Conservative Fields#

Conservative Fields and Potentials#

If for a vector field \(\overrightarrow{\boldsymbol{F}}(x, y, z)\) there exists a function \(f(x, y, z)\) satisfying \(\overrightarrow{\boldsymbol{F}}=\overrightarrow{\boldsymbol{\nabla}} f\) then \(\overrightarrow{\boldsymbol{F}}\) is called a conservative vector field, and \(f\) is called a potential function for vector field \(\overrightarrow{\boldsymbol{F}}\).

  • Not all vector fields are conservative, but many important fields in physics are conservative.

  • For a conservative field, field lines are orthogonal to equipotential surfaces.

The Fundamental Theorem for Line Integrals#

For a smooth curve \(C\) given by \(\overrightarrow{\boldsymbol{r}}(t), a \leq t \leq b\), and a scalar field \(f(x, y, z)\) such that \(\vec{\nabla} f\) exists

\[ \int_{C} \vec{\nabla} f \cdot d \overrightarrow{\boldsymbol{r}}=f(\mathbf{r}(b))-f(\mathbf{r}(a)) . \]

and is continuous on \(C\),

Hence for a conservative field \(\overrightarrow{\boldsymbol{F}}=\overrightarrow{\boldsymbol{\nabla}} f\)

  • the line integral of \(\overrightarrow{\boldsymbol{F}}\) is equal to the net change in the potential function, \(\Delta f\)

  • the line integral of \(\overrightarrow{\boldsymbol{F}}\) is independent of path

  • around any closed path, \(\oint_{C} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{r}}=0\).

The following converse statements can also be proved to be true (see textbooks):

  • \(\int_{C} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{r}}\) is independent of path if and only if \(\overrightarrow{\boldsymbol{F}}\) is a conservative vector field (that is, if and only if there exists a scalar function \(f\) such that \(\overrightarrow{\boldsymbol{F}}=\overrightarrow{\boldsymbol{\nabla}} f\).)

  • \(\int_{C} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{r}}\) is independent of path in a region \(D\) if and only if \(\oint_{C} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{r}}=0\) for every closed path \(C\) in \(D\).

Application

This theorem is the mathematical basis of the principle of conservation of energy. For a conservative force field \(\overrightarrow{\boldsymbol{F}}\), the potential energy, \(E_{p}\), of a body due to the field is defined to be minus the associated potential, \(\overrightarrow{\boldsymbol{F}}=-\overrightarrow{\boldsymbol{\nabla}} E_{p}\). So if we move a body along a path \(C\) from point \(A\) to point \(B\) against the field, the increase in potential energy is equal to the work done against the field and is independent of path: \(\quad-\int_{C} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{r}}=\int_{A}^{B} \overrightarrow{\boldsymbol{\nabla}} E_{p} \cdot d \overrightarrow{\boldsymbol{r}}=E_{p}(B)-E_{p}(A)\).

(Component Test) Theorem#

  1. If \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) is a conservative vector field on an open, simply connected” region \(D\), and \(M\) and \(N\) have continuous first order partial derivatives in \(D\), then \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\) throughout \(D\).

  2. Conversely, if \(P\) and \(Q\) have continuous first order partial derivatives in \(D\) and satisfy \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\) then \(\mathbf{F}\) is conservative.

  • open means it does not contain any boundary points

  • simply connected means it consists of just one piece, without any holes

img

Note:

  • Part ii) is what we use to determine whether or not a given vector field is conservative, and is sometimes known as the component test. The equivalent result in \(\boldsymbol{R}^{3}\) is (from Thomas p.994):

Component Test for Conservative Fields#

Let \(\mathbf{F}=M(x, y, z) \mathbf{i}+N(x, y, z) \mathbf{j}+P(x, y, z) \mathbf{k}\) be a field on an open simply connected domain whose component functions have continuous first partial derivatives. Then, \(\mathbf{F}\) is conservative if and only if

\[ \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}, \quad \frac{\partial M}{\partial z}=\frac{\partial P}{\partial x}, \quad \text { and } \quad \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y} . \]