Vector and scalar potentials
The final part of our module concerns special types of vector fields.
We call a field \({\mathbf{A}}\) solenoidal if \[ \nabla\cdot {\mathbf{A}} = 0 \] and we call a field \({\mathbf{B}}\) irrotational if \[ \nabla\times{\mathbf{B}} = {\mathbf{0}}. \]
A field can be both solenoidal and irrotational, for example any constant field. But in general they are different things.
Irrotational fields
These are sometimes called curl free fields, for obvious reasons. The key theorem is:
\[\nabla\times{\mathbf{A}} = {\mathbf{0}} \text{ if and only if }{\mathbf{A}} = \nabla \phi\text{ for some scalar field }\phi.\]
We call \(\phi\) a scalar potential for \({\mathbf{A}}\).
Proving \(\impliedby\) is easy: If \({\mathbf{A}} = \nabla \phi\) then \[ \nabla\times {\mathbf{A}} = \nabla \times \nabla \phi = 0, \] this was one of our second derivative identities.
The proof in the other direction is much more complicated, but for any given example you can make it work.
Example
Find a scalar potential for \[ {\mathbf{A}} = [x^2,y^2,0]. \]
First check that \(\nabla\times{\mathbf{A}}={\mathbf{0}}\), otherwise it doesn’t work.
Then we need to find \(\phi\) such that \(\nabla \phi = [x^2,y^2,0]\).
Maybe you can spot it directly, but otherwise we can break it down. We need to find \(\phi(x,y,z)\) such that \[ \begin{aligned} {\frac{\partial \phi}{\partial x}} &= x^2,\qquad\text{(1)} \\ {\frac{\partial \phi}{\partial y}} &= y^2,\qquad\text{(2)}\\ {\frac{\partial \phi}{\partial z}} &= 0.\qquad\text{(3)} \end{aligned} \]
Integrating the (1), we have \[\phi(x,y,z) = \frac{1}{3} x^3 + C_1(y,z)\] Here \(C_1\) is just a normal constant of integration, but because we integrated a partial derivative, this constant can depend on \(y\) and \(z\).
Then (2) becomes \[{\frac{\partial C_1}{\partial y}} = y^2\] so \[ C_1 = \frac{1}{3} y^3 + C_2(z).\] \(C_1\) depends only on \(y\) and \(z\), so the new constant of integration \(C_2\) depends only on \(z\). So we have \[ \phi = \frac{1}{3}x^3 + \frac{1}{3}y^3 + C_2(z), \] so (3) gives \[ \frac{dC_2}{dz} = 0 \] so \(C_2\) is just a constant.
So finally we have \[ \phi(x,y,z) = \frac{1}{3}x^3 + \frac{1}{3}y^3 + C_2, \] where \(C_2\) is any constant.
Scalar potentials are not unique - you can add any constant to \(\phi\) and it’s still valid.
Solenoidal fields
These are also called incompressible or divergence free1. The key theorem here is:
\[\nabla\cdot{\mathbf{A}} = {\mathbf{0}} \text{ if and only if }{\mathbf{A}} = \nabla\times {\mathbf{B}}\text{ for some vector field }{\mathbf{B}}.\]
We call \({\mathbf{B}}\) a vector potential for \({\mathbf{A}}\). Again, you can easily prove that \(\nabla\times{\mathbf{B}}\) is solenoidal, but the other way is harder. You can also easily prove this:
Vector potentials are not unique - If \({\mathbf{B}}\) is a vector potential for \({\mathbf{A}}\) then so is \({\mathbf{B}}+\nabla\phi\) for any scalar field \(\phi\).2
This non-uniqueness is very important, because it means we can choose properties we want \({\mathbf{B}}\) to have. One convenient choice is often \[B_3=0,\] no component in the \(z\) direction.
Example
Find a vector potential for \[ {\mathbf{A}} = [x+y^2, -y+z^2, y^2] \]
This works the same as for scalar potentials, but it gets much messier. I strongly recommend staring at it for 30 seconds to see if you can guess a potential that works, before trying as follows:
First, you could check it’s actually solenoidal. That’s quite quick in this case.
We need to find \({\mathbf{B}}=[B_1,B_2,B_3]\) such that \[ [x+y^2, -y+z^2, y^2] = \nabla\times{\mathbf{B}} = [{\frac{\partial B_3}{\partial y}}-{\frac{\partial B_2}{\partial z}},\,{\frac{\partial B_1}{\partial z}}-{\frac{\partial B_3}{\partial x}},\,{\frac{\partial B_2}{\partial x}}-{\frac{\partial B_1}{\partial y}}]. \]
Let’s make the gauge choice that \(B_3=0\), so then we get \[x+y^2 = -{\frac{\partial B_2}{\partial z}},\] \[-y+z^2 = {\frac{\partial B_1}{\partial z}},\] \[y^2 = {\frac{\partial B_2}{\partial x}} - {\frac{\partial B_1}{\partial y}}.\]
Integrating the first two of these, we have: \[ B_2 = -x z - y^2 z + f(x, y) \] and \[ B_1 = -y z + \frac{1}{3} z^3 + g(x, y) \] where \(f\) and \(g\) are some unknown integration constant functions.
Now, substitute \(B_1\) and \(B_2\) into the third equation: \[ y^2 = (-z + {\frac{\partial f}{\partial x}}) - (-z + {\frac{\partial g}{\partial y}}) = {\frac{\partial f}{\partial x}} - {\frac{\partial g}{\partial y}} \]
We still have some freedom here. A simple choice is \(f(x, y) = 0\) and \(g(x, y) = -\frac{1}{3} y^3\): \[ {\frac{\partial f}{\partial x}} = 0,\quad {\frac{\partial g}{\partial y}} = -y^2 \] So, \[ y^2 = 0 - (-y^2) = y^2 \] This works.
Therefore, a vector potential is: \[ {\mathbf{B}} = \left[ -y z + \frac{1}{3} z^3 - \frac{1}{3} y^3,\; -x z - y^2 z,\; 0. \right] \]
As with many things in maths, finding the potentials is slow but checking they are valid is quick. Always check!