# Abstract

By introducing a new form of metric tensor the same derivation for the electromanetic tensor $$F_{\mu \nu}$$ from potentials $$A_{\mu}$$ leads to the dual space (Hodge Dual) of the regular $$F_{\mu \nu}$$ tensor. There are additional components in the $$i,j,k$$ planes, however if after the derivation only the real part is considered a physically consistent electromagnetic theory is recovered with a relabelling of $$\vec{E}$$ fields to $$\vec{B}$$ fields and vice versa.

# Introduction

A prominent feature in relativistic physics is the Minkowski metric tensor $\eta = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{bmatrix}$

On probing where this comes from it was postulated that the matrix could be the ’Real’ (non-quaternion) part of the outer product of two unit quaternions $$Q=1+i+j+k$$, $\eta_{\mu \nu} = Q \otimes Q =Re \left( \begin{bmatrix} 1 & i & j & k \\ i & -1 & k & -j\\ j & -k & -1 & i\\ k & j & -i & -1 \end{bmatrix} \right)$

The implications of carrying through the physics made with this tensor without taking the real part were considered. The creation of an electromagnetic tensor is considered.

When being used for a metric in the form $ds^2= \begin{bmatrix} dt & dx & dy & dz \end{bmatrix} \begin{bmatrix} 1 & i & j & k \\ i & -1 & k & -j \\ j & -k & -1 & i \\ k & j & -i & -1 \end{bmatrix} \begin{bmatrix} dt \\ dx \\ dy \\ dz \end{bmatrix}$

Then from explicit calculation it can be shown that an equaivalent matrix gives $ds^2= \begin{bmatrix} dt & dx & dy & dz \end{bmatrix} \begin{bmatrix} 1 & i & j & k \\ i & -1 & 0 & 0 \\ j & 0 & -1 & 0 \\ k & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} dt \\ dx \\ dy \\ dz \end{bmatrix}$

This would be equaivalent to a new type of number with the rules $i \cdot i =-1 \\ j \cdot j =-1 \\ k \cdot k =-1 \\ i\cdot j=j\cdot i=0 \\ i\cdot k=k\cdot i=0 \\ j\cdot k=k\cdot j=0$

This is similar to having 3 independant imaginary numbers or basis vectors. If this were an inner product then they are orthoganal but antiparallell with themselves. They form a NON-ASSOCIATIVE ’group’ under a product with elements $${0,1,i,j,k,-1,-i,-j,-k}$$, this is an Abelian relationship as the non-Abelian properties of the quaternions was removed with the cross interactions. For example, $(i \cdot i) \cdot j = -1 \cdot j = -j \\ i \cdot ( i \cdot j ) = i \cdot 0 = 0 \\ (a \cdot b) \cdot c \ne a \cdot (b \cdot c)$

# Electromagnetism

This formulation would require $\eta^{\mu \nu}=\frac{1}{1+i^2+j^2+k^2} \begin{bmatrix} 1 & i & j & k \\ i & -1 & -k & j\\ j & k & -1 & -i\\ k & -j & i & -1 \end{bmatrix}$

This would require perhaps a normalisation of $$\frac{1}{\sqrt{2}}$$ on each matrix.

For the potential 4 vector $A^{\mu}=(\varphi/c,A_x,A_y,A_z)$

Then $A_{\mu}=\eta_{\mu \nu}A^{\nu}$

Which gives $A_\mu = \begin{bmatrix} \frac{\varphi}{c} +A_xi+A_yj+A_zk \\ \frac{\varphi i}{c} -A_x +A_yk -A_zj \\ \frac{\varphi j}{c} -A_xk -A_y +A_zi \\ \frac{\varphi k}{c} +A_xj +A_yi-A_z \end{bmatrix} \\ = \begin{bmatrix} \frac{\varphi}{c} \\ -A_x \\ -A_y \\ -A_z \end{bmatrix} + \begin{bmatrix} A_x \\ \frac{\varphi}{c} \\ A_z \\ -A_y \end{bmatrix} i + \begin{bmatrix} A_y \\ -A_z \\ \frac{\varphi}{c} \\ A_x \end{bmatrix} j + \begin{bmatrix} A_z \\ A_y \\ -A_x \\ \frac{\varphi}{c} \end{bmatrix} k$

There are now hypercomplex four potentials from the real potential such that $$A_\mu \to A_\mu + B_\mu i + \Gamma_\mu j + \Delta_\mu k$$. Now, as usually defined we can create the tensor $H_{\mu \nu} = (dA)_{\mu \nu}= \partial_\mu A_\nu - \partial_\nu A_\mu$

But with the hypercomplex components it is clear that $$H_{\mu \nu} = F_{\mu \nu} +I_{\mu \nu}i + J_{\mu \nu}j + K_{\mu \nu}k$$ where $F_{\mu \nu}=Re[H_{\mu \nu}]=\partial_\mu A_\nu - \partial_\nu A_\mu \\ I_{\mu \nu}=Im_i[H_{\mu \nu}]=\partial_\mu B_\nu - \partial_\nu B_\mu \\ J_{\mu \nu}=Im_j[H_{\mu \nu}]=\partial_\mu \Gamma_\nu - \partial_\nu \Gamma_\mu \\ K_{\mu \nu}=Im_k[H_{\mu \nu}]=\partial_\mu \Delta_\nu - \partial_\nu \Delta_\mu$

This means that the real component $$F_{\mu \nu}$$ is equal to th regular EM tensor $F_{\mu \nu} = \frac{1}{c} \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -cB_z & cB_y \\ -E_y & cB_z & 0 & -cB_x \\ -Ez & -cB_y & cB_x & 0 \end{bmatrix}$

However, $F^{\alpha \beta}=\eta^{\alpha \gamma}\eta^{\beta \delta}F_{\gamma \delta}$

Through explicit calculation this results in $Re[F^{\mu \nu}]=\frac{1}{2c} \begin{bmatrix} 0 & cB_x & cB_y & cB_z \\ -cB_x & 0 & E_z & -E_y \\ -cB_y & -E_z & 0 & E_x \\ -cB_z & E_y & -E_x & 0 \end{bmatrix} \\ =\frac{1}{2}G_{\gamma \delta} = \frac{1}{4} \varepsilon_{\alpha \beta \gamma \delta}F^{\alpha \beta}$

# Maxwell’s Equations

In SI units we have the Faraday-Gauss Law $\partial_\alpha F^{ \alpha \beta} = \mu_0 J_{e}^{\beta}$

The Ampere-Gauss Law $\partial_\alpha \star F^{\alpha \beta} = \frac{\mu_0}{c}J_{m}^{\beta}$

and the Lorentz force law $\frac{dp_\alpha}{d\tau} =q_eF_{\alpha \beta}v^{\beta}+q_m\star F_{\alpha \beta}v^{\beta}$

But we now have the different form for the Hodge Dual $$\star F$$