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Black Holes, how much energy do they give off?

  1. Jun 9, 2008 #1
    Sorry for the dumb question but...

    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. jcsd
  3. Jun 9, 2008 #2
    I'd start with wikipedia's article on Hawking Radiation:

    It goes on to give the formula and more info.
     
  4. Jun 10, 2008 #3
    Thanks alot, this really helps. :smile:
     
  5. Oct 28, 2008 #4
    A black hole gets hotter as it radiates and gets smaller, so a smaller black hole dissipates more quickly than a massive one...but the lives in general are extremely long...I believe on roughly the order of the age of the universe....or at least ,really, really slowly...in other words, you'll have to watch really carefully to see one die....
     
  6. Oct 28, 2008 #5

    George Jones

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  7. Nov 3, 2008 #6
    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 (Mir) 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-Mir) 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
     
    Last edited by a moderator: May 3, 2017
  8. Nov 3, 2008 #7

    [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
     
    Last edited: Nov 3, 2008
  9. Nov 4, 2008 #8
    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
     
    Last edited by a moderator: May 3, 2017
  10. Nov 6, 2008 #9

    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
     
    Last edited: Nov 6, 2008
  11. Nov 6, 2008 #10
    Last edited: Nov 6, 2008
  12. Nov 6, 2008 #11
    Kerr metric...


    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
     
    Last edited: Nov 7, 2008
  13. Nov 7, 2008 #12
    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.
     
    Last edited: Nov 7, 2008
  14. Nov 8, 2008 #13

    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]

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

    Reference:
    Photon sphere - Wikipedia
    Inside a black hole - the Kerr black hole
     

    Attached Files:

    Last edited: Nov 8, 2008
  15. Nov 10, 2008 #14
    Retrograde photon orbit-

    [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/
     
  16. Nov 10, 2008 #15

    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]
     
    Last edited by a moderator: May 3, 2017
  17. Nov 11, 2008 #16
    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
     
    Last edited: Nov 12, 2008
  18. Nov 11, 2008 #17

    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.
     
    Last edited: Nov 11, 2008
  19. Nov 11, 2008 #18
    That seems to be the case.
     
    Last edited: Nov 12, 2008
  20. Nov 11, 2008 #19
    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
     

    Attached Files:

    Last edited by a moderator: May 3, 2017
  21. Nov 11, 2008 #20
    Kerr metric quantum black hole evaporation time...


    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]


    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]
     

    Attached Files:

    Last edited by a moderator: May 3, 2017
  22. Jan 14, 2010 #21
    The "Figure 3.14" referenced in post No. 19 relates specifically to the Alexander Burinskii electron model that includes a naked ring singularity. The electron angular momentum is so great that the electron cannot collapse to its event horizon radius and so its ring singularity is naked. Electron mass energy relationships imply that electron gravitational collapse is halted at the equatorial photon sphere radius which is larger than the projected event horizon. In the equatorial plane, the static limit is (4/2) times the projected event horizon while the photon sphere radius is (3/2) times the projected event horizon. See:
    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.
     
  23. Jan 21, 2010 #22
    A paper referenced in post #20 "Quantum Black Holes: the Event Horizon as a Fuzzy Sphere" is interesting because the author finds (page 3) the smallest possible mass for a black hole in this scheme is (1/2) (Planck mass). This is confirmed when the angular momentum value h/4pi is included as a requirement.

    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.
     
  24. Jul 3, 2011 #23
    We can add to the Alexander Burinskii electron model by introducing the concept of a "limit gravitational time dilation factor" as shown:
    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.
     
  25. Nov 13, 2011 #24
    We can now see that the electron mass, muon mass and tau mass are each related to the Planck mass by three factors. The first two factors are (1/2) and (2/3)^1/2. The third factor is the square root of the ratio, 4pi (3Gm/c^2) divided by (Compton wavelength of the particle). This third factor is the gravitational time dilation factor that applies at the collapsed particle radius, 3Gm/c^2. In the special (electron) case, this time dilation value is equal to the square root of the proposed limit time dilation value. These equations are either precisely correct, or so very close that we are unable to define their error. Many theorists expected that the Planck mass should relate to the electron mass. A serious effort is now needed to determine if these equations are just due to coincidence or if these equations can lead to an improved understanding of particles. I suggest that Orion 1 has worked in this area and so he could provide help with this.
     
  26. Feb 23, 2012 #25
    We find that photons have ever greater energy density as a photon wavelength is reduced. An interesting (limit) short wavelength value is found when photon energy density is so large that photon energy is correctly specified by either the Planck constant h, or the gravitational constant G. Photon energy at this limit is h times frequency or (hc/ wavelength) and also equal to (wavelength) (c^4/3pi G).
    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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