MajinVegeta Posted March 2, 2003 Share Posted March 2, 2003 How do you find the wave function of a particle in a black hole? Do you just use Schrodinger's Equation? Link to comment Share on other sites More sharing options...
Raider Posted March 2, 2003 Share Posted March 2, 2003 Don't our formulas break down at singularities? Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 2, 2003 Author Share Posted March 2, 2003 I heard that my aforementioned question is still contrevertial. I thought maybe we could specualte or something. I don't know for sure, do they breakdown at singularities? Link to comment Share on other sites More sharing options...
Deslaar Posted March 2, 2003 Share Posted March 2, 2003 Originally posted by MajinVegeta I heard that my aforementioned question is still contrevertial. I thought maybe we could specualte or something. I don't know for sure, do they breakdown at singularities? I can't conceive of how a wave function for a particle attribute could be preserved in a singularity. Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 3, 2003 Author Share Posted March 3, 2003 Why can't you just use the schrodinger equation? If you new the particle's original velocity, and position, couldn't you know the probable position and velocity inside a singularity/black hole? Link to comment Share on other sites More sharing options...
Deslaar Posted March 3, 2003 Share Posted March 3, 2003 Originally posted by MajinVegeta Why can't you just use the schrodinger equation? If you new the particle's original velocity, and position, couldn't you know the probable position and velocity inside a singularity/black hole? No, because neither position or momentum exist in a singularity. Remember that space is infinitely warped in a singularity. Link to comment Share on other sites More sharing options...
fafalone Posted March 3, 2003 Share Posted March 3, 2003 The warp in space-time approaches infinity as you approach the center of density; like an asymtote on a graph. Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 3, 2003 Author Share Posted March 3, 2003 Originally posted by Deslaar No, because neither position or momentum exist in a singularity. Remember that space is infinitely warped in a singularity. Oh, I get it! So like in a singularity, particles are "piled" on top of one another...?(not rhetorical) Link to comment Share on other sites More sharing options...
Deslaar Posted March 4, 2003 Share Posted March 4, 2003 Originally posted by MajinVegeta Oh, I get it! So like in a singularity, particles are "piled" on top of one another...?(not rhetorical) They occupy 0 space. Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 5, 2003 Author Share Posted March 5, 2003 0 space? Link to comment Share on other sites More sharing options...
Radical Edward Posted March 5, 2003 Share Posted March 5, 2003 no space, they are a point, on account of the fact that nothing can stop them from collapsing any more. Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 14, 2003 Author Share Posted March 14, 2003 What would happen if a singualrity were (hypothetically of course) to collapse? Link to comment Share on other sites More sharing options...
Deslaar Posted March 14, 2003 Share Posted March 14, 2003 Originally posted by MajinVegeta What would happen if a singualrity were (hypothetically of course) to collapse? A singularity is already collapsed, that's why it's a singularity. Link to comment Share on other sites More sharing options...
JaKiri Posted March 14, 2003 Share Posted March 14, 2003 It's been proven that you can't have an exposed singularity. Link to comment Share on other sites More sharing options...
Deslaar Posted March 14, 2003 Share Posted March 14, 2003 Originally posted by MrL_JaKiri It's been proven that you can't have an exposed singularity. What's an "exposed singularity"? Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 14, 2003 Author Share Posted March 14, 2003 Originally posted by Deslaar A singularity is already collapsed, that's why it's a singularity. What is the radius of a singularity? Link to comment Share on other sites More sharing options...
Deslaar Posted March 15, 2003 Share Posted March 15, 2003 Originally posted by MajinVegeta What is the radius of a singularity? A singularity doesn't have radius (unless you say it has a radius of 0) . That's what makes it a singularity. A singularity is infinitely dense because the space it occupies is 0. Are you getting confused with singularities and black holes? Black are products of singularities. A black hole does have a radius. Link to comment Share on other sites More sharing options...
JaKiri Posted March 16, 2003 Share Posted March 16, 2003 Originally posted by Deslaar What's an "exposed singularity"? A singularity without an event horizon. Link to comment Share on other sites More sharing options...
Deslaar Posted March 16, 2003 Share Posted March 16, 2003 Originally posted by MrL_JaKiri A singularity without an event horizon. I thought so, yes that's an impossibility. Link to comment Share on other sites More sharing options...
MajinVegeta Posted March 17, 2003 Author Share Posted March 17, 2003 Are you getting confused with singularities and black holes? Black are products of singularities. A black hole does have a radius. heh, yes, I usually do get singularities and black holes confused. Anyway, what you said leads me to think that singularities DO emit schwartzchild radiation, and therefore evaporate. Has anyone heard of something called a "naked singularity"? It is still strictly hypothesis, as it is still being debated over. (At least that's how its depicted in Stephen Hawking's "The Universe in a Nutshell"). Link to comment Share on other sites More sharing options...
Tom Mattson Posted April 2, 2003 Share Posted April 2, 2003 Originally posted by MajinVegeta Why can't you just use the schrodinger equation? If you new the particle's original velocity, and position, couldn't you know the probable position and velocity inside a singularity/black hole? Hi, it's me Tom from Physics Forums. (I get around a lot ) The curvature of spacetime around a black hole demands a relativistic treatment. Since the Schrodinger equation is nonrelativistic, it cannot be used. You can, however, use relativistic quantum mechanics (RQM), which is the quantized Hamiltonian: H2=(pc)2+(mc2)2 Amazingly, it is quantized according to the same rules as the nonrelativistic case, namely: H-->i(hbar)d/dt p-->-i(hbar)grad RQM in curved spacetime has been worked out. I will come up with some references when I get home from work. Tom Link to comment Share on other sites More sharing options...
Radical Edward Posted April 3, 2003 Share Posted April 3, 2003 Originally posted by MajinVegeta Stephen Hawking don't mention hawking... MrL gets all upset. Link to comment Share on other sites More sharing options...
Tom Mattson Posted April 4, 2003 Share Posted April 4, 2003 As promised... http://xxx.lanl.gov/PS_cache/gr-qc/pdf/0212/0212076.pdf That should keep you busy for a while! Tom Link to comment Share on other sites More sharing options...
MajinVegeta Posted April 26, 2003 Author Share Posted April 26, 2003 Originally posted by Tom Hi, it's me Tom from Physics Forums. (I get around a lot ) Hi! *waves* The curvature of spacetime around a black hole demands a relativistic treatment. Since the Schrodinger equation is nonrelativistic, it cannot be used. You can, however, use relativistic quantum mechanics (RQM), which is the quantized Hamiltonian: H2=(pc)2+(mc2)2 Amazingly, it is quantized according to the same rules as the nonrelativistic case, namely: H-->i(hbar)d/dt p-->-i(hbar)grad RQM in curved spacetime has been worked out. I will come up with some references when I get home from work. Tom I don't understand the use of the hamiltonian constant. is "p" poise or momentum? Link to comment Share on other sites More sharing options...
Tom Mattson Posted April 26, 2003 Share Posted April 26, 2003 Originally posted by MajinVegeta I don't understand the use of the hamiltonian constant. In classical mechanics, the Hamiltonian is the total energy. In quantum mechanics, it becomes an operator. When that operator acts on the wavefunction, you get the Schrodinger equation. is "p" poise or momentum? p=momentum Tom Link to comment Share on other sites More sharing options...
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