File:Kerr Newman De Sitter (KNdS) Horizons vs Ergospheres.png

Summary[edit]

DescriptionKerr Newman De Sitter (KNdS) Horizons vs Ergospheres.png	English: KNdS: horizons & escape velocity vs ergospheres & frame dragging velocity
Date	24 August 2023
Source	Own work, Code: knds.yukterez.net
Author	Yukterez (Simon Tyran, Vienna)
Other versions

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Equations[edit]

The frame-dragging angular velocity is

${\displaystyle \omega ={\frac {\rm {d\phi }}{\rm {dt}}}=-{\frac {g_{\rm {t\phi }}}{g_{\rm {\phi \phi }}}}={\rm {\frac {a\ [a^{2}\ \Lambda \left(a^{2}+r^{2}\right)\cos ^{2}\theta +a^{2}\ \Lambda \ r^{2}+6\ r+\Lambda \ r^{4}-3\ \mho ^{2}]}{a^{2}\ \sin ^{2}\theta \ [a^{2}\left(\Lambda \ r^{2}-3\right)+6\ r+\Lambda \ r^{4}-3\ r^{2}-3\ \mho ^{2}]+a^{2}\ \Lambda \ \left(a^{2}+r^{2}\right)^{2}\cos ^{2}\theta +3\ \left(a^{2}+r^{2}\right)^{2}}}}}$

and the local frame-dragging velocity relative to constant ${\displaystyle {\rm {\{r,\theta ,\phi \}}}}$ positions (the speed of light at the ergospheres)

${\displaystyle \nu ={\sqrt {g_{\rm {t\phi }}\ g^{\rm {t\phi }}}}={\rm {\sqrt {-{\frac {a^{2}\ \sin ^{2}\theta \ [a^{2}\ \Lambda \left(a^{2}+r^{2}\right)\cos ^{2}\theta +a^{2}\Lambda \ r^{2}+6\ r+\Lambda \ r^{4}-3\ \mho ^{2}]^{2}}{\left(a^{2}\ \Lambda \ \cos ^{2}\theta +3\right)\left(a^{2}+r^{2}-a^{2}\sin ^{2}\theta \right)^{2}[a^{2}\left(\Lambda \ r^{2}-3\right)+6\ r+\Lambda \ r^{4}-3\ r^{2}-3\ \mho ^{2}]}}}}}}$

The escape velocity (the speed of light at the horizons) relative to the local corotating ZAMO (zero angular momentum observer) is

${\displaystyle {\rm {v}}={\sqrt {1-1/g^{\rm {tt}}}}={\rm {\sqrt {{\frac {3\left(a^{2}\Lambda \cos ^{2}\theta +3\right)\left(a^{2}+r^{2}-a^{2}\sin ^{2}\theta \right)^{2}\left[a^{2}\left(\Lambda r^{2}-3\right)+\Lambda r^{4}-3r^{2}+6r-3\mho ^{2}\right]}{\left(a^{2}\Lambda +3\right)^{2}\left(a^{2}\cos ^{2}\theta +r^{2}\right)\{a^{2}\Lambda \left(a^{2}+r^{2}\right)^{2}\cos ^{2}\theta +3\left(a^{2}+r^{2}\right)^{2}+a^{2}\sin ^{2}\theta \left[a^{2}\left(\Lambda r^{2}-3\right)+\Lambda r^{4}-3r^{2}+6r-3\mho ^{2}\right]\}}}+1}}}}$

Setting ${\displaystyle \nu =1}$ or equivalently ${\displaystyle g_{\rm {tt}}=0}$ and solving for ${\displaystyle {\rm {r}}}$ gives the Boyer Lindquist radii of the ergospheres and setting ${\displaystyle g^{\rm {rr}}=0}$ the horizon radii, for more information see here and here