Simplifying the previous adjoint matrix results.

We previously found determinant expressions for the matrix elements of the adjoint for 2D and 3D matrices $$M$$. However, we can extract additional structure from each of those results.

2D case.

Given a matrix expressed in block matrix form in terms of it’s columns
M =
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix},

we found that the adjoint $$A$$ satisfying $$M A = \Abs{M} I$$ had the structure
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
& \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{bmatrix}.

We initially had wedge product expressions for each of these matrix elements, and can discover our structure by putting back those wedge products. Modulo sign, each of these matrix elemens has the form
\begin{aligned}
\begin{vmatrix} \Be_i & \Bm_j \end{vmatrix}
&=
\lr{ \Be_i \wedge \Bm_j } i^{-1} \\
&=
\lr{ \Be_i \wedge \Bm_j } i^{-1}
} \\
&=
\lr{ \Be_i \Bm_j – \Be_i \cdot \Bm_j } i^{-1}
} \\
&=
\Be_i \Bm_j i^{-1}
} \\
&=
\Be_i \cdot \lr{ \Bm_j i^{-1} },
\end{aligned}

where $$i = \Be_{12}$$. The adjoint matrix is
A =
\begin{bmatrix}
-\lr{ \Bm_2 i } \cdot \Be_1 & -\lr{ \Bm_2 i } \cdot \Be_2 \\
\lr{ \Bm_1 i } \cdot \Be_1 & \lr{ \Bm_1 i } \cdot \Be_2 \\
\end{bmatrix}.

If we use a column vector representation of the vectors $$\Bm_j i^{-1}$$, we can write the adjoint in a compact hybrid geometric-algebra matrix form
A =
\begin{bmatrix}
-\lr{ \Bm_2 i }^\T \\
\lr{ \Bm_1 i }^\T
\end{bmatrix}.

Check:

Let’s see if this works, by multiplying with $$M$$
\begin{aligned}
A M &=
\begin{bmatrix}
-\lr{ \Bm_2 i }^\T \\
\lr{ \Bm_1 i }^\T
\end{bmatrix}
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix} \\
&=
\begin{bmatrix}
-\lr{ \Bm_2 i }^\T \Bm_1 & -\lr{ \Bm_2 i }^\T \Bm_2 \\
\lr{ \Bm_1 i }^\T \Bm_1 & \lr{ \Bm_1 i }^\T \Bm_2
\end{bmatrix}.
\end{aligned}

Those dot products have the form
\begin{aligned}
\lr{ \Bm_j i }^\T \Bm_i
&=
\lr{ \Bm_j i } \cdot \Bm_i \\
&=
\gpgradezero{ \lr{ \Bm_j i } \Bm_i } \\
&=
\gpgradezero{ -i \Bm_j \Bm_i } \\
&=
-i \lr{ \Bm_j \wedge \Bm_i },
\end{aligned}

so
\begin{aligned}
A M &=
\begin{bmatrix}
i \lr{ \Bm_2 \wedge \Bm_1 } & 0 \\
0 & -i \lr { \Bm_1 \wedge \Bm_2 }
\end{bmatrix} \\
&=
\Abs{M} I.
\end{aligned}

We find the determinant weighted identity that we expected. Our methods are a bit schizophrenic, switching fluidly between matrix and geometric algebra representations, but provided we are careful enough, this isn’t problematic.

3D case.

Now, let’s look at the 3D case, where we assume a column vector representation of the matrix of interest
M =
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix},

and try to simplify the expression we found for the adjoint
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_3 & \Bm_1 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_1 & \Bm_2 \end{vmatrix}
\end{bmatrix}.

As with the 2D case, let’s re-express these determinants in wedge product form. We’ll write $$I = \Be_{123}$$, and find
\begin{aligned}
\begin{vmatrix} \Be_i & \Bm_j & \Bm_k \end{vmatrix}
&=
\lr{ \Be_i \wedge \Bm_j \wedge \Bm_k } I^{-1} \\
&=
\gpgradezero{ \lr{ \Be_i \wedge \Bm_j \wedge \Bm_k } I^{-1} } \\
&=
\Be_i \lr{ \Bm_j \wedge \Bm_k }
\Be_i \cdot \lr{ \Bm_j \wedge \Bm_k }
} I^{-1} } \\
&=
\Be_i \lr{ \Bm_j \wedge \Bm_k }
I^{-1} } \\
&=
\Be_i \lr{ \Bm_j \cross \Bm_k } I
I^{-1} } \\
&=
\Be_i \cdot \lr{ \Bm_j \cross \Bm_k }.
\end{aligned}

We see that we can put the adjoint in block matrix form
A =
\begin{bmatrix}
\lr{ \Bm_2 \cross \Bm_3 }^\T \\
\lr{ \Bm_3 \cross \Bm_1 }^\T \\
\lr{ \Bm_1 \cross \Bm_2 }^\T \\
\end{bmatrix}.

Check:

\begin{aligned}
A M
&=
\begin{bmatrix}
\lr{ \Bm_2 \cross \Bm_3 }^\T \\
\lr{ \Bm_3 \cross \Bm_1 }^\T \\
\lr{ \Bm_1 \cross \Bm_2 }^\T \\
\end{bmatrix}
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix} \\
&=
\begin{bmatrix}
\lr{ \Bm_2 \cross \Bm_3 }^\T \Bm_1 & \lr{ \Bm_2 \cross \Bm_3 }^\T \Bm_2 & \lr{ \Bm_2 \cross \Bm_3 }^\T \Bm_3 \\
\lr{ \Bm_3 \cross \Bm_1 }^\T \Bm_1 & \lr{ \Bm_3 \cross \Bm_1 }^\T \Bm_2 & \lr{ \Bm_3 \cross \Bm_1 }^\T \Bm_3 \\
\lr{ \Bm_1 \cross \Bm_2 }^\T \Bm_1 & \lr{ \Bm_1 \cross \Bm_2 }^\T \Bm_2 & \lr{ \Bm_1 \cross \Bm_2 }^\T \Bm_3
\end{bmatrix} \\
&=
\Abs{M} I.
\end{aligned}

Essentially, we found that the rows of the adjoint matrix are each parallel to the reciprocal frame vectors of the columns of $$M$$. This makes sense, as the reciprocal frame encodes a generalized inverse of sorts.

January 16, 2024 math and physics play , , , , ,

I started reviewing a book draft that mentions the adjoint in passing, but I’ve forgotten what I knew about the adjoint (not counting self-adjoint operators, which is different.) I do recall that adjoint matrices were covered in high school linear algebra (now 30+ years ago!), but never really used after that.

It appears that the basic property of the adjoint $$A$$ of a matrix $$M$$, when it exists, is
M A = \Abs{M} I,

so it’s proportional to the inverse, where the numerical factor is the determinant of that matrix. Let’s try to compute this beastie for 1D, 2D, and 3D cases.

Simplest case: $$1 \times 1$$ matrix.

For a one by one matrix, say
M =
\begin{bmatrix}
m_{11}
\end{bmatrix},

the determinant is just $$\Abs{M} = m_11$$, so our adjoint is the identity matrix
A =
\begin{bmatrix}
1
\end{bmatrix}.

Not too interesting. Let’s try the 2D case.

Less trivial case: $$2 \times 2$$ matrix.

For the 2D case, let’s define our matrix as a pair of column vectors
M =
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix},

and let’s write the adjoint out in full in coordinates as
A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}.

We seek solutions to a pair of vector equations
\begin{aligned}
\Bm_1 a_{11} + \Bm_2 a_{21} &= \Abs{M} \Be_1 \\
\Bm_1 a_{12} + \Bm_2 a_{22} &= \Abs{M} \Be_2.
\end{aligned}

We can immediately solve either of these, by taking wedge products, yielding
\begin{aligned}
\lr{ \Bm_1 \wedge \Bm_2 } a_{11} + \lr{ \Bm_2 \wedge \Bm_2 } a_{21} &= \Abs{M} \lr{ \Be_1 \wedge \Bm_2 } \\
\lr{ \Bm_1 \wedge \Bm_1 } a_{11} + \lr{ \Bm_1 \wedge \Bm_2 } a_{21} &= \Abs{M} \lr{ \Bm_1 \wedge \Be_1 } \\
\lr{ \Bm_1 \wedge \Bm_2 } a_{12} + \lr{ \Bm_2 \wedge \Bm_2 } a_{22} &= \Abs{M} \lr{ \Be_2 \wedge \Bm_2 } \\
\lr{ \Bm_1 \wedge \Bm_1 } a_{12} + \lr{ \Bm_1 \wedge \Bm_2 } a_{22} &= \Abs{M} \lr{ \Bm_1 \wedge \Be_2}.
\end{aligned}

Any wedge with a repeated vector is zero.
Provided the determinant is non-zero, we can divide both sides by $$\Bm_1 \wedge \Bm_2 = \Abs{M} \Be_{12}$$ to find a single determinant for each element in the adjoint
\begin{aligned}
a_{11} &= \begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} \\
a_{21} &= \begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} \\
a_{12} &= \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
a_{22} &= \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{aligned}

or
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
& \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{bmatrix},

or
A_{ij} =
\epsilon_{ir}
\begin{vmatrix}
\Be_j & \Bm_r
\end{vmatrix},

where $$\epsilon_{ir}$$ is the completely antisymmetric tensor, and the Einstein summation convention is in effect (summation implied over any repeated indexes.)

Check:

We should verify that expanding these determinants explicitly reproduces the usual representation of the 2D adjoint:
\begin{aligned}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} &= \begin{vmatrix} 1 & m_{12} \\ 0 & m_{22} \end{vmatrix} = m_{22} \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} &= \begin{vmatrix} m_{11} & 1 \\ m_{21} & 0 \end{vmatrix} = -m_{21} \\
\begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} &= \begin{vmatrix} 0 & m_{12} \\ 1 & m_{22} \end{vmatrix} = -m_{12} \\
\begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix} &= \begin{vmatrix} m_{11} & 0 \\ m_{21} & 1 \end{vmatrix} = m_{11},
\end{aligned}

or
A =
\begin{bmatrix}
m_{22} & -m_{12} \\
-m_{21} & m_{11}
\end{bmatrix}.

Multiplying everything out should give us determinant weighted identity
\begin{aligned}
M A
&=
\begin{bmatrix}
m_{11} & m_{12} \\
m_{21} & m_{22}
\end{bmatrix}
\begin{bmatrix}
m_{22} & -m_{12} \\
-m_{21} & m_{11}
\end{bmatrix} \\
&=
\lr{ m_{11} m_{22} – m_{12} m_{21} }
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
&= \Abs{M} I,
\end{aligned}

as expected.

3D case: $$3 \times 3$$ matrix.

For the 3D case, let’s also define our matrix as column vectors
M =
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix},

and let’s write the adjoint out in full in coordinates as
A =
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}.

This time, we seek solutions to three vector equations
\begin{aligned}
\Bm_1 a_{11} + \Bm_2 a_{21} + \Bm_3 a_{31} &= \Abs{M} \Be_1 \\
\Bm_1 a_{12} + \Bm_2 a_{22} + \Bm_3 a_{32} &= \Abs{M} \Be_2 \\
\Bm_1 a_{13} + \Bm_2 a_{23} + \Bm_3 a_{33} &= \Abs{M} \Be_3,
\end{aligned}

and can immediately solve, once again, by taking wedge products, yielding
\begin{aligned}
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{11} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{21} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{31} &= \Abs{M} \Be_1 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{11} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{21} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{31} &= \Abs{M} \Bm_1 \wedge \Be_1 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{11} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{21} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{31} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_1 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{12} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{22} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{32} &= \Abs{M} \Be_2 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{12} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{22} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{32} &= \Abs{M} \Bm_1 \wedge \Be_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{12} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{22} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{32} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_2 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{13} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{23} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{33} &= \Abs{M} \Be_3 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{13} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{23} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{33} &= \Abs{M} \Bm_1 \wedge \Be_3 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{13} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{23} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{33} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_3,
\end{aligned}

Any wedge with a repeated vector is zero.
Like before, provided the determinant is non-zero, we can divide both sides by $$\Bm_1 \wedge \Bm_2 \wedge \Bm_3 = \Abs{M} \Be_{123}$$ to find a single determinant for each element in the adjoint
\begin{aligned}
A &=
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Bm_1 & \Be_1 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_3 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Bm_1 & \Bm_2 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Bm_2 & \Be_2 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Bm_2 & \Be_3 \end{vmatrix}
\end{bmatrix} \\
&=
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_3 & \Bm_1 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_1 & \Bm_2 \end{vmatrix}
\end{bmatrix},
\end{aligned}

or
A_{ij} = \frac{\epsilon_{irs}}{2!} \begin{vmatrix} \Be_j & \Bm_r & \Bm_s \end{vmatrix}.

Observe that the inclusion of the $$\Be_j$$ column vector in this determinant, means that we really need only compute a $$2 \times 2$$ determinant for each adjoint matrix element. That is

A_{ij} = \frac{(-1)^j \epsilon_{irs}\epsilon_{jab}}{(2!)^2}
\begin{vmatrix}
m_{ar} & m_{as} \\
m_{br} & m_{bs}
\end{vmatrix}
.

This looks a lot like the usual minor/cofactor recipe, but written out explicitly for each element, using the antisymmetric tensor to encode the index alternation. It’s worth noting that there may be an error or subtle difference from the usual in my formulation, since wikipedia defines the adjoint as the transpose of the cofactor matrix, see: [1].

General case: $$n \times n$$ matrix.

It appears that if we wanted an induction hypotheses for the general $$n > 1$$ case, the $$ij$$ element of the adjoint matrix is likely
\begin{aligned}
A_{ij} &= \frac{\epsilon_{i s_1 s_2 \cdots s_{n-1}}}{(n-1)!} \begin{vmatrix} \Be_j & \Bm_{s_1} & \Bm_{s_2} & \cdots & \Bm_{s_{n-1}} \end{vmatrix} \\
&= \frac{(-1)^j \epsilon_{i r_1 r_2 \cdots r_{n-1}} \epsilon_{j s_1 s_2 \cdots s_{n-1}} }{\lr{(n-1)!}^2}
\begin{vmatrix}
m_{r_1 s_1} & \cdots & m_{r_1 s_{n-1}} \\
\vdots & & \vdots \\
m_{r_{n-1} s_{1}} & \cdots & m_{r_{n-1} s_{n-1}}
\end{vmatrix}.
\end{aligned}

I’m not going to try to prove this, inductively or otherwise.

References

[1] Wikipedia contributors. Minor (linear algebra) — Wikipedia, the free encyclopedia, 2023. URL https://en.wikipedia.org/w/index.php?title=Minor_(linear_algebra)&oldid=1182988311. [Online; accessed 16-January-2024].

Motivation.

This revisits my last blog post where I covered this content in a meandering fashion. This is an attempt to re-express this in a more compact form. In particular, in a form that is amenable to include in my book. When I wrote the potential section of my book, I cheated, and didn’t try to motivate the results. My cheat was figuring out the multivector potential representation starting with STA where things are simpler, and then translating it back to a multivector representation, instead of figuring out a reasonable way to motivate things from the foundation already laid.

I’d like to eventually have a less rushed treatment of potentials in my book, where the results are not pulled out of a magic hat. Here is an attempted step in that direction. I’ve opted to put some of the motivational material in problems (with solutions at the chapter end.)

Multivector potentials.

We know from conventional electromagnetism (given no fictitious magnetic sources) that we can represent the six components of the electric and magnetic fields in terms of four scalar fields
\label{eqn:mvpotentials:80}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} \\
\BH &= \inv{\mu} \spacegrad \cross \BA.
\end{aligned}

The conventional way of constructing these potentials makes use of the identities
\label{eqn:mvpotentials:60}
\begin{aligned}
\end{aligned}

applying those to the source free Maxwell’s equations to find representations of $$\BE, \BH$$ that automatically satisfy those equations. For that conventional analysis, see section 18-6 [2] (available online), or section 10.1 [3], or section 6.4 [4]. We can also find such a potential representation using geometric algebra methods that are cross product free (problem 1.)

For Maxwell’s equations with fictitious magnetic sources, it can be shown that a potential representation of the field
\label{eqn:mvpotentials:100}
\begin{aligned}
\BH &= -\spacegrad \phi_m – \PD{t}{\BF} \\
\BE &= -\inv{\epsilon} \spacegrad \cross \BF.
\end{aligned}

satisfies the source-free grades of Maxwell’s equation.
See [1], and [5] for such derivations. As with the conventional source potentials, we can also apply our geometric algebra toolbox to easily find these results (problem 2.)

We have a mix of time partials and curls that is reminiscent of Maxwell’s equation itself. It’s obvious to wonder whether there is a more coherent integrated form for the potential. This is in fact the case.

Lemma 1.1: Multivector potentials.

For Maxwell’s equation with electric sources, the total field $$F$$ can be expressed in multivector potential form
\label{eqn:mvpotentials:520}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.

For Maxwell’s equation with only fictitious magnetic sources, the total field $$F$$ can be expressed in multivector form
\label{eqn:mvpotentials:540}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } I \eta \lr{ -\phi_m + c \BF } }{1,2}.

The reader should try to verify this themselves (problem 3.)

Using superposition, we can form a multivector potential that includes all grades.

Definition 1.1: Multivector potential.

We call $$A$$, a multivector with all grades, the multivector potential, defining the total field as
\label{eqn:mvpotentials:600}
\begin{aligned}
F
&=
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } A

\end{aligned}

Imposition of the constraint
\label{eqn:mvpotentials:680}

is called the Lorentz gauge condition, and allows us to express $$F$$ in terms of the potential without any grade selection filters.

Lemma 1.2: Conventional multivector potential.

Let
\label{eqn:mvpotentials:620}
A = -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }.

This results in the conventional potential representation of the electric and magnetic fields
\label{eqn:mvpotentials:640}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} – \inv{\epsilon} \spacegrad \cross \BF \\
\end{aligned}

In terms of potentials, the Lorentz gauge condition \ref{eqn:mvpotentials:680} takes the form
\label{eqn:mvpotentials:660}
\begin{aligned}
0 &= \inv{c} \PD{t}{\phi} + \spacegrad \cdot (c \BA) \\
0 &= \inv{c} \PD{t}{\phi_m} + \spacegrad \cdot (c \BF).
\end{aligned}

See problem 4.

Problem 1: Potentials for no-fictitious sources.

Starting with Maxwell’s equation with only conventional electric sources
\label{eqn:mvpotentials:120}

Show that this may be split by grade into three equations
\label{eqn:mvpotentials:140}
\begin{aligned}
\spacegrad \wedge \BE + \inv{c}\PD{t}{} \lr{ I \eta \BH } &= 0 \\
\spacegrad \wedge \lr{ I \eta \BH } &= 0.
\end{aligned}

Then use the identities $$\spacegrad \wedge \spacegrad \wedge \BA = 0$$, for vector $$\BA$$ and $$\spacegrad \wedge \spacegrad \phi = 0$$, for scalar $$\phi$$ to find the potential representation.

Taking grade(0,1) and (2,3) selections of Maxwell’s equation, we split our equations into source dependent and source free equations
\label{eqn:mvpotentials:200}

\label{eqn:mvpotentials:220}

In terms of $$F = \BE + I \eta \BH$$, the source free equation expands to
\label{eqn:mvpotentials:240}
\begin{aligned}
0
&=
\lr{ \spacegrad + \inv{c} \PD{t}{} } \lr{ \BE + I \eta \BH }
}{2,3} \\
&=
&=
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \inv{c} \PD{t}{\BH},
\end{aligned}

which can be further split into a bivector and trivector equation
\label{eqn:mvpotentials:260}
0 = \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH}

\label{eqn:mvpotentials:280}
0 = \spacegrad \wedge \lr{ I \eta \BH }.

It’s clear that we want to write the magnetic field as a (bivector) curl, so we let
\label{eqn:mvpotentials:300}
I \eta \BH = I c \BB = c \spacegrad \wedge \BA,

or
\label{eqn:mvpotentials:301}
\BH = \inv{\mu} \spacegrad \cross \BA.

\Cref{eqn:mvpotentials:260} is reduced to
\label{eqn:mvpotentials:320}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH} \\
&= \spacegrad \wedge \BE + \inv{c} \PD{t}{} \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}

We can now let
\label{eqn:mvpotentials:340}
\BE + \PD{t}{\BA} = -\spacegrad \phi.

We sneakily adjust the sign of the gradient so that the result matches the conventional representation.

Problem 2: Potentials for fictitious sources.

Starting with Maxwell’s equation with only fictitious magnetic sources
\label{eqn:mvpotentials:160}

show that this may be split by grade into three equations
\label{eqn:mvpotentials:180}
\begin{aligned}
-\eta \spacegrad \wedge \BH + \inv{c}\PD{t}{(I \BE)} &= 0 \\
\spacegrad \wedge \lr{ I \BE } &= 0.
\end{aligned}

Then use the identities $$\spacegrad \wedge \spacegrad \wedge \BF = 0$$, for vector $$\BF$$ and $$\spacegrad \wedge \spacegrad \phi_m = 0$$, for scalar $$\phi_m$$ to find the potential representation \ref{eqn:mvpotentials:100}.

We multiply \ref{eqn:mvpotentials:160} by $$I$$ to find
\label{eqn:mvpotentials:360}

which can be split into
\label{eqn:mvpotentials:380}
\begin{aligned}
\end{aligned}

We expand the source free equation in terms of $$I F = I \BE – \eta \BH$$, to find
\label{eqn:mvpotentials:400}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } \lr{ I \BE – \eta \BH } }{0,3} \\
&= \spacegrad \wedge \lr{ I \BE } + \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH,
\end{aligned}

which has the respective bivector and trivector grades
\label{eqn:mvpotentials:420}
0 = \spacegrad \wedge \lr{ I \BE }

\label{eqn:mvpotentials:440}
0 = \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH.

We can clearly satisfy \ref{eqn:mvpotentials:420} by setting
\label{eqn:mvpotentials:460}
I \BE = -\inv{\epsilon} \spacegrad \wedge \BF,

or
\label{eqn:mvpotentials:461}
\BE = -\inv{\epsilon} \spacegrad \cross \BF.

Here, once again, the sneaky inclusion of a constant factor $$-1/\epsilon$$ is to make the result match the conventional. Inserting this value for $$I \BE$$ into our bivector equation yields
\label{eqn:mvpotentials:480}
\begin{aligned}
0
&= -\inv{\epsilon} \inv{c} \PD{t}{} (\spacegrad \wedge \BF) – \eta \spacegrad \wedge \BH \\
&= -\eta \spacegrad \wedge \lr{ \PD{t}{\BF} + \BH },
\end{aligned}

so we set
\label{eqn:mvpotentials:500}
\PD{t}{\BF} + \BH = -\spacegrad \phi_m,

and have a field representation that automatically satisfies the source free equations.

Problem 3: Total field in terms of potentials.

Prove lemma 1.1, either by direct expansion, or by trying to discover the multivector form of the field by construction.

Proof by expansion is straightforward, and left to the reader. We form the respective total electromagnetic fields $$F = \BE + I \eta H$$ for each case.

We find
\label{eqn:mvpotentials:560}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\spacegrad \phi – \PD{t}{\BA} + I \frac{\eta}{\mu} \spacegrad \cross \BA \\
&= -\spacegrad \phi – \inv{c} \PD{t}{(c \BA)} + \spacegrad \wedge (c\BA) \\
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – \inv{c} \PD{t}{(c \BA)} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}

For the field for the fictitious source case, we compute the result in the same way, inserting a no-op grade selection to allow us to simplify, finding
\label{eqn:mvpotentials:580}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\inv{\epsilon} \spacegrad \cross \BF + I \eta \lr{ -\spacegrad \phi_m – \PD{t}{\BF} } \\
&= \inv{\epsilon c} I \lr{ \spacegrad \wedge (c \BF)} + I \eta \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } \\
&= I \eta \lr{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } } \\
&= I \eta \gpgrade{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } }{1,2} \\
&= I \eta \gpgrade{ \spacegrad (-\phi_m + c \BF) – \inv{c} \PD{t}{(c \BF)} }{1,2} \\
&= I \eta \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } (-\phi_m + c \BF) }{1,2}.
\end{aligned}

Problem 4: Fields in terms of potentials.

Prove lemma 1.2.

Let’s expand and then group by grade
\label{eqn:mvpotentials:n}
\begin{aligned}
\lr{ \spacegrad – \inv{c} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }} \\
&=
-\inv{c} \PD{t}{\phi} + c \inv{c} \PD{t}{ \BA } + I \eta \lr{ -\inv{c} \PD{t}{\phi_m} + c \inv{c} \PD{t}{\BF} } \\
&=
+ I \eta c \spacegrad \wedge \BF
– c \inv{c} \PD{t}{\BA}
– c I \eta \inv{c} \PD{t}{\BF} \\
+\inv{c} \PD{t}{\phi}
+ \inv{c} \PD{t}{\phi_m} } \\
&=
– \PD{t}{\BA}
– \PD{t}{\BF}
} \\
+\inv{c} \PD{t}{\phi}
+ \inv{c} \PD{t}{\phi_m} }.
\end{aligned}

Observing that $$F = \gpgrade{ \lr{ \spacegrad -(1/c) \partial_t } A }{1,2} = \BE + I \eta \BH$$, completes the problem. If the Lorentz gauge condition is assumed, the scalar and pseudoscalar components above are obliterated, leaving just
$$F = \lr{ \spacegrad -(1/c) \partial_t } A$$.

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.

[2] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics, Volume II.[Lectures on physics], chapter The Maxwell Equations. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963. URL https://www.feynmanlectures.caltech.edu/II_18.html.

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[4] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[5] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

Conventional formulation.

The idea behind introducing the scalar potential $$\phi$$ and vector potential $$\BA$$ is that we can impose a constraint on the form of our observable fields $$\BE, \BB$$, (or $$\BD, \BH$$), that reduces the complexity and coupling of Maxwell’s equations. These potentials are not unique, but the types of allowed variations in those potentials (gauge transformations) do not change the observable fields.

The basic idea is that we are looking for representations of the fields that automatically satisfy the pair of source free Maxwell’s equations
\label{eqn:gapotentials:40}
\begin{aligned}
\spacegrad \cdot \BB &= 0 \\
c \partial_0 \BB + \spacegrad \cross \BE &= 0,
\end{aligned}

so that the problem is reduced to solving just the remaining source dependent Maxwell’s equations.

The conventional way of constructing these potentials makes use of the identities
\label{eqn:gapotentials:60}
\begin{aligned}
\end{aligned}

where $$\Bf$$ is a vector, and $$\chi$$ is a scalar. This approach is straightforward. Instead of replicating it, here are a few well known references where such a treatment can be found

1. section 18-6 potentials and the wave equation in [2] (available online),
2. section 10.1 The potential formulation in [3], and
3. section 6.4 Vector and Scalar Potentials, in [4],

Multivector potentials in geometric algebra.

The multivector form of Maxwell’s equation is
\label{eqn:gapotentials:820}
\lr{ \spacegrad + \partial_0 } F = J,

where $$\partial_0 = (1/c)\partial/\partial t$$, the electromagnetic field $$F = \BE + I c \BB = \BE + I \eta H$$ has grades(1,2), and a multivector charge and current density $$J$$. Grades(0,1) of the current are the charge and current densities respectively, and if desired, the grade(2,3) portion of the current has the fictitious magnetic charge and current densities (used in microwave and antenna engineering.)

It’s best to consider the case of electric sources, separately from the case of (fictitious) magnetic sources, and then use superposition to construct a potential representation that includes both.

We require a tool, that generalizes the $$\mathbb{R}^3$$ cross product curl identities above.

Lemma 1.1: Curl of curl.

Let $$A \in \bigwedge^k$$ be a blade of grade $$k$$. Then
\begin{equation*}
\nabla \wedge \nabla \wedge A = 0.
\end{equation*}

Observe that for scalar $$A$$, this reduces to
\label{eqn:gapotentials:1740}
\nabla \wedge \nabla A = 0.

We’ve recently proved this, so we won’t do it again now.

Now we are ready to figure out the structure of the potentials.

Case I. No (fictitious) magnetic sources.

Without magnetic sources, Maxwell’s equation is
\label{eqn:gapotentials:840}

This can be split into two equations, one that has just the sources, and one that is source free
\label{eqn:gapotentials:860}

\label{eqn:gapotentials:880}

If you are clever, or have the benefit of having worked out the answer already, you can look directly at \ref{eqn:gapotentials:880} and guess the multivector form for the potential. Hint: you want something closely related to $$F = \lr{ \spacegrad – \partial_0 } A$$, where $$A$$ has grades(0,1).

If you aren’t that clever, or don’t have a time machine that let’s you look that clever, you’ll have to work it out systematically like the rest of us. We can start by breaking down $$F$$ into it’s constituent observer dependent fields. That means that we want to find values for $$\BE, \BH$$ that satisfy
\label{eqn:gapotentials:900}
\gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{ \BE + I \eta \BH } }{2,3} = 0.

Expanding the multivector factors gives us
\label{eqn:gapotentials:920}
\begin{aligned}
\gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{ \BE + I \eta \BH } }{2,3}
&=
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \partial_0 \BH.
\end{aligned}

Splitting this into one equation for each grade, leaves us with
\label{eqn:gapotentials:940}
0 = \spacegrad \wedge \BE + I \eta \partial_0 \BH

\label{eqn:gapotentials:960}
0 = \spacegrad \wedge \lr{ I \eta \BH }.

Observe that we could have also written \ref{eqn:gapotentials:960} as $$0 = I \eta \lr{ \spacegrad \cdot \BH }$$, which is the starting point of the conventional non-GA approach.
It’s clear that we want to write $$I \eta \BH = I c \BB$$ as a (bivector) curl, and let
\label{eqn:gapotentials:980}
I \eta \BH = c \spacegrad \wedge \BA.

It’s a bit sneaky to toss that factor of $$c$$ in here, but that’s done to make the units of $$\BA$$ turn out in a way that matches the conventional vector potential. If it makes you feel better, you can think of this as an undetermined constant multiplicative undetermined factor that will be used to adjust the dimensions of $$\BA$$ down the line.

Having made that choice, \ref{eqn:gapotentials:960} is automatically satisfied, and \ref{eqn:gapotentials:940} is reduced to
\label{eqn:gapotentials:1000}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \partial_0 \BH \\
&= \spacegrad \wedge \BE + \partial_0 \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + c \partial_0 \BA }.
\end{aligned}

We can now let
\label{eqn:gapotentials:1020}
\BE + \partial_0 c \BA = -\spacegrad \phi.

Again, we had the option of including an arbitrary multiplicative constant, but this time, we managed to find the right switch for our time machine, and look ahead to see that we want that constant to be $$-1$$ in order to have agreement with the conventional result.

We are left with a potential construction for our individual field components
\label{eqn:gapotentials:1040}
\begin{aligned}
\BE &= -\spacegrad \phi – c \partial_0 \BA \\
I \eta \BH &= c \spacegrad \wedge \BA,
\end{aligned}

or
\label{eqn:gapotentials:1060}
F = -\spacegrad \phi – c \partial_0 \BA + c \spacegrad \wedge \BA.

This automatically satisfies the grades of Maxwell’s equation that are source free, leaving us to solve just
\label{eqn:gapotentials:1080}

Multivector potential.

It’s natural to wonder if there is a more structured form for $$F$$ than \ref{eqn:gapotentials:1060}, just as we found a GA structure for Maxwell’s equation that eliminated the crazy mix of divs and curls that we had in the original Gibbs representation. Let’s find that structure. To do so, we can enclose $$F$$ in a no-op grade selection operation
\label{eqn:gapotentials:1100}
\begin{aligned}
F
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – c \partial_0 \BA + \lr{ \partial_0 \phi – \partial_0 \phi } }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}

We can now introduce a multivector potential, and express the remaining non-zero grades of Maxwell’s equation in terms of this potential
\label{eqn:gapotentials:1120}
\begin{aligned}
A &= -\phi + c \BA \\
\end{aligned}

Lorentz gauge.

The grade selection in our representation of $$F$$ is a bit annoying, and can be eliminated if we impose additional constraints on the potential. We can write
\label{eqn:gapotentials:1140}
F =
\lr{ \spacegrad – \partial_0 } A

and then ask what conditions are required for this grade(0,3) selection to be zero. In terms of our constituent potentials, that is
\label{eqn:gapotentials:1160}
\begin{aligned}
0 &=
&=
\gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA } }{0,3} \\
&=
c \spacegrad \cdot \BA + \partial_0 \phi,
\end{aligned}

This is the Lorentz gauge condition, recognized a bit more easily if written out in terms of the time partials explicitly
\label{eqn:gapotentials:1180}
\inv{c^2} \PD{t}{\phi} + \spacegrad \cdot \BA = 0.

We can now write Maxwell’s equations, in the potential formulation, as
\label{eqn:gapotentials:1200}
\begin{aligned}
A &= -\phi + c \BA \\
F &= \lr{ \spacegrad – \partial_0 } A \\
0 &= \inv{c} \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} = \inv{c^2} \PD{t}{\phi} + \spacegrad \cdot \BA \\
\end{aligned}

This is quite nice. We have a one to one decoupled relationship between the potential and the current, and are free to use the well known techniques for solving the wave equation (using convolution and a superposition of advanced and retarded Green’s functions for the wave equation operator.)

Gauge transformation.

There’s one more thing that we should look at before moving on to the magnetic sources case, and that’s the question of gauge freedom. We’ve said that the potentials are not unique, but this non-uniqueness has a very specific form.

Since we’ve constructed $$F$$ with a grade selection as
\label{eqn:gapotentials:1220}

so it’s clear that any transformation
\label{eqn:gapotentials:1240}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \psi_{0,3},

where $$\psi_{0,3}$$ is any multivector with grades(0,3) components, will leave $$F$$ invariant. That is
\label{eqn:gapotentials:1260}
\begin{aligned}
A &= -\phi + c \BA \\
&\rightarrow
-\phi + c \BA + \lr{ \spacegrad + \partial_0 } \psi_{0,3} \\
&=
-\phi + c \BA + \lr{ \spacegrad + \partial_0 } \lr{ c \psi + I \bar{\psi} } \\
&=
\lr{ -\phi + c \partial_0 \psi }
+ c \lr{ \BA + \spacegrad \psi }
+ I \partial_0 \bar{\psi}.
\end{aligned}

We see that the contributions of $$\bar{\psi}$$ result in grade(2,3) terms, which are not of interest, and we find that a paired transformation of the potentials
\label{eqn:gapotentials:1280}
\begin{aligned}
\phi &\rightarrow \phi – \PD{t}{\psi} \\
\BA &\rightarrow \BA + \spacegrad \psi,
\end{aligned}

called a gauge transformation, leaves the field $$F$$ unchanged. This can be expressed slightly more compactly as
\label{eqn:gapotentials:1300}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } c \psi,

where, once again, the multiplicative constant $$c$$ is included so for consistency with the conventional expression for potential gauge transformation.

Case II. With (fictitious) magnetic sources.

With magnetic sources, Maxwell’s equation is
\label{eqn:gapotentials:1500}

We put this in dual form
\label{eqn:gapotentials:1520}

which now has the sources all with grades (0,1) as we just analyzed. The dual vector $$I F$$, like $$F$$, has only grade(1,2) components.

Expanding the source free Maxwell’s equations in terms of $$\BE, \BH$$, we have
\label{eqn:gapotentials:1340}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{I \BE – \eta \BH } }{2,3} \\
&= \gpgrade{ I \spacegrad \BE – \eta \spacegrad \BH + I \partial_0 \BE – \eta \partial_0 \BH }{2,3} \\
&= \spacegrad \wedge \lr{ I \BE } – \eta \spacegrad \wedge \BH + I \partial_0 \BE,
\end{aligned}

\label{eqn:gapotentials:1360}
0 = \spacegrad \wedge \lr{ I \BE },

\label{eqn:gapotentials:1361}
0 = – \eta \spacegrad \wedge \BH + I \partial_0 \BE.

We see that the dual electric field needs to be a curl to satisfy \ref{eqn:gapotentials:1360}
\label{eqn:gapotentials:1400}
I \BE = -\eta \spacegrad \wedge c \BF,

and after substitution into \ref{eqn:gapotentials:1361} we are left with
\label{eqn:gapotentials:1540}
\begin{aligned}
0
&= – \eta \spacegrad \wedge \BH + \partial_0 \lr{ – \eta \spacegrad \wedge c \BF } \\
&= \eta \spacegrad \wedge \lr{ -\BH – \partial_0 c \BF } \\
\end{aligned}

We set
\label{eqn:gapotentials:1420}
-\BH – \partial_0 c \BF = \spacegrad \phi_m,

Our fields are
\label{eqn:gapotentials:1440}
\begin{aligned}
\BE &= – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= -\spacegrad \phi_m – \PD{t}{\BF}.
\end{aligned}

This has the structure that matches the potential conventions from antenna theory, for example as stated in [1].

Multivector potential.

As with the electrical sources, we expect that we can write this as something like
\label{eqn:gapotentials:1460}

Let’s verify that this is the case.
\label{eqn:gapotentials:1480}
\begin{aligned}
F
&= I \eta \spacegrad \wedge (c \BF) -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF \\
&= \gpgrade{ I \eta \spacegrad \wedge (c \BF) -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF }{1,2} \\
&= \gpgrade{ I \eta \spacegrad c \BF -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF }{1,2} \\
&= \gpgrade{ I \eta \lr{ \spacegrad \lr{ – \phi_m + c \BF } – \partial_0 c \BF + \partial_0 \phi_m – \partial_0 \phi_m} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF } }{1,2}.
\end{aligned}

Lorentz gauge.

Let’s see what constraints we need to write our field in terms of a potential without a grade selection, that is
\label{eqn:gapotentials:1560}
F = \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF }.

We need the grade(0,3) components of this multivector to be zero. Those components are
\label{eqn:gapotentials:1580}
\begin{aligned}
0 &=
\gpgrade{ \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF }}{0,3} \\
&=
\gpgrade{-\spacegrad I \eta \phi_m+\spacegrad I \eta c \BF+ \partial_0 I \eta \phi_m – \partial_0 I \eta c \BF }{0,3} \\
&=
+ \partial_0 I \eta \phi_m \\
&=
I \eta \lr{ c \lr{ \spacegrad \cdot \BF} + \partial_0 \phi_m },
\end{aligned}

or
\label{eqn:gapotentials:1600}
0 = \inv{c^2} \PD{t}{\phi_m} + \spacegrad \cdot \BF.

This is the Lorentz gauge condition. With this condition we can we can express Maxwell’s equation with magnetic sources, as a forced wave equation
\label{eqn:gapotentials:1620}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
F &= \lr{ \spacegrad – \partial_0 } A \\
0 &= \inv{c} \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} = \inv{c^2} \PD{t}{\phi_m} + \spacegrad \cdot \BF \\
\end{aligned}

Gauge transformation.

Without the Lorentz gauge assumption, our potential representation for the field is
\label{eqn:gapotentials:1640}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
\end{aligned}

It’s clear that any transformation of the form
\label{eqn:gapotentials:1660}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \psi_{0,3},

leaves the field unchanged.
\label{eqn:gapotentials:1680}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
&\rightarrow
I \eta \lr{ -\phi + c \BF } + \lr{ \spacegrad + \partial_0 } \psi_{0,3} \\
&=
I \eta \lr{ -\phi_m + c \BF } + \lr{ \spacegrad + \partial_0 } \lr{ \psi + I \eta c \bar{\psi} } \\
&=
I \eta \lr{
-\phi_m
+ c \partial_0 \bar{\psi}
+ c \BF
}
+ \lr{ \spacegrad + \partial_0 } \psi.
\end{aligned}

We can drop the $$\psi$$ contributions, since this time we want only grades(2,3) in our potential, and find that the
desired form of the gauge transformation, for scalar $$\bar{\psi}$$, is
\label{eqn:gapotentials:1700}
\begin{aligned}
\phi_m &\rightarrow \phi_m – \PD{t}{\bar{\psi}} \\
\BF &\rightarrow \BF + \spacegrad \bar{\psi}.
\end{aligned}

The multivector form of this is
\label{eqn:gapotentials:1720}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } I \eta c \bar{\psi}.

Superposition.

We can now use superposition to construct a potential representation that works for both conventional electric and fictitious magnetic charges and currents.

Without a Lorentz gauge assumption, that is
\label{eqn:gapotentials:1760}
\begin{aligned}
A &= -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } \\
J &= \lr{ \spacegrad + \partial_0 } F,
\end{aligned}

where, given scalar functions $$\psi, \bar{\psi}$$, we are free to make gauge transformations of the multivector potential that satisfy
\label{eqn:gapotentials:1800}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \lr{ c \psi + I \eta c \bar{\psi} },

With a Lorentz gauge constraint, we have a wave equation operator acting on $$A$$, with the multivector current as a forcing term.
\label{eqn:gapotentials:1780}
\begin{aligned}
A &= -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } \\
F &= \lr{ \spacegrad – \partial_0 } A \\
J &= \lr{ \spacegrad^2 – \partial_{00} } A.
\end{aligned}

Check.

It’s worth expansion to verify that we got all the dimensional constants write, and compare the results to Maxwell’s equations in their Gibbs form.

Let’s start with an expansion of $$F$$ in terms of the potentials
\label{eqn:gapotentials:1820}
\begin{aligned}
F &=
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } }{1,2} \\
&=
\gpgrade{ \spacegrad \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } -\partial_0 \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } }{1,2} \\
&=
\gpgrade{ \spacegrad \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } -\partial_0 \lr{ c \BA + I \eta c \BF } }{1,2} \\
&=
-\partial_0 \lr{ c \BA + I \eta c \BF }.
\end{aligned}

That is
\label{eqn:gapotentials:1840}
\begin{aligned}
\BE &= -\spacegrad \phi + I \eta c \spacegrad \wedge \BF -c \partial_0 \BA \\
I \eta \BH &= c \spacegrad \wedge \BA – I \eta \spacegrad \phi_m – I \eta c \partial_0 \BF,
\end{aligned}

or
\label{eqn:gapotentials:1860}
\begin{aligned}
\BE &= – \spacegrad \phi -\partial_t \BA – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= – \spacegrad \phi_m – \partial_t \BF + \inv{\mu} \spacegrad \cross \BA.
\end{aligned}

All is good. This is exactly the form that we expect.

Let’s expand out Maxwell’s equation in terms of this potential representation and see what we get.

Let’s write the total field without the grade(1,2) selection, by subtracting off any grade(0,3) contributions
\label{eqn:gapotentials:1880}
F = \lr{ \spacegrad – \partial_0 } A – \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3}.

That difference term is
\label{eqn:gapotentials:1900}
\begin{aligned}
&=
– \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA – I \eta \phi_m + I \eta c \BF } }{0,3} \\
&=
– c \spacegrad \cdot \BA – I \eta c \spacegrad \cdot \BF – \partial_0 \phi – I \eta \partial_0 \phi_m.
\end{aligned}

The field is nicely split into a multivector term that depends directly on the full multivector potential $$A$$, and a difference term that wipes out any scalar and pseudoscalar terms
\label{eqn:gapotentials:1920}
F
=
\lr{ \spacegrad – \partial_0 } A
– \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } – I \eta \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF }.

Maxwell’s equations are now reduced to
\label{eqn:gapotentials:1940}
\lr{ \spacegrad^2 – \partial_{00} } A

\lr{ \partial_0 \phi + c \spacegrad \cdot \BA }

I \eta \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF }
= J.

This splits nicely into a single equation for each grade of $$A, J$$ respectively. We write
\label{eqn:gapotentials:1960}
J = \eta\lr{ c \rho – \BJ } + I \lr{ c \phi_m – \BM },

so
\label{eqn:gapotentials:1980}
\begin{aligned}
\lr{ \spacegrad^2 – \partial_{00} } (-\phi) – \partial_0 \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } &= \eta c \rho \\
\lr{ \spacegrad^2 – \partial_{00} } (c \BA) – \spacegrad \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } &= -\eta \BJ \\
\lr{ \spacegrad^2 – \partial_{00} } (I \eta c \BF) – I \eta \partial_0 \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF } &= -I \BM \\
\lr{ \spacegrad^2 – \partial_{00} } (-I \eta \phi_m) – I \eta \spacegrad \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF } &= I c \rho_m.
\end{aligned}

If we choose the Lorentz gauge conditions
\label{eqn:gapotentials:2000}
0 = \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } = \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF },

all of these equations decouple nicely, leaving us with 8 (scalar) equations in 8 unknowns
\label{eqn:gapotentials:2020}
\begin{aligned}
\lr{ \spacegrad^2 – \partial_{00} } \phi &= -\frac{\rho}{\epsilon} \\
\lr{ \spacegrad^2 – \partial_{00} } \BA &= -\mu \BJ \\
\lr{ \spacegrad^2 – \partial_{00} } \BF &= -\epsilon \BM \\
\lr{ \spacegrad^2 – \partial_{00} } \phi_m &= – \frac{\rho_m}{\mu}.
\end{aligned}

Potentials in STA (space time algebra).

All of this was very convoluted. Maxwell’s equation in STA form is considerably simpler, as is the potential formulation.

STA form of Maxwell’s equation.

We identify
\label{eqn:gapotentials:2040}
\begin{aligned}
\Be_k &= \gamma_k \gamma_0 \\
I &= \Be_1 \Be_2 \Be_3 = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \\
\gamma^\mu \cdot \gamma_\nu &= {\delta^\mu}_\nu.
\end{aligned}

Our field multivector
\label{eqn:gapotentials:2060}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= \gamma_{k0} E^k + \eta \gamma_{0123k0} H^k \\
&= \gamma_{k0} E^k + \eta \gamma_{123k} H^k,
\end{aligned}

now has a pure bivector representation in STA (since $$k$$ will always clobber one of the $$1,2,3$$ indexes.) To find the STA representation of Maxwell’s equation, we simply multiply both sides of our multivector representation, from the left, by $$\gamma_0$$.
\label{eqn:gapotentials:2080}
\gamma_0 \lr{ \spacegrad + \partial_0 } F = \gamma_0 \lr{ \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM } }.

The LHS is just the spacetime gradient of $$F$$, which we can see by expanding the product
\label{eqn:gapotentials:2100}
\begin{aligned}
\gamma_0 \lr{ \spacegrad + \partial_0 }
&=
\gamma_0 \lr{ \gamma_{k0} \PD{x^k}{} + \PD{x^0}{} } \\
&=
-\gamma_{k} \PD{x^k}{} + \gamma_0 \PD{x^0}{}.
\end{aligned}

\label{eqn:gapotentials:2120}
\grad \equiv \gamma^k \PD{x^k}{} + \gamma^0 \PD{x^0}{} = \gamma^\mu \partial_\mu.

Our RHS is
\label{eqn:gapotentials:2140}
\begin{aligned}
\gamma_0 \lr{ \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM } }
&=
\gamma_0 \frac{\rho}{\epsilon} – \gamma_{0k0} \eta (\BJ \cdot \Be_k)
– I \lr{ c \rho_m \gamma_0 – \gamma_{0k0} (\BM \cdot \Be_k) } \\
&=
\gamma_0 \frac{\rho}{\epsilon} + \gamma_k \eta (\BJ \cdot \Be_k)
– I \lr{ c \rho_m \gamma_0 + \gamma_{k} (\BM \cdot \Be_k) }.
\end{aligned}

If we let
\label{eqn:gapotentials:2160}
\begin{aligned}
J_e^0 &= \frac{\rho}{\epsilon} \\
J_e^k &= \eta (\BJ \cdot \Be_k) \\
J_m^0 &= c \rho_m \\
J_m^k &= (\BM \cdot \Be_k) \\
J_e &= J_e^\mu \gamma_\mu \\
J_m &= J_m^\mu \gamma_\mu,
\end{aligned}

then we are left with
\label{eqn:gapotentials:2180}
\grad F = J_e – I J_m,

or just
\label{eqn:gapotentials:2640}

where we now give a different meaning to $$J$$ than we had in the multivector formulation. This $$J$$ is now a multivector with grade(1,3) components.

Case I: potential formulation for conventional sources.

Much like we did with to find the potential formulation for the multivector form of Maxwell’s equation, we use superposition, and tackle the conventional sources, and fictitious magnetic sources separately.

With no fictitious sources, Maxwell’s equation is
\label{eqn:gapotentials:2200}

which we may split into vector and trivector components
\label{eqn:gapotentials:2220}
\begin{aligned}
\grad \cdot F &= J_e \\
\end{aligned}

Clearly, the trivector equation can be satified by setting
\label{eqn:gapotentials:2240}

for some vector $$A$$. We may also make gauge transformations of $$A$$ of the form
\label{eqn:gapotentials:2260}
A \rightarrow A + \grad \psi,

without changing $$F$$, showing that $$A$$ is not uniquely determined. With such a representation, Maxwell’s equation is now reduced to
\label{eqn:gapotentials:2280}

or
\label{eqn:gapotentials:2300}
\begin{aligned}
J_e
&=
&=
\end{aligned}

Clearly the equivalent of the Lorentz gauge condition is now just
\label{eqn:gapotentials:2320}

so the Lorentz gauge potential form of Maxwell’s equation is just
\label{eqn:gapotentials:n}S

Case II: potential formulation for fictitious sources.

If we have only fictious sources, Maxwell’s equation is
\label{eqn:gapotentials:2340}

or after left multiplication by $$I$$ we have
\label{eqn:gapotentials:2360}

Let $$G = I F$$, for the dual field, which is still a bivector. As before, we can split Maxwell’s equations into vector and trivector compoents
\label{eqn:gapotentials:2380}
\begin{aligned}
\grad \cdot G &= J_m \\
\end{aligned}

We may set
\label{eqn:gapotentials:2400}

for vector $$K$$. Maxwell’s equation is now reduced to
\label{eqn:gapotentials:2420}

or
\label{eqn:gapotentials:2440}
\begin{aligned}
J_m
&=
&=
\end{aligned}

As before we may make gauge transformations by adding gradient to our potential
\label{eqn:gapotentials:2460}
K \rightarrow K + \grad \bar{\psi},

which will not change $$G$$. For such sources, the Lorentz gauge condition is $$\grad \cdot K = 0$$. With the Lorentz gauge, Maxwell’s equation is reduced to
\label{eqn:gapotentials:2480}

Superposition.

For non-fictious sources, we have
\label{eqn:gapotentials:2500}

and for fictious sources, we have
\label{eqn:gapotentials:2520}
I F = G = \grad \wedge K,

or
\label{eqn:gapotentials:2540}
F = -I G = -I \lr{ \grad \wedge K }.

Combining these results, we have
\label{eqn:gapotentials:2560}
\begin{aligned}
F
\end{aligned}

or
\label{eqn:gapotentials:2580}
F = \grad \lr{ A + I K } – \gpgrade{ \grad \lr{ A + I K } }{0,4}.

Maxwell’s equation is
\label{eqn:gapotentials:2600}

With the Lorentz gauge, this splits nicely into one forced wave equation for each vector potential
\label{eqn:gapotentials:2620}
\begin{aligned}
\end{aligned}

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.

[2] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics, Volume II.[Lectures on physics], chapter The Maxwell Equations. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963. URL https://www.feynmanlectures.caltech.edu/II_18.html.

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[4] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

Hodge star vs. pseudoscalar multiplication.

We find a definition of the hodge star for basic k-forms in [2].

Definition 1.7: Hodge star.

Let $$\omega$$ be a basic k-form on $$\mathbb{R}^n$$. The hodge star of $$\omega$$, denoted by $${*} \omega$$ is the unique $$n-k$$-form with the property
\begin{equation*}
\omega \wedge {*} \omega = dx_1 \wedge \cdots \wedge dx_n.
\end{equation*}

I find it interesting that this duality definition is completely free of any notion of metric or inner product. That isn’t the case with the hodge star definition from [3]. This is certainly an easier definition to understand.

Let’s calculate all the duals for the basic forms from $$\mathbb{R}^3$$. We let $$I = dx_1 \wedge dx_2 \wedge dx_3$$, and then by inspection find all the duals satisfying
\label{eqn:formAndCurl:1110}
\begin{aligned}
I &= 1 \wedge {*} 1 \\
I &= dx \wedge {*} dx \\
I &= dy \wedge {*} dy \\
I &= dz \wedge {*} dz \\
I &= (dx dy) \wedge {*} (dx dy) \\
I &= (dy dz) \wedge {*} (dy dz) \\
I &= (dz dx) \wedge {*} (dz dx) \\
I &= dx dy dz \wedge {*} (dx dy dz).
\end{aligned}

Those are
\label{eqn:formAndCurl:1130}
\begin{aligned}
{*} 1 &= dx dy dz \\
{*} dx &= dy dz \\
{*} dy &= dz dx \\
{*} dz &= dx dy \\
{*} (dx dy) &= dz \\
{*} (dy dz) &= dx \\
{*} (dz dx) &= dy \\
{*} (dx dy dz) &= 1.
\end{aligned}

Now let’s compare this to multiplication of the $$\mathbb{R}^3$$ basis vectors with the pseudoscalar $$I = \Be_1 \Be_2 \Be_3$$. We have
\label{eqn:formAndCurl:1140}
\begin{aligned}
1 I &= I \\
\Be_1 I &= \Be_{1123} = \Be_{23} \\
\Be_2 I &= \Be_{2123} = \Be_{31} \\
\Be_3 I &= \Be_{3123} = \Be_{12} \\
\Be_{23} I &= \Be_{23123} = – \Be_1 \\
\Be_{31} I &= \Be_{31123} = – \Be_2 \\
\Be_{12} I &= \Be_{12123} = – \Be_3 \\
\Be_{123} I &= \Be_{123123} = -1.
\end{aligned}

With differential forms, the duals of the duals of all our basic forms recovered the original, that is $$** \omega = \omega$$, but that isn’t the case if we use pseudoscalar multiplication to define duality. We see that to model the Hodge dual, we need to multiply by a grade specific pseudoscalar.

Definition 1.8: Hodge dual of an $$\mathbb{R}^3$$ multivector

Let $$M$$ be a $$\mathbb{R}^3$$ multivector. The Hodge dual $${*} M$$ of that multivector is
\begin{equation*}
{*} M
=
\end{equation*}

In particular, if $$A$$ is a k-blade in $$\mathbb{R}^3$$, a round trip requires multiplication with different signed unit pseudoscalars.

Let’s step back and consider the $$\mathbb{R}^2$$ case as well. This time we let $$i = dx_1 \wedge dx_2$$. We seek all the duals satisfying
\label{eqn:formAndCurl:1180}
\begin{aligned}
i &= 1 \wedge {*} 1 \\
i &= dx \wedge {*} dx \\
i &= dy \wedge {*} dy \\
i &= (dx dy) \wedge {*} (dx dy).
\end{aligned}

Those duals are
\label{eqn:formAndCurl:1200}
\begin{aligned}
{*} 1 &= dx dy \\
{*} dx &= dy \\
{*} dy &= -dx \\
{*} (dx dy) &= 1 \\
\end{aligned}

Now let’s compare this to multiplication of the $$\mathbb{R}^2$$ basis vectors with the pseudoscalar $$i = \Be_1 \Be_2$$. We have
\label{eqn:formAndCurl:1220}
\begin{aligned}
1 i &= i \\
\Be_1 i &= \Be_{112} = \Be_{2} \\
\Be_2 i &= \Be_{212} = -\Be_{1} \\
\Be_{12} i &= \Be_{1212} = -1 \\
\end{aligned}

Definition 1.9: Hodge dual of $$\mathbb{R}^2$$ multivector

Let $$M$$ be a $$\mathbb{R}^2$$ multivector. The Hodge dual $${*} M$$ of that multivector is
\begin{equation*}
{*} M
=
\end{equation*}

Neither of these grade specific duality operations are as nice as simply multiplying by a unit pseudoscalar, but if we care about correspondence with the Hodge dual (at least according to the definition in the article), then this is what we need.

Having done that, let’s now look at the Hodge dual that produces the divergence operation.

Lemma 1.13: Divergence relation to the exterior derivative.

Let $$\omega = f dx + g dy + h dz$$ be a one-form in $$\mathbb{R}^3$$. The exterior derivative of the Hodge dual of $$\omega$$ is a divergence three-form
\begin{equation*}
d({*} \omega) = \lr{ \PD{x}{f} + \PD{y}{g} + \PD{z}{h} } dx \wedge dy \wedge dz.
\end{equation*}
The GA equivalent of this, for a vector corresponding to this one-form $$\Bf = f \Be_1 + g \Be_2 + h \Be_3 \in \mathbb{R}^3$$, is
\begin{equation*}
\end{equation*}

Start proof:

The dual of the one form is
\label{eqn:formAndCurl:1280}
{*} \omega =
f dy \wedge dz
+ g dz \wedge dx
+ h dx \wedge dy,

so the exterior derivative is
\label{eqn:formAndCurl:1300}
\begin{aligned}
d({*} \omega) &=
\lr{
\PD{x}{f} dx +
\PD{y}{f} dy +
\PD{z}{f} dz
}
\wedge dy \wedge dz \\
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dz \wedge dx \\
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dx \wedge dy \\
&=
\lr{
\PD{x}{f} +
\PD{y}{g} +
\PD{z}{h}
}
dx \wedge dy \wedge dz.
\end{aligned}

We expect that the GA equivalent of this is $$\spacegrad \wedge ({*} \Bf) = \lr{ \spacegrad \cdot \Bf} I$$. Let’s check that this is the case. The dual, for a vector, is
\label{eqn:formAndCurl:1320}
{*} \Bf
= \Bf I,

so
\label{eqn:formAndCurl:1340}
\begin{aligned}