[Click here for a PDF version of this post]

I’ve been enjoying XylyXylyX’s QED Prerequisites Geometric Algebra: Spacetime YouTube series, which is doing a thorough walk through of [1], filling in missing details. The last episode QED Prerequisites Geometric Algebra 15: Complex Structure, left things with a bit of a cliff hanger, mentioning a “canonical” form for STA bivectors that was intriguing.

The idea is that STA bivectors, like spacetime vectors can be spacelike, timelike, or lightlike (i.e.: positive, negative, or zero square), but can also have a complex signature (squaring to a 0,4-multivector.)

The only context that I knew of that one wanted to square an STA bivector is for the electrodynamic field Lagrangian, which has an \( F^2 \) term. In no other context, was the signature of \( F \), the electrodynamic field, of interest that I knew of, so I’d never considered this “Canonical form” representation.

Here are some examples:

\begin{equation}\label{eqn:canonicalbivectors:20}

\begin{aligned}

F &= \gamma_{10}, \quad F^2 = 1 \\

F &= \gamma_{23}, \quad F^2 = -1 \\

F &= 4 \gamma_{10} + \gamma_{13}, \quad F^2 = 15 \\

F &= \gamma_{10} + \gamma_{13}, \quad F^2 = 0 \\

F &= \gamma_{10} + 4 \gamma_{13}, \quad F^2 = -15 \\

F &= \gamma_{10} + \gamma_{23}, \quad F^2 = 2 I \\

F &= \gamma_{10} – 2 \gamma_{23}, \quad F^2 = -3 + 4 I.

\end{aligned}

\end{equation}

You can see in this table that all the \( F \)’s that are purely electric, have a positive signature, and all the purely magnetic fields have a negative signature, but when there is a mix, anything goes. The idea behind the canonical representation in the paper is to write

\begin{equation}\label{eqn:canonicalbivectors:40}

F = f e^{I \phi},

\end{equation}

where \( f^2 \) is real and positive, assuming that \( F \) is not lightlike.

The paper gives a formula for computing \( f \) and \( \phi\), but let’s do this by example, putting all the \( F^2 \)’s above into their complex polar form representation, like so

\begin{equation}\label{eqn:canonicalbivectors:60}

\begin{aligned}

F &= \gamma_{10}, \quad F^2 = 1 \\

F &= \gamma_{23}, \quad F^2 = 1 e^{\pi I} \\

F &= 4 \gamma_{10} + \gamma_{13}, \quad F^2 = 15 \\

F &= \gamma_{10} + \gamma_{13}, \quad F^2 = 0 \\

F &= \gamma_{10} + 4 \gamma_{13}, \quad F^2 = 15 e^{\pi I} \\

F &= \gamma_{10} + \gamma_{23}, \quad F^2 = 2 e^{(\pi/2) I} \\

F &= \gamma_{10} – 2 \gamma_{23}, \quad F^2 = 5 e^{ (\pi – \arctan(4/3)) I}

\end{aligned}

\end{equation}

Since we can put \( F^2 \) in polar form, we can factor out half of that phase angle, so that we are left with a bivector that has a positive square. If we write

\begin{equation}\label{eqn:canonicalbivectors:80}

F^2 = \Abs{F^2} e^{2 \phi I},

\end{equation}

we can then form

\begin{equation}\label{eqn:canonicalbivectors:100}

f = F e^{-\phi I}.

\end{equation}

If we want an equation for \( \phi \), we can just write

\begin{equation}\label{eqn:canonicalbivectors:120}

2 \phi = \mathrm{Arg}( F^2 ).

\end{equation}

This is a bit better (I think) than the form given in the paper, since it will uniformly rotate \( F^2 \) toward the positive region of the real axis, whereas the paper’s formula sometimes rotates towards the negative reals, which is a strange seeming polar form to use.

Let’s compute \( f \) for \( F = \gamma_{10} – 2 \gamma_{23} \), using

\begin{equation}\label{eqn:canonicalbivectors:140}

2 \phi = \pi – \arctan(4/3).

\end{equation}

The exponential expands to

\begin{equation}\label{eqn:canonicalbivectors:160}

e^{-\phi I} = \inv{\sqrt{5}} \lr{ 1 – 2 I }.

\end{equation}

Multiplying each of the bivector components by \(1 – 2 I\), we find

\begin{equation}\label{eqn:canonicalbivectors:180}

\begin{aligned}

\gamma_{10} \lr{ 1 – 2 I}

&=

\gamma_{10} – 2 \gamma_{100123} \\

&=

\gamma_{10} – 2 \gamma_{1123} \\

&=

\gamma_{10} + 2 \gamma_{23},

\end{aligned}

\end{equation}

and

\begin{equation}\label{eqn:canonicalbivectors:200}

\begin{aligned}

– 2 \gamma_{23} \lr{ 1 – 2 I}

&=

– 2 \gamma_{23}

+ 4 \gamma_{230123} \\

&=

– 2 \gamma_{23}

+ 4 \gamma_{23}^2 \gamma_{01} \\

&=

– 2 \gamma_{23}

+ 4 \gamma_{10},

\end{aligned}

\end{equation}

leaving

\begin{equation}\label{eqn:canonicalbivectors:220}

f = \sqrt{5} \gamma_{10},

\end{equation}

so the canonical form is

\begin{equation}\label{eqn:canonicalbivectors:240}

F = \gamma_{10} – 2 \gamma_{23} = \sqrt{5} \gamma_{10} \frac{1 + 2 I}{\sqrt{5}}.

\end{equation}

It’s interesting here that \( f \), in this case, is a spatial bivector (i.e.: pure electric field), but that clearly isn’t always going to be the case, since we can have a case like,

\begin{equation}\label{eqn:canonicalbivectors:260}

F = 4 \gamma_{10} + \gamma_{13} = 4 \gamma_{10} + \gamma_{20} I,

\end{equation}

from the table above, that has both electric and magnetic field components, yet is already in the canonical form, with \( F^2 = 15 \). The canonical \( f \), despite having a positive square, is not necessarily a spatial bivector (as it may have both grades 1,2 in the spatial representation, not just the electric field, spatial grade-1 component.)

# References

[1] Justin Dressel, Konstantin Y Bliokh, and Franco Nori. Spacetime algebra as a powerful tool for electromagnetism. *Physics Reports*, 589:1–71, 2015.