# 17. Divergence, Curl and Potentials

## g. Vector Potentials

Given a vector field \(\vec F\), a vector potential for \(\vec F\) is any vector field \(\vec A\) whose curl is \(\vec F\): \[ \vec\nabla\times\vec A=\vec F \]

## 2. Non-Existence of a Vector Potential

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

Find a vector potential for \(\vec F=\left\langle -x-y,2z,\underline{\large 2}\,z-y\right\rangle\) or show it does not exist.

Notice that this is almost the same vector field as in the example on the previous page except for the extra factor of \(\underline{\large 2}\). So watch the differences in the solutions, all of which are underlined in a larger font.

We attempt to find a vector field \(\vec A\) satisfying \(\vec\nabla\times\vec A=\vec F\). Assuming \(A_3=0\), the equations reduce to \[\begin{aligned} -\partial_z A_2&=-x-y\quad&&\text{(1)} \\ \partial_z A_1&=2z\quad&&\text{(2)} \\ \partial_x A_2-\partial_y A_1&=\underline{\large 2}\,z-y\quad&&\text{(3)} \end{aligned}\] We solve these successively.

- Equation (1) says \(\partial_z A_2=x+y\). So \[ A_2=xz+yz+f(x,y) \]
- Equation (2) says \(\partial_z A_1=2z\). So \[ A_1=z^2+g(x,y) \]
- Using what we know about \(A_1\) and \(A_2\), equation (3) becomes \[\begin{aligned} \partial_x A_2-\partial_y A_1&=\underline{\large 2}\,z-y \\ (z+\partial_x f)-(\partial_y g)&=\underline{\large 2}\,z-y \end{aligned}\] or \[ \partial_x f-\partial_y g=\underline{\large z}-y \] which we need to solve for \(f\) and \(g\). However, this is a contradiction because \(f\) and \(g\) are functions of only \(x\) and \(y\). So their derivatives cannot depend on \(z\). Consequently, there is no vector potential.

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

### Using the Divergence of a Curl Identity

There is usually an easier way to show a vector potential does not exist using the Divergence of a Curl Identity: \[ \vec\nabla\cdot\vec\nabla\times\vec A=0 \] If a vector field \(\vec F\) has a vector potential, so that \(\vec F=\vec\nabla\times\vec A\), then \[ \vec\nabla\cdot\vec F=\vec\nabla\cdot\vec\nabla\times\vec A=0 \] Turning this around, we have

If \(\vec\nabla\cdot F\ne 0\) then \(\vec F\) does not have a vector potential.

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

We compute the divergence of \(\vec F\): \[ \vec\nabla\cdot F =\partial_x(xy)+\partial_y(yz)+\partial_z(zx) =y+z+x\ne0 \] Since this is not \(0\), there is no vector potential. (Finding the divergence is a really easy thing to do before going through the effort of finding a vector potential.)

Caution: The proposition does NOT say that if \(\vec\nabla\cdot\vec F=0\) then \(\vec F\) has a vector potential. However, \(99\%\) of the time, there is a vector potential. In fact, a therorem from higher math says, if \(\vec\nabla\cdot\vec F=0\) in a region then \(\vec F\) has a vector potential in that region. So for a vector potential not to exist, even though \(\vec\nabla\cdot\vec F=0\), we would have to be looking for a potential on a region with holes, like Swiss cheese. This does happen, but certainly not in this class.

A region is contractable if the region can be shrunk to a point while still staying in the region. Thus a ball is contractable while a donut or Swiss cheese is not.

Find a vector potential for the vector field
\(\vec F=\left\langle x^2,y^2,z^2\right\rangle\)
or show it does not exist.

Be sure to check your work by computing \(\vec\nabla\times\vec A\).

There is no vector potential.

We first check if the vector potential exists by computing its divergence: \[ \vec\nabla\cdot\vec F=\partial_x(x^2)+\partial_y(y^2)+\partial_z(z^2) =2x+2y+2z\ne0 \] Since \(\vec\nabla\cdot\vec F\ne0\), there is no vector potential \(\vec A\).

There is nothing to check, since there is no vector potential.

Find a vector potential for the vector field
\(\vec F=\left\langle -x,-2x,z-x\right\rangle\)
or show it does not exist.

Be sure to check your work by computing \(\vec\nabla\times\vec A\).

\(\vec A=\left\langle -2xz+xy,xz,0\right\rangle\)

There are many other correct solutions.

We first check if the vector potential exists by computing its divergence:
\[
\vec\nabla\cdot\vec F=\partial_x(-x)+\partial_y(-2x)+\partial_z(z-x)=-1+0+1=0
\]
Since \(\vec\nabla\cdot\vec F=0\) the vector field most likely has a
vector potential \(\vec A\). We need to solve the equation
\(\vec\nabla\times\vec A=\vec F\). Assuming \(A_3=0\), the equations reduce to
\[\begin{aligned}
-\partial_z A_2&=-x\quad&&\text{(1)} \\
\partial_z A_1&=-2x\quad&&\text{(2)} \\
\partial_x A_2-\partial_y A_1&=z-x\quad&&\text{(3)}
\end{aligned}\]
Equation (1) says \(A_2=xz+f(x,y)\).

Equation (2) says \(A_1=-2xz+g(x,y)\).

Equation (3) reduces to
\[
(z+\partial_x f)-(\partial_y g)=z-x
\]
or
\[
\partial_x f-\partial_y g=-x
\]
We take \(f=0\) and \(g=xy\). So a vector potential is
\[
\vec A=\left\langle A_1,A_2,A_3\right\rangle
=\left\langle -2xz+xy,xz,0\right\rangle
\]
There are many other correct solutions.

We check our answer by computing the curl of \(\vec A\): \[\begin{aligned} \text{curl}\vec A&=\vec\nabla\times\vec A =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -2xz+xy & xz & \quad0\quad \end{vmatrix} \\ &=\hat\imath(0-x)-\hat\jmath(0--2x)+\hat k(z-x)\\ &=\left\langle -x,-2x,z-x\right\rangle \end{aligned}\] which is our original vector field \(\vec F\).

There is some terminology to describe this distinction between \(\vec\nabla\cdot\vec F=0\) and \(\vec F=\vec\nabla\times\vec A\).

A vector field \(\vec F\) is called incompressible or divergence-free if \(\vec\nabla\cdot\vec F=0\).

A vector field \(\vec F\) is called solenoidal or a curl field if \(\vec F=\vec\nabla\times\vec A\) for some vector potential, \(\vec A\).

Consequently, a curl field is divergence-free. Equivalently, a solenoidal field is incompressible. But the converse is not necessarily true, except that most often, a divergence-free field is a curl.

In fluid mechanics, a fluid is called incompressible if its fluid velocity is divergence-free. In electromagnetism, the magnetic field is always divergence-free and usually found by writing it as the curl of a vector potential.

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