Conservative Fields and Potential Functions

From a Scalar Function to a Vector Field

A vector field assigns a vector to each point in its domain. In two dimensions, write such a field as \(\mathbf{f}(x,y)=\langle P(x,y),Q(x,y)\rangle\). In three dimensions, write \(\mathbf{f}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\). The component functions describe the vector assigned at each point.

A scalar function assigns a number to each point. A differentiable scalar function can also generate a vector field: take its partial derivative with respect to each coordinate and place those derivatives in a vector. This vector is the function’s gradient. When a vector field is the gradient of a scalar function, the scalar function is called a potential for the field.

In two dimensions, the definition means that if \(\mathbf{f}(x,y)=\langle P(x,y),Q(x,y)\rangle\), then \(\phi\) is a potential precisely when

In three dimensions, there are three partial derivatives to check:

The equality in the definition is an equality of vector fields: each component must agree at every point of the domain. Checking that the components agree at a few selected points is not enough. For polynomial fields and proposed polynomial potentials, the derivatives can be calculated directly and compared as functions.

Verifying a Potential Component by Component

A potential is a candidate that must be tested against the definition. The test is direct: differentiate the candidate with respect to each coordinate, assemble the gradient, and compare the result with the given field. The same order of coordinates must be used on both sides.

The example illustrates why partial derivatives, rather than ordinary one-variable derivatives, appear in the definition. When differentiating with respect to one coordinate, the other coordinate is treated as constant. Each resulting partial derivative supplies the matching component of the vector field.

Why the Gradient Structure Matters

The definition links the field at a point to how a scalar function changes near that point. This link can be seen by following any differentiable curve through the domain. Let \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\) be a differentiable curve and suppose \(\mathbf{f}=\nabla\phi\). The ordinary chain rule gives

Since \(\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle\) and \(\nabla\phi=\mathbf{f}\), the right-hand side is the dot product of the field with the curve’s velocity:

This identity explains why potentials are useful: along a curve, the rate of change of the scalar potential is determined by the field and the direction of motion. The argument uses only the chain rule and the definition of a potential. It does not depend on a special choice of curve.

The same reasoning works in three dimensions. If \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), then the chain rule gives

When the field equals \(\nabla\phi\), this rate is also \(\mathbf{f}(\mathbf{r}(t))\mathbin{\cdot}\mathbf{r}'(t)\). This is the local calculus connection between a conservative field and its potential.

Verifying a Three-Dimensional Field

In three dimensions, a valid potential must reproduce all three components, not just two. A field that matches a proposed gradient in two coordinates but not in the third does not satisfy the definition.

A Common Mistake: Matching the Wrong Components

A proposed function can look plausible and still fail. For example, consider the field \(\mathbf{f}(x,y)=\langle x,y\rangle\) and the proposed function \(\psi(x,y)=xy\). A mistaken check might say that the \(x\)-derivative is \(x\) and the \(y\)-derivative is \(y\), then conclude that the gradient matches the field. But those derivatives have been interchanged: the product rule for partial derivatives gives

Unless \(x=y\), this is not \(\langle x,y\rangle\), so \(\psi=xy\) is not a potential for this field on \(\mathbb{R}^2\). The correction is to test a candidate whose partial derivatives match in the right order. For \(\phi(x,y)=\tfrac12x^2+\tfrac12y^2\),

This verifies that \(\phi\) is a potential for the field. The important practice is to compare the complete gradient with the complete field, coordinate by coordinate, rather than relying on the appearance of the candidate.

Potentials Are Not Numerically Unique

If \(\phi\) is a potential for \(\mathbf{f}\), adding a constant does not change its gradient. For any constant \(C\),

because each partial derivative of a constant is zero. Thus \(\phi+C\) is also a potential for the same field. For instance, \(x^2+y^2+7\) and \(x^2+y^2\) both generate the field \(\langle2x,2y\rangle\). When verifying a potential, the definition asks whether its gradient equals the field; it does not prescribe a particular constant term.