What Is a Divergence-Free Vector Field?

A divergence-free vector field is a vector field whose divergence is zero throughout the region being considered. The condition describes the field locally: around every sufficiently small volume, the amount flowing in balances the amount flowing out.

For a vector field in three-dimensional Cartesian coordinates,

$$\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle,$$

its divergence is

$$\operatorname{div}\mathbf{F}=\nabla\cdot\mathbf{F} =\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.$$

Therefore, $\mathbf{F}$ is divergence-free on a region $D$ precisely when

$$\nabla\cdot\mathbf{F}=0$$

at every point of $D$.

Physical meaning of zero divergence

Think of a vector field as the velocity field of a fluid. Positive divergence at a point indicates local expansion: more fluid leaves a tiny region than enters it. Negative divergence indicates local compression or a sink. Zero divergence means there is no net creation, destruction, expansion, or compression of fluid at that point.

This is why divergence-free fields are often called solenoidal fields. In fluid mechanics, the velocity field of an incompressible fluid is commonly modeled as divergence-free. In electromagnetism, the magnetic field satisfies

$$\nabla\cdot\mathbf{B}=0,$$

which expresses the absence of magnetic monopoles in the standard theory.

Zero divergence does not mean that the vector field itself is zero. A field can have large, rapidly varying vectors and still have zero divergence if the component derivatives cancel.

How to test a vector field

Use this procedure:

1. Identify the three components $P$, $Q$, and $R$.
2. Differentiate $P$ with respect to $x$, $Q$ with respect to $y$, and $R$ with respect to $z$.
3. Add the three results.
4. Check whether the sum is zero throughout the specified domain.

Worked example

Consider

$$\mathbf{F}(x,y,z)=\langle 2xy+z,\ x^2-3y,\ 4z-2x\rangle.$$

The relevant derivatives are

$$\frac{\partial P}{\partial x}=2y, \qquad \frac{\partial Q}{\partial y}=-3, \qquad \frac{\partial R}{\partial z}=4.$$

Thus,

$$\nabla\cdot\mathbf{F}=2y-3+4=2y+1.$$

This is not zero everywhere, so the field is not divergence-free on $\mathbb{R}^3$. It happens to have zero divergence on the plane $y=-\tfrac12$, but that does not make it divergence-free on a three-dimensional region.

Now consider

$$\mathbf{G}(x,y,z)=\langle yz,\ xz,\ xy\rangle.$$

Then

$$\begin{aligned} \nabla\cdot\mathbf{G} &=\frac{\partial (yz)}{\partial x} +\frac{\partial (xz)}{\partial y} +\frac{\partial (xy)}{\partial z}\\ &=0+0+0\\ &=0. \end{aligned}$$

Therefore, $\mathbf{G}$ is divergence-free everywhere in $\mathbb{R}^3$.

Divergence-free fields and flux

The divergence theorem states that, for a sufficiently smooth vector field and a solid region $V$ with outward-oriented boundary surface $\partial V$,

$$\iint_{\partial V}\mathbf{F}\cdot\mathbf{n}\,dS =\iiint_V \nabla\cdot\mathbf{F}\,dV.$$

If $\nabla\cdot\mathbf{F}=0$ throughout $V$, then

$$\iint_{\partial V}\mathbf{F}\cdot\mathbf{n}\,dS=0.$$

So the total outward flux through any closed surface surrounding a region is zero. This does not imply that the flux through each part of the surface is zero. One part may have positive outward flux while another has an equal negative flux.

For example, the constant field

$$\mathbf{F}=\langle 1,0,0\rangle$$

is divergence-free. Through a closed box, the flux entering through the face with outward normal $\langle-1,0,0\rangle$ exactly cancels the flux leaving through the opposite face. The other four faces contribute zero.

Constructing divergence-free vector fields

A useful way to produce divergence-free fields is to take the curl of a sufficiently smooth vector potential $\mathbf{A}$:

$$\mathbf{F}=\nabla\times\mathbf{A}.$$

The identity

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

guarantees that $\mathbf{F}$ is divergence-free wherever the required derivatives are continuous.

For instance, let

$$\mathbf{A}(x,y,z)=\langle 0,0,x^2+y^2\rangle.$$

Then

$$\nabla\times\mathbf{A} =\left\langle 2y,-2x,0\right\rangle.$$

Checking directly,

$$\nabla\cdot\langle 2y,-2x,0\rangle =\frac{\partial(2y)}{\partial x} +\frac{\partial(-2x)}{\partial y} +\frac{\partial 0}{\partial z} =0.$$

In two dimensions, a similar construction uses a stream function. If $\psi(x,y)$ is twice continuously differentiable, then

$$\mathbf{F}=\left\langle \frac{\partial\psi}{\partial y},-\frac{\partial\psi}{\partial x}\right\rangle$$

is divergence-free because

$$\nabla\cdot\mathbf{F} =\frac{\partial^2\psi}{\partial x\partial y} -\frac{\partial^2\psi}{\partial y\partial x}=0.$$

The equality of the mixed partial derivatives is the reason for the cancellation.

Common mistakes

Differentiating with respect to the wrong variable

For $\mathbf{F}=\langle P,Q,R\rangle$, divergence uses $P_x+Q_y+R_z$. It does not use all nine first partial derivatives; that collection belongs to the Jacobian matrix.

Confusing divergence with curl

Divergence is a scalar:

$$\nabla\cdot\mathbf{F}.$$

Curl is a vector:

$$\nabla\times\mathbf{F}.$$

A divergence-free field need not have zero curl, and a curl-free field need not be divergence-free. For example, $\langle-y,x,0\rangle$ has zero divergence but nonzero curl.

Checking only one point

A field is divergence-free on a region only if its divergence vanishes at every point in that region. Finding one point where the divergence is zero is not enough.

Applying the flux conclusion to an open surface

Zero total flux is guaranteed for closed surfaces under the divergence theorem. An open surface can have nonzero flux even when the field is divergence-free.