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19 Feb 2015: Employing the one-particleidentity. ik1B1ik. 1B1, 5.24. we see that with no sum over a.

5 Feb 2015: It is readily verified that Ka = K2a = a, or, as an operator equation,. ... Ka = Ka = (a) K. (4.52). It follows thatKej = ej, Kej = ej,.

14 Feb 2015: The most important quantities associated with a single 2-spinor κA are its flag-pole Ka = κAκA′, and the flagplane determined by the bivector Pab = κAκBA′B′ ABκA. ... P = 12ψσ1ψ̃ = κ(γ1(γ0 γ3))κ̃. (24). Since σ1 anticommutes with

5 Feb 2015: where the ka are a set of three Lagrangian multipliers,. R1 = κAκ̇A κ̄A′ ˙̄κA′. ... J. Math. Phys., 33(5):1831, 1992. [24] W.I. Fushchich and R.Z. Zhdanov.

14 Feb 2015: Theessential techniques are described in Chapter 5, which introduces the concept of themultivector derivative [18, 24]. ... ArBs 〈AB〉|sr| (1.23)ArBs 〈AB〉sr, (1.24). which we call the interior and exterior products respectively.

18 Feb 2015: Anumber of other helpful introductory articles can be found, including those byHestenes [35, 36], Vold [24, 28], and Doran & Lasenby [37]. ... 3.24). Hence µRR̃ is equal to minus its reverse and so must be a bivector.