DescriptionBlack.Hole,Extremal.Kerr.Newman,Raytracing.png	English: Extremal Kerr-Newman black hole with spin a/M=√½ and charge Q/M=√½, in natural units of G=K=c=1. Therefore a²+Q²=M². The observer is at r=50M and views the black hole from the equatorial plane (edge on). FOV: 77.4°×38.7°. The equations that were used to raytrace the image can be found here.
Date	10 avril 2018
Source	Travail personnel
Auteur	Yukterez (Simon Tyran, Vienna). Source material for the Milky way background (also available on Commons): ESO/S.Brunier. Code for the relativistic raytracer: yukterez.net
Autres versions

Line-element in Boyer-Lindquist-coordinates:

${\displaystyle {\rm {d\tau ^{2}\ =\ \left(1-{\frac {2r-\mho ^{2}}{\Sigma }}\right)\mathrm {d} t^{2}\ -\ {\frac {\Sigma }{\Delta }}\ \mathrm {d} r^{2}\ -\ \Sigma \ d\theta ^{2}\ -\ {\frac {\chi }{\Sigma }}\ \sin ^{2}\theta \ d\phi ^{2}\ +\ 2\ {\frac {\Lambda }{\Sigma }}\ dt\ d\phi }}}$

Shorthand terms:

${\displaystyle {\rm {\Delta =r^{2}-2r+a^{2}+\mho ^{2}\ ,\ \Sigma =r^{2}+a^{2}\ \cos ^{2}\theta \ ,\ \chi =(a^{2}+r^{2})^{2}-a^{2}\ \sin ^{2}\theta \ \Delta \ ,\ \ \Lambda =a\ (2r-\mho ^{2})\ \sin ^{2}\theta }}}$

with the dimensionless spin parameter a=Jc/G/M² and the dimensionless electric charge parameter ℧=Qₑ/M·√(K/G). Here G=M=c=K=1 so that a=J und ℧=Qₑ, with lengths in GM/c² and times in GM/c³.

Co- and contravariant metric:

${\displaystyle {g_{\mu \nu }={\rm {\left({\begin{array}{cccc}{\rm {1-{\frac {2r-\mho ^{2}}{\Sigma }}}}&0&0&{\frac {\Lambda }{\Sigma }}\\0&{\rm {-{\frac {\Sigma }{\Delta }}}}&0&0\\0&0&{\rm {-\Sigma }}&0\\{\frac {\Lambda }{\Sigma }}&0&0&-{\frac {\chi \sin ^{2}\theta }{\Sigma }}\ \end{array}}\right)}}\ \to \ g^{\mu \nu }={\rm {\left({\begin{array}{cccc}{\rm {\frac {\chi }{\Delta \Sigma }}}&0&0&{\rm {-{\frac {a\left({\rm {\mho ^{2}-2r}}\right)\Sigma }{\rm {\left({\rm {\mho ^{2}-2r+\Sigma }}\right)\chi -a\Lambda }}}}}\\0&{\rm {-{\frac {\Delta }{\Sigma }}}}&0&0\\0&0&{\rm {-{\frac {1}{\Sigma }}}}&0\\{\rm {-{\frac {a\left({\rm {\mho ^{2}-2r}}\right)\Sigma }{\rm {\left({\rm {\mho ^{2}-2r+\Sigma }}\right)\chi -a\Lambda }}}}}&0&0&{\rm {-{\frac {\Delta -a^{2}\sin ^{2}\theta }{\Delta \Sigma \sin ^{2}\theta }}}}\\\end{array}}\right)}}}}$

Contravariant Maxwell tensor:

${\displaystyle {\rm {F}}^{\mu \nu }=\left({\begin{array}{cccc}0&-{\frac {\rm {4(a^{2}+r^{2})\ \mho \ (\cos(2\theta )\ a^{2}+a^{2}-2r^{2})}}{\rm {(\cos(2\theta )\ a^{2}+a^{2}+2r^{2})^{3}}}}&-{\frac {\rm {8a^{2}r\ \mho \sin(2\theta )}}{\rm {(\cos(2\theta )\ a^{2}+a^{2}+2r^{2})^{3}}}}&0\\{\frac {\rm {4(a^{2}+r^{2})\ \mho \ (\cos(2\theta )\ a^{2}+a^{2}-2r^{2})}}{\rm {(\cos(2\theta )\ a^{2}+a^{2}+2r^{2})^{3}}}}&0&0&{\frac {a\ \mho \ (a^{2}\cos ^{2}\theta -r^{2})}{(r^{2}+a^{2}\cos ^{2}\theta )^{3}}}\\{\frac {\rm {8a^{2}r\ \mho \sin(2\theta )}}{\rm {(\cos(2\theta )\ a^{2}+a^{2}+2r^{2})^{3}}}}&0&0&{\frac {\rm {16a\ r\ \mho \cot \theta }}{\rm {(\cos(2\theta )\ a^{2}+a^{2}+2r^{2})^{3}}}}\\0&{\frac {\rm {a\ \mho \ (r^{2}-a^{2}\cos ^{2}\theta )}}{\rm {(r^{2}+a^{2}\cos ^{2}\theta )^{3}}}}&-{\frac {\rm {16a\ r\ \mho \cot \theta }}{\rm {(\cos(2\theta )\ a^{2}+a^{2}+2r^{2})^{3}}}}&0\\\end{array}}\right)}$

The coordinate acceleration of a test-particle with the specific charge q is given by

${\displaystyle {\rm {{\ddot {x}}^{i}=-\sum _{j=1}^{4}\sum _{k=1}^{4}{\dot {x}}^{j}\ {\dot {x}}^{k}\ \Gamma _{jk}^{i}+q\ {F^{ik}}\ {{\dot {x}}^{j}}}}\ {g_{\rm {jk}}}}$

with the Christoffel-symbols

${\displaystyle \Gamma _{\rm {jk}}^{\rm {i}}=\sum _{\rm {s=1}}^{4}{\frac {g^{\rm {is}}}{2}}\left({\frac {\partial {g}_{\rm {sj}}}{\partial {\rm {x^{k}}}}}+{\frac {\partial {g}_{\rm {sk}}}{\partial {\rm {x^{j}}}}}-{\frac {\partial {g}_{\rm {jk}}}{\partial {\rm {x^{s}}}}}\right)}$

So the second proper time derivatives are

${\displaystyle {\rm {{\ddot {t}}=-(a^{2}\ {\dot {\theta }}\ (\sin(2\theta )(q\ \mho \ r+(\mho ^{2}-2r)\ {\dot {t}})-2a\sin ^{3}\theta \cos \theta \ (\mho ^{2}-2r)\ {\dot {\phi }})+}}}$ ${\displaystyle {\rm {({\dot {r}}\ ((a^{2}+r^{2})(a^{2}\cos ^{2}\theta \ (q\mho -2{\dot {t}})+r(2\ (r-\mho ^{2}){\dot {t}}-q\ \mho \ r))+a\sin ^{2}\theta \ {\dot {\phi }}\ (2a^{4}\cos ^{2}\theta +}}}$ ${\displaystyle {\rm {a^{2}\mho ^{2}r\ (\cos(2\theta )+3)-a^{2}r^{2}(\cos(2\theta )+3)+4\mho ^{2}r^{3}-6r^{4})))/(a^{2}+(r-2)r+\mho ^{2}))/((a^{2}\cos ^{2}\theta +r^{2})^{2})}}}$

for the time component,

${\displaystyle {\rm {{\ddot {r}}=(a^{2}{\dot {\theta }}\sin(2\theta )\ {\dot {r}})/(a^{2}\cos ^{2}\theta +r^{2})+{\dot {r}}^{2}((r-1)/(a^{2}+(r-2)\ r+\mho ^{2})-r/(a^{2}\cos ^{2}\theta +r^{2}))+}}}$ ${\displaystyle {\rm {((a^{2}+(r-2)\ r+\mho ^{2})(8a\sin ^{2}\theta \ {\dot {\phi }}\ (a^{2}\cos ^{2}\theta \ (q\ \mho -2{\dot {t}})+r(2(r-\mho ^{2}){\dot {t}}-q\ \mho \ r))+}}}$ ${\displaystyle {\rm {8{\dot {t}}\ (a^{2}\cos ^{2}\theta \ ({\dot {t}}-q\ \mho )+r\ (q\ \mho \ r+(\mho ^{2}-r)\ {\dot {t}}))+8r\ {\dot {\theta }}^{2}\ (a^{2}\cos ^{2}\theta +r^{2})^{2}+}}}$ ${\displaystyle {\rm {\sin ^{2}\theta \ {\dot {\phi }}^{2}\ (2a^{4}\sin ^{2}(2\theta )+r\ (a^{2}(a^{2}\cos(4\theta )+3a^{2}+4\ (a-\mho )(a+\mho )\cos(2\theta )+4\mho ^{2})+}}}$ ${\displaystyle {\rm {8r\ (-a^{2}\sin ^{2}\theta +2a^{2}r\cos ^{2}\theta +r^{3})))))/(8\ (a^{2}\cos ^{2}\theta +r^{2})^{3})}}}$

for the radial component,

${\displaystyle {\rm {{\ddot {\theta }}=-(2r\ {\dot {\theta }}\ {\dot {r}})/(a^{2}\cos ^{2}\theta +r^{2})-(a^{2}\sin \theta \cos \theta \ {\dot {r}}^{2})/((a^{2}+(r-2)\ r+}}}$ ${\displaystyle {\rm {\mho ^{2})(a^{2}\cos ^{2}\theta +r^{2}))+(\sin(2\theta )(a^{2}(8{\dot {\theta }}^{2}(a^{2}\cos ^{2}\theta +r^{2})^{2}-8{\dot {t}}(2q\ \mho \ r+}}}$ ${\displaystyle {\rm {(\mho ^{2}-2r)\ {\dot {t}}))+16a\ (a^{2}+r^{2})\ {\dot {\phi }}(q\ \mho \ r+(\mho ^{2}-2r)\ {\dot {t}})+{\dot {\phi }}^{2}(3a^{6}+11a^{4}r^{2}+10a^{4}r-}}}$ ${\displaystyle {\rm {5a^{4}\mho ^{2}+4a^{2}(a^{2}+2r^{2})\cos(2\theta )(a^{2}+(r-2)r+\mho ^{2})-8a^{2}\mho ^{2}r^{2}+16a^{2}r^{4}+16a^{2}r^{3}+a^{4}\cos(4\theta )(a^{2}+}}}$ ${\displaystyle {\rm {(r-2)r+\mho ^{2})+8r^{6})))/(16(a^{2}\cos ^{2}\theta +r^{2})^{3})}}}$

the poloidial component and

${\displaystyle {\rm {{\ddot {\phi }}=-(({\dot {r}}(4a\ q\ \mho \ (a^{2}\cos ^{2}\theta -r^{2})-8a\ {\dot {t}}(a^{2}\cos ^{2}\theta +r\ (\mho ^{2}-r))+{\dot {\phi }}\ (2a^{4}\sin ^{2}(2\theta )+}}}$ ${\displaystyle {\rm {8r^{3}(a^{2}\cos(2\theta )+a^{2}+\mho ^{2})+a^{2}r\ (a^{2}(4\cos(2\theta )+\cos(4\theta ))+3a^{2}+8\mho ^{2})-4a^{2}r^{2}(\cos(2\theta )+3)+8r^{5}-16r^{4})))/(a^{2}+}}}$ ${\displaystyle {\rm {(r-2)\ r+\mho ^{2})+{\dot {\theta }}\ ({\dot {\phi }}\ (a^{4}(-\sin(4\theta ))-2a^{2}\sin(2\theta )(3a^{2}+4(r-1)r+2\mho ^{2})+8\ (a^{2}+r^{2})^{2}\cot \theta )+}}}$ ${\displaystyle {\rm {8a\cot \theta \ (q\ \mho \ r+(\mho ^{2}-2r)\ {\dot {t}})))/(4(a^{2}\cos ^{2}\theta +r^{2})^{2})}}}$

for the axial component of the 4-acceleration. The total time dilation is

${\displaystyle {\rm {\dot {t}}}}$${\displaystyle {\rm {={\frac {\csc ^{2}\theta \ ({L_{z}}(a\ \Delta \sin ^{2}\theta -a\ (a^{2}+r^{2})\sin ^{2}\theta )-q\ \mho \ r\ (a^{2}+r^{2})\sin ^{2}\theta +E((a^{2}+r^{2})^{2}\sin ^{2}\theta -a^{2}\Delta \sin ^{4}\theta ))}{\Delta \Sigma }}}}}$${\displaystyle {\rm {={\frac {a(L_{z}-aE\sin ^{2}\theta )+(r^{2}+a^{2})P/\Delta }{\Sigma }}}}}$

where the differentiation goes by the proper time τ for charged (q≠0) and neutral (q=0) particles (μ=-1, v<1), and for massless particles (μ=0, v=1) by the spatial affine parameter ŝ. The relation between the first proper time derivatives and the local three-velocity components relative to a ZAMO is

${\displaystyle {\rm {{\dot {r}}={\frac {v_{r}{\sqrt {\Delta }}}{\sqrt {\Sigma (1-\mu ^{2}v^{2})}}}}}={\frac {{\rm {Sign}}({\rm {v_{r}){\sqrt {\rm {V_{r}}}}}}}{\Sigma }}\ \ \ \ \ \ \ \ {\rm {{\dot {\theta }}={\frac {v_{\theta }}{\sqrt {\Sigma (1-\mu ^{2}v^{2})}}}={\frac {\rm {Sign(v_{\theta }){\sqrt {\rm {V_{\theta }}}}}}{\Sigma }}}}}$

${\displaystyle {\dot {\phi }}{\rm {={\frac {a\left(a^{2}E-aL_{z}-\Delta E-qr\mho +Er^{2}\right)+\Delta L_{z}\csc ^{2}\theta }{\Delta \Sigma }}}}}$

The local three-velocity in terms of the position and the constants of motion is

${\displaystyle {\rm {v=\left|{\frac {\sqrt {-a^{2}L_{z}^{2}\Sigma ^{2}\left(\mho ^{2}-2r\right)^{2}+2aL_{z}\Sigma \chi \left(2r-\mho ^{2}\right)(E\Sigma -qr\mho )+\chi \left(\Delta \Sigma ^{3}-\chi (E\Sigma -qr\mho )^{2}\right)}}{aL_{z}\Sigma \left(\mho ^{2}-2r\right)+\chi (E\Sigma -qr\mho )}}\right|}}}$

which reduces to

${\displaystyle {\rm {v={\sqrt {\frac {\chi \ (E-L_{z}\ \Omega )^{2}-\Delta \ \Sigma }{\chi \ (E-L_{z}\ \Omega )^{2}}}}={\frac {\sqrt {{\dot {t}}^{2}-\varsigma ^{2}}}{\dot {t}}}}}}$

if the charge of the test particle is q=0. The escape velocity of a charged particle with zero orbital angular momentum is

${\displaystyle {\rm {v_{esc}=\left|{\frac {\sqrt {a^{4}\cos ^{4}\theta (\Delta \Sigma -\chi )+2a^{2}r\cos ^{2}\theta (q\chi \mho +\Delta r\Sigma -r\chi )+r^{2}\left(-q^{2}\chi \mho ^{2}+2qr\chi \mho +r^{2}(\Delta \Sigma -\chi )\right)}}{{\sqrt {\chi }}\left(a^{2}\cos ^{2}\theta +r(r-q\mho )\right)}}\right|}}}$

which for a neutral test particle with q=0 reduces to

${\displaystyle {\rm {v_{esc}}}={\frac {\sqrt {\varsigma ^{2}-1}}{\varsigma }}}$

with the gravitational time dilation of a locally stationary ZAMO

${\displaystyle \varsigma ={\frac {\rm {dt}}{\rm {d\tau }}}={\sqrt {|g^{\rm {tt}}|}}={\sqrt {\frac {\chi }{\Delta \ \Sigma }}}}$

which is infinite at the horizon. The time dilation of a globally stationary particle (with respect to the fixed stars) is

${\displaystyle \sigma ={\frac {\rm {dt}}{\rm {d\tau }}}={\sqrt {|1/g_{\rm {tt}}|}}={\frac {1}{\sqrt {1-{\frac {\rm {2r-\mho ^{2}}}{\Sigma }}}}}}$

which is infinite at the ergosphere. The Frame-Dragging angular velocity observed at infinity is

${\displaystyle \omega =\left|{\frac {g_{\rm {t\phi }}}{g_{\phi \phi }}}\right|={\rm {a\left(2r-\mho ^{2}\right)/\chi }}}$

The local frame dragging velocity with respect to the fixed stars is therefore

${\displaystyle v_{\phi }={\sqrt {g_{\rm {t\phi }}\ g^{\rm {t\phi }}}}={\sqrt {1-g_{\rm {tt}}\ g^{\rm {tt}}}}=|{\frac {g_{\rm {t\phi }}}{g_{\rm {\phi \phi }}}}{\sqrt {g^{\rm {tt}}}}\ {\sqrt {g_{\rm {\phi \phi }}}}|=\omega {\bar {\rm {R}}}_{\phi }\varsigma }$

which is c at the ergosphere. The axial radius of gyration is

${\displaystyle {\bar {\rm {R}}}_{\phi }={\sqrt {|g_{\phi \phi }|}}={\sqrt {\frac {\chi }{\Sigma }}}\ \sin \theta }$

The 3 conserved quantities are 1) the total energy:

${\displaystyle {{\rm {E}}=g_{\rm {tt}}\ {\dot {\rm {t}}}+g_{\rm {t\phi }}\ {\rm {{\dot {\phi }}+{\rm {q\ A_{t}={\dot {t}}\left(1-{\frac {2r-\mho ^{2}}{\Sigma }}\right)+{\dot {\phi }}{\frac {a\sin ^{2}\theta \left(2r-\mho ^{2}\right)}{\Sigma }}+{\frac {\mho \ q\ r}{\Sigma }}={\rm {{\sqrt {\frac {\Delta \ \Sigma }{(1-\mu ^{2}v^{2})\ \chi }}}+\omega \ L_{z}+{\frac {\mho \ q\ r}{\Sigma }}}}}}}}}}$

2) the axial angular momentum:

${\displaystyle {\rm {L_{z}}}=-g_{\phi \phi }\ {\dot {\phi }}-g_{\rm {t\phi }}\ {\rm {{\dot {t}}-q\ A_{\phi }={\rm {{\frac {{\dot {\phi }}\ \chi \sin ^{2}\theta }{\Sigma }}-{\frac {{\dot {t}}\ a\ \sin ^{2}\theta \left(2r-Q^{2}\right)}{\Sigma }}+{\frac {a\ r\ \mho \ q\ \sin ^{2}\theta }{\Sigma }}}}={\frac {v_{\phi }\ {\bar {R}}_{\phi }}{\sqrt {1-\mu ^{2}\ v^{2}}}}+{\frac {(1-\mu ^{2}v^{2})\ a\ r\ \mho \ q\ \sin ^{2}\theta }{\Sigma }}}}}$

3) the Carter constant:

${\displaystyle {\rm {Q={\frac {{v_{\theta }}^{2}\ \Sigma }{1-\mu ^{2}v^{2}}}+\cos ^{2}\theta \left(a^{2}(\mu ^{2}-E^{2})+{\frac {L_{z}^{2}}{\sin ^{2}\theta }}\right)}}}$

The effective radial potential whose zero roots define the turning points is

${\displaystyle {\rm {V_{r}=P^{2}-\Delta \left((L_{z}-aE)^{2}+Q+\mu ^{2}r^{2}\right)}}}$

and the poloidial potential

${\displaystyle {\rm {V_{\theta }={\frac {{v_{\theta }}^{2}\ \Sigma }{1-\mu ^{2}v^{2}}}=Q-\cos ^{2}\theta \left(a^{2}\left(\mu ^{2}-E^{2}\right)+{\frac {\rm {L_{z}^{2}}}{\sin ^{2}\theta }}\right)}}}$

with the parameter

${\displaystyle {\rm {P=E\left(a^{2}+r^{2}\right)-aL_{z}+qr\mho }}}$

The azimutal and latitudinal impact parameters are

${\displaystyle {\rm {b_{\phi }={\frac {L_{z}}{E}}\ ,\ \ b_{\theta }={\sqrt {\frac {Q}{E^{2}}}}}}}$

The horizons and ergospheres have the Boyer-Lindquist-radius

${\displaystyle {\rm {r_{H}^{\pm }=1\pm {\sqrt {1-a^{2}-\mho ^{2}}}}}\ ,\ \ {\rm {r_{E}^{\pm }=1\pm {\sqrt {\rm {1-a^{2}\cos ^{2}\theta -\mho ^{2}}}}}}}$

In this article the total mass equivalent M, which also contains the rotational and the electrical field energy, is set to 1; the relation of M with the irreducible mass is

${\displaystyle {\rm {M_{\rm {irr}}={\frac {\sqrt {2M^{2}-\mho ^{2}+2M{\sqrt {M^{2}-\mho ^{2}-a^{2}}}}}{2}}\ \to \ M={\sqrt {\frac {16M_{\rm {irr}}^{4}+8M_{\rm {irr}}^{2}\ \mho ^{2}+\mho ^{4}}{16M_{\rm {irr}}^{2}-4a^{2}}}}}}}$

where a is in units of M.

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