- #1
- 19,773
- 10,726
Definition/Summary
Hawking radiation is black body thermal radiation predicted to be emitted by black holes due to quantum effects-
'The vacuum in quantum field theory is not really empty; it's filled with "virtual pairs" of particles and antiparticles that pop in and out of existence, with lifetimes determined by the Heisenberg uncertainty principle. When such pairs forms near the event horizon of a black hole, though, they are pulled apart by the tidal forces of gravity. Sometimes one member of a pair crosses the horizon, and can no longer recombine with its partner. The partner can then escape to infinity, and since it carries off positive energy, the energy (and thus the mass) of the black hole must decrease.'
Source- Steve Carlip
Equations
Close to the event horizon of a black hole, a local observer must accelerate to keep from falling in. This implies that the Unruh effect (the prediction that an accelerating observer will observe black-body radiation where an inertial observer would observe none) takes place near the event horizon. An accelerating observer would see a thermal bath of particles pop out of the local acceleration horizon, turn around, and free-fall back in. Some of the particles emitted by the horizon are not reabsorbed and in the case of a BH, become outgoing Hawking radiation.
The equation for Unruh temperature as observed by an accelerating observer is-
T=\frac{a \hbar}{2\pi k_B c}
where a is the local acceleration, \hbar is the reduced Planck constant, k_B is the Boltzmann constant and c is the speed of light.
For a black hole, acceleration is replaced by surface gravity (\kappa). Black-body radiation for a static black hole-
T_H=\frac{\kappa \hbar}{2\pi k_B c}
for a static BH-
\kappa=\frac{c^4}{4Gm}
where G is the gravitational constant and m is mass. T_H can be rewritten-
T_H=\frac{\hbar c^3}{8\pi G k_Bm}
The power emitted by a static BH, according to the Stefan-Boltzmann-Schwarzschild-Hawking power law-
P=A\epsilon \sigma T^4
where A is surface area of the BH, \epsilon is emissivity (for a true black body, \epsilon = 1), \sigma is the Stefan-Boltzmann constant and T is temperature.
for a static black hole-
A=4\pi r_s^2=\frac{16 \pi G^2 m^2}{c^4}
where r_s is the Schwarzschild radius. P can be rewritten -
P=\frac{\hbar c^6}{15360 \pi G^2 m^2}
where P is the energy output in watts.
Evaporation time via Hawking radiation for a static BH-
t_{ev}=\frac{5120 \pi G^2 m^3}{\hbar c^4}
where t_{ev} is the evaporation time of the BH in seconds based on an empty universe (i.e. no matter, photons, CMB radiation or any other type of energy falling into the BH). For derivation, see Hr (wiki) link in 'See Also'.
'..since the universe contains the cosmic microwave background radiation, in order for a black hole to dissipate, it must have a temperature greater than that of the present-day black-body radiation of the universe of 2.7 K = 2.3e-4 eV. This implies that m must be less than 0.8% of the mass of the Earth.' The smaller the BH, the quicker it evaporates.
Extended explanation
For a rotating BH, the equation for TH can be rewritten-
T_H=\frac{\hbar\kappa}{2\pi k_B c}=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}<br />
where M is the gravitational radius (M=Gm/c2) and a is the spin parameter (a=J/mc)
For a charged BH, the equation can be rewritten as-
T_H=\frac{\hbar\kappa}{2\pi k_B c}=\left(1-\frac{Q^4}{r_+^4}\right)^{-1} \frac{\hbar c^3}{8\pi Gk_bm}<\frac{\hbar c^3}{8\pi Gk_bm}<br />
where Q is charge and r+ is the outer event horizon.
The above implies that spin and charge reduce Hawking radiation though a/M or Q/M would have to virtually equal 1 for HR to be significantly reduced.
If both spin and charge are present then-
\kappa=c^2\ \frac{(r_+-r_-)}{2\left(r_+^2+a^2\right)}
where r_{\pm}=M \pm \sqrt(M^2-Q^2-a^2) and a^2+Q^2\leq M^2 applies.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
Hawking radiation is black body thermal radiation predicted to be emitted by black holes due to quantum effects-
'The vacuum in quantum field theory is not really empty; it's filled with "virtual pairs" of particles and antiparticles that pop in and out of existence, with lifetimes determined by the Heisenberg uncertainty principle. When such pairs forms near the event horizon of a black hole, though, they are pulled apart by the tidal forces of gravity. Sometimes one member of a pair crosses the horizon, and can no longer recombine with its partner. The partner can then escape to infinity, and since it carries off positive energy, the energy (and thus the mass) of the black hole must decrease.'
Source- Steve Carlip
Equations
Close to the event horizon of a black hole, a local observer must accelerate to keep from falling in. This implies that the Unruh effect (the prediction that an accelerating observer will observe black-body radiation where an inertial observer would observe none) takes place near the event horizon. An accelerating observer would see a thermal bath of particles pop out of the local acceleration horizon, turn around, and free-fall back in. Some of the particles emitted by the horizon are not reabsorbed and in the case of a BH, become outgoing Hawking radiation.
The equation for Unruh temperature as observed by an accelerating observer is-
T=\frac{a \hbar}{2\pi k_B c}
where a is the local acceleration, \hbar is the reduced Planck constant, k_B is the Boltzmann constant and c is the speed of light.
For a black hole, acceleration is replaced by surface gravity (\kappa). Black-body radiation for a static black hole-
T_H=\frac{\kappa \hbar}{2\pi k_B c}
for a static BH-
\kappa=\frac{c^4}{4Gm}
where G is the gravitational constant and m is mass. T_H can be rewritten-
T_H=\frac{\hbar c^3}{8\pi G k_Bm}
The power emitted by a static BH, according to the Stefan-Boltzmann-Schwarzschild-Hawking power law-
P=A\epsilon \sigma T^4
where A is surface area of the BH, \epsilon is emissivity (for a true black body, \epsilon = 1), \sigma is the Stefan-Boltzmann constant and T is temperature.
for a static black hole-
A=4\pi r_s^2=\frac{16 \pi G^2 m^2}{c^4}
where r_s is the Schwarzschild radius. P can be rewritten -
P=\frac{\hbar c^6}{15360 \pi G^2 m^2}
where P is the energy output in watts.
Evaporation time via Hawking radiation for a static BH-
t_{ev}=\frac{5120 \pi G^2 m^3}{\hbar c^4}
where t_{ev} is the evaporation time of the BH in seconds based on an empty universe (i.e. no matter, photons, CMB radiation or any other type of energy falling into the BH). For derivation, see Hr (wiki) link in 'See Also'.
'..since the universe contains the cosmic microwave background radiation, in order for a black hole to dissipate, it must have a temperature greater than that of the present-day black-body radiation of the universe of 2.7 K = 2.3e-4 eV. This implies that m must be less than 0.8% of the mass of the Earth.' The smaller the BH, the quicker it evaporates.
Extended explanation
For a rotating BH, the equation for TH can be rewritten-
T_H=\frac{\hbar\kappa}{2\pi k_B c}=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}<br />
where M is the gravitational radius (M=Gm/c2) and a is the spin parameter (a=J/mc)
For a charged BH, the equation can be rewritten as-
T_H=\frac{\hbar\kappa}{2\pi k_B c}=\left(1-\frac{Q^4}{r_+^4}\right)^{-1} \frac{\hbar c^3}{8\pi Gk_bm}<\frac{\hbar c^3}{8\pi Gk_bm}<br />
where Q is charge and r+ is the outer event horizon.
The above implies that spin and charge reduce Hawking radiation though a/M or Q/M would have to virtually equal 1 for HR to be significantly reduced.
If both spin and charge are present then-
\kappa=c^2\ \frac{(r_+-r_-)}{2\left(r_+^2+a^2\right)}
where r_{\pm}=M \pm \sqrt(M^2-Q^2-a^2) and a^2+Q^2\leq M^2 applies.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
