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- Thread starter TimeRip496
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- Since it's a point-mass, you might expect that it would be a black hole. However, the sort of black hole that is studied most extensively, one with an "event horizon", is only possible with the constraint: [itex]Q^2 + \frac{J^2}{M^2} \leq M^2[/itex], where [itex]Q[/itex] is the charge, [itex]J[/itex] is the angular momentum, and [itex]M[/itex] is the mass (in some units). An electron violates this constraint, because its mass is so small compared with its charge and angular momentum. GR doesn't say that nothing can violate this constraint, but something that violates it will not have an event horizon. It will be a "naked singularity" I guess.

- As I understand it (which is not very well), standard General Relativity does not allow for point particles with intrinsic spin. This is a technical point that I really don't understand very well, but GR assumes that the stress-energy tensor, which is the source of spacetime curvature, is symmetric. If you have particles with nonzero intrinsic spin, the stress-energy tensor becomes nonsymmetric, and a generalization of GR is required.

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Nugatory

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That's not a dumb question, and the answer is pretty much as you suspect: for many reasons like the ones that stevendaryl gave above, we can't ignore the quantum effects so the EFE doesn't work. In practice, this is not a problem because the gravitational forces are so small compared with the other forces at work here (try calculating the numerical values of the constraint that stevendaryl posted to see what I mean) that they can be ignored and we can solve the problem using just quantum mechanics.

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GR doesn't say that nothing can violate this constraint, but something that violates it will not have an event horizon. It will be a "naked singularity" I guess.

Yes. The specific solution of the EFE that would apply to a nonspinning charged particle (as you note, there are additional issues that arise for particles with spin, like the electron) is the Reissner-Nordstrom geometry with ##Q > M##, and this geometry has a naked singularity at ##r = 0## and no event horizon. There is a conjecture (called "cosmic censorship") in GR that says that such a solution cannot be physically realized because of the naked singularity, but this has never been proven.

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Do you mind telling me more about that event horizon? As far as I know, it is a border of a black hole at which light cannot escape upon crossing it. But how do you obtain that equation?

- Since it's a point-mass, you might expect that it would be a black hole. However, the sort of black hole that is studied most extensively, one with an "event horizon", is only possible with the constraint: [itex]Q^2 + \frac{J^2}{M^2} \leq M^2[/itex], where [itex]Q[/itex] is the charge, [itex]J[/itex] is the angular momentum, and [itex]M[/itex] is the mass (in some units). An electron violates this constraint, because its mass is so small compared with its charge and angular momentum. GR doesn't say that nothing can violate this constraint, but something that violates it will not have an event horizon. It will be a "naked singularity" I guess.

- As I understand it (which is not very well), standard General Relativity does not allow for point particles with intrinsic spin. This is a technical point that I really don't understand very well, but GR assumes that the stress-energy tensor, which is the source of spacetime curvature, is symmetric. If you have particles with nonzero intrinsic spin, the stress-energy tensor becomes nonsymmetric, and a generalization of GR is required.

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Do you mind telling me more about that event horizon? As far as I know, it is a border of a black hole at which light cannot escape upon crossing it.

Yes.

how do you obtain that equation?

From the Kerr-Newman solution of the Einstein Field Equation.

- #7

stevebd1

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Do you mind telling me more about that event horizon? As far as I know, it is a border of a black hole at which light cannot escape upon crossing it. But how do you obtain that equation?

The equation for the event horizon(s) for a Kerr-Newman black hole is-

[tex]r_{\pm}=M\pm\sqrt{M^2-Q^2-a^2}[/tex]

where [itex]\pm[/itex] denotes the outer (+) and inner (-) horizon, [itex]M=Gm/c^2[/itex], [itex]a=J/mc[/itex] and [itex]Q=C\sqrt(Gk_e)/c^2[/itex] where [itex]k_e=1/(4\pi\varepsilon_0)[/itex] and [itex]C[/itex] is charge.

If you plug in the units for mass and charge of an electron, you'll see there is already an issue with the cosmic censorship law (as stated in post #2) before you've even consider spin, which causes issues with black hole thermodynamics.

- #8

stevebd1

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http://universeinproblems.com/index...etry_of_the_stationary_limit_surfaces_in_Kerr

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what would happen to the ergosphere in the case of an over-extremal black hole

There isn't one; in fact there isn't even a black hole any more in the super-extremal case, because there isn't an event horizon.

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Do you mean Maxwell's equations?

The OP said "EFE", which means the Einstein Field Equation.

Doesn't electromagnetic shielding prove gravity is not an electromagnetic phenomenon.

Yes, but that's not the question the OP was asking.

- #12

stevebd1

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There isn't one; in fact there isn't even a black hole any more in the super-extremal case, because there isn't an event horizon.

I'm aware that there is no event horizon and technically no black hole with the extremal and over-extremal cases in Kerr metric but according to the link, it appears that the ergoregion still exists. It's not as recognised as the outer ergosurface but there is an ergosurface within the inner (Cauchy) horizon (sometimes denoted as [itex]r_{e+}[/itex] for the outer ergosurface and [itex]r_{e-}[/itex] for the inner ergosurface, see equations here). Below are links to the images '..Solid lines denote ergosurfaces, the dashed ones are horizons. The thin circle shows the value of a, equal to its radius'.

a=0.97 (generic astrophysical)

http://universeinproblems.com/index.php/File:BHfig-Kerr2-09generic.png

a=1.0 (extremal)

http://universeinproblems.com/index.php/File:BHfig-Kerr3-10extremal.png

a=1.1 (naked singularity)

http://universeinproblems.com/index.php/File:BHfig-Kerr4-11naked.png

You can see in the case of a=1.0, the inner and outer ergosurfaces make contact at the poles and in the case of a=1.1, the ergoregion disappears at the poles but still exists off pole.

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