【相対論】Newman-Penrose formalism 1

$g(k,l)=(u^c,{\gamma }^{a}u)(v^c,{\gamma }_{a}v)=-8|(u,v){|}^2$なので、$g(k,l)=-1$となるようにrescaleすればよい。
さらにこのとき
$$\begin{array}{rl}g(m,\overline{m})& =(v^c,{\gamma }^{a}u)(u^c,{\gamma }_{a}v)=(v^c,{\gamma }^{a}u)(v,{\gamma }_a{u}^c)\\ & =(v^c,e_A{u}^c)(v,{\gamma }_a{e}^A{\gamma }^{a}u)(\because {\gamma }_a{e}^{bc}{\gamma }^a=0)\\ & =(v^c,u^c)(v,{\gamma }_a{\gamma }^{a}u)-(v^c,z{u}^c)(v,{\gamma }_{a}z{\gamma }^{a}u)\\ & =4(v,u{)}^{\ast }(v,u)+(v^c,z{u}^c)(v,{\gamma }_a{\gamma }^{a}zu)\\ & =4|(v,u){|}^2-4(v^c,-i{u}^c)(v,iu)\\ & =4|(v,u){|}^2-4(v^c,u^c)(v,u)\\ & =4|(v,u){|}^2+4(v,u{)}^{\ast }(v,u)\\ & =8|(v,u){|}^2=1\end{array}$$
となる。
また
$$\begin{array}{rl}g(k,m)& =(u^c,{\gamma }^{a}u)(v,{\gamma }_a{u}^c)=(u^c,e_A{u}^c)(v^c,{\gamma }_a{e}^A{\gamma }^{a}u)=0\\ g(m,m)& =(v^c,{\gamma }^{a}u)(v^c,{\gamma }_{a}u)=(v^c,{\gamma }^{a}u)(u,{\gamma }_a{v}^c)\\ & =(v^c,e_A{v}^c)(u,{\gamma }_a{e}^A{\gamma }^{a}u)(\because {\gamma }_a{e}^{bc}{\gamma }^a=0)\\ & =(v^c,v^c)(v,{\gamma }_a{\gamma }^{a}u)+(v^c,z{v}^c)(u,{\gamma }_{a}z{\gamma }^{a}u)=0\end{array}$$
となる。

以下の計算から従う。
$$\begin{array}{rl}{\mathrm{\nabla }}_k{k}_a& =({\mathrm{\nabla }}_k{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a{\mathrm{\nabla }}_{k}u)=(\overline{\kappa }{v}^c-\overline{ϵ}{u}^c,{\gamma }_{a}u)+(u^c,\kappa {\gamma }_{a}v-ϵ{\gamma }_{a}u)\\ & =\overline{\kappa }(v^c,{\gamma }_{a}u)-\overline{ϵ}(u^c,{\gamma }_{a}u)+\kappa (u^c,{\gamma }_{a}v)-ϵ(u^c,{\gamma }_{a}u)\\ & =\overline{\kappa }{m}_a-\overline{ϵ}{k}_a-\kappa {\overline{m}}_a-ϵk_a\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_{k}m& =({\mathrm{\nabla }}_k{v}^c,{\gamma }_{a}u)+(v^c,{\gamma }_a{\mathrm{\nabla }}_{k}u)=(\overline{ϵ}{v}^c-\overline{\pi }{u}^c,{\gamma }_{a}u)+(v^c,{\gamma }_a(\kappa v-ϵu))\\ & =\overline{ϵ}m-\overline{\pi }k+\kappa l-ϵm\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_l{l}_a& =({\mathrm{\nabla }}_l{v}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a{\mathrm{\nabla }}_{l}v)=(\overline{\gamma }{v}^c-\overline{\nu }{u}^c,{\gamma }_{a}v)+(v^c,\gamma v-\nu u)\\ & =\overline{\gamma }{l}_a-\overline{\nu }\overline{m}+\gamma l-\nu m\\ {\mathrm{\nabla }}_l\overline{m}& =({\mathrm{\nabla }}_l{u}^c,{\gamma }_{a}v)+(u^c,{\gamma }_a{\mathrm{\nabla }}_{l}v)=(\overline{\tau }{v}^c-\overline{\gamma }{u}^c,{\gamma }_{a}v)+(u^c,{\gamma }_a(\gamma v-\nu u))\\ & =\overline{\tau }{l}_a-\overline{\gamma }\overline{m}+\gamma \overline{m}-\nu k_a\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_{\overline{m}}k& =({\mathrm{\nabla }}_{\overline{m}}{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a{\mathrm{\nabla }}_{\overline{m}}u)=(\overline{\rho }{v}^c-\overline{\alpha }{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a(\rho v-\alpha u))\\ & =\overline{\rho }{m}_a-\overline{\alpha }{k}_a+\rho {\overline{m}}_a-\alpha k_a\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_m{l}_a& =({\mathrm{\nabla }}_m{v}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a{\mathrm{\nabla }}_{m}v)=(\overline{\beta }{v}^c-\overline{\mu }{u}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a(\beta v-\mu u))\\ =& \overline{\beta }l-\overline{\mu }\overline{m}+\beta l_a-\mu m_a\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_m{k}_a& =({\mathrm{\nabla }}_m{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a{\mathrm{\nabla }}_{m}u)=(\overline{\sigma }{v}^c-\overline{\beta }{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a(\sigma v-\beta u))\\ & =\overline{\sigma }{m}_a-\overline{\beta }{k}_a+\sigma {\overline{m}}_a-\beta k\end{array}$$

$$\begin{array}{rl}{\mathrm{\nabla }}_{m}m& =({\mathrm{\nabla }}_m{v}^c,{\gamma }_{a}u)+(v^c,{\gamma }_a{\mathrm{\nabla }}_{m}u)=(\overline{\beta }{v}^c-\overline{\mu }{u}^c,{\gamma }_{a}u)+(v^c,{\gamma }_a(\sigma v-\beta u))\\ & =\overline{\beta }m-\overline{\mu }k+\sigma l-\beta m\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_{\overline{m}}{l}_a& =({\mathrm{\nabla }}_{\overline{m}}{v}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a{\mathrm{\nabla }}_{\overline{m}}v)=(\overline{\alpha }{v}^c-\overline{\lambda }{u}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a(\alpha v-\lambda u))\\ & =\overline{\alpha }{l}_a-\overline{\lambda }\overline{m}+\alpha l_a-\lambda m_a\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_{\overline{m}}\overline{m}& =({\mathrm{\nabla }}_{\overline{m}}{u}^c,{\gamma }_{a}v)+(u^c,{\gamma }_a{\mathrm{\nabla }}_{\overline{m}}v)=(\overline{\rho }{v}^c-\overline{\alpha }{u}^c,{\gamma }_{a}v)+(u^c,{\gamma }_a(\alpha v-\lambda u))\\ & =\overline{\rho }l-\overline{\alpha }\overline{m}+\alpha \overline{m}-\lambda k\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_l{k}_a& =({\mathrm{\nabla }}_l{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a{\mathrm{\nabla }}_{l}u)=(\overline{\tau }{v}^c-\overline{\gamma }{u}^c,{\gamma }_{a}u)+(u^c,{\gamma }_a(\tau v-\gamma u))\\ & =\overline{\tau }m-\overline{\gamma }{k}_a+\tau \overline{m}-\gamma k_a\end{array}$$
$$\begin{array}{rl}{\mathrm{\nabla }}_{k}l& =({\mathrm{\nabla }}_k{v}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a{\mathrm{\nabla }}_{k}v)=(\overline{ϵ}{v}^c-\overline{\pi }{u}^c,{\gamma }_{a}v)+(v^c,{\gamma }_a(ϵv-\pi u))\\ & =\overline{ϵ}{l}_a-\overline{\pi }\overline{m}+ϵl_a-\pi m_a\end{array}$$