I can't imagine the forces involved in black holes' lives. So please, help me to find out, if it is possible or not to destroy black hole in this specific way.
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$\begingroup$ I am not sure, but I think that in order to increase its spin efficiently, you need to throw stuff at it, and if you do you will also increase its mass; in the strong field limit of GR, the corresponding mass increase prevents centrifugal force from breaking it apart. At best it becomes a Kerr black hole with 'a' asymptotically close to 1. For black holes, gravity always win! (if you want to torque it up without throwing things, you ll have to be fairly far away and it won't be very efficient) $\endgroup$– chrisMay 25, 2014 at 0:50
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6 Answers
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Can we (theoretically) spin the black hole so strong so it will be broken apart by centrifugal force?
For a Kerr-Newman (rotating, charged, isolated) black hole of mass $M$, angular momentum $J$, and charge $Q$, the surface area of the event horizon is given by $$A = 8M\left[M^2 + (M^2-a^2-Q^2)^{1/2} - Q^2/2\right]\text{,}$$ where $a = J/M$. An extremal black hole occurs when $M^2 = a^2 + Q^2$. Beyond that, if the black hole is even more overspun or overcharged, is an "overextremal" Kerr-Newman spacetime, which wouldn't really be a black hole at all, but rather a naked singularity.
Thus, I interpret your question as asking whether or not a black hole can be be spun up to the extremal limit and beyond, so as to destroy the event horizon. It's very probable that it can't be done.
Wald proved in 1974 that as one flings matter into a black hole to try to increase its angular momentum, the nearer to an extremal black hole it is, the harder it is to continue this process: a fast-spinning black hole will repel matter that would take it beyond the extremal limit. There are other schemes, and though I'm not aware of any completely general proof within classical general relativity, the continual failure of schemes like this is well-motivated by the connection between black hole dynamics and thermodynamics.
For example, the Hawking temperature of the black hole is $T_\text{H} = \kappa/2\pi$, where $$\kappa = \frac{\sqrt{M^2-a^2-Q^2}}{2M\left(M+\sqrt{M^2-a^2-Q^2}\right)-Q^2}$$ is the black hole's surface gravity. Thus, even reaching the extremal limit is thermodynamically equivalent to cooling a system to absolute zero.
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I don't have all of the math off the top of my head, but from my conceptual understanding, it is not possible.
Black holes have a large enough gravitational attraction that even light cannot escape even from well beyond the "surface" (that is if the black hole has a low enough mass that it still has a surface and has not collapsed into a singularity). That would mean that it would have to spin fast enough that the surface is moving significantly faster than the speed of light in order to have enough linear momentum (often colloquially called "centrifugal force" in a circular reference frame) to escape, which according to the theory of relativity is not possible.
Hawking radiation is only possible because the electromagnetic radiation is moving very nearly to orthogonally to the "surface" of the black hole and light can only be "bent" by gravity, it can't be pulled to a stop.
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2$\begingroup$ There are several conceptual problems with this answer. It suggests that the "surface" is something other than the event horizon--then what is it? It suggests that Hawking radiation is limited to massless radiation like electromagnetic--not true. It suggests that black holes can't stop light--the horizon is a lightlike surface that does that. $\endgroup$ May 25, 2014 at 8:44
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Black holes can evaporate through a quantum process known as Hawking Radiation and that's it.
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As far as we know there is nothing that can stop a black hole. For this notion to make sense you must first look at what is currently known about black holes. Once you get a grasp of it, then you will see that due to our current understanding of the Cosmos there is nothing we can do to black holes.
It is true that Hawking Radiation can affect a black hole, but that is only for very very small black holes.
By the way, in Physics there is no centrifugal force -- this is actually a misconception many people have. However, there is centripetal force.
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3$\begingroup$ @IlmariKaronen Thanks for the down-vote, however, you do realize that the cartoon doesn't prove anything right? There is a difference between Centripetal force and centrifugal. $\endgroup$– apxcodeMay 24, 2014 at 22:01
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2$\begingroup$ To be fair, the cartoon points out that centrifugal force is as inertial force, which is true. Arguably it's sensible to interpret inertial forces as "not real", but I'm not sure what bearing that has on this question, since gravitational force is an inertial force as well. $\endgroup$ May 26, 2014 at 3:54
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$\begingroup$ This "answer" is pretty vague, makes questionable statements, and has no real bearing on the original query. $\endgroup$ Jun 3, 2015 at 19:48
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Interesting. This process could impact the black hole formation in the first place. Consider a rotating star that dies and starts to shrink due to gravitational forces. As it shrinks, all its mass will be more and more compacted in a smaller radius. This will have two consequences: 1) the gravitational force atracting different parts of the body will will grow up with the inverse of squared radius and 2) its rotational speed will increase due to angular momentum conservation and the expanding force, due to the rotation, will grow up with the inverse of radius cubed. This means the expanding force will grow more quickly than the contracting one and, at least in Newtonian view, the expanding force will win. From this point of view, it looks like a rotating star would never form a black hole...
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$\begingroup$ You might like this article. slate.com/blogs/quora/2014/02/10/… $\endgroup$– userLTKOct 5, 2017 at 16:21
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Let's try the following:
Equate the forces:
$$\begin{eqnarray*} F_c &=& \frac{mv^2}R\\ F_g &=& \frac{GMm}{R^2}\\ \newline mv^2/R &=& \frac{GMm}{R^2}\\ v^2 &=& \frac{GM}R \end{eqnarray*}$$
Since the radius is $R_s = 2GM/c^2$
$$\begin{eqnarray*} v^2 &=& \frac{GM}{2GM/c^2}\\ v &=& \sqrt{c^2/2}\\ v &=& \frac{c}{\sqrt2}\\ v &=& 0,707c \end{eqnarray*}$$
I don't know if I made all the considerations, but as far as my knowledge goes, if the blackhole is spinning faster than $0,707c$, the event horizon won't be able to maintain itself on the current radius.
However, when the radius expands, the rotation will slow down by conservation of angular momentum... so I don't think it will rip apart... perhaps become a "grey hole"?
Please forgive me if any fundamental mistake was made, I'm new to all this... :P
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3$\begingroup$ This looks Newtonian. Did you consider the curvature of spacetime near a Black Hole? See also "Kerr solution" for Black Holes. $\endgroup$– GeraldOct 7, 2014 at 14:32
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$\begingroup$ This is an old answer, but I thought I'd drop this here because it addresses orbital velocity of black holes. you'd need to have the theoretical edge of the black hole at the event horizon spinning at C to escape, which is obviously impossible. physics.stackexchange.com/questions/207816/… $\endgroup$– userLTKOct 5, 2017 at 15:10