$ \begin{aligned} \nabla\cdot(\nabla\times\mathbf{F}) &= \frac{\partial}{\partial x}(R_y-Q_z) + \frac{\partial}{\partial y}(P_z-R_x) + \frac{\partial}{\partial z}(Q_x-P_y)\\ &= R_{yx}-Q_{zx}+P_{zy}-R_{xy}+Q_{xz}-P_{yz}. \end{aligned} $$

Because the second partial derivatives are continuous, the order of differentiation can be exchanged. Thus \(R_{yx}=R_{xy}\), \(Q_{zx}=Q_{xz}\), and \(P_{zy}=P_{yz}\). Substituting those equalities gives

$$ (R_{xy}-R_{xy})+(-Q_{xz}+Q_{xz})+(P_{yz}-P_{yz})=0. $$

Every term cancels with its matching mixed partial. This proves the identity wherever the field is \(C^2\). The result is a pointwise statement; it does not require an integral theorem or any assumptions about the shape of a region.

Theorem: If \(\mathbf{F}\) has continuous second partial derivatives, then \(\nabla\times\mathbf{F}\) has zero divergence. The vanishing comes from cancellation of mixed partial derivatives, not from a special choice of \(\mathbf{F}\).

Worked Example: Curl of a Curl

Worked Example: Curl of a Curl

For \(\mathbf{F}=(xy,yz,zx)\), compute \(\nabla\times(\nabla\times\mathbf{F})\) directly and compare it with the identity.

First compute the curl. Its components are \(R_y-Q_z=0-y=-y\), \(P_z-R_x=0-z=-z\), and \(Q_x-P_y=0-x=-x\). Therefore

$$ \nabla\times\mathbf{F}=(-y,-z,-x). $$

Take the curl of this new vector field. Its first component is \(\partial(-x)/\partial y-\partial(-z)/\partial z=0-(-1)=1\). Its second component is \(\partial(-y)/\partial z-\partial(-x)/\partial x=0-(-1)=1\). Its third component is \(\partial(-z)/\partial x-\partial(-y)/\partial y=0-(-1)=1\). Hence

$$ \nabla\times(\nabla\times\mathbf{F})=(1,1,1). $$

Now check using the identity. The divergence of \(\mathbf{F}\) is \(y+z+x\), so its gradient is \(\nabla(\nabla\cdot\mathbf{F})=(1,1,1)\). Each component \(xy\), \(yz\), and \(zx\) has zero scalar Laplacian, since its second derivative with respect to each of \(x,y,z\) is zero. Thus \(\nabla^2\mathbf{F}=(0,0,0)\), and

$$ \nabla(\nabla\cdot\mathbf{F})-\nabla^2\mathbf{F} = (1,1,1)-(0,0,0) = (1,1,1). $$

The identity agrees with the direct calculation.

Deriving the Curl-of-Curl Identity

For \(\mathbf{F}=(P,Q,R)\), the curl is \((R_y-Q_z,\ P_z-R_x,\ Q_x-P_y)\). Taking its curl again gives the following first component:

$$ \begin{aligned} \bigl(\nabla\times(\nabla\times\mathbf{F})\bigr)_1 &= \frac{\partial}{\partial y}(Q_x-P_y) - \frac{\partial}{\partial z}(P_z-R_x)\\ &= Q_{xy}-P_{yy}-P_{zz}+R_{xz}. \end{aligned} $$

On the other hand, the first component of \(\nabla(\nabla\cdot\mathbf{F})-\nabla^2\mathbf{F}\) is

$$ \begin{aligned} \frac{\partial}{\partial x}(P_x+Q_y+R_z) - (P_{xx}+P_{yy}+P_{zz}) &= Q_{yx}+R_{zx}-P_{yy}-P_{zz}. \end{aligned} $$

Since \(Q_{xy}=Q_{yx}\) and \(R_{xz}=R_{zx}\), these first components agree. The same expansion for the second and third components is

$$ \begin{aligned} \bigl(\nabla\times(\nabla\times\mathbf{F})\bigr)_2 &= R_{yz}-Q_{zz}-Q_{xx}+P_{yx}\\ &= \frac{\partial}{\partial y}(P_x+Q_y+R_z) - (Q_{xx}+Q_{yy}+Q_{zz}),\\[4pt] \bigl(\nabla\times(\nabla\times\mathbf{F})\bigr)_3 &= P_{zx}-R_{xx}-R_{yy}+Q_{zy}\\ &= \frac{\partial}{\partial z}(P_x+Q_y+R_z) - (R_{xx}+R_{yy}+R_{zz}). \end{aligned} $$

In the second line, equality uses \(P_{yx}=P_{xy}\) and \(R_{yz}=R_{zy}\); in the third, it uses \(P_{zx}=P_{xz}\) and \(Q_{zy}=Q_{yz}\). The three component equalities establish the vector identity:

$$ \nabla\times(\nabla\times\mathbf{F}) = \nabla(\nabla\cdot\mathbf{F})-\nabla^2\mathbf{F}. $$

The minus sign before the vector Laplacian is essential. The gradient of divergence is a vector, as is the componentwise Laplacian, so the right-hand side is a vector difference.

Worked Example: Simplifying a Vector-Field Expression

Worked Example: Simplifying with the Curl-of-Curl Identity

Let \(\mathbf{F}=(x^2y,y^2z,z^2x)\). Simplify \(\nabla\times(\nabla\times\mathbf{F})+\nabla^2\mathbf{F}\) using the identity, and check the result componentwise.

From the earlier component definitions, the divergence is

$$ \nabla\cdot\mathbf{F} = \frac{\partial(x^2y)}{\partial x} + \frac{\partial(y^2z)}{\partial y} + \frac{\partial(z^2x)}{\partial z} = 2xy+2yz+2zx. $$

Its gradient is \((2y+2z,\ 2x+2z,\ 2x+2y)\). The component Laplacians are \(\nabla^2(x^2y)=2y\), \(\nabla^2(y^2z)=2z\), and \(\nabla^2(z^2x)=2x\). Therefore

$$ \nabla^2\mathbf{F}=(2y,2z,2x). $$

Adding \(\nabla^2\mathbf{F}\) to both sides of the curl-of-curl identity gives \(\nabla\times(\nabla\times\mathbf{F})+\nabla^2\mathbf{F}=\nabla(\nabla\cdot\mathbf{F})\). Thus the expression simplifies to

$$ (2y+2z,2x+2z,2x+2y). $$

To check, the curl of \(\mathbf{F}\) is \((-y^2,-z^2,-x^2)\), so the curl of that field is \((2z,2x,2y)\). Adding the previously computed vector Laplacian gives \((2z,2x,2y)+(2y,2z,2x)=(2y+2z,2x+2z,2x+2y)\), as required.

A Common Sign Error

A frequent mistake is to write the curl-of-curl identity with a plus sign before the vector Laplacian. For example, the incorrect rearrangement

$$ \nabla\times(\nabla\times\mathbf{F}) = \nabla(\nabla\cdot\mathbf{F})+\nabla^2\mathbf{F} $$

fails for \(\mathbf{F}=(x^2,0,0)\). Its divergence is \(2x\), so \(\nabla(\nabla\cdot\mathbf{F})=(2,0,0)\), while \(\nabla^2\mathbf{F}=(2,0,0)\). Directly, \(\nabla\times\mathbf{F}=(0,0,0)\), so its curl is also zero. The correct identity gives \((2,0,0)-(2,0,0)=(0,0,0)\). The incorrect plus sign gives \((4,0,0)\), contradicting the direct calculation.

For a quick sign check, remember that the vector Laplacian is subtracted when expressing curl of curl as gradient of divergence minus Laplacian. If instead solving for the gradient of divergence, rearrange carefully: \(\nabla(\nabla\cdot\mathbf{F})=\nabla\times(\nabla\times\mathbf{F})+\nabla^2\mathbf{F}\).

Using the Identities Carefully

These identities are particularly useful when an expression contains repeated differential operations. They can replace a nested curl by a gradient and a componentwise Laplacian, or show immediately that a divergence-of-curl term contributes zero. Before applying them, check that the field is defined on an open region and has continuous second partial derivatives there. They are Cartesian identities in this form; the displayed component Laplacian should not be transferred unchanged to another coordinate system.

The condition is sufficient for the identities as stated. If a field has singularities or lacks the required smoothness, the componentwise argument may not apply at those points. In such cases, do not infer the identity there without separately examining the field and its domain.

Check Your Understanding

Try each question by writing out the relevant Cartesian components.

  1. For \(\mathbf{F}=(P,Q,R)\), which mixed partial derivatives cancel when computing \(\nabla\cdot(\nabla\times\mathbf{F})\)?
  2. State the \(C^2\) identity for \(\nabla\times(\nabla\times\mathbf{F})\), including the sign of the vector Laplacian term.
  3. For \(\mathbf{F}=(x^2,0,0)\), compute its divergence, the gradient of its divergence, and its vector Laplacian. How do these confirm curl of curl?
  4. Why is continuity of the second partial derivatives relevant to the derivations?
  5. Rewrite \(\nabla(\nabla\cdot\mathbf{F})\) in terms of \(\nabla\times(\nabla\times\mathbf{F})\) and \(\nabla^2\mathbf{F}\).

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