# On black hole ground states

Wheeler suggested a ground state black hole suggesting the name Geons, a play on the word for a particle trapped in its own gravitational field. Later, Motz suggested that pair black holes could form a system which is not a gravitational black hole in itself. He likened this to the bound state of two quarks, also known as confinement phenomenon. I later found after some of my own investigations into how to model this, that Holeums where in 2002 and wrongly credited as an idea to originating with Chavda and Abhijit Chavda in 2002, though the credit they should get is providing the first strong mathematical foundation for the theory.

Motz suggested that a ground state black hole may not radiate because like a hydrogen atom, it could possess an infinite lifetime so that it cannot decay into anything below this. Here is a summary of my own investigation and calculations:

Please be patient as this takes some time to explain and I want to demonstrate that this was a serious look into the hypothetical stable black holes.

Black holes, even the microscopic kind, cannot be mathematically ruled out from physics - and if the work of Hawking is to be taken seriously, including analysis provided by Motz, then a black hole system really is deduced from a discrete set of quantum processes - those discrete processes always lead to an increase in the entropy of a system like a black hole .I have taken this recently to mean, that even a Planck particle (aka. black hole particles, or the Uniton as Motz named them) also must increase in entropy. How you actually interpret this with a particle with an infinite lifetime like a hypothetical class of particle black holes, is actually uncertain. We cannot of course rule out that entropic phenomenon take place in an increasing fashion since the third law of thermodynamics insists that the entropy of a system is only zero when the temperature is zero - we know this cannot ever be the case in science, since a zero temperature would correspond to zero motion inside the field. It’s an interesting question what happens to entropy in the ground state of a black hole particle since it has been shown the overall entropy of a black hole cannot decrease… only increase in time.

The earliest model of a black hole stated that the entropy of a black hole was exactly zero - Hawking challenged this and stated if it has an entropy, then a black hole must possess a temperature and later we came to understand it as a thermal property of the black body radiation of the black hole. It may still be true, at least for the ground state black hole, that entropy may be essentially constant.

So how can we have an increasing energy but a system with an infinite lifetime? Something needs to give. The intuitive answer is that somewhere in the ground state, our usual understanding of entropy must break down. Entropy is a measure of disorder, and if a system is infinitely stable, then the disorder remains a constant. It could be that the black hole particle at near zero temperatures exhibits a behaviour like a zeno effect. The zeno effect is when an atom is incapable of giving off energy suspended in an otherwise, infinite animation and it will remain in a ground state because its wave function is incapable of evolving from the ground state. It is therefore similar to how an atom, ready to give up energy can be suspended in infinitely in time - and the reason why it cannot give up radiation is essentially the same for the micro black hole, since the zeno effect is all about keeping an atom in the ground state. To put in a summary for clarity I have came to some conclusions:

1. The Black Hole entropy rule dictating it always increases may apply only to macroscopic black holes

2. A ground state black hole has stability owed to it from two phenomenon: There is no fundamental particle it can decay into, coupled with a mechanism brought on about by a zeno effect (which halts the wave function evolution).

I did find a way to represent the energy levels of a microblack hole in those discrete processes spoken about before above. That took the idea of taking the ordinary transition equation for atomic spectral emission or gain of energy and reinterpret the fine structure as the gravitational fine structure.

In natural units,of $c = G = \hbar = 1$ the gravitational fine structure constant is equal to the square of the mass of a particle

$$\alpha_G = m^2$$

And the quantization of a mass depends on that spectral property:

$$n\hbar = m^2$$

The hope or immediate realization of this are attempts to find quantization of mass depending on factors of $n\hbar$. The $Gm^2$ can be thought of as the gravitational charge of the system analogous to the electric charge $e^2$. If we define the Rydberg constant in terms of the gravitational coupling constant we get:

$$\mathbf{R} = \frac{\alpha_G}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$$

Even though the Rydberg constant was first applied to hydrogen atoms, it could be derived from fundamental concepts (according to Bohr). In which case we may hypothesize energy levels:

$$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$$

Plugging in the last expressions we get an energy equation:

$$\Delta E_G = \frac{n\hbar c}{\Delta \lambda} = \frac{1}{4 \pi }\frac{m_0v^2}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{4 \pi \lambda_0}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2}) = \frac{p}{4 \pi \hbar}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})$$

The relativistic gamma appears from the definition of the deBroglie wavelength [1] and it suggests a relationship between Einstein's relativistic mass and the gravitational charge.

[1] - The deBroglie relationship used was: $$\frac{1}{\lambda} = \frac{\gamma m_0 v}{\hbar} = \frac{1}{\hbar} \frac{m_0v}{(1 - \frac{v^2}{c^2})}$$

The entropy of the black hole is calculated in the following way:

The SI value for the Boltzmann constant is given in units of the Hartree energy divided by the Kelvin

$$\frac{E}{K} = k_B = 3.1668114(29)\times {10}^{−6}$$

In which the Hartree energy is simply

$$E = 2\mathbf{R}\hbar c$$

and so we can write it with temperature $T$ which is also measured in Kelvin,

$$k = \frac{E}{T} = \mathbf{R}\frac{\hbar c}{T}$$

This can be written in such a way to introduce those quantized atomic levels ~ by making use of the formula

$$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$$

giving

$$k_B = \frac{1}{T} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1T} - \frac{Gm^2}{n^2_2T})$$

In which $\mathbf{R}$ is the usual Rydberg constant. The reason why this is interesting for the case of a black hole, is that the entropy of a system can be deduced entirely from Planck constants

$$S = \frac{m_PL^2_P}{T_P t^2_P} = \frac{\sqrt{\frac{\hbar c}{G}} \frac{\hbar G}{c^3}}{\sqrt{\frac{\hbar c^5}{Gk^2_B}} \frac{\hbar G}{c^5}}$$

---- The questions: This is to help me understand or if I have missed anything in the theories I have been reading. If I understand degenerate systems properly all this ''wiki'' talk about them possessing a non-zero entropy at zero kelvin, has to be physically rubbish. Quantum forbids experimentally any case that satisfies T=0 and by the third law of thermodynamics, when a system temperature is zero, it must also have a zero entropy. I think I am right to take all this nonsense talk about systems at absolute zero is really only an approximation towards this non-reachable limit?

second question: Am I right in thinking it is not a true analogy in my case given above in regards to a ground state black hole and the concept of the zeno effect in quantum behaviour of atoms because there are also anti-zeno effects - there is also the issue of when observing the atom ceases, the unstable atom will radiate, whereas, the ground state hydrogen has an infnite lifetime and thus stable. Regardless, there are similarities since the zeno effect is about reducing the system to the ground state and so I am not entirely incorrect to draw upon these possible connections to the physics? Or have I missed something?

Thank you community!

• "Quantum forbids experimentally any case that satisfies T=0 and by the third law of thermodynamics, when a system temperature is zero, it must also have a zero entropy". The third law of thermodynamics has been expressed in different, inequivalent ways. The most accepted (and useful) version is the "unattainability principle", which says that the derivative of the entropy goes to zero as $T\to 0$. This doesn't rule out that the entropy itself is non-vanishing at zero temperature. See Masanes & Oppenheim. – Mark Mitchison May 29 '18 at 12:30
• This question got several arguments for nonzero entropy at T=0: physics.stackexchange.com/q/109941 – Anders Sandberg May 29 '18 at 22:45
• After much consideration, I have found the links provided and material interesting. I read the Masane-Oppenheim paper. I understood most of it. I still have some issues here. Take a black hole, it would take an infinite amount of steps for it to reach zero kelvin. It is often we must take this an an unphysical limit. A true system at absolute zero, must have zero entropy. In much the same sense, this is how Hawking discovered black holes where radiators- because he knew if it had a non-vanishing entropy, it must have a temperature. They seem to be codependent. – Gareth Meredith May 31 '18 at 10:56
• Of course I could be wrong, but this is just how I understand the physics. Temperature translates to motion and it seems to be an experimental fact of nature that all particles do possess motion and even there is no such thing as a true Newtonian vacuum (ie zero point fields). – Gareth Meredith May 31 '18 at 10:58
• The only way you could demonstrate it seems, that entropy vanishes with temperature (is impossible to prove) for the very reasons given above. Though it is accepted things are never at zero Kelvin, it seems like motion and disorder and the so-called passage of time is really a measure of the entropy of something. – Gareth Meredith May 31 '18 at 11:02