# Connection between quaternions and electromagnetism

Consider the multiplication of two quaternions, which can be written as a matrix equation: $$xy = \begin{pmatrix} x_0 & -x_1 & -x_2 & -x_3 \\ x_1 & x_0 & -x_3 & x_2 \\ x_2 & x_3 & x_0 & -x_1 \\ x_3 & -x_2 & x_1 & x_0 \end{pmatrix}\begin{pmatrix} y_0\\y_1\\y_2\\y_3 \end{pmatrix} = x_0y + \begin{pmatrix} 0 & -x_1 & -x_2 & -x_3 \\ x_1 & 0 & -x_3 & x_2 \\ x_2 & x_3 & 0 & -x_1 \\ x_3 & -x_2 & x_1 & 0 \end{pmatrix}\begin{pmatrix} y_0\\y_1\\y_2\\y_3 \end{pmatrix}$$ if we put the four components of a quaternion inside a four-component vector. Let's compare this to the electromagnetic field strength tensor, which is given in terms of the electric and magnetic fields as: $$F^{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}$$ The structure of these matrices seems oddly similar. Why could this be?

I assume this has to do with $$F^{\mu\nu}$$ being a differential 2-form, at least that's where I think the antisymmetry comes from.

• IMO there is no similarity, unless you suppose that $\,\mathbf E\equiv(c)\mathbf B$. Commented Jan 17, 2022 at 14:51
• I have to agree with @Frobenius. There are just too many DOFs here. What you really want is complex quaternions.
– J.G.
Commented Jan 17, 2022 at 20:25
• Commented Jan 22, 2022 at 20:32

The field strength tensor can be written as $$\tag{1} F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu=2\,\partial_{[\mu}A_{\nu]}$$ where $$A_\mu$$ is the vector potential. This is antisymmetric in all four indices $$\mu=0,1,2,3$$ (and $$\nu=0,1,2,3$$ of course).
The multiplication of two quaternions \begin{align} p&=x_0+\boldsymbol{i}x_1+\boldsymbol{j}x_2+\boldsymbol{k}x_3\,,\\ q&=y_0+\boldsymbol{i}y_1+\boldsymbol{j}y_2+\boldsymbol{k}y_3\, \end{align} can be written as \begin{align} pq&=\underbrace{x_0y_0-x_1y_1-x_2y_2-x_3y_3}_{\text{scalar}}\\ &+\underbrace{\boldsymbol{i}(x_0y_1+x_1y_0)+\boldsymbol{j}(x_0y_2+x_2y_0)+\boldsymbol{k}(x_0y_3+x_3y_0)}_{\text{symmetric}}\\ &+\underbrace{\boldsymbol{i}(x_2y_3-x_3y_2)+\boldsymbol{j}(x_3y_1-x_1y_3)+\boldsymbol{k}(x_1y_2-x_2y_1)}_{\text{anti symmetric}}\,. \end{align} Even if we take from this only the antisymmetric part it is only in three indices $$1,2,3$$ which makes it pretty hopless to formulate (1) only with quaternions.
A viable approach though is Space time algebra that uses the Clifford algebra $$\mathcal{Cl}_{1,3}(\mathbb R)$$ which is built on Dirac matrices. In a sense this is a space of higher "quaternions" that has the right signature for special relativity.
• Clifford algebras have one relation that distinguishes them from "all" matrices: $\gamma_i\gamma_j+\gamma_j\gamma_i=2\eta_{ij}$ where $\eta$ is a pseudo metric with some signature. To me it is pretty obvious that ${\cal Cl}_{1,3}(\mathbb R)$ is the right algebra when SR plays a role. Why we bother to construct all other ${\cal Cl}_{p,q}(\mathbb K)$ is a far reaching question that I cannot answer in such a comment. Commented Jan 18, 2022 at 9:36