Is a black hole a perfect sphere?

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Is a black hole a perfect sphere?
No matter how perfect we make a ball of iron it'll never be a mathematically perfect sphere because it's basically made out of atoms. But how about black holes?
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Re: Is a black hole a perfect sphere?
My 2 cents. A black hole's jet makes at least one nonmaterial hole in the mathematical sphere, thus if radiation is mathematically described, the sphere it's not perfect...
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Re: Is a black hole a perfect sphere?
Just my curiosity here...
Aside from the jets expelling radiation, how could a black hole NOT
be a sphere?
This extreme gravity pulls in all directions at once, no? What could
possibly cause the shape to be other than spherical?
Again, just curiosity from the BrainDead.
Aside from the jets expelling radiation, how could a black hole NOT
be a sphere?
This extreme gravity pulls in all directions at once, no? What could
possibly cause the shape to be other than spherical?
Again, just curiosity from the BrainDead.
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 Hungry4info
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Re: Is a black hole a perfect sphere?
I think he's talking about the singularity itself. Neutron stars have small mmscale topographic variations preventing them from being perfect spheres. Do black hole singularities have perfectly spherical shapes?BobHegwood wrote:Aside from the jets expelling radiation, how could a black hole NOT be a sphere?
This, I think, is what the OP is asking.
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 t00fri
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Re: Is a black hole a perfect sphere?
BobHegwood wrote:Just my curiosity here...
Aside from the jets expelling radiation, how could a black hole NOT
be a sphere?
Angular momentum!
Rotating black holes (socalled Kerr black holes) do have nonvanishing angular momentum and the gravitational singularity correspondingly becomes a ring (rather than a point!).
NB: There exist four types of BH solutions of Einstein's equations of GR: two of these are rotating (Kerr (uncharged) and KerrNewman (charged) BHs), respectively. The best known (uncharged) Schwarzschild BH has vanishing angular momentum and hence doesn't rotate.
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Re: Is a black hole a perfect sphere?
Perhaps the OP is thinking about the event horizon of the black hole? The event horizon is a mathematical surface, and you wouldn't necessarily notice very much as you fell through it, but it's certainly a significant surface, especially if you hoped to visit anywhere but the singularity in future.
The event horizon of an isolated nonrotating black hole is perfectly spherical, but such Schwarzschild black holes are vanishingly unlikely to occur in nature. Realworld black holes will have angular momentum, and therefore nonspherical event horizons to go with the ring singularity Fridger mentioned.
Grant
The event horizon of an isolated nonrotating black hole is perfectly spherical, but such Schwarzschild black holes are vanishingly unlikely to occur in nature. Realworld black holes will have angular momentum, and therefore nonspherical event horizons to go with the ring singularity Fridger mentioned.
Grant
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Re: Is a black hole a perfect sphere?
 edit 
Last edited by John Van Vliet on 19.10.2013, 04:03, edited 1 time in total.
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Re: Is a black hole a perfect sphere?
A long time ago, when my wife and I were young PostDocs in theoretical physics, we investigated the effects of angular momentum in case of a (Kerr) black hole as to modifying the BH's familiar gravitational attraction.
Those of you who ever attended a lecture about Quantum Mechanics will remember the important effect of angular momentum (J) in the Schroedinger equation! It is well known to generate an additive repulsive term of the form J(J+1)/(2*mass * r^2) in the potential: this is the socalled angular momentum barrier. There are many important physical implications of this term that I can hardly review here...
We found at the time that this is quite similar with a (Kerr) black hole carrying nonvanishing angular momentum in the Einstein equations. Again one finds a repulsive angular momentum barrier that tends to compensate the gravitational attraction of the BH! The surprising result are certain stable orbits around such a rotating BH, such that the object orbiting the BH is NOT falling into the BH as happens invariably for a Schwarzschild BH with J=0.
Fridger
Those of you who ever attended a lecture about Quantum Mechanics will remember the important effect of angular momentum (J) in the Schroedinger equation! It is well known to generate an additive repulsive term of the form J(J+1)/(2*mass * r^2) in the potential: this is the socalled angular momentum barrier. There are many important physical implications of this term that I can hardly review here...
We found at the time that this is quite similar with a (Kerr) black hole carrying nonvanishing angular momentum in the Einstein equations. Again one finds a repulsive angular momentum barrier that tends to compensate the gravitational attraction of the BH! The surprising result are certain stable orbits around such a rotating BH, such that the object orbiting the BH is NOT falling into the BH as happens invariably for a Schwarzschild BH with J=0.
Fridger
 Hungry4info
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Re: Is a black hole a perfect sphere?
Very interesting! Can you give us an idea of what the orbital radii of these orbits would be? (like in units of the event horizon radius for example).
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Re: Is a black hole a perfect sphere?
Hungry4info wrote:I think he's talking about the singularity itself. Neutron stars have small mmscale topographic variations preventing them from being perfect spheres. Do black hole singularities have perfectly spherical shapes?BobHegwood wrote:Aside from the jets expelling radiation, how could a black hole NOT be a sphere?
This, I think, is what the OP is asking.
This is should be more interesting, because the "perfection" of the sphere was assumed mathematically whereas nothing was pointed out about its geometric perfection. If a singularity is often referred as "point", it is not extended (points have no parts); there is no geometrical place of the points equidistant from another central point to get a circle and in which its diameter can rotate about itself to produce a (perfect) sphere. Moreover, like within a singularity the common physics is said to cease of exist, even the math describing that physics would cease of exist.
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Re: Is a black hole a perfect sphere?
Well, thank you ALL for once again educating the BrainDead.
Much appreciated.
Much appreciated.
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 Hungry4info
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Re: Is a black hole a perfect sphere?
Sure that makes sense. However I think there should be a time between being a point, and being an extended object, where the star core is shrinking and collapses to within its schwarzchild radius. So I'd expect that at least at some brief time, there's a spherical extended singularity within the event horizon.Fenerit wrote:If a singularity is often referred as "point", it is not extended (points have no parts); there is no geometrical place of the points equidistant from another central point to get a circle and in which its diameter can rotate about itself to produce a (perfect) sphere.
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Re: Is a black hole a perfect sphere?
IMHO, the horizon sphere is a perfect sphere, because it's not really physical, it's a geometric concept.
As far as the black hole itself (the matter it is made of, whatever form it has), we don't know and we can not know since it is inside the "noreturn of information" zone. We might as well consider that it does not exist, as if the universe had holes whereever there is a black hole. Like a swiss cheese.
As far as the black hole itself (the matter it is made of, whatever form it has), we don't know and we can not know since it is inside the "noreturn of information" zone. We might as well consider that it does not exist, as if the universe had holes whereever there is a black hole. Like a swiss cheese.
Re: Is a black hole a perfect sphere?
Hungry4info wrote:Sure that makes sense. However I think there should be a time between being a point, and being an extended object, where the star core is shrinking and collapses to within its schwarzchild radius. So I'd expect that at least at some brief time, there's a spherical extended singularity within the event horizon.Fenerit wrote:If a singularity is often referred as "point", it is not extended (points have no parts); there is no geometrical place of the points equidistant from another central point to get a circle and in which its diameter can rotate about itself to produce a (perfect) sphere.
Seem that the angular momentum makes a string, like that of the relevant theories, of which a component's direction is emitted as radiation, escaping from the black hole because the balanced gravitational pull of this latter with the string's curvature strenght at the circumference of the circle. Inside the black hole, is the lenght of the "time" through the diameter the same of the "time" through the long run string? How much space (lenght) there is between the singularity and the event horizon?
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Re: Is a black hole a perfect sphere?
I may be misunderstanding what Fridger intended, but I think we're talking about ISCO  the innermost stable circular orbit. This lies at three Schwarzschild radii from a nonrotating black hole. For rotating black holes, ISCO splits into two  prograde orbits are stable nearer to the event horizon, while retrograde orbits have to be farther out to be stable. For nearextremal Kerr black holes, the prograde ISCO is pulled right down into the ergosphere; for the extremal case, ISCO is at the event horizon (at the equator).Hungry4info wrote:Very interesting! Can you give us an idea of what the orbital radii of these orbits would be? (like in units of the event horizon radius for example).
This has potentially observable implications for the extent of the accretion disc of real black holes, IIRC.
The Schwarzschild radius quoted for nonrotating black holes is actually a bit of a fudge: it's the "reduced circumference" of the event horizon. You measure the circumference "at" the event horizon, divide by two pi, and call that the radius of the black hole. This gives you a consistent set of coordinates you can work in (GR is all about choosing useful coordinates).Fenerit wrote:How much space (lenght) there is between the singularity and the event horizon?
But spacetime coordinates get switched around below the event horizon. In a Schwarzschild black hole, that means the radial coordinate is "timelike" everywhere inside the event horizon. So once below the horizon, you can't help but travel radially until you reach the singularity  the singularity is in your future, rather than in a place you can point to. So maybe we should measure the radius of a black hole in seconds rather than metres!
There's a nice paper about survival time below the event horizon at http://arxiv.org/abs/0705.1029, and the sums are easy to do if anyone is interested.
(Rotating black holes have very complicated insides, with an extra horizon and an inner region which has a spacelike radial coordinate.)
Grant
Re: Is a black hole a perfect sphere?
granthutchison wrote:So maybe we should measure the radius of a black hole in seconds rather than metres!
Moreover, is the use of the mass to measure the time, the fact more "funny" of the general relativity. What time is it? It's two kg o' clock. Thank you for pointing me to that paper.
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Re: Is a black hole a perfect sphere?
The paper uses normalized units in which G = c = 1. If you want to calculate times with the formulae provided, wherever m (the mass of the black hole) appears, replace it with Gm/c?. That will give you time in "lightmetres"; divide by c to convert to seconds.Fenerit wrote:granthutchison wrote:So maybe we should measure the radius of a black hole in seconds rather than metres!
Moreover, is the use of the mass to measure the time, the fact more "funny" of the general relativity. What time is it? It's two kg o' clock. Thank you for pointing me to that paper.
Grant
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Re: Is a black hole a perfect sphere?
granthutchison wrote:The paper uses normalized units in which G = c = 1. If you want to calculate times with the formulae provided, wherever m (the mass of the black hole) appears, replace it with Gm/c?. That will give you time in "lightmetres"; divide by c to convert to seconds.Fenerit wrote:granthutchison wrote:So maybe we should measure the radius of a black hole in seconds rather than metres!
Moreover, is the use of the mass to measure the time, the fact more "funny" of the general relativity. What time is it? It's two kg o' clock. Thank you for pointing me to that paper.
Grant
In my field of research (theoretical (Astro)Particle Physics) the following units are ALWAYS used:
? = c =1
where ? aka hbar is the socalled reduced Planck constant with ? = h/2?
In this case, time and space coordinates have the same dimension, since generally a 4D spacetime vector in Minkowski space reads (x,y,z,ict), as required by the validity of special Relativity. For c=1: [x] = [y] = [z] = [t], since the components of vectors must have all the same dimensions.
There are many further unfamiliar dimensions ...Like momenta and energies get the same dimension.
Due to the famous Heisenberg Uncertainty Relation
?x ?p > ?/2 =1/2;
?E ?t > ?/2 =1/2;
configuration space coordinates carry the inverse dimensions as momenta and energies do and time carries the dimension of an inverse energy!
We measure energies in eV (electron Volt), hence times are measured in [1/eV].
In practice, 1 eV is a tiny scale, hence we usually employ GeV = Giga electron Volt or more recently TeV = Tera electron Volt (think of the operation energy of the LHC!)
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Re: Is a black hole a perfect sphere?
For some odd reason I found "momenergy" (as Wheeler liked to call it) harder to get my head around than spacetime. Why I should resist merging two "bookkeeping" quantities like momentum and energy, while breezily accepting the merger of such intuitively different concepts as time and distance, I don't know.t00fri wrote:In this case, time and space coordinates have the same dimension, since generally a 4D spacetime vector in Minkowski space reads (x,y,z,ict), as required by the validity of special Relativity. For c=1: [x] = [y] = [z] = [t], since the components of vectors must have all the same dimensions.
There are many further unfamiliar dimensions ...Like momenta and energies get the same dimension.
Grant
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Re: Is a black hole a perfect sphere?
granthutchison wrote:For some odd reason I found "momenergy" (as Wheeler liked to call it) harder to get my head around than spacetime. Why I should resist merging two "bookkeeping" quantities like momentum and energy, while breezily accepting the merger of such intuitively different concepts as time and distance, I don't know.t00fri wrote:In this case, time and space coordinates have the same dimension, since generally a 4D spacetime vector in Minkowski space reads (x,y,z,ict), as required by the validity of special Relativity. For c=1: [x] = [y] = [z] = [t], since the components of vectors must have all the same dimensions.
There are many further unfamiliar dimensions ...Like momenta and energies get the same dimension.
Grant
Grant,
the reason for merging space and time and correspondingly in Fourier space energy and momentum into 4vectors is simply relativistic covariance in 4d Minkowski space (Unlike 4d Euclidean space the latter is characterized by a mixed metric (+) or (+++)).
[@relativistic covariance:
The Minkowski spacetime of special relativity is a socalled 4d real vector space, meaning that all elements of that space are (built of) 4vectors, such as to transform the same way as the position 4vector does under any Poincar? transformation. Correspondingly, all relativistically invariant quantities are found as the scalar products ("dot" products) of these various 4vectors in Minkowski space.]
I am sure you have already seen somewhere, how appealingly elegant the Maxwell equations of classical Electrodynamics look in relativistically covariant notation. They furnish the r?le model for the relativistic gauge field theories that we rely on in everyday's research work...
Fridger
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