New Theoretical Considerations For The Black Hole System

[Shustaire]

Can you reference a link to that please. Spin 2 would apply to the graviton. Seeing as how gravity displays a spin 2 quadrupole 4 polarity states. As opposed to spin 1 2 polarity states. This is critical for the amplitudes of a function.

 

scalar naturally S=0

Edited July 10, 2018 by Shustaire

[Super Polymath]

Why would you think this? Have you noticed that your describing the ground state of the applicable field ? ie spin 1.... Ths field also lies outside the EH with the EH itself being a boundary. Ie infinite redshift....

 

I also hope your looking at information loss as per degrees of freedom...

 

 

Your theory subjects every particle to being a Singularity of some sort but the Kerr metric would apply to any BH that is spinning in actuality. If an anti-proton is a singularity then the photon sphere would apply to its electron's orbital too.

Again this only applies to a number to the negative  power of 54 meter (sub-planckian singular graviton [as in a singularity] metric) that doesn't represent the inversion of the anti protons singular spherical rindler coordinate but an actual micro black hole smaller than planck particle. These are only in a newly produced proton, while it's in it's initial negative charge before turning into a neutron (evaporation point where only the quasar is left), and then a proton (the scattering of the quasar remnant of that black hole), and then back into an anti-proton.

 

There is only an actual singularity in the anti proton, no other particle has this feature other than the proton/neutron/anti proton except for a planck particle or strangelets which are just heavier particles that are made from the fusion of protons. The lighter particles are all made from compacting photons with innumerable actual infinitesimal black holes all within their inner structures that have incalculable coordinates because they evaporate too quickly, even in electrons. These are the gravitons in your pilot wave. We don't calculate them mathematically, they don't exist long enough, they are pixy dust. 

 

So the photons in the photon sphere of the anti proton aren't the photons of our macroscopic cosmic scale, they are photons in a microverse a miniature reality with an exponentially special relativity

Edited July 10, 2018 by Super Polymath

[Super Polymath]

The whole physics seems to be missed on everyone here. The fact is the physics will not satisfy a stable black hole, it has nothing to do with extending it for creation and annihilation operators, the underlying discovery is that a ground state stable black hole violates the third law of thermodynamics. Indeed, if one wanted to describe the black hole with an actual evaporation time then it would make sense to wish to talk about this. But it had no value to what I was investigating. 

It can exist if the anti-proton (made out of the quasar of the micro bh) was surrounded by positrons. This is an anti-hydrogen atom. 

Edited July 10, 2018 by Super Polymath

[Super Polymath]

Your equation dealt with it being surrounded by electrons as with any hydrogen atom.

[Shustaire]

The whole physics seems to be missed on everyone here. The fact is the physics will not satisfy a stable black hole, it has nothing to do with extending it for creation and annihilation operators, the underlying discovery is that a ground state stable black hole violates the third law of thermodynamics. Indeed, if one wanted to describe the black hole with an actual evaporation time then it would make sense to wish to talk about this. But it had no value to what I was investigating. 

 

Then I am clearly misreading your article in terms of your connection to mass in terms of the relevant section on your reference to Bose-Einstein statistics as well as your entropy to mass relations.

[Vmedvil2]

Can you reference a link to that please. Spin 2 would apply to the graviton. Seeing as how gravity displays a spin 2 quadrupole 4 polarity states. As opposed to spin 1 2 polarity states. This is critical for the amplitudes of a function.

 

scalar naturally S=0

 

Shustaire here is your link for the citation, all of my links are from creditable sources. https://ncatlab.org/nlab/show/string+theory+FAQ

[Vmedvil2]

The third law states that as the entropy approaches a constant value, it must also approach zero point temperatures. A true black hole particle with no Hawking radiation is equivalent to the system having no entropy. 

 

And... if a black hole has no entropy, it cannot have a temperature. And that sort of rings alarm bells in my head because, it was for the same reasons Hawking envisioned the entropy of a black hole not to be zero, and found as a result, it had a corresponding temperature.

 

Yes, BH do have a non zero temperature because of emission of photons basically, because of hawking radiation is why hawking thought that.

Edited July 10, 2018 by VictorMedvil

[Shustaire]

Shustaire here is your link for the citation, all of my links are from creditable sources. https://ncatlab.org/nlab/show/string+theory+FAQ

thanks

[Super Polymath]

A true black hole particle with no Hawking radiation is equivalent to the system having no entropy. 

Not necessarily. As I said a while ago Hawing radiation under the alt to QM (the deleterious di-brane) interpretation I offered up is the result of a de sitter space-matter-time-energy medium made of spherical qubits contacting an anti desitter space-matter-time-energy medium made of spherical qubits (black hole gravitons). The absorption of gravitons may not be enough to counteract the Hawking radiation if they're too small.

 

It would however, be enough for the ground state BH of the schwarzschild radius of a proton being 2.484 to the negative power of 54 meters, that's the EH of the BH in an anti-proton according to what Vmedvil referred to as my 9 dimensional Yau Manifold differential geometry, if that anti-proton is surrounded by positrons

 

The reason I say that if surrounded by positrons it would be hit by heavier gravitons to feed its growth & delay Hawking radiation is because all the bosonic graviton particles composing a pilot wave are being pushed towards the atomic nucleus in the anti-hydrogen element as opposed to being pulled away from it as you would expect in a normal hydrogen atom, allowing for the convergence of mass to fuel the BH at the heart of the anti-proton in the anti-atom's nucleus. 

Edited July 10, 2018 by Super Polymath

[Super Polymath]

And I've told you loads of times before, stop peddling you own pet theories in my threads, I do not read them, and they are always ignored. 

You can't ignore the fact that one is needed in order to distinguish a two state vector formalism from the standard LCDM's interpretation of quantum mechanics which ignores a quintessential differential geometry with probability statistics. 

 

If you can't determine how a system behaves you might as well turn to a luxon metric. But you won't find the answers there, you weren't meant to, QM was designed to be a matter of chance not offering the same level of predictability inherent in GR. 

Edited July 10, 2018 by Super Polymath

[Shustaire]

Interesting article Dubbleosix.

[Shustaire]

The last post is in essence the effect of relativity on redshift for part one. For part two the surface area as it gets smaller follows the thermodynamic laws via temperaature over volume relations.

 

If I understand it correctly the zero entropy is due to observer limits on the EH condition however in the same reference frame as the EH the BH isn't a true zero entropy state. It may appear that way to an observer at infinity

[Super Polymath]

The only class of black hole that could survive my conditions is a new class of dark matter which is incapable of absorbing energy and it's entropy has an origin at the horizon, but its source is not allowed to come from the interior. I had to think about this hard, so if there is such a particle be rest assured, I'll try and prove it.This way the system resists being in zero temperatures while still having an entropy. I'd do this by assuming that the black hole has to be treated like any quantum mechanical system including a Planck-Einstein corrective zero point energy term of [math]\frac{1}{2} \hbar \omega[/math].

The bolded is the paramount in understanding needed to take on the milestone physics has set out for itself in explaining qm with the same precision that it does classically

 

The holy grail is a specific differential geometry you'll have make & plot all starting with the photon because light's inertial reference is quintessential & how it moves by the drag of an innate internal composition of motion-structure

Edited July 10, 2018 by Super Polymath

[Shustaire]

The only class of black hole that could survive my conditions is a new class of dark matter which is incapable of absorbing energy and it's entropy has an origin at the horizon, but its source is not allowed to come from the interior. I had to think about this hard, so if there is such a particle be rest assured, I'll try and prove it.This way the system resists being in zero temperatures while still having an entropy. I'd do this by assuming that the black hole has to be treated like any quantum mechanical system including a Planck-Einstein corrective zero point energy term of [math]\frac{1}{2} \hbar \omega[/math].

 

 

Well under QM a absolute zero is impossible under that formula so that would certainly apply. The point I was trying to make about relativity is that according the an infalling observer, he wouldn't even know he crossed the EH until he is already inside and looking out.

 

If one were to take a potential box ie used in QM analogies for describing waves of particles or strings. The temperature in the same frame as the EH doesn't change temperature it is only to the outside observer that the temperature varies via redshift.

 

Now the physicists involved all know this, so why the zero entropy. If I recall it was based on the detail that due to infinite redshift much like Unruh radiation at the cosmological event horizon which is also a horizon dependent on observer the information concerning entropy is lost from the observer point of view.

 

I've been looking through my spouses collection he used to have a copy of the Unruh paper regarding this

Edited July 10, 2018 by Shustaire

[Vmedvil2]

Right but let us not jump the gun, let me get this clear first, do you understand what I meant in the last post? If not, I will try and articulate this clearer, because the relativistic analogies I have brought forth are no more moot than Einstein's elevator thought experiment. Better yet, these are pure relativistic concepts which seem to have an unusual relationship with each other, in a relativistic sense. 

 

A shorter way to articulate this:

 

There is both a length contraction and increase of heat for:

 

1) The rest observer moving at relativistic speeds, will measure the black hole to be larger than the person at rest, and conditionally, measures it to be cooler. This is a direct analogue of the larger the black hole, the cooler it appears. 

 

2) Alternatively, the black hole moving at relativistic speeds will be space dilated, meaning it corresponds to a smaller black hole and so hotter (in accordance with theory).

 

Have you ever considered that the reason the BH seems to be a a higher temperature with less mass is the fact that more radiation escapes the less massive gravitational well, more particles and virtual photons escape the gravity when the gravity has less mass feeding it, which goes back to the Newtonian principal of F_g=M₂M₁G/R² , The greater the mass the greater the gravitational attraction and the further out V = C , thus more particles are sucked directly in rather than pair producing into photons, thus since less matter escapes it seems to be cooler at higher masses.

[Vmedvil2]

The only class of black hole that could survive my conditions is a new class of dark matter which is incapable of absorbing energy and it's entropy has an origin at the horizon, but its source is not allowed to come from the interior. I had to think about this hard, so if there is such a particle be rest assured, I'll try and prove it.This way the system resists being in zero temperatures while still having an entropy. I'd do this by assuming that the black hole has to be treated like any quantum mechanical system including a Planck-Einstein corrective zero point energy term of [math]\frac{1}{2} \hbar \omega[/math].

That constant is the effect of gravitational zero point energy in my theory I tend to agree that, it is a correct term dubbel, that term appears twice right next to the Einstein Tensor/Energy-Stress Tensor.  ℏω that in any case, is a equivalence constant between GR and QM for Gravity. It is actually that expression plus the Einstein field Equation several times. That can be expressed as the Energy of a Graviton or gravitational wave depending on the ω used as well. That expression, there is definitely a particle that escapes the singularity that causes the photon pair.

 

Edited July 11, 2018 by VictorMedvil

[Shustaire]

I couldn't locate his full article but here is a link covering notes from it.

 

http://blogs.umass.edu/grqft/files/2015/01/Unruh-black-hole-evaporation.pdf

 

also pertinent

 

https://arxiv.org/abs/0710.5373

 

I know your not model building per se, however it also helps me recall numerous pertinent details that went into the thermodynamics. I find the second link a good supplement to understand the first better.

 

I assume you used Naturals units throughout ?

 

[math]c=\hbar=G=k=1[/math]

Edited July 11, 2018 by Shustaire