Talk:Black hole electron

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Help with this template Comments: edit – history – watch – refresh Contradicting evidence relating to the electron size presents a challenge to physicists. Arguments have been presented that favor a Compton-sized electron, while scattering properties indicate a much smaller radius. Under probing with electromagnetic fields, the electron appears to be a point without extension. The conception of the electron as a point particle has dominated particle physics. Other ideas have not been seriously considered. It is known that negative electric charge elements repel one another, so that a large force is required in order to hold the electron together. No combination of electric and magnetic forces can produce a stable state. Malcolm MacGregor and other writers have noted that gravitational forces could solve the stability problem if the electron is very small (approaching Schwarzschild radius size). Feynman wrote (1961) "--- We have been computing terms like a blind man exploring a new room, but soon we must develop some concept of this room as a whole, and to have some general idea of what is contained in it ---". The challenge by Feynman is to devise an electron model that accounts for electron properties and the diagrams used in QED. The "Black hole electron" article is significant because it provides information that relates to the electron model challenge. DonJStevens 20:57, 29 December 2006 (UTC)

Archives Archive 1

Contents 1 Planck length/ Compton wavelength 2 Superfluid 3He 3 "In need of an expert tag" 4 Matter-wave quantum 5 Stability / Infinite density 6 super-extremal? 7 Source; m squared requirement 8 Balanced pattern 9 Quantum - gravitational effect

[edit] Planck length/ Compton wavelength

A clear size relationship is specified by an equation that compares length values as shown below. The length labeled (L2) is defined as 1/2 of the electron Compton wavelength. The value used for the Planck length is 1.616x10 exp-35 meters.

(L2)squared=(L3)(2pi)(Planck length)(3/2)exponent 1/2

When this equation is solved for (L3) the value found is 1.1835x10 exp 10 meters. This is a unique value because it is the wavelength of a photon that has energy equal to the Planck constant divided by the constant (2pi) squared. The implied value for the Planck length is 1.6159455x10 exp-35. I believe this relationship will become useful as the Bh electron article is filled in with information from other contributors.DonJStevens 15:43, 20 August 2006 (UTC)

The (L3) value is extremely close to the wavelength of a photon that has the energy value h divided by (2pi) squared. When the implied value for the Planck length is 1.6159455x10 exp-35, the implied value for the gravitational constant is 6.6717456x10 exp-11. It is an interesting surprise to find that the G value implied by electron properties is close to and even overlapped by G values that have been previously determined from laboratory measurements.DonJStevens 17:37, 16 September 2006 (UTC)

Don, I'm not sure I entirely follow what you're trying to do here. You've defined

and then you say that this is very close to "the wavelength of a photon that has the energy value h divided by (2pi) squared". But h has the dimensions of action (energy × time). How are you getting from h to a photon energy? -- Jheald 17:09, 21 September 2006 (UTC).

Answer; The L3 equation as you have written it is correct. The numerical value for L3 is 1.1834933x10 exp 10 meters when the Planck length value is 1.616x10 exp -35 meters. The L3 value found is very close to one light second (or 2.9979246x10 exp 8 meters) multiplied by the constant (2pi) squared. A photon with wavelength equal to one light second (and frequency of one cycle per second) will have (E=h times one cycle /sec) while a photon with the wavelength (2pi) squared times one light second will have [E=h times (1/2pi) squared cycle /sec]. It is correct to say L3 is approximately equal to (2pi) squared times one light second however K-N black hole properties imply that L3 should be precisely equal to (2pi) squared times one light second.DonJStevens 14:43, 22 September 2006 (UTC)

More facts; The approximate electron Schwarzschild radius can be specified as shown. The Le value is the electron Compton wavelength.

rs = 2/3 (Le/4pi)(Le/2L3) squared

With L3 equal to (2pi) squared times one light second, the radius is 1.3524368x10 exp-57 meter. This is approximately equal to the well known value 2Gm/c squared. The approximate radius and the well known radius values will be numerically equal when the implied G value (6.6717456x10 exp-11) is used. The two radius values are approximately equal when any of the (other) published G values is used.

The following equations can be easily verified to be approximately correct. They will be numerically correct when the implied G value is used.

3Gm/c squared = (Le/4pi)(Le/2L3) squared

m = (c squared/3G)(Le/4pi)(Le/2L3) squared

m = (1/3G)(Le/4pi) cubed, times (1/2pi) squared

m c squared = (c exp 4)(1/3G)(Le/4pi)(Le/2L3) squared

hc/Le = (c exp 4)(1/3G)(Le/4pi)(Le/2L3) squared

m = (h/4pi c)(c/3pi hG) exponent 1/4

The values h and G are shown to be closely related. DonJStevens 16:24, 17 October 2006 (UTC)

[edit] Superfluid 3He

I'm reverting the most recent edit by User:Quasarq again:

1. Micro black hole is already linked
2. superfluid 3He stuff:
   1. It's not written well
   2. It's original research (reference to primary sources, I'm going to check out the SA article)
   3. It doesn't follow WP:MoS
3. unification: M-Theory claims unification. M-Theory is unverified, but is definitely (not really arguable) adequate. Also, non-"M-theory" string theory also has a good explanation of unification (I'm not familiar with loop quantum gravity), so unless you're willing to list all the exceptions (which I can't say how many there are), it should remain closer to how it is.

It should be noted that I'm not opposed to this content in general, but it's going to have to be incorporated better. McKay 23:10, 13 September 2006 (UTC)

I read the relevant article in Scientific American. Black hole analogies, are merely that -- analogies of black holes. They are not black hole electrons with a new name. They do not belong in this article. For more information see User_talk:Quasarq. McKay 04:25, 14 September 2006 (UTC)

[edit] "In need of an expert tag"

This page looks pretty good now, do we still need it? X [Mac Davis] (SUPERDESK|Help me improve) 07:47, 24 October 2006 (UTC)

[edit] Matter-wave quantum

The electron mass energy is related to a specific photon wavelength so its mass energy is related to the Planck constant. In the book "The Enigmatic Electron", by Malcolm MacGregor (page 25) the electron is referred to as "the discrete quantum of matter-wave systems". Later (page 67) the author suggests that electron properties may be a guide to bridge the quantum and classical domains. From the electron mass and light velocity, a relationship between the (quantum) Planck constant and the (classical) gravitational constant is specified.

Le = 2(3Gm)exp 1/3, times (2pi)exp 5/3

h/2mc = (3Gm)exp 1/3, times (2pi)exp 5/3

h = 2mc(3Gm)exp 1/3, times (2pi)exp 5/3

This provides a partial bridge between the two domains.

DonJStevens 19:33, 3 January 2007 (UTC)

[edit] Stability / Infinite density

When an electron is infinitely small, its electric field will be infinitely large. In his book "The Quantum World", author Kenneth Ford writes, "Physicists will be happy if the things we call particles and that seem to exist at points are eventually shown to occupy some region - albeit an incredibly small region - of space" (page 241).

Theorists are not satisfied with words like infinitely small and infinitely large to describe electron properties. It is known that negative electric charge elements repel one another so that a significant force is required to hold the electron together. No combination of electric and magnetic forces has been proposed that can produce a stable state. Author, Malcolm MacGregor noted that gravitational forces could solve the stability problem if the electron is incredibly small. See book, "The Enigmatic Electron" page 72. The density required for stability is clearly achieved before the electron radius is reduced to zero meters (a point). The electron must occupy a small region of space so that its electric field and density are not required to be infinitely large.

DonJStevens 18:58, 19 January 2007 (UTC) DonJStevens 15:58, 3 February 2007 (UTC)

[edit] super-extremal?

A black hole must satisfy |Q| ≤ M for there to be an event horizon, while |Q_e| >> M_e. Without an event horizon, how can an object meaningfully be termed a black hole? —The preceding unsigned comment was added by 209.250.133.53 (talk) 17:40, 17 February 2007 (UTC).

Hmm, I made reference to a problem with the page you referenced. Take a look at the talk page. But if I understand you correctly, it's actually the whole point behind this article. McKay 01:37, 18 February 2007 (UTC)

Added reply: The Russian physicist, Alexander Burinskii describes the Kerr-Newman black hole electron as a particle without a horizon. In his paper "The Dirac-Kerr electron" he writes (page 2) "Recall, that angular momentum J=h bar/2 for parameters of electron is so high that black hole horizons disappear and the source of the spinning particle represents a naked singular ring". DonJStevens 18:02, 18 February 2007 (UTC)

Wow, I think that's just a tad bit over my head. I guess I'm not sure how that applies to a comparison of Charge and Mass? McKay 05:09, 19 February 2007 (UTC)

Improved reply: The charge and mass comparison discussion should begin with relationships that all can agree on. The electron charge energy (electric field plus magnetic field) will be equal to the electron mass energy and this energy will be equal to 1/2 of the energy of a photon with wavelength 1.213x10 exp-12 meters (approx). This wavelength photon will have the energy density to convert to a pair of black hole particles when subjected to appropriate gravitational blue shift (time dilation) and an equal amount of gravitational space contraction. If this wavelength photon is absorbed by a black hole, its electromagnetic energy becomes mass. If the black hole is unstable after absorption, it may emit a pair of electron mass particles. Each mass particle will have a size closely related to its Schwarzschild radius value. In order to be extremal and stable, the electron must have the minimum energy required for it to be confined by its own gravitational field. DonJStevens 18:09, 19 February 2007 (UTC)

Ahh, so we're comparing the energy of the charge and the energy of the mass, if so, then you've basically determined that the electron is the basis for comparison, then saying that the electron is a denerate case of the black hole because q=m doesn't really say much. Where did the comparisons of Q and M come from? McKay 04:36, 20 February 2007 (UTC)

Be careful here: There are three parameters to consider here: mass, charge, and angular momentum (actually angular momentum per unit mass). The last one is usually represented by a. For the case of m² > q² + a², the Kerr-Newman horizins do vanish. To compare all three, geometrized units are used where c=G=1. More specifically, m = MG / c², (in cgs units), and a = A / c, where the capital letters are in "normal" units and the small letters are in geometrized units. Note that all geometrized units are lengths. I regret that I don't have the time to do the calculations for an electron right now. --EMS | Talk 06:18, 20 February 2007 (UTC)

The electron mass and the electromagnetic energy released during particle annihilation are "given" values that require explanation. The dimensionless ratio (electron Compton wavelength) divided by 4pi(3Gm/c squared) is required to be a function of the gravitational potential at the radius 3Gm/c squared. When this ratio is explained we will be close to explaining why the electron mass is quantized. DonJStevens 16:33, 20 February 2007 (UTC)

Okay, I can help with the calculations for an electron. For an electron:

• m = 6.76345536e-58 meters
• q = 1.38057828e-36 meters
• a = ...

  I can't quite get the units to work out right. you say it should be a = A / c, but A is in units of J * s, or kg * m^2 / s. Dividing by c gives us units in kg * m, which gets us off by a factor of kilograms, the conversion to eliminate the kilograms (G/c²), results in a units of m² which is also wrong, can I just take the square root? That doesn't quite seem right, I'm not sure. In any case, what I got was 3.04639314e-43 m kg.

Where do we go from here? I still don't feel right about applying "m² > q² + a²" because my units aren't right yet. McKay 16:44, 20 February 2007 (UTC)

Angular momentum has geometric dimension of area. Angular momentum per unit mass will be meters.DonJStevens 18:52, 21 February 2007 (UTC)

Ahh, I missed the parenthetical remark "(actually angular momentum per unit mass)" so, the numbers are:

• m = 6.76345536e-58 meters
• q = 1.38057828e-36 meters
• a = 3.34423694e-13 meters

For this case, the mass is much smaller than the combined charge and angular momentum (when geometrized with G and c), so there isn't an event horizon (according to this metric). Perhaps this should be incorporated into the article? McKay 19:42, 21 February 2007 (UTC)

I'm glad that you figured out the per-unit-mass business. Sorry that I was not around to help earlier with that. As for incorporating this into the article: I would first double-check the numbers, but if they are right the next issue may be seeking a reliable source for this. This looks like something that should be known, but without a source I cannot vouch that it is. (BTW, the m and q values do look right to me, but I have never computed a before.) --EMS | Talk 05:32, 22 February 2007 (UTC)

Yeah, we can't add it until we have a source. I'm merely adding information to (and learning about :) ) a claim that someone made on the talk page. It does appear as if the claim has merit, I would imagine that the person who made the intial edit (User:209.250.133.53) might have a source, Feel free to check my work. I could have made a mistake. I feel confident about my work, it was done on google calculator. I usually use units to check my work, and they came out clean. Feel free to check it, errors could be anywhere.McKay 06:02, 22 February 2007 (UTC)

I don't doubt your math and have fewer doubts on the conclusion. I have run across the electron being super-extremal before. My only concern here is the integrity of the encyclopedia, but even then I see this as an important and relevant finding which almost certainly has been noticed before. Perhaps using the {{fact}} template may be sufficient to permit the inclusion of this interesting finding in the article. At the least, this is not a controverial finding for those of us who are familiar with GR and see the math. --EMS | Talk 15:54, 22 February 2007 (UTC)

I need a math check so we can all be on the same page. I have angular momentum value (h/4pi)(G/c cubed) or 1.306x10exp -70. This value divided by 6.763x10exp -58 is 1.931x10exp -13. The "a" value would then be close to 1.931x10exp -13 meters unless I missed something. Let me know if I made an error. DonJStevens 18:16, 23 February 2007 (UTC)

Hmm, at first, I thought your units were messed up, but they're not.

So I reduced yours to elminate duplicates, and only then did I check that it was off by a factor of two. hbar is h/2pi, not h/4pi. so I think my original number is correct.

For my result:Check this out Google calculator McKay 05:11, 24 February 2007 (UTC)

Hi McKay. Thanks for the math check. In any case the horizons vanish. This is significant. DonJStevens 15:12, 24 February 2007 (UTC)

[edit] Source; m squared requirement

In his book "The Road To Reality" page 832, Roger Penrose writes, regarding electrons and the K-N gyromagnetic ratio 2,"It could only apply if an electron could be regarded as being, in some sense a black hole". He then notes that the electron does not meet the requirement that (m)exp 2 must be equal to or greater than (a)exp 2 plus (e)exp 2 in order that the "--Kerr-Newman metric can represent a black hole".

Many theorists agree however, with the statement by Chris Isham; "One of the major predictions of Einstein's theory is the phenomenon of gravitational collapse in which ---, matter that is compressed to more than a critical density will inevitably collapse under its self-gravitational attraction until it becomes a --- gravitational singularity".

The critical density is clearly reached before the electron radius is reduced to a point. DonJStevens 17:25, 5 March 2007 (UTC)

Regarding the problem of cosmic censorship, Roger Penrose has noted that "cosmic censorship" is a mathematical conjecture that is " -- as yet neither proved or refuted -- concerning general solutions of the Einstein equation". See pages 768 and 769 in his book "The Road To Reality". DonJStevens 18:41, 10 March 2007 (UTC)

Many readers will agree that the electron is, in some sense a black hole particle if it has the (black hole) mass density required for gravitational confinement. Theorist A. Burinskii included a "naked singular ring" in his electron particle description. The absence of horizons was fully expected by Burinskii. DonJStevens 22:42, 10 March 2007 (UTC)

[edit] Balanced pattern

When lengths labeled (L1) and (L4) are compared to lengths previously labeled (L2) and (L3) an interesting ratio pattern appears.

L1 = 2pi (Planck length)(3/2)exp 1/2

L2 = 1/2 (Le)

L3 = (2pi)squared times (one light second)

L4 = 2pi (3Gm/c squared)

A balanced ratio pattern follows.

L1/L2 = L2/L3 = L4/L1 = 2(L1)/Le = Le/2(L3)

When the implied G value is used, these dimensionless ratios are all equal. In the following equation, the G value will cancel so that the equation is correct without regard to a true G value.

L4/L2 = (L1/L2)squared

The above equation can be solved for (L1) to show that (L1) squared is (3pi hG/c cubed). This (L1) value is 2pi (Planck length) times (3/2) exponent 1/2. The equation below is derived from the balanced ratio pattern.

2(L4)/Le = [Le/2(L3)] squared

The balanced pattern results from relationships that are readily verifiable.---DonJStevens 19:57, 16 March 2007 (UTC)

[edit] Quantum - gravitational effect

At the extreme high energy photon wavelength (L1), a critical energy density is reached. This wavelength photon has energy equal to the mass energy of two black holes. Each of these black holes has a photon capture circumference [2pi (3Gm/c squared)] equal to the photon wavelength.

E = hc/L1 = 2E black hole = L1 (c exp 4)(1/3pi G)

hc/L1 = L1 (c exp 4)(1/3pi G)

L1 = (3pi hG/c cubed) exp 1/2

The (L1) wavelength is an effective model for all photons because photon energy is inversely proportional to wavelength. The (L1) wavelength provides a useful conversion factor to relate the Planck constant to the gravitational constant.

(L1) squared, (c cubed/3pi) = hG

(L1) squared, (c cubed/3pi h) = G

The Planck length to electron Compton wavelength relationship (shown earlier) is known to be either precisely correct, as implied or extremely close to correct, as demonstreted.

(L2) squared = (L3)(L1)

Full confirmation of this relationship would allow the electron to be correctly defined as a "quantum - gravitational" mass particle.

m = (h/4pi c)(c/3pi hG) exp 1/4

DonJStevens 15:37, 24 March 2007 (UTC)