Conservative Vector Fields

Section 2.3 Conservative Vector Fields

Not all vector fields are created equal. In particular, some vector fields are easier to work with than others. One important class of vector fields that are relatively easy to work with, at least sometimes, but that still arise in many applications are “conservative vector fields”.

Definition 2.3.1.

The vector field $\vF$ is said to be conservative if there exists a function $\varphi$ such that $\vF = \vnabla\varphi\text{.}$ Then $\varphi$ is called a potential for $\vF\text{.}$ Note that if $\varphi$ is a potential for $\vF$ and if $C$ is a constant, then $\varphi+C$ is also a potential for $\vF\text{.}$
If $\vF=\vnabla\varphi$ is a conservative field with potential $\varphi$ and if $C$ is a constant, then the set of points that obey $\varphi(x,y,z)=C$ is called an equipotential surface. Similarly, in two dimensions, the set of points that obey $\varphi(x,y)=C$ is called an equipotential curve.

Warning 2.3.2.

In physics, when a vector field is of the form $\vF=-\vnabla\varphi\text{,}$ then $\varphi$ is called a potential for $\vF\text{.}$ Note the minus

Physicists introduce this minus sign in order to eliminate the minus sign in the next footnote.

Example 2.3.3. Potential energy.

The “conservative” in “conservative vector field” has nothing to do with politics. It comes from “conservation of energy”. Here is how. Suppose that you have a particle of mass $m$ moving in a force field $\vF$ that happens to be of the form $\vF=\vnabla\varphi$ for some function $\varphi\text{.}$ If the position of the particle a time $t$ is $\big(x(t),y(t),z(t)\big)\text{,}$ then, by Newton’s law of motion,

\begin{alignat*}{2} m\va = \vF \qquad &\implies & m\diff{\vv}{t}(t) &= \vF\big(x(t),y(t),z(t)\big)\\ &\implies & m\diff{\vv}{t}(t) &= \vnabla\varphi\big(x(t),y(t),z(t)\big)\\ \end{alignat*}

Now dot both sides with $\vv(t)\text{.}$

\begin{alignat*}{2} &\implies & m\,\vv(t)\cdot\diff{\vv}{t}(t) &= \vv(t)\cdot\vnabla\varphi\big(x(t),y(t),z(t)\big)\\ & & &=x'(t)\frac{\partial\varphi}{\partial x}\big(x(t),y(t),z(t)\big) +y'(t)\frac{\partial\varphi}{\partial y}\big(x(t),y(t),z(t)\big)\\ & & & \hskip1in +z'(t)\frac{\partial\varphi}{\partial z}\big(x(t),y(t),z(t)\big)\\ \end{alignat*}

Next use $\diff{}{t}\vv\cdot\vv=2\vv\cdot\diff{\vv}{t}$ on the left hand side and the chain rule on the right hand side.

\begin{alignat*}{2} &\implies &\ \ \diff{\ }{t}\Big(\frac{1}{2} m \vv(t)\cdot\vv(t)\Big) &= \diff{\ }{t}\Big(\varphi\big(x(t),y(t),z(t)\big)\Big)\\ &\implies & &\hskip-1.3in\diff{\ }{t}\Big(\frac{1}{2} m \vv(t)\cdot\vv(t) -\varphi\big(x(t),y(t),z(t)\big)\Big) =0\\ &\implies & &\hskip-1.0in\frac{1}{2} m |\vv(t)|^2 -\varphi\big(x(t),y(t),z(t)\big) = \text{const} \end{alignat*}

So $\frac{1}{2} m |\vv(t)|^2 -\varphi\big(x(t),y(t),z(t)\big)\text{,}$ which is called the energy

$\frac{1}{2} m |\vv(t)|^2$ is the kinetic energy and $-\varphi$ is the potential energy. See Warning 2.3.2.

of the particle at time $t\text{,}$ does not actually depend on time — it is conserved. Let’s call the initial energy $E\text{.}$ That is, $E=\frac{1}{2} m |\vv(0)|^2 -\varphi\big(x(0),y(0),z(0)\big)\text{.}$ Then $\frac{1}{2} m |\vv(t)|^2 -\varphi\big(x(t),y(t),z(t)\big)=E$ for all $t$ and, in particular

\begin{equation*} \varphi\big(x(t),y(t),z(t)\big) = \frac{1}{2} m |\vv(t)|^2 -E \ge -E \end{equation*}

So even without having to find $\big(x(t)\,,\,y(t)\,,\,z(t)\big)\text{,}$ we know that our particle can never escape the region $\Set{(x,y,z)}{\varphi(x,y,z)\ge -E}\text{.}$

Example 2.3.4. Gravity.

The gravitational force that a body of mass $M$ at the origin exerts on a body of mass $m$ at $\vr=(x,y,z)$ is

\begin{equation*} \vF(\vr) = -\frac{GMm}{r^3}\vr \end{equation*}

where $r=|\vr|=\sqrt{x^2+y^2+z^2}$ and $G$ is the gravitational constant. This force is conservative, with potential $\varphi(\vr) = \frac{GMm}{r}\text{.}$ To verify that this is correct, observe that

\begin{alignat*}{2} \frac{\partial\ }{\partial x}\varphi(\vr) &=\frac{\partial\ }{\partial x}\frac{GMm}{\sqrt{x^2+y^2+z^2}} &=-\frac{1}{2}\frac{GMm(2x)}{[x^2+y^2+z^2]^{3/2}} &=-\frac{GMm}{r^3}x\\ \frac{\partial\ }{\partial y}\varphi(\vr) &=\frac{\partial\ }{\partial y}\frac{GMm}{\sqrt{x^2+y^2+z^2}} &=-\frac{1}{2}\frac{GMm(2y)}{[x^2+y^2+z^2]^{3/2}} &=-\frac{GMm}{r^3}y\\ \frac{\partial\ }{\partial z}\varphi(\vr) &=\frac{\partial\ }{\partial z}\frac{GMm}{\sqrt{x^2+y^2+z^2}} &=-\frac{1}{2}\frac{GMm(2z)}{[x^2+y^2+z^2]^{3/2}} &=-\frac{GMm}{r^3}z \end{alignat*}

We have already found conservation of energy very helpful a couple of times in Section 1.7 (Sliding on a Curve). So, at this point, there are probably several questions gnawing away at you.

Is every vector field conservative?
If not, is there an easy way to tell whether or not a vector field is conservative?
If we know that a given vector field is conservative, how do you find a potential for it?

Have no fear. We will consider those questions in some detail shortly. But first, a couple of more examples.

Example 2.3.5.

In this example we will show that the vector field $\vF(x,y) = x\,\hi-y\,\hj$ is conservative and find both its potential and its field lines.

The potential: Our vector field $\vF(x,y) = x\,\hi-y\,\hj$ is conservative if we can find a $\varphi$ obeying
\begin{align*} \frac{\partial \varphi}{\partial x}(x,y) &= x\\ \frac{\partial \varphi}{\partial y}(x,y) &= -y \end{align*}
Recall that, when taking the partial derivative $\frac{\partial\ }{\partial x}$ the coordinate $y$ is viewed as a constant. So the first of these equations is satisfied if and only if there is a $\psi(y)\text{,}$ which does not depend on $x\text{,}$ so that
\begin{equation*} \varphi(x,y) = \frac{x^2}{2} +\psi(y) \end{equation*}
For this to also satisfy the second equation, we need
\begin{gather*} -y=\frac{\partial \varphi}{\partial y}(x,y) =\frac{\partial\ }{\partial y}\Big(\frac{x^2}{2} +\psi(y)\Big) =\psi'(y) \end{gather*}
which is the case if and only if there is a constant $C$ with
\begin{equation*} \psi(y) =-\frac{y^2}{2} +C \end{equation*}
So, for any choice of the constant $C\text{,}$
\begin{equation*} \frac{x^2}{2} - \frac{y^2}{2} +C \end{equation*}
is a potential. In particular, taking $C=0\text{,}$ one possible potential is
\begin{equation*} \varphi(x,y) = \frac{x^2}{2} - \frac{y^2}{2} \end{equation*}
Some equipotential curves for this potential are sketched in (c) below. They are the blue curves.
The field lines (Optional): Recalling (2.2.5), the field lines of the vector field $\vF(x,y) = x\,\hi-y\,\hj$ are determined by
\begin{align*} \frac{\dee{x}}{x} = \frac{\dee y}{-y} &\iff -y\dee x = x\dee{y}\\ &\iff x\dee{y} + y\dee x = 0\\ &\iff \dee{(xy)} = 0 \qquad\text{by the product rule}\\ &\iff xy = C \end{align*}
for some constant $C\text{.}$ If you are not comfortable with the use of the product rule above, here is another way to write the same computation.
\begin{align*} \diff{y}{x}=-\frac{y}{x} &\iff x\diff{y}{x} = -y\\ &\iff x\diff{y}{x} +y = 0\\ &\iff \diff{}{x}(xy) = 0 \qquad\text{by the product rule}\\ &\iff xy = C \end{align*}
Some field lines are sketched in (c) below. They are the red curves. Note that they appear to cross the equipotential curves, the blue curves, at right angles. We shall see in Lemma 2.3.6, below, that this is not a coincidence. Also note that, while the above computation tells what the field lines are, it does not give us the direction of motion along the field lines. We determine the direction of motion next.
Direction of motion (Optional): The sign data

\begin{align*} &\hi\cdot\vF(x,y) = x \left.\begin{cases} \gt 0 &\text{if } x \gt 0 \\ =0 &\text{if } x=0 \\ \lt 0 &\text{if } x \lt 0 \end{cases}\right\}\\ &\hj\cdot\vF(x,y) = -y \left.\begin{cases} \gt 0 &\text{if } y \lt 0 \\ =0 &\text{if } y=0 \\ \lt 0 &\text{if } y \gt 0 \end{cases}\right\}\qquad \end{align*}

is visually displayed in the figure on the left below. The arrows in the figure on the left gives us the direction of motion along the field lines (in red) in the figure on the right below. Some equipotential curves are also sketched (in blue) in the figure on the right below.

We have just seen one example of a conservative vector field for which the field lines appear to cross the equipotential curves at right angles. Here is a result which says that that was no accident. The field lines of conservative fields always cross the equipotential surfaces at right angles.

Lemma 2.3.6. Optional.

If $\vF$ is a conservative vector field, then the field lines of $\vF$ are perpendicular to the equipotential surfaces of $\vF\text{.}$

Proof.

Let $\vF=\vnabla\varphi\text{.}$ Pick any point $\vr_0$ and any nonzero vector $\vT$ that is tangent to the equipotential surface at $\vr_0\text{.}$ That equipotential surface is $\varphi\big(x,y,z\big)=\varphi(\vr_0)\text{.}$ Consider any curve $\vr(t)=\big(x(t),y(t),z(t)\big)$ that

lies in the equipotential surface of $\vF$ through $\vr_0\text{,}$ so that $\varphi\big(\vr(t)\big)=\varphi(\vr_0)$ for all $t\text{,}$ and also obeys
$\vr(0)=\vr_0$ and
$\diff{\vr}{t}(0) = \vT\text{.}$

Differentiating $\varphi\big(\vr(t)\big)=\varphi(\vr_0)$ with respect to $t$ and applying the chain rule gives

\begin{align*} \diff{\ }{t}\big[\varphi\big(x(t),y(t),z(t)\big)\big] &=0 \end{align*}

\begin{align*} \frac{\partial\varphi}{\partial x}\big(x(t),y(t),z(t)\big)\diff{x}{t}(t) &+\frac{\partial\varphi}{\partial y}\big(x(t),y(t),z(t)\big)\diff{y}{t}(t)\\ &+\frac{\partial\varphi}{\partial z}\big(x(t),y(t),z(t)\big)\diff{z}{t}(t) =0 \end{align*}

Notice that the left hand side is exactly the dot product of $\big(\frac{\partial\varphi}{\partial x} \,,\,\frac{\partial\varphi}{\partial y} \,,\,\frac{\partial\varphi}{\partial z}\big)=\vnabla\varphi$ with $\big(\diff{x}{t} \,,\,\diff{y}{t} \,,\,\diff{z}{t}\big) =\diff{\vr}{t}\text{.}$ So

\begin{align*} \vnabla\varphi\big(\vr(t)\big)\cdot\diff{\vr}{t}(t) &= 0\\ \vF\big(\vr(t)\big)\cdot\diff{\vr}{t}(t) &= 0 \end{align*}

Then set $t=0$ to get

\begin{equation*} \vF\big(\vr_0\big)\cdot\vT = 0 \end{equation*}

This says that the vector $\vT\text{,}$ which is tangent to the equipotential surface at $\vr_0\text{,}$ is perpendicular to the vector field at $\vr_0\text{,}$ which is a tangent vector to the field line of $\vF$ through $\vr_0\text{.}$

Here is another example in which we try to find a potential for a vector field.

Example 2.3.7.

Let’s try to find a potential for the vortex vector field $\vv(x,y) = \Om\big(-y\hi +x\hj\big)$ of Example 2.1.4. The potential would have to obey

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y) &= -\Om y\\ \frac{\partial \varphi}{\partial y}(x,y) &= \Om x \end{align*}

We proceed just as we did in Example 2.3.5. The first of these equations is satisfied if and only if there is a $\psi(y)\text{,}$ which does not depend on $x\text{,}$ so that

\begin{equation*} \varphi(x,y) = -\Om xy +\psi(y) \end{equation*}

For this to also satisfy the second equation, we need

\begin{gather*} \Om x=\frac{\partial \varphi}{\partial y}(x,y) =\frac{\partial\ }{\partial y}\Big(-\Om xy +\psi(y)\Big) =-\Om x +\psi'(y) \iff \psi'(y) = 2\Om x \end{gather*}

If $\Om\ne 0\text{,}$ the right hand side of this equation depends on $x$ while the left hand side in independent of $x\text{,}$ no matter what $\psi$ is. So no $\psi$ can work, and $\vv(x,y) = \Om\big(-y\hi +x\hj\big)$ is not conservative.

The previous example shows that not all vector fields are conservative. That answers the first of the questions that we posed just before Example 2.3.5. The next theorem provides a simple screening test for conservativeness, which partially answers the second question. The easy way to remember the screening test uses the curl, which we now define.

Definition 2.3.8.

The curl of a vector field $\vF(x,y,z)$ is denoted by $\vnabla\times\vF(x,y,z)$ and is defined by

\begin{align*} \vnabla\times\vF &=\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} \Big)\hi +\Big(\frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x} \Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \Big)\hk\\ &=\det\left[\begin{matrix} \hi &\hj & \hk \\ \frac{\partial\ }{\partial x} & \frac{\partial\ }{\partial y} & \frac{\partial\ }{\partial z} \\ F_1(x,y,z) & F_2(x,y,z) & F_3(x,y,z) \end{matrix}\right] \end{align*}

The determinant in the second row is really just a mnemonic device used to make it easy to remember the expression after the equals sign in the first row. One must be careful about the signs in this definition — the determinant helps with that.

Theorem 2.3.9. Screening test for conservative vector fields..

Assume that $F_1(x,y)$ and $F_2(x,y)$ are continuously differentiable. If the vector field $F_1(x,y)\hi + F_2(x,y)\hj$ is conservative, then we must have
\begin{equation*} \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x} \end{equation*}
Assume that $F_1(x,y,z)\text{,}$ $F_2(x,y,z)$ and $F_3(x,y,z)$ are continuously differentiable. If the vector field $F_1(x,y,z)\hi + F_2(x,y,z)\hj + F_3(x,y,z)\hk$ is conservative, then
\begin{equation*} \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\qquad \frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}\qquad \frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y} \end{equation*}
Equivalently,
\begin{equation*} \vnabla\times\vF =\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} \Big)\hi +\Big(\frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x} \Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \Big)\hk =\vZero \end{equation*}
That is, $\vF$ is curl free.

Proof.

(a) If the vector field $F_1(x,y)\hi + F_2(x,y)\hj$ is conservative, then there is a potential $\varphi(x,y)$ such that

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y) &= F_1(x,y)\\ \frac{\partial \varphi}{\partial y}(x,y) &= F_2(x,y) \end{align*}

Applying $\frac{\partial\ }{\partial y}$ to the first equation and $\frac{\partial\ }{\partial x}$ to the second equation gives

\begin{align*} \frac{\partial^2 \varphi}{\partial y\partial x} &= \frac{\partial F_1}{\partial y}\\ \frac{\partial^2 \varphi}{\partial x\partial y} &= \frac{\partial F_2}{\partial x} \end{align*}

Recall that, for any twice continuously differentiable function, $\frac{\partial^2 \varphi}{\partial y\partial x} = \frac{\partial^2 \varphi}{\partial x\partial y}\text{.}$ So the two left hand sides are equal, and the two right hand sides must also be equal.

(b) If the vector field $F_1(x,y,z)\hi + F_2(x,y,z)\hj + F_3(x,y,z)\hk$ is conservative, then there is a potential $\varphi(x,y,z)$ such that

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= F_1(x,y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= F_2(x,y,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= F_3(x,y,z) \end{align*}

We proceed just as in (a).

Applying $\frac{\partial\ }{\partial y}$ to the first equation and $\frac{\partial\ }{\partial x}$ to the second equation gives
\begin{gather*} \left\{\atp{ \frac{\partial^2 \varphi}{\partial y\partial x} = \frac{\partial F_1}{\partial y} } { \frac{\partial^2 \varphi}{\partial x\partial y} = \frac{\partial F_2}{\partial x} }\right\} \implies \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x} \end{gather*}
Applying $\frac{\partial\ }{\partial z}$ to the first equation and $\frac{\partial\ }{\partial x}$ to the third equation gives
\begin{gather*} \left\{\atp{ \frac{\partial^2 \varphi}{\partial z\partial x} = \frac{\partial F_1}{\partial z} } { \frac{\partial^2 \varphi}{\partial x\partial z} = \frac{\partial F_3}{\partial x} }\right\} \implies \frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x} \end{gather*}
Applying $\frac{\partial\ }{\partial z}$ to the second equation and $\frac{\partial\ }{\partial y}$ to the third equation gives
\begin{gather*} \left\{\atp{ \frac{\partial^2 \varphi}{\partial z\partial y} = \frac{\partial F_2}{\partial z} } { \frac{\partial^2 \varphi}{\partial y\partial z} = \frac{\partial F_3}{\partial y} }\right\} \implies \frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y} \end{gather*}

Combining the three bullet points gives $\vnabla\times\vF=\vZero\text{.}$

Warning 2.3.10.

As always, we have to be careful with the flow of logic

Use your favourite search engine to look up a list of common logical errors. One is “affirming the consequent”. An example would be concluding that because Shakespeare is dead, Elvis, who is also dead, must also be Shakespeare.

. The screening test in Theorem 2.3.9 is a one-way test. If, for example, $\frac{\partial F_1}{\partial y} \ne \frac{\partial F_2}{\partial x}$ then the vector field $\vF$ cannot be conservative. But if $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ Theorem 2.3.9 does not guarantee that $\vF$ is conservative. In fact there are fields that are not conservative but do obey $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}\text{.}$ We’ll see one in Example 2.3.14, below. We shall later find some additional regularity conditions which, when combined with $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}\text{,}$ do imply conservativeness. See Theorem 2.4.8, below.

Example 2.3.11. Example 2.3.7 revisited.

In Example 2.3.7, we attempted to find a potential for the vector field

\begin{equation*} \vv(x,y) = \Om\big(-y\hi +x\hj\big) \end{equation*}

In the end we showed that, if $\Om\ne 0\text{,}$ no potential could exist, i.e. $\vv(x,y)$ is not conservative. Had we known the screening test of Theorem 2.3.9.a, we could have concluded that $\vv(x,y)$ is not conservative by simply observing that

\begin{alignat*}{2} \frac{\partial \vv_1}{\partial y}&= \frac{\partial \ }{\partial y}\big(-\Om y) &= -\Om\\ \frac{\partial \vv_2}{\partial x}&= \ \ \frac{\partial \ }{\partial x}\big(\Om x) &= +\Om \end{alignat*}

are not equal, unless $\Om=0\text{.}$ But $\Om=0$ makes a rather boring vector field.

Example 2.3.12.

Determine whether or not the vector field

\begin{equation*} \vF(x,y,z) = y\hi -z\hj +x\hk \end{equation*}

is conservative. If it is conservative, find a potential.

Solution.

Let’s start by applying the screening test Theorem 2.3.9.b. Since

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj & \hk\\ \frac{\partial\ }{\partial x} & \frac{\partial\ }{\partial y} & \frac{\partial\ }{\partial z}\\ y & -z & x \end{matrix}\right] =\hi-\hj-\hk \end{align*}

is not $\vZero\text{,}$ the vector field $\vF$ cannot be conservative.

Example 2.3.13.

Determine whether or not the vector field

\begin{equation*} \vF(x,y,z) = (y^2+2xz^2-1)\hi +(2x+1)y\,\hj +(2x^2z+z^3)\hk \end{equation*}

is conservative. If it is conservative, find a potential.

Solution.

Again start by applying the screening test Theorem 2.3.9.b. This time

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj & \hk \\ \frac{\partial\ }{\partial x} & \frac{\partial\ }{\partial y} & \frac{\partial\ }{\partial z} \\ y^2+2xz^2-1 & (2x+1)y & 2x^2z+z^3 \end{matrix}\right]\\ &=0\hi-(4xz-4xz)\hj+(2y-2y)\hk\\ &=\vZero \end{align*}

So $\vF$ passes the screening test. Let’s look for a function $\varphi(x,y,z)$ obeying

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= y^2+2xz^2-1 \\ \frac{\partial \varphi}{\partial y}(x,y,z) &= (2x+1)y \tag{$*$}\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 2x^2z+z^3 \end{align*}

The partial derivative $\frac{\partial \ }{\partial x}$ treats $y$ and $z$ as constants. So $\varphi(x,y,z)$ obeys the first equation if and only if there is a function $\psi(y,z)$ with

\begin{equation*} \varphi(x,y,z) =xy^2+x^2z^2-x +\psi(y,z) \end{equation*}

This $\varphi(x,y,z)$ will also obey the second equation if and only if

\begin{align*} &\frac{\partial \ }{\partial y}\big(xy^2+x^2z^2-x+\psi(y,z)\big) = (2x+1)y\\ &\hskip1in\iff 2xy +\frac{\partial\psi}{\partial y}(y,z) = (2x+1)y\\ &\hskip1in\iff \frac{\partial\psi}{\partial y}(y,z) = y\\ &\hskip1in\iff \psi(y,z) = \frac{y^2}{2} +\zeta(z) \end{align*}

for some function $\zeta(z)$ which depends only on $z\text{.}$ At this stage we know that

\begin{equation*} \varphi(x,y,z) =xy^2+x^2z^2-x+\psi(y,z) =xy^2+x^2z^2-x+\frac{y^2}{2}+\zeta(z) \end{equation*}

obeys the first two equations of ($*$), for any function $\zeta(z)\text{.}$ Finally to have the third equation of ($*$) also satisfied, we also need to chose $\zeta(z)$ to obey

\begin{align*} &\frac{\partial \ }{\partial z}\left(xy^2+x^2z^2-x+\frac{y^2}{2}+\zeta(z)\right) = 2x^2z+z^3\\ &\hskip1in\iff 2x^2z +\zeta'(z) = 2x^2z + z^3\\ &\hskip1in\iff \zeta'(z) = z^3\\ &\hskip1in\iff \zeta(z) = \frac{z^4}{4} + C \end{align*}

for any constant $C\text{.}$ So one possible potential, namely that with $C=0\text{,}$ is

\begin{equation*} \varphi(x,y,z) =xy^2+x^2z^2-x+\frac{y^2}{2}+\frac{z^4}{4} \end{equation*}

Note, as a check

⁴

It is always worth doing this check.

, that

\begin{equation*} \vnabla \varphi(x,y,z) =\big(y^2+2xz^2-1\big)\hi+\big(2xy+y)\hj+\big(2x^2z+z^3\big)\hk \end{equation*}

as desired.

Example 2.3.14. Optional: First look at $-\frac{y}{x^2+y^2}\hi + \frac{x}{x^2+y^2}\hj$.

Now is a good time to reread Warning 2.3.10. In this example we will show that the vector field

\begin{align*} &\vF(x,y) = -\frac{y}{x^2+y^2}\hi + \frac{x}{x^2+y^2}\hj\\ &\qquad \text{defined for all $(x,y)$ in $\bbbr^2$ except $(x,y)=(0,0)$} \end{align*}

passes the screening test of Theorem 2.3.9.a. We will also begin to see why it is not conservative on the domain $\bbbr^2\setminus\{(0,0)\}\text{.}$ To verify the screening test, we compute

\begin{alignat*}{2} \frac{\partial\ }{\partial y}\Big(-\frac{y}{x^2+y^2}\Big) &= -\frac{(x^2+y^2) - y(2y)}{{(x^2+y^2)}^2} &&= \frac{y^2-x^2}{{(x^2+y^2)}^2}\\ \frac{\partial\ }{\partial x}\Big(\frac{x}{x^2+y^2}\Big) &=\phantom{-} \frac{(x^2+y^2) - x(2x)}{{(x^2+y^2)}^2} &&= \frac{y^2-x^2}{{(x^2+y^2)}^2} \end{alignat*}

and observe that the two right hand sides are identical. So the screening test is passed.

In order for $\vF$ to be conservative on the domain $\bbbr^2\setminus\{(0,0\}\text{,}$ there must exist a function $\varphi(x,y)\text{,}$ that, together with both partial derivatives $\frac{\partial \varphi}{\partial x}(x,y)$ and $\frac{\partial \varphi}{\partial y}(x,y)\text{,}$ is defined for all $(x,y)$ in $\bbbr^2$ except $(x,y)=(0,0)\text{,}$ and obeys

\begin{alignat*}{3} \frac{\partial \varphi}{\partial x}(x,y) &= -\frac{y}{x^2+y^2} &&= \frac{-\frac{y}{x^2}}{1+\big(\frac{y}{x}\big)^2} &&=\frac{\partial\ }{\partial x}\Big(\arctan\frac{y}{x}\Big)\\ \frac{\partial \varphi}{\partial y}(x,y) &= \phantom{-} \frac{x}{x^2+y^2} &&= \frac{\frac{1}{x}}{1+\big(\frac{y}{x}\big)^2} &&=\frac{\partial\ }{\partial y}\Big(\arctan\frac{y}{x}\Big) \end{alignat*}

by the chain rule, because

\begin{equation*} \frac{\partial\ }{\partial x}\Big(\frac{y}{x}\Big) =-\frac{y}{x^2} \qquad \frac{\partial\ }{\partial y}\Big(\frac{y}{x}\Big)=\frac{1}{x} \end{equation*}

It looks like we have found a potential, namely $\arctan\frac{y}{x}\text{.}$ But there is a problem. Recall that, by definition, $\arctan\frac{y}{x}$ is an angle $\theta(x,y)$ that obeys $\tan\theta(x,y)= \arctan\frac{y}{x}\text{;}$ but for any $(x,y)\in\bbbr^2\setminus\{(0,0\}$ there are infinitely many angles having the tangent $\frac{y}{x}\text{.}$ To define $\varphi(x,y)$ we have to select exactly one such angle. It is impossible to do so in such a way that $\varphi(x,y)$ is continuous on all of $\bbbr^2\setminus\{(0,0\}\text{.}$

To see why, fix any $r \gt 0\text{,}$ and imagine that you are walking on the circle $x^2+y^2=r^2$ in the $xy$-plane. At time $\theta\text{,}$ you are at $x=r\cos\theta\text{,}$ $y=r\sin\theta$ and then $\frac{y}{x} = \tan\theta$ and you are allowed to define $\varphi(x,y)=\theta+k\pi\text{,}$ for any integer $k\text{.}$

Suppose that at time $\theta=0$ you choose $k=0\text{.}$ That is, you choose $\varphi(r,0)=0\text{.}$ Now start walking, choosing an allowed $\varphi(x,y)\text{,}$ i.e. choosing a $k\text{,}$ for each point $(x,y)$ that you cross. Because $\varphi(x,y)$ has to vary continuously

⁵

If $\varphi(x,y)$ is not continuous, its gradient does not exist, and $\varphi$ cannot be a potential.

with $(x,y)\text{,}$ you have to continue choosing $k=0\text{.}$ But you run off a cliff as $\theta$ approaches $2\pi\text{,}$ because then

you are approaching $(r,0)$ from below, as in the figure below, and
because you are choosing $k=0\text{,}$ $\varphi(x,y)$ is just a little less than $2\pi\text{,}$ but
you have already chosen $\varphi(r,0)=0\text{,}$ not $2\pi\text{.}$ So $\varphi(x,y)$ has a jump discontinuity
⁶
Those who have taken some complex analysis may recognize this as the branch cut in $\ln z\text{.}$
along the positive $x$-axis.

If you are having trouble following this argument, don’t worry about it. We will return with a less hand-wavy argument later.

Exercises Exercises

Exercise Group.

Exercises — Stage 1

1.

We’ve seen two calculations of the energy $E$ of a system. Equation 1.7.1 told us $E=\frac{1}{2}m|\vv|^2+mgy\text{,}$ while Example 2.3.3 says $\frac{1}{2} m |\vv(t)|^2 -\varphi\big(x(t),y(t),z(t)\big)=E\text{.}$

Consider a force given by $\vF = \vnabla \varphi$ for some differentiable function $\varphi:\mathbb R^3 \to \mathbb R\text{.}$ A particle of mass $m$ is being acted on by $\vF$ and no other forces, and its position at time $t$ is given by $(x(t),y(t),0)\text{.}$

True or false: $mgy(t)=-\varphi(x(t),y(t),0)\text{.}$

2.

For each of the following fields, decide which of the following holds:

The screening test for conservative vector fields tells us $\vF$ is conservative.
The screening test for conservative vector fields tells us $\vF$ is not conservative.
The screening test for conservative vector fields does not tell us whether $\vF$ is conservative or not.

(The screening test is Theorem 2.3.9.)

$\displaystyle \vF=x\hi + z\hj + y\hk$
$\displaystyle \vF=y^2z\hi + x^2z\hj + x^2y\hk$
$\displaystyle \vF=(ye^{xy}+1)\hi + (xe^{xy}+z)\hj + \left( \frac1z+y\right)\hk$
$\displaystyle \vF=y\cos(xy)\hi + x\sin(xy)\hj $

3.

Suppose $\vF$ is conservative and let $a\text{,}$ $b\text{,}$ and $c$ be constants. Find a potential for $\vF+(a,b,c)\text{,}$ OR give a conservative field $\vF$ and constants $a\text{,}$ $b\text{,}$ and $c$ for which $\vF+(a,b,c)$ is not conservative.

4.

Prove, or find a counterexample to, each of the following statements.

If $\vF$ is a conservative field and $\vG$ is a non-conservative field, then $\vF+\vG$ is non-conservative.
If $\vF$ and $\vG$ are both non-conservative fields, then $\vF+\vG$ is non-conservative.
If $\vF$ and $\vG$ are both conservative fields, then $\vF+\vG$ is conservative.

Exercise Group.

Exercises — Stage 2

5. (✳).

Let $D$ be the domain consisting of all $(x,y)$ such that $x \gt 1\text{,}$ and let $\vF$ be the vector field

\begin{gather*} \vF = -\frac{y}{x^2+y^2}\,\hi + \frac{x}{x^2+y^2}\,\hj \end{gather*}

Is $\vF$ conservative on $D\text{?}$ Give reasons for your answer.

6.

Find a potential $\varphi$ for $\vF(x,y)=(x+y)\hi+(x-y)\hj\text{,}$ or prove none exists.

7.

Find a potential $\varphi$ for $\vF(x,y)=\left( \frac{1}{x}-\frac{1}{y}\right)\hi+\left(\frac{x}{y^2}\right)\hj\text{,}$ or prove none exists.

8.

Find a potential $\varphi$ for $\vF(x,y,z)=\left(x^2yz+xz\right)\hi+\left( \frac13x^3z+y \right)\hj+\left(\frac13x^3y+\frac12x^2+y\right)\hk\text{,}$ or prove none exists.

9.

Find a potential $\varphi(x,y,z)$ for

\begin{equation*} \vF(x,y,z)=\left( \frac{x}{x^2+y^2+z^2}\right)\hi+\left( \frac{y}{x^2+y^2+z^2}\right)\hj+\left( \frac{z}{x^2+y^2+z^2}\right)\hk, \end{equation*}

or prove none exists.

10.

Determine whether or not each of the following vector fields are conservative. Find the potential if it is.

$\displaystyle \vF(x,y,z)=x\hi-2y\hj+3z\hk$
$\displaystyle \vF(x,y)=\frac{x\hi-y\hj}{x^2+y^2}$

11.

Let $\vF= e^{(z^2)}\,\hi+2Byz^3\,\hj +\big(Axze^{(z^2)}+3By^2z^2\big)\,\hk\text{.}$

For what values of the constants $A$ and $B$ is the vector field $\vF$ conservative on $\bbbr^3\text{?}$
If $A$ and $B$ have values found in (a), find a potential function for $\vF\text{.}$

Exercise Group.

Exercises — Stage 3

12.

Find the velocity field for a two dimensional incompressible fluid when there is a point source of strength $m$ at the origin. That is, fluid is emitted from the origin at area rate $2\pi m$ ${\rm cm}^2$/sec. Show that this velocity field is conservative and find its potential.

13.

A particle of mass $10$ kg moves in the force field $\vF=\vnabla\varphi\text{,}$ where $\varphi(x,y,z)=-(x^2+y^2+z^2)\text{.}$ When its potential energy is 0, the particle is at the origin, and it moves with a velocity $2$ m/s.

Following Example 2.3.3, give a region the particle can never escape.

14.

A particle with constant mass $m=1/2$ moves under a force field $\vF=\hj+3\sqrt[3]{z}\,\hk\text{.}$ At position $(0,0,0)\text{,}$ its speed is $1\text{.}$ What is its speed at $(1,1,1)\text{?}$

(You may assume without proof that the particle does indeed reach the point $(1,1,1)\text{.}$)

15.

For some differentiable, real-valued functions $f,g,h:\mathbb R \to \mathbb R\text{,}$ we define

\begin{equation*} \vF=2f(x)f'(x)\hi+g'(y)h(z)\hj+g(y)h'(z)\hk. \end{equation*}

Verify that $\vF$ is conservative.

16.

Describe the region in $\mathbb R^3$ where the field

\begin{equation*} \vF=\left \lt xy, xz,y^2+z \right \gt \end{equation*}

has curl $\mathbf0\text{.}$

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CLP-4 Vector Calculus

Section 2.3 Conservative Vector Fields

Definition 2.3.1.

Warning 2.3.2.

Example 2.3.3. Potential energy.

Example 2.3.4. Gravity.

Example 2.3.5.

Lemma 2.3.6. Optional.

Proof.

Example 2.3.7.

Definition 2.3.8.

Theorem 2.3.9. Screening test for conservative vector fields..

Proof.

Warning 2.3.10.

Example 2.3.11. Example 2.3.7 revisited.

Example 2.3.12.

Example 2.3.13.

Example 2.3.14. Optional: First look at \(-\frac{y}{x^2+y^2}\hi + \frac{x}{x^2+y^2}\hj\).

Exercises Exercises

Exercise Group.

1.

2.

3.

4.

Exercise Group.

5. (✳).

6.

7.

8.

9.

10.

11.

Exercise Group.

12.

13.

14.

15.

16.