Gravity for a stellar black hole

  • Thread starter stevebd1
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Thanks for the response. It appears the equation deals with a rotating black hole only. While it's likely that most black holes are rotating (Kerr black holes) due to that fact that most stars rotate, I still feel that what ever occurs with gravity at the event horizon and beyond happens to both a rotating (Kerr) black hole and static (Schwarzschild) black hole regardless of rotation. There may be some way of factoring in retardation for a rotating black hole but there would be an equation, in principle, that would apply to both in respect of gravity.

regards
Steve
 
  • #27
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Gravity well for a 3 sol mass rotating black hole, a = 0.95

I kept the following basic in order to keep the latex to a minimum. Hopefully it makes sense (it appears to make sense on the graph attached). I've kept my justification to a minimum but will probably elaberate on this later.


Equation for gravitational acceleration up to the event horizon of a static black hole-

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{2GM}{rc^{2}}\right)^{-\frac{1}{2}}[/tex]

which can be expressed as

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{R_{s}}{R}\right)^{-\frac{1}{2}}[/tex]

where Rs is the Schawartzschild radius- [tex]R_{s}=2GM/c^{2}[/tex]


Following equations established to obtain a simple estimation of gravity acceleration-


Gravitational acceleration past the event horizon of a static black hole (based on divergence at EH and gravity and space swapping properies)-

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{rc^{2}}{2GM}\right)^{-\frac{1}{2}}[/tex]

which can be expressed as

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{R}{R_{s}}\right)^{-\frac{1}{2}}[/tex]


Attached is a graph for a 3 sol mass rotating black hole (a = 0.95) put together using the equations below-


Gravitational acceleration up to the event horizon for a rotating black hole (where a represents angular momentum, 0-1)-

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{GM(1+\sqrt{(1-a^{2})})}{rc^{2}}\right)^{-\frac{1}{2}}[/tex]

which can be expressed as-

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{R_{+}}{R}\right)^{-\frac{1}{2}}[/tex]

where R+ is the event horizon- [tex]R_{+}=(GM/c^{2}) (1+\sqrt{(1-a^{2})})[/tex]


Gravitational acceleration between cauchy horizon and event horizon of a rotating black hole-

[tex]g=\frac{GM}{r^{2}}\left(\left(1-\frac{R}{R_{+}}\right)^{-\frac{1}{2}}+\left(1-\frac{R_{-}}{R}\right)^{-\frac{1}{2}}\right)[/tex]

where R_ is the Cauchy horizon- [tex]R_{-}=(GM/c^{2}) (1-\sqrt{(1-a^{2})})[/tex]


Gravitational acceleration within Cauchy horizon up to ring singularity-

[tex]g=\frac{GM}{r^{2}}\left(1-\frac{R}{R_{-}}\right)^{-\frac{1}{2}}[/tex]


It could be said that the gravity acceleration peaks of infinity signify a 'flip' as the properties of gravity and space swap properties.

Steve
 

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