QM + GR = black holes cannot exist

[OnlyMe]

Prof. Mitch Begelman:]

I think the key assumption is that you are somehow able to remain stationary at the surface of the planet as it becomes a black hole. At the instant that the horizon engulfs the surface of the planet you actually would feel infinite gravitational force --- you only feel zero gravity if you are in free fall. The beam from the laser pointer then undergoes infinite gravitational redshift while rising an infinitesimal distance. This means that a photon carries no energy in an upward direction.[/B]

Sounds almost as if the atomic/subatomic processes that lead to photon emission, cease at or within an event horizon. Does a photon, that carries no energy exist?

Note that this thread began with the title, QM + GR ... and yet,

Most of this discussion and the expert responses tashja has been getting are limited to examining the problem(s) exclusively from GR. Most of those same experts would likely agree that the implied mathematical singularity does not exist in reality and that whatever does exist within the event horizon will have to be explained by some future rendition of quantum theory.

Several times through this discussion I have remembered a past discussion about the core composition of neutron stars. I think it focused on a paper suggesting cubic neutronium at the core, or something to that effect. What continues to come to mind for me is that if that case (cubic neutron), is even close or possible as a state of matter/mass that precedes a colapse to a black hole, the thought follows, that what ever exists as the mass of a black hole likely does not not allow for any photon emission. You would not expect that a core composed of nothing but neutrons, as might be the case with a neutron star, arranged in a manner and under such gravitational pressures that even zitterbewegung motions may be dampened, that photon emission would even occur. In the case of a black hole perhaps photon emission would not even be present.

While it might be reasonable to imagine photons falling into a black hole, is it really reasonable to imagine that there are any photons emitted from the mass of a black hole? Once within the event horizon are there even any photons being emitted? Most of the GR answers including that above, would remain the same for any event occurring at or outside the event horizon.., a photon emitted at the event horizon might be infinitely red shifted to the point of zero energy content, since some form of atomic/subatomic process which could result in photon emissions may be possible... But inside the event horizon is it even reasonable to expect that what we think of as matter and atomic and subatomic processes even continue to exist?

 

[paddoboy]

You would not expect that a core composed of nothing but neutrons, as might be the case with a neutron star

It has been theorised that deep within a Neutron/Pulsar could exist Quark matter.

While it might be reasonable to imagine photons falling into a black hole, is it really reasonable to imagine that there are any photons emitted from the mass of a black hole? Once within the event horizon are there even any photons being emitted?

100% Categorically no. Once inside the EH, all paths lead to the Singularity. Plus the fact that all the mass of the BH, is at the Singularity...The rest is just critically curved spacetime.[/QUOTE]

 

[OnlyMe]

It has been theorised that deep within a Neutron/Pulsar could exist Quark matter.



100% Categorically no. Once inside the EH, all paths lead to the Singularity. Plus the fact that all the mass of the BH, is at the Singularity...The rest is just critically curved spacetime.

The word Singularity is too often associated with "point singularity". Not likely something that exists in reality. So for that last comment above to have any connection with reality, the word Singuarity would require a clarification of definition.

Beyond that I was suggesting that other than as an exercise in logic or extrapolation of the mathematics of GR, it may not be reasonable to expect that photons originating from inside of the event horizon exist at all. That was where I was using the neutron star core as an example. Those processes we associate with the emission of photons may not even exist, inside the event horizon.

GR treats even planets, moons and stars as point masses.., or at the very least as objects of uniform mass distribution, which then leads to the same results as if they were point masses.

Yes the path of any photon falling across the event horizon toward the black hole would be a straight path toward the black hole's center of mass, for a non rotating black hole. There should still be some angular curvature for a rotating black hole, or a real black hole...

 

[paddoboy]

The word Singularity is too often associated with "point singularity". Not likely something that exists in reality. So for that last comment above to have any connection with reality, the word Singuarity would require a clarification of definition.

I like to think of the Singularity as simply the region of spacetime at the quantum/Planck level, where the laws of physics and GR do not apply.

 

[tashja]

OnlyMe, I sent your post (#564) to Prof. Begelman. Here's what he said:

Prof. Begelman:

My colleague Prof. Andrew Hamilton has thought much more about such questions than I have, so perhaps could give more satisfying answers.

According to GR, any observer falling through the horizon notices nothing special about physics, so atomic processes should appear the same. Relative to local spacetime, nothing can travel faster than c, but there is nothing to preclude space itself from being pulled into the black hole at faster than c, relative to any external observer. This is what GR predicts inside the event horizon. The result is that a photon can travel outwards at c relative to local space inside the black hole, yet still be dragged inward so no escape is possible.

If GR is wrong about this and there is a "firewall" or something else bounding the BH instead of a singularity then all bets are off, but I have no expertise to assess these arguments. Hamilton has thought about this and has been arguing with the string theorists about the firewall argument. He has also analyzed the recent papers claiming that BHs can't exist and can explain what he thinks is the fallacy in this argument.

I also contacted Prof. Hamilton and he recommends that you visit this page:

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

And that if you have any questions, to ''please feel free to ask.''

 

[Beaconator]

I like to think of the Singularity as simply the region of spacetime at the quantum/Planck level, where the laws of physics and GR do not apply.

T_uv still contains physical values of density where an infinite mass and zero volume is considered. In other words it does not reach an "undefined" physical value even if every number in the tensor happens to be zero.

 

[paddoboy]

T_uv still contains physical values of density where an infinite mass and zero volume is considered. In other words it does not reach an "undefined" physical value even if every number in the tensor happens to be zero.

???? Infinite mass???
Would you like to give that some proper thought?

 

[Beaconator]

Not when the derivative of Pi*D is still pi and the integration of 0 leaves the possability of a variable, no.

 

[OnlyMe]

OnlyMe, I sent your post (#564) to Prof. Begelman. Here's what he said:


I also contacted Prof. Hamilton and he recommends that you visit this page:

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

And that if you have any questions, to ''please feel free to ask.''

Tashja, Prof. Begelman responce, except for reference to firewalls in the last paragraph seems limited to a description of spacetime as per GR. The point I was attempting to make by emphasizing, the QM + GR, portion of the title was and is that within the event horizon QM effects and conditions would become more important, and may not resemble anything we expect from QM observations outside the event horizon. The GR thought experiment is valid where the conditions are restricted to GR.

Again this is why I tried to use as reference, the core of a neutron star, where matter is not thought to be organized in the same manner as in everyday experience.., and the emission of photons is already, at least changed if not non existent. The QM conditions within a black hole and perhaps even an event horizon, would seem to be even more exotic than those we might expect at the core of a neutron star. My comments were not intended as restricted to a GR thought experiment and I did repeatedly phrase the QM aspect of what might be occurring within an event horizon as, it it reasonable to...., a question. This was not intended as a comment on the implications of the GR thought experiment. It was intended to question whether the matter of a black hole or in at least some cases within the event horizon, would from the context of QM be such that photon emission would occur, at all.

At first glance, Prof. Hamilton's link also appears to be limiting its focus to GR...

I have no issues with the GR thought experiments. My comments were an attempt to at least reintroduce the general implications of the thread title and early posts... It seems that what happens within the event horizon, must involve both the affects of GR and QM... And the QM aspect has not yet been settled... but it seems to me from the neutron star reference, that the emission of photons is at least altered, if not, once again non existent.

 

[QuarkHead]

T_uv still contains physical values of density where an infinite mass and zero volume is considered. In other words it does not reach an "undefined" physical value even if every number in the tensor happens to be zero.

Yes, let me expand on that

Suppose a volume V of spacetime where the mass-energy is nowhere zero i.e. $$T_{\mu \nu}\ne 0 \in V$$. Then given the important caveat that provided the distance that separates the observer from the "boundary "of this volume significantly exceeds its radius, then one may define a mass-energy density function $$D(x,y,z,t)$$ such that the integral

$$\int_V \,D(x,y,z,t}\,dV = \text{constant}$$

Um, something wrong with LaTex rendering

Then by letting $$V \to 0$$ the integral $$\int_V D(x,y,z,t}\,dV \to \infty$$, a singularity - mass-energy density is infinite at that point

 

[river]

Yes, let me expand on that

Suppose a volume V of spacetime where the mass-energy is nowhere zero i.e. $$T_{\mu \nu}\ne 0 \in V$$. Then given the important caveat that provided the distance that separates the observer from the "boundary "of this volume significantly exceeds its radius, then one may define a mass-energy density function $$D(x,y,z,t)$$ such that the integral

$$\int_V \,D(x,y,z,t}\,dV = \text{constant}$$

Um, something wrong with LaTex rendering

Then by letting $$V \to 0$$ the integral $$\int_V D(x,y,z,t}\,dV \to \infty$$, a singularity - mass-energy density is infinite at that point

I don' t see

 

[paddoboy]

I don' t see

I'm actually in agreement.
A BH Singularity need not be infinite...It just has that probability to lead to infinite quantities of spacetime curvature and density.

 

[Beaconator]

Yes, let me expand on that

Suppose a volume V of spacetime where the mass-energy is nowhere zero i.e. $$T_{\mu \nu}\ne 0 \in V$$. Then given the important caveat that provided the distance that separates the observer from the "boundary "of this volume significantly exceeds its radius, then one may define a mass-energy density function $$D(x,y,z,t)$$ such that the integral

$$\int_V \,D(x,y,z,t}\,dV = \text{constant}$$

Um, something wrong with LaTex rendering

Then by letting $$V \to 0$$ the integral $$\int_V D(x,y,z,t}\,dV \to \infty$$, a singularity - mass-energy density is infinite at that point

Very well said sir.

I don' t see

With a radius of zero we immediately imply there is a volume of zero. This is not quite true. The rules for integration ensure that all constants keep their value through the process making the density of a singularity infinite (or nearly infinite as I suspect) as opposed to zero or undefined.

 

[Motor Daddy]

With a radius of zero we immediately imply there is a volume of zero. This is not quite true. The rules for integration ensure that all constants keep their value through the process making the density of a singularity infinite (or nearly infinite as I suspect) as opposed to zero or undefined.

Infinity is not a number like zero. Zero is finite whereas infinity is infinite, two different deals. Zero is not changing, but infinity is. So you have a contradiction with infinity and zero, one changing and the other not, at the same point in space.

 

[QuarkHead]

So zero is different from infinity? Breaking news!!!

Look, when it comes to mathematics, 95% of the time I know exactly what I am talking about. It very looks to me as though Beaconator does too.

So if there is something you don't understand, just ASK. But if you dispute an assertion from a stand-point of ignorance, it doesn't reflect well on you.

For explanation..... consider the function $$f(x)=\frac{1}{x}$$. One may perfectly legitimately say " as $$x \to 0,\,\,f(x) \to \infty$$", even though the world, his wife and their dog knows this function is undefined at $$x =0$$

 

[Motor Daddy]

Look, when it comes to mathematics, 95% of the time I know exactly what I am talking about. It very looks to me as though Beaconator does too.

So 5 out of 100 times (1 out of 20) you don't know what you're talking about when it comes to mathematics? Beaconator too??

 

[QuarkHead]

Cheap shot from a small mind

 

[Motor Daddy]

Cheap shot from a small mind

Well? 1 out of 20? Is that on your resume??

 

[Motor Daddy]

"Dear Sir, I submit to you that I am wrong 1 out of 20 times!"

 

[Motor Daddy]

Cheap shot from a small mind

Funny as hell nonetheless!! Your words, not mine. Bwahahahahaahaha