Why Conservative Fields Have Zero Curl

Introduction

A conservative vector field is a gradient field: its components come from the first partial derivatives of a scalar potential. Curl, by contrast, measures a field’s local tendency to circulate. The connection between these ideas is precise: when a potential has continuous second partial derivatives, the mixed partial derivatives agree, and the curl of its gradient vanishes.

The regularity condition matters. The conclusion uses equality of mixed partial derivatives, which is guaranteed when the relevant second partial derivatives are continuous. The result is a necessary consequence of having such a potential; by itself, it does not establish that every field with zero curl has a potential.

Main result: If a vector field is the gradient of a twice continuously differentiable potential on an open region, then its curl is zero throughout that region.

Definitions and conditions

Let \(D\) be an open subset of \(\mathbb{R}^3\). A vector field \(\mathbf{F}:D\to\mathbb{R}^3\) is conservative if there is a scalar function \(\phi:D\to\mathbb{R}\), called a potential, such that \(\mathbf{F}=\nabla\phi\). In coordinates, for \(\mathbf{F}=\langle P,Q,R\rangle\), this means

P=\frac{\partial \phi}{\partial x},\qquad Q=\frac{\partial \phi}{\partial y},\qquad R=\frac{\partial \phi}{\partial z}.

The curl of a continuously differentiable field \(\mathbf{F}=\langle P,Q,R\rangle\) is the vector field

\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.

In the plane, a field \(\mathbf{F}=\langle P,Q\rangle\) has scalar curl \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\). It is the \(z\)-component of the three-dimensional curl when the planar field is viewed as \(\langle P,Q,0\rangle\), with no dependence on \(z\).

Regularity condition: The theorem below applies on an open region where a potential \(\phi\) has continuous second partial derivatives. This condition is written \(\phi\in C^2(D)\). It ensures that the mixed partial derivatives used in the curl are equal.

Why the mixed partial derivatives agree

The equality of mixed partial derivatives is the key step. Here is a justification under the continuity condition used in the theorem. Suppose \(g\) has continuous second partial derivatives near \((a,b)\). For sufficiently small positive \(h\) and \(k\), form the rectangular difference

\Delta = g(a+h,b+k)-g(a+h,b)-g(a,b+k)+g(a,b).

Apply the one-variable Fundamental Theorem of Calculus first in the \(x\)-direction, and then in the \(y\)-direction. The result is

\Delta = \int_b^{b+k}\int_a^{a+h}g_{xy}(x,y)\,dx\,dy.

Alternatively, apply it first in the \(y\)-direction and then in the \(x\)-direction. This gives

\Delta = \int_a^{a+h}\int_b^{b+k}g_{yx}(x,y)\,dy\,dx.

Divide each expression by \(hk\). As \(h\) and \(k\) approach zero through positive values, continuity of the second partial derivatives ensures that each average approaches the value of its integrand at \((a,b)\). Both averages are the same number, \(\Delta/(hk)\), so their limits are equal. Therefore

g_{xy}(a,b)=g_{yx}(a,b).

Since \((a,b)\) was arbitrary, the two mixed partial derivatives agree throughout any region where the stated continuity condition holds. The same argument applies to each pair of coordinates in three variables.

The zero-curl theorem

Theorem: Let \(D\) be open in \(\mathbb{R}^3\), and suppose \(\phi\in C^2(D)\). If \(\mathbf{F}=\nabla\phi\), then \(\nabla\times\mathbf{F}=\mathbf{0}\) at every point of \(D\). In two dimensions, if \(\mathbf{F}=\langle\phi_x,\phi_y\rangle\), its scalar curl is zero throughout \(D\).

To see why, write the components of the gradient as \(P=\phi_x\), \(Q=\phi_y\), and \(R=\phi_z\). Substitute these into the definition of curl. Each component becomes a difference of mixed partial derivatives:

\nabla\times\mathbf{F} = \left\langle \phi_{zy}-\phi_{yz}, \phi_{xz}-\phi_{zx}, \phi_{yx}-\phi_{xy} \right\rangle.

Because \(\phi\in C^2(D)\), equality of mixed partial derivatives gives \(\phi_{zy}=\phi_{yz}\), \(\phi_{xz}=\phi_{zx}\), and \(\phi_{yx}=\phi_{xy}\). Every component is consequently zero. In the plane the same reasoning reads

\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = \phi_{yx}-\phi_{xy} = 0.

This proves the result: the zero curl follows from the gradient representation and the equality of mixed partial derivatives. The proof does not require calculating any line integrals.

Worked examples

Worked Example: A conservative field in the plane

Let \(\phi(x,y)=x^2y+3x+2y^2\), and define \(\mathbf{F}=\nabla\phi=\langle P,Q\rangle\). Find the field and verify that its scalar curl is zero.

Differentiate the potential with respect to each coordinate:

P=\phi_x=2xy+3,\qquad Q=\phi_y=x^2+4y.

For the scalar curl, differentiate \(Q\) with respect to \(x\) and \(P\) with respect to \(y\):

Q_x=2x,\qquad P_y=2x,\qquad Q_x-P_y=2x-2x=0.

Check the result against the potential: differentiating the stated components gives \(\phi_x=2xy+3=P\) and \(\phi_y=x^2+4y=Q\). Thus the field really is the gradient of \(\phi\), and its computed curl is zero.

Worked Example: A three-dimensional polynomial potential

Take \(\phi(x,y,z)=xyz+x^2+\frac{3}{2}y^2+2z^2\). Compute its gradient and check all three components of the curl.

The first partial derivatives are

P=\phi_x=yz+2x,\qquad Q=\phi_y=xz+3y,\qquad R=\phi_z=xy+4z.

Now calculate the three differences in the curl formula separately:

R_y=x,\qquad Q_z=x,\qquad R_y-Q_z=x-x=0,

P_z=y,\qquad R_x=y,\qquad P_z-R_x=y-y=0,

Q_x=z,\qquad P_y=z,\qquad Q_x-P_y=z-z=0.

Therefore \(\nabla\times\mathbf{F}=\langle 0,0,0\rangle\). As a check on the field itself, differentiating \(\phi\) gives exactly \(yz+2x\), \(xz+3y\), and \(xy+4z\), the three stated components.

Worked Example: A potential with exponential and trigonometric terms

Let \(\phi(x,y,z)=e^{xy}+z\cos x+y^3\). Find the components of \(\nabla\phi\), then verify that each component of the curl vanishes.

Using the chain rule for \(e^{xy}\), the gradient components are

P=\phi_x=ye^{xy}-z\sin x,\qquad Q=\phi_y=xe^{xy}+3y^2,\qquad R=\phi_z=\cos x.

For the first curl component, \(R_y=0\) and \(Q_z=0\), so \(R_y-Q_z=0\). For the second, \(P_z=-\sin x\) and \(R_x=-\sin x\), so \(P_z-R_x=0\). For the third, the product rule gives

Q_x=e^{xy}+xye^{xy},\qquad P_y=e^{xy}+xye^{xy},

and hence \(Q_x-P_y=0\). Thus the full curl is \(\langle 0,0,0\rangle\). To check the gradient, these three components were obtained directly by differentiating the given \(\phi\) with respect to \(x\), \(y\), and \(z\), respectively.

A common mistake: differentiating the wrong component

The planar curl uses \(Q_x-P_y\): differentiate the second component with respect to \(x\), then subtract the derivative of the first component with respect to \(y\). Reversing one of these coordinate choices can produce a false nonzero result, even for a gradient field.

For example, let \(\phi(x,y)=x^2y\). Then \(P=\phi_x=2xy\) and \(Q=\phi_y=x^2\). An incorrect calculation might use \(Q_x-P_x\), giving \(2x-2y\), which is \(2\) at \((1,0)\). But \(P_x\) is not the derivative required in the curl. The correct calculation is

$$ Q_x-P_y=2x-2x=0. $$

Always match each derivative to the formula before evaluating it. The equality responsible for cancellation is \(Q_x=\phi_{yx}\) and \(P_y=\phi_{xy}\), not an equality involving \(P_x\).

What the result tells us

Zero curl is a necessary local consequence of having a \(C^2\) potential. It can therefore be used as a consistency check: if a field known to come from such a potential has a nonzero curl at a point, there is an error in the field, the differentiation, or the stated assumptions.

The implication proved here runs from a sufficiently smooth potential to zero curl. The reverse direction is a separate question and requires attention to the field’s domain. A calculation of zero curl alone should not be presented as a proof that a potential exists.

Takeaway: For \(\mathbf{F}=\nabla\phi\), the components of \(\nabla\times\mathbf{F}\) are differences of mixed second partial derivatives of \(\phi\). If those derivatives are continuous, the mixed partials agree and each difference is zero.

Check Your Understanding

Use the curl formulas and the role of the potential’s regularity to answer these questions.

For a planar gradient field \(\mathbf{F}=\langle\phi_x,\phi_y\rangle\), write its scalar curl in terms of mixed partial derivatives of \(\phi\).
Why does the theorem require continuity of the second partial derivatives of the potential?
If \(\mathbf{F}=\langle P,Q,R\rangle=\nabla\phi\), write the second component of \(\nabla\times\mathbf{F}\) and identify the mixed partial derivatives that cancel.
For \(\phi(x,y)=x^2y+xy^2\), calculate \(P\), \(Q\), and \(Q_x-P_y\).
Does proving \(\nabla\times\mathbf{F}=\mathbf{0}\) by itself establish that \(\mathbf{F}\) is conservative? Explain what direction the theorem proves.