# 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 \(\vec F\): \[ \vec\nabla\times\vec A=\vec F \]

## 1. Finding Vector Potentials

It is quite difficult to find a vector potential by inspection. So we will need to do some work. However, most problems only ask us to find "a" vector potential, not "all" vector potentials. We can use this to simplify the process of finding the vector potential.

Find a vector potential for \(\vec F=\left\langle -x-y,2z,z-y\right\rangle\).

We need to find a vector field \(\vec A\) satisfying \(\vec\nabla\times\vec A=\vec F\). For a general vector field \(\vec A\), the curl is \[\begin{aligned} \vec\nabla\times\vec A &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ A_1 & A_2 & A_3 \end{vmatrix} \\ &=\left\langle \partial_y A_3-\partial_z A_2,\partial_z A_1-\partial_x A_3, \partial_x A_2-\partial_y A_1 \right\rangle \end{aligned}\] So we need to solve the equations \[\begin{aligned} \partial_y A_3-\partial_z A_2&=F_1=-x-y\quad&&\text{(1)} \\ \partial_z A_1-\partial_x A_3&=F_2=2z\quad&&\text{(2)} \\ \partial_x A_2-\partial_y A_1&=F_3=z-y\quad&&\text{(3)} \end{aligned}\] for \(A_1\), \(A_2\) and \(A_3\).

Since we only want *some* vector potential, and not *all*
vector potentials, we can make assumptions about the solution and see if
they work. It turns out that it is always possible to find a vector
potential with \(A_3=0\). (Or any one component is \(0\). This will be
justified at the bottom of this page.) So the equations reduce to
\[\begin{aligned}
-\partial_z A_2&=-x-y\quad&&\text{(1)} \\
\partial_z A_1&=2z\quad&&\text{(2)} \\
\partial_x A_2-\partial_y A_1&=z-y\quad&&\text{(3)}
\end{aligned}\]
We solve these successively.

- Equation (1) says \(\partial_z A_2=x+y\). So \[ A_2=xz+yz+f(x,y) \] where the "constant" of integration, \(f(x,y)\), can be a function of \(x\) and \(y\) because we took the \(z\)-antiderivative.
- Equation (2) says \(\partial_z A_1=2z\). So \[ A_1=z^2+g(x,y) \] where the "constant" of integration, \(g(x,y)\), can again be a function of \(x\) and \(y\).
- Using what we know about \(A_1\) and \(A_2\), equation (3) becomes \[\begin{aligned} \partial_x A_2-\partial_y A_1&=z-y \\ (z+\partial_x f)-(\partial_y g)&=z-y \end{aligned}\] or \[ \partial_x f-\partial_y g=-y \] which we need to solve for \(f\) and \(g\). Since we are only looking for one solution, we can pick \(g=0\) and solve for \[ f=-xy \]
- Reassembling the solution, we find the vector potential is \[ \vec A=\left\langle A_1,A_2,A_3\right\rangle =\left\langle z^2,xz+yz-xy,0\right\rangle \]

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

Notice that there are many correct answers. We did not necessarily need to pick \(A_3=0\) and could have solved for \(f\) and \(g\) differently.

Find a vector potential for
\(\vec F=\left\langle -\sin(x),\cos(y),z\cos(x)+z\sin(y)\right\rangle\).

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

\(\vec A=\left\langle z\cos(y),z\sin(x),0\right\rangle\)

There are many other correct solutions.

We need to find a vector field \(\vec A\) satisfying \(\vec\nabla\times\vec A=\vec F\). So we need to solve \[\begin{aligned} \partial_y A_3-\partial_z A_2&=F_1=-\sin(x)\quad&&\text{(1)} \\ \partial_z A_1-\partial_x A_3&=F_2=\cos(y)\quad&&\text{(2)} \\ \partial_x A_2-\partial_y A_1&=F_2=z\cos(x)+z\sin(y)\quad&&\text{(3)} \end{aligned}\] for \(A_1\), \(A_2\) and \(A_3\). We assume \(A_3=0\). So the equations reduce to \[\begin{aligned} -\partial_z A_2&=-\sin(x)\quad&&\text{(1)} \\ \partial_z A_1&=\cos(y)\quad&&\text{(2)}\\ \partial_x A_2-\partial_y A_1&=z\cos(x)+z\sin(y)\quad&&\text{(3)} \end{aligned}\] We solve these successively.

- Equation (1) says \(\partial_z A_2=\sin(x)\). So \[ A_2=z\sin(x)+f(x,y) \]
- Equation (2) says \(\partial_z A_1=\cos(y)\). So \[ A_1=z\cos(y)+g(x,y) \]
- Using what we know about \(A_1\) and \(A_2\), equation (3) becomes \[\begin{aligned} \partial_x A_2\qquad-\qquad\partial_y A_1\qquad &=z\cos(x)+z\sin(y) \\ (z\cos(x)+\partial_x f)-(-z\sin(y)+\partial_y g) &=z\cos(x)+z\sin(y) \end{aligned}\] or \[ \partial_x f-\partial_y g=0 \] which we need to solve for \(f\) and \(g\). This is trivial. We take \(f=g=0\)
- Reassembling the solution, we find the vector potential is \[ \vec A=\left\langle A_1,A_2,A_3\right\rangle =\left\langle z\cos(y),z\sin(x),0\right\rangle \]

There are many other correct solutions.

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

### Non-Uniqueness of the Vector Potential

We have said that there are many vector potentials for a given vector field, but how much non-uniqueness is there?

If \(\vec A\) and \(\vec B\) are both vector potentials for a vector field
\(\vec F\), then \(\vec A=\vec B+\vec C\) for some curl-free vector field
\(\vec C\).

Since \(\vec F=\vec\nabla\times\vec A=\vec\nabla\times\vec B\), we have \(\vec\nabla\times(\vec A-\vec B)=\vec0\). Consequently, \(\vec C=\vec A-\vec B\) is curl free.

Since every gradient field is curl-free and most curl-free fields are gradient fields, one frequently says you can add an arbitrary gradient, \(\vec\nabla f\), to a vector potential, \(\vec B\), and get another vector potential, \(\vec A\): \[ \vec A=\vec B+\vec\nabla f \]

We can now fulfill a promise from earlier on this page.

If a vector field \(\vec F\) has a vector potential, \(\vec B\), then
there is another vector potential, \(\vec A\), for which \(A_3=0\).

Since \(\vec B\) is a vector potential for \(\vec F\), so is \(\vec A=\vec B+\vec\nabla f\) for any function \(f\). We want to find \(f\) so that \(A_3=0\). The equation for \(A_3\) is: \[ A_3=B_3+\partial_z f \] So we pick \(f\) so that \(\partial_z f=-B_3\). In other words, \(f\) can be any \(z\)-antiderivative of \(-B_3\).

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