Electromagnetism according to geometric algebra: An appropriate and comprehensive formulation

Abstract

While the Maxwell’s equations describe the electric and magnetic fields developed by electrically charged particles and/or currents, the Lorentz force law completes the picture of classical electromagnetic theory by defining the force acting on a localised charge (or charge distribution) moving in the field. However, the standard formulation using vector algebra suffers from several inadequacies and unwarranted features. Clifford’s geometric algebra or more specifically space–time algebra, i.e geometric algebra in 4D Minkowski space–time, provides an elegant, compactified and comprehensive description by removing the discrepancies of the earlier formulation. It provides an invariant description, in the appropriate space–time setting, in terms of the combined electromagnetic field without reference to any inertial system. Moreover, using elementary geometric calculus, it facilitates direct analytical introduction of the putative concept of magnetic monopole and renders the equations for both the constituent fields, symmetric and inhomogeneous. In terms of the single space–time force equation, space–time algebra also encapsulates both the Lorentz force equation and the electromagnetic power equation.

Appendices

Appendix I: Derivation of ${\mathbf {F}}^{2}$ ($\langle {\mathbf {F}}^{2}\rangle _{0}, \langle {\mathbf {F}}^{2}\rangle _{2}$ and $\langle {\mathbf {F}}^{2}\rangle _{4}$)

Since ${\hat{\alpha }}_{j}\,{\hat{\alpha }}_{0}: {\hat{\alpha }}_{k}\,{\hat{\alpha }}_{0}=\delta _{jk}$; ${\hat{\alpha }}_{j}\,{\hat{\alpha }}_{0}: {\hat{\alpha }}_{k}\,{\hat{\alpha }}_{l}=0$ and ${\hat{\alpha }}_{j}\,{\hat{\alpha }}_{k}: {\hat{\alpha }}_{l}\,\hat{\mathbf {\alpha }}_{m}=\delta _{jm}\, \delta _{kl}-\delta _{jl}\,\delta _{km}$, from eq. (24) we get

$$\begin{aligned} \langle {\mathbf {F}}^{2}\rangle _{0}= & {} {\mathbf {F}}:{\mathbf {F}}=(F^{10})^{2}+(F^{20})^{2}\\&+(F^{30})^{2}-\{(F^{23})^{2}+(F^{31})^{2}+(F^{12})^{2}\}\\= & {} {\mathbf {e}}^{2} -c^{2}{\mathbf {b}}^{2}\equiv {\mathbf {e}}^{2} +c^{2}{\mathbf {B}}^{2}. \end{aligned}$$

Again, since ${\hat{\alpha }}_{j}\,{\hat{\alpha }}_{0}\cdot {\hat{\alpha }}_{k}\,{\hat{\alpha }}_{0}={\hat{\alpha }}_{j} \,{\hat{\alpha }}_{k}=-{\hat{\alpha }}_{k}\,{\hat{\alpha }}_{0} \cdot {\hat{\alpha }}_{j}\,{\hat{\alpha }}_{0};\;{\hat{\alpha }}_{j}\, {\hat{\alpha }}_{0}\cdot {\hat{\alpha }}_{k}\,{\hat{\alpha }}_{l}= {\hat{\alpha }}_{l}\,{\hat{\alpha }}_{0}\,\delta _{jk}- {\hat{\alpha }}_{k}\,{\hat{\alpha }}_{0}\,\delta _{jl}=- {\hat{\alpha }}_{k}\,{\hat{\alpha }}_{l}.{\hat{\alpha }}_{j}\, {\hat{\alpha }}_{0}$ and ${\hat{\alpha }}_{j}\,{\hat{\alpha }}_{k}. {\hat{\alpha }}_{l}\,\hat{\mathbf {\alpha }}_{m}={\hat{\alpha }}_{j}\, \hat{\mathbf {\alpha }}_{m}\delta _{kl}-{\hat{\alpha }}_{k}\, \hat{\mathbf {\alpha }}_{m}\delta _{jl}=0$, it follows

$$\begin{aligned} \langle {\mathbf {F}}^{2}\rangle _{2}= & {} {\mathbf {F}}\cdot {\mathbf {F}}\\= & {} F^{10}\,F^{20}\cdot 0+ F^{10}\,F^{30}\cdot 0+F^{20}\,F^{30}\cdot 0\\&+F^{10}\,F^{23}\cdot 0+F^{10}\,F^{31}\cdot 0+F^{10}\, F^{12}\cdot 0\\&+F^{20}\,F^{23}\cdot 0+F^{20}\,F^{31}\cdot 0+F^{20}\,F^{12}\cdot 0\\&+F^{30}\,F^{23}\cdot 0 +F^{30}\,F^{31}\cdot 0+F^{30}\,F^{12}\cdot 0\\&+0+ \cdots +0=0. \end{aligned}$$

Finally,

$$\begin{aligned}&\langle {\mathbf {F}}^{2}\rangle _{4}&\\= & {} {\mathbf {F}} \wedge {\mathbf {F}}\\= & {} F^{10}\, F^{23}(\hat{\mathbf {\alpha }}_{1}\,{\hat{\alpha }}_{0}\,\hat{\mathbf {\alpha }}_{2} \,\hat{\mathbf {\alpha }}_{3}+\hat{\mathbf {\alpha }}_{2}\,\hat{\mathbf {\alpha }} _{3}\,\hat{\mathbf {\alpha }}_{1}\,{\hat{\alpha }}_{0})\\&+F^{20}\,F^{31} (\hat{\mathbf {\alpha }}_{2}\,{\hat{\alpha }}_{0}\,\hat{\mathbf {\alpha }}_{3}\, \hat{\mathbf {\alpha }}_{1}+\hat{\mathbf {\alpha }}_{3}\,\hat{\mathbf {\alpha }}_{1}\, \hat{\mathbf {\alpha }}_{2}\,{\hat{\alpha }}_{0})\\&+F^{30}\,F^{12} (\hat{\mathbf {\alpha }}_{3}\,{\hat{\alpha }}_{0}\,\hat{\mathbf {\alpha }}_{1}\, \hat{\mathbf {\alpha }}_{2}+\hat{\mathbf {\alpha }}_{1}\,\hat{\mathbf {\alpha }}_{2}\, \hat{\mathbf {\alpha }}_{3}\,{\hat{\alpha }}_{0})\\= & {} -2(F^{10}\,F^{23} +F^{20}\,F^{31}+F^{30}\,F^{12}){\hat{\alpha }}_{0}\,\hat{\mathbf {\alpha }}_{1} \,\hat{\mathbf {\alpha }}_{2}\,\hat{\mathbf {\alpha }}_{3}\\= & {} 2c\,{\mathbf {e}}\cdot {\mathbf {b}} \,I_{4}\equiv 2c\,{\hat{\alpha }}_{0}\, {\mathbf {e}} \wedge {\mathbf {B}} . \end{aligned}$$

Appendix II: Derivation of ($\Box \cdot {\mathbf {F}}$)$\cdot {\mathbf {F}}$

$$\begin{aligned}&(\Box \cdot {\mathbf {F}})\cdot {\mathbf {F}}\\&\quad =\{(-{\hat{\alpha }}_{0} c^{-1}\partial _{t}+\nabla )\cdot ({\mathbf {e}} \wedge {\hat{\alpha }}_{0}-c\,{\mathbf {B}})\}\\&\qquad \cdot ({\mathbf {e}} \wedge {\hat{\alpha }}_{0}-c\,{\mathbf {B}})=(-c^{-1}\partial _{t} {\mathbf {e}}+{\hat{\alpha }}_{0}\, \nabla \cdot {\mathbf {e}}\\&\qquad -c\nabla \cdot {\mathbf {B}})\cdot ({\mathbf {e}} \wedge {\hat{\alpha }}_{0}-c\, {\mathbf {B}})\\&\quad =-{\hat{\alpha }}_{0} c^{-1}(\partial _{t} {\mathbf {e}}).{\mathbf {e}}+ (\partial _{t} {\mathbf {e}}).{\mathbf {B}}\\&\qquad +(\nabla \cdot {\mathbf {e}}){\mathbf {e}}+0- {\hat{\alpha }}_{0}\,c\,(\nabla \cdot {\mathbf {B}}).{\mathbf {e}}\\&\qquad +c^{2}(\nabla \cdot {\mathbf {B}}). {\mathbf {B}}=-{\hat{\alpha }}_{0} c^{-1}\{(\partial _{t} {\mathbf {e}}).{\mathbf {e}}\\&\qquad +c^{2}(\nabla \cdot {\mathbf {B}}).{\mathbf {e}}\}+(\partial _{t} {\mathbf {e}}).{\mathbf {B}}\\&\qquad +(\nabla \cdot {\mathbf {e}}){\mathbf {e}}+c^{2}(\nabla \cdot {\mathbf {B}}).{\mathbf {B}}\\&\quad =-{\hat{\alpha }}_{0}c^{-1}\{2^{-1}\partial _{t}({\mathbf {e}}^{2}-c^{2}\,{\mathbf {B}}^{2})\\&\qquad +c^{2}\nabla \cdot ({\mathbf {B}}\cdot {\mathbf {e}})\}-\partial _{t}({\mathbf {B}}\cdot {\mathbf {e}}) -2^{-1}\nabla {\mathbf {e}}^{2}\\&\qquad +({\mathbf {e}}.\nabla ){\mathbf {e}}+(\nabla \cdot {\mathbf {e}}) {\mathbf {e}}+2^{-1}c^{2}\nabla {\mathbf {B}}^{2}-c^{2}({\mathbf {B}} \wedge \nabla ): {\mathbf {B}}\\&\Rightarrow \bar{\mathbf {f}}=\epsilon _{0}\langle (\Box {\mathbf {F}}){\mathbf {F}}\rangle _{1} \equiv \epsilon _{0}\,(\Box \cdot {\mathbf {F}})\cdot {\mathbf {F}}\\&\quad =-c^{-1}\,{\hat{\alpha }}_{0} (\partial _{t} u_{em}+\nabla \cdot {\mathbf {s}})+\epsilon _{0}[(\nabla \cdot {\mathbf {e}}){\mathbf {e}}+ ({\mathbf {e}}\cdot \nabla ){\mathbf {e}}]\\&\quad -\mu _{0}^{-1}[(\nabla \wedge {\mathbf {B}}):{\mathbf {B}}+ ({\mathbf {B}} \wedge \nabla ):{\mathbf {B}}]-\nabla {\mathrm {u}}_{em}\\&\quad -c^{-2}\,\partial _{t} {\mathbf {s}}, \end{aligned}$$

subtracting a null term involving $\nabla \wedge {\mathbf {B}}$. Since

• (i) $\nabla \cdot ({\mathbf {B}}\cdot {\mathbf {e}})=(\nabla \wedge {\mathbf {e}}):{\mathbf {B}} -(\nabla \cdot {\mathbf {B}})\cdot {\mathbf {e}} \Rightarrow (\nabla \cdot {\mathbf {B}})\cdot {\mathbf {e}}=(\nabla \wedge {\mathbf {e}}):{\mathbf {B}}-\nabla \cdot ({\mathbf {B}}\cdot {\mathbf {e}})$ $=-\partial _{t} {\mathbf {B}}:{\mathbf {B}}-\nabla \cdot ({\mathbf {B}}\cdot {\mathbf {e}})$, using eq. (19) ($(\nabla \wedge {\mathbf {e}}):{\mathbf {B}}+\partial _{t} {\mathbf {B}}=0$),

• (ii) $\partial _{t}({\mathbf {B}}\cdot {\mathbf {e}})=(\partial _{t}{\mathbf {B}})\cdot {\mathbf {e}}+ {\mathbf {B}}\cdot \partial _{t}{\mathbf {e}} \Rightarrow (\partial _{t}{\mathbf {e}})\cdot {\mathbf {B}} =-\partial _{t}({\mathbf {B}}\cdot {\mathbf {e}})+(\partial _{t}{\mathbf {B}})\cdot {\mathbf {e}}$ $=-\partial _{t}({\mathbf {B}}\cdot {\mathbf {e}})-(\nabla \wedge {\mathbf {e}})\cdot {\mathbf {e}} =-\partial _{t}({\mathbf {B}}\cdot {\mathbf {e}})-2^{-1}\nabla {\mathbf {e}}^{2}+({\mathbf {e}}\cdot \nabla ) {\mathbf {e}}]$, using eq. (19) and the identity: $2^{-1}\nabla {\mathbf {e}}^{2}=({\mathbf {e}}\cdot \nabla )){\mathbf {e}}+(\nabla \wedge {\mathbf {e}})\cdot {\mathbf {e}}$ and

• (iii) $2^{-1}\nabla ({\mathbf {B}}:{\mathbf {B}})=(\nabla \cdot {\mathbf {B}})\cdot {\mathbf {B}} +({\mathbf {B}} \wedge \nabla ):{\mathbf {B}}$. (Note ${\mathbf {B}}^{2}={\mathbf {B}}:{\mathbf {B}} =-{\mathbf {b}}^{2},\;({\mathbf {B}} \wedge \nabla ):{\mathbf {B}}=-({\mathbf {b}}\cdot \nabla ){\mathbf {b}}$ and $(\nabla \wedge {\mathbf {B}}):{\mathbf {B}}=-(\nabla \cdot {\mathbf {b}}){\mathbf {b}}$ in VA).

Appendix III: Expressions for the transformed (reverse) bivector bases

The transformed basis vectors with respect to the space–time ‘rotor’ ${\mathcal {R}}$ are given by

$$\begin{aligned} \hat{\mathbf {\alpha }}'_{0}= & {} {\mathcal {R}} {\hat{\alpha }}_{0} {\mathcal {R}}^{\dag }=\exp (\hat{\mathbf {R}}\,\omega /2) {\hat{\alpha }}_{0} \exp (-\hat{\mathbf {R}}\,\omega /2) \nonumber \\= & {} (\cosh \omega /2+{\hat{\alpha }}_{0} \hat{\mathbf {v}} \sinh \omega /2) {\hat{\alpha }}_{0}(\cosh \omega /2\nonumber \\&-{\hat{\alpha }}_{0} \hat{\mathbf {v}} \sinh \omega /2) \nonumber \\= & {} (\cosh \omega /2+ {\hat{\alpha }}_{0} \hat{\mathbf {v}}\sinh \omega /2)({\hat{\alpha }}_{0} \cosh \omega /2\nonumber \\&+\hat{\mathbf {v}} \sinh \omega /2) \nonumber \\= & {} {\hat{\alpha }}_{0}(\cosh ^{2}\omega /2+ \sinh ^{2}\omega /2)\nonumber \\&+\hat{\mathbf {v}} (\sinh \omega /2 \cosh \omega /2+\sinh \omega /2 \cosh \omega /2) \nonumber \\= & {} {\hat{\alpha }}_{0}\cosh \omega + \hat{\mathbf {v}}\sinh \omega \nonumber \\= & {} \gamma ({\hat{\alpha }}_{0}+\beta \hat{\mathbf {v}})\Rightarrow \hat{\mathbf {\alpha }}'_{0}=c^{-1}\gamma \bar{\mathbf {v}} \end{aligned}$$

(52)

and obviously, $\hat{\mathbf {\alpha }}'_{0}\cdot \hat{\mathbf {\alpha }}'_{0}=-1$. In STA, therefore, the inertial rest frame of the observer is defined by the future-pointing unit time-like basis vector ($\hat{\mathbf {\alpha }}'_{0}$) along its space–time velocity $\bar{\mathbf {v}}$, i.e. along the tangent to the world line of the observer. Putting ${\mathbf {v}}={\hat{\alpha }}_{k}$ we get directly, $\hat{\mathbf {\alpha }}'_{0}=\gamma ({\hat{\alpha }}_{0}+\beta \hat{\mathbf {\alpha }} _{k})$. Similarly,

$$\begin{aligned} \hat{\mathbf {\alpha }}'_{j}= & {} {\mathcal {R}} {\hat{\alpha }}_{j} {\mathcal {R}}^{\dag }=\exp (\hat{\mathbf {R}}\,\omega /2) {\hat{\alpha }}_{j} \exp (-\hat{\mathbf {R}}\,\omega /2)\\= & {} (\cosh \omega /2+\hat{\mathbf {R}} \sinh \omega /2)({\hat{\alpha }}_{j} \cosh \omega /2\\&-{\hat{\alpha }}_{j} \hat{\mathbf {R}} \sinh \omega /2)\\= & {} {\hat{\alpha }}_{j} \cosh ^{2} \omega /2- {\hat{\alpha }}_{j} \hat{\mathbf {R}} \cosh \omega /2 \,\sinh \omega /2\\&+\hat{\mathbf {R}} {\hat{\alpha }}_{j} \cosh \omega /2\, \sinh \omega /2\\&-\hat{\mathbf {R}} ({\hat{\alpha }}_{j} \hat{\mathbf {R}}) \sinh ^{2} \omega /2\\= & {} {\hat{\alpha }}_{j} \cosh ^{2} \omega /2+(\hat{\mathbf {R}} {\hat{\alpha }}_{j}-{\hat{\alpha }}_{j}\hat{\mathbf {R}}) \cosh \omega /2\, \sinh \omega /2\\&-\hat{\mathbf {R}}({\hat{\alpha }}_{j}\cdot \hat{\mathbf {R}}+{\hat{\alpha }}_{j} \wedge \hat{\mathbf {R}}) \sinh ^{2} \omega /2\\= & {} {\hat{\alpha }}_{j}+\hat{\mathbf {R}}\cdot {\hat{\alpha }}_{j} \sinh \omega -\hat{\mathbf {R}}({\hat{\alpha }}_{j}\cdot \hat{\mathbf {R}}) (\cosh \omega -1). \end{aligned}$$

Since $\hat{\mathbf {R}}{\hat{\alpha }}_{j}-{\hat{\alpha }}_{j} \hat{\mathbf {R}}=2 \hat{\mathbf {R}}\cdot {\hat{\alpha }}_{j}$ and for the unit simple bivector $\hat{\mathbf {R}}$ we can write

$$\begin{aligned}&{\hat{\alpha }}_{j}=(\hat{\mathbf {R}}\hat{\mathbf {R}}) {\hat{\alpha }}_{j} \equiv \hat{\mathbf {R}}(\hat{\mathbf {R}} {\hat{\alpha }}_{j})=\hat{\mathbf {R}}(\hat{\mathbf {R}}. {\hat{\alpha }}_{j}+\hat{\mathbf {R}} \wedge {\hat{\alpha }}_{j})\\&\quad \,\,\, =\hat{\mathbf {R}}(-{\hat{\alpha }}_{j}. \hat{\mathbf {R}}+{\hat{\alpha }}_{j} \wedge \hat{\mathbf {R}})\\&\Rightarrow \hat{\mathbf {R}}({\hat{\alpha }}_{j}\cdot \hat{\mathbf {R}}+{\hat{\alpha }}_{j} \wedge \hat{\mathbf {R}}) ={\hat{\alpha }}_{j}+2 \hat{\mathbf {R}}({\hat{\alpha }}_{j}\cdot \hat{\mathbf {R}}). \end{aligned}$$

Finally, we get

$$\begin{aligned}&\hat{\mathbf {\alpha }}'_{j} ={\hat{\alpha }}_{j}+\gamma \,\beta ({\hat{\alpha }}_{0}\,\hat{\mathbf {v}})\cdot {\hat{\alpha }}_{j}\nonumber \\&-(\gamma -1)({\hat{\alpha }}_{0}\,\hat{\mathbf {v}})\cdot \{{\hat{\alpha }}_{j}\cdot ({\hat{\alpha }}_{0}\, \hat{\mathbf {v}})\} \nonumber \\&={\hat{\alpha }}_{j}+\gamma \,\beta |{\mathbf {v}}|^{-1} v_{j} {\hat{\alpha }}_{0}+(\gamma -1)|{\mathbf {v}}|^{-1} v_{j}\,\hat{\mathbf {v}}. \end{aligned}$$

(53)

Note that, with $\hat{\mathbf {v}}={\hat{\alpha }}_{k};\; |{\mathbf {v}}|^{-1} v_{k}=1$ and $v_{j}=0,\; j \ne k \Rightarrow \hat{\mathbf {\alpha }}'_{k}=\gamma ({\hat{\alpha }}_{k}+\beta \hat{\mathbf {\alpha }} _{0})$ and for all $j \ne k,\;\hat{\mathbf {\alpha }}'_{j}= {\hat{\alpha }}_{j}$.

$$\begin{aligned}&\Rightarrow \hat{\mathbf {\alpha }}'_{j}\cdot \hat{\mathbf {\alpha }}'_{k}= [|{\mathbf {v}}|^{-1} v_{j}\{\gamma (\hat{\mathbf {v}}+\beta {\hat{\alpha }}_{0})-\hat{\mathbf {v}}\}+ {\hat{\alpha }}_{j}]\\&\cdot [|{\mathbf {v}}|^{-1} v_{k}\{\gamma (\hat{\mathbf {v}}+\beta {\hat{\alpha }}_{0})- \hat{\mathbf {v}}\}+{\hat{\alpha }}_{k}]\\&=|{\mathbf {v}}|^{-2} v_{j}v_{k}\{\gamma ^{2} (1-\beta ^{2})+1-2\gamma \}\\&\quad +2|{\mathbf {v}}|^{-2} v_{j}v_{k} (\gamma -1)+{\delta }_{j\,k}\\&=|{\mathbf {v}}|^{-2} v_{j}v_{k}\{1+1-2\gamma +2(\gamma -1)\}+{\delta }_{j\,k}\\&={\delta }_{j\,k}. \end{aligned}$$

Also,

$$\begin{aligned} \hat{\mathbf {\alpha }}'_{0}\cdot \hat{\mathbf {\alpha }}'_{j}= & {} \gamma ({\hat{\alpha }}_{0}\\&+\beta \hat{\mathbf {v}})\cdot [|{\mathbf {v}}|^{-1} v_{j}\{\gamma (\hat{\mathbf {v}}+\beta {\hat{\alpha }}_{0})- \hat{\mathbf {v}}\}\\&+ {\hat{\alpha }}_{j}] =|{\mathbf {v}}|^{-1} v_{j}\{-\gamma ^{2}\beta \\&+ \gamma ^{2}\beta -\gamma \beta +\gamma \beta \}=0. \end{aligned}$$

Thus, $\hat{\mathbf {\alpha }}'_{0}$ and $\hat{\mathbf {\alpha }}'_{j}$ are the new orthogonal basis vectors under the Lorentz boost along any arbitrary direction of $\hat{\mathbf {v}}$. Now, for the same electromagnetic field bivector ${\mathbf {F}}$ expressed in two different coordinate systems, we can write,

$$\begin{aligned} {\mathbf {F}}= & {} F^{\mu \nu }\,\hat{\mathbf {\alpha }}_{\mu }\, \hat{\mathbf {\alpha }}_{\nu }=F'^{\mu \nu }\,\hat{\mathbf {\alpha }}'_{\mu }\, \hat{\mathbf {\alpha }}'_{\nu }\nonumber \\= & {} F'^{\mu \nu }\, {\mathcal {R}} \hat{\mathbf {\alpha }}_{\mu } {\mathcal {R}}^{\dag }{\mathcal {R}} \hat{\mathbf {\alpha }}_{\nu }{\mathcal {R}}^{\dag }\nonumber \\= & {} F'^{\mu \nu }\,{\mathcal {R}}\hat{\mathbf {\alpha }}_{\mu }\,\hat{\mathbf {\alpha }}_{\nu } {\mathcal {R}}^{\dag } \end{aligned}$$

(54)

and using the reverse transformation of ${\mathbf {F}}$, we get

$$\begin{aligned} {\mathcal {R}}^{\dag }{\mathbf {F}}{\mathcal {R}}= & {} F^{\mu \nu }\, {\mathcal {R}}^{\dag }\hat{\mathbf {\alpha }}_{\mu }\,\hat{\mathbf {\alpha }}_{\nu } {\mathcal {R}}=F'^{\mu \nu }\,\hat{\mathbf {\alpha }}_{\mu }\, \hat{\mathbf {\alpha }}_{\nu }\nonumber \\= & {} {\mathbf {F}}' (\hbox {say}). \end{aligned}$$

(55)

Now, the reverse transformations of the basis vectors produce ${\mathcal {R}}^{\dag }{\hat{\alpha }}_{0} {\mathcal {R}}=\gamma ({\hat{\alpha }}_{0}-\beta \hat{\mathbf {v}})$ and ${\mathcal {R}}^{\dag }{\hat{\alpha }}_{j} {\mathcal {R}}={\hat{\alpha }}_{j} -\gamma \,\beta |{\mathbf {v}}|^{-1} v_{j} {\hat{\alpha }}_{0}+(\gamma -1) |{\mathbf {v}}|^{-1} v_{j}\,\hat{\mathbf {v}} \Rightarrow {\mathcal {R}}^{\dag }{\hat{\alpha }}_{j}\, {\hat{\alpha }}_{0}{\mathcal {R}}=\{{\hat{\alpha }}_{j} -\gamma \,\beta |{\mathbf {v}}|^{-1} v_{j}{\hat{\alpha }}_{0} +(\gamma -1)|{\mathbf {v}}|^{-1} v_{j}\,\hat{\mathbf {v}}\} \wedge \gamma ({\hat{\alpha }}_{0}-\beta \hat{\mathbf {v}}) =\gamma {\hat{\alpha }}_{j} \wedge {\hat{\alpha }}_{0} -\gamma \beta |{\mathbf {v}}|^{-1} v_{k} {\hat{\alpha }}_{j} \wedge {\hat{\alpha }}_{k}+ (1-\gamma )|{\mathbf {v}}|^{-1} v_{j} \hat{\mathbf {v}} \wedge {\hat{\alpha }}_{0},\;j \ne k$, from which we get the explicit expressions for ${\mathcal {R}}^{\dag } {\hat{\alpha }}_{1}\,{\hat{\alpha }}_{0}{\mathcal {R}}, \;{\mathcal {R}}^{\dag }{\hat{\alpha }}_{2}\,{\hat{\alpha }}_{0} {\mathcal {R}}$ and ${\mathcal {R}}^{\dag }\hat{\mathbf {\alpha }}_{3}\,{\hat{\alpha }}_{0}{\mathcal {R}}$. Also, ${\mathcal {R}}^{\dag }{\hat{\alpha }}_{j}\,{\hat{\alpha }}_{k}{\mathcal {R}}= \{{\hat{\alpha }}_{j}-\gamma \,\beta |{\mathbf {v}}|^{-1} v_{j}{\hat{\alpha }}_{0}+ (\gamma -1)|{\mathbf {v}}|^{-1}v_{j}\,\hat{\mathbf {v}}\} \wedge \{{\hat{\alpha }}_{k} -\gamma \,\beta |{\mathbf {v}}|^{-1}v_{k} {\hat{\alpha }}_{0}+(\gamma -1) |{\mathbf {v}}|^{-1}v_{k}\,\hat{\mathbf {v}}\} =\gamma {\hat{\alpha }}_{j} \wedge {\hat{\alpha }}_{k}+\gamma \beta |{\mathbf {v}}|^{-1}(v_{j} {\hat{\alpha }}_{k} \wedge {\hat{\alpha }}_{0}-v_{k} {\hat{\alpha }}_{j} \wedge {\hat{\alpha }}_{0})+ (1-\gamma )|{\mathbf {v}}|^{-2} \{v_{l} v_{l} {\hat{\alpha }}_{j} \wedge {\hat{\alpha }}_{k}+v_{j} v_{l} {\hat{\alpha }}_{k} \wedge {\hat{\alpha }}_{l} -v_{k} v_{l} {\hat{\alpha }}_{j} \wedge {\hat{\alpha }}_{l}\}$, which gives expressions for ${\mathcal {R}}^{\dag } \hat{\mathbf {\alpha }}_{2}\, \hat{\mathbf {\alpha }}_{3}{\mathcal {R}},\;{\mathcal {R}}^{\dag } \hat{\mathbf {\alpha }}_{3}\,\hat{\mathbf {\alpha }}_{1}{\mathcal {R}}$ and ${\mathcal {R}}^{\dag }\hat{\mathbf {\alpha }}_{1}\,\hat{\mathbf {\alpha }}_{2} {\mathcal {R}}$. Using these expressions in eq. (44) one easily gets all the expressions from eqs (45) to (50).

Alternatively, we can also get the above expressions by resolving ${\mathbf {F}}$ with respect to the space–time unit bivector $\hat{\mathbf {R}}={\hat{\alpha }}_{0}\,\hat{\mathbf {v}}$ ($\hat{\mathbf {v}}=|{\mathbf {v}}|^{-1}{\mathbf {v}}$ and ${\mathbf {v}}= v_{l}\,{\hat{\alpha }}_{l}$) as follows:

$$\begin{aligned} {\mathbf {F}}={\mathbf {F}}(\hat{\mathbf {R}}\,\hat{\mathbf {R}})= & {} ({\mathbf {F}} \,\hat{\mathbf {R}})\hat{\mathbf {R}}\;(\hbox {since}\; \hat{\mathbf {R}}\,\hat{\mathbf {R}}=1)\\= & {} ({\mathbf {F}}:\hat{\mathbf {R}}+{\mathbf {F}} \cdot \hat{\mathbf {R}}+{\mathbf {F}} \wedge \hat{\mathbf {R}}) \hat{\mathbf {R}}\\= & {} ({\mathbf {F}}:\hat{\mathbf {R}})\hat{\mathbf {R}} +({\mathbf {F}} \cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}+({\mathbf {F}} \wedge \hat{\mathbf {R}}): \hat{\mathbf {R}}\\= & {} {\mathbf {F}}_{\Vert \hat{\mathbf {R}}}+{\mathbf {F}}_{\perp \hat{\mathbf {R}}}. \end{aligned}$$

The first term $({\mathbf {F}}:\hat{\mathbf {R}}) \hat{\mathbf {R}}$ on the right-hand side lies in the plane of $\hat{\mathbf {R}}$ (denoted by ${\mathbf {F}}_{\Vert \hat{\mathbf {R}}}$) and the second term consists of parts of ${\mathbf {F}}_{\perp \hat{\mathbf {R}}}$ orthogonal to the plane of $\hat{\mathbf {R}}$ and also having a common basis vector with $\hat{\mathbf {R}}$. The final term $({\mathbf {F}} \wedge \hat{\mathbf {R}}):\hat{\mathbf {R}}$ (also a part of ${\mathbf {F}}_{\perp \hat{\mathbf {R}}}$) exists for dimensions higher than 3 and lies in plane touching the orthogonal plane of $\hat{\mathbf {R}}$ only at a single point. The inner products of $\hat{\mathbf {R}}$ with both the first and the final terms vanish and simple working out with geometrical calculus produce

$$\begin{aligned}&({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}\\&={\mathbf {F}}-|{\mathbf {v}}|^{-2}[\{v_{1}^{2}\,\hat{\mathbf {\alpha }}_{1} {\hat{\alpha }}_{0}+(v_{1}v_{2}\,\hat{\mathbf {\alpha }}_{2} {\hat{\alpha }}_{0}\\&\quad +v_{1}v_{3}\,\hat{\mathbf {\alpha }}_{3} {\hat{\alpha }}_{0})\}F^{10}\\&\quad +\{v_{2}^{2}\,\hat{\mathbf {\alpha }}_{2} {\hat{\alpha }}_{0}+(v_{1}v_{2}\,\hat{\mathbf {\alpha }}_{1} {\hat{\alpha }}_{0}+v_{2}v_{3}\,\hat{\mathbf {\alpha }}_{3} {\hat{\alpha }}_{0})\}F^{20}\\&\quad +\{v_{3}^{2}\,\hat{\mathbf {\alpha }}_{3} {\hat{\alpha }}_{0}+(v_{2}v_{3}\,\hat{\mathbf {\alpha }}_{2} {\hat{\alpha }}_{0}\\&\quad +v_{1}v_{3}\,\hat{\mathbf {\alpha }}_{1} {\hat{\alpha }}_{0})\}F^{30}+\{v_{1}^{2}\, \hat{\mathbf {\alpha }}_{2}\,\hat{\mathbf {\alpha }}_{3}+(v_{1}\, v_{2}\,\hat{\mathbf {\alpha }}_{3}\,\hat{\mathbf {\alpha }}_{1}\\&\quad +v_{1}\,v_{3}\,\hat{\mathbf {\alpha }}_{1}\, \hat{\mathbf {\alpha }}_{2})\} F^{23}\\&\quad +\{v_{2}^{2}\,\hat{\mathbf {\alpha }}_{3}\, \hat{\mathbf {\alpha }}_{1}+(v_{2}\,v_{3}\,\hat{\mathbf {\alpha }}_{1}\, \hat{\mathbf {\alpha }}_{2}+v_{1}\,v_{2}\,\hat{\mathbf {\alpha }}_{2}\, \hat{\mathbf {\alpha }}_{3})\}F^{31}\\&\quad +\{v_{3}^{2}\,\hat{\mathbf {\alpha }}_{1}\, \hat{\mathbf {\alpha }}_{2}+(v_{1}\,v_{3}\,\hat{\mathbf {\alpha }}_{2}\, \hat{\mathbf {\alpha }}_{3}+v_{2}\,v_{3}\,\hat{\mathbf {\alpha }}_{3}\, \hat{\mathbf {\alpha }}_{1})\}F^{12}]. \end{aligned}$$

Now, for example, if we consider the velocity boost along the x-axis, i.e. $\hat{\mathbf {v}}=\hat{\mathbf {\alpha }}_{1}$, we simply get:

$$\begin{aligned}&{\mathbf {F}}_{\Vert \hat{\mathbf {R}}}=({\mathbf {F}}:\hat{\mathbf {R}}) \hat{\mathbf {R}}=F^{10}\,\hat{\mathbf {\alpha }}_{1}\, {\hat{\alpha }}_{0},\;\\&({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}=F^{20}\,\hat{\mathbf {\alpha }}_{2}\, {\hat{\alpha }}_{0}+F^{30}\, \hat{\mathbf {\alpha }}_{3}\, {\hat{\alpha }}_{0}\\&\qquad \qquad \qquad +F^{31}\, \hat{\mathbf {\alpha }}_{3}\, \hat{\mathbf {\alpha }}_{1}+F^{12}\, \hat{\mathbf {\alpha }}_{1}\, \hat{\mathbf {\alpha }}_{2}\\&\mathrm{and} \\&({\mathbf {F}} \wedge \hat{\mathbf {R}}):\hat{\mathbf {R}}=F^{23}\, \hat{\mathbf {\alpha }}_{2}\,\hat{\mathbf {\alpha }}_{3}. \end{aligned}$$

The boost, generated by $\hat{\mathbf {R}}$, operates only on the second term $({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}$, leaving $({\mathbf {F}}:\hat{\mathbf {R}})\hat{\mathbf {R}}$ and $({\mathbf {F}} \wedge \hat{\mathbf {R}}): \hat{\mathbf {R}}$ unchanged.

$$\begin{aligned} \Rightarrow {\mathbf {F}}'= & {} {\mathcal {R}}^{\dag } {\mathbf {F}}{\mathcal {R}}\nonumber \\= & {} {\mathbf {F}}-({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}+{\mathcal {R}}^{\dag }\{({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}\} {\mathcal {R.}} \end{aligned}$$

(56)

Noting that, ${\mathcal {R}}^{\dag }\{({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}\}{\mathcal {R}}=({\mathcal {R}}^{\dag })^{2}\{({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}\}=\{({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}\}{\mathcal {R}}^{2}=\{({\mathbf {F}}\cdot \hat{\mathbf {R}})\cdot \hat{\mathbf {R}}\}\,\gamma (1+\beta \,{\hat{\alpha }}_{0}\, \hat{\mathbf {v}})$, we get all the expressions, eqs (45)–(50) from eqs (42) and (51).