Nano Posted June 12, 2010 Share Posted June 12, 2010 As far as I've understood, the whole supernova process starts with that the core is collapsing when the gravitational force is grater than the electron degeneracy pressure. This collapse happens in a very brief amount of time, and it reach to a sudden halt when the neutron-pressure and the weak force is holding back gravity for collapsing even further. Apparently it is this "halt" which is triggering the outer layers to explode. My question is how do the core and the outer layers interract? Is it particles which is carrying the kinetic energy from the core to the outer layers? Link to comment Share on other sites More sharing options...

Martin Posted June 14, 2010 Share Posted June 14, 2010 (edited) It is a good thing to be asking about. Many people here can help by giving some more details. I can only start us off. There are several kinds of supernova. You are talking about the core-collapse type. There is another sort called "Type Ia" which involves a binary star, a red giant with a small white dwarf companion. It is the small companion which explodes. I think you don't want to learn about all types of supernovas, and the different mechanisms by which they explode. You just want to look at the typical core-collapse case. In that case you have a big star with has done all the fusion it can, meaning that it's largely IRON at the center. Iron is the final endpoint of the series of energy-producing fusion reactions. HERE'S the section on core collapse in the Wikipedia article on supernovae ==quote== The core collapses in on itself with velocities reaching 70,000 km/s (0.23c),[62] resulting in a rapid increase in temperature and density. The energy loss processes operating in the core cease to be in equilibrium. Through photodisintegration, gamma rays decompose iron into helium nuclei and free neutrons, absorbing energy, whilst electrons and protons merge via electron capture, producing neutrons and electron neutrinos, which escape. In a typical Type II supernova the newly formed neutron core has an initial temperature of about 100 billion kelvin (100 GK), 6000 times the temperature of the sun's core. A further release of neutrinos carries away much of the thermal energy, allowing a stable neutron star to form (the neutrons would "boil away" if this cooling did not occur).[63] These 'thermal' neutrinos form as neutrino-antineutrino pairs of all flavors, and total several times the number of electron-capture neutrinos.[64] About 10^46 joules of gravitational energy—approximately 10% of the star's rest mass—is converted into a ten-second burst of neutrinos, which is the main output of the event.[57][65] These carry away energy from the core and accelerate the collapse, while some neutrinos may later be absorbed by the star's outer layers to provide energy to the supernova explosion.[66] The inner core eventually reaches typically 30 km diameter,[57] and a density comparable to that of an atomic nucleus, and further collapse is abruptly stopped by strong force interactions and by degeneracy pressure of neutrons. The infalling matter, suddenly halted, rebounds, producing a shock wave that propagates outward. Computer simulations indicate that this expanding shock does not directly cause the supernova explosion;[57] rather, it stalls within milliseconds[67] in the outer core as energy is lost through the dissociation of heavy elements, and a process that is not clearly understood is necessary to allow the outer layers of the core to reabsorb around 10^44 joules of energy, producing the visible explosion.[68] Current research focuses upon a combination of neutrino reheating, rotational and magnetic effects as the basis for this process.[57] ==endquote== Elements lighter than iron can fuse and release energy. Iron is the complete dead end of fusion. Once you get up to iron you can't get any more fusion energy. A star needs to be producing heat at the core in order to support its outer layers. It is a war of heat pressure against the squeeze of gravity. Running out of fusion fuel in the core is dangerous. You mentioned the electron degeneracy pressure. So you are thinking about the Chandrasekhar limit. Around 1.4 solar masses. As long as the core has fusion fuel (elements lighter than iron) it can make heat and the heat pressure can support the outer layers. But finally there is just this big inert mass of iron in the center of the core. Approaching the fatal Chandrasekhar limit. Suddenly it is collapsing in free fall! Suddenly the iron is all converting to neutron matter which occupies a billionth of the volume. That's effectively nothing! So the whole ball is in free fall towards its own center. Falling not with the Earth acceleration we are used to but with enormously greater acceleration. Now what energy could "rebound" from that huge collapse and blow off the outer layers? One possible mechanism is a wind of neutrinos. When protons swallow electrons--to become neutron matter--they BURP neutrinos. And in this collapse to neutron matter such a huge wind of neutrinos is produced that, as I recall, it is able to blow away stuff. Even though neutrinos in ordinary numbers are not very interactive and pass thru stuff without exerting any force, in such large numbers they act like a blast of wind. That's how I was taught, and Wikipedia seems to confirm it. But probably the whole core-collapse scenario is not perfectly understood. There may be other mechanisms besides this neutrino blastwave. Edited June 14, 2010 by Martin Link to comment Share on other sites More sharing options...

Nano Posted June 14, 2010 Author Share Posted June 14, 2010 Thanks for a detailed explanation of the SN type II. Now something "rebounds" from that huge collapse and blows off the outer layers. One possible mechanism is a wind of neutrinos. When protons swallow electrons--to become neutron matter--they BURP neutrinos. And in this collapse to neutron matter such a huge wind of neutrinos is produced that, as I recall, it is able to blow away stuff. Even though neutrinos in ordinary numbers are not very interactive and pass thru stuff without exerting any force, in very large numbers they act like a blast of wind. The neutrinos are light, but do indeed have high velocity therefore also momentum. I can understand that their kinetic energy would blast the outer layers of, remembering that it is a vast amount of those small bastards coming from the core. The problem, as you also pointed out, is that neutrinos hardly interact with ordinary matter, which the outer layers consist of. In this very second 50 trillion solar neutrinos passes through your body (Wikipedia) Link to comment Share on other sites More sharing options...

Johanluus Posted June 17, 2010 Share Posted June 17, 2010 "In a typical Type II supernova the newly formed neutron core has an initial temperature of about 100 billion kelvin (100 GK), 6000 times the temperature of the sun's core. A further release of neutrinos carries away much of the thermal energy, allowing a stable neutron star to form (the neutrons would "boil away" if this cooling did not occur)." The above explanation raises further questions for me: Would I be right to deduce that a collapsing typical Type II supernova forms a stable Neutron Star IF the initial mass of the star ~ Chandrasekhar limit. Around 1.4 solar masses. If that is the case what is the critical mass limit required by a star , if it were to collapes into a black hole ? Furthemore what is the final degeneracy force( particle) that we know of that the accelleration would need to overcome(collapse-degenerate)? Finally what would this accelleration be in terms of c? Im not sure if my questions make sense , but there seems to be some fundamental correlations with the Accelleration, mass of star, and Degeneracy Force, to what the star ultimately produces when it 'dies', which I have not yet understood. Can anybody point me into the right direction to get further reading material on these questions? Link to comment Share on other sites More sharing options...

Spyman Posted June 17, 2010 Share Posted June 17, 2010 Can anybody point me into the right direction to get further reading material on these questions? Did you read the Wikipedia article with corresponding links? A type II supernova belongs to a sub-category of cataclysmic variable star known as a core-collapse supernova, which results from the internal collapse and violent explosion of a massive star. The presence of hydrogen in its spectrum is what distinguishes a type II supernova from other classes of supernova explosions. A star must have at least 9 times the mass of the Sun in order to undergo this type of core-collapse. http://en.wikipedia.org/wiki/Type_II_supernova Link to comment Share on other sites More sharing options...

Johanluus Posted June 18, 2010 Share Posted June 18, 2010 yes I have but I am still not clear on some of their their interpretations. To be more specific: Wikipedia's "Classification by Schwarzschild radius:" "if one accumulates matter at normal density (1 g/cm³, for example, the density of water) up to about 150,000,000 times the mass of the Sun, such an accumulation will fall inside its own Schwarzschild radius and thus it would be a supermassive black hole of 150,000,000 solar masses" I can understand why this would happen due to gravity.But why does it end at the Schwarzschild radius. But Calculating force of accelleration just outside the schwarzschild radius of different stellar masses gives enourmous values. Each value differes by many orders of magnitude for different values of M. What is the correlation of the "Schwarzschild radius" and the degeneracy of all matter. And what is the critical factor that determines the raduis. From the formula the Schwarzschild radius is proportional to the mass with a proportionality constant involving the gravitational constant and the speed of light: I do not understand the significance of this radius. To formulate my question better , why would an object need different amounts of energy( using newtons law) to escape a mass inside its "Schwarzschild radius" if the object were just outside the radius? Why is the accelleration different just outside ,but just inside everything breaks down to a singularity.(if that is true)? Link to comment Share on other sites More sharing options...

Spyman Posted June 18, 2010 Share Posted June 18, 2010 I do not understand the significance of this radius. To formulate my question better , why would an object need different amounts of energy( using newtons law) to escape a mass inside its "Schwarzschild radius" if the object were just outside the radius? Why is the accelleration different just outside ,but just inside everything breaks down to a singularity.(if that is true)? Since a Black hole is an object with an Escape Velocity greater than the speed of light, the Schwarzschild radius with Newton gravity is where the Escape Velocity is exactly the speed of light, with General Relativity it's the boundary where all future trajectory paths, for an object placed there, points towards the center. In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero. It is commonly described as the speed needed to "break free" from a gravitational field. http://en.wikipedia.org/wiki/Escape_velocity Formula for the Schwarzschild radius [math] r_{s}=\frac{2Gm}{c^2} [/math] where: [math]r_{s}[/math] is the Schwarzschild radius, [math]m[/math] is the mass of the gravitating object, [math]G[/math] is the gravitational constant = 6.67428×10^{-11} m^{3}kg^{-1}s^{-2}, [math]c[/math] is the speed of light in vacuum = 299792458 m/s. Formula for surface acceleration [math] g=\frac{Gm}{r^2} [/math] where: [math]g[/math] is the acceleration due to gravity, [math]m[/math] is the mass of the gravitating object, [math]G[/math] is the gravitational constant = 6.67428×10^{-11} m^{3}kg^{-1}s^{-2}, [math]r[/math] is the radius of the gravitating object. Formula for Escape velocity [math] v_{e}=\sqrt{\frac{2GM}{r}} [/math] where: [math]v_{e}[/math] is the Escape velocity, [math]M[/math] is the mass of the gravitating object, [math]G[/math] is the gravitational constant = 6.67428×10^{-11} m^{3}kg^{-1}s^{-2}, [math]r[/math] is the distance between the center of the body and the point of calculation. Link to comment Share on other sites More sharing options...

Johanluus Posted June 18, 2010 Share Posted June 18, 2010 [math] r_{s}=\frac{2Gm}{c^2} [/math] - EQ1 math]g=\frac{Gm}{r^2}[/math] - EQ2 Thanks Spyman I am trying to digest all the info . Would it be correct to say that when the radius in EQ2 is equal to its Schwarzschild radius in EQ1 then the accelleration g just outside the radius would have a value of ( c^4/4)? This would be the "minimum" value that g may be at the surface(horizon) of the black hole? I deduce this from the fact that if the gravitating body's raduis were greater than its Schwarzschild radius photons would have enough escape velocity to reach us. (with the assumption that the body is not rotating). Link to comment Share on other sites More sharing options...

Spyman Posted June 18, 2010 Share Posted June 18, 2010 Would it be correct to say that when the radius in EQ2 is equal to its Schwarzschild radius in EQ1 then the accelleration g just outside the radius would have a value of ( c^4/4)? Wouldn't the acceleration at the Event Horizon be: [math] a_{EH}=\frac{c^4}{4Gm} [/math] This would be the "minimum" value that g may be at the surface(horizon) of the black hole? I don't think there is a "minimum" value since the surface acceleration decreases when mass increases and there is no known "maximum" value of mass for Black Holes. I deduce this from the fact that if the gravitating body's raduis were greater than its Schwarzschild radius photons would have enough escape velocity to reach us.(with the assumption that the body is not rotating). This is correct, we can see the Sun because its current radius is greater than its Schwarzschild radius. Link to comment Share on other sites More sharing options...

Johanluus Posted June 21, 2010 Share Posted June 21, 2010 [math] a_{EH}=\frac{c^4}{4Gm} [/math] yes of course , my apologies. "surface acceleration decreases when mass increases " This is suprising , because with Newtons gravitational law , force(hence accelleration) between objects increases with increased mass. [math] F_{EH}=\frac{G M_{1}m_{2}}{r^2} [/math] Is this perhaps because the distance between the "centre" of the black hole and the event horizon i.e the Schwarzschild radius, has also increased due to the mass increase and this affects the force ( accelleration) in Newtons gravitational law, with a inverse square relationship? Reducing the accelleration. Link to comment Share on other sites More sharing options...

Spyman Posted June 21, 2010 Share Posted June 21, 2010 Is this perhaps because the distance between the "centre" of the black hole and the event horizon i.e the Schwarzschild radius, has also increased due to the mass increase and this affects the force ( accelleration) in Newtons gravitational law, with a inverse square relationship? Reducing the accelleration. Yes, thats how I interpret it. Link to comment Share on other sites More sharing options...

Johanluus Posted June 22, 2010 Share Posted June 22, 2010 To summarize then [math] a_{EH}=\frac{c^4}{4Gm} [/math] Given the initial mass M of the black hole already there. The accelleration [math]a_{EH}[/math] can only decrease with time , as matter (energy) is sucked into the EH , with the EH growing proportionately relative to its Schwarzschild radius.(Assuming Hawkin radiation does not exist). One could possibly deduce that a black holes "maximum" [math]a_{EH}[/math] would be at the instant of creation i.e at the collapse of the large star forming the black hole. And as it sucks in matter, the space curvature outside the event horizon would try and "renormalize" as it tends back to flat ecludian space. [math]a_{EH}[/math] tends back to zero. Surely then, there would be a point in time where photons and matter would , then have enough escape velocity to reach outside observers, and not fall into the EH?? What am i missing here? Link to comment Share on other sites More sharing options...

Spyman Posted June 22, 2010 Share Posted June 22, 2010 Surely then, there would be a point in time where photons and matter would , then have enough escape velocity to reach outside observers, and not fall into the EH?? What am i missing here? Maybe the missing piece will be revealed if you try to show at which amount of mass in a Black Hole the calculated escape velocity at its Event Horizon would go below c. Link to comment Share on other sites More sharing options...

Johanluus Posted June 22, 2010 Share Posted June 22, 2010 " at which amount of mass in a Black Hole, the calculated escape velocity at its Event Horizon would go below c" That point is always the Schwarzschild radius no matter what the mass is inside it , not so? My assumption in (post #8) was "when the radius in eq2 is equal to its Schwarzschild radius (eq1)" [math]r_{s}=\frac{2Gm}{c^2}[/math] -eq1 [math]g=\frac{Gm}{r^2}[/math] - eq2 Link to comment Share on other sites More sharing options...

Spyman Posted June 22, 2010 Share Posted June 22, 2010 " at which amount of mass in a Black Hole, the calculated escape velocity at its Event Horizon would go below c" That point is always the Schwarzschild radius no matter what the mass is inside it , not so? Yes, at that radius the escape velocity is always c according to eq.3 in post#7. So why do you think that "there would be a point in time where photons and matter would , then have enough escape velocity to reach outside observers" if the Black Hole is allowed to grow sufficient enough? Assuming the same radius for eq.1 and eq.2 does reveal the acceleration where the escape velocity is c for a BH but inserting any mass into the resulting equation do not change the value of the escape velocity itself at that specific radius. Merged post follows: Consecutive posts mergedMaybe I have misunderstod you... There is huge difference between Newton and Einstein gravity concerning Black Holes. With Newtonian Mechanics it is fully possible to lower an object down below the calculated radius of escape velocity and then pull it right up again, if the rope and puller is strong enough. According to the modern theory of General Relativity it is not possible to force something out of a Black Hole since the geometry inside the Event Horizon don't have any path to the outside. Is that the missing information you are asking about? A black hole, according to the general theory of relativity, is a region of space from which nothing, including light, can escape. It is the result of the deformation of spacetime caused by a very compact mass. Around a black hole there is an undetectable surface which marks the point of no return, called an event horizon. The defining feature of a black hole is the appearance of an event horizon—a boundary in spacetime through which matter and light can only pass inward towards the mass of the black hole. Nothing, including light, can escape from inside the event horizon. The event horizon is referred to as such because if an event occurs within the boundary, light from that event cannot reach an outside observer, making it impossible to determine if such an event occurred. As predicted by general relativity, the presence of a large mass deforms spacetime in such a way that the paths particles take bend towards the mass. At the event horizon of a black hole, this deformation becomes so strong that there are no paths that lead away from the black hole. http://en.wikipedia.org/wiki/Black_hole A dark star is a theoretical object compatible with Newtonian mechanics that, due to its large mass, has a surface escape velocity that equals or exceeds the speed of light. Whether light is affected by gravity under Newtonian mechanics is questionable but if it were, any light emitted at the surface of a dark star would be trapped by the star’s gravity rendering it dark, hence the name. Einstein’s general theory of relativity has yielded more insight into the nature of objects of extraordinary mass. Such objects by modern understanding would be described in more modern terms as black holes. Unlike a modern black hole, the object behind the horizon is assumed to be stable against collapse. http://en.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics) Planetary or lunar escape velocity is sometimes misunderstood to be the speed a powered vehicle (such as a rocket) must reach to leave orbit; however, this is not the case, as the quoted number is typically the escape velocity at the body's surface, and vehicles need never achieve that speed. This barycentric escape velocity is the speed required for an object to leave the planet if the object is simply projected from the surface of the planet and then left without any more kinetic energy input: in practice the vehicle's propulsion system will continue to provide energy after it has left the surface. In fact a vehicle can leave the Earth's gravity at any speed. At higher altitudes, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at that position. As is obvious from the equation, at sufficiently high altitudes this speed approaches 0 as r becomes large. http://en.wikipedia.org/wiki/Escape_velocity Link to comment Share on other sites More sharing options...

Johanluus Posted June 23, 2010 Share Posted June 23, 2010 "So why do you think that "there would be a point in time where photons and matter would , then have enough escape velocity to reach outside observers" if the Black Hole is allowed to grow sufficient enough?" Thanks for your input spyman: The above statment is the crux of my question. The equation for [math]a_{EH}}[/math] (the accelleration at the event horizon (which we made equal to its Schwarzschild radius for simplicity) is given by:(post #12) [math]a_{EH}=\frac{c^4}{4Gm}[/math] We also deduced that "surface acceleration decreases when mass increases " for this equation , the reason being(post #10): "Is this perhaps because the distance between the "centre" of the black hole and the event horizon i.e the Schwarzschild radius, has also increased due to the mass increase and this affects the force ( accelleration) in Newtons gravitational law, with a inverse square relationship? Reducing the accelleration." So my logic says that: As time goes by and the Black hole swallows more and more mass , the accelletation [math]a_{EH}}[/math] which is just outside our Schwarzschild radius from our assumption, will tend to zero. So at some point in time , this accelleration [math]a_{EH}}[/math] will eventually be comparable to say earths gravitational pull on its surface ( 9.8 ms^2) at which photons can easily escape, which implies no more black hole? Link to comment Share on other sites More sharing options...

Spyman Posted June 23, 2010 Share Posted June 23, 2010 (edited) So at some point in time , this accelleration ... will eventually be comparable to say earths gravitational pull on its surface ( 9.8 ms^2) at which photons can easily escape, which implies no more black hole? Earth radius = 6371000 m. and mass = 5.9736×10^{24} kg. which reveals an surface acceleration of ~9.82 m/s^{2}. Mass of a hypothetical Black Hole when assumed that a_{EH} = g_{Earth} is ~3.08×10^{42} kg. (~375000 times heavier than the estimates of Milky Ways Supermassive Black Hole.) Radius for the hypothetical Black Hole's Event Horizon is then ~4.575×10^{15} m. (~250 times wider than the estimates of Milky Ways Supermassive Black Hole.) Escape velocity at Event Horizon is then ~299792458 m/s. which clearly shows that photons won't be able to escape there, even though the gravitational acceleration is as low as Earth levels. ---------- What you seem to be asking is how there can be an Event Horizon with lower surface acceleration than we have here on Earth and still have an escape velocity higher than speed of light. The gravitational acceleration at a distance from the center of a body is the slope of the gravitational field that the massive object imposes on the space surrounding it. Plot of a two-dimensional slice of the gravitational potential in and around a uniform spherical body. http://en.wikipedia.org/wiki/Gravitational_potential If the gravitational well gets bigger, it does not only get deeper it gets wider too. When we enlarge such a field by increasing the mass we can still find a distance from the center where the slope has a equal rate, (as compared to the lower mass), but by changing the displacement in the field to a higher altitude the acceleration itself will also be subjected to a slower rate of change. Essentially, even if the acceleration at the Event Horizon of a extremly massive Black Hole, is as low as on Earth, the acceleration will decrease much slower than the acceleration above Earth does, if we continue to take measurements more and more distant from them. A lightray that leaves Earth don't have to struggle against an acceleration of 9.82 m/s^{2} very far, but a lightray escaping from close above the hypothetical Black Hole's Event Horizon will have to struggle against that acceleration during a very long distance. ---------- Lets take the example above with the hypothetical Black Hole and see what happens when we increase the distance and calculate the acceleration one Earth radius above the surfaces. Acceleration at 6371000 meters above Earth surface is lowered to ~2.46 m/s^{2}. Acceleration at 6371000 meters above BH EH surface is still at ~9.82 m/s^{2}. Conclusion: Black Holes remains Black Holes even if the surface acceleration at the Event Horizon tends to zero when the mass approaches infinity. Edited June 23, 2010 by Spyman LOL Link to comment Share on other sites More sharing options...

Johanluus Posted June 23, 2010 Share Posted June 23, 2010 "What you seem to be asking is how there can be an Event Horizon with lower surface acceleration than we have here on Earth and still have an escape velocity higher than speed of light." Yes that is exactly right "A lightray that leaves Earth don't have to struggle against an acceleration of 9.82 m/s2 very far, but a lightray escaping from close above the hypothetical Black Hole's Event Horizon will have to struggle against that acceleration during a very long distance" Agreed. "If the gravitational well gets bigger, it does not only get deeper it gets wider too." This was the information that i was missing !! -{1} I believe the area under such a (slope/function) would be the total amount of energy needed to escape the gravity well. But what determines the shape of such a slope, or topology. What information / equation relates the "DEPTH" of the gravitational well to its "WIDTH". Conclusion: Black Holes remains Black Holes even if the surface acceleration at the Event Horizon tends to zero when the mass approaches infinity. Based one {1} i can finally understand this. Thanks spyman for your persistance. Link to comment Share on other sites More sharing options...

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