Is the Universe / an electron a Black Hole?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by Reiku, Sep 18, 2007.

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  1. Nasor Valued Senior Member

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    I'd still like to see some actual math indicating that the electron has any meaningful surface gravity - because as far as I'm able to calculate, it doesn't. Of course I could be wrong, but I'd like to see some math to back up all the nice, pretty words that people are throwing around here.
     
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  3. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    only quickly skimed this page and no one seems to have noted the obvious proof that electron is NOT a black hole (no need to show surface gravity too small. etc.) Just note that all electrons have the exactly the same mass but black holes can have any mass. Thread is dead.
     
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  5. fmonroy Registered Senior Member

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    I've no idea of it is or not, but your thought doesn't exclude black holes of exactly the same mass, so they can be a subset of black holes, right?
     
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  7. DonJStevens Registered Member

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    This is probably the time to examine some facts that all can agree are valid. The length 2pi (Planck length) times (3/2) exp 1/2 is also equal to (3pi hG/c cubed) exp 1/2. The energy of a photon with this wavelength can be specified either by the Planck constant and light velocity or by the gravitational constant and light velocity.

    E = hc/wavelength

    E = (c) exp 4, times (wavelength)/(3pi G)

    hc/wavelength = (c) exp 4, times (wavelength)/(3pi G)

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

    This wavelength labeled (L1) is clearly a very special wavelength. All longer wavelength photons are stretched copies of this photon. This photon has energy equal to the mass energy of two black holes, each with a photon capture radius (3Gm/c squared) equal to the photon wavelength divided by two pi. A simple equation shows how (L1) relates to the wavelength labeled (L2). The (L2) value is defined as (electron Compton wavelength/2). This wavelength photon has energy equal to the mass energy of two electrons. The simple length equation below shows how (L1) and (L2) are related to a derived length labeled (L3).

    L1/L2 = L2/L3

    When this equation is solved for (L3) the value is found to be (2pi) squared times one light second. The (L2) value is then required to be equal to the square root of (L1) (L3). The electron Compton wavelength will then be 4pi times (3pi hG/c) exp 1/4. The electron mass is then required to be (h/4pi c) times (c/3pi hG) exp 1/4. If this equation is precisely correct, as implied, the applicable G value must be very close to 6.6717456x10 exp -11.

    These relationships are easily verified. It is not so easy to know their full meaning.
     
  8. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Perhaps, but no reason for this "new class of black holes" to exists.

    This new class would also not be alowed to have the normal properties of black holes. I.e. they could not "swallow" any more mass and could not "evaporate" via Hawkins radiation and would all be constrained to have a fixed ratio of charge to mass, when all other black hole permit this ratio to vary and their angular momentum needs also to be fixed, not variable as is all other black holes. Summary: they share no property in common with other black holes.

    So what you are suggestion is a new class of objects that have none of the properties of black holes (and all of the properties of electrons) but you want to call them "black holes."

    OK it is a free country, but that is a strange terminology/ name for them since in no way do they resemble other black holes, but go ahead and call them that if you wish. However, if you do, I am equally free to call horses another "new class of black holes" also as they too have none of the properties of black holes (and all of the properties of horses).- an exact parallel to yours and equally silly.
     
    Last edited by a moderator: Sep 21, 2007
  9. Reiku Banned Banned

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    Yes, well... I'd have to admit i agree with all of this. I never believed in my heart they where black holes anyway, so, the general opinion only heightens m initial thoughts.

    Reiku :m:
     
  10. DonJStevens Registered Member

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    It is correct to describe electrons as special case black holes. Here are some words from Brian Greene: Electrons are in the class called "extremal" black holes. These black holes have charge and have the minimal possible mass consistant with the charge they carry. They do not evaporate or emit Hawking radiation.

    Many theorists are trying to resolve this and debates will continue. We gather data, analyze it and try many ideas that don't work before finding the one that is correct. Since we know with full confidence, that what is correct will eventually prevail, we can respect the Brian Greene opinion and still disagree. As you know, I personally side with Greene and Burinskii. These people are good company.
     
  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Very much like that other special class of black holes called "horses" (not "electrons").

    I.e.:
    Horse black holes also "do not evaporate or emit Hawking radiation."
    Horse black holes also are "extremal" black holes (in that they carry the maximal mass consistent with only four index finger nails (sometimes called hoofs).

    Neither "electron" nor "horse" black holes shair any property with conventional black holes (or with each other, except that both are "extremal special class" black holes).

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    More details about horse black holes in my prior post 17.
     
  12. DonJStevens Registered Member

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    I have one other equation that I would like to make readers aware of, as shown below:

    electron Schwarzschild radius = 2/3 (Le/4pi) times (Le/2L3) squared

    The Le value is the electron Compton wavelength and the L3 value is (2pi) squared times one light second. If there are questions about why this is correct, I will try to answer them. If the only response is from the land of Houyhnhnms, I may not determine that any reply is appropriate.
     
  13. James R Just this guy, you know? Staff Member

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    DonJStevens:

    Welcome to sciforums. You can post math using TeX tags in this forum, which might make your posts more readable. Quote this post to see how to do this.

    For example, from your post above:

    \(r = \frac{2}{3} \left(\frac{L_e}{4\pi}\right) \times \left(\frac{L_e}{2L_3}\right)^2\)
     
  14. Nasor Valued Senior Member

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    Just doing it in my head, it looks like the Schwarzschild radius for the mass of an electron would be around 10E-58 meters. I don't think electron are that small...
     
  15. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Nothing is, if memory serves (less than the Planck length, I think)
     
  16. DonJStevens Registered Member

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    \( \frac{3 G_m}{c^2} = (\frac{L_e}{4\pi}) \times\left(\frac{L_e}{2L_3}\right)^2\)

    James R:

    Thank you for the suggestion. I just gave it a try, but I am very slow at this. The 3Gm/c squared equation is easily solved for an implied G value. When the implied G value is used, the electron mass and Compton wavelength equations are precisely correct. If any of the (other) published values for G are used, the mass and wavelength equations are extremely close to correct. With the values I used for mass and wavelength, the implied G value is 6.6717456x10 exp -11. The average of three gravitational constant values measured in 1942, 1972, and 1982 is 6.672x10 exp -11. All of these have significant uncertainty. The electron mass equation is either precisely correct, as implied or extremely close to correct as demonstrated.

    Note that the electron mass formula, like the black hole entropy formula, contains the Planck constant as well as the gravitational constant, indicating that the electron mass value is the result of a quantum-gravitational effect. See page 715 in "The Road To Reality" by Penrose.

    The Schwarzschild radius is (approx.) 1.35x10 exp -57 meter. We usually define the electron as a zero radius particle but if it is gravitationally collapsed, as proposed by physicist John Wheeler, it will have a photon capture radius value 3Gm/c squared. This is small but not infinitely small (like zero).
     
    Last edited: Sep 23, 2007
  17. Nasor Valued Senior Member

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    So how fast does it have to be going before its Schwarzschild radius is more than its de Broglie wavelength?
     
  18. DonJStevens Registered Member

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    Hi Nasor:

    As an exchange, I will provide an answer to your question, regarding de Broglie wavelength but I will request an answer in return. Do you agree or disagree that the energy of the photon with wavelength labeled L1 has electromagnetic energy that can be sprcified by either the Planck constant and light velocity or the gravitational constant and light velocity? Also tell me why if you disagree. Is that fair?
     
    Last edited: Sep 24, 2007
  19. DonJStevens Registered Member

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    Matter wavelength limit

    Hi Nasor:

    Some theories come with a high-energy cutoff, beyond which we do not expect that the theory provides a good description of nature. Gravity appears to be valid up to its cutoff at the Planck scale.

    As an electron approaches ever closer to light velocity, a high-energy cutoff is expected. A limit gamma factor is expected to apply when the electron mass reaches the value (Planck mass/2) times (2/3) exponent 1/2. This is 8.887x10 exp -9 kg (gamma factor 9.756x10 exp 21).

    The electron de Broglie wavelength with this limit gamma factor will not be less than 2.49x10 exp -34 meter. My answer to your question is "The electron cannot gain enough mass from high veocity to reach a condition where the de Broglie wavelength is less than 1.35x10 exp -57 meter."
     
  20. Reiku Banned Banned

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    ''Some theories come with a high-energy cutoff, beyond which we do not expect that the theory provides a good description of nature. Gravity appears to be valid up to its cutoff at the Planck scale.

    As an electron approaches ever closer to light velocity, a high-energy cutoff is expected. A limit gamma factor is expected to apply when the electron mass reaches the value (Planck mass/2) times (2/3) exponent 1/2. This is 8.887x10 exp -9 kg (gamma factor 9.756x10 exp 21).

    The electron de Broglie wavelength with this limit gamma factor will not be less than 2.49x10 exp -34 meter. My answer to your question is "The electron cannot gain enough mass from high veocity to reach a condition where the de Broglie wavelength is less than 1.35x10 exp -57 meter.''

    Excellent obervation made here. Worthy of recognition.
     
  21. Reiku Banned Banned

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    Would anybody know what i was talking about, if i said, ''could we apply the Special Relativistic Gamma Factor with consciousness?''

    Reiku - Peace
     
  22. quadraphonics Bloodthirsty Barbarian Valued Senior Member

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    Nope.
     
  23. Reiku Banned Banned

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    I would like to proclaim against Spammers? Is this npt the main goal...
     
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