What is the Wavelength of Hawking Radiation?

In summary, Hawking Radiation is a type of radiation emitted by black holes, specifically the Kerr Black Hole which is a rotating black hole. It is caused by the Kerr Temperature which is responsible for the Kerr Particle Energy Spectrum and is considered the genesis of Thermodynamic Quantum Gravitation. This is the combination of thermodynamics and quantum gravitation. When a black hole falls below the Planck radius, it becomes a Thermodynamic Quantum-Schwarzschild Black Hole and its radiation diminishes instead of increasing, resulting in a flux that does not violate conservation of energy/momentum. The wavelength of Hawking radiation is still unknown, as it requires a full theory of quantum gravity.
  • #1
Orion1
973
3
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
 
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  • #2
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?
 
  • #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.

 
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  • #4
Originally posted by Orion1
A Schwarzschild-Planck Black Hole would evaporate instantly.

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.
 
  • #5
http://www.th.physik.uni-frankfurt.de/~lxd/English/bhs_e.html
http://relativity.livingreviews.org/Articles/lrr-2001-6/node7.html

"Imaginary Time" (which theoretically exists at right angles to ordinary time) may hold the key to this one. However, we’re yet to probe THAT dimension…
 
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  • #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.
 
  • #7
Planck Probability...

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.

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 equation 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 equations are described in 'slow motion'.
 
  • #8
NateTG said:
Let's say that I have an extremely small black hole -- i.e. the Schwartzschild radius is smaller than the Planck length.

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.
 
  • #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
 

1. What is Hawking Radiation?

Hawking Radiation is a theoretical phenomenon proposed by physicist Stephen Hawking in which black holes emit particles and energy over time, eventually leading to their complete evaporation.

2. How does Hawking Radiation occur?

Hawking Radiation occurs due to a quantum effect known as vacuum fluctuations. These fluctuations cause virtual particle-antiparticle pairs to appear near the event horizon of a black hole, with one particle falling into the black hole and the other escaping as radiation.

3. What types of particles are emitted in Hawking Radiation?

Hawking Radiation is theorized to emit all types of particles, including photons, neutrinos, and even heavier particles such as protons and electrons. However, the rate at which these particles are emitted is incredibly slow, making it difficult to detect.

4. Can Hawking Radiation be observed?

Currently, there is no direct observational evidence for Hawking Radiation as it is incredibly faint and difficult to detect. However, scientists are working on developing new technologies and experiments to potentially observe this phenomenon in the future.

5. What are the implications of Hawking Radiation for black hole physics?

Hawking Radiation is a crucial concept in understanding the behavior and evolution of black holes. It suggests that black holes have a finite lifespan and will eventually evaporate completely, leading to important implications for the study of the universe and its evolution.

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