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Hawking Radiation

  1. Dec 29, 2003 #1
    One of the real important aspects of Hawking Radiation is the Kerr Temperature.

    Note that the Kerr Temperature is responsible for the Kerr Particle Energy Spectrum and represents the genesis of Thermodynamic Quantum Gravitation.

    Thermodynamic Quantum Gravitation is the combination of Thermodynamics and Quantum Gravitation:

    A Kerr Black Hole is a rotating Black Hole.

    [tex]T_k = \frac{\hbar c^3}{K_o G M}[/tex]

    Ko = Boltzmann's Thermal Constant
     
  2. jcsd
  3. Dec 29, 2003 #2

    NateTG

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    Since you brought it up, I have this rather silly question about Hawking radiation:

    Let's say that I have an extremely small black hole -- i.e. the Schwartzschild radius is smaller than the Planck length.

    Wouldn't a black hole like this have a tendency to radiate light that had more mass/momentum than the black hole does?
     
  4. Dec 29, 2003 #3


    Schwarzschild radius:
    [tex]r_s = \frac{2 G M}{c^2}[/tex]

    Planck Length:
    [tex]r_p = \sqrt{ \frac{ \hbar G}{c^3}}[/tex]

    Schwarzschild Temperature:
    [tex]T_s = \frac{ \hbar c^3}{4 K_o G M}[/tex]

    [tex]r_s = r_p[/tex]

    [tex]\frac{2 G M}{c^2} = \sqrt { \frac{ \hbar G}{c^3}}[/tex]

    Schwarzschild-Planck Mass:
    [tex]M_s = \frac{1}{2} \sqrt { \frac{ \hbar c}{G}}[/tex]

    Integral:
    [tex]M_s = \frac{1}{2} \sqrt { \frac{ \hbar c}{G}} = \frac{ \hbar c^3}{4 K_o G T_s}[/tex]

    Schwarzschild-Planck Temperature:
    [tex]T_s = \frac {1}{2K_o} \sqrt { \frac{\hbar c^5}{G}}}[/tex]

    [tex]T_s = 7.084E+31 K[/tex]
    T_s = 7.084*10^31 Kelvin


    Schwartzschild radius is smaller than the Planck length.

    Wouldn't a black hole like this have a tendency to radiate light that had more mass/momentum than the black hole does?

    if the black hole were smaller than that, then would it be able to radiate?

    [tex]r_s \ll r_p[/tex]

    [tex]\Delta = \frac {\hbar c}{ \lambda K_o T_s} = \frac{2}{ \lambda} \sqrt { \frac {\hbar G}{c^3}}[/tex]

    [tex]\Delta = \frac{2}{ \lambda} \sqrt { \frac {\hbar G}{c^3}}[/tex]

    [tex]T_q = \frac {\hbar c^3}{4 K_o G M_s ( e^\Delta - 1)}[/tex]

    [tex]I(\lambda) = \frac { 2 \pi h c^2}{ \lambda^5 (e^\Delta - 1)}[/tex]

    [tex]I_q = \sigma T_q^4[/tex]



    The Schwarzschild-Planck Radius is Mass dependent.

    As a Thermodynamic Schwarzschild-Planck Black Hole radius falls below the Planck Radius [tex]r_s \ll r_p[/tex], it becomes a Thermodynamic Quantum-Schwarzschild Black Hole, the resulting radiation diminishes instead of increasing.

    The resulting radiation flux becomes less energetic than the mass equivalency.

    A Thermodynamic Schwarzschild-Planck Black Hole would evaporate instantly however a Thermodynamic Quantum-Schwarzschild Black Hole diminishes more gradually, however still relatively instantaneous. The relative flux intensity also diminishes.

    A Quantum-Schwarzschild Black Hole Thermodynamic Temperature is quantumized below the Planck Radius.

     
    Last edited: Dec 30, 2003
  5. Dec 30, 2003 #4

    NateTG

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    Right, but if the black hole were smaller than that, then would it be able to radiate? Conservation of energy/mass would be grossly violated if the black hole could produce radiation with more energy than it's equivalent mass. Similarly, there might be problems with conservation of linear momentum.
     
  6. Jan 27, 2004 #5
  7. Apr 24, 2004 #6
    Conservation of energy/mass would be grossly violated if the black hole could produce radiation with more energy than it's equivalent mass

    Not if the black hole can absorb as much energy from the vacuum as it radiates and there is theoretically 10^120 Joules per cubic metre in the vacuum.
     
  8. Apr 25, 2004 #7
    Planck Probability...

    What if a Schwarzschild-Planck Black Hole is capable of absorbing more radiation than its Schwarzschild-Planck Temperature vacuum, does this also violate Conservation of energy/mass?

    I presume that a Schwarzschild-Planck Black Hole which is capable of absorbing more radiation than its Schwarzschild-Planck Temperature vacuum would momentarily increase in mass, then still evaporate instantly.

    To my understanding, a 'perfect radiation absorber' is not possible, but then again it was once thought that black holes were a 'zero radiation emitter'.

    Is there an equasion that exists that determines how much radiation a Schwarzschild-Planck Black Hole is capable of absorbing?

    According to my integrations above, the amount of Schwarzschild-Planck Black Hole radiation flux produced below the Planck Radius [tex]r_s \ll r_p[/tex] is no longer determined by mass thermodynamics, but by probability and radiation wavelength similar to a blackbody radiator, given here: [tex]P = (e^\Delta - 1)[/tex].

    Thereby, when Schwarzschild-Planck Black Hole falls below Planck radius [tex]r_s \ll r_p[/tex], the radiation flux probabilisticly diminishes, resulting in a radiation flux that no longer violates Conservation of energy/momemtum. However note that the evaporation is still relatively instantaneous, as such equasions are described in 'slow motion'.
     
  9. Apr 25, 2004 #8

    Stingray

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    In that case, the derivation of Hawking radiation is no longer valid. You'd need a fully theory of quantum gravity, whereas Hawking radiation is derived by formulating quantum field theory on top of a fixed classical spacetime.
     
  10. Feb 27, 2008 #9
    Wavelength

    I am Doing a project on Hawking Radiation and I am wondering if anyone actually Knows the wavelength of it?? It would be great if someone could email me cos I don't always have time to do much more than check my emails

    --
    a.random.persona@gmail.com
    --
    Thanks
     
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