Timeline for Does conservation of energy make black holes impossible?
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22 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2016 at 6:05 | history | tweeted | twitter.com/StackAstronomy/status/709984027885871104 | ||
| Mar 15, 2016 at 7:13 | comment | added | ProfRob | To understand the energetics of black holes it is necessary to understand the rudiments of GR. They certainly cannot be understood in terms of Newtonian physics and "escape velocities". | |
| Mar 14, 2016 at 23:02 | answer | added | Stan Liou | timeline score: 5 | |
| Mar 14, 2016 at 21:41 | comment | added | James K | Paper on gravitational collapse aei.mpg.de/~rezzolla/lnotes/mondragone/collapse.pdf Suffice to say the maths is hard. | |
| Mar 14, 2016 at 21:04 | comment | added | ckersch | Let us continue this discussion in chat. | |
| Mar 14, 2016 at 20:57 | comment | added | Era | @ckersch I don't know what it means to apply kinetic energy slowly. My point is that the energy differential is the energy required to move between the points. It doesn't matter whether that energy comes from kinetic energy or from somewhere else (as in the case with the bowling ball example: I put energy in by doing the work of pulling the rope, but the bowling ball's KE remains low because it is moving slowly). | |
| Mar 14, 2016 at 20:45 | comment | added | ckersch | @Era Why would the potential energy be finite? The potential energy differential between two points will always exactly equal the kinetic energy required to move between them, regardless of whether you apply that kinetic energy quickly or slowly. | |
| Mar 14, 2016 at 20:40 | comment | added | Era | @ckersch The kinetic energy of the escape velocity at the event horizon is infinite. The point is that you don't need to reach the kinetic energy of the escape velocity in order to escape, you just need as much energy as you gained by accelerating. The "escape energy" is just the potential energy, which is finite. (Edit: minor point- the kinetic energy also depends on the mass of the object, not just the velocity.) | |
| Mar 14, 2016 at 20:36 | comment | added | ckersch | @Era The question, though, is how much energy it gains by falling. Assuming there is no friction, the kinetic energy of your bowling ball will be exactly equal to the potential energy it loses during the fall. For something approaching the event horizon, where escape energy should be infinite, kinetic energy should asymptotically approach infinity. The asymptotic nature of that approach should mean we need to impart arbitrarily large amounts of energy over an arbitrarily small distance to escape. | |
| Mar 14, 2016 at 20:35 | answer | added | Era | timeline score: 0 | |
| Mar 14, 2016 at 20:28 | comment | added | Era | @ckersch I see what you're getting at now. Remember that it's not necessary to reach escape velocity in order to return to your original position after falling. Imagine that I'm some distance away from the earth and release a bowling ball on a rope: after it hits the ground, I can slowly pull it back up, putting in at least as much work (energy) as was gained in kinetic energy by its falling. Escape velocity does not take into account any power source that might be exerting a force on you: it's the velocity at which you would escape if there were no other forces acting on you. | |
| Mar 14, 2016 at 20:25 | comment | added | ckersch | @userLTK Yes, the shell would increase in radius as more matter falls in, which makes sense if the innermost particles are gaining energy (and therefore velocity) from gravitational interaction with the infalling particles. | |
| Mar 14, 2016 at 20:23 | comment | added | ckersch | Interestingly, I think if we assume the spherical black hole is a fermi liquid like a quark-gluon plasma with all possible states being filled, the exclusion principle should make it impossible for incoming fermions to enter or cross the matter shell closest to the event horizon, assuming that shell is filled. | |
| Mar 14, 2016 at 20:22 | comment | added | ckersch | @Era Why not? If a particle falls from arbitrarily far away, all of its potential energy should convert to kinetic energy. The kinetic energy of the particle should then be equivalent to the amount of energy required for it to return to its original position. Reaching the event horizon should require shedding an infinite amount of potential energy, which would accelerate a particle to the 'speed of light', which should be impossible. | |
| Mar 14, 2016 at 20:16 | comment | added | Era | @ckersch I think you're mixing up the escape velocity with the velocity at which matter falls towards the black hole. The matter accelerating towards the black hole can increase in speed immensely, but it doesn't approach the speed of light as it approaches the event horizon. | |
| Mar 14, 2016 at 20:13 | comment | added | ckersch | @Era it seems like, as I get arbitrarily close to the event horizon, I can choose an arbitrarily small distance across which the change in potential energy of a particle increases by an arbitrarily large amount. As escape velocity from a specific depth approaches the speed of light, the kinetic energy of the particle would then approach infinity. | |
| Mar 14, 2016 at 20:12 | comment | added | userLTK | @ckersch So the matter outside the event horizon would move outwards, against gravity as the black hole grows? | |
| Mar 14, 2016 at 20:03 | comment | added | Era | @ckersch But why would it require infinite energy? Accelerating matter to sub-lightspeed velocity requires only finite energy, and the black hole does not accelerate matter to light speed. | |
| Mar 14, 2016 at 19:40 | comment | added | ckersch | @userLTK Sort of. My conclusion was that reaching the event horizon would require infinite energy, so it must be impossible to cross the event horizon, and as such nothing can fall into the black hole and all accumulated matter would collect in a sphere outside the event horizon, which would increase in radius as the event horizon grows. | |
| Mar 14, 2016 at 19:11 | comment | added | userLTK | You're assuming the object falling in actually reaches the speed of light. It doesn't and obviously can't, as that would require an infinite amount of energy. The in-falling velocity is quite a bit less than the speed of light, though you're correct on it taking an infinite amount of energy to pull a particle from the event horizon. This question is related. physics.stackexchange.com/questions/24319/… | |
| Mar 14, 2016 at 18:39 | review | First posts | |||
| Mar 14, 2016 at 18:41 | |||||
| Mar 14, 2016 at 18:38 | history | asked | ckersch | CC BY-SA 3.0 |