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.

1. Equation (1) says \(\partial_z A_2=x+y\). So \[ A_2=xz+yz+f(x,y) \]
2. Equation (2) says \(\partial_z A_1=2z\). So \[ A_1=z^2+g(x,y) \]
3. 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 contractable 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.

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

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.