[Bored chemist]

This is incorrect

Yes; what you say is incorrect.

On the other hand, the point about the 1/r^3 dependence of force on distance from a magnetic dipole is correct.
https://en.wikipedia.org/wiki/Dipole#Magnitude

Well, you had confused me with that message about magnetic filed.

You're still confused.

[Dave Lev (OP)]

On the other hand, the point about the 1/r^3 dependence of force on distance from a magnetic dipole is correct.
https://en.wikipedia.org/wiki/Dipole#Magnitude

Is it real?
Do you have an understanding which formula we need to use?

Based on Kryptid message:

Magnetic field strength is related to the strength of the electric current used to produce it.

So, we need to look for a formula which is based on current.
The following formula is the ultimate solution for that:

http://wikipremed.com/01physicscards600/371a.gif
B is affected by 1/r, while the gravity force is affected by 1/r^2

B = μ * I /  ( 2 π r)

Therefore
B is affected by 1/r

In any case, let's look at your formula.
We discuss hear on Dipole which consists of two equal and opposite point charges:
https://en.wikipedia.org/wiki/Dipole#Magnitude
"A physical dipole consists of two equal and opposite point charges: in the literal sense, two poles. Its field at large distances (i.e., distances large in comparison to the separation of the poles) depends almost entirely on the dipole moment as defined above"
This doesn't fully represent our case
However, even in this case, the formula is:

B =  ( μ/4π) * (3(m r̂ ) r̂  - m)/R^3
where

B is the field
R is the vector from the position of the dipole to the position where the field is being measured
r is the absolute value of R: the distance from the dipole
r̂ = R / r is the unit vector parallel to R
m is the (vector) dipole moment
μ0 is the permeability of free space

if we assume that r̂ ^ 2  is bigger than m we get

B = ( μ/4π) * (3(m r̂ ) r̂ /R^3

If we assume that any r = r̂  = R represents absolute value of R
r = r̂  = R
we actually get

B= ( μ/4π) * 3m/r
So again B is affected by 1/r

In any case, the message from Kryptid is very clear:

Magnetic field strength is related to the strength of the electric current used to produce it.

So, we must use:

B = μ * I /  ( 2 π r)

Hence

B is affected by 1/r

I think that at this stage the science community should offer me a reward for my discovery!

[Bored chemist]

The force produced by a magnetic dipole falls as 1/r^3

This will remain true, no matter how often you pretend otherwise.

[Kryptid]

Do you agree that the dynamo at the Sun is more massive than the one on earth and therefore, the Electric current there is higher so it can generate Higher magnetic field?

Yes, but the fact that it has more mass isn't what causes the field strength to be higher.

The idea that our scientists don't know how the SMBH works, doesn't mean that it has no core.
We need to look around it.
If we see flares next to the SMBH and we know for sure that similar flares around the Sun are due to the Sun' magnetic field, than it is clear that those flares also there due to the SMBH' magnetic field.
As the accretion disc is located at the equatorial  "where the polarity of the SMBH's magnetic field changes from north to south it proves that it is there due to the SMBH' magnetic field.
You have also agreed that the molecular jet stream is due to Magnetic field. As its velocity (0.8c) and its length is 27,000Ly above/below the SMBH poles.

And... those didn't answer my questions at all. So let me ask again: How do you know that a SMBH even has a core? If it does, how do you know it has electric currents in it? What is it made of?

it is clear that it is due to the SMBH' magnetic field

Not clear at all, actually. You have to show that the magnetic field is due to the black hole and not due to the accretion disk. We know for a fact, based on the laws of physics, that the accretion disk must generate a magnetic field. So saying that the black hole makes one too is redundant unless you can provide some actually evidence for there being one there.

Well, it seems to me that you have a mistake.

That, again, did absolutely nothing to answer my question. So tell me, how can the core withstand the black hole's gravity without collapsing?

Hence, at the accretion disc near the SMBH, the gravity force is much stronger and therefore it can hold the new created particles at a circular orbiting cycle around the SMBH.

Those particles form at the event horizon, which you say is at "short range" and thus the particles must have a decaying orbit, not an expanding one.

Hence, the plasma there is so hot 10^9K. The conditions there transform those new created particles into real atoms and Molecular.

Where did the plasma come from? Remember, there is no accretion disk when we start things off. If you have to have a hot accretion disk in order to create atoms and molecules for the accretion disk to exist, then, well, you've created a catch-22.

Now that we know how it really works, we do understand how it had been created at the first phase.

Then tell me how, because you haven't explained it yet.

I think that at this stage the science community should offer me a reward for my discovery!

You haven't discovered a single thing.

[Dave Lev (OP)]

How do you know that a SMBH even has a core? If it does, how do you know it has electric currents in it? What is it made of?

All of those questions are none relevant as we can't see it from inside and we would never ever see it.
We can just see what is going around it.
So please try to answer the following:
1. Do you agree that the activity of the Sun' magnetic field could give us some indication on the SMBH' magnetic filed activity?
2. Solar flares
Do you agree that the Flares around the Sun are due to the Sun magnetic Field?
https://www.space.com/11506-space-weather-sunspots-solar-flares-coronal-mass-ejections.html
"The high magnetic fields in the sunspot-producing active regions also give rise to explosions known as solar flares. When the twisted field lines cross and reconnect, energy explodes outward with a force exceeding that of millions of hydrogen bombs. [The Sun's Wrath: Worst Solar Storms in History]"
So, Why the flares around the SMBH couldn't be due to the SMBH' magnetic filed?
3. Hot Plasma of 100MK at the Sun' Corona
Do you argree that the Hot plasma at the Corona around the Sun is due to the Sun magnetic field?
https://www.space.com/11506-space-weather-sunspots-solar-flares-coronal-mass-ejections.html
Temperatures in the outer layer of the sun, known as the corona, typically fall around a few million kelvin. As solar flares push through the corona, they heat its gas to anywhere from 10 to 20 million K, occasionally reaching as high as 100 million K. According to NASA, the energy released in a solar flare "is the equivalent of millions of 100-megaton hydrogen bombs exploding at the same time."
So, why the hot plasma in the accretion disc (which is estimated at a range of 100MK to 10^9K) around the SMBH couldn't be due to the SMBH' magnetic filed?

[Bored chemist]

All of those questions are none relevant as we can't see it from inside and we would never ever see it.
We can just see what is going around it.

They are relevant because you say it's the core that generates the magnetic field.
To say that, you need to show that there's a core.

1. Do you agree that the activity of the Sun' magnetic field could give us some indication on the SMBH' magnetic filed activity?

No
Because the Sun isn't a point, but a black hole is.
.

So, why the hot plasma in the accretion disc (which is estimated at a range of 100MK to 10^9K) around the SMBH couldn't be due to the SMBH' magnetic filed?

Well, for a start, there's no credible mechanism here for a BH to have a field.

Do you agree that the Flares around the Sun are due to the Sun magnetic Field?

So what?

So, why the hot plasma in the accretion disc (which is estimated at a range of 100MK to 10^9K) around the SMBH couldn't be due to the SMBH' magnetic filed?

Why not just follow sensible physics and saty it's due to stuff falling in?

[Kryptid]

All of those questions are none relevant as we can't see it from inside and we would never ever see it.

If you want the black hole to have a magnetic field, it is very much relevant.

The laws of physics actually won't let such a core exist. Nothing can travel away from a black hole's center if it is already inside of the hole, not even if the signal is moving at the speed of light. Matter's ability to support itself against pressure is due to the fact that each particle can "feel" the other particles around it using one of the four fundamental forces. However, those particles can only transmit forces between each other at the speed of light at the very most. So particles in the center of the core therefore cannot transmit a force to those into the outer region of the core because such a signal cannot move away from the black hole's center. If the force cannot be transmitted, then there will be nothing supporting those particles in the outer layer against collapse.

1. Do you agree that the activity of the Sun' magnetic field could give us some indication on the SMBH' magnetic filed activity?

No. Black holes are not made of plasma like the Sun is.

Do you agree that the Flares around the Sun are due to the Sun magnetic Field?

Yes.

Do you argree that the Hot plasma at the Corona around the Sun is due to the Sun magnetic field?

Last time I checked, the reason that the Sun's corona is so much hotter than its surface is still a mystery. That might have changed recently, but if so, I haven't heard about it.

So, why the hot plasma in the accretion disc (which is estimated at a range of 100MK to 10^9K) around the SMBH couldn't be due to the SMBH' magnetic filed?

It's not that it couldn't be due to the black hole's magnetic field so much as there is no need to invoke one in order to explain what we observe. A magnetic field generated by the accretion disk is enough. We know from the laws of physics that an accretion disk must create a magnetic field. By contrast, the laws of physics do not require a black hole to have a magnetic field.

I also noticed you still haven't addressed the fact that there is no way for a black hole to create an accretion disk in your model. Is the magnetic field impassible? If so, then particles can't get out in order to form an accretion disk. If it isn't impassible, then matter can fall into the black hole from inside.

[Dave Lev (OP)]

Matter's ability to support itself against pressure is due to the fact that each particle can "feel" the other particles around it using one of the four fundamental forces

Based on Theory D the BH/SMBH is full with Particles (without a single atom).
Those particles can feel each other. Therefore, their "shell" actually touch each other back to back.

However, those particles can only transmit forces between each other at the speed of light at the very most. So particles in the center of the core therefore cannot transmit a force to those into the outer region of the core because such a signal cannot move away from the black hole's center. If the force cannot be transmitted, then there will be nothing supporting those particles in the outer layer against collapse.

Any particle and any atom is a pure cell of energy.
That energy represents its mass.
If you break the particle shell, its energy is lost and therefore it should also lose its total mass.
Hence, if particles at the core of the BH would collapse, they would lose their energy and their mass would be ZERO.
Therefore, the assumption that the BH or a SMBH is a point is a severe mistake.

Because the Sun isn't a point, but a black hole is.

So, that statement is clearly incorrect.

Black holes are not made of plasma like the Sun is.

The BH is made out of Particles while the Sun is made out of atoms (mainly hydrogen)
This is the main difference between the two objects.
So, what could be the difference in their size?
Let's assume that the BH is full with nucleus and verify the difference:
https://en.wikipedia.org/wiki/Atomic_radius
Under most definitions the radii of isolated neutral atoms range between 30 and 300 pm (trillionths of a meter), or between 0.3 and 3 ångströms. Therefore, the radius of an atom is more than 10,000 times the radius of its nucleus (1–10 fm),[2] and less than 1/1000 of the wavelength of visible light (400–700 nm).
So, the difference between the nucleus to its atom is one to 10,000.
Hence, if we would set the entire nucleus in the BH in a row, their total length should be 1/10,000= 1/10^4 with regards to the same quantity of atoms.
However, as we discuss on a sphere, the ratio should be (1/10^4)^3 = 1/10^7.
Hence, the size of our SMBH with its 4*10^6 sun mass is
4/10 = 0.4 Sun size
So simple and clear.
Do you agree with that?

[Kryptid]

Based on Theory D the BH/SMBH is full with Particles (without a single atom).
Those particles can feel each other. Therefore, their "shell" actually touch each other back to back.

Then your model violates relativity (yet again). That's the only way that force can propagate outward in a black hole: by moving faster than light.

If you break the particle shell, its energy is lost and therefore it should also lose its total mass.

There is no evidence that particles have shells, so that statement doesn't make any sense.

Any particle and any atom is a pure cell of energy.

An unevidenced statement. Particles are not made of energy. That's no more valid than saying that particles are made of charge or spin. Energy is a property of particles.

Let's assume that the BH is full with nucleus

Let's not, because that isn't possible. Nuclear degeneracy pressure is insufficient to support a nucleus against collapse inside of a black hole. That's why neutron stars can only become so massive (that limit is a bit more than twice the mass of the Sun). If they are any heavier, nuclear degeneracy pressure is overwhelmed and the neutron star is forced to collapse into a black hole.

Do you agree with that?

No. The radius of a black hole is defined as the distance at which its escape velocity reaches the speed of light (obviously, since light can't escape black holes by definition). That is also called the Schwarzschild radius. For a black hole with a mass 4 million times the mass of the Sun, the escape velocity reaches the speed of light at a radius of about 12 million kilometers. That's significantly larger than the Sun's radius (which is only about 695,700 kilometers). So when you say:

Hence, the size of our SMBH with its 4*10^6 sun mass is
4/10 = 0.4 Sun size

It is wrong by a large margin.

So simple and clear.

So simple and wrong, just like your "Model D" (I'm not going to call it "Theory D", because scientific theories don't invoke literal miracles like you did when said your model requires a miracle for a black hole to generate an accretion disk).

[Bored chemist]

So simple and clear.

And so wrong.
You really need to look up what black holes are like.
Since there is nothing which is strong enough to counter their gravity, they collapse down to point sized objects.
Maybe they are as big as the planck length, but they certainly are not bigger.

You really should have listened when I pointed this out here

Because it hasn't got a core; it's a point.

Then you wouldn't have wasted further time on it.

[Bored chemist]

No. The radius of a black hole is defined as the distance at which its escape velocity reaches the speed of light (obviously, since light can't escape black holes by definition). That is also called the Schwarzschild radius.

That's the size of the event horizon of the hole, not of the hole itself, which is a singularity.

[Dave Lev (OP)]

There is no evidence that particles have shells, so that statement doesn't make any sense.

Any particle has some sort of volume/size. It is a key element in its chemical properties:
https://en.wikipedia.org/wiki/Particle
In the physical sciences, a particle (or corpuscule in older texts) is a small localized object or entity to which can be ascribed several physical or chemical properties such as volume, density or mass.[1][2] They vary greatly in size or quantity, from subatomic particles like the electron, to microscopic particles like atoms and molecules, to macroscopic particles like powders and other granular materials.
Today we can even estimate the size of electron:
https://arxiv.org/pdf/0709.3041.pdf
"Experimentally, we know (now) that the “size” of the electron is small, Re <∼ 10^−17 cm"
So, if it has a volume it also must have size or some sort of shell.
You can't squeeze it to zero.
If you try to do so, you would break down the Particle/subatomic particle

With regards to proton:
It is considered as subatomic particle and it has a size:
https://en.wikipedia.org/wiki/Proton
"Because protons are not fundamental particles, they possess a measurable size; the root mean square charge radius of a proton is about 0.84–0.87 fm (or 0.84×10−15 to 0.87×10−15 m)"
https://en.wikipedia.org/wiki/Quark
A proton is composed of two up quarks, one down quark, and the gluons that mediate the forces "binding" them together.
The mass of the up quark is - 2.2 Mev/c^2
The mass of the down quark is - 4.7 Mev/c^2
Therefore, the total mass in the three quarks is - 2.2 +2.2 + 4.7 = 9.1 Mev/c^2
However, the mass of a proton is - 938 Mev/c^2
So, how could it be that those 3 quarks could set a proton with mass of  938 Mev/c^2?
The extra mass is coming by gluons
https://en.wikipedia.org/wiki/Proton#/media/File:Quark_structure_proton.svg
"The quark structure of the proton. There are two up quarks in it and one down quark. The strong force is mediated by gluons (wavey)."
So, the total mass of the gluons is a proton is:
938-9.1 / 938 * 100% = 99%
Hence,
The root mean square charge radius of a proton is about 0.84–0.87 fm (or 0.84×10−15 to 0.87×10−15 m)
if that proton collapse, we might get there quarks out of it at total mass of 9.1 Mev/c^2, but it must lose the gluons which represents 9% of its total mass.
Therefore, the assumption that particles/protons could be collapsed and still maintain their original mass is a pure fiction.
The ONLY way to keep the proton' mass is by keeping its "charged radius".
Therefore, any object (even if you call it - BH) with real radius of Zero would carry a mass of Zero.

You really need to look up what black holes are like.
Since there is nothing which is strong enough to counter their gravity, they collapse down to point sized objects.
Maybe they are as big as the planck length, but they certainly are not bigger.

If BH is at the size of Planck length, then you need to verify how many Particles/quarks you could fit in this size.
This will give you the total mass in that Planck length - which should be close to zero.

No. The radius of a black hole is defined as the distance at which its escape velocity reaches the speed of light (obviously, since light can't escape black holes by definition). That is also called the Schwarzschild radius. For a black hole with a mass 4 million times the mass of the Sun, the escape velocity reaches the speed of light at a radius of about 12 million kilometers. That's significantly larger than the Sun's radius (which is only about 695,700 kilometers). So when you say:

Quote from: Dave Lev on Today at 05:58:34
Hence, the size of our SMBH with its 4*10^6 sun mass is
4/10 = 0.4 Sun size

It is wrong by a large margin.

So do you claim that the radius of the SMBH is 12 MK?
How could it be that a radius of Planck length had been shifted to 12MK?

. The radius of a black hole is defined as the distance at which its escape velocity reaches the speed of light (obviously, since light can't escape black holes by definition). That is also called the Schwarzschild radius.

That's the size of the event horizon of the hole,

Yes, I agree with that.

[Kryptid]

Any particle has some sort of volume/size. It is a key element in its chemical properties:

Atoms have a size. The fundamental particles that make them up (quarks and electrons), not necessarily so.

"Experimentally, we know (now) that the “size” of the electron is small, Re <∼ 10^−17 cm"

That represents an upper limit of the electron's size, not its actual size. Electrons could be point particles for all we know.

So, if it has a volume it also must have size or some sort of shell.

Non-sequitur. A shell is not implied by there being a size.

You can't squeeze it to zero.

If electrons are point particles, then they already have a size of zero.

if that proton collapse, we might get there quarks out of it at total mass of 9.1 Mev/c^2, but it must lose the gluons which represents 9% of its total mass.

You can't have free quarks, so this is wrong.

Therefore, the assumption that particles/protons could be collapsed and still maintain their original mass is a pure fiction.

Non-sequitur.

How could it be that a radius of Planck length had been shifted to 12MK?

Nobody said that it was. You are mistaking the singularity for the event horizon.

[Dave Lev (OP)]

Quote from: Dave Lev on Yesterday at 19:35:42
"Experimentally, we know (now) that the “size” of the electron is small, Re <∼ 10^−17 cm"

That represents an upper limit of the electron's size, not its actual size. Electrons could be point particles for all we know.

Even if this value represents its upper limit, don't you agree that there must be also lower limit?
If the electron collapse below that lower limit how can we still consider it as electron?
Could it be that your assumption that Electrons could be a zero point particle is incorrect?

Quote from: Dave Lev on Yesterday at 19:35:42
Any particle has some sort of volume/size. It is a key element in its chemical properties:

Atoms have a size. The fundamental particles that make them up (quarks and electrons), not necessarily so.

Our scientists claim that the size of the Atom is bigger by 10,000 times with regards to the size of its nucleus.
So, as you confirm that atom has a size then you also must confirm that its nucleus MUST has a size.
A proton is located in that Nucleus.
I have offered full explanation about Proton:

https://en.wikipedia.org/wiki/Quark
A proton is composed of two up quarks, one down quark, and the gluons that mediate the forces "binding" them together.
The mass of the up quark is - 2.2 Mev/c^2
The mass of the down quark is - 4.7 Mev/c^2
Therefore, the total mass in the three quarks is - 2.2 +2.2 + 4.7 = 9.1 Mev/c^2
However, the mass of a proton is - 938 Mev/c^2
So, how could it be that those 3 quarks could set a proton with mass of  938 Mev/c^2?
The extra mass is coming by gluons
https://en.wikipedia.org/wiki/Proton#/media/File:Quark_structure_proton.svg
"The quark structure of the proton. There are two up quarks in it and one down quark. The strong force is mediated by gluons (wavey)."
So, the total mass of the gluons is a proton is:
938-9.1 / 938 * 100% = 99%
Hence,
The root mean square charge radius of a proton is about 0.84–0.87 fm (or 0.84×10−15 to 0.87×10−15 m)
if that proton collapse, we might get there quarks out of it at total mass of 9.1 Mev/c^2, but it must lose the gluons which represents 9% of its total mass.
Therefore, the assumption that particles/protons could be collapsed and still maintain their original mass is a pure fiction.
The ONLY way to keep the proton' mass is by keeping its "charged radius".
Therefore, any object (even if you call it - BH) with real radius of Zero would carry a mass of Zero.

It is stated clearly that "The root mean square charge radius of a proton is about 0.84–0.87 fm (or 0.84×10−15 to 0.87×10−15 m)"?
We also understand from the article that the Gluons is a pure energy which holds the quarks in order to form a proton.
Hence, what could be the outcome if it collapses much below its minimal size?
How the Gluons could still exist if we disconnect it from the three quarks?
Don't you agree that at the moment that we break the structure of the proton, the Gluons is lost forever?
At that moment 99% of the proton mass (938 Mev/c^2) had been lost.
Therefore, the left over from that proton are the three quarks with only 9.1 Mev/c^2.
However, you claim:

You can't have free quarks, so this is wrong.

So, there is no way to break the proton.
In any case, if you believe that a proton could carry mass at a size of zero, then why atom couldn't carry mass at size of zero?
Please remember, 10,000 times of zero is still zero.
So, do you agree that a proton MUST have a minimal size?
Please - yes or no?

[Bored chemist]

don't you agree that there must be also lower limit?

Yes, zero.

If the electron collapse below that lower limit how can we still consider it as electron?

Because.if that's how big it is then  that's how big it is.

Could it be that your assumption that Electrons could be a zero point particle is incorrect?

All the evidence supports the idea.
It could be wrong, but there's no reason to think it is.

So, as you confirm that atom has a size then you also must confirm that its nucleus MUST has a size.

Nobody has disputed this; why raise it?

I have offered full explanation about Proton:

No you have not; mainly you quoted WIKI.

So, there is no way to break the proton.

And yet we do, at places like CERN.

Again it's the familiar idea that, when reality  disagrees with you. it's not reality which is mistaken

[Kryptid]

Even if this value represents its upper limit, don't you agree that there must be also lower limit?

Yes, that lower limit is zero.

If the electron collapse below that lower limit how can we still consider it as electron?

You can't collapse below zero.

Could it be that your assumption that Electrons could be a zero point particle is incorrect?

Yes. Superstring theory posits that electrons are actually tiny strings with a length on the order of the Planck length. If that is true, then they could not collapse below that length and we would expect black holes not to collapse below that length either.

So, as you confirm that atom has a size then you also must confirm that its nucleus MUST has a size.
A proton is located in that Nucleus.

Yes, the nucleus and proton do have a size. However, the proton is made up of quarks, which have no known size.

Hence, what could be the outcome if it collapses much below its minimal size?

You get a positively-charged singularity, presumably.

How the Gluons could still exist if we disconnect it from the three quarks?

Who said anything about disconnecting gluons from quarks?

Don't you agree that at the moment that we break the structure of the proton, the Gluons is lost forever?

No. Any mass-energy represented by the gluons would simply become a part of the mass-energy of the resulting singularity. That's due to conservation of energy.

At that moment 99% of the proton mass (938 Mev/c^2) had been lost.

Nope.

So, there is no way to break the proton.

Protons are broken all the time in particle accelerators. It's just that new quarks and anti-quarks are created in the process so that no single quark is left by itself.

In any case, if you believe that a proton could carry mass at a size of zero, then why atom couldn't carry mass at size of zero?

It's not that an atom couldn't have mass with a size of zero. Rather, it's that an object with a size of zero is no longer an atom.

So, do you agree that a proton MUST have a minimal size?

Yes. If all three quarks are crushed into a singularity, then you no longer have a proton but a black hole instead. That singularity will have the charge and mass of the proton so that conservation laws are not violated.

[Dave Lev (OP)]

Quote from: Dave Lev on Today at 05:49:22
So, do you agree that a proton MUST have a minimal size?

Yes.

Thanks
So, you confirm that it has a minimal size. However, why don't you agree that if a proton is crushed, we actually split it to its basic element as three separated quarks and gluons?
However, why do you claim that::

Who said anything about disconnecting gluons from quarks?

How can you crush a proton without breaking it to its basic elements as quarks and Gluons?
In any case, what is the real function of the gluons?:
https://www.quantamagazine.org/what-goes-on-in-a-proton-quark-math-still-conflicts-with-experiments-20200506/
Three particles called quarks ricochet back and forth at nearly the speed of light, snapped back by interconnected strings of particles called gluons. Bizarrely, the proton’s mass must somehow arise from the energy of the stretchy gluon strings, since quarks weigh very little and gluons nothing at all.
So, it is stated clearly:
"The proton’s mass must somehow arise from the energy of the stretchy gluon strings, since quarks weigh very little and gluons nothing at all."
Therefore, the gluon is an energy that acts as some sort of stretchy strings (or if you wish a glue) that holds the proton structure.
It is clearly stated:
“We know absolutely that quarks and gluons interact with each other, but we can’t calculate” the result."
So, if you crush/break that interact between the quarks and the Gluons, we actually break the proton. In this case, we get three quarks and gluons.
However, as the Gluons is energy (not mass) then why that energy is not released at the same moment that we break the proton.
Please remember - Gluons is energy. Not mass.
It is considered as mass as long as it is in the proton.
So, if we break it from the internal interact in the proton we clearly get the three quarks mass and the energy of the gluons.

If all three quarks are crushed into a singularity, then you no longer have a proton but a black hole instead. That singularity will have the charge and mass of the proton so that conservation laws are not violated.

Ok - you consider the gluons as some sort of charge.
Our scientists prefer to call it energy.
Therefore, this gluons charge/energy can't represent any sort of mass as the proton is crashed.
Why the gluons wouldn't be transformed into some sort heat or flare and lost forever without violating the energy conservation law?
Don't forget that you have stated that:

You can't have free quarks,

So, at the same moment that you crush/break the proton, the gluons is lost as energy (heat or flare) while the quarks also ended their life as it is impossible to have free quarks.
In other words - at the same moment that the proton is crushed into a singularity, the gluon is transformed into pure energy (as it is energy) and as we can't have free quarks, those quarks also should transform their mass into energy.

Protons are broken all the time in particle accelerators. It's just that new quarks and anti-quarks are created in the process so that no single quark is left by itself.

Well, in the accelerator the particles orbits at ultra high velocity under ultra high magnetic force.
Based on your theory, there is no magnetic field at the core of BH and the particles aren't moving as they crushed to zero point.
So, how can you compare a BH core to accelerator?
If you wish, you can compare accelerator to accretion disc.

In any case, can you please offer the article that proves that you can crush a proton to zero without breaking it to its basic elements or losing most/all of its mass into pure energy.

[Bored chemist]

So, you confirm that it has a minimal size. However, why don't you agree that if a proton is crushed, we actually split it to its basic element as three separated quarks and gluons?

A car has a minimum size- big enough to put a person in.
If you crush it then it is no longer big enough to put a person in, and thus no longer a car.
It is also not separated into its component parts.

So, as well as thinking things fall upwards, you think squeezing things together makes them spread apart.

Do you see why we are not taking your ideas very seriously?

[Bored chemist]

Ok - you consider the gluons as some sort of charge.

Nobody said that.

Our scientists prefer to call it energy.

Based on past evidence you have little or no idea what scientists think or do.

Therefore, this gluons charge/energy can't represent any sort of mass as the proton is crashed.

Non sequitur

Why the gluons wouldn't be transformed into some sort heat or flare and lost f

because that wouldn't be compressing stuff, it would be expanding it.

[Bored chemist]

So, how can you compare a BH core to accelerator?

Because, from the PoV of the proton the process is the same.
Too many protons in too small a space.