# 17. Divergence, Curl and Potentials

## f. Scalar Potentials

Given a vector field \(\vec F\), a scalar potential for \(\vec F\) is any scalar field \(f\) whose gradient is \(\vec F\): \[ \vec\nabla f=\vec F \]

## 2. Non-Existence of a Scalar Potential

Not every vector field has a scalar potential. What can go wrong in the process of finding a scalar potential?

Find a scalar potential for \(\vec F =\left\langle 2xy^3+z^3,3x^2y^2-\underline{\large 6}\,y^2z,3xz^2+4z^3-y^3\right\rangle\) or show it does not exist.

Notice that this is almost the same vector field as in the last exercise on the previous page except for the \(\underline{\large 6}\) replacing a \(3\). So watch the differences in the solutions, all of which are underlined in a larger font.

We attempt to find a function \(f\) satisfying \(\vec\nabla f=\vec F\), or \[\begin{aligned} \partial_x f&=2xy^3+z^3\quad&&\text{(1)} \\ \partial_y f&=3x^2y^2-\underline{\large 6}\,y^2z\quad&&\text{(2)} \\ \partial_z f&=3xz^2+4z^3-y^3\quad&&\text{(3)} \end{aligned}\] We solve these successively.

- We first take the \(x\)-antiderivative of (1) to get \[ f=x^2y^3+xz^3+g(y,z) \]
- We take the \(y\)-derivative of this: \[ \partial_y f=3x^2y^2+\partial_y g \] and compare it to (2) to get \[ \partial_y g=-\underline{\large 6}\,y^2z \] We take the \(y\)-antiderivative of this to get \[ g=-\underline{\large 2}\,y^3z+h(z) \] and substitute this \(g\) into \(f\) to get \[ f=x^2y^3+xz^3-\underline{\large 2}\,y^3z+h(z) \]
- Finally, we take the \(z\)-derivative: \[ \partial_z f=3xz^2-\underline{\large 2}\,y^3+\dfrac{dh}{dz} \] and compare it to (3) to get \[ \dfrac{dh}{dz}=4z^3+\underline{\large y^3} \] This is a contradiction because \(h\) is only a function of \(z\) and so its derivative cannot depend on \(y\). Consequently, there is no scalar potential.

If we try to find a scalar potential and derive a contradiction,
then no scalar potential exists.

### Using the Curl of a Gradient Identity

There is usually an easier way to show a scalar potential does not exist using the Curl of a Gradient Identity: \[ \vec\nabla\times\vec\nabla f=\vec0 \] If a vector field \(\vec F\) has a scalar potential, so that \(\vec F=\vec\nabla f\), then \[ \vec\nabla\times\vec F=\vec\nabla\times\vec\nabla f=\vec0 \] Turning this around, we have

If \(\vec\nabla\times\vec F\ne\vec 0\), then \(\vec F\) does not have a scalar potential.

Find a scalar potential for \(\vec F=\left\langle yz,2xz,3xy\right\rangle\) or show it does not exist.

We compute the curl of \(\vec F\): \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz & 2xz & 3xy \end{vmatrix} \\ &=\hat\imath(3x-2x)-\hat\jmath(3y-y)+ \hat k(2z-z) \\ &=\left\langle x,-2y,z\right\rangle\ne\vec0 \end{aligned}\] Since this is not \(\vec0\), there is no scalar potential.

Caution: The proposition does NOT say that if \(\vec\nabla\times\vec F=\vec0\) then \(\vec F\) has a scalar potential. However, \(99\%\) of the time, there is still a scalar potential. In fact, a therorem from higher math says, if \(\vec\nabla\times\vec F=\vec0\) in a region then \(\vec F\) has a scalar potential in that region. So for a scalar potential not to exist, even though \(\vec\nabla\times\vec F=\vec0\), we would have to be looking for a potential on a region with holes, like a donut. This does happen, for example, in studying the theory of magnetic monopoles, but certainly not in this class.

A region is simply connected if every circle in the region can be shrunk to a point while still staying in the region. Thus, a ball is simply connected while a donut is not.

Find a scalar potential for \(\vec F=\left\langle xz,yz,z^2\right\rangle\) or show it does not exist.

There is no scalar potential.

We compute the curl of \(\vec F\): \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xz & yz & z^2 \end{vmatrix} \\ &=\hat\imath(-y)-\hat\jmath(-x)+\hat k(0) \\ &=\left\langle -y,x,0\right\rangle\ne\vec0 \end{aligned}\] Since this is not \(\vec0\), there is no scalar potential.

Find a scalar potential for \(\vec F=\left\langle y+z,x+z,x+y\right\rangle\) or show it does not exist.

\(f=xy+xz+yz+C\)

We compute the curl of \(\vec F\): \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ y+z & x+z & x+y \end{vmatrix} \\ &=\hat\imath(1-1)-\hat\jmath(1-1)+\hat k(1-1) \\ &=\vec0 \end{aligned}\] Since this is \(\vec0\), we expect there is a scalar potential. To find it, we solve: \[ \partial_x f=y+z \qquad \partial_y f=x+z \qquad \partial_z f=x+y \] By inspection, the scalar potential is: \[ f=xy+xz+yz+C \]

There is some terminology to describe this distinction between \(\vec\nabla\times\vec F=\vec0\) and \(\vec F=\vec\nabla f\).

A vector field \(\vec F\) is called irrotational or curl-free if \(\vec\nabla\times\vec F=\vec0\).

A vector field \(\vec F\) is called conservative or a gradient field if \(\vec F=\vec\nabla f\) for some scalar potential, \(f\).

Consequently, a gradient field is curl-free. Equivalently, a conservative field is irrotational. But the converse is not necessarily true, except that most often, a curl-free field is a gradient.

In physics, if a force is conservative, then its scalar potential is the negative of its potential energy (hence the word "potential"). Further, the work done by the force will equal the change in potential energy, guaranteeing the conservation of energy (hence the word "conservative").

Optional: When can it happen that there is NO scalar potential even though \(\vec\nabla\times\vec F=\vec0\). As an exmple, consider the region which is all of \(\mathbb R^3\) except for the \(z\)-axis. This region is not simply connected because a circle which goes around the \(z\)-axis cannot be shrunk to a point.

Notice the vector field, \(\vec F=\left\langle \dfrac{2x}{x^2+y^2},\dfrac{2y}{x^2+y^2},0\right\rangle\) is defined throughout \(\mathbb R^3\) except for the \(z\)-axis. Find a scalar potential for \(\vec F\) defined on all of \(\mathbb R^3\) except for the \(z\)-axis, or show it does not exist.

A scalar potential is \(f=\ln(x^2+y^2)\) which is defined on all of \(\mathbb R^3\) except for the \(z\)-axis.

We compute the curl of \(\vec F\): \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \dfrac{2x}{x^2+y^2} & \dfrac{2y}{x^2+y^2} & 0 \end{vmatrix} \\ &=\hat\imath(0)-\hat\jmath(0) \\ &\quad+\hat k\left(\dfrac{-4xy}{(x^2+y^2)^2} -\dfrac{-4xy}{(x^2+y^2)^2}\right) \\ &=\langle0,0,0\rangle \end{aligned}\] Since this is \(\vec0\), we expect there is a scalar potential. To find it, we solve: \[ \partial_x f=\dfrac{2x}{x^2+y^2} \qquad \partial_y f=\dfrac{2y}{x^2+y^2} \qquad \partial_z f=0 \] By inspection, a scalar potential is: \[ f=\ln(x^2+y^2) \] (Check this by computing its gradient.) Further, this function is defined on all of \(\mathbb R^3\) except where \(x^2+y^2=0\) which is the \(z\)-axis.

Notice the vector field, \(\vec F=\left\langle \dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2},0\right\rangle\) is defined throughout \(\mathbb R^3\) except for the \(z\)-axis. Find a scalar potential for \(\vec F\) defined on all of \(\mathbb R^3\) except for the \(z\)-axis, or show it does not exist.

There is no scalar potential on the entire plane missing the origin. The closest we can get is the function \[ f=\arctan\left(\dfrac{y}{x}\right)=\theta \] where \(\theta\) is the angular coordinate of cylindrical coordinates. \(\theta\) is defined everywhere except on some one radial ray starting from the origin, because the angle jumps by \(2\pi\) across this line.

We compute the curl of \(\vec F\): \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \dfrac{-y}{x^2+y^2} & \dfrac{x}{x^2+y^2} & 0 \end{vmatrix} \\ &=\hat\imath(0)-\hat\jmath(0) \\ &\quad+\hat k\left(\dfrac{x^2+y^2-x(2x)}{(x^2+y^2)^2} -\dfrac{(x^2+y^2)(-1)-(-y)(2y)}{(x^2+y^2)^2}\right) \\ &=\langle0,0,0\rangle \end{aligned}\] Since this is \(\vec0\), we expect there is a scalar potential. To find it, we solve: \[ \partial_x f=\dfrac{-y}{x^2+y^2} \qquad \partial_y f=\dfrac{x}{x^2+y^2} \qquad \partial_z f=0 \] By inspection, a scalar potential is: \[ f=\arctan\left(\dfrac{y}{x}\right) \] (Check this by computing its gradient.) Unfortunately, \(f\) is not defined on all of \(\mathbb R^3\) except for the \(z\)-axis. In fact, it is not defined when \(x=0\) which is the entire \(yz\)-plane. Can we fix this?

Notice, that \(\arctan\left(\dfrac{y}{x}\right)\) is the cylindrical coordinate \(\theta\). So we rewrite the scalar potential as \[ f=\theta \] which appears to be well-defined and continuous throughout \(\mathbb R^3\) except for the \(z\)-axis. The problem is that \(\theta\) is not single valued. When you go around the \(z\)-axis, \(\theta\) jumps by \(2\pi\). So it does not actually define a function throughout \(\mathbb R^3\) except for the \(z\)-axis. It cannot be a scalar potential. Further, any other possible scalar potential would have to differ by a constant. So there is no scalar potential.

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