6.5 Divergence and Curl

Big idea

Two particular combinations of first partial derivatives kept appearing in the previous section. One was $P_x + Q_y$, which governed flux across a closed curve. The other was $Q_x - P_y$, which governed circulation around it. This section gives them names, extends both to three dimensions, and explains what each one measures. After this they stop being bookkeeping and become the operators that the remaining theorems are stated in.

Divergence is a scalar. At a point it reports the rate at which the field is spreading out there, measured per unit volume. If you imagine the field as the velocity of a fluid, positive divergence marks a source, where material is appearing, and negative divergence marks a sink. Zero divergence everywhere describes an incompressible flow: whatever enters a small box leaves it.

Curl is a vector. At a point it reports how much the field twists there, with the direction of the vector giving the axis of the twist by the right-hand rule and its length giving the strength. A paddle wheel dropped into the flow spins fastest when its axis is lined up with the curl, and the angular speed it reaches is half the length of the curl vector. Zero curl everywhere describes an irrotational flow, which is the three dimensional version of the cross-partials test.

Two identities tie the operators together and to the gradient: the curl of any gradient field is zero, and the divergence of any curl is zero. They are both consequences of the equality of mixed partial derivatives, and both are used constantly, once as a quick disqualification test and once as a structural fact about the last two theorems in the chapter.

Decoder

Divergence is the dot product of the del operator with the field and is a scalar; curl is the cross product of the del operator with the field and is a vector.

Treating $\nabla = \langle \partial/\partial x, \partial/\partial y, \partial/\partial z\rangle$ as a vector of operators is a memory device, not a definition, but it predicts everything correctly: a dot product of two vectors is a scalar and a cross product is a vector, which is exactly the type of each answer. It also explains why the cross product determinant is the fastest way to compute a curl by hand.

The order matters in a way that ordinary dot products do not prepare you for. Writing $\mathbf{F}\cdot\nabla$ produces an operator, not a number, so $\nabla\cdot\mathbf{F}$ cannot be reversed. The same holds for the cross product, where the usual antisymmetry does not apply because one factor is not a vector.

The classic mistake is a type error: reporting a vector for divergence or a scalar for curl. Check the shape of your answer before anything else. A second frequent slip is the middle sign in the curl determinant, which is negative; expanding along the first row with the alternating signs $+, -, +$ prevents it.

Definitions and results

Divergence. For $\mathbf{F} = \langle P, Q, R\rangle$ with differentiable components, $$ \operatorname{div}\mathbf{F} = \nabla\cdot\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $$ In the plane the third term is absent. The result is a scalar function.

Curl. For the same field, $$ \operatorname{curl}\mathbf{F} = \nabla\times\mathbf{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \;\; \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \;\; \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right\rangle $$ equivalently the formal determinant with rows $\mathbf{i},\mathbf{j},\mathbf{k}$, then the three partial operators, then $P, Q, R$. The result is a vector field.

Planar fields. A field $\langle P, Q\rangle$ in the plane can be read as $\langle P, Q, 0\rangle$ with no $z$ dependence. Its curl is then $\langle 0, 0, Q_x - P_y\rangle$, pointing straight out of the plane, so the scalar $Q_x - P_y$ in the circulation form of Green’s theorem is the only surviving component.

Physical reading of divergence. The flux of $\mathbf{F}$ outward across the boundary of a small region, divided by the volume of that region, tends to $\operatorname{div}\mathbf{F}$ at the point as the region shrinks. Positive means a source, negative a sink, zero means balanced.

Physical reading of curl. The circulation of $\mathbf{F}$ around a small closed curve with unit normal $\mathbf{n}$, divided by the area enclosed, tends to $(\operatorname{curl}\mathbf{F})\cdot\mathbf{n}$. So curl is a circulation density, and the component you get depends on the plane you measure in.

Curl of a gradient is zero. If $f$ has continuous second partials, $\nabla\times(\nabla f) = \mathbf{0}$. Each component is a difference of mixed second partials taken in the two orders, and those agree. Consequently a field with nonzero curl cannot be conservative.

Divergence of a curl is zero. If $\mathbf{F}$ has continuous second partials, $\nabla\cdot(\nabla\times\mathbf{F}) = 0$, again by cancellation of mixed partials in pairs. So a field with nonzero divergence is not the curl of anything.

Test for conservative fields in space. On a simply connected open region, a field with continuous partials is conservative if and only if its curl is zero. This restates the three cross-partial equations as a single vector condition.

Irrotational, incompressible, harmonic. A field is irrotational where its curl vanishes and incompressible, or solenoidal, where its divergence vanishes. Applying divergence to a gradient gives the Laplacian $\nabla\cdot\nabla f = f_{xx} + f_{yy} + f_{zz}$, written $\nabla^2 f$; its solutions of $\nabla^2 f = 0$ are the harmonic functions. A conservative field that is also incompressible has a harmonic potential.

Worked examples

Computing both for one field

Let $\mathbf{F} = \langle xy, \; yz, \; zx\rangle$.

The divergence is the sum of the three matching partials: $$ \operatorname{div}\mathbf{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(zx) = y + z + x $$

For the curl, take the three differences in order. The first component is $\partial_y(zx) - \partial_z(yz) = 0 - y = -y$. The second is $\partial_z(xy) - \partial_x(zx) = 0 - z = -z$. The third is $\partial_x(yz) - \partial_y(xy) = 0 - x = -x$. So $$ \operatorname{curl}\mathbf{F} = \langle -y, \; -z, \; -x\rangle $$ Check the second identity on this answer: the divergence of $\langle -y, -z, -x\rangle$ is $\partial_x(-y) + \partial_y(-z) + \partial_z(-x) = 0 + 0 + 0 = 0$, as it must be for any curl.

Rotation and radial flow

The field $\mathbf{F} = \langle -y, \; x, \; 0\rangle$ is the rigid rotation from the first section. Its divergence is $0 + 0 + 0 = 0$, so the flow neither creates nor destroys material, which matches the picture of a disk spinning without stretching. Its curl has first two components zero and third component $\partial_x(x) - \partial_y(-y) = 1 + 1 = 2$, so $\operatorname{curl}\mathbf{F} = \langle 0,0,2\rangle$. The axis is the $z$-axis, and the length $2$ is twice the angular speed $1$, exactly as the paddle wheel reading predicts.

The radial field $\mathbf{G} = \langle x, y, z\rangle$ behaves oppositely. Its divergence is $1+1+1 = 3$, a uniform source, and its curl is $\langle 0-0,\; 0-0,\; 0-0\rangle = \mathbf{0}$, since each component of $\mathbf{G}$ involves only its own variable. Spreading without spinning.

Using curl as a conservative test

Let $\mathbf{F} = \langle y+z, \; x+z, \; x+y\rangle$. Compute the curl: $$ \operatorname{curl}\mathbf{F} = \langle 1 - 1, \;\; 1 - 1, \;\; 1 - 1\rangle = \mathbf{0} $$ The domain is all of space, which is simply connected, so the field is conservative. Recovering a potential, antidifferentiating the first component in $x$ gives $f = xy + xz + g(y,z)$; matching $f_y = x + g_y$ to $x + z$ forces $g_y = z$, so $g = yz + h(z)$; matching $f_z = x + y + h'(z)$ to $x+y$ forces $h' = 0$. Hence $f = xy + yz + xz$, and differentiating confirms $\nabla f = \langle y+z, x+z, x+y\rangle$.

Confirming the gradient identity

Take $f = x^2 y z^3$, so $\nabla f = \langle 2xyz^3, \; x^2 z^3, \; 3x^2 y z^2\rangle$. The three curl components are $$ \partial_y(3x^2yz^2) - \partial_z(x^2z^3) = 3x^2z^2 - 3x^2z^2 = 0 $$ $$ \partial_z(2xyz^3) - \partial_x(3x^2yz^2) = 6xyz^2 - 6xyz^2 = 0 $$ $$ \partial_x(x^2z^3) - \partial_y(2xyz^3) = 2xz^3 - 2xz^3 = 0 $$ Every component cancels in a pair, which is the whole proof of the identity written out for one function. Nothing about this particular $f$ mattered except that its second partials are continuous.

Practice

These problems ask for divergence and curl of given fields in two and three dimensions, for the physical reading of the results, for the use of curl as a test for a conservative field, and for the two identities applied as shortcuts. Check the type of each answer before checking the algebra.

Quiz

Five items on computing divergence and curl, interpreting them, and applying the curl of a gradient and divergence of a curl identities.