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The event horizon for a rotating black hole does not only depend on the mass but also on the angular momentum.

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Bill_K

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No. The most general black hole is represented by the Kerr-Newman solution, and it has this property.Specifically, I'm interested in whether it's possible for a blackhole to have an event horizon which is not of the form: constant * mass.

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Also Kerr-Newman is not the most general black hole. It is only the most general stationary solution to the Einstein-Maxwell equations in four dimensions. There are obviously many more general solutions like for example time dependent ones with matter falling to the horizon.

The Schwarzschild solution is a very very special case.

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WannabeNewton

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The OP asked if there are black holes which do not have a schwarzchild radius i.e. r = 2GM. The kerr - newman solution is the most general stationary solution that DOES have this property.Kerr-Newman don't have the property that the radius is proportional to the mass times a constant. It depends on both the angular momentum(as passionflower said) and the charge.

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No, for a kerr black holeThe OP asked if there are black holes which do not have a schwarzchild radius i.e. r = 2GM. The kerr - newman solution is the most general stationary solution that DOES have this property.

[itex]r = M + \sqrt{M^2-a^2}[/itex]

in units G=1 where a = J/M and J is angular momentum.

Go read a book!

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WannabeNewton

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That is a definition of r to find the location of the horizon in the kerr geometry i.e. at [tex]g_{rr} = \infty [/tex]. This has nothing to with the fact that r = 2M IS the schwarzchild radius for a kerr black hole.No, for a kerr black hole

[itex]r = M + \sqrt{M^2-a^2}[/itex]

in units G=1 where a = J/M and J is angular momentum.

Go read a book!

Last edited:

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Bill_K

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The point is that a is restricted to the range 0 < a < M, so M < r < 2M. The inner horizon is no longer spherical, but its size is proportional to M. In fact this must be the case, since M is the only parameter in the solution that sets the scale.No, for a kerr black hole r = M + √(M^{2}- a^{2})

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r=2M has nothing to do with the event horizon of a Kerr black hole.That is a definition of r to find the location of the horizon in the kerr geometry i.e. at [tex]g_{rr} = \infty [/tex]. This has nothing to with the fact that r = 2M IS the schwarzchild radius for a kerr black hole.

The OP asked whether the event horizon of a black hole is proportional to the mass times a constant. This is only the case for schwarzschild black hole.

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The inner horizon radius is not proportional to the mass either. Take fix J and take M large and the inner horizon shrinks to zero as J/M goes to zero.The point is that a is restricted to the range 0 < a < M, so M < r < 2M. The inner horizon is no longer spherical, but its size is proportional to M. In fact this must be the case, since M is the only parameter in the solution that sets the scale.

a is not restricted to a<M. a>M is a naked singularity.

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WannabeNewton

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Yes excuse me I see what you are saying. [tex]g_{rr} = \infty[/tex] and [tex]g_{tt} = 0[/tex] for a schwarzchild black hole is r = 2m but for a kerr black hole the actual horizon (not the ergosphere) [tex]g_{rr} = \infty [/tex] is when [tex]\Delta = 0[/tex]. Thanks mate; indeed I was mixing up the terms for the event horizon and the definition of schwarzchild radius.r=2M has nothing to do with the event horizon of a Kerr black hole.

The OP asked whether the event horizon of a black hole is proportional to the mass times a constant. This is only the case for schwarzschild black hole.

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George Jones

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When "mass" is suitably defined, there are dynamic (non-stationary; no exterior timelike killing vector) black hole spacetimes that don't have this property. For an example that uses the Vaidya metric, see section 5.1.8 on page 132 (pdf page 148) of Eric Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of black hole Mechanics.

In this example, the apparent horizon has this property, but (a portion of) the event doesn't.

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edguy99

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I would appreciate if you had any good links to the subject "Kerr-Newman is not the most general black hole"

Also Kerr-Newman is not the most general black hole. It is only the most general stationary solution to the Einstein-Maxwell equations in four dimensions. There are obviously many more general solutions like for example time dependent ones with matter falling to the horizon.

The Schwarzschild solution is a very very special case.

I am interested in black holes where a lot of material is falling in from close to the top and bottem of the spinning black hole. Mostly in 3D aspects as I find many references and drawings on the internet really only seem to be 2D in nature (the axis of rotation is not precessing).

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George Jones

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Also Kerr-Newman is not the most general black hole. It is only the most general stationary solution to the Einstein-Maxwell equations in four dimensions. There are obviously many more general solutions like for example time dependent ones with matter falling to the horizon.

I gave an example for non-spinning black holes in post #12.I would appreciate if you had any good links to the subject "Kerr-Newman is not the most general black hole"

I am interested in black holes where a lot of material is falling in from close to the top and bottem of the spinning black hole. Mostly in 3D aspects as I find many references and drawings on the internet really only seem to be 2D in nature (the axis of rotation is not precessing).

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