Pierre Laplace: The Evolution Of Stars
BLACK HOLE
Introduction:
"A luminous star, of the same density as the Earth, and whose diameter should be two hundred and fifty times larger than that of the Sun, would not, in consequence of its attraction, allow any of its rays to arrive at us; it is therefore possible that the largest luminous bodies in the universe may, through this cause, be invisible." -- Pierre Laplace, The System of the World, Book 5, Chapter VI (1798).
Evolution of Stars
1. Clouds of Hydrogen begin condensing into more dense clusters due to gravitation. 2. Eventually the density gets high enough that the Hydrogen begins fusing into Helium. This fusion releases energy, mostly in the form of electromagnetic radiation. Our sun is currently in this phase. …show more content…
Even though the collapse takes a finite amount of time from the reference frame of infalling matter, a distant observer sees the infalling material slow and halt just above the event horizon, due to gravitational time dilation. Light from the collapsing material takes longer and longer to reach the observer, with the light emitted just before the event horizon forms delayed an infinite amount of time. Thus the external observer never sees the formation of the event horizon; instead, the collapsing material seems to become dimmer and increasingly red-shifted, eventually fading …show more content…
He got this result by applying quantum field theory in a static black hole background. The result of his calculations is that a black hole should emit particles in a perfect black body spectrum. This effect has become known as Hawking radiation. Since Hawking's result, many others have verified the effect through various methods. If his theory of black hole radiation is correct, then black holes are expected to emit a thermal spectrum of radiation, and thereby lose mass, because according to the theory of relativity mass is a form of energy (E = mc2). Black holes will shrink and evaporate over time. The temperature of this spectrum (Hawking temperature) is proportional to the surface gravity of the black hole, which for a Schwarzschild black hole is inversely proportional to the mass. Large black holes, therefore, emit less radiation than small black holes.
A stellar black hole of one solar mass has a Hawking temperature of about 100 nanokelvins. This is far less than the 2.7 K temperature of the cosmic microwave background. Stellar mass (and larger) black holes receive more mass from the cosmic microwave background than they emit through Hawking radiation and will thus grow instead of shrink. To have a Hawking temperature larger than 2.7 K (and be able to evaporate), a black hole needs to be lighter than the Moon (and therefore a diameter of less than a tenth of a
Introduction:
"A luminous star, of the same density as the Earth, and whose diameter should be two hundred and fifty times larger than that of the Sun, would not, in consequence of its attraction, allow any of its rays to arrive at us; it is therefore possible that the largest luminous bodies in the universe may, through this cause, be invisible." -- Pierre Laplace, The System of the World, Book 5, Chapter VI (1798).
Evolution of Stars
1. Clouds of Hydrogen begin condensing into more dense clusters due to gravitation. 2. Eventually the density gets high enough that the Hydrogen begins fusing into Helium. This fusion releases energy, mostly in the form of electromagnetic radiation. Our sun is currently in this phase. …show more content…
Even though the collapse takes a finite amount of time from the reference frame of infalling matter, a distant observer sees the infalling material slow and halt just above the event horizon, due to gravitational time dilation. Light from the collapsing material takes longer and longer to reach the observer, with the light emitted just before the event horizon forms delayed an infinite amount of time. Thus the external observer never sees the formation of the event horizon; instead, the collapsing material seems to become dimmer and increasingly red-shifted, eventually fading …show more content…
He got this result by applying quantum field theory in a static black hole background. The result of his calculations is that a black hole should emit particles in a perfect black body spectrum. This effect has become known as Hawking radiation. Since Hawking's result, many others have verified the effect through various methods. If his theory of black hole radiation is correct, then black holes are expected to emit a thermal spectrum of radiation, and thereby lose mass, because according to the theory of relativity mass is a form of energy (E = mc2). Black holes will shrink and evaporate over time. The temperature of this spectrum (Hawking temperature) is proportional to the surface gravity of the black hole, which for a Schwarzschild black hole is inversely proportional to the mass. Large black holes, therefore, emit less radiation than small black holes.
A stellar black hole of one solar mass has a Hawking temperature of about 100 nanokelvins. This is far less than the 2.7 K temperature of the cosmic microwave background. Stellar mass (and larger) black holes receive more mass from the cosmic microwave background than they emit through Hawking radiation and will thus grow instead of shrink. To have a Hawking temperature larger than 2.7 K (and be able to evaporate), a black hole needs to be lighter than the Moon (and therefore a diameter of less than a tenth of a