# An Ultra-compact Object from Semi-classical Gravity

I. Prasetyo<EMAIL_ADDRESS>H. S. Ramadhan<EMAIL_ADDRESS>A. Sulaksono<EMAIL_ADDRESS>Departemen Fisika, FMIPA,
Universitas Indonesia, Depok 16424, Indonesia.

###### Abstract

In a recent report, Carballo-Rubio Carballo-Rubio:2017tlh  utilizes the semi-
classical theory of gravity to obtain a generalized Tolman-Oppenheimer-Volkoff
(TOV) equation. This model has a new coupling constant $l_{p}$ implying two
different modified TOV equation forms characterized by the sign of
$p^{\prime}$. The negative branch reduces to the ordinary GR-TOV in the limit
of $l_{p}\to 0$, while the positive one does not. In the positive branch,
Carballo-Rubio was able to find the exact solutions using the
constant-$\lambda$ trick. In this work, we investigate whether this theory’s
negative branch can also provide a different feature of the ultra-compact
object compared to those obtained from the GR-TOV equation. We use a numerical
method to calculate the properties of an ultra-compact object. We tested our
code by calculating numerically the pressure profile of the object in a
positive branch with the star’s radius $R$ in a macroscopic unit by
integrating from surface to center. The result reproduces the analytical
result of Carballo-Rubio Carballo-Rubio:2017tlh . While in a negative branch,
we reproduce the pressure profile results of Ho and Matsuo Ho:2017vgi  using
constant energy density $\rho=\rho_{0}$ not only by integrating from surface
to center but also by integrating from center to surface, respectively. In
this work, we study ultra-compact objects with an isotropically ideal fluid
matter where we use a simple but physically motivated equation of state
$\rho=p/w+\rho_{0}$ with $w$=1 and $w$=1/3. In general, we obtain that the
range of $l_{p}$ is very restricted and must not be equal to $r_{c}$. Here
$r_{c}$ is the starting point of integration located at the center of the
star. While $l_{p}$ should be set to be much larger than Planck length
$L_{\text{Pl}}$. Consequently, the mass-radius curves for the various value of
$l_{p}$ for both $w$ cases are still indistinguishable from the standard GR-
TOV results. Hence from the negative branch of $p^{\prime}(r)$, the additional
free parameter $l_{p}$ does not provide a significant effect compared to the
standard GR-TOV equation results, even though $l_{p}$ is not in the limit of
$l_{p}\to 0$ anymore. Therefore, similar to the conclusion in Ref.
Urbano:2018nrs  with GR theory that the ultra-compact objects from negative
branch of semi-classical gravity with a linear equation of state are unable to
generate demanding gravitational echoes.

##  I Introduction

The recent excitement over gravitational wave astronomy by LIGO and Virgo
GBM:2017lvd  gave rise to many discussions over what kind of massive bodies
produce the waves. From the astronomical point of view, the curiosity of
synthesizing heavy metals such as silver and platinum from a collision of two
neutron stars also triggered excitement. From the theoretical point of view,
many attempts have been made to justify the existence of horizonless ultra-
compact objects other than known compact objects such as neutron stars or
white dwarfs when interacting with other massive bodies by probing its
behavior through its waves. There are many proposals given, including
gravastars Mazur:2001fv ; Mazur:2004fk  where inside the surface of the star
lies another de-Sitter space connected to the usual Schwarzschild space-time
by a thin shell of ultrarelativistic matter. Interestingly recently, Carballo-
Rubio Carballo-Rubio:2017tlh  has claimed to predict exotic stars that have
many similar features with gravastar using semiclassical gravity theory.

The semiclassical theory of gravity is probably one of the earliest attempts
by physicists to reconcile quantum field theory and gravity. It began with the
remarkable result on black hole entropy and temperature and the Unruh effect
(see the famous classic monograph Birrell:1982ix  and for more recent one see
Parker:2009uva ). In this theory, the concept of particle creation in general
space-time is different from the usual one in Minkowski space-time. This
ambiguity is mainly caused by the Bogolyubov transformation, which is similar
to coordinate transformation. Although the problem seems unsolvable, it turns
out that the energy-momentum tensor is indifferent to what “coordinate system”
we chose. Even though there are infinities inside them, physicists had
identified what kind of infinities leads to the physical quantum state,
resulting in the so-called Hadamard state, i.e., a state whose two-point
function has infinities whose form is the same as the one calculated in
Minkowskian space-time. Another state exists, i.e., Boulware state, which had
been extensively studied again since it describes a Schwarzschild black hole
vacuum. Since the study of $n$-point functions are related to regularization
of infinities from expectation value of energy-momentum tensor,
$\langle\hat{T}_{\mu\nu}\rangle$, hence the main interest in the semiclassical
theory of gravity is mainly around renormalization of the energy-momentum
tensor which is related to the Einstein tensor by

$G_{\mu\nu}=8\pi G\langle\hat{T}_{\mu\nu}\rangle.$

The resulting renormalization should satisfy Wald’s renormalization axioms
(see page 89 of Wald:1995yp ). It has been known that the methods of
renormalization only can apply for some specific cases DeWitt:1975ys , and in
the case of conformal field theory, these methods give rise to the so-called
trace anomaly Birrell:1982ix  which is a still unsolvable problem in semi-
classical gravity theory. The value of the cosmological constant is restricted
in a nontrivial way Juarez-Aubry:2019jbw ; hence one cannot treat this
constant as a free parameter in this theory.

Here we provide a short discussion related to Boulware and Unruh vacuums. The
definition of Boulware vacuum is a quantum state of the exterior of any
massive body but singular at the horizon. In contrast, the Unruh vacuum is the
quantum state of gravitational collapse, and it features Hawking radiation at
large distances Visser1997 ; Barbado2011 . In general, these two vacuums are
inequivalent even though they could be interlinked classically by coordinate
transformation. Please see Ref. Shapiro2008  for a detailed discussion about
how to relate both quantum vacuums. The authors of Ref. Barcelo2008
investigated the dynamical evolution of a collapsed star within semiclassical
gravity. They found that in some cases, the trapping horizon is prevented from
forming in a semiclassical approach, and the new collapsed objects could
exist. These objects make the confrontation with information paradox and run-
away endpoint problem unnecessary. They also found that both quantum vacuums
can describe the same exterior compact object locally as long as the horizon
is unformed. If the horizon is formed, then the Boulware vacuum could no
longer be used, and one should use Unruh vacuum to describe the object. Please
see detail technical discussions of this matter in Ref. Barcelo2008 . A recent
study by Ho and Matsuo Ho:2017vgi  has shown that if the compact objects have
no singularity nor horizons implies that arbitrary heavy objects can have a
physical state in the Boulware vacuum as long as it has stationary state.
Furthermore, Ho and Matsuo in other paper HoMa2018  have shown that it is
physically sensible to consider the Boulware vacuum for any macroscopic radius
of a compact object from nonperturbative analysis of the semiclassical
Einstein equation. These results Ho:2017vgi ; HoMa2018  are contrary to the
common belief that Boulware vacuum becomes unphysical if the radius is smaller
than the Schwarzchild radius since the energy-momentum tensor is divergent at
the Schwarzchild radius. In this work, we study the compact object without
horizon, and the radius is larger than the Schwarzchild radius. Therefore,
using a Boulware vacuum is relatively safe.

Meanwhile, on gravitational wave astronomy, there are some indications that
all black holes cannot be observed except by their intense gravity effects.
Therefore, there exists the possibility of the existence of the so-called
black hole mimickers. These are compact horizonless astronomical objects,
usually assumed to have spherical symmetry, whose mass is enormous, yet their
size is relatively small. As an illustration, a neutron star’s mass $M$ had
been observed to be around $1.4-2$ solar mass, and its radius $R$ is roughly
within the range of 10 to 20 km. These compact objects are mainly featured by
their compactness $C=GM/R$, a dimensionless quantity due to Newton’s constant
$G$. Those objects can be categorized as follows Cardoso:2019rvt :

  1. 1.

compact objects, if $2C>1/3$,

  2. 2.

ultra-compact objects, if $2C>2/3$,

  3. 3.

objects violating Buchdhal limit, if $2C>8/9$,

  4. 4.

clean-photon sphere objects, if $2C>1/(1+0.019)$, and

  5. 5.

near-horizon quantum structures, if $2C>1/(1+10^{-40})$.

A black hole’s compactness is $2C=1$, so these exotic compact objects should
have compactness less than the black hole’s ($2C<1$). Suppose that any two
massive objects had collided and formed another massive object in the so-
called ring-down phase, this final object, in general, can be anything,
including a black hole. Due to the gravitational field’s intensity, these
final objects will most likely be a black hole, but if it is not, it will be
at least an ultra-compact object. By many analyses on a geodesic of light rays
propagating near the object, perturbating its exterior gravitational field,
people knew that the effective potential $V_{eff}$ from its gravity would
affect its ring-down gravitational wave pulse Urbano:2018nrs . After two
massive objects are combined into either a black hole or an ultra-compact
object, its gravitational wave pulse will stop or produce echoes as the final
object forms. This echo is the leaking of those gravitational waves that are
trapped by the effective potential barrier located at radius $r=3GM$
($V_{eff}^{\prime}(3GM)=0$ and $V_{eff}^{\prime\prime}(3GM)<0$), which is the
location of the so-called photon sphere. To have the photon sphere, then the
object’s radius should satisfy $R<3GM$, which implies compactness $2C>2/3$. It
is worthy to note that the photon sphere’s existence is due to the geodesic of
light rays propagating near the massive object, but its existence does not
imply gravitational wave echo, which came from the so-called TT-tensor part of
metric perturbation. The potential wall location at $r=3GM$ is in coincidence
with the photon sphere and is a byproduct of the spherical symmetry
assumption, which is generally not true for rotating objects that obey axial
symmetry. In the rest of this paper, we shall only assume spherical symmetry.
The gravitational wave echo frequency $f_{echo}$ is in one-to-one
correspondence on the object’s compactness $C$ by Cardoso:2019rvt

$\tau_{echo}^{(approx)}\sim 4M|\log\epsilon|,$ (1)

with $\epsilon={1/2C}-1$ and $f_{echo}=\pi/\tau_{echo}$. This echo time delay
can also be estimated by a calculation of its metric component $g_{tt}$ and
$g_{rr}$ both inside and outside the star Mannarelli:2018pjb

$\tau_{echo}^{(num)}=\int_{0}^{3GM}\sqrt{-{g_{rr}\over g_{tt}}}~{}dr.$ (2)

The integration means that the trapped waves propagated from $r=3GM$ to $r=0$
and back to $r=3GM$ by assuming that the gravitational wave penetrates the
ultra-compact object. (In fact, Eq. (1) came from Eq. (2) but integrated from
$r=R$ to $r=3GM$.) It had been calculated that the binary neutron stars merger
GW170817 produced an approximation to a “tentative” echo frequency
$f_{echo}\simeq 72$ Hz from a 2.6-2.7 solar mass ”black hole” remnant with
dimensionless spin $0.84-0.87$ Abedi:2018npz . This observation had been
predicted to be compatible with a toy model of an incompressible star with
mass $2-3$ solar mass Pani:2018flj . There are varieties of the proposed
compact objects (see the compilation of the properties of those objects and
the corresponding discussions in Ref. Cardoso:2019rvt ). These proposed ultra-
compact object studies are motivated by the hope that at least the echo from
one of those objects will eventually be detected in the near future.

Recently, Carballo-Rubio Carballo-Rubio:2017tlh  proposed a modified TOV
equation generated by semi-classical gravity theory. Solutions obeying
boundary conditions are found. They arise from the pressure equation
$p^{\prime}$ with a positive sign and are identified with a nontrivial
combination of the black stars and gravastars. This calculation is done by
defining a suitably new constant $\lambda>1$ that relates all pressure $p$,
mass $m$, and energy density $\rho$, hence the EoS are unique. The form of
pressure equation $p^{\prime}$ with the negative sign has also been analyzed
by Ho and Matsuo Ho:2017vgi  to show, using constant energy density, that the
Buchdahl limit can be violated without making the pressure goes to infinity.

In this work, we investigate the form of pressure equation $p^{\prime}$ with
negative sign much further. This paper is organized as follows. In section II,
we briefly revisit the models (Refs. Carballo-Rubio:2017tlh  and Ho:2017vgi ).
We discuss the numerical scheme in Appendix (page A), which is crucial to
justify our results in the following section III. In section III part A, we
discuss the model in more detail with non-negative pressure and energy density
assumption, and we use a simple linear EoS. We look at the effect from the
semi-classical terms compared to the usual TOV equation. In section III part
B, we discuss the numerical results. Finally, we summarize our work in Section
IV.

##  II COMPACT STARS IN THE THEORY OF SEMI-CLASSICAL GRAVITY

In ref. Carballo-Rubio:2017tlh  a new type of TOV equations can obtained by
using the renormalized energy-momentum tensor and solving the modified
Einstein field equation (EFE)

$G_{\mu\nu}=\frac{8\pi
G}{c^{4}}T_{\mu\nu}+96\pi^{2}l_{p}^{2}\langle\hat{T}_{\mu\nu}\rangle,$ (3)

where $l_{p}$ is the so-called renormalized Planck length. This quantity is
related to the usual Planck length $L_{\text{Pl}}=\sqrt{\hbar G/c^{3}}$ by

$l_{p}=\sqrt{N\over 12\pi}L_{\text{Pl}},$

where $N\gg 1$ is the number of particle fields. From now on, we shall use the
natural units ($c=1$). The renormalized stress-energy tensor (RSET) describes
the quantum vacuum polarization of $N\gg 1$ matter fields. We shall symbolize
the usual Planck length as $L_{\text{Pl}}$ so that there is no ambiguity with
the normalized one $l_{p}$, which will be treated as an adjustable coupling
constant.

Similar to the derivation of the usual TOV equation, the space-time metric
$g_{\mu\nu}$ is the static spherically symmetric metric

$\displaystyle ds^{2}$ $\displaystyle=$ $\displaystyle
ds^{2}_{(2)}+r^{2}(d\theta^{2}+\sin^{2}\theta d\varphi^{2}),$ (4)
$\displaystyle ds^{2}_{(2)}$ $\displaystyle=$
$\displaystyle-e^{\nu(r)}dt^{2}+{dr^{2}\over 1-2Gm(r)/r},$ (5)

which has a time symmetry denoted mathematically by a timelike Killing vector
$\xi=\partial_{t}$, and the energy-momentum tensor can be expressed by the
usual perfect fluid

$T^{\mu\nu}=(\rho+p)u^{\mu}u^{\nu}+pg^{\mu\nu},$ (6)

where $u^{\mu}$ is the 4-velocity of the fluid, which is timelike and
normalized ($u^{\mu}u_{\mu}=-1$), and $g^{\mu\nu}$ is the inverse of the
metric. The RSET is given by the so-called s-wave Polyakov approximation (see
Ref. Fabbri:2005  page 216)

$\langle\hat{T}_{\mu\nu}\rangle={\delta^{a}_{\mu}\delta^{b}_{\nu}\over 4\pi
r^{2}}\langle\hat{T}_{ab}^{(2)}\rangle,$ (7)

where indices $a$ and $b$ denote 2-dimensional space-time coordinates
$ds^{2}_{(2)}$ where $\langle\hat{T}_{ab}^{(2)}\rangle=\langle
0|\hat{T}_{ab}^{(2)}|0\rangle$ is evaluated. The Boulware state $|0\rangle$ is
associated with the Killing vector $\xi$ from the metric, i.e.,
$\xi^{\mu}=\delta^{\mu}_{t}$ which implies $\xi_{\mu}=-e^{\nu}\delta_{\mu t}$.
It is a usual practice to use null coordinate to obtain
$\langle\hat{T}_{ab}^{(2)}\rangle$ and then transform back to $(t,r)$
coordinate. It was shown in Barcelo:2011bb  that there is a shortcut
equivalent to the usual practice, i.e.,

$\langle\hat{T}_{ab}^{(2)}\rangle={1\over
48\pi}\left(R^{(2)}g_{ab}+A_{ab}-{1\over 2}g_{ab}A,\right)$ (8)

where $R^{(2)}$ is the Ricci scalar from the 2-dimensional metric and $A_{ab}$
is related to the Killing vector $\xi$ (in this case $|\xi|=e^{\nu/2}$) by

$A_{ab}={4\over|\xi|}\nabla_{a}\nabla_{b}|\xi|.$ (9)

From Wald’s axioms Wald:1995yp , the contracted Bianchi identity
$\nabla_{\mu}\langle\hat{T}^{\mu\nu}\rangle=0$ should be satisfied. This can
be checked by substituting the components of the RSET the identity.

Then from the contracted Bianchi identity $\nabla_{\mu}T^{\mu\nu}=0$ and the
above modified EFE (3) we obtain three equations of motion

$\displaystyle-p^{\prime}-{(p+\rho)\over 2}{\nu^{\prime}}$ $\displaystyle=$
$\displaystyle 0,$ (10) $\displaystyle{\nu^{\prime}\over r}-{2Gm\over
r^{3}(1-2Gm/r)}$ $\displaystyle=$ $\displaystyle{8\pi Gp\over
1-2Gm/r}-{l_{P}^{2}\over 4}\left({\nu^{\prime}\over r}\right)^{2},$ (11)
$\displaystyle{2Gm^{\prime}\over r^{2}}$ $\displaystyle=$ $\displaystyle 8\pi
G\rho+{l_{P}^{2}\over r^{2}}\left[\left(1-{2Gm\over
r}\right)\left(\nu^{\prime\prime}+\left(\nu^{\prime}\right)^{2}\right)\right.$
(12) $\displaystyle\left.-\left({Gm^{\prime}\over r}-{Gm\over
r^{2}}\right){\nu^{\prime}}-{3\over 4}\left(1-{2Gm\over
r}\right)\left(\nu^{\prime}\right)^{2}\right].$

Here the prime denotes differentiation with respect to $r$.

There are two roots of the metric solution $\nu^{\prime}$ from Eq. (11):

$\nu^{\prime}=-{2r\over l_{P}^{2}}\left(1\pm\sqrt{1+{l_{p}^{2}\over
r^{2}}{2Gm\over r}{(1+4\pi r^{3}p/m)\over(1-2Gm/r)}}\right).$ (13)

It leads to two different expressions for both of $p^{\prime}$ and
$m^{\prime}$. The negative and positive signs on the right-hand side will be
named as _negative_ and (_positive_) branches. In the limit of $l_{p}\to 0$,
the negative branch approaches the standard GR-TOV equation while the positive
branch does not. In the next section, we focus on this negative branch.
Regarding the validity of our numerical calculation, we provide the detailed
discussions in Appendix in page A.

##  III RESULTS AND DISCUSSION

###  III.1 Modified TOV Equations from Semi-classical Gravity

Here we investigate the model in Ref. Carballo-Rubio:2017tlh  with
phenomenological point of view, i.e, we only consider non-negative $p$. Since
in general $(l_{p}/r)^{2}X$ is not constant, where

$X={2Gm\over r}{(1+4\pi r^{3}p/m)\over(1-2Gm/r)},$ (14)

there was a hope for a more general EoS than (85), that is by relaxing the
requirement $\lambda=$ constant so that $\lambda$ is a function of $r$. By
inspecting equation

$p^{\prime}={(p+\rho)r\over l_{P}^{2}}\left(1+k\sqrt{1+{l_{p}^{2}\over
r^{2}}X}\right)$ (15)

where $k=\pm 1$, we discuss it in three cases: (1) ${l_{p}^{2}\over
r^{2}}X<1$, (2) ${l_{p}^{2}\over r^{2}}X>1$ and (3) ${l_{p}^{2}\over
r^{2}}X=1$. Clearly $X>0$, since $X\leq 0$ must be from $2Gm/r\geq 1$. From
case (1), by binomial expansion with respect to ${l_{p}^{2}\over r^{2}}X$ to
only a leading order, we obtain

$p^{\prime}=\begin{cases}{\rho+p\over 2}\left({4r\over l_{p}^{2}X}+{X\over
r}\right),&k=+1,\\\ -{\rho+p\over 2}{X\over r},&k=-1.\end{cases}$ (16)

Since for any reasonable perfect fluid demand positive pressure and surface of
the star demand zero pressure, we need $p^{\prime}<0$, which is not satisfied
by $p^{\prime}$ from $k=+1$ while $k=-1$ is just standard TOV equation. From
case (2), again by binomial expansion with respect to ${r^{2}/(l_{p}^{2}X)}$,
this leads us to

$p^{\prime}=k{(\rho+p)\sqrt{X}\over l_{p}}.$ (17)

This expression eliminates $k=+1$. From the case (3), this leads us to

$p^{\prime}={(p+\rho)r\over l_{P}^{2}}\left(1+k\sqrt{2}\right),$ (18)

which again imply that we should choose $k=-1$. Hence from now we shall only
investigate $p^{\prime}$ generated from $k=-1$ so that both $p_{c}>0$ and
$p(R)=0$ are satisfied.

Notice that Eq. (12) has $m^{\prime}$ on both sides, the full expression of
$m^{\prime}$ is rather lengthy. Since we choose $k=-1$, we use

$p^{\prime}={(p+\rho)r\over l_{P}^{2}}\left(1-\sqrt{1+{l_{p}^{2}\over
r^{2}}X}\right),$ (19)

then we have

$m^{\prime}={4\pi\rho r^{2}}\left({1+\sum_{i=1}^{7}A_{i}\over
1+\sum_{i=1}^{4}B_{i}}\right),$ (20)

where the nominator consists of

$\displaystyle A_{1}$ $\displaystyle=$ $\displaystyle\frac{3l_{p}^{2}p}{\rho
r^{2}\sqrt{1+{l_{p}^{2}\over r^{2}}X}},$ (21) $\displaystyle A_{2}$
$\displaystyle=$ $\displaystyle-\frac{3l_{p}^{2}m(1+4\pi r^{3}p/m)}{4\pi\rho
r^{5}\sqrt{1+{l_{p}^{2}\over r^{2}}X}},$ (22) $\displaystyle A_{3}$
$\displaystyle=$ $\displaystyle-\frac{Gl_{p}^{2}m^{2}(1+4\pi
r^{3}p/m)}{2\pi\rho r^{6}(1-2Gm/r)\sqrt{1+{l_{p}^{2}\over r^{2}}X}},$ (23)
$\displaystyle A_{4}$ $\displaystyle=$
$\displaystyle-\frac{m\left(1-\sqrt{1+{l_{p}^{2}\over
r^{2}}X}\right)}{4\pi\rho r^{3}},$ (24) $\displaystyle A_{5}$ $\displaystyle=$
$\displaystyle-\frac{\left(1-2Gm/r\right)\left(1-\sqrt{1+{l_{p}^{2}\over
r^{2}}X}\right)}{4\pi G\rho r^{2}},$ (25) $\displaystyle A_{6}$
$\displaystyle=$
$\displaystyle\frac{\left(1-2Gm/r\right)\left(1-\sqrt{1+{l_{p}^{2}\over
r^{2}}X}\right)^{2}}{8\pi G\rho l_{p}^{2}},$ (26) $\displaystyle A_{7}$
$\displaystyle=$ $\displaystyle\frac{l_{p}^{2}}{\rho r\sqrt{1+{l_{p}^{2}\over
r^{2}}X}}p^{\prime},$ (27)

and the denominator consists of

$\displaystyle B_{1}$ $\displaystyle=$ $\displaystyle\frac{4\pi
l_{p}^{2}pr}{m\sqrt{1+{l_{p}^{2}\over r^{2}}X}},$ (28) $\displaystyle B_{2}$
$\displaystyle=$ $\displaystyle-\frac{l_{p}^{2}\left(1+4\pi
r^{3}p/m\right)}{r^{2}\sqrt{1+{l_{p}^{2}\over r^{2}}X}},$ (29) $\displaystyle
B_{3}$ $\displaystyle=$ $\displaystyle-\frac{2Gl_{p}^{2}m\left(1+4\pi
r^{3}p/m\right)}{r^{3}\left(1-2Gm/r\right)\sqrt{1+{l_{p}^{2}\over r^{2}}X}},$
(30) $\displaystyle B_{4}$ $\displaystyle=$
$\displaystyle-{\left(1-\sqrt{1+{l_{p}^{2}\over r^{2}}X}\right)}.$ (31)

Since $m^{\prime}$ is complicated, we need to be careful of fixing the
constants and the initial data of pressure.

Before solving those equations numerically, let us investigate both
$p^{\prime}$ and $m^{\prime}$ at limit $l_{p}\to 0$. Expanding $p^{\prime}$
with respect to $l_{p}$, we have

$p^{\prime}=-\frac{G\left(m+4\pi
r^{3}p\right)(\rho+p)}{r(r-2Gm)}+\frac{G^{2}l_{p}^{2}\left(m+4\pi
r^{3}p\right)^{2}(\rho+p)}{2r^{3}(r-2Gm)^{2}}+\mathcal{O}(l_{p}^{3}).$ (32)

Let us expand these $A_{i}$ and $B_{i}$ w.r.t. $l_{p}$:

$\displaystyle A_{1}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\frac{3p}{\rho}+\mathcal{O}(l_{p}^{3})>0,$
(33) $\displaystyle A_{2}$ $\displaystyle=$
$\displaystyle-\frac{l_{p}^{2}}{r^{2}}\frac{3m\left(1+4\pi
r^{3}p/m\right)}{4\pi r^{3}\rho}+\mathcal{O}(l_{p}^{3})<0,$ (34)
$\displaystyle A_{3}$ $\displaystyle=$
$\displaystyle-\frac{l_{p}^{2}}{r^{2}}\frac{Gm^{2}\left(1+4\pi
r^{3}p/m\right)}{2\pi r^{4}\rho(1-2Gm/r)}+\mathcal{O}(l_{p}^{3})<0,$ (35)
$\displaystyle A_{4}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\frac{Gm^{2}\left(1+4\pi
r^{3}p/m\right)}{4\pi r^{4}\rho(1-2Gm/r)}+\mathcal{O}(l_{p}^{3})>0,$ (36)
$\displaystyle A_{5}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\frac{m\left(1+4\pi r^{3}p/m\right)}{4\pi
r^{3}\rho}+\mathcal{O}(l_{p}^{3})>0,$ (37) $\displaystyle A_{6}$
$\displaystyle=$ $\displaystyle\frac{l_{p}^{2}}{r^{2}}\frac{Gm^{2}\left(1+4\pi
r^{3}p/m\right)^{2}}{8\pi r^{4}\rho(1-2Gm/r)}+\mathcal{O}(l_{p}^{3})>0,$ (38)
$\displaystyle A_{7}$ $\displaystyle=$
$\displaystyle-\frac{l_{p}^{2}}{r^{2}}\frac{Gm(1+p/\rho)\left(1+4\pi
r^{3}p/m\right)}{r(1-2Gm/r)}+\mathcal{O}(l_{p}^{3})<0,$ (39)

and

$\displaystyle B_{1}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\frac{4\pi
r^{3}p}{m}+\mathcal{O}(l_{p}^{3})>0,$ (40) $\displaystyle B_{2}$
$\displaystyle=$ $\displaystyle-\frac{l_{p}^{2}}{r^{2}}\left(1+\frac{4\pi
r^{3}p}{m}\right)+\mathcal{O}(l_{p}^{3})<0,$ (41) $\displaystyle B_{3}$
$\displaystyle=$ $\displaystyle-\frac{l_{p}^{2}}{r^{2}}\frac{2Gm\left(1+4\pi
r^{3}p/m\right)}{r(1-2Gm/r)}+\mathcal{O}(l_{p}^{3})<0,$ (42) $\displaystyle
B_{4}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\frac{Gm\left(1+4\pi
r^{3}p/m\right)}{r(1-2Gm/r)}+\mathcal{O}(l_{p}^{3})>0,$ (43)

Since we know that $(Gm/r),(r^{3}p/m),(r^{3}\rho/m)$ and $(p/\rho)$ are
dimensionless, notice that no singularity occur on each $A_{i}$ and $B_{i}$ as
$r=r_{c}\sim 0$. Summing them up,

$\displaystyle\sum_{i=1}^{7}A_{i}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\left[\frac{3p}{\rho}-2\frac{m\left(1+4\pi
r^{3}p/m\right)}{4\pi r^{3}\rho}-\frac{Gm^{2}\left(1+4\pi
r^{3}p/m\right)}{4\pi r^{4}\rho(1-2Gm/r)}\right.$ (44)
$\displaystyle\left.+\frac{Gm^{2}\left(1+4\pi r^{3}p/m\right)^{2}}{8\pi
r^{4}\rho(1-2Gm/r)}-\frac{Gm(1+p/\rho)\left(1+4\pi
r^{3}p/m\right)}{r(1-2Gm/r)}\right]+\mathcal{O}(l_{p}^{3}),$
$\displaystyle\sum_{i=1}^{4}B_{i}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r^{2}}\left[-1-\frac{Gm\left(1+4\pi
r^{3}p/m\right)}{r(1-2Gm/r)}\right]+\mathcal{O}(l_{p}^{3}).$ (45)

The value of $\rho_{0}$ plays a significant role. On the surface of the
object, $p=0$, $\rho=\rho_{0}$, $r=R$ and $m=M$, so we have

$p^{\prime}(R)=-\frac{GM\rho_{0}}{R^{2}(1-2GM/R)}+\frac{G^{2}l_{p}^{2}M^{2}\rho_{0}}{2R^{5}(1-2GM/R)^{2}}+\mathcal{O}(l_{p}^{3})$
(46)

and

$\displaystyle\sum_{i=1}^{7}A_{i}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{R^{2}}\left[-2\frac{M}{4\pi
R^{3}\rho_{0}}-\frac{GM^{2}}{4\pi R^{4}\rho_{0}(1-2GM/R)}\right.$ (47)
$\displaystyle\left.+\frac{GM^{2}}{8\pi
R^{4}\rho_{0}(1-2GM/R)}-\frac{GM}{R(1-2GM/R)}\right]+\mathcal{O}(l_{p}^{3}),$
$\displaystyle\sum_{i=1}^{4}B_{i}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{R^{2}}\left[-1-\frac{GM}{R(1-2GM/R)}\right]+\mathcal{O}(l_{p}^{3}),$
(48)

hence $m^{\prime}(R)$ finite since $\rho_{0}\neq 0$. It seems that if the
value of $\rho_{0}$ goes to zero, the changes of both $R$ and $M$ might be
significant.

Now in the center $r=r_{c}\sim 0$ we have

$\displaystyle\sum_{i=1}^{7}A_{i}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r_{c}^{2}}\left[\frac{3p_{c}}{\rho_{c}}-2\frac{\left(1+3p_{c}/\rho_{c}\right)}{3}-\frac{Gm_{c}^{2}\left(1+3p_{c}/\rho_{c}\right)}{4\pi
r_{c}^{4}\rho_{c}(1-2Gm_{c}/r_{c})}\right.$ (49)
$\displaystyle\left.+\frac{Gm_{c}^{2}\left(1+3p_{c}/\rho_{c}\right)^{2}}{8\pi
r_{c}^{4}\rho_{c}(1-2Gm_{c}/r_{c})}-\frac{Gm_{c}(1+p_{c}/\rho_{c})\left(1+3p_{c}/\rho_{c}\right)}{r_{c}(1-2Gm_{c}/r_{c})}\right]+\mathcal{O}(l_{p}^{3})$
$\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r_{c}^{2}}\left[\frac{p_{c}}{\rho_{c}}-\frac{2}{3}\right]+\mathcal{O}(r_{c}^{2}),$
$\displaystyle\sum_{i=1}^{4}B_{i}$ $\displaystyle=$
$\displaystyle\frac{l_{p}^{2}}{r_{c}^{2}}\left[-1-\frac{Gm_{c}\left(1+3p_{c}/\rho_{c}\right)}{r_{c}(1-2Gm_{c}/r_{c})}\right]+\mathcal{O}(l_{p}^{3})=-\frac{l_{p}^{2}}{r_{c}^{2}}+\mathcal{O}(r_{c}^{2}),$
(50)

where we have substitute the mass in this limit
$m_{c}=(4/3)\pi\rho_{c}r_{c}^{3}$. Then we have

$p^{\prime}(r_{c})\sim-\frac{4}{3}\pi
Gr_{c}(\rho_{c}+p_{c})(\rho_{c}+3p_{c})+\frac{8}{9}\pi^{2}G^{2}l_{p}^{2}r_{c}(\rho_{c}+p_{c})(\rho_{c}+3p_{c})^{2}$
(51)

and

$m^{\prime}(r_{c})\sim 4\pi
r_{c}^{2}\rho_{c}\left(\frac{1+(l_{p}/r_{c})^{2}(3p_{c}/\rho_{c}-2/3)}{1-(l_{p}/r_{c})^{2}}\right).$
(52)

These equations are valid only when $r$ near $r_{c}$ and clearly
$m^{\prime}(r_{c})$ is singular when $r_{c}=l_{p}$.

Suppose that we have at the center $r_{c}=\alpha l_{p}$ where $\alpha>0$. Then
this means that $\alpha\neq 1$. Notice that Eq. (51) can be rewritten as

$p^{\prime}(r_{c})\simeq-\frac{4}{3}\pi
G\rho_{c}^{2}\left(1+\frac{p_{c}}{\rho_{c}}\right)\left(1+3\frac{p_{c}}{\rho_{c}}\right)\alpha
l_{p}\left[1-\frac{2}{3}\pi
G\rho_{c}\left(1+3\frac{p_{c}}{\rho_{c}}\right)l_{p}^{2}\right].$ (53)

The second term in the square bracket can be bigger than the first term if
$\alpha l_{p}^{3}$ is large enough than $\alpha l_{p}$. Hence, we conclude
that $\alpha l_{p}^{3}$ should be sufficiently small such that
$p^{\prime}(r)<0$. Looking at Eq. (53), then the higher-order terms should
have the pattern $\mathcal{O}\left(\alpha l_{p}^{2n+1}\right)$, with $n$ an
integer. This means that we should hold $l_{p}$ first then adjusting $\alpha$
such that the second term in Eq. (53) does not dominate the first one, i.e.,

$l_{p}<\sqrt{\frac{3}{2\pi G(\rho_{c}+3p_{c})}}\text{, or equivalently,
}N<\frac{18}{L_{\text{Pl}}^{2}G(\rho_{c}+3p_{c})}.$ (54)

Hence the upper bound of $l_{p}$ depends on both $p_{c}$ and the EoS, which
should satisfy the strong energy condition.

By defining an arbitrary positive valued constant $\alpha=r_{c}/l_{p}$, Eq.
(52) becomes

$m^{\prime}(r_{c})\sim 4\pi
r_{c}^{2}\rho_{c}\left[1-\frac{3p_{c}/\rho_{c}+1/3}{1-\alpha^{2}}\right].$
(55)

Logically the mass should grow from center to the surface, so
$m^{\prime}(r)>0$ ($0<r\leq R$). Notice that if $\alpha>1$ then
$m^{\prime}(r_{c})>0$ is trivially satisfied. On the other hand, if $\alpha<1$
then $m^{\prime}(r_{c})>0$ can happen if the second term inside the square
bracket is less than unity. This implies

$3p_{c}<(2/3-\alpha^{2})\rho_{c}.$ (56)

If $\alpha=\sqrt{2/3}$ then $p_{c}<0$. This negative center pressure is
contradicting with our previous assumption that $p>0$ and $\rho>0$ inside the
star. If $\alpha>\sqrt{2/3}$ then

$\rho_{c}<-\left({3p_{c}\over\alpha^{2}-2/3}\right)<0.$ (57)

This expression is also a contradiction with our previous assumption. If
$\alpha<\sqrt{2/3}$ then

$\rho_{c}>\left({3p_{c}\over 2/3-\alpha^{2}}\right)>4.5p_{c},$ (58)

which imply speed of sound squared $dp/d\rho=w<2/9$, lower than the upper
bound from QCD and causality, where $w\leq 1/3$ and $w\leq 1$, respectively
Moustakidis:2016sab . According to Ref. Urbano:2018nrs , the maximum
compactness produced by a linear equation of state (EoS) $\rho=p/w+\rho_{0}$
is

$\left({2GM\over
R}\right)_{\text{max}}\sim\frac{8}{9\left(\frac{0.51w+0.77}{w(w+4.18)}+1\right)},$
(59)

which is a monotonically increasing function of $w$ as $w$ grow but cannot go
beyond the Buchdahl limit $2GM/R=8/9$. So $w<2/9$ will produce less compact
stars than $w=1/3$. Also notice that this condition $m^{\prime}(r)$ applies
only at $r=r_{c}$, so it may not be true for $r>r_{c}$.

The case of $\alpha<1$ may still be investigated with negative valued $m_{c}$,
which is possible according to Ref Anastopoulos:2020mrt . From Eq. (45) then
we can see that singularity of $m^{\prime}(r)$ may not be there anymore for
$\alpha=1$. If we demand $\sum_{i}B_{i}>0$ then

$m_{c}<\frac{\rho_{c}r_{c}}{G\rho_{c}-3Gp_{c}}\equiv m_{c,max}$ (60)

and since $m_{c}<0$ we obtain another restriction for the EoS

$w>1/3.$ (61)

Notice that $m_{c,max}$ can be made close to zero so we can use limit
$m_{c}\to 0^{-}$. Eq. (32) in the limit of $m_{c}\to 0^{-}$ becomes
$p^{\prime}(r_{c})<0$ so $p$ will decrease from $p_{c}$. In this limit, Eq.
(44) becomes

$\sum_{i}A_{i}\sim\alpha^{-2}\left[{p_{c}\over\rho_{c}}-{2\over 3}\right].$
(62)

Since we demand $\sum_{i}B_{i}>0$ and $m^{\prime}(r)>0$, then

$w>2/3.$ (63)

This is from considering $m_{c}\to 0^{-}$. If on the other hand we consider
$m_{c}\to-\infty$, notice that Eq. (32) set the minimum value of $m_{c}$,
i.e.,

$0>m_{c}>-4\pi r_{c}^{3}p_{c}\equiv m_{c,min}.$ (64)

Notice that $m_{c,min}$ can be very close to zero so the range of $m_{c}$ is
quite small. Thus for the case of $m_{c}<0$, we obtain that both $w$ and
$m_{c}$ are restricted to $w>2/3$ and $m_{c,min}<m_{c}<m_{c,max}$. Due to the
tight restriction of $m_{c}<0$, we do not focus on $m_{c}<0$ case in the
following sections but rather on the usual positive valued $m_{c}=4\pi
r_{c}^{3}\rho_{c}/3$.

Figure 1: Here we have profiles of $p$ and $m$ in the upper panels from
$l_{p}=1$ mm and $\alpha=20$ and we vary the speed of sound squared $w$. The
metric function $\exp(\nu)$ and $1-2Gm/r$ are also shown for each $w$ in the
lower panels. Figure 2: Here we show the M-R curve from varying $w$. The semi-
classical correction does not produce significant discrepancy compared to TOV
from GR. Figure 3: Here we show the M-R curve from varying $B$. Decreasing $B$
produces higher mass and larger radius. Figure 4: Here we show the M-R curve
from varying $\alpha$. The choice of $\alpha$ does not show any difference in
the result. Figure 5: Here we show the M-R curve from varying $l_{p}$.
Increasing $l_{p}$ also does not significantly shift the M-R curve.

###  III.2 Numerical Results

The EoS used here have the following linear form

$\rho(p)=p/w+\rho_{0},$ (65)

where $w$ is the speed of sound squared from thermodynamics and $\rho_{0}$ is
a positive valued constant. The constant $\rho_{0}$ cannot be zero since it
implies no solution and cannot be negative, or it will violate weak energy
condition Urbano:2018nrs . The constant $w$ is restricted to at least two
conditions Moustakidis:2016sab : (1) from causality, we have $0<w\leq 1$ and
(2) from QCD and other known theories, we have $0<w\leq 1/3$. Here we focus on
compact stars with largest compactness possible, so we shall choose $1/3\leq
w\leq 1$ and $\alpha>1$.

To integrate both $p^{\prime}$ and $m^{\prime}$ numerically we need to fix
$l_{p},\alpha,\rho_{0}$, and $p_{c}$ carefully such that all boundary
conditions are satisfied. Here we shall use natural units for those four
constants following Ref. Diener:2008bj . In this set of units called “NS”
units, $r$ is in metres, both $p$ and $\rho$ are in MeV/fm3 and $m$ is in MeV
m3/fm3. The compactness is defined as $C=GM/R$. We use $\rho_{0}=4B$, where
$B$ is the so-called bag constant from the MIT bag model Mannarelli:2018pjb .

The upper bound of $l_{p}$ can now be determined. It is usual to have both
$p_{c}$ and $\rho_{0}$ at most $\sim 10^{3}$ MeV/fm3, then by Eq. (54), we
have $l_{p}\lesssim 10$ km. Since the observed neutron stars have $R\sim 10$
km, we have $l_{p}<R$, which is trivial if we consider $l_{p}$ to be related
to the Planck length. This finding is equivalent to fixing the upper bound for
the dimensionless parameter $N$, which is $N<10^{79}$. This number is a huge
quantity since the RSET should satisfy (in SI units using $\hbar c=3.162\times
10^{-27}$ kg m3 s-2)

$|\langle T_{\mu\nu}\rangle|<\frac{|T_{\mu\nu}|}{3.162\times 10^{52}\text{ kg
m${}^{3}$ s${}^{-2}$}}.$ (66)

Since the smallness of $l_{p}$ often make calculators cannot detect the second
term in the square root in Eq. (19), this form tends to make $p^{\prime}=0$ in
some $r$. To evade this, we use an equivalent form of Eq. (19), i.e.,

$p^{\prime}=-\frac{(p+\rho)X}{r\left(1+\sqrt{1+(l_{p}/r)^{2}X}\right)}.$ (67)

The following numerical results show that the contribution of RSET does not
significantly affect the maximum mass. An example for profiles from varying
$w$ are shown in Fig. 1. We see that the contribution of $w$ does make higher
mass and larger radius resulting in an ultra-compact star, but we will not
discuss its echo property since this compactness can also be obtained from the
standard TOV equation in GR.

Varying $B$, $\alpha$ and $l_{p}$ lead to similar profiles. The M-R curves
from varying $w$, $B$, $\alpha$ and $l_{p}$ are shown in Figs. 2, 3, 4 and 5,
respectively. Here the semi-classical correction does not produce significant
discrepancy from arbitrary choice of $p_{c}$ compared to TOV from GR.

We also had tried using $\alpha<1$, but we cannot find the numerical solution.
The mass always goes down to a negative value, even though it increases a
little in the beginning. It is interesting to note that for the similar case
considered in Ref. Ho:2017vgi  their integration method breaks down at
$r=l_{p}$. They use constant energy density, and they integrate the equations
from the surface to the core.

##  IV Conclusions

In this work, we analyze the semi-classical theory of gravity proposed in Ref.
Carballo-Rubio:2017tlh . Note that the theory has two different sets of
equations characterized by the sign in $p^{\prime}$. This happened since the
Einstein equation for metric solution $\exp(\nu)$ in (13) has two roots and
$p^{\prime}$ is related to $\nu^{\prime}$ by Eq. (10).

From the positive branch, defining a constant $\lambda$ is done to obtain
analytic solutions. By definition of $\lambda$, $p$ is related to energy
density $\rho$ and mass $m$. The solution’s character $p$ is as follows:
starting from the negative value of $p$ at $r\to 0$ and increase the $p$ to
zero. This solution resembles a combination of the known gravastars or black
star models. We show that the numerical solution can be obtained by
integrating from its surface to its core.

From the negative branch, which goes to the usual TOV equation at the formal
limit $l_{p}\to 0$, we had reproduced the results from Ho:2017vgi  for the
case of constant energy density. In this paper, we analyze this negative
branch further by applying linear EoS $\rho=p/w+\rho_{0}$ with $w$=1/3 and
$w$=1, respectively.

The range of $l_{p}$ is dependent on the choice of $r_{c}$. The reason is as
follows. Since the equation $m^{\prime}(r)$ is much more complicated than
usual TOV equation (see Eqs.(20)-(31)), it has terms that can make
$m^{\prime}(r)<0$, which may imply negative $m$ since at the center $m_{c}\sim
0$. These terms are dependent on the values of both $l_{p}$ and $r_{c}$ such
that $m^{\prime}(r_{c})$ is singular when $l_{p}=r_{c}$.

To investigate $m^{\prime}(r)$ around this singularity, we fix a relation
$r_{c}=\alpha l_{p}$ with $\alpha\neq 1$. We also demand $m^{\prime}(r)>0$ for
$0<r<R$ since it is usual practice to expect the mass increase from the center
to surface. We found that, with some approximation steps,
$0<\alpha<\sqrt{2/3}$ demands a very restricted set of EoS with a speed of
sound squared $w<2/9$. This value implies the compactness lowered, so we are
not discussing this aspect. For $\sqrt{2/3}\leq\alpha<1$, this demands either
$p_{c}<0$ or $\rho_{c}<0$ which contradict our starting assumption that both
pressure and energy density should not be negative. For the case of
$\alpha>1$, we can use any EoS. Thus our analysis agrees with Ref. Ho:2017vgi
, i.e., their integration breaks down at $r=l_{p}$.

The value of $l_{p}$ has an upper bound-constrained by the pressure at the
center $p_{c}$. From our approximation, we found from the dimensional analysis
that $l_{p}<10$ km, which is of order $R$. This upper bound implies
$N<10^{79}$. This fact implies that $l_{p}$ can be much larger than Planck
length.

Through varying all four parameters $w$, $B$, $\alpha$, and $l_{p}$, our
numerical results show that the M-R curves are indistinguishable compared to
the TOV equation results in GR. Hence the parameter $l_{p}$, though not in the
limit $l_{p}\to L_{\text{Pl}}$, has no significant signature compared to the
standard TOV equation.

Compactness reaching the ultra-compact range can be achieved by adjusting both
$w$ and $B$ in the TOV equation system with and without semi-classical
correction. Moreover, the authors in Ref. Ho:2017vgi  showed that the Buchdahl
limit could be violated using constant energy density, which means setting $w$
equal to infinity. Hence, it seems that we need to reach somehow much higher
compactness larger than the Buchdahl limit to see the significance of the
semi-classical correction. This way might be done either by adjusting
decreasing $B$ to near zero or increasing $w$ such that $w>1$, which violates
causality.

Hence we conclude this section with the following points. (1) The structure of
the Eqs. (19)-(31) restricts the range of $\alpha$ and $w$ in some ways that
turn out to give results not significant if compared to the usual TOV equation
system. Therefore, similar to the conclusion in Ref. Urbano:2018nrs  with GR
theory that the ultra-compact objects from the negative branch of semi-
classical gravity with a linear equation of state are unable to generate
demanding gravitational echoes. (2) Our numerical results are consistent with
both Ref. Urbano:2018nrs  and Ref. Ho:2017vgi . See the detailed discussion in
Appendix. (3) The positive branch is still open for more detailed numerical
analysis. We expect that the quantum effect within semi-classical
approximation is only significant for exotic compact objects, considered the
solution related to the positive branch.

###### Acknowledgements.

This work is funded by Publikasi Terindeks Internasional (PUTI) Doktor 2020,
No. NKB-614/UN2.RST/HKP.05.00/2020.

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##  Appendix A Discussions Regarding Numerical Method

In this appendix, we discuss the integration schemes and their results. As a
disclaimer, in this paper, all the calculations are done using Mathematica
10.0. The schemes are named as _forward_ (_backward_ , resp.) integration,
corresponding to integrating from center to surface (surface to center). It is
known that for the standard TOV equation, we can do both forward and backward
integration. Here we discuss both forward, and backward integration to obtain
numerical solutions for (1) negative branch obtained previously by Ho and
Matsuo Ho:2017vgi  in the case of constant energy density $\rho$ and (2) the
negative branch obtained previously by Carballo-Rubio Carballo-Rubio:2017tlh
using constant $\lambda$ trick.

We emphasize these forward and backward integration schemes because we use
forward integration in the next section, but both Carballo-Rubio:2017tlh  and
Ho:2017vgi  imply backward integration. We show that for the negative branch,
we get consistent results from both backward and forward integration. This
evidence justifies our numerical method for the negative branch system that we
discussed in the section III onwards.

####  A.0.1 Forward and backward integration solutions from negative branch

Suppose we redefine the $rr$ component of the metric by

$\displaystyle F(r)$ $\displaystyle=$
$\displaystyle\sqrt{C(r)\left(1-{2Gm(r)\over r}\right)},$ (68) $\displaystyle
C(r)$ $\displaystyle=$ $\displaystyle e^{\nu(r)},$ (69)

and the constants by $\alpha=l_{p}^{2}$ and $\kappa=8\pi G$. Then the
equations (11) and (12) are equivalent to equations (5.7) and (5.8) in
Ho:2017vgi ,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{8r^{2}C(r)^{2}}\left[-3\alpha
F(r)^{2}C^{\prime}(r)^{2}+4rC(r)^{2}F(r)F^{\prime}(r)\right.$ (70)
$\displaystyle\left.+2C(r)F(r)\left(\alpha
C^{\prime}(r)F^{\prime}(r)+F(r)\left(\alpha
C^{\prime\prime}(r)-2rC^{\prime}(r)\right)\right)\right.$
$\displaystyle\left.+2\kappa r^{2}C(r)^{3}(p+\rho)\right],$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle\frac{1}{4r^{2}C(r)^{2}}\left[\alpha
F(r)^{2}C^{\prime}(r)^{2}-\alpha
C(r)F(r)\left(F(r)C^{\prime\prime}(r)+C^{\prime}(r)F^{\prime}(r)\right)\right.$
(71)
$\displaystyle\left.-2C(r)^{2}F(r)\left(rF^{\prime}(r)+F(r)\right)+C(r)^{3}\left(2-\kappa
r^{2}(\rho-p)\right)\right]$

To avoid confusion, we keep our symbol for energy density and metric function
as $\rho$ and $\nu$, respectively, while in Ho:2017vgi  the authors use $m$
and $\rho$. In this paper, we assign $m$ as the mass function. Setting the
energy density as constant $\rho=\rho_{0}$, we can see from (10) that the
pressure becomes

$p(r)=-\rho_{0}+p_{0}e^{-\nu(r)/2}.$ (72)

Here $p_{0}$ is a constant that should satisfy the boundary condition at the
surface $p(R)=0$, i.e., $p_{0}=\rho_{0}\sqrt{1-2GM/R}$ with $M=m(R)$.

Only in this subsection, the integrations are done for the tortoise coordinate
$r_{*}$ which is defined as

$r_{*}=\int{dr\over F(r)}.$ (73)

This then give us the equations for $r(r_{*})$ and $\nu(r_{*})$ as

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{8r(r_{*})^{2}}\left[2\left\\{\alpha\nu^{\prime\prime}(r_{*})-2r(r_{*})\nu^{\prime}(r_{*})r^{\prime}(r_{*})\right\\}\right.$
(74) $\displaystyle\left.+4r(r_{*})r^{\prime\prime}(r_{*})+2\kappa
r(r_{*})^{2}p_{0}e^{\nu(r_{*})/2}-\alpha[\nu^{\prime}(r_{*})]^{2}\right],$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{1}{4r(r_{*})^{2}}\left[-\alpha\nu^{\prime\prime}(r_{*})-2(r(r_{*})r^{\prime\prime}(r_{*})+[r^{\prime}(r_{*})]^{2})\right.$
(75) $\displaystyle\left.+e^{\nu(r_{*})}[2-\kappa
r(r_{*})^{2}(2\rho_{0}-p_{0}e^{-\nu(r_{*})/2})]\right].$

From substracting (74) with (75) we can see there are two roots of
$\eta^{\prime}(r_{*})$, corresponding to the negative and positive branch we
had discussed before. Since in this subsection we focus on the negative
branch, which goes to TOV equation in the limit $\alpha\to 0$

$\nu^{\prime}(r_{*})=-\frac{2}{\alpha}\left(r(r_{*})r^{\prime}(r_{*})-\sqrt{r(r_{*})^{2}\left(\alpha\kappa
p_{0}e^{\frac{\nu(r_{*})}{2}}+r^{\prime}(r_{*})^{2}-\alpha\kappa\rho_{0}e^{\nu(r_{*})}\right)+\alpha\left(e^{\nu(r_{*})}-r^{\prime}(r_{*})^{2}\right)}\right).$
(76)

This corresponds to the second equation

$\displaystyle r^{\prime\prime}(r_{*})$ $\displaystyle=$
$\displaystyle\frac{1}{4\left(\alpha-r(r_{*})^{2}\right)r^{\prime}(r_{*})}\left[\kappa
r(r_{*})^{2}e^{\frac{\nu(r_{*})}{2}}\left(2\rho_{0}e^{\frac{\nu(r_{*})}{2}}-p_{0}\right)\right.$
(77) $\displaystyle\left.\times\left(2\sqrt{r(r_{*})^{2}\left(\alpha\kappa
p_{0}e^{\frac{\nu(r_{*})}{2}}+r^{\prime}(r_{*})^{2}-\alpha\kappa\rho_{0}e^{\nu(r_{*})}\right)+\alpha\left(e^{\nu(r_{*})}-r^{\prime}(r_{*})^{2}\right)}-\alpha\nu^{\prime}(r_{*})\right)\right.$
$\displaystyle\left.-\left(2\sqrt{r(r_{*})^{2}\left(\alpha\kappa
p_{0}e^{\frac{\nu(r_{*})}{2}}+r^{\prime}(r_{*})^{2}-\alpha\kappa\rho_{0}e^{\nu(r_{*})}\right)+\alpha\left(e^{\nu(r_{*})}-r^{\prime}(r_{*})^{2}\right)}-\alpha\nu^{\prime}(r_{*})\right)\right.$
$\displaystyle\left.\times
2e^{\nu(r_{*})}+4r(r_{*})r^{\prime}(r_{*})\left(\alpha\kappa
p_{0}e^{\frac{\nu(r_{*})}{2}}+r^{\prime}(r_{*})^{2}-\alpha\kappa\rho_{0}e^{\nu(r_{*})}\right)\right]$

So we have a first order equation for $\nu(r_{*})$ and second order equation
for $r(r_{*})$. These depends on three parameters $M$, $R_{*}=r_{*}(R)$,
$\rho_{0}$ and two coupling constants $\alpha$ and $\kappa$. The initial
conditions at $r_{*}=R_{*}$ are

$\displaystyle R_{*}$ $\displaystyle=$ $\displaystyle R+2GM\ln(R/2GM-1),$ (78)
$\displaystyle\nu(R_{*})$ $\displaystyle=$ $\displaystyle\ln(\sqrt{1-2GM/R}),$
(79) $\displaystyle r(R_{*})$ $\displaystyle=$ $\displaystyle R,$ (80)
$\displaystyle r^{\prime}(R_{*})$ $\displaystyle=$ $\displaystyle 1-2GM/R.$
(81)

Then we do the backward integration. To fix the three input parameters $R,M,$
and $\rho_{0}$, we follow recipe in Ho:2017vgi : fix $M$ first, then fix $R$
such that $R>2GM$, then explore the value of $\rho_{0}$.

To every value of $\rho_{0}$, we have different shooting parameter values $k$
related to $R$ by $R=2GM/k$, so it is restricted to $0<k<1$. It is also
carefully chosen so that at the center the slope of $p$ is not steep
($|p^{\prime}(r_{*}\to-\infty)|<\infty$). This result has two reasons. First,
assigning $p^{\prime}(r_{*}\to-\infty)\sim 0$ is done so that when we
integrate the equation in the opposite direction (forward), we can get the
same curves. Second, the negative branch goes to the TOV equation in the small
$\alpha$ limit and it is known that, in the TOV system, $p^{\prime}(r=0)\sim
0$ so the negative branch should also has this property for sufficiently small
$\alpha$. We assume that $r_{*}$ is linear to $r$ in the region near $r=0$ so
that $p^{\prime}(r_{*}\to-\infty)\sim 0$ imply $|p^{\prime}(r\to 0)|<\infty$.

Figure 6: The backward and forward integration on the negative branch gives
same numerical curves. Here we integrate them with respect to tortoise
coordinate $r_{*}$. Here $k$ is the shooting parameter to determine $R$ by
$R=2GM/k$ so that $p^{\prime}(r_{*}=0)\sim 0$. Here $e^{(\nu)}$ goes really
close to zero as $r_{*}\to-\infty$ but never reach zero.

The results from doing backward integration and calculating again using
forward integration are shown in Fig. 6. We obtain that the forward
integration produces the same curves as the backward one. The pressure, in the
region $r_{*}\to-\infty$, is indeed does not go to Planck scale for the
macroscopic size of $R$.

####  A.0.2 Forward and backward integration solutions from positive branch

The author in Ref. Carballo-Rubio:2017tlh  had shown that the positive branch
have exact solutions whose properties are very similar to a mixture of
gravastars and black stars, i.e., the negative pressure in the interior

$p={-1+\mathcal{O}(l_{p}^{2}/r^{2})\over 8\pi
Gr^{2}R^{2}}\left[R^{2}-r^{2}e^{\frac{(\lambda+1)(r^{2}-R^{2})}{l_{p}^{2}}}\right],$
(82)

with $\lambda>1$ a constant, but its energy density is still positive valued

$\rho(r)={1+\mathcal{O}(l_{p}^{2}/r^{2})\over 8\pi
Gr^{2}R^{2}}\left[R^{2}+r^{2}e^{\frac{(\lambda+1)(r^{2}-R^{2})}{l_{p}^{2}}}\right].$
(83)

Notice that $p(0)<0$ and $\rho(0)>0$. Moreover, both satisfy weak, dominant,
and null energy conditions. The strong energy conditions are assumed
unnecessary since the existence of Casimir energy in experiments violate it.
The key point of the derivation of Eqs. (82) and (83) is the introduction of a
constant $\lambda>1$ that is defined as

$\lambda\equiv\sqrt{1+{l_{p}^{2}\over r^{2}}{2Gm\over r}{(1+4\pi
r^{3}p/m)\over(1-2Gm/r)}},$ (84)

where it implies a fixed form of EoS and mass profile

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle-p+\frac{l_{p}^{2}}{(\lambda+1)r}p^{\prime},$ (85)
$\displaystyle m$ $\displaystyle=$ $\displaystyle\frac{r^{3}\left(-8\pi
Gl_{p}^{2}p+\lambda^{2}-1\right)}{2G\left(l_{p}^{2}+\left(\lambda^{2}-1\right)r^{2}\right)}.$
(86)

Since $p=0$ at $r=R$, we have the compactness determined by $R,l_{p}$ and
$\lambda$ as

$2C={2GM\over
R}=\frac{1}{\left(l_{p}^{2}/\left[R^{2}\left(\lambda^{2}-1\right)\right]+1\right)},$
(87)

with $M=m(R)$. The resulting exact solution for metric is

$\nu(r)=\nu(R)+{\frac{(\lambda+1)(R^{2}-r^{2})}{l_{p}^{2}}},\\\ $ (88)

with $\nu(R)=\ln\left(1-2GM/R+\mathcal{O}(l_{p}^{2}/R^{2})\right)$. The
solutions can lead to arbitrary compactness depending on the value of
$\lambda$ and are stable by curvature and boundary conditions arguments.

After some algebra, one can arrive at the following equation of motion

$\displaystyle p^{\prime}(r)$ $\displaystyle=$ $\displaystyle g(r)+p(r)h(r),$
(89) $\displaystyle g(r)$ $\displaystyle=$
$\displaystyle\frac{(\lambda+1)^{2}\left(l_{p}^{4}+(\lambda-2)l_{p}^{2}r^{2}-\left(\lambda^{2}-1\right)r^{4}\right)}{4\pi
Gl_{p}^{2}r\left(l_{p}^{2}-(\lambda+1)r^{2}\right)\left(l_{p}^{2}+\left(\lambda^{2}-1\right)r^{2}\right)},$
(90) $\displaystyle h(r)$ $\displaystyle=$
$\displaystyle\frac{2(\lambda+1)r\left((2\lambda+1)l_{p}^{4}+\left(\lambda^{3}+\lambda^{2}-2\lambda-2\right)l_{p}^{2}r^{2}-(\lambda-1)(\lambda+1)^{2}r^{4}\right)}{l_{p}^{2}\left(l_{p}^{2}-(\lambda+1)r^{2}\right)\left(l_{p}^{2}+\left(\lambda^{2}-1\right)r^{2}\right)}.$
(91)

We integrate by backward integration starting from $r=R$ and $p(R)=0$ inward.
Since this is a first order equation for $p$, we can see that $R$ is already
tied with $M$ so there is no shooting parameter $k$ like in the previous
subsection, where $R$ can be set at an arbitrary value as long as $R>2GM$ and
$p^{\prime}(r=0)\sim 0$. From $g(r)$, we can see that the equation has
singularity at $r=0$ so we expect uncontrolled behavior of $p$ near $r=0$.

Figure 7: Profiles of $p(r)$ from the positive branch from backward
integration (top panel), forward integration (middle panel), and the
comparison between each numerical results (bottom panel).Here we calculate
first using backward integration, then the values from backward integration is
used for forward integration. In the most bottom panel, we show the real and
imaginary part of $\log(p/p_{c})$ since for the forward integration,
$p/p_{c}=0$ at $r=r_{f}\sim 5$ m and after this point $p/p_{c}<0$ when
$r>r_{f}$ and thus $\log(p/p_{c})\in\mathbb{C}$. Hence both scheme does not
give similar result.

We show the backward integration solutions in top panel of Fig. 7 and it is
clear that it violates $|p^{\prime}(r\to 0)|<\infty$. We tried to use the
smallest possible $r=r_{c}$ as the initial position and its corresponding
$p(r_{c})$ for starting the forward integration. The resulting curve in the
middle panel, while the comparison is in the bottom panel. The results from
backward and forward integration are different. The real and imaginary words
in the most bottom panel is the real and imaginary part of $\log(p/p_{c})$
from both backward and forward integration. We show them here since the
pressure $p/p_{c}$ from forward integration suddenly goes to zero at $r=r_{f}$
much earlier than $R$ from the backward integration, where after this point
$p/p_{c}<0$ and thus $\log(p/p_{c})\in\mathbb{C}$ for forward integration. In
Fig. 7, $r_{f}\sim 5$ m while $R=10$ km. We still do not yet know how to
remedy this, but the numerical result from backward integration perfectly fits
the analytic solution (82).

####  A.0.3 Comparison of our numerical results with the density for regular
geometry part of Ho-Matsuo model

In this subsubsection, we intend to check the validity of our numerical
calculation by comparing the typical result of section 6.1.3, density for
regular geometry in Ref. Ho:2017vgi  with the same EOS (constant energy
density EOS) input but calculating by using our code where we restricted the
calculation only for the interior of the star.Note that we perform a
calculation by using backward and forward integrations to check the
consistency of the numerical calculation. It means that the geometry is
regular for 0 $\leq$ r $\leq$ R, and pressure is finite and positive inside
the star. Note that we do not investigate the too small density and too large
density cases of the Ho-Matsuo model wherein both latter cases $r_{*}$ do not
always monotonically increase when $r$ increased. We had reproduced relatively
similar plots with Fig. 14 and Fig. 15 of Ref. Ho:2017vgi , and the trend of
the plot satisfies the condition discussed in section 6.1.3 of Ref. Ho:2017vgi
. The results are shown in Fig. 6. However, we unable to reproduce exactly the
quantitative result as shown in Fig. 14 and Fig. 15 of Ref. Ho:2017vgi  since
the authors of Ref. Ho:2017vgi  did not specify the units they are using.
Therefore, we only guess the initial conditions until we obtain similar
behavior of both $r(r_{*})$ and $\nu(r_{*})$. We choose $R$ and $\rho_{0}$
suitably (with $2GM$ fixed) such that the profile $p(r_{*})$ decrease
monotonically as $r_{*}$ increases. It turns out that this choice gives us $r$
that is increasing monotonically as $r_{*}$ increases as the one shows in Ref.
Ho:2017vgi , so in principle, we can interchange $r$ and $r_{*}$. If we replot
all functions in Fig. 6 as functions of $r$ instead of $r_{*}$, we obtain Fig.
8. In Fig. 8, we show the numerical calculation result by integrating the
equations (76)-(77) from center to surface. We compare it with the usual TOV
equation in GR. We can notice that $p$ becomes very steep as $r\to 0$ since
$r$ (as a function of $r_{*}$) drop really quick in that area. We also notice
that $m_{c}$ (mass near the center) is negative valued and quite large in
magnitude. We suspect that this negative $m_{c}$ may be why the compactness
can be very close to the black hole limit ($2GM/R=1$). We suspect that this
$m_{c}<0$ is the reason for such high compactness. However, this value is
indeed outside the range from our lower and upper bound ($m_{c,min}$ and
$m_{c,max}$ from Eq. (64) and Eq. (60)) which is $-10^{-3}M_{\odot}\lesssim
m_{c}\lesssim-10^{-7}M_{\odot}$. Therefore, it seems that our analytical upper
and lower bound estimation of $m_{c}<0$ may not be to justify for constant
energy density EOS.

Figure 8: The solid line in these plots are the same as in Fig. 6 but the
dashed line is from TOV GR system. Both data are integrated from center to
surface and notice the large compactness $2GM/R$ at the right bottom panel.
Notice that $p$ increases dramatically as $r\to 0$. Since in SCGrav $r>l_{p}$,
then $p$ will never go to infinity. But in TOV GR, $r$ can be arbitrarily
close to zero so $p$ will go to infinity as $r\to 0$. Notice also that the
mass at the center ($m_{c}$) is negative valued. This may be the reason why
the compactness can be very close to black hole limit ($2GM/R=1$) for the case
of constant energy density EOS.

To obtain such high compactness, we also observe that the system should have
very large density $\rho_{0}$ and central pressure $p_{c}$. In our units, we
estimate that $\rho_{0}\sim 10^{12}$ MeV fm-3. This is about nine orders of
magnitude larger than the bag constants $B$ that we used. Notice also from
Fig. 8 that $p_{c}\sim 10^{16}$ MeV fm-3. Now we input $\rho_{0}$ and $p_{c}$
around this estimation, but with $m_{c}=0$, into our equations (20)-(31) and
integrate them from $r=\alpha l_{p}$ to $r=R$. The result is shown in Fig. 9.
Note here that $r_{*}$ is also monotonically increasing with respect to $r$.
We also have the compactness slightly above the Buchdahl limit (BL)
$2GM/R=8/9$.

Figure 9: Here we use linear EOS with $w=\infty$, $B\sim 10^{12}$ MeV fm-3,
and $p_{c}\sim 10^{16}$ MeV fm-3 while $m_{c}=0$. Unlike Fig. 8, the
compactness is still far from black hole limit, although is larger than
Buchdahl limit (BL).

To verify our expectation that a very high uniform density and central
pressure can give us a star with compactness beyond BL, we calculate similar
profiles as Fig. 9 with variations of $w$, $B$, and $p_{c}$. We show the
results in Table 1. In the top three rows, we vary $w$. In the middle three
rows, we vary $p_{c}$. Moreover, in the top three rows, we vary $B$. We can
see that increasing $w$ will increase the compactness. However, to obtain
compactness over the Buchdahl limit, we need to increase $B$ to at least eight
orders of magnitude and the pressure to at least thirteen orders of magnitude.
The entries with bold fonts are from Fig. 9. We also compare them with the
result from TOV GR and find the compactness quite close to SCGrav. Of course,
the results from TOV GR that violate the Buchdahl limit will have $p(r\to
0)\to\infty$. Of course, the results from TOV GR that violate the Buchdahl
limit will have $p(r\to 0)\to\infty$, which is not correct for the
semiclassical model since it should satisfy $r>l_{p}>0$ condition. We verify
this expectation by showing Table 1.

Table 1: Here is the comparison of compactness with variations of $w$, $B$, and $p_{c}$, with $m_{c}=0$ in our semiclassical model (SCGrav) using constant density EOS. $l_{p}\text{(m)}$ | $\alpha$ | $\left.p_{c}\text{(MeV }\text{fm}^{-3}\right)$ | $w$ | $\left.B(\text{MeV}^{4}\right)$ | $2GM/R\text{(SCGrav)}$ | $2GM/R\text{(TOV GR)}$
---|---|---|---|---|---|---
0.001 | 10 | 230 | 1/3 | $(145)^{4}$ | $0.531671<\text{BL}$ | $0.528683<\text{BL}$
0.001 | 10 | 230 | 1 | $(145)^{4}$ | $0.662569<\text{BL}$ | $0.658682<\text{BL}$
0.001 | 10 | 230 | $\infty$ | $(145)^{4}$ | $0.754262<\text{BL}$ | $0.74993<\text{BL}$
0.001 | 10 | $230\times 10^{12}$ | $\infty$ | $(145)^{4}\times 10^{9}$ | $0.888742<\text{BL}$ | $0.888769<\text{BL}$
0.001 | 10 | $230\times 10^{13}$ | $\infty$ | $(145)^{4}\times 10^{9}$ | $0.888961>\text{BL}$ | $0.888902>\text{BL}$
0.001 | 10 | $230*10^{14}$ | $\infty$ | $(145)^{4}*10^{9}$ | $0.890161>\text{BL}$ | $0.888916>\text{BL}$
0.001 | 10 | $230*10^{14}$ | $\infty$ | $(145)^{4}*10^{8}$ | $0.889297>\text{BL}$ | $0.888892>\text{BL}$
0.001 | 10 | $230*10^{14}$ | $\infty$ | $(145)^{4}*10^{7}$ | $0.889019>\text{BL}$ | $0.888889=\text{BL}$
0.001 | 10 | $230*10^{14}$ | $\infty$ | $(145)^{4}*10^{6}$ | $0.88893>\text{BL}$ | $0.888889=\text{BL}$


$\displaystyle-(m-1)\left(\sum_{j\in\pi^{t_{0}+p-l}(d)}r_{d}(j)-\sum_{j\in\pi^{t_{0}+p-l-1}(d)}r_{d}(j)\right)$
$\displaystyle=u_{d}^{t_{0}+p-l-1}(\pi^{t_{0}+p-l})-u_{d}^{t_{0}+p-l-1}(\pi^{t_{0}+p-l-1})$
$\displaystyle-(m-1)\left(\sum_{j\in\sigma(d)}r_{d}(j)-\sum_{j\in\sigma^{\prime}(d)}r_{d}(j)\right)\text{.}$

Therefore,

$\displaystyle(m-1)\left(\sum_{j\in\sigma(d)}r_{d}(j)-\sum_{j\in\sigma^{\prime}(d)}r_{d}(j)\right)$
$\displaystyle<u_{d}^{t_{0}+p-l-1}(\pi^{t_{0}+p-l})-u_{d}^{t_{0}+p-l-1}(\pi^{t_{0}+p-l-1})\text{.}$

However, the left hand side of this equation is unbounded while the right hand
side is fixed. Hence, this cannot be true for all $m\in\mathbb{N}_{0}$, a
contradiction. We have thus established Equation 1.

We can compute that the deviation is indeed a NS deviation in every period:

$\displaystyle\bar{u}_{d}^{kp+l}(\sigma^{\prime})-\bar{u}_{d}^{kp+l}(\sigma)$
$\displaystyle=\bar{u}_{d}^{l}(\sigma^{\prime})-\bar{u}_{d}^{l}(\sigma)$
$\displaystyle+k\left(\sum_{j\in\sigma^{\prime}(d)}r_{d}(j)-\sum_{j\in\sigma(d)}r_{d}(j)\right)$
$\displaystyle\overset{\mathit{\eqref{eq:builtresent}}}{\geq}\bar{u}_{d}^{l}(\sigma^{\prime})-\bar{u}_{d}^{l}(\sigma)$
$\displaystyle=\bar{u}_{d}^{0}(\sigma^{\prime})+\sum_{j\in\sigma^{\prime}(d)}(r_{d}(j)-r_{d}^{p-l}(j))$
$\displaystyle-\bar{u}_{d}^{0}(\sigma)-\sum_{j\in\sigma(d)}(r_{d}(j)-r_{d}^{p-l}(j))$
$\displaystyle=-u_{d}^{t_{0}}(\pi^{t_{0}+p-l-1})+\sum_{j\in\pi^{t_{0}+p-l-1}(d)}(r_{d}(j)-r_{d}^{p-l}(j))$
$\displaystyle+u_{d}^{t_{0}}(\pi^{t_{0}+p-l})-\sum_{j\in\pi^{t_{0}+p-l}(d)}(r_{d}(j)-r_{d}^{p-l}(j))$
$\displaystyle=-u_{d}^{t_{0}+p-l}(\pi^{t_{0}+p-l-1})+\sum_{j\in\pi^{t_{0}+p-l-1}(d)}r_{d}(j)$
$\displaystyle+u_{d}^{t_{0}+p-l}(\pi^{t_{0}+p-l})-\sum_{j\in\pi^{t_{0}+p-l}(d)}r_{d}(j)$
$\displaystyle\overset{\mathit{\eqref{eq:builtresent}}}{\geq}u_{d}^{t_{0}+p-l}(\pi^{t_{0}+p-l})-u_{d}^{t_{0}+p-l}(\pi^{t_{0}+p-l-1})>0\text{.}$

The strict inequality in the end holds because the deviation performed by $d$
in the sequence $(\pi^{t})_{t\geq 0}$ is a NS deviation. Hence, the sequence
$(\sigma^{t})_{t\geq 0}$ evolves through NS deviations. ∎

Table 4: Example from Theorem 4.7. Utilities after $x$ cycles of deviations. Each row depicts the utility that one agent has for the other agents. | $a$ | $a^{\prime}$ | $b$ | $b^{\prime}$ | $c$ | $c^{\prime}$
---|---|---|---|---|---|---
$a$ | $-$ | $110+x$ | $120+x$ | $-100+x$ | $130+x$ | $-100+x$
$a^{\prime}$ | $20+x$ | $-$ | $100+x$ | $10+x$ | $100+x$ | $30+x$
$b$ | $130+x$ | $-100+x$ | $-$ | $110+x$ | $120+x$ | $-100+x$
$b^{\prime}$ | $100+x$ | $30+x$ | $20+x$ | $-$ | $100+x$ | $10+x$
$c$ | $120+x$ | $-100+x$ | $130+x$ | $-100+x$ | $-$ | $110+x$
$c^{\prime}$ | $100+x$ | $10+x$ | $100+x$ | $30+x$ | $20+x$ | $-$

See 4.7

###### Proof.

We now describe an involved example of an MFHG together with an infinite
periodic sequence of NS deviations for appreciative agents. Our example is
very similar to the one presented for resentful agents in Theorem 3.4. In
fact, following the general idea of Theorem 4.1, we reverse the deviation
sequence and appropriately adjust the initial utilities. As in Theorem 3.4,
each agent joins each other agent exactly once in each cycle. This establishes
the same invariant: if a deviating agent prefers a joined coalition $C_{1}$ to
an abandoned coalition $C_{2}$ before (and during) the _first_ execution of
the cycle, then it still prefers $C_{1}$ to $C_{2}$ before (and during) _each_
execution of the cycle.

Consider the game with agent set
$N=\\{a,a^{\prime},b,b^{\prime},c,c^{\prime}\\}$ and utilities as depicted in
Table 4. The initial utilities result from setting $x=0$ in Table 4. We now
present an infinite sequence $(\pi^{t})_{t\geq 0}$ of partitions, always
consisting of three coalitions. For each partition, we refer to the first
listed coalition as $C_{1}$, to the second as $C_{2}$, and the third as
$C_{3}$. For the sake of clarity, for each partition, we also specify which
agent deviates to which coalition in the next step. Specifically, for $n\geq
0$, we have

  * •

$\pi^{18n+1}=\\{\\{b^{\prime}\\},\\{a,a^{\prime}\\},\\{b,c,c^{\prime}\\}\\}$
with agent $b$ deviating to $C_{2}$,

  * •

$\pi^{18n+2}=\\{\\{b^{\prime}\\},\\{b,a,a^{\prime}\\},\\{c,c^{\prime}\\}\\}$
with agent $c^{\prime}$ deviating to $C_{1}$,

  * •

$\pi^{18n+3}=\\{\\{b^{\prime},c^{\prime}\\},\\{b,a,a^{\prime}\\},\\{c\\}\\}$
with agent $c^{\prime}$ deviating to $C_{2}$,

  * •

$\pi^{18n+4}=\\{\\{b^{\prime}\\},\\{b,a,a^{\prime},c^{\prime}\\},\\{c\\}\\}$
with agent $b$ deviating to $C_{1}$,

  * •

$\pi^{18n+5}=\\{\\{b^{\prime},b\\},\\{a,a^{\prime},c^{\prime}\\},\\{c\\}\\}$
with agent $a$ deviating to $C_{1}$,

  * •

$\pi^{18n+6}=\\{\\{b^{\prime},b,a\\},\\{a^{\prime},c^{\prime}\\},\\{c\\}\\}$
with agent $c^{\prime}$ deviating to $C_{3}$,

  * •

$\pi^{18n+7}=\\{\\{b^{\prime},b,a\\},\\{a^{\prime}\\},\\{c,c^{\prime}\\}\\}$
with agent $a$ deviating to $C_{3}$,

  * •

$\pi^{18n+8}=\\{\\{b^{\prime},b\\},\\{a^{\prime}\\},\\{c,c^{\prime},a\\}\\}$
with agent $b^{\prime}$ deviating to $C_{2}$,

  * •

$\pi^{18n+9}=\\{\\{b\\},\\{a^{\prime},b^{\prime}\\},\\{c,c^{\prime},a\\}\\}$
with agent $b^{\prime}$ deviating to $C_{3}$,

  * •

$\pi^{18n+10}=\\{\\{b\\},\\{a^{\prime}\\},\\{c,c^{\prime},a,b^{\prime}\\}\\}$
with agent $a$ deviating to $C_{2}$,

  * •

$\pi^{18n+11}=\\{\\{b\\},\\{a^{\prime},a\\},\\{c,c^{\prime},b^{\prime}\\}\\}$
with agent $c$ deviating to $C_{2}$,

  * •

$\pi^{18n+12}=\\{\\{b\\},\\{a^{\prime},a,c\\},\\{c^{\prime},b^{\prime}\\}\\}$
with agent $b^{\prime}$ deviating to $C_{1}$,

  * •

$\pi^{18n+13}=\\{\\{b,b^{\prime}\\},\\{a^{\prime},a,c\\},\\{c^{\prime}\\}\\}$
with agent $c$ deviating to $C_{1}$,

  * •

$\pi^{18n+14}=\\{\\{b,b^{\prime},c\\},\\{a^{\prime},a\\},\\{c^{\prime}\\}\\}$
with agent $a^{\prime}$ deviating to $C_{3}$,

  * •

$\pi^{18n+15}=\\{\\{b,b^{\prime},c\\},\\{a\\},\\{c^{\prime},a^{\prime}\\}\\}$
with agent $a^{\prime}$ deviating to $C_{1}$,

  * •

$\pi^{18n+16}=\\{\\{b,b^{\prime},c,a^{\prime}\\},\\{a\\},\\{c^{\prime}\\}\\}$
with agent $c$ deviating to $C_{3}$,

  * •

$\pi^{18n+17}=\\{\\{b,b^{\prime},a^{\prime}\\},\\{a\\},\\{c^{\prime},c\\}\\}$
with agent $b$ deviating to $C_{3}$,

  * •

$\pi^{18n+18}=\\{\\{b^{\prime},a^{\prime}\\},\\{a\\},\\{c^{\prime},c,b\\}\\}$
with agent $a^{\prime}$ deviating to $C_{2}$.

Then, it is possible to verify that for $k\geq 1$, $\pi^{k-1}$ leads to
$\pi^{k}$ by means of an NS deviation. Hence, we have presented an MFHG with
an infinite sequence of NS deviations for appreciative agents. ∎

##  Appendix C Additional Material for Section 7

The goal of this section is to complete the proof of Theorem 7.1. Since we
have already considered NS and IS in Lemma 7.2, it remains to consider CNS and
CS. We split the remaining proof into two more lemmas.

######  Lemma C.1.

resentful- and appreciative-ashg-CNS-Sequence are $\NP$-hard.

###### Proof.

We first show the statement for resentful agents. We again reduce from RX3C.

Construction. From an instance of RX3C, given by a universe
$U=\\{x_{1},\dots,x_{3t}\\}$ and a family
$\mathcal{S}=\\{S_{1},\dots,S_{3t}\\}$ of $3$-subsets of $U$, we create an
additively separable hedonic game as follows. We create the set of agents
$N=U\cup\mathcal{S}\cup F\cup Q$ with element agents $U$, set agents
$\mathcal{S}$, filling agents $F=\\{f_{1},\ldots,f_{2t}\\}$, and penalizing
agents $Q=\\{p_{0},\ldots,p_{4}\\}$. The initial utilities of the agents are
illustrated in Figure 8:

  * •

Each element agent has utility $1$ for each other element agent.

  * •

Each set agent $S\in\mathcal{S}$ has utility $20t$ for the three element
agents $x\in S$, utility $60t$ for each filling agent, utility $60t$ for
$p_{0}$, and utility zero for $p_{1}$.

  * •

$p_{0}$ has utility $10t$ for each set agent and utility zero for $p_{1}$.

  * •

$p_{1}$ has utility $1$ for $p_{0}$.

  * •

$p_{2}$, $p_{3}$, and $p_{4}$ have utility $200t$ for $p_{1}$.

  * •

$p_{2}$ has utility $-100t$ for $p_{3}$ and utility $-400t$ for $p_{4}$.

  * •

$p_{3}$ has utility $-100t$ for $p_{4}$ and utility $-400t$ for $p_{2}$.

  * •

$p_{4}$ has utility $-100t$ for $p_{2}$ and utility $-400t$ for $p_{3}$.

  * •

All not explicitly mentioned utilities are $-1000t$.

The penalizing gadget consisting of agents $p_{1},\ldots,p_{4}$ is a variant
of an ASHG by Sung and Dimitrov (2007, Example 2) for which no CNS partition
exists (in normal ASHGs).

We set $k$ to $5t+1$ and show that there is an exact cover of $U$ if and only
if there is a sequence starting from the singleton partition of at most $k$
CNS deviations leading to an CNS partition, where agents are resentful.

${x_{1}}$$\vdots$${x_{3t}}$$S_{1}$$\vdots$$S_{3t}$$f_{1}$$\vdots$$f_{2t}$$p_{0}$$p_{1}$$p_{2}$$p_{3}$$p_{4}$$1$$20t$$x\in
S$$60t$$60t$$10t$$0$$1$$0$$200t$$200t$$200t$$-400t$$-100t$$-400t$$-100t$$-400t$$-100t$
Figure 8: Additively separable hedonic games in Lemma C.1. All omitted
utilities are $-1000t$.

($\Rightarrow$) Given an exact cover
$\mathcal{S}^{\prime}\subseteq\mathcal{S}$, consider the following sequence of
$k=5t+1$ deviations. First, for each set $S\in\mathcal{S}^{\prime}$ with
$S=\\{x_{i},x_{j},x_{l}\\}$, $x_{i}$ and $x_{j}$ join $x_{l}$ and afterwards
the set agent $S$ joins $\\{x_{i},x_{j},x_{l}\\}$ ($3t$ deviations). Second,
all set agents $S$ with $S\notin\mathcal{S}^{\prime}$ join a filling agent
each ($2t$ deviations). Last, $p_{1}$ joins $p_{0}$.

All these deviations are NS deviations as each of these deviations is
performed from a singleton to a coalition with positive utility for the
deviator. Moreover, all these NS deviations are CNS deviations because they
are performed from singletons; so there are no other agents who could veto the
deviation.

Next, we show that the resulting partition $\pi$ is CNS. First, each set agent
has utility $60t$ for $\pi(S)$. Since every other coalition either contains an
agent for which she has utility $-1000t$ or is the coalition
$\\{p_{0},p_{1}\\}$, for which she has utility $60t$, she does not want to
deviate. Second, each element and filling agent do not have any CNS deviation
because they have a set agent in their coalition who would veto their
deviation. Third, $p_{1}$ would veto a deviation of $p_{0}$, so $p_{0}$ has no
CNS deviation. Fourth, $p_{1}$ does not want to deviate to another coalition
as she has utility $1$ for $\pi(p_{1})=\\{p_{0},p_{1}\\}$ and every other
coalition in $\pi$ contains an agent for which she has utility $-1000t$. Last,
$p_{2}$, $p_{3}$, and $p_{4}$ do not want to deviate as they only have a
positive utility (of $200t$) for agent $p_{1}$ but $p_{1}$ is together with
$p_{0}$ for which they have utility $-1000t$.

($\Leftarrow$) Assume that we have a sequence of $\ell\leq k=5t+1$ CNS
deviations leading from the singleton partition $\pi^{0}$ to a CNS partition
$\pi^{\ell}$. In this sequence, no two agents who mutually have utility
$-1000t$ for each other can ever be in the same coalition because they would
never join each other. Furthermore, as we start from the singleton partition,
the resent an agent has accumulated for the other agents after $\ell$
deviations sums up to at most $\frac{\ell}{2}\leq 3t$.

We will now show that no set agent is together with $p_{0}$ in $\pi^{\ell}$.
For the sake of a contradiction, assume that $S\in\pi^{\ell}(p_{0})$ for some
$S\in\mathcal{S}$. By the first observation, this implies that
$\pi^{\ell}(p_{0})\subseteq\\{S,p_{0},p_{1}\\}$. It further holds that
$p_{1}\notin\pi^{\ell}(p_{0})$ because, otherwise, $p_{1}$ would have a CNS
deviation to a singleton coalition. Hence, $\pi^{\ell}(p_{0})=\\{S,p_{0}\\}$.
Next, consider coalition $\pi^{\ell}(p_{1})$. By the first observation, we
have $\pi^{\ell}(p_{1})\subseteq\mathcal{S}\cup Q$. Furthermore, there is no
set agent in $\pi^{\ell}(p_{1})$: Indeed, assume for contradiction that there
is some set agent $S^{\prime}$ in $\pi^{\ell}(p_{1})$. Then, by the first
observation, neither $p_{2}$, $p_{3}$, nor $p_{4}$ are in $\pi^{\ell}(p_{1})$.
Furthermore, we already now that $p_{0}$ is not in $\pi^{\ell}(p_{1})$. It
follows that $\pi^{\ell}(p_{1})=\\{S^{\prime},p_{1}\\}$ which means that
$p_{1}$ has a CNS deviation to a singleton coalition, a contradiction. Hence,
$\pi^{\ell}(p_{1})\subseteq\\{p_{1},\ldots,p_{4}\\}$. For $i\in\\{2,3,4\\}$,
it also holds by the first observation that
$\pi^{\ell}(p_{i})\subseteq\\{p_{1},\ldots,p_{4}\\}$. We will now show that
$\\{p_{1},\ldots,p_{4}\\}$ can not be divided into a CNS partition (cf. Sung
and Dimitrov, 2007, Example 2). Note that the arguments also hold when
lowering any utilities among this agents by at most $3t$ and therefore
$\pi^{\ell}$ is no CNS partition, which is a contradiction.

First, none of $p_{2}$, $p_{3}$, and $p_{4}$ can be in the same coalition
because the agent with a initial utility of $-400t$ for another agent in her
coalition would have a CNS deviation to a singleton. Hence, the remaining
possible partitions of $\\{p_{1},\ldots,p_{4}\\}$ are
$\pi^{\prime}=\\{\\{p_{1}\\},\\{p_{2}\\},\\{p_{3}\\},\\{p_{4}\\}\\}$,
$\pi^{\prime\prime}=\\{\\{p_{1},p_{2}\\},\\{p_{3}\\},\\{p_{4}\\}\\}$,
$\pi^{\prime\prime\prime}=\\{\\{p_{2}\\},\\{p_{1},p_{3}\\},\\{p_{4}\\}\\}$,
and
$\pi^{\prime\prime\prime\prime}=\\{\\{p_{2}\\},\\{p_{3}\\},\\{p_{1},p_{4}\\}\\}$.
But all these partitions are not CNS: $\pi^{\prime}$ is not stable because
$p_{2}$ has a CNS deviation to $\\{p_{1}\\}$; $\pi^{\prime\prime}$ is not
stable because $p_{4}$ has a CNS deviation to $\\{p_{1},p_{2}\\}$;
$\pi^{\prime\prime\prime}$ is not stable because $p_{2}$ has a CNS deviation
to $\\{p_{1},p_{3}\\}$; $\pi^{\prime\prime\prime\prime}$ is not stable because
$p_{3}$ has a CNS deviation to $\\{p_{1},p_{4}\\}$. Thus, we have shown that
$\pi^{\ell}$ is not CNS. This is a contradiction and therefore the initial
assumption was wrong, i.e., no set agent is together with $p_{0}$ in
$\pi^{\ell}$.

Since $\pi^{\ell}$ is CNS, no set agent has a CNS deviation to $p_{0}$. Note
that $p_{0}$ is the only agent who has positive utilities for the set agents.
So, for any set agent $S\in\mathcal{S}$, there is no agent in $\pi^{\ell}(S)$
who would veto a deviation of $S$. Therefore, no set agent has an NS deviation
to $p_{0}$. Furthermore, we know that
$\pi^{\ell}(p_{0})\subseteq\\{p_{0},p_{1}\\}$ because of the first observation
and because no set agent is with $p_{0}$. So, for every set agent
$S\in\mathcal{S}$, we have $u_{S}^{\ell}(\pi^{\ell})\geq
u_{S}^{\ell}(\pi^{\ell}(p_{0})\cup\\{S\\})\geq 60t-3t=57t$ (where $60t$ is
$S$’s initial utility for $p_{0}$ and $-3t$ is a bound for the resent that $S$
might have build for $p_{0}$ or $p_{1}$). Therefore, $\pi^{\ell}(S)$ contains
at least one filling agent or three element agents $x$ with $x\in S$. Since no
two set agents are in the same coalition, this implies the existence of an
exact cover for $U$.

For appreciative agents, we can use almost the same construction. We just set
$p_{0}$’s initial utility for $p_{1}$ to $-1$ while all other utilities stay
the same. Then, the proof of correctness is very similar to the proof for
resentful agents. ∎

######  Lemma C.2.

resentful- and appreciative-ashg-CS-Sequence are $\NP$-hard.

###### Proof.

We first show the statement for resentful agents and reduce from RX3C.

Construction. Let an instance of RX3C $(U,\mathcal{S})$ with a universe
$U=\\{x_{1},\dots,x_{3t}\\}$ and a family
$\mathcal{S}=\\{S_{1},\dots,S_{3t}\\}$ of $3$-subsets of $U$ be given. We set
$k=3t+2$ and create the following additively separable hedonic game. The set
of agents is $N=U\cup\mathcal{S}\cup F\cup Q$ with element agents $U$, set
agents $\mathcal{S}$, filling agents $F=\\{f_{1},\ldots,f_{2t}\\}$, and
penalizing agents $Q=\\{p_{0},\ldots,p_{6}\\}$. The initial utilities of the
agents are illustrated in Figure 9:

${x_{1}}$$\vdots$${x_{3t}}$${x_{3t}}$$S_{1}$$\vdots$$S_{3t}$$S_{3t}$$S_{3t}$$f_{1}$$f_{1}$$\vdots$$f_{2t}$$f_{2t}$$p_{0}$$p_{1}$$p_{2}$$p_{3}$$p_{4}$$p_{5}$$p_{6}$$p_{6}$$0$$10t$$300t$$60t$$60t$$60t$$50t$$50t$$50t$$40t$$40t$$40t$$60t$$20t$$1$$20t$$x\in
S$$1$$60t$ Figure 9: Additively separable hedonic games in Lemma C.2. All
omitted utilities are $-1000t$.

  * •

Each set agent $S$ has utility $60t$ for each filling agent, utility $20t$ for
element agents $x$ with $x\in S$, and utility $60t$ for the penalizing agent
$p_{0}$.

  * •

Each element agent has utility $1$ for each set agent and utility $0$ for all
other element agents.

  * •

Each filling agent has utility $1$ for each set agent.

  * •

$p_{0}$ has utility $20t$ for all set agents and utility $10t$ for $p_{1}$.

  * •

Agents $p_{1}$, $p_{3}$, and $p_{5}$ have utility $40t$ for each other.

  * •

For $i\in\\{1,3,5\\}$, agent $p_{i}$ has utility $60t$ for agent $p_{i+1}$ and
vice versa.

  * •

For $i\in\\{2,4,6\\}$, agent $p_{i}$ has utility $50t$ for agent $p_{i+1}$ and
vice versa (where $p_{7}=p_{1}$).

  * •

All not explicitly mentioned utilities are $-1000t$.

Note that the penalizing gadget consisting of agents $p_{1},\ldots,p_{6}$ is a
variant of an ASHG by Aziz et al. (2013, Example 1) for which no CS partition
exists (for normal ASHGs).

($\Rightarrow$) Given an exact cover
$\mathcal{S}^{\prime}\subseteq\mathcal{S}$, consider the following sequence of
group deviations. For each $S\in\mathcal{S}^{\prime}$, $S$ and the three $x\in
S$ deviate together ($t$ deviations). For each
$S\in\mathcal{S}\setminus\mathcal{S}^{\prime}$, $S$ deviates with one filling
agent that is still in a singleton ($2t$ deviations). Now, $p_{0}$ deviates
with $p_{1}$. Lastly, $p_{3}$, $p_{4}$, and $p_{5}$ deviate together. Hence,
we have $3t+2$ deviations in total.

All these deviations are CS deviations as all these deviations are performed
from singletons to coalitions with positive utilities for all deviators.

Next, we show that the resulting partition $\pi$ is CS. First note that no two
agents who have utility $-1000t$ for each other want to deviate together. Each
element agent $x$ only has a positive utility of $1$ for all set agents and is
in a coalition with one set agent. As no two set agents will ever deviate
together, there is no coalition that $x$ could deviate with to improve her
utility. The same holds for all filling agents. Each set agent $S$ has utility
$60t$ for $\pi(S)$. As no element or filling agent wants to deviate, the only
remaining agent for which $S$ has positive utility is $p_{0}$. Yet, $S$ only
has utility $60t$ for $p_{0}$ which is no improvement compared to $\pi(S)$. So
no set agent has a CS deviation. Therefore, also $p_{0}$ has no CS deviation
as she would only like to deviate with some set agents. $p_{1}$ has utility
$300t$ for $\pi(p_{1})=\\{p_{0},p_{1}\\}$ and would thus not deviate without
$p_{0}$. Similarly, the remaining agents $p_{2},\ldots,p_{6}$ have no CS
deviation without $p_{1}$. Thus, the whole partition $\pi$ is CS.

($\Leftarrow$) Assume that we have a sequence of $\ell\leq k=3t+2$ CS
deviations leading from the singleton partition $\pi^{0}$ to a CS partition
$\pi^{\ell}$. In this sequence, no agent is ever in a coalition with an agent
for which she has utility $-1000t$ because she would never deviate with this
agent. Therefore, the largest coalitions in this sequence have size at most
four (containing one set agent and three element agents). It follows that the
sum of the resent that a single agent accumulates for the other agents during
$\ell\leq 3t+2$ deviations is bounded by $\frac{\ell}{2}\cdot
3\leq\frac{3t+2}{2}\cdot 3=\frac{9}{2}t+3$ because she can be left by at most
three agents every second deviation.

We will now show that no set agent is in a joint coalition with $p_{0}$ in
$\pi^{\ell}$. For the sake of a contradiction, assume that
$S\in\pi^{\ell}(p_{0})$ for some $S\in\mathcal{S}$. By the first observation,
this implies that $\pi^{\ell}(p_{0})=\\{S,p_{0}\\}$. This means that
$p_{1},\ldots,p_{6}$ have formed coalitions among each other. Yet, as argued
by Aziz et al. (2013, Example 1), there is no CS partition for these six
agents. Note that resent, which influences each player by a total utility
change of at most $\frac{9}{2}t+3$, does not facilitate stability because the
differences between the utilities in this group are at least $10t$. It follows
that $\pi^{\ell}$ is not CS which is a contradiction. So no set agent is
together with $p_{0}$ in $\pi^{\ell}$.

Next, since $\pi^{\ell}$ is CS, no coalition $\\{S,p_{0}\\}$ with
$S\in\mathcal{S}$ has a CS deviation. Since $p_{0}$ is together with no set
agent in $\pi^{\ell}$, $p_{0}$ can have a utility of at most $10t$ for
$\pi^{\ell}$ and wants to deviate to $\\{S,p_{0}\\}$. So, $S$ does not want to
deviate to $\\{S,p_{0}\\}$. Therefore, $u_{S}^{\ell}(\pi^{\ell})\geq
u_{S}^{\ell}(\\{S,p_{0}\\})\geq 60t-(\frac{9}{2}t+3)=\frac{111}{2}t-3$. Hence,
$\pi^{\ell}(S)$ contains at least one filling agent or three element agents
$x$ with $x\in S$. Since no two set agents are in the same coalition, this
implies the existence of an exact cover for $U$.

For appreciative agents, we can use the same construction and the proof is
very similar. ∎