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My friend and I were wondering how much energy is given off by an average black hole relative to its mass. (like X black hole gives off Y% of its mass as energy per second)

Thanks in advance for your help.

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My friend and I were wondering how much energy is given off by an average black hole relative to its mass. (like X black hole gives off Y% of its mass as energy per second)

Thanks in advance for your help.

- #2

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Wikipedia said:When particles escape, the black hole loses a small amount of its energy and therefore of its mass (mass and energy are related by Einstein's equation E = mc²).

The power emitted by a black hole in the form of Hawking radiation can easily be estimated for the simplest case of a nonrotating, non-charged Schwarzschild black hole of mass M. Combining the formulae for the Schwarzschild radius of the black hole, the Stefan-Boltzmann law of black-body radiation, the above formula for the temperature of the radiation, and the formula for the surface area of a sphere (the black hole's event horizon) we get:

It goes on to give the formula and more info.

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Thanks alot, this really helps.

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George Jones

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stevebd1

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It might be worth noting that black holes (particulaly rotating ones) can also loose energy via other processes-

According to one source, the total mass-energy of a black hole is-

[tex]\tag{1}M^2=\frac{J^2}{4M_{ir}^2}+\left(\frac{Q^2}{4M_{ir}}+M_{ir}\right)^2[/tex]

where

[tex]\tag{2}M_{ir}=\frac{1}{2}\sqrt{\left(M+\sqrt{M^2-Q^2-a^2}\right)^2+a^2}[/tex]

The first term (J) is rotational energy, the second term (Q) is coulomb energy and the third term (M_{ir}) is irreducible energy *(in all cases, M is the gravitational radius, a is the spin parameter in metres and J is angular momentum)*.

The first and second are extractable by physical means, such as the Penrose process, superradiance or electrodynamical process.

The irreducible part cannot be lowered by classical (e.g. non-quantum) processes and can only be lost through Hawking radiation. As high as 29% of a black holes total mass can be extracted by the first process and up to 50% for the second process (but realistically, charged black holes probably only exist in theory or are very short lived as they would probably neutralise quite quickly after forming).

Interestingly, the above makes it appear that kinetic rotational energy does contribute to the overall total mass of a black hole (or any rotating object for that matter) as the rest mass would be less than the rotating mass.

Another equation to calculate the rotational energy (which equals M-M_{ir}) is-

[tex]\tag{3}E_{rot}=M-\sqrt{\frac{1}{2}M\left(M+\sqrt{M^2-a^2\right)}}[/tex]

Also, the effects of rotation on Hawking radiation is equal to-

[tex]\tag{4}T_H=\frac{\hbar\kappa}{2\pi k_bc}=2\left(1+\frac{M}{\sqrt{M^2-a^2}\right)^{-1}} \frac{\hbar c^3}{8\pi Gk_bm}<\frac{\hbar c^3}{8\pi Gk_bm}[/tex]

where [itex]\kappa[/itex] is the Killing surface gravity of the black hole, [itex]\hbar[/tex] is the reduced Planck constant and [itex]k_b[/itex] is the Boltzmann constant.

The above effects of rotation on Hawking radiation imply that a maximal Kerr black hole (a/M=1) would give off no radiation.

(1),(2)

'Black Holes: A General Introduction' by Jeane-pierre Luminet

http://www.ece.uic.edu/~tsarkar/Goodies/Black Hole.pdf

pages 12 & 13

(3)

'Compact Objects in Astrophysics' by Max Camenzind

http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf [Broken]

page 271

(4)

'Black Hole Thermodynamics' by Narit Pidokrajt

http://www.physto.se/~narit/bh.pdf [Broken]

page 10

According to one source, the total mass-energy of a black hole is-

[tex]\tag{1}M^2=\frac{J^2}{4M_{ir}^2}+\left(\frac{Q^2}{4M_{ir}}+M_{ir}\right)^2[/tex]

where

[tex]\tag{2}M_{ir}=\frac{1}{2}\sqrt{\left(M+\sqrt{M^2-Q^2-a^2}\right)^2+a^2}[/tex]

The first term (J) is rotational energy, the second term (Q) is coulomb energy and the third term (M

The first and second are extractable by physical means, such as the Penrose process, superradiance or electrodynamical process.

The irreducible part cannot be lowered by classical (e.g. non-quantum) processes and can only be lost through Hawking radiation. As high as 29% of a black holes total mass can be extracted by the first process and up to 50% for the second process (but realistically, charged black holes probably only exist in theory or are very short lived as they would probably neutralise quite quickly after forming).

Interestingly, the above makes it appear that kinetic rotational energy does contribute to the overall total mass of a black hole (or any rotating object for that matter) as the rest mass would be less than the rotating mass.

Another equation to calculate the rotational energy (which equals M-M

[tex]\tag{3}E_{rot}=M-\sqrt{\frac{1}{2}M\left(M+\sqrt{M^2-a^2\right)}}[/tex]

Also, the effects of rotation on Hawking radiation is equal to-

[tex]\tag{4}T_H=\frac{\hbar\kappa}{2\pi k_bc}=2\left(1+\frac{M}{\sqrt{M^2-a^2}\right)^{-1}} \frac{\hbar c^3}{8\pi Gk_bm}<\frac{\hbar c^3}{8\pi Gk_bm}[/tex]

where [itex]\kappa[/itex] is the Killing surface gravity of the black hole, [itex]\hbar[/tex] is the reduced Planck constant and [itex]k_b[/itex] is the Boltzmann constant.

The above effects of rotation on Hawking radiation imply that a maximal Kerr black hole (a/M=1) would give off no radiation.

(1),(2)

'Black Holes: A General Introduction' by Jeane-pierre Luminet

http://www.ece.uic.edu/~tsarkar/Goodies/Black Hole.pdf

pages 12 & 13

(3)

'Compact Objects in Astrophysics' by Max Camenzind

http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf [Broken]

page 271

(4)

'Black Hole Thermodynamics' by Narit Pidokrajt

http://www.physto.se/~narit/bh.pdf [Broken]

page 10

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stevebd1 said:The above effects of rotation on Hawking radiation imply that a maximal Kerr black hole (a/M=1) would give off no radiation.

[tex]T_H = 2 \left(1 + \frac{M}{\sqrt{M^2 - a^2} \right)^{-1}} \frac{\hbar c^3}{8 \pi G k_b m} \leq \frac{\hbar c^3}{8 \pi G k_b m}[/tex]

If that equation is correct under zeroth law, what would prevent a maximal Kerr quantum black hole from becoming completely stable?

Reference:

'Black Hole Thermodynamics' by Narit Pidokrajt - pg. 10

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- #8

stevebd1

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Initially I would say cosmic censorship but I found this in the black hole electron section of wiki-

The equation for the photon sphere that applies to both static and rotating black holes is-

[tex]\tag{1}R_{ph}=2M\left[1+cos\left(\frac{2}{3}cos^{-1}\frac{-a}{M}\right)\right][/tex]

which reduces to (what was) the event horizon radius for a maximal Kerr black hole. If the above statement is right, then the reduction of the photon sphere radius would appear to have some impact on the stability of a maximal Kerr quantum black hole (or maybe the photon sphere works differently at quantum level).

(1)

'Compact Objects in Astrophysics' by Max Camenzind

http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf [Broken]

page 259

Some theorists, including Stephen Hawking and Kip Thorne, have recently concluded that the laws of nature do permit the formation of a naked singularity during gravitational collapse. In a later paper (2007) titled, Kerr Geometry as Space-Time Structure of the Dirac Electron, Burinskii writes:"In this work we obtain an exact correspondence between the wave function of the Dirac equation and the spinor (twistorial) structure of the Kerr geometry. It allows us to assume that the Kerr-Newman geometry reflects the specific space-time structure of electron, and electron contains really the Kerr-Newman circular string of Compton size". The Burinskii papers describe an electron as a gravitationally confined ring singularity without an event horizon. It has some, but not all of the predicted properties of a black hole. A new name other than "Black hole electron" is needed for this model.

At the radius [itex]3Gm/c^2[/itex], a special space curvature condition is found. An electromagnetic wave has a 50 percent probability of either orbiting and spiraling inward or spiraling away to infinity due to the gravitational space curvature at [itex]3Gm/c^2[/itex]. The radius [itex]3Gm/c^2[/itex] is the gravitational photon orbit radius or photon sphere radius. This radius is critical if self-gravitational attraction is required to produce a stable state.

The equation for the photon sphere that applies to both static and rotating black holes is-

[tex]\tag{1}R_{ph}=2M\left[1+cos\left(\frac{2}{3}cos^{-1}\frac{-a}{M}\right)\right][/tex]

which reduces to (what was) the event horizon radius for a maximal Kerr black hole. If the above statement is right, then the reduction of the photon sphere radius would appear to have some impact on the stability of a maximal Kerr quantum black hole (or maybe the photon sphere works differently at quantum level).

(1)

'Compact Objects in Astrophysics' by Max Camenzind

http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf [Broken]

page 259

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In reference 1 - pg. 253, the equation stated in relativistic units for the dual outer and inner horizons:

[tex]r_{\pm} = M_H \pm \sqrt{M_H^2 - a_H^2}[/tex]

Where the key is:

[tex]M_H = r_g = \frac{Gm}{c^2}[/tex]

[tex]a_H = a \cdot r_g[/tex]

Therefore, the equation for the dual outer and inner horizons in physical natural units (S.I.) is:

[tex]\boxed{r_{\pm} = r_g \pm \sqrt{r_g^2(1 - a^2)}}[/tex]

Is this equation translation from relativistic to natural units correct?

Reference:

Compact Objects in Astrophysics - Max Camenzind

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- #10

stevebd1

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One other source I know of provides the following solution-

[tex]{r_{\pm} = r_g \left[1 \pm \sqrt{(1 - a^2)}\right][/tex]

Source-

'Inside Black Holes' by G. H. George

http://www.engr.mun.ca/~ggeorge/astron/blackholes.html

[tex]{r_{\pm} = r_g \left[1 \pm \sqrt{(1 - a^2)}\right][/tex]

Source-

'Inside Black Holes' by G. H. George

http://www.engr.mun.ca/~ggeorge/astron/blackholes.html

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The effects of angular momentum on Hawking radiation in physical units (S.I.):

[tex]\boxed{T_H = 2 \left(1 + \frac{1}{\sqrt{(1 - a^2)} \right)^{-1}} \frac{\hbar c^3}{8 \pi G k_b m} \leq \frac{\hbar c^3}{8 \pi G k_b m}}[/tex]

Kerr metric angular momentum:

[tex]\alpha = \frac{J}{mc}[/tex]

Radii for the dual outer and inner horizons in physical units (S.I.):

[tex]{r_{\pm} = r_g \left[1 \pm \sqrt{(1 - a^2)}\right][/tex]

Ergosphere radius:

[tex]r_{e} = r_g \left[1 + \sqrt{(1 - a^2 \cos^{2} \theta)}\right][/tex]

Radii for the dual outer and inner horizons in physical units (S.I.):

[tex]r_{\pm} = \frac{r_{s} \pm \sqrt{r_{s}^{2} - 4\alpha^{2}}}{2}[/tex]

Ergosphere radius:

[tex]r_{e} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4\alpha^{2} \cos^{2}\theta}}{2}[/tex]

[tex]r_s = 2 \cdot r_g[/tex]

Establishing the equations between the Kerr metric angular momentum terms [tex]\alpha[/tex] and [tex]a[/tex] for the inner horizon radius:

[tex]r_{-} = \frac{1}{2} \left(2 r_g - \sqrt{(2 r_g)^{2} - 4 \alpha^{2}} \right) = r_g \left[1 - \sqrt{(1 - a^2)}\right][/tex]

Solving for [tex]a[/tex] for the inner horizon radius:

[tex]\boxed{a = \frac{\alpha}{r_g}}[/tex]

Reference:

Kerr Metric - Important surfaces - Wikipedia

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- #12

stevebd1

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For some reason, wikipedia do not recognise the inner Cauchy horizon of Kerr black holes, so when they talk about the outer horizon, they refer to the static limit or the ergosphere and when they talk about the inner horizon, they refer to the outer event horizon (I don't know if it's still recorded under discussion but there was a request to remove any mention of the inner Cauchy horizon which went ahead). This would mean the following would apply-

ergosphere-

[tex]r_{e} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4\alpha^{2} cos^{2}}}{2} = r_g \left[1 + \sqrt{(1 - a^2 cos^{2} \theta)}\right][/tex]

(wiki has this down as the outer event horizon)

outer event horizon-

[tex]r_{+} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4\alpha^{2}}}{2} = r_g \left[1 + \sqrt{(1 - a^2)}\right][/tex]

(wiki has this down as the inner event horizon)

And to extrapolate-

inner event horizon-

[tex]r_{-} = \frac{r_{s} - \sqrt{r_{s}^{2} - 4\alpha^{2}}}{2} = r_g \left[1 - \sqrt{(1 - a^2)}\right][/tex]

(which wiki has ignored)

In regard of your solution expressing [itex]a[/itex] as a unitless spin parameter between 0 and 1, the following would apply-

[tex]a=\frac{\alpha}{r_g}[/tex]

It's worth noting that while wikipedia use [itex]\alpha[/itex] (alpha) to represent the spin parameter in metres, I have seen [itex]\alpha[/itex] represent redshift in frame-dragging equations (which derive from Kerr metric) and [itex]a[/itex] is used to represent the spin parameter in metres. To represent the spin parameter as a unitless number between 0 and 1 (or in natural units), I've seen the following terms used- [itex]|a|[/itex], [itex]a_\ast[/itex] or even simply [itex]a/M[/itex] as in the 'black hole parameters' section on scholarpedia.

ergosphere-

[tex]r_{e} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4\alpha^{2} cos^{2}}}{2} = r_g \left[1 + \sqrt{(1 - a^2 cos^{2} \theta)}\right][/tex]

(wiki has this down as the outer event horizon)

outer event horizon-

[tex]r_{+} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4\alpha^{2}}}{2} = r_g \left[1 + \sqrt{(1 - a^2)}\right][/tex]

(wiki has this down as the inner event horizon)

And to extrapolate-

inner event horizon-

[tex]r_{-} = \frac{r_{s} - \sqrt{r_{s}^{2} - 4\alpha^{2}}}{2} = r_g \left[1 - \sqrt{(1 - a^2)}\right][/tex]

(which wiki has ignored)

In regard of your solution expressing [itex]a[/itex] as a unitless spin parameter between 0 and 1, the following would apply-

[tex]a=\frac{\alpha}{r_g}[/tex]

It's worth noting that while wikipedia use [itex]\alpha[/itex] (alpha) to represent the spin parameter in metres, I have seen [itex]\alpha[/itex] represent redshift in frame-dragging equations (which derive from Kerr metric) and [itex]a[/itex] is used to represent the spin parameter in metres. To represent the spin parameter as a unitless number between 0 and 1 (or in natural units), I've seen the following terms used- [itex]|a|[/itex], [itex]a_\ast[/itex] or even simply [itex]a/M[/itex] as in the 'black hole parameters' section on scholarpedia.

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Wikipedia said:A rotating black hole has two photon spheres. As a black hole rotates it drags space with it. The photon sphere that is closer to the black hole is moving in the same direction as the rotation, whereas the photon sphere further away is moving against it. The greater the angular velocity of the rotation of a black hole the greater distance between the two photon spheres. Because the black hole has an axis of rotation this only holds true if approaching the black hole in the direction of the equator. If approaching at a different angle, such as one from the poles of the black hole, to the equator there is only one photon sphere. This is because approaching at this angle the possibility of traveling with or against the rotation does not exist.

The effects of angular momentum on photon sphere radius in physical S.I. units:

[tex]r_{ps} = 2 r_g \left[1 + \cos \left(\frac{2}{3} \cos^{-1} -a \right) \right][/tex]

[tex]\boxed{r_{ps} = 3 r_g \; \; \; a = 0}[/tex]

[tex]\boxed{r_{ps} = r_g \; \; \; \; a = 1}[/tex]

Wikipedia said:The greater the angular velocity of the rotation of a black hole the greater distance between the two photon spheres.

Does anyone have the radius equations for the two photon spheres?

Reference:

Photon sphere - Wikipedia

Inside a black hole - the Kerr black hole

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- #14

stevebd1

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[tex]R_{ph2}=2M\left[1+cos\left(\frac{2}{3}cos^{-1}\frac{a}{M}\right)\right][/tex]

which puts it at 3M at a/M=0 and 4M at a/M=1

'Spherical Photon Orbits Around a Kerr Black Hole'

http://www.physics.nus.edu.sg/~phyteoe/kerr/

- #15

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The effects of angular momentum on photon sphere radius in physical S.I. units:

[tex]r_{ps \mp} = 2 r_g \left[1 + \cos \left(\frac{2}{3} \cos^{-1} \mp |a| \right) \right][/tex]

[tex]-|a|[/tex] - prograde photon orbit

[tex]+|a|[/tex] - retrograde photon orbit

[tex]\boxed{r_{ps-} = 3 r_g \; \; \; a = 0}[/tex]

[tex]\boxed{r_{ps+} = 3 r_g \; \; \; \; a = 0}[/tex]

[tex]\boxed{r_{ps-} = r_g \; \; \; \; \; a = 1}[/tex]

[tex]\boxed{r_{ps+} = 4 r_g \; \; \; \; a = 1}[/tex]

Reference:

http://www.physics.nus.edu.sg/~phyteoe/kerr/" [Broken]

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- #16

stevebd1

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There seems to be a problem with the equation at the poles. As *a* increases, the prograde orbit passes through the event horizon. (EDIT: Equation in post #15 removed)

Below is a link to images from the 'Spherical Photon Orbits' website of example orbits (which look fairly spherical)-

http://www.physics.nus.edu.sg/~phyteoe/kerr/table.html

Below is a link to images from the 'Spherical Photon Orbits' website of example orbits (which look fairly spherical)-

http://www.physics.nus.edu.sg/~phyteoe/kerr/table.html

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- #17

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The graphic model shown in attachment #1 of post #13, displays the dual photon spheres as oblate spheroids, this graphic model appears to be incorrect.

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- #18

stevebd1

Gold Member

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That seems to be the case.

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- #19

stevebd1

Gold Member

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Regarding the OP's question of how much energy do black holes give off, there is also a process called Penrose Pair Production (PPP)-

key to Fig. 3.14-

'Magnetohydrodynamics on the Kerr Geometry' by A Mueller

http://www.wissenschaft-online.de/astrowissen/downloads/PhD/PhD_AMueller.pdf

pages 47 & 48

A variant of the classical http://en.wikipedia.org/wiki/Penrose_process" [Broken] is the Penrose Pair Production. The mechanism is based on another ingredient, the photonsphere. Photons are instably trapped in the photon sphere. Other photons may now infall on radial null geodesics and hit these trapped photons. If the energy of the quanta exceeds the rest frame energy of about one MeV, gamma photons produce pairs of leptons. For rather large values of the black hole spin, a > 0.7, the spherically symmetric photon sphere plunges into the oblate ergosphere. Then, the PPP is expected to occur dominantly. PPP is sketched in Fig. 3.14. (R.K.) Williams applied this model to explain the populations of ultrarelativistic electrons in the quasars 3C 279 and 3C 273

key to Fig. 3.14-

Illustration of the production of a pair plasma via Penrose processes in

the ergosphere (blue) of a Kerr black hole (violet). The gamma photons

annihilate in the photon sphere (yellow) to produce electrons (cyan) and

positrons (green).

'Magnetohydrodynamics on the Kerr Geometry' by A Mueller

http://www.wissenschaft-online.de/astrowissen/downloads/PhD/PhD_AMueller.pdf

pages 47 & 48

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- #20

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Effects of angular momentum on a (4+n)-dimensional Kerr metric quantum black hole Hawking radiation evaporation time:

[tex]\boxed{t(n, a)_{ev} = \frac{320 G^2 m_p^3}{\hbar c^4 \sqrt{\pi}} \left[ \frac{E_{BH}}{E_p} \left( \frac{8 \Gamma\left(\frac{n+3}{2} \right)}{n+2} \right) \right]^{\frac{3}{n+1}} \left(1 + \frac{1}{\sqrt{(1 - a^2)} \right)^{4}}} \; \; \; a < 1[/tex]

[tex]\boxed{t_0 = \frac{320 G^2 m_p^3}{\hbar c^4 \sqrt{\pi}}}[/tex]

[tex]\boxed{t_0 = 9.733 \cdot 10^{-42} \; \text{s}}[/tex]

[tex]E_{BH} = 14 \; \text{Tev}[/tex]

[tex]\boxed{t(10, 0.999)_{ev} = 3.275 \cdot 10^{-40} \; \text{s}}[/tex]

Note:

[tex]\boxed{\lim_{a \to 1} t(n, a)_{ev} = \infty}[/tex]

Wikipedia said:Computer simulations of the collapse of a disk of dust have indicated that these objects can exist, and thus the cosmic censorship hypothesis (stating that singularities are always hidden) does not hold.

Computer models of gravitational collapse have shown that naked singularities can arise, but these models rely on very special circumstances (such as spherical symmetry).

Some theorists, including Stephen Hawking and Kip Thorne, have recently concluded that the laws of nature do permit the formation of a naked singularity during gravitational collapse.

Reference:

http://www.youtube.com/watch?v=kVsZdgz5oFM"

http://en.wikipedia.org/wiki/Hawking_radiation#Black_hole_evaporation"

http://www.physto.se/~narit/bh.pdf" [Broken]

https://www.physicsforums.com/showpost.php?p=1946387&postcount=11"

http://en.wikipedia.org/wiki/Micro_black_hole#Stable_micro_black_holes"

http://eprints.may.ie/archive/00000240/01/fuzzyBH-4.pdf" [Broken]

https://www.physicsforums.com/showpost.php?p=1891354&postcount=454"

http://en.wikipedia.org/wiki/Naked_singularity" [Broken]

http://en.wikipedia.org/wiki/Cosmic_censorship_hypothesis" [Broken]

https://www.physicsforums.com/showpost.php?p=1877224&postcount=428"

http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf" [Broken]

http://www.wissensnavigator.ch/documents/OTTOROESSLERMINIBLACKHOLE.pdf" [Broken]

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- #21

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http://www.absoluteastronomy.com/discussionpost/Electron_as_a_ring_singularity_56595

In the Burinskii electron model rotational energy accounts for all of the electron mass energy. The ingrediants of the electron are an electromagnetic field and rotating space-time.

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mcr = h/4pi

mc (2Gm/c^2) = h/4pi = angular momentum

(m)^2 = (h/4pi) (c^2/2G) (1/c) = (1/4) (hc/2pi G)

m = (1/2) (hc/2pi G)^1/2 = (1/2) (Planck mass)

The mass value that relates to electron mass, is collapsed to its photon orbit radius where forces are balanced. When space-time spin is canceled (by merging with a positron) all of its mass energy is radiated away.

mc (3Gm/c^2) = h/4pi = angular momentum

(m)^2 = (h/4pi) (c^2/3G) (1/c) = (1/4) (hc/3pi G) = (hc/12pi G)

m = (hc/12pi G)^1/2 = (1/2) (2/3)^1/2 (hc/2pi G)^1/2 = (1/2) (2/3)^1/2 (planck mass)

In this model, the electron is not fully collapsed to produce an event horizon. Without an event horizon, it has some but not all of the properties predicted for a black hole.

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Limit time factor = (3/2)^1/2 (Planck time) divided by (2pi seconds)

Limit time factor = (3hG /4pi c^5)^1/2 (1/2pi) seconds per second

(Limit time factor)^1/2 = (3hG /4pi c^5)^1/4 (1/2pi)^1/2 seconds per second

In this ratio equation, one second is the time required for light to travel 299792458 meters. The quantized mass value noted in post #22 was (hc/ 12 pi G)^1/2. We find that this mass value, when degraded (reduced) by the (limit time factor) exponent 1/2 is equal the electron mass.

electron mass = (hc/ 12pi G)^1/2 (3hG /4pi c^5)^1/4 (1/2pi)^1/2 kg

This equation implies that the true gravitational constant value G, must be very close to 6.67174557x10 exponent -11.

electron mass = 9.10938215x10 exponent -31 kg

Ths implies that the time (3/2) exponent 1/2 times (Planck time) is the smallest amount of time that has meaning.

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- #25

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Energy = hc/ wavelength = (wavelength) (c^4/3pi G)

We find, (wavelength)^2 = (hc) (3pi G/c^4). Then the wavelength is:

wavelength = (3pi hG/c^3)^1/2

This wavelength is (2pi) (3/2)^1/2 (Planck length). This photon wavelength has energy that is specified by light velocity and the gravitational constant. Photon energy is inversely proportional to wavelength and so the energy of any longer wavelength photon can be specified by light velocity and the gravitational constant. The photon wavelength with energy equal to two electron mass particles or (h/2mc) meters can also be specified by light velocity and the gravitational constant. The rato (2pi) (3/2)^1/2(Planck length) divided by (h/2mc) meters becomes very significant when we try to understand how the electron mass is quantized. This ratio relates to gravitational blueshift and gravitational time dilation.

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