How do you find the wave function of a particle in a black hole?

Do you just use Schrodinger's Equation?

Don't our formulas break down at singularities?

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?

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.

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?

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.

The warp in space-time approaches infinity as you approach the center of density; like an asymtote on a graph.

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)

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.

0 space?

no space, they are a point, on account of the fact that nothing can stop them from collapsing any more.

What would happen if a singualrity were (hypothetically of course) to collapse?

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.

It's been proven that you can't have an exposed singularity.

Originally posted by MrL_JaKiri

It's been proven that you can't have an exposed singularity.

 

What's an "exposed singularity"?

Originally posted by Deslaar

A singularity is already collapsed, that's why it's a singularity.

 

What is the radius of a singularity?

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.

Originally posted by Deslaar

What's an "exposed singularity"?

 

A singularity without an event horizon.

Originally posted by MrL_JaKiri

A singularity without an event horizon.

 

I thought so, yes that's an impossibility.

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").

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

Originally posted by MajinVegeta

Stephen Hawking

 

don't mention hawking... MrL gets all upset.

As promised...

 

http://xxx.lanl.gov/PS_cache/gr-qc/pdf/0212/0212076.pdf

 

That should keep you busy for a while!

 

Tom

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?

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