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The power emitted by a black hole in the form of Hawking radiation can be estimated for the simplest case of a non-rotating, non-charged Schwarzschild black hole of mass M. Combining the formulas for the black hole Schwarzschild radius, the Stefan Boltzmann law of blackbody radiation, the black hole Bekenstein Hawking luminosity surface radiation temperature, and the sphere surface area formula, which is also the black holes event horizon surface area, several equations can be derived:
$\;$
Stefan Boltzmann constant: (ref. 1, ref. 2)
$\sigma = \frac{\pi^{2} k_{B}^{4}}{60 \hbar^{3} c^{2}}$
$\;$
Schwarzschild radius: (ref. 3, ref. 4)
$r_{s} = \frac{2 G M}{c^{2}}$
$\;$
Gravitational acceleration at the Schwarzschild event horizon:
$g = \frac{G M}{r_{s}^{2}} = G M \left(\frac{c^{2}}{2GM} \right)^{2} = \frac{c^{4}}{4 G M}$
$\boxed{g = \frac{c^{4}}{4 G M}}$
$\;$
$E = k_{B} T = \frac{\hbar g}{2 \pi c} = \frac{\hbar}{2 \pi c} \left(\frac{c^{4}}{4 G M} \right) = \frac{\hbar c^{3}}{8 \pi G M}$
$\;$
Hawking peak radiation temperature: (ref. 6, ref. 7)
$\boxed{T_{H} = \frac{\hbar c^{3}}{8 \pi G M k_{B}}}$
$\;$
For a one solar mass black hole, the Hawking peak radiation temperature is:
$T_{H} = \frac{\hbar c^3}{8 \pi G M_{\odot} k_{B}} = 6.170 \times 10^{-8} \; \text{K}$
$\boxed{T_{H} = \frac{\hbar c^3}{8 \pi G M_{\odot} k_{B}}}$
$\boxed{T_{H} = 6.170 \times 10^{-8} \; \text{K}}$
$\;$
Wiens displacement law: (ref. 8)
$\lambda_\mathrm{peak} = \frac{h c}{4.965 k_{B} T_{H}} = \frac{h c}{4.965 k_{B}} \left(\frac{8 \pi G M_{\odot} k_{B}}{\hbar c^{3}} \right) = \frac{h c}{4.965 k_{B}} \left(\frac{8 \pi^{2} k_{B}}{h c} \right)\left(\frac{2 G M}{c^{2}} \right) = \frac{16 G M \pi^{2}}{4.965 c^{2}} = \frac{8 \pi^{2}}{4.965} \; r_{s} = 15.902 \; r_{s}$
$\boxed{\lambda_\mathrm{peak} = \frac{16 G M \pi^{2}}{4.965 c^{2}}}$
$\boxed{\lambda_\mathrm{peak} = 15.902 \; r_{s}}$
$\;$
The radiation peak wavelength is nearly 16 times the black hole Schwarzschild radius.
$\;$
Schwarzschild sphere surface area with Schwarzschild radius integration via substitution:
$A_{s} = 4 \pi r_{s}^{2} = 4 \pi \left(\frac{2 G M}{c^{2}} \right)^{2} = \frac{16 \pi G^{2} M^{2}}{c^{4}}$
$\;$
Schwarzschild sphere surface area with Schwarzschild radius:
$\boxed{A_{s} = \frac{16 \pi G^{2} M^{2}}{c^{4}}}$
$\;$
Stefan Boltzmann power law: (ref. 9)
$P = A_{s} j^{\star} = A_{s} \epsilon \sigma T^{4}$
$\;$
A black hole is a perfect blackbody:
$\boxed{\epsilon = 1}$
$\;$
Stefan Boltzmann Schwarzschild Hawking black hole radiation power law derivation:
$P = A_{s} \epsilon \sigma T_{H}^{4} = \left(\frac{16 \pi G^{2} M^{2}}{c^{4}} \right)\left(\frac{\pi^{2} k_{B}^{4}}{60 \hbar^{3} c^{2}} \right)\left(\frac{\hbar c^{3}}{8 \pi G M k_{B}} \right)^{4} = \frac{\hbar c^{6}}{15360 \pi G^{2} M^{2}}$
$\;$
This yields the Bekenstein Hawking luminosity of a black hole, for pure photon emission and no other emitted particles, and under the assumption that the horizon is the radiating surface:
$\boxed{P = \frac{\hbar c^{6}}{15360 \pi G^{2} M^{2}}}$
$\;$
$P$ - radiated power luminosity
$\hbar$ - reduced Planck constant
$c$ - speed of light
$G$ - gravitational constant
$M$ - black hole mass
$\;$
It is worth mentioning that the above formula has not yet been derived in the framework of semiclassical gravity.
$\;$
Substituting the numerical values of the physical constants in the formula for luminosity, the Hawking radiation power from a solar mass black hole turns out to be minuscule:
$P = \frac{\hbar c^{6}}{15360 \pi G^{2} M_{\odot}^{2}} = 9.007 \cdot 10^{-29} \; \text{W}$
$\boxed{P = \frac{\hbar c^{6}}{15360 \pi G^{2} M_{\odot}^{2}}}$
$\boxed{P = 9.007 \cdot 10^{-29} \; \text{W}}$
$\;$
It is indeed an extremely good approximation to call such an object black. Under the assumption of an otherwise empty universe, so that no matter, cosmic microwave background radiation, or other radiation falls into the black hole, it is possible to calculate how long it would take for the black hole to dissipate.
Bekenstein Hawking Schwarzschild evaporation constant:

$K_{ev} = \frac{\hbar c^{6}}{15360 \pi G^{2}} = 3.562 \times 10^{32} \; \text{W} \; \text{kg}^{2}$
$\boxed{K_{ev} = \frac{\hbar c^{6}}{15360 \pi G^{2}}}$
$\boxed{K_{ev} = 3.562 \times 10^{32} \; \text{W} \; \text{kg}^{2}}$
$\;$
Given that the power of the Hawking radiation is the rate of evaporation energy loss of the black hole:
$P = - \frac{dE}{dt} = \frac{K_{ev}}{M^{2}}$
$\;$
Since the black hole total energy E is related to the black hole mass M by the Einstein mass energy formula:
$E = Mc^{2}$
$\;$
$P = - \frac{dE}{dt} = -\left(\frac{d}{dt} \right) M c^2 = -c^2 \frac{dM}{dt}$
$\;$
It is possible to equate this with the above expression for the power equation:
$-c^{2} \frac{dM}{dt} = \frac{K_\mathrm{ev}}{M^{2}}$
$\;$
The differential equation variables can be seperated:
$M^{2} \; dM = - \frac{K_{ev}}{c^{2}} \; dt$
$\;$
The black holes mass is now a function $M(t)$ of time $t$. Integrating over $M$ from $M_{0}$, the initial mass of the black hole to zero, complete evaporation, and over $t$ from zero to $t_{ev}$:
$\int_{M_{0}}^{0} M^{2} \; dM = - \frac{K_{ev}}{c^{2}} \int_{0}^{t_{ev}} \; dt$
$\;$
Schwarzschild black hole evaporation time integration via substitution:
$t_{ev} = \frac{c^{2} M_{0}^{3}}{3 K_{ev}} = \left(\frac{c^{2} M_{0}^{3}}{3} \right)\left(\frac{15360 \pi G^{2}}{\hbar c^{6}} \right) = \frac{5120 \pi G^{2} M_{0}^{3}}{\hbar c^{4}}$
$\;$
Schwarzschild black hole evaporation time:
$\boxed{t_{ev} = \frac{5120 \pi G^{2} M_{0}^{3}}{\hbar c^{4}}}$
$\;$
Evaporation time For a one solar mass black hole:
$t_{ev} = \frac{5120 \pi G^{2} M_\odot^{3}}{\hbar c^{4}} = 6.617 \times 10^{74} \; \text{s} = 2.098 \; \cdot \; 10^{67} \; \text{years}$
$\;$
$\boxed{t_{ev} = \frac{5120 \pi G^{2} M_\odot^{3}}{\hbar c^{4}}}$
$\boxed{t_{ev} = 2.098 \; \cdot \; 10^{67} \; \text{years}}$
$\;$
Universe age:
$t_{u} = 13.799 \; \cdot \; 10^{9} \; \text{years}$
$\;$
The evaporation time for a one solar mass black hole is much greater than the universe age.
$\boxed{t_{ev} \gg t_{u}}$
$\;$
The lower classical quantum limit for mass for this equation is equivalent to the Planck mass $m_{P}$:
$\;$
Hawking radiation evaporation time for a Planck mass quantum black hole:
$t_{ev} = \frac{5120 \pi G^{2} m_{P}^{3}}{\hbar c^{4}} = 5120 \pi t_{P} = 5120 \pi \sqrt{\frac{\hbar G}{c^{5}}} = 8.671 \times 10^{-40} \; \text{s}$
$\;$
$\boxed{t_{ev} = 5120 \pi \sqrt{\frac{\hbar G}{c^{5}}}}$
$\boxed{t_{ev} = 8.671 \cdot 10^{-40} \; \text{s}}$
$\;$
Where $t_{P}$ is the Planck time. (ref. 10)
$\;$
However, since the universe contains the cosmic microwave background radiation, in order for the black hole to dissipate, it must have a temperature greater than that of the present day blackbody radiation of the universe of 2.7 K. This implies that M must be less than 0.8 percent of the mass of the Earth.
Cosmic microwave background radiation universe temperature: (ref. 11)

$T_{u} = 2.725 \; \text{K}$
$\;$
Schwarzschild Hawking total black hole mass:
$M_{H} \leq \frac{\hbar c^{3}}{8 \pi G k_{B} T_{u}} \leq 4.503 \; \cdot \; 10^{22} \; \text{kg}$
$\;$
$\boxed{M_{H} \leq \frac{\hbar c^{3}}{8 \pi G k_{B} T_{u}}}$
$\boxed{M_{H} \leq 4.503 \; \cdot \; 10^{22} \; \text{kg}}$
$\;$
$\frac{M_{H}}{M_\oplus} = 7.539 \; \cdot \; 10^{-3} = 0.754 \; \%$
$\;$
$M_{\oplus}$ - total Earth mass
$\;$
$\;$
Reference:
Wikipedia - Stefan-Boltzmann law: (ref. 1)
https://en.wikipedia.org/wiki/Stefan-Boltzmann_law
Wikipedia - Stefan-Boltzmann_constant: (ref. 2)
https://en.wikipedia.org/wiki/Stefan-Boltzmann_constant
Wikipedia - Schwarzschild metric: (ref. 3)
https://en.wikipedia.org/wiki/Schwarzschild_metric
Wikipedia - Schwarzschild radius: (ref. 4)
Wikipedia - Blackbody radiation: (ref. 5)
Wikipedia - Hawking radiation: (ref. 6)
Wikipedia - Black hole thermodynamics: (ref. 7)
https://en.wikipedia.org/wiki/Black_hole_thermodynamics
Wikipedia - Wien's_displacement law: (ref. 8)
https://en.wikipedia.org/wiki/Wien's_displacement_law
Wikipedia - Stefan-Boltzmann law: (ref. 9)
https://en.wikipedia.org/wiki/Stefan-Boltzmann_law
Wikipedia - Planck time: (ref. 10)
https://en.wikipedia.org/wiki/Planck_time
Wikipedia - Cosmic microwave background radiation: (ref. 11)

Edited by Orion1
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Other than the fact that the CMB hasn't always been at 2.7 deg., but has come down from about 3500 deg. , that seems a pretty good approximation.
Something slightly less massive, to account for decreased radiation prior to the present era, could not have resulted from stellar collapse.

Since only 'primordial' Black Holes of about 0.7 % Earth's mass would be completely evaporating by now ( in a final shedding of the Event Horizon, and a massive Gamma Ray burst ), the fact that none are observed might mean that the energy density of the early universe was homogenous and isotropic enough to prevent the formation of 'primordial' BHs.

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On 2/10/2020 at 11:07 PM, MigL said:

Something slightly less massive, to account for decreased radiation prior to the present era, could not have resulted from stellar collapse.

Affirmative, stellar black hole masses typically range between 2.27 to 16 solar masses, and are generated from stellar supernovas. (ref. 1)
$2.27 \cdot M_{\odot} \leq M_{bh} \leq 16 \cdot M_{\odot}$
$\;$
According to this model, any Schwarzschild black hole capable of evaporating completely from Hawking radiation within the universe age, the Schwarzschild black hole must have an initial primordial mass less than or equal to this Hawking evaporation mass. This Hawking evaporation mass is similar to the mass of a small asteroid.
$\;$
Schwarzschild black hole Bekenstein-Hawking total power luminosity:
$\boxed{P_{H} = \frac{\hbar c^{6}}{15360 \pi G^{2} M^{2}}}$
$\;$
Universe age:
$t_{u} = 13.799 \cdot 10^{9} \; \text{years} = 4.355 \cdot 10^{17} \; \text{s}$
$\;$
Schwarzschild black hole total Hawking evaporation mass in universe lifetime integration:
$P_{H} = - \frac{dE}{dt} = -\left(\frac{d}{dt} \right) M c^2 = -c^2 \frac{dM}{dt} = \frac{\hbar c^{6}}{15360 \pi G^{2} M^{2}}$
$\;$
The differential variables are seperable, and the integrals can be written as:
$- \int_{M_{ev}}^{0} M^{2} dM = \frac{\hbar c^{4}}{15360 \pi G^{2}} \; \int_{0}^{t_{u}} dt = \frac{M_{ev}^{3}}{3} = \frac{t_{u} \hbar c^{4}}{15360 \pi G^{2}}$
$\;$
Schwarzschild black hole total Hawking evaporation mass in universe lifetime:
$\boxed{M_{ev} \leq \left(\frac{t_{u} \hbar c^{4}}{5120 \pi G^{2}} \right)^{\frac{1}{3}}}$
$\;$
$\boxed{M_{ev} \leq 1.730 \cdot 10^{11} \; \text{kg}}$
$\;$
Asteroid Castalia mass: (ref. 2)
$M = 5.000 \cdot 10^{11} \; \text{kg}$
$\;$
$\;$
Reference:
Science Forums - toy model black holes per galaxy average number - Orion1: (ref. 1)
https://www.scienceforums.net/topic/120012-toy-model-black-holes-per-galaxy-average-number/
Asteroid Fact Sheet - NASA: (ref. 2)
https://nssdc.gsfc.nasa.gov/planetary/factsheet/asteroidfact.html

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Posted (edited)
On 2/10/2020 at 11:07 PM, MigL said:

( in a final shedding of the Event Horizon, and a massive Gamma Ray burst )

According to this model, a result which occurs during Schwarzschild black hole Hawking radiation evaporation, when Hawking radiation evaporation lifetime is set at one second remaining.
$\;$
Schwarzschild black hole total Hawking evaporation time with one second lifetime remaining:
$\boxed{t_{1} = 1 \; \text{s}}$
$\;$
Schwarzschild black hole total Hawking evaporation mass with one second lifetime remaining:
$\boxed{M_{1} = \left(\frac{t_{1} \hbar c^{4}}{5120 \pi G^{2}} \right)^{\frac{1}{3}} }$
$\;$
$\boxed{M_{1} = 2.282 \cdot 10^{5} \; \text{kg}}$
$\;$
Schwarzschild black hole Bekenstein-Hawking total power luminosity integration via substitution:
$P_{1} = \frac{\hbar c^{6}}{15360 \pi G^{2} M_{1}^{2}} = \frac{\hbar c^{6}}{15360 \pi G^{2}} \left[\left(\frac{5120 \pi G^{2}}{t_{1} \hbar c^{4}} \right)^{\frac{1}{3}} \right]^{2} = \frac{}{24} \left(\frac{\hbar c^{10}}{10 \pi \left(G t_{1} \right)^{2}} \right)^{\frac{1}{3}}$
$\;$
Schwarzschild black hole Bekenstein-Hawking total power luminosity at one second lifetime remaining:
$\boxed{P_{1} = \frac{}{24} \left(\frac{\hbar c^{10}}{10 \pi \left(G t_{1} \right)^{2}} \right)^{\frac{1}{3}}}$
$\;$
$\boxed{P_{1} = 6.838 \cdot 10^{21} \; \text{W}}$
$\;$
$1 \; \text{megaton TNT} = 4.184 \cdot 10^{15} \; \text{j}$
$\;$
$\boxed{P_{1} = 1.634 \cdot 10^{6} \; \frac{\text{megatons TNT}}{\text{s}}}$
$\;$
Schwarzschild black hole peak Hawking radiation temperature at one second lifetime remaining integration via substitution:
$T_{1} = \frac{\hbar c^{3}}{8 \pi G M_{1} k_{B}} = \frac{\hbar c^{3}}{8 \pi G k_{B}} \left(\frac{5120 \pi G^{2}}{t_{1} \hbar c^{4}} \right)^{\frac{1}{3}} = \frac{}{k_{B}} \left(\frac{10 \hbar^{2} c^{5}}{G t_{1} \pi^{2}} \right)^{\frac{1}{3}}$
$\;$
Schwarzschild black hole peak Hawking radiation temperature at one second lifetime remaining:
$\boxed{T_{1} = \frac{}{k_{B}} \left(\frac{10 \hbar^{2} c^{5}}{G t_{1} \pi^{2}} \right)^{\frac{1}{3}}}$
$\;$
$\boxed{T_{1} = 5.376 \cdot 10^{17} \; \text{K}}$
$\;$
Schwarzschild black hole total Hawking energy radiation emission after one second lifetime remaining integration via substitution:
$M_{1} = \frac{E_{1}}{c^{2}} = \left(\frac{t_{1} \hbar c^{4}}{5120 \pi G^{2}} \right)^{\frac{1}{3}} \rightarrow E_{1} = c^{2} \left(\frac{t_{1} \hbar c^{4}}{5120 \pi G^{2}} \right)^{\frac{1}{3}} = \left(\frac{t_{1} \hbar c^{10}}{5120 \pi G^{2}} \right)^{\frac{1}{3}}$
$\;$
Schwarzschild black hole total Hawking energy radiation emission after one second lifetime remaining:
$\boxed{E_{1} = \left(\frac{t_{1} \hbar c^{10}}{5120 \pi G^{2}} \right)^{\frac{1}{3}}}$
$\;$
$\boxed{E_{1} = 2.051 \cdot 10^{22} \; \text{j}}$
$\;$
$1 \; \text{megaton TNT} = 4.184 \cdot 10^{15} \; \text{j}$
$\;$
$\boxed{E_{1} = 4.903 \cdot 10^{6} \; \text{megatons TNT}}$
$\;$
Is this the eventual fate for every black hole in the entire universe?
$\;$
$\;$

Edited by Orion1
source code correction...

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Posted (edited)
On 2/10/2020 at 11:07 PM, MigL said:

the fact that none are observed might mean that the energy density of the early universe was homogenous and isotropic enough to prevent the formation of 'primordial' BHs.

Affirmative, during the inflationary epoch, the universe was homogeneous and isotropic, every particle point location in space-time was inflating and expanding from every other particle point location in space-time faster than invariant luminous velocity. Any primordial quantum black holes generated as bosonic radiation when the strong force broke symmetry from gravity, would have evaporated instantly from Hawking blackbody radiation.
$\;$
Hawking blackbody radiation evaporation time for a Planck mass quantum black hole:
$t_{ev} = \frac{5120 \pi G^{2} m_{P}^{3}}{\hbar c^{4}} = 5120 \pi t_{P} = 5120 \pi \sqrt{\frac{\hbar G}{c^{5}}} = 8.671 \cdot 10^{-40} \; \text{s}$
$\;$
$\boxed{t_{ev} = 5120 \pi \sqrt{\frac{\hbar G}{c^{5}}}}$
$\boxed{t_{ev} = 8.671 \cdot 10^{-40} \; \text{s}}$
$\;$
This next presented Hawking blackbody radiation model is based on quantum physics and classical Newtonian gravitation.
$\;$
$g = \frac{G M}{r^{2}}$
$\;$
$v_{e} = \sqrt{\frac{2 G M}{r}}$
$\;$
$\boxed{\frac{2 \pi E}{\hbar} = \frac{g}{v_{e}}}$
$\;$
Hawking blackbody radiation peak radiation energy initial model condition integration via substitution:
$\frac{2 \pi E}{\hbar} = \frac{g}{v_{e}} = \left(\frac{G M}{r^{2}} \right)\left(\sqrt{\frac{r}{2 G M}} \right) = \sqrt{\frac{G M}{2 r^{3}}}$
$\;$
$\boxed{\frac{2 \pi E}{\hbar} = \sqrt{\frac{G M}{2 r^{3}}}}$
$\;$
$E = k_{B} T_{H} = \frac{\hbar}{2 \pi} \sqrt{\frac{G M}{2 r^{3}}}$
$\;$
$\boxed{T_{H} = \frac{\hbar}{2 \pi k_{B}} \sqrt{\frac{G M}{2 r^{3}}}}$
$\;$
Stellar model radiation source is a perfect blackbody:
$\boxed{\epsilon = 1}$
$\;$
Stefan-Boltzmann Hawking blackbody radiation power law derivation integration via substitution:
$P_{H} = A_{s} \epsilon \sigma T_{H}^{4} = \left(4 \pi r^{2} \right) \left(\frac{\pi^{2} k_{B}^{4}}{60 c^{2} \hbar^{3}} \right) \left(\frac{\hbar}{2 \pi k_{B}} \sqrt{\frac{G M}{2 r^{3}}} \right)^{4} = \frac{\hbar G^{2} M^{2}}{960 \pi c^{2} r^{4}}$
$\;$
Stefan-Boltzmann Hawking blackbody radiation power law:
$\boxed{P_{H} = \frac{\hbar G^{2} M^{2}}{960 \pi c^{2} r^{4}}}$
$\;$
One solar mass neutron star model total stellar radius: (ref. 1)
$\boxed{R_{ns} = 1 \cdot 10^{4} \; \text{m}}$
$\;$
One solar mass neutron star model Hawking blackbody radiation peak radiation temperature:
$T_{H} = \frac{\hbar}{2 \pi k_{B}} \sqrt{\frac{G M_{\odot}}{2 R_{ns}^{3}}} = 9.903 \cdot 10^{-9} \; \text{K}$
$\boxed{T_{H} = \frac{\hbar}{2 \pi k_{B}} \sqrt{\frac{G M_{\odot}}{2 R_{ns}^{3}}}}$
$\boxed{T_{H} = 9.903 \cdot 10^{-9} \; \text{K}}$
$\;$
One solar mass neutron star model Hawking blackbody radiation power law:
$P_{H} = \frac{\hbar G^{2} M_{\odot}^{2}}{960 \pi c^{2} R_{ns}^{4}} = 6.852 \cdot 10^{-31} \; \text{W}$
$\boxed{P_{H} = \frac{\hbar G^{2} M_{\odot}^{2}}{960 \pi c^{2} R_{ns}^{4}}}$
$\boxed{P_{H} = 6.852 \cdot 10^{-31} \; \text{W}}$
$\;$
Can thermodynamic neutron stars generate Hawking blackbody radiation?
$\;$
$\;$
Reference:
Wikipedia - neutron star: (ref. 1)
https://en.wikipedia.org/wiki/Neutron_star
Wikipedia - Hawking radiation: (ref. 2)

Edited by Orion1
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20 hours ago, Orion1 said:

Can thermodynamic neutron stars generate Hawking blackbody radiation?

No neutron stars do not emit blackbody radiation, though they do emit other forms of radiation such as electromagnetic.

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1 hour ago, Mordred said:

No neutron stars do not emit blackbody radiation, though they do emit other forms of radiation such as electromagnetic.

I'm confused. Blackbody radiation is electromagnetic. Alpha, beta, EM are forms of radiation. Blackbody describes the spectrum and something about the process (i.e. it's thermal).

If neutron stars emit non-thermal radiation, it would not be blackbody.

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Anything that has a temperature, has to emit black body radiation.

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9 minutes ago, MigL said:

Anything that has a temperature, has to emit black body radiation.

But that's not a type of radiation. As I said, it describes the spectrum, and indicates its origin is thermal. The distinction between blackbody and electromagnetic is, AFAIK, not correct. Blackbody radiation from the sun, for example, is EM radiation.

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31 minutes ago, MigL said:

Anything that has a temperature, has to emit black body radiation.

Only if it approximates a black body

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I agree, a neutron star would emit EM radiation, based on its temperature, in accordance with Black body radiation spectrum.
As to whether it would also emit non-thermal ( non-lack body, but consisting if alpha, beta, gamma and other particles ) radiation also, I'm sure there is some, but would have to look into it.

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39 minutes ago, MigL said:

I agree, a neutron star would emit EM radiation, based on its temperature, in accordance with Black body radiation spectrum.
As to whether it would also emit non-thermal ( non-lack body, but consisting if alpha, beta, gamma and other particles ) radiation also, I'm sure there is some, but would have to look into it.

I was thinking more of the pulsar aspects of a rotating neutron star. EM, but not thermal.

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That would be true for rotation powered pulsars or magnetars.
But most accretion powered pulsars emit x-rays in relation to the 'temperature' of the accretion disk.
Also , pulsars are thought to be one of the sources for high energy cosmic rays ( non-EM )

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Posted (edited)
7 hours ago, swansont said:

I'm confused. Blackbody radiation is electromagnetic. Alpha, beta, EM are forms of radiation. Blackbody describes the spectrum and something about the process (i.e. it's thermal).

If neutron stars emit non-thermal radiation, it would not be blackbody.

I should have been clearer, whether or not one can accurately describe the surface temperature as a blackbody is a topic under considerable debate. The restriction I'm aware of is the Geminga effect. Here is a relevant paper

It was 5 am when I posted on the way to work so was a bit rushed. Anyways I've seen numerous papers over the years that indicate treating the surface temperature of a neutron star as a blackbody is unrealistic.

A black hole though is a near perfect blackbody while it's questionable if a neutron star can be accurately approximated with a blackbody. As as I understand the debate.

Edited by Mordred

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