うべの時空代数

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\def\ou#1#2#3{\overset{#2}{\underset{#3}{#1}}}\def\os#1#2{\overset{#2}{#1}}\def\diff{\mathrm{d}}\def\biff{\mathrm{b}}\def\D{\mathrm{D}}\def\p{\mathrm{p}}\def\pu#1{\underset{#1}{\mathrm{p}}}\def\G#1#2{\overset{#1}{\underset{#2}{\Gamma}}}\def\abs#1{|#1|}電磁場のラグランジアンを幾何代数で書いたら
F:=A\wedge\D\\L=-\frac12F\vee F+A\vee J
である.確かめよう
A\vee J=\ou{A}{\mu}{}\ou{J}{\nu}{}\ou{\gamma}{}{\mu}\vee\ou{\gamma}{}{\nu}=\ou{A}{}{\nu}\ou{J}{\nu}{}=\ou{A}{}{\mu}\ou{J}{\mu}{}
\ou{F}{}{\mu\nu}:=\ou{A}{}{\mu}\ou{\p}{}{\nu}-\ou{A}{}{\nu}\ou{\p}{}{\mu}\\F=A\wedge\D=\ou{A}{}{\mu}\ou{\p}{}{\nu}\ou{\gamma}{\mu}{}\wedge\ou{\gamma}{\nu}{}=\frac12\ou{F}{}{\mu\nu}\ou{\gamma}{\mu}{}\wedge\ou{\gamma}{\nu}{}
\frac12F\vee F=\frac18E_0((\ou{F}{}{\mu\nu}\ou{\gamma}{\mu}{}\wedge\ou{\gamma}{\nu}{})^2)
直交座標系(\os{x'}{\iota})(基底\ou{\gamma'}{}{\mu}偏微分\ou{p'}{}{\mu})を使って計算を進める.
\ou{F}{}{\mu\nu}\ou{\gamma}{\mu}{}\wedge\ou{\gamma}{\nu}{}=\ou{F}{}{\mu\nu}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\os{x}{c}\ou{\p'}{}{a}\os{x}{d}\ou{\p'}{}{b}\ou{\gamma'}{}{c}\wedge\ou{\gamma'}{}{d}
\ou{\gamma'}{}{c}\ou{\gamma'}{}{d}\ (c\neq d)の成分は
2\ou{F}{}{\mu\nu}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\os{x}{c}\ou{\p'}{}{a}\os{x}{d}\ou{\p'}{}{b}
時空代数2次同士の幾何積の0次の部分は
(i,\ j)=(2,\ 3),\ (3,\ 1),\ (1,\ 2)\\a:=\displaystyle\sum_{k=1}^3a_{0k}\gamma_0\gamma_k+\displaystyle\sum_{ij}a_{ij}\gamma_i\gamma_j\\b:=\displaystyle\sum_{k=1}^3b_{0k}\gamma_0\gamma_k+\displaystyle\sum_{ij}b_{ij}\gamma_i\gamma_j\\E_0(ab)=\displaystyle\sum_{k=1}^3a_{0k}b_{0k}-\displaystyle\sum_{ij}a_{ij}b_{ij}
であるため
\frac18E_0((\ou{F}{}{\mu\nu}\ou{\gamma}{\mu}{}\wedge\ou{\gamma}{\nu}{})^2)=\frac12\{\displaystyle\sum_{k=1}^3(\ou{F}{}{\mu\nu}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\os{x}{0}\ou{\p'}{}{a}\os{x}{k}\ou{\p'}{}{b})^2-\displaystyle\sum_{ij}(\ou{F}{}{\mu\nu}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\os{x}{i}\ou{\p'}{}{a}\os{x}{j}\ou{\p'}{}{b})^2\}\\=\frac12\{\displaystyle\sum_{k=1}^3\ou{F}{}{\mu\nu}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\os{x}{0}\ou{\p'}{}{a}\os{x}{(k)}\ou{\p'}{}{b}\ou{F}{}{\rho\sigma}\ou{g}{\rho e}{}\ou{g}{\sigma f}{}\os{x}{0}\ou{\p'}{}{e}\os{x}{(k)}\ou{\p'}{}{f}-\displaystyle\sum_{ij}\ou{F}{}{\mu\nu}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\os{x}{(i)}\ou{\p'}{}{a}\os{x}{(j)}\ou{\p'}{}{b}\ou{F}{}{\rho\sigma}\ou{g}{\rho e}{}\ou{g}{\sigma f}{}\os{x}{(i)}\ou{\p'}{}{e}\os{x}{(j)}\ou{\p'}{}{f}\\=\frac12\ou{F}{}{\mu\nu}\ou{F}{}{\rho\sigma}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\ou{g}{\rho e}{}\ou{g}{\sigma f}{}\{\displaystyle\sum_{k=1}^3\os{x}{0}\ou{\p'}{}{a}\os{x}{(k)}\ou{\p'}{}{b}\os{x}{0}\ou{\p'}{}{e}\os{x}{(k)}\ou{\p'}{}{f}-\os{x}{(i)}\ou{\p'}{}{a}\os{x}{(j)}\ou{\p'}{}{b}\os{x}{(i)}\ou{\p'}{}{e}\os{x}{(j)}\ou{\p'}{}{f}\}
\ou{g}{}{ea}\ou{g}{}{fb}=\os{x}{h}\ou{\p'}{}{a}\os{x}{m}\ou{\p'}{}{b}\os{x}{g}\ou{\p'}{}{e}\os{x}{l}\ou{\p'}{}{f}\ou{\eta}{}{gh}\ou{\eta}{}{lm}\\=\{\displaystyle\sum_{k=1}^3\{\os{x}{0}\ou{\p'}{}{a}\os{x}{(k)}\ou{\p'}{}{b}\os{x}{0}\ou{\p'}{}{e}\os{x}{(k)}\ou{\p'}{}{f}+\os{x}{(k)}\ou{\p'}{}{a}\os{x}{0}\ou{\p'}{}{b}\os{x}{(k)}\ou{\p'}{}{e}\os{x}{0}\ou{\p'}{}{f}\}-\displaystyle\sum_{ij}\{\os{x}{(i)}\ou{\p'}{}{a}\os{x}{(j)}\ou{\p'}{}{b}\os{x}{(i)}\ou{\p'}{}{e}\os{x}{(j)}\ou{\p'}{}{f}+\os{x}{(j)}\ou{\p'}{}{a}\os{x}{(i)}\ou{\p'}{}{b}\os{x}{(j)}\ou{\p'}{}{e}\os{x}{(i)}\ou{\p'}{}{f}\}\}
よって
\frac12F\vee F=\frac18E_0((\ou{F}{}{\mu\nu}\ou{\gamma}{\mu}{}\wedge\ou{\gamma}{\nu}{})^2)=\frac14\ou{F}{}{\mu\nu}\ou{F}{}{\rho\sigma}\ou{g}{\mu a}{}\ou{g}{\nu b}{}\ou{g}{\rho e}{}\ou{g}{\sigma f}{}\ou{g}{}{ea}\ou{g}{}{fb}\\=\frac14\ou{F}{}{\mu\nu}\ou{F}{}{\rho\sigma}\ou{g}{\mu\rho}{}\ou{g}{\nu\sigma}{}