# 4.1 – Constructing a Macro Holeum from the inside out

Let us consider a Holeum that is a bound state of two microscopic black holes of masses $m_{1}$ and $m_{2}$ revolving around their center of mass under the action of gravity. From [1] and section 1.2 – A stable Holeum, the energy values of the Holeum are given by

 $E_{2}=-\frac{m_{1}m_{2}c^{2}\alpha_{2}^{2}}{2\left( m_{1}+m_{2}\right)n_{2}^{2}}$ (eq. 4.1)

where $\alpha_{2}$ is the gravitational coupling constant for the interaction between masses $m_{1}$ and $m_{2}$, given by

 $\alpha_{2}=\frac{m_{1}m_{2}}{m_{p}^{2}}$ (eq. 4.2)

where

 $m_{p}=\left( \frac{\hbar c}{G}\right) ^{\frac{1}{2}}$ (eq. 4.3)

is the Planck mass, $\hbar$ is the Planck’s constant divided by $2\pi$, $c$ is the speed of light in vacuum, and $G$ is Newton’s universal constant of gravity. $n_{2}$ is the principal quantum number of the bound state. From [1], the most probable radius of the Holeum is given by

 $r_{2}=\frac{\left( R_{1}+R_{2}\right) \pi^{2}n_{2}^{2}}{16\alpha_{2}^{2}}$ (eq. 4.4)

where

 $R_{i}=\frac{2m_{i}G}{c^{2}}$ (eq. 4.5)

is the Schwarzschild radius of the microscopic black hole of mass $m_{i}$. The mass of the bound state is given by

 $M_{2}=m_{1}+m_{2}+\frac{E_{2}}{c^{2}}$ (eq. 4.6)

We shall call this bound state of two microscopic black holes a di-Holeum. Now consider a bound state of the di-Holeum and a microscopic black hole of mass $m_{3}$. We call it a tri-Holeum. Its energy eigenvalues are given by

 $E_{3}=-\frac{m_{3}M_{2}c^{2}\alpha_{3}^{2}}{2\left( m_{3}+M_{2}\right)n_{3}^{2}}$ (eq. 4.7)

where

 $\alpha_{3}=\frac{m_{3}M_{2}}{m_{p}^{2}}$ (eq. 4.8)

The bound state radius of the tri-Holeum is given by

 $r_{3}=\frac{\left( m_{3}+M_{2}\right) G\pi^{2}n_{3}^{2}}{8c^{2}\alpha_{3}^{2}}$ (eq. 4.9)

With the help of equation (4.6) this may be written as

 $r_{3}=\frac{\left( R_{1}+R_{2}+R_{3}+\frac{E_{2}}{c^{2}}\right) \pi^{2}n_{3}^{2}}{16\alpha_{3}^{2}}$ (eq. 4.10)

The mass of the tri-Holeum is given by

 $M_{3}=m_{3}+M_{2}+\frac{E_{3}}{c^{2}}$ (eq. 4.11)

Continuing in this manner we arrive at a $k$-Holeum, also called a Macro Holeum, consisting of $k$ microscopic black holes. Its energy values, bound state radius, and bound state mass are given, respectively, by

 $E_{k}=-\frac{m_{k}^{3}c^{2}\left[ m_{1}+m_{2}+\ldots+m_{k-1}+\frac{\left(E_{2}+E_{3}+\ldots+E_{k-1}\right) }{c^{2}}\right] ^{3}}{2m_{p}^{4}n_{k}^{2}\left[ m_{1}+m_{2}+\ldots+m_{k}+\frac{\left( E_{2}+E_{3}+\ldots+E_{k-1}\right) }{c^{2}}\right] }$ (eq. 4.12)

 $r_{k}=\frac{\left( R_{1}+R_{2}+\ldots+R_{k}\right) \left[ 1+\frac{E_{2}+E_{3}+\ldots+E_{k-1}}{\left( m_{1}+m_{2}+\ldots+m_{k}\right) c^{2}}\right]\pi^{2}n_{k}^{2}}{\left( \frac{m_{k}}{m_{p}}\right) ^{2}\left[ \frac{m_{1}+m_{2}+\ldots+m_{k-1}}{m_{p}}\right] ^{2}\left[ 1+\frac{E_{2} +E_{3}+\ldots+E_{k-1}}{\left( m_{1}+m_{2}+\ldots+m_{k-1}\right) c^{2}}\right] ^{2}}$ (eq. 4.13)

 $M_{k}=\left( m_{1}+m_{2}+\ldots+m_{k}+\frac{E_{2}+E_{3}+\ldots+E_{k}}{c^{2}}\right)$ (eq. 4.14)

The gravitational coupling constant $\alpha_{k}$ for the interaction between a Macro Holeum containing $k-1$ microscopic black holes and the microscopic black hole of mass $m_{k}$ is given by

 $\alpha_{k}=\frac{m_{k}M_{k-1}}{m_{p}^{2}}$ (eq. 4.15)

Using equation (4.14), this may be rewritten as

 $\alpha_{k}=\frac{m_{k}\left( m_{1}+m_{2}+\ldots+m_{k-1}+\frac{\left(E_{2}+E_{3}+\ldots+E_{k-1}\right) }{c^{2}}\right) }{m_{p}^{2}}$ (eq. 4.16)

For the sake of simplicity we now consider an equal-mass case: $m_{1}=m_{2}=\ldots=m_{k}=m$, say. Then equations (4.12) – (4.16) reduce, respectively, to

 $E_{k}=-\frac{\left( k-1\right) ^{3}mc^{2}\alpha_{g}^{2}}{2kn_{k}^{2}}\left[\frac{g_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) ^{3}}{f_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) }\right]$ (eq. 4.17)

 $r_{k}=\frac{kR\pi^{2}n_{k}^{2}}{16\alpha_{g}^{2}\left( k-1\right) ^{2}}\left[ \frac{f_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right)}{g_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) ^{2}}\right]$ (eq. 4.18)

 $M_{k}=kmg_{k}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k}\right)$ (eq. 4.19)

 $\alpha_{k}=\left( k-1\right) \alpha_{g}g_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right)$ (eq. 4.20)

where

 $\alpha_{g}=\left( \frac{m}{m_{p}}\right) ^{2}$ (eq. 4.21)

Here $n_{j}$ is the principal quantum number of the microscopic black hole of mass $m_{j}$; $j=2,3,\ldots k$. From equations (4.12) and (4.13) as well as from equations (4.17) and (4.18) we see that whereas $E_{k}$ and $r_{k}$ depend strongly upon $n_{k}$, the principal quantum number of the outermost microscopic black hole, their dependence on the rest of the principal quantum numbers is quite symmetric. None of the latter is singled out the way $n_{k}$ is.

Hence for the sake of convenience we shall describe the $k$-Holeum in terms of a “valence” microscopic black hole, namely the outermost one and and the “core” microscopic black holes, namely, the rest of them, $j=2,3,\ldots k-1$.

$f_{k-1}$ and $g_{k-1}$ are given by

 $f_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) =1+\frac{E_{2}+E_{3}+\ldots+E_{k-1}}{kmc^{2}}$ (eq. 4.22)

 $g_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) =1+\frac{E_{2}+E_{3}+\ldots+E_{k-1}}{\left( k-1\right) mc^{2}}$ (eq. 4.23)

Note that we must have at least $k=2$ for a bound state formation. Thus in equations (4.22) and (4.23), we must define $f_{1}=g_{1}=1$. Note that the Schwarzschild radius, $R_{k}$, of the $k$-Holeum containing $k$ microscopic black holes is given by

 $R_{k}=\frac{2M_{k}G}{c^{2}}=kRg_{k}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k}\right)$ (eq. 4.24)

where $M_{k}$ is given by equation (4.19). From equations (4.18) and (4.24) we get the ratio of the Schwarzschild radius $R_{k}$ of the $k$-Holeum to its bound state radius $r_{k}$ as follows:

 $\frac{R_{k}}{r_{k}}=\frac{16\alpha_{g}^{2}\left( k-1\right) ^{2}}{\pi^{2}n_{k}^{2}}\frac{g_{k}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k}\right) g_{k-1}\left( \alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) ^{2}}{f_{k-1}\left(\alpha_{g},n_{2},n_{3},\ldots,n_{k-1}\right) }$ (eq. 4.25)

Substituting for $E_{j}$ from equation (4.17) into equations (4.22) and (4.23) we obtain

 $f_{k-1}=1-\frac{\alpha_{g}^{2}}{2k}\sum_{j=2}^{k-1}\frac{\left( j-1\right)^{3}}{jn_{j}^{2}}\frac{g_{j-1}^{3}}{f_{j-1}}$ (eq. 4.26)

 $g_{k-1}=1-\frac{\alpha_{g}^{2}}{2\left( k-1\right) }\sum_{j=2}^{k-1}\frac{\left( j-1\right) ^{3}}{jn_{j}^{2}}\frac{g_{j-1}^{3}}{f_{j-1}}$ (eq. 4.27)

Since $r_{k}>0$ and $M_{k}>0$, for all $k$, we conclude from equations (4.18) and (4.19) that

 $f_{k-1}>0\text{, }g_{k-1}>0$ (eq. 4.28)

From equations (4.26) – (4.28) we have

 $f_{k-1}\leq1\text{, }g_{k-1}\leq1$ (eq. 4.29)

for all $k$ and $\alpha_{g}$. In this equation and subsequent ones the equality holds if either $\alpha_{g}=0$ or $n_{2}=n_{3}=\ldots=n_{k-1}=\infty$. From equations (4.26) and (4.27) it trivially follows that

 $f_{k-1}\geq g_{k-1}$ (eq. 4.30)

From equation (4.29) it follows that

 $g_{k-1}\geq g_{k-1}^{2}\geq g_{k-1}^{3}$ (eq. 4.31)

From equations (4.30) – (4.31) we have

 $f_{k-1}\geq g_{k-1}\geq g_{k-1}^{2}\geq g_{k-1}^{3}$ (eq. 4.32)

From the latter we conclude that

 $\frac{g_{j-1}^{3}}{f_{j-1}}\leq1$ (eq. 4.33)

From equations (4.33) and (4.17) we have

 $E_{k}\geq-\frac{\left( k-1\right) ^{3}}{2k}\frac{mc^{2}\alpha_{g}^{2}}{n_{k}^{2}}$ (eq. 4.34)

Letting $p=k\alpha_{g}$ and $k\gg2$ in equation (4.34) we may rewrite it as

 $E_{k}\geq-\frac{p^{2}mc^{2}}{2n_{k}^{2}}$ (eq. 4.35)

If $p\rightarrow\infty$ as $k\rightarrow\infty$ in equation (4.35), then the energy of the system has no lower bound. Therefore the system will go on losing energy during interactions until it becomes a part of the infinite energy of the vacuum state of the universe. In this case, the system cannot have an independent existence. This is averted if $p would be in the astronomical range, say, $k=10^{50}$ to $k=10^{100}$. The huge mass of this system gives it a huge inertia. It will be very difficult to perturb such a system. Thus, we may assume that most of the system will remain unperturbed and only the outermost one or two microscopic black holes will be affected by the interactions of the system with its environment. In order to reduce this many-body problem to a two-body one, we will make the extreme simplification that all the microscopic black holes in the core are in the same quantum state described by $n_{2}=n_{3}=\ldots=n_{k-1}=n$, say, and that the outermost microscopic black hole is in an arbitrary quantum state described by the principal quantum number $n_{k}$. The first assumption embodies the great inertia of the system.

We recall here that in the nuclear many-body problem, the nucleus is assumed to be an infinite medium in which every nucleon moves in the same average potential produced by the rest of them. This reduces the many-body problem to a one-body problem. This, plus a heuristic spin-orbit potential with an arbitrarily chosen sign, fetches us good quantitative agreement with nuclear data in the nuclear shell model. Great complexity often yields to extreme simplification; provided the simplification embodies the physics of the system correctly. In our problem, if there is a greater excitation, we may describe it in terms of $n,n_{k-1}$ and $n_{k}$; wherein the core now consists of $k-2$ microscopic black holes all in the same state of excitation $n$, and the two outer-most microscopic black holes are in arbitrary states of excitation described by $n_{k-1}$ and $n_{k}$. But in the first approximation, in the following, we describe the system in terms of $n$ and $n_{k}$ only, apart from the other parameters.

Now we make a Taylor series expansion of $f_{k-1}$, equation (4.26), in powers of $x=\alpha_{g}^{2}$. Then we get

 $f_{k-1}(x)=1+xf_{k-1}^{\prime}(0)+\frac{x^{2}}{2!}f_{k-1}^{\prime\prime}(0)+\ldots$ (eq. 4.36)

From equation (4.26) we have

 $f_{k-1}^{\prime}(x)=-\frac{1}{2k}\sum_{j=2}^{k-1}\frac{\left( j-1\right)^{3}}{jn_{j}^{2}}\frac{g_{j-1}^{3}}{f_{j-1}}-\frac{x}{2k}\sum_{j=2}^{k-1} \frac{\left( j-1\right) ^{3}}{jn_{j}^{2}}\left[ \frac{3g_{j-1}^{2}g_{j-1}^{\prime}}{f_{j-1}}-\frac{g_{j-1}^{3}f_{j-1}^{\prime}}{f_{j-1}^{2}}\right]$ (eq. 4.37)

From equation (4.37) it follows that

 $f_{k-1}^{\prime}(0)=-\frac{S_{k-1}}{2kn^{2}}$ (eq. 4.38)

In these equations we have taken $n_{2}=n_{3}=\ldots=n_{k-1}=n$. Similarly from equation (4.27) it follows that

 $g_{k-1}^{\prime}(0)=-\frac{S_{k-1}}{2(k-1)n^{2}}$ (eq. 4.39)

where

 $S_{k-1} =\sum_{j=2}^{k-1}\frac{\left( j-1\right) ^{3}}{j} =\frac{\left( k-1\right) k\left( 2k-1\right) }{6}-\frac{3k\left(k-1\right) }{2}+3\left( k-1\right) -C-\ln\left( k-1\right) -\frac{1}{2\left( k-1\right) }+\sum_{j=2}^{\infty}\frac{A_{j}}{\left( k-1\right) k\left( k+1\right) \ldots \left( k+j-2\right) }$ (eq. 4.40)

where $A_{2}=A_{3}=\frac{1}{12}$, $A_{4}=\frac{19}{80}$, etc. and $C=0.577216$ is the Euler constant. Now substituting for $S_{k-1}$ from equation (4.40) into equation (4.38) and keeping only the two leading terms we get

 $f_{k-1}^{\prime}(0)=-\frac{1}{2kn^{2}}\left[ \frac{k^{3}}{3}-\frac{3k^{2}}{2}+O(k)\right] \ldots$ (eq. 4.41)

 $xf^{\prime}(0) =-\frac{\alpha_{g}^{2}k^{2}}{6n^{2}}+\frac{3\alpha_{g}^{2}k}{4n^{2}}+O(\alpha_{g}^{2}k^{0})=-\frac{p^{2}}{6n^{2}}+\frac{3p^{2}}{4kn^{2}}+O(\frac{p^{2}}{k^{2}})=-\frac{p^{2}}{6n^{2}}+O(k^{-1})$ (eq. 4.42)

where $p=k\alpha_{g}$ is taken as a constant. Differentiating equation (4.37) again with respect to $x$ and letting $x=0$ we obtain

 $f_{k-1}^{\prime\prime}(0)=-\frac{1}{kn^{2}}\sum_{j=2}^{k-1}\frac{\left(j-1\right) ^{3}}{j}\left[ 3g_{k-1}^{\prime}(0)-f_{k-1}^{\prime}(0)\right]$ (eq. 4.43)

Substituting from equations (4.38) and (4.39) into equation (4.43), we obtain

 $f_{k-1}^{\prime\prime}(0)=-\frac{1}{2kn^{4}}\sum_{j=2}^{k-1}\frac{(2j+1)\left( j-1\right) ^{2}S_{j-1}}{j^{2}}$ (eq. 4.44)

Since we need only the leading order terms for $k\gg2$, we substitute only the first term of $\ S_{j-1}$ from equation (4.40) into equation (4.44) to obtain

 $f_{k-1}^{\prime\prime}(0) =-\frac{1}{2kn^{4}}\sum_{j=2}^{k-1}\frac {(4j^{2}-1)\left( j-1\right) ^{3}}{6j}+ \text{ l.o.t.} =-\frac{1}{2kn^{4}}\sum_{j=2}^{k-1}\left[ \frac{2}{3}j^{4}-2j^{3}+\frac{11}{6}j^{2}-\frac{1}{6}j-\frac{1}{2}+\frac{1}{6j}\right] +\text{ l.o.t.}$ (eq. 4.45)

where $l.o.t.$ stands for lower order terms. Now we use the identity

 $\sum_{i=1}^{m}i^{4}=\frac{m^{5}}{5}+\frac{m^{4}}{2}+\frac{m^{3}}{3}-\frac{m}{30}$ (eq. 4.46)

where $m=k-1\gg1$. Therefore substituting only the leading term from equation (4.46) into the leading term in equation (4.45) we get

 $f_{k-1}^{\prime\prime}(0)=-\frac{k^{4}}{15n^{4}}+O(k^{3})$ (eq. 4.47)

 $\frac{x^{2}}{2!}f_{k-1}^{\prime\prime}(0) =-\frac{k^{4}\alpha_{g}^{4}}{30n^{4}}+O(\frac{p^{4}}{k})=-\frac{p^{4}}{30n^{4}}+O(k^{-1})$ (eq. 4.48)

Substituting equations (4.42) and (4.48) into equation (4.36), we obtain

 $f_{k-1}(x)=1-\frac{p^{2}}{6n^{2}}-\frac{p^{4}}{30n^{4}}+O(k^{-1})\ldots$ (eq. 4.49)

Now we further assume $n\gg1$. This allows us to keep only the first two terms in equation (4.49). Therefore we get

 $f_{k-1}(x)\simeq1-\frac{p^{2}}{6n^{2}}+O(k^{-1})\ldots$ (eq. 4.50)

Now for $k\gg2$, $f_{k-1}$ $\simeq g_{k-1}$. Henceforth, for simplicity, we will drop the notation $O\left(k^{-1}\right)$ and we will replace the symbol $\simeq$ by $=$ in the following. Therefore, for $k\gg2$ and $n\gg1$ we have

 $f_{k-1}=g_{k-1}=1-\frac{p^{2}}{6n^{2}}$ (eq. 4.51)

Note that $1\geq f_{k-1}(x)\geq0$. Therefore from equation (4.51) we have

 $0\leq p^{2}\leq6.$ (eq. 4.52)

Finally, substituting from equation (4.51) into equations (4.17) - (4.20) and equations (4.24) and (4.25), we have

 $E_{k}=-\frac{p^{2}mc^{2}}{2n_{k}^{2}}\left( 1-\frac{p^{2}}{6n^{2}}\right)^{2}$ (eq. 4.53)

 $r_{k}=\frac{\pi^{2}kRn_{k}^{2}}{16p^{2}\left( 1-\frac{p^{2}}{6n^{2}}\right)}$ (eq. 4.54)

 $M_{k}=mk\left( 1-\frac{p^{2}}{6n^{2}}\right)$ (eq. 4.55)

 $R_{k}=kR\left( 1-\frac{p^{2}}{6n^{2}}\right)$ (eq. 4.56)

 $\alpha_{k}=p\left( 1-\frac{p^{2}}{6n^{2}}\right)$ (eq. 4.57)

 $\frac{R_{k}}{r_{k}}=\frac{16p^{2}}{\pi^{2}n_{k}^{2}}\left( 1-\frac{p^{2}}{6n^{2}}\right) ^{2}$ (eq. 4.58)

The density of the Macro Holeum is given by

 $\rho_{k}=\frac{3072m_{p}p^{5}\left( 1-\frac{p^{2}}{6n^{2}}\right) ^{4}}{\pi^{7}kR_{p}^{3}n_{k}^{6}}$ (eq. 4.59)

In equations (4.53) - (4.59), $p=k\alpha_{g}$ satisfies equation (4.52); and $R_{p}$ given by

 $R_{p}=\frac{2m_{p}G}{c^{2}}$ (eq. 4.60)

is the Schwarzschild radius of a microscopic black hole of Planck mass. Apart from the integers $n$ and $n_{k}$ the seven properties listed in equations (4.53) - (4.59) depend, in general, on two parameters $m$ and $k$ or their two suitable combinations. But $\alpha_{k}$ and $\frac{R_{k}}{r_{k}}$ depend only upon the combination $p=k\alpha_{g}$. As mentioned earlier, $p$ may be regarded as an effective coupling strength that determines the formation of the Macro Holeum.

### The Ground State of the Macro Holeum

The ground state of Macro Holeums is characterized by $n=\infty$ and $n_{k}=1$. The Macro Holeum has maximum binding energy, minimum physical radius, maximum Schwarzschild radius and maximum mass in this state. Such a system can be thought of as consisting of a gas of $k-1$ free $(n=\infty)$ micro black holes that is bounded and therefore isolated from the outside world by a solitary outermost micro black hole whose principal quantum number is $n_{k}=1$.

### The Stability of the Macro Holeum

It can be seen from the equations (4.53) - (4.57) that the condition for the stability of Holeums is given by

\frac{p^{2}}{6n^{2}}<1[/latex] (eq. 4.61)
Substituting the relations [latex size="1"]p=k\alpha _{g}
and $\alpha _{g}=\frac{m^{2}}{m_{P}^{2}}$ into this inequality, we see that the condition for the stability of Holeums can be expressed as

m (eq. 4.62)
The ground state of Holeums is characterized by [latex size="1"]n=\infty
, which gives us

 m<\infty[/latex] (eq. 4.63)

as the condition for stability. Thus, the ground state of Holeums is guaranteed to be always stable.

### References

[1] L. K. Chavda and Abhijit L. Chavda, Dark matter and stable bound states of primordial black holes, arXiv:gr-qc/0308054 (2002).

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