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Black hole electron

  1. Nov 7, 2008 #1
    There seems to be a strong theory that the electron could be a 'naked' singularity or maximal Kerr black hole (maximal meaning it wouldn't give off any Hawking radiation and therefore wouldn't evaporate instantly). While technically there would be no event horizon due to the fact that the inner and outer event horizons meet at the gravitational radius (M), the Schwarzschild radius would still apply as it is always the reduced circumference (R) of r+ and r- regardless of spin so for a maximal rotating black hole where the two horizons meet at the gravitational radius, the reduced circumference (R) at the gravitational radius is Rs. Based on the mass of the electron, 9.109e-31 kg, Rs would be 1.353e-57 m, which is way smaller than the current accepted minimum of Planck length, 1.616e-35 m, of which anything smaller is considered to be just quantum foam. Does the theory for the black hole electron pose a solution to this or is it one of the problems posed by the theory?

    Black hole electron - wikipedia
    Last edited: Nov 7, 2008
  2. jcsd
  3. Feb 2, 2009 #2
    The gravitational collapse electron model was at first, considered by most physicists, to be unacceptable because the model includes a "naked singularity". Now a number of physicists have concluded that the laws of nature do permit a naked singularity to be formed. See February 2009 issue Scientific American article "Naked Singularities" by Pankaj S. Joshi. A quote from this article follows.

    "General relativity does not account for the quantum effects that become important for microscopic objects, and those effects presumably intervene to prevent the strength of gravity from becoming truly infinite. But physicists are still struggling to develop the quantum theory of gravity they need to explain singularities."

    When a force becomes infinite, we need to realize "we don't have it right". This is the first step toward getting it right. I was the first author to publish the electron mass equation in a small book (copyright 2001).

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

    The mass equation though numerically correct, was considered by many to be a numerical coincidence rather than a guideline to a deeper electron model. With the recent work by Burinskii, Williamson, van der Mark, Greene, Joshi and others, a new evaluation of the gravitationally confined electron concept is in progress.

    The "numerical coincidence" explanation for the electron mass equation (and related equations) is becoming ever more difficult to defend.

    Don Stevens
    Last edited: Feb 2, 2009
  4. Mar 7, 2009 #3
    Re: Black hole proton?

    Hi Don,

    If one is going to model elementary particles as relativistic objects, the proton radius is a critical test.

    Have you tried to apply your methods to retrodicting the proton radius?

    I am having a closely related discussion [which mentioned your efforts] at Advanced Physics Forum [Planck Scale Redux] in the "lounge" forum.

    Yours in science,
  5. Mar 7, 2009 #4
    quantum black hole lepton mass spectrum...

    According to QCD, a proton is not a elementary or fundamental particle.

    The solution I derived for the quantum black hole electron mass:
    [tex]\boxed{m_e = \frac{1}{2 \sqrt{\pi} (6G)^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}}}[/tex]

    The solution I derived for the quantum black hole mass:
    [tex]\boxed{m = C_1 \cdot m_L}[/tex]

    [tex]C_1[/tex] - quantized dimensionless constant

    Quantum black hole quantum lepton mass:
    [tex]\boxed{m_L = \frac{1}{3 \sqrt{2 \pi} G^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}}}[/tex]

    Quantum black hole quantum lepton mass theorem:
    [tex]\boxed{m = C_1 \left[ \frac{1}{3 \sqrt{2 \pi} G^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}} \right]}[/tex]

    Quantum black hole Lepton quantum mass spectrum:

    Quantum black hole Electron mass:
    [tex]\boxed{m_e = \left( \frac{3}{2} \right)^{\frac{3}{4}} \left[ \frac{1}{3 \sqrt{2 \pi} G^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}} \right]}[/tex]

    Quantum black hole Muon mass:
    [tex]\boxed{m_{\mu} = \left( \frac{7}{2} \right)^{\frac{9}{2}} \left[ \frac{1}{3 \sqrt{2 \pi} G^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}} \right]}[/tex]

    Quantum black hole Tauon mass:
    [tex]\boxed{m_{\tau} = \left( \frac{8}{5} \right)^{18} \left[ \frac{1}{3 \sqrt{2 \pi} G^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}} \right]}[/tex]

    Quantum black hole Proton mass:
    [tex]\boxed{m_p = \left( \frac{7}{4} \right)^{14} \left[ \frac{1}{3 \sqrt{2 \pi} G^{\frac{1}{4}}} \left( \frac{\hbar}{c} \right)^{\frac{3}{4}} \right]}[/tex]

    http://en.wikipedia.org/wiki/Compton_wavelength" [Broken]
    http://en.wikipedia.org/wiki/Electron" [Broken]
    http://en.wikipedia.org/wiki/Muon" [Broken]
    http://en.wikipedia.org/wiki/Tau_lepton" [Broken]
    http://en.wikipedia.org/wiki/Proton" [Broken]
    Last edited by a moderator: May 4, 2017
  6. Mar 7, 2009 #5
    Proton is elementary

    Hi Orion,

    I contend that the proton is most definitely an elementary particle.

    Impossible, you say?

    See: http://arxiv.org/ftp/astro-ph/papers/0701/0701006.pdf and you will find a fairly strong case for seriously considering both the proton and the electron as elementary particles that can be best modeled as Kerr-Newman solutions of the Einstein-Maxwell equations. The proton satisfies the M^2 > Q^2 + a^2 requirement for an event horizon, and so it is a black hole solution. The electron has M^2 << Q^2 + a^2, and so it is represents a naked singularity solution (note: naked singularities are not really totally naked, but rather have a compact spherical halo of lower scale particles surrounding them).

    Does this mean that there is serious trouble in the land of High Energy Physics? Isn't that completely obvious? Just look at the horrendous vacuum energy density fiasco: http://arxiv.org/ftp/arxiv/papers/0901/0901.3381.pdf [Broken] . It's a whole new ballgame, my friend!

    Yours in science,
    Last edited by a moderator: May 4, 2017
  7. Mar 8, 2009 #6
    Hello Orion and Knecht,

    I believe the proton must also be gravitationally confined. I have concentrated specifically on the electron but a better understanding of electrons will surely lead to a better understanding of other particles as well.

    Orion, I confirm your quantum black hole electron mass equation.

    m = (3/2)^3/4 [ 6.720792346x10^-31 ] = 9.109382144x10^-31

    We must keep in mind the electron mass equation that uses only G and the electron Compton wavelength, labeled (Le).

    m = (1/3G) (1/2pi)^2 (Le/4pi)^3 = 9.109382143x10^-31

    Orion, your work to define the mass of other particles is very interesting. I obtain a muon mass value, 1.886803515x10^-28 kg. I believe we are about to see "a whole new ballgame".

    Don Stevens
  8. Mar 8, 2009 #7

    Vanadium 50

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    Re: Proton is elementary

    Nonsense. First, I outline some (of many) of the errors your paper is making https://www.physicsforums.com/showthread.php?p=2108022" Second, your paper ignores the vast quantity of high quality data on deeply elastic scattering and proton structure (one paper of which received the Nobel prize.

    Physics Forums is here for, among other reasons, to help students learn the current status of physics as practiced by the scientific community. Posting non-mainstream theories does not help with that.

    Orion, I am having trouble getting dimensions to work out right. [itex] \hbar / c [/itex] has units of (Js)/(m/s) or Js2/m. Since a Joule is a kgm2/s2, then [itex] \hbar / c [/itex] has units of kg-m. G has units of m3/kgs2, so [itex] (\hbar / c)^3 / G [/itex] must have units (kgm)3 / (m3/kgs2), or kg4s2. So the mass of the electron you give doesn't have units of kg, but rather kgs1/2. Is there something missing somewhere?
    Last edited by a moderator: Apr 24, 2017
  9. Mar 8, 2009 #8
    Hi Vanadium 50,

    You may find this is interesting. The Le length is the electron Compton wavelength.

    2pi (Planck length) (3/2)^1/2 (c) (one second) = (Le/4pi)^2

    (3pi hG/c^3)^1/2 (c) (one second) = (Le/4pi)^2

    (4pi)^2 (3pi hG/c^3)^1/2 (c) (one second) = (Le)^2

    (length) (length) = (length)^2

    The equation is correct when the G value is 6.671745578x10^-11. From the Le value the electron mass is h/c (1/Le).

    Don Stevens
  10. Mar 8, 2009 #9
    Proton Radius Without Mathematical Gymnastics

    A very good test of Don's approach is : Can he retrodict the proton radius?

    The paradigm I am proposing can do this very simply, as follows. The Schwarzschild radius relation is: R = 2GM/c^2. (A good 1st approximation for the proton radius.)

    If you use the proton mass and my G' candidate for the correct value of G within atomic scale systems (= 2.31 x 10^31 cgs), then R = 0.8 x 10^-13 cm. Spot on!

    See how easy and accurate that is. No weird scaling factors, no mathematical gymnastics. No "strings " attached! I'll bet that Don cannot retrodict the proton radius without a lots of numerology, and that the numerology will mathematically reduce to what I am proposing.

    For a discussion of how I derive G', see: http://arxiv.org/ftp/physics/papers/0701/0701132.pdf .

    And to Vanadium 50: when one seeks a new and more unifying paradigm one has to start somewhere and gradually develop that new paradigm. When generations of physicists have worked to hammer the Standard Model into conformity with observations, obviously the fit is going to be fairly good. The new discrete scale invariant paradigm is worth considering, even in its present state and especially if you are an open-minded young student, because it can unify a badly schizophrenic physics (disconnect between QFT/QM and GR), it can solve the VED crisis, it can explain the fine structure constant, it can elucidate the real meaning of Planck's constant, it can finally give us an reasonable Planck Scale, it generates a definitive prediction for the dark matter (not some vague arm-waving about hypothetical "unicornons"), ...

    Yours in science,
  11. Mar 8, 2009 #10

    Vanadium 50

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    I don't find it interesting at all. I find it numerological claptrap. How can an electron possibly know how long one second is? How can 1/86400 of the Earth's rotation possibly have any impact on a fundamental property of an elementary particle.

    You wrote an "equation" with mismatched units, and then tried to fix it by sticking in "one second". That's incorrect.
  12. Mar 8, 2009 #11

    Vanadium 50

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    Accepting "equations" with mismatched units is not a sign of "open-mindedness". It's a mistake.
  13. Mar 8, 2009 #12

    George Jones

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    This thread has degenerated into personal speculation as opposed to a discussion of mainstream science, so I'm shutting it down.

    A reminder: Physics Forums rules,


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