# Dynamical Property of Black Hole Matter

C.S. Lam  Department of Physics, McGill University
Montreal, Q.C., Canada H3A 2T8
Department of Physics and Astronomy, University of British Columbia,
Vancouver, BC, Canada V6T 1Z1
Email<EMAIL_ADDRESS>
###### Abstract

Matter loses its original characteristics after entering a black hole, thus
becoming a new kind of (black hole) matter. The property of this new matter
cannot be measured experimentally, but some of it can be deduced theoretically
from the Einstein equations and the conservation laws which it must still
satisfy. In a previous paper, this matter is modelled by an ideal fluid, with
an equation of state $p(r)=-\xi\rho(r)$ between the pressure $p(r)$ and the
density $\rho(r)$. In order for this matter to fill the inside of a black hole
so that its property can be teased out from the Einstein and conservation
equations, it must possess a negative pressure ($\xi>0$) to counter the
gravitation attraction which draws all matter to the center. In that case a
solution of the Einstein and conservation equations exists if and only if the
constant $\xi$ is confined within a narrow range, between 0.1429 and 0.1716.
In the present paper, we try to find out its dynamical response by injecting
additional matter into the black hole over a period of time. The resulting
solutions of the six time-dependent Einstein equations and conservation laws
are presented in perturbation theory, valid if the total amount of injection
is small. Even in perturbation, the solutions can be obtained only with a
special trick. The result shows that the equation of state
$p(r,t)=-\xi\rho(r,t)$ remains unchanged with the same $\xi$ when the
injection rate is constant. When the rate changes with time, $\xi$ requires a
correction, $\xi\to\xi+\xi_{1}(r,t)$, where $\xi_{1}(r,t)$ appears to be
correlated with the acceleration of the injected matter in a way to be shown
in the text.

##  I Introduction

According to the no-hair theorem MTW73 ; sI67 ; sI68 , a stationary black hole
is completely characterized by its mass, its electric charge, and its spin.
For neutral and non-rotating black holes, the only thing that tells them apart
is their mass. In the Standard Model, other than mass and electric charge, a
particle is characterized also by its baryonic number, isotopic spin,
hypercharge, and statistics, etc. With the no-hair theorem, all these are lost
once matter enters into the black hole, so what is inside a black hole must be
a new kind of matter, to be referred hereafter as the ‘black hole matter’.
Very little is known about it because there is no way to get inside the black
hole to measure. Nevertheless, it must obey the Einstein equations and the
conservation laws, from which some information of the new matter can be teased
out. There are some studies along this line MM15 ; BMN19 ; MM23  but to my
knowledge not many. The most popular and well known model is the Schwarzschild
solution, which assumes all black hole matter sink to the singularity at the
center of the black hole, leaving a vacuum prevailing everywhere else inside
the black hole. This solution unfortunately tells us nothing about the nature
of the new matter as it is nowhere to be found except at the singularity.

In a recent paper Lam , we studied the black hole matter inside a neutral,
spherical, and non-rotational black hole modelled by an ideal fluid, whose
density $\rho(r)$ and pressure $p(r)$ are related by the equation of state
$p(r)=-\xi\rho(r)$. In order for matter to fill the black hole so that its
property can be probed, a negative pressure ($\xi>0$) is required to
counteract the gravitational attraction which tends to draw everything to the
singularity at the center. In order to satisfy the Einstein equations and the
conservation laws, we found that the constant parameter $\xi$ must be confined
to a narrow range, between $\xi=1/7\simeq 0.1429$ and $\xi=3-2\sqrt{2}\simeq
0.1716$. A singularity of $\rho(r)$ still appears at the origin as demanded by
the singularity theorem aK56 ; aR57 ; rP65 ; SG14 ; kL22 , but $\rho(r)$ is
also non-zero elsewhere inside the black hole, except at the event horizon
where it must vanish to match the vacuum condition outside the black hole.

In this paper, we try to find out more property of the new matter by probing
its dynamical response. About the only thing one can do to probe its reaction
is to inject more matter into the black hole at different rates. In this way,
starting from a static black hole of mass $M_{0}$, additional mass is injected
to increase it from $M_{0}$ to $M(t)=M_{0}(1+\sigma(t))$ at time $t$. By
calculating $\rho(r,t)$ and $p(r,t)$ from the Einstein equations and the
conservation laws for an arbitrary $\sigma(t)$, one can in principle deduce
how black hole responds to such a dynamical probing.

Unfortunately, the resulting Einstein and conservation equations (Appendix A)
are so complicated and so nonlinear that there is no hope to find an exact
solution. For that reason we turn to perturbative solutions by assuming
$\sigma(t)\ll 1$ at all times. With that caveat and a trick to be explained,
the equations can then be solved to obtain $\rho(r,t)$ and $p(r,t)$ and other
quantities.

We find that the static equation of state $p(r,t)=-\xi\rho(r,t)$ remains
unchanged with the original $\xi$ if the second time derivative
$\ddot{\sigma}(t)=0$. If $\ddot{\sigma}(t)\not=0$, then the constant $\xi$
receives a correction, $\xi\to\xi+\xi_{1}(r,t)$, with
$\xi_{1}(r,t)=\hat{\xi}_{1}(r)\ddot{\sigma}(t)$.

That the equation of state can be altered by external conditions is also seen
in ordinary matter. A non-relativistic gas has an equation of state with
$\xi=0$, but if heat is applied to that system to make the movement
relativistic, then $\xi=-{1\over 3}$. Similarly, in the present case changes
of equation of state occurs if the injected matter has an acceleration
$\ddot{\sigma}(t)\not=0$.

Sec. II reviews the black hole matter model proposed in Lam . It is the zeroth
order solution of the time-dependent Einstein equations and conservation laws
to be discussed in Sec. III. The first-order equation are shown in Eq.(26) to
Eq.(31), but it is not easy solve them directly. However, their time-
independent version can be solved using a trick discussed in Sec. IV, which
then serves as an effective stepping stone to yield the solutions of the time-
dependent equations to be presented in Sec. V. These solutions are summarized
in Sec. VIA, their validity and interpretations discussed in Sec. VIB, and the
solutions are illustrated numerically in Sec. VIC.

##  II The Static Model

The matter distribution inside a neutral, non-rotating, and spherical black
hole is reviewed here. In this model Lam , black hole matter is described by
an ideal fluid with the energy-momentum tensor

$\displaystyle T_{\mu\nu}=(p+\rho)U_{\mu}U_{\nu}+pg_{\mu\nu},$ (1)

where as usual the four-velocity is normalized to $U^{\mu}U_{\mu}=-1$. The
energy density $\rho$ is positive, and the pressure $p$ obeying the equation
of state $p=-\xi\rho$ is negative. The constant $\xi$ has to be confined to a
narrow range $0.1429<\xi<0.1716$ in order to satisfy the Einstein equation and
the covariant conservation laws.

The spacetime metric

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle-e^{2\alpha(r)}dt^{2}+\left(1-2Gm(r)\over
r\right)^{-1}dr^{2}+r^{2}d\Omega^{2}$ (2)

is Schwarzschild outside the black hole, where
$e^{2\alpha(r)}=\left(1-{2GM\over r}\right)$ and $m(r)=M$, with $G$ being the
gravitational constant and $M$ the black hole mass with an event horizon
located at $r=R=2GM$. Inside the black hole, the metric is no longer
Schwarzschild because it is now filled with matter given by Eq.(1). Einstein
equations of this system lead to the Tolman-Oppenheimer-Volkoff equation for
the pressure gradient sC04 ; OV39

$\displaystyle{dp\over dr}$ $\displaystyle=$
$\displaystyle-{(\rho+p)[Gm(r)+4\pi Gr^{3}p]\over r[r-2Gm(r)]},$ (3)

and a relation between $m(r)$ and $\rho(r)$,

$\displaystyle m(r)$ $\displaystyle=$ $\displaystyle
4\pi\int_{0}^{r}\rho(r^{\prime}){r^{\prime}}^{2}dr^{\prime},$ (4)

so $m(r)$ can be interpreted as the mass of a ball of black-fluid with a
radius $r\leq R$. The metric function $\alpha$ inside the black hole satisfies
the equation

$\displaystyle{d\alpha\over dr}$ $\displaystyle=$
$\displaystyle-{1\over(\rho+p)}{dp\over dr}.$ (5)

Continuity across the black hole surface is imposed as the boundary condition
to match the inside and the outside of the black hole:

$\displaystyle p(R)=\rho(R)=0,\quad m(R)=M,\quad e^{2\alpha(R)}=0.$ (6)

For $R-r$ small and positive, Eq.(3) can be approximated by

$\displaystyle{dp\over dr}=-{(\rho+p)GM\over R[r-2GM]}={\xi-1\over
2\xi}{p\over R-r},\quad(R-r\ll R),$ (7)

whose solution for small and positive $R-r$ is

$\displaystyle
p(r)\simeq-\tilde{c}(R-r)^{(1-\xi)/2\xi}\equiv-\tilde{c}(R-r)^{\gamma}$ (8)

for some $\tilde{c}>0$. Thus $0<\xi<1$ is required in order to have $p(R)=0$,
so the black-hole matter must carry a negative pressure.

Using Eq.(5) and $p=-\xi\rho$, one gets

$\displaystyle e^{2\alpha(r)}=c_{1}\rho(r)^{1/\gamma}$ (9)

for some $c_{1}>0$. By rescaling time $t$ in Eq.(2), one can make $c_{1}=1$.
When $r$ approaches $R$, Eq.(8) shows that $e^{2\alpha(r)}\to c_{2}(R-r)$ for
some $c_{2}>0$, matching the value 0 for this metric component outside the
black hole.

These equations can be written in a dimensionless form by letting let $x=r/R$,
$\bar{m}(x)=m(r)/M$, $\bar{\rho}(x)=\rho(r)(R^{3}/M)$,
$\bar{p}(x)=p(r)(R^{3}/M)=-\xi\bar{\rho}(x)$, and $\bar{\alpha}(x)=\alpha(r)$.
Then Eq.(3) in the dimensionless form becomes

$\displaystyle{d\bar{\rho}(x)\over dx}$ $\displaystyle=$
$\displaystyle-{\xi-1\over 2\xi}{\bar{\rho}(x)\over x}{\bar{m}(x)-4\pi\xi
x^{3}\bar{\rho}(x)\over x-\bar{m}(x)},$ (10) $\displaystyle{d\bar{m}(x)\over
dx}$ $\displaystyle=$ $\displaystyle 4\pi x^{2}\bar{\rho}(x),$ (11)

and the boundary conditions are $\bar{\rho}(1)=0$ and $\bar{m}(1)=1$. Eq.(5)
can be written as

$\displaystyle{d\bar{\alpha}(x)\over dx}={\xi\over
1-\xi}\left({1\over\bar{\rho}(x)}{d\bar{\rho}(x)\over dx}\right),$ (12)

whose solution from Eq.(9) is

$\displaystyle e^{2\bar{\alpha}(x)}=\bar{\rho}(x)^{1\over\gamma}.$ (13)

The interior solution equivalent to Eq.(8) near $x=1$ is then

$\displaystyle\bar{\rho}(x)$ $\displaystyle\simeq$ $\displaystyle
c_{3}(1-x)^{(1-\xi)/2\xi}\equiv c_{3}(1-x)^{\gamma},$
$\displaystyle\bar{m}(x)$ $\displaystyle\simeq$ $\displaystyle 1-{8\pi\xi
c_{3}\over\xi+1}(1-x)^{(\xi+1)/2\xi}=1-{8\pi\xi
c_{3}\over\xi+1}(1-x)^{\gamma+1},$ $\displaystyle e^{2\bar{\alpha}(x)}$
$\displaystyle\simeq$ $\displaystyle c_{3}^{1/\gamma}(1-x)\qquad(1-x\ll 1)$
(14)

for some $c_{3}>0$.

Near $x=0$, $\bar{m}(x)=\mu_{0}x^{\beta}$ for some $\mu_{0}>0$, and it is
necessary to impose $0<\beta<1$ in order for Eq.(10) and Eq.(11) to have a
solution. It follows from Eq.(10) that
$\bar{\rho}(x)=(\mu_{0}\beta/4\pi)x^{\beta-3}$, and $\beta$ is related to
$\xi$ by the relation

$\displaystyle\beta=(\\!-\\!1\\!+\\!7\xi)/\xi(1\\!+\\!\xi).$ (15)

The requirement for $\beta$ to be between 0 and 1 forces $\xi$ to be in a
narrow range between $\xi=1/7\simeq 0.1429$ (when $\beta=0$) and
$\xi=3-2\sqrt{2}\simeq 0.1716$ (when $\beta=1$).

The corresponding values for $\gamma=(1-\xi)/2\xi$, the power of
$\bar{\rho}(x)$ near $x=1$, are $\gamma=3$ when $\xi=1/7$, and $\gamma=2.414$
when $\xi=3-2\sqrt{2}$.

A numerical solution for $\xi=0.16$ is shown in Fig. 1.

Fig. 1. Scaled mass and density distribution in the black-hole interior for
$\xi=0.16$

This diagram illustrates the genereal behavior of functions true for any $\xi$
within the allowed range. $\bar{\rho}(x)$ decreases steadily from $\sim
x^{\beta-3}\ (\beta<1)$ at small $x$ towards $\sim(1-x)^{\gamma}\ (\gamma>0)$
near $x=1$. In the mean time, $\bar{m}(x)$ increases steadily from $\sim
x^{\beta}$ at small $x$ towards 1, keeping $\bar{m}(x)-x\geq 0$. The steady
increase of $\bar{m}(x)$ is a consequence of Eq.(11) and $\bar{\rho}(x)\geq
0$, and the positivity of $\bar{m}(x)-x$ except at the boundaries is needed to
avoid a singularity appearing in Eq.(10).

##  III Time-dependent injection

We found in the last section that black hole matter can be represented by an
ideal fluid with an equation of state $p(r)=-\xi\rho(r)$, as long as
$0.1429<\xi<0.1716$. In the rest of this article, we shall investigate the
dynamical property of this matter, deduced from its response to an injection
of additional matter into the black hole. As the response cannot be measured
experimentally, we must resort to the solution of time-dependent Einstein
equations to find out what it is.

Intuition might tell us that the internal adjustment to such an input would
start at the surface of the black hole, then propagating inward at the speed
of sound. However, such an intuition based on physics of compressible fluid in
a spacetime with a signature $(-,+,+,+)$ may not be applicable inside a
blackhole whose signature is $(-,-,+,+)$. After all, with this change of
signature, a hyperbolic wave equation in $r$ and $t$ becomes an elliptic
equation, suggesting that the internal adjustment might actually be
instantaneous.

The only way to find out the truth is to solve the time-dependent Einstein
equations, but unfortunately that proves to be too difficult a task. To make
some headway, we shall consider the idealized situation where matter is
accreted spherically, with a total amount so small that perturbation theory
can be applied to solve the Einstein equations.

Time-dependent density $\rho$ and pressure $p$ require time-dependent metric
functions $\alpha$ and $m$ in Eq.(16). The continuous injection of matter
brings about a fluid flow with a non-zero momentum density $T_{tr}$, which in
turn requires a non-diagonal metric element $g_{tr}$. The appropriate metric
to be used inside the black hole is therefore

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle-e^{2\alpha(r,t)}dt^{2}+\left(1-{2Gm(r,t)\over
r}\right)^{-1}dr^{2}+2\epsilon(r,t)drdt+r^{2}d\Omega^{2}\equiv
g_{\mu\nu}dx^{\mu}dx^{\nu}.$ (16)

As in the stationary case, $m(r,t)$ can be interpreted as the mass of a ball
of fluid with radius $r$ at time $t$. The event horizon $R(t)$ at time $t$ is
related to the black hole mass $M(t)$ by $R(t)=2GM(t)$, where
$M(t)=m(R(t),t)$.

The time-dependent energy-momentum tensor of an ideal fluid is now

$\displaystyle
T_{\mu\nu}=\big{(}\rho(r,t)+p(r,t)\big{)}U_{\mu}U_{\nu}+p(r,t)g_{\mu\nu},$
(17)

where the four-velocity $U^{\mu}=(u^{0}(r,t),u(r,t),0,0)$ [for the components
$\mu=(t,r,\theta,\phi)$] is as usual normalized to
$U^{\mu}U_{\mu}=g_{\mu\nu}U^{\mu}U^{\nu}=-1$. This implies

$\displaystyle
u^{0}(r,t)=e^{-2\alpha(r,t)}\Bigg{(}u(r,t)\epsilon(r,t)+\sqrt{e^{2\alpha(r,t)}\left(\frac{ru(r,t)^{2}}{r-2Gm(r,t)}+1\right)+u(r,t)^{2}\epsilon(r,t)^{2}}\Bigg{)}.$
(18)

The computed expressions for the conservation law $\nabla_{\mu}T^{\mu}_{\
\nu}=0$ [to be referred to as $CO_{\nu}$] and the Einstein equation
$G_{\mu\nu}-8\pi GT_{\mu\nu}=0$ [to be referred to as $EQ_{\mu\nu}$] are given
in Appendix A. There are only four independent Einstein equations,
$EQ_{tt},EQ_{rr},EQ_{tr}$, and $EQ_{\theta\theta}$, because $EQ_{\phi\phi}$ is
the same as $EQ_{\theta\theta}$, and all other components are zero. There are
two non-zero conservation laws, $CO_{t}$ and $CO_{r}$. In principle, these six
equations would be used to solve the unknown variables in the metric and the
momentum tensor. However, a glance at Appendix A makes it obvious that these
equations are so complicated and so non-linear that there is no hope to solve
them exactly, so we resort to perturbation theory. Starting from a static
black hole of mass $M_{0}$ and radius $R_{0}$, as described in Sec. II,
additional matter is injected over a period of time so that the black hole
mass at any given time becomes $M(t)=M_{0}(1+\sigma(t/R_{0}))$, where
$\sigma(t/R_{0})$ is a non-negative and monotonically increasing function with
$\sigma(t/R_{0})\ll 1$ at all time. Accordingly, we make the expansions

$\displaystyle\alpha(r,t)$ $\displaystyle=$
$\displaystyle\alpha_{0}(r)+\alpha_{1}(r,t),$ (19) $\displaystyle p(r,t)$
$\displaystyle=$ $\displaystyle p_{0}(r)+p_{1}(r,t),$ (20)
$\displaystyle\rho(r,t)$ $\displaystyle=$
$\displaystyle\rho_{0}(r)+\rho_{1}(r,t),$ (21) $\displaystyle m(r,t)$
$\displaystyle=$ $\displaystyle m_{0}(r)+m_{1}(r,t),$ (22)

and keep only the 0th and 1st order terms of
$\epsilon(r,t),u(r,t),\alpha_{1}(r,t),p_{1}(r,t),\rho_{1}(r,t),m_{1}(r,t)$ in
the equations. To this order, it follows from Eq.(18) that
$u^{0}=e^{-\alpha_{0}(r)}\left(1-\alpha_{1}(r,t)\right)$.

###  III.1 Zeroth order

The 0th order equations describe a static black hole that is already
considered in Sec. II. In that case $EQ_{tr}$ is identically zero and the
metric is diagonal. It turns out that $CO_{t}$ is also zero, and that
$EQ_{\theta\theta}=0$ if the following three equations are satisfied. The
remaining equations can be extracted from the expressions in Appendix A to be

$\displaystyle CO_{r}$ $\displaystyle 0={dp_{0}(r)\over
dr}+\big{(}p_{0}(r)+\rho_{0}(r)\big{)}{d\alpha_{0}(r)\over dr}$ (23)
$\displaystyle EQ_{tt}$ $\displaystyle 0={dm_{0}(r)\over dr}-4\pi
r^{2}\rho_{0}(r)$ (24) $\displaystyle EQ_{rr}$ $\displaystyle
0={d\alpha_{0}(r)\over dr}r\big{(}r-2Gm_{0}(r)\big{)}-G\big{(}m_{0}(r)-4\pi
r^{3}p_{0}(r)\big{)}.$ (25)

These are equivalent to Eq.(5), Eq.(4), Eq.(3) used to calculate the static
black hole if we insert the equation of state $p_{0}(r)=-\xi\rho_{0}(r)$,
except that a subscript 0 is now added to the functions to emphasize its
static nature.

###  III.2 First order

To save writing, arguments of functions will not be shown below. All first-
order functions ($\epsilon,u$ and those with subscript 1) depend on $r$ and
$t$, and all the zeroth order functions (those with subscript 0) depend only
on $r$. Time derivative $\partial/\partial t$ is indicated by a dot on top,
and spatial derivative $\partial/\partial r$ is indicated by a prime.

The first-order relations of the six equations can be extracted from the
expression in Appendix A. After using Eq.(23) to Eq.(25) to simplify, the
result is

$\displaystyle\noindent CO_{t}$ (26) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\xi
r\bigg{(}e^{\alpha_{0}}(r-2Gm_{0})\dot{\rho}_{1}-(\xi-1)\rho_{0}\Big{(}e^{2\alpha_{0}}(r-2Gm_{0})u^{\prime}+Ge^{\alpha_{0}}\dot{m}_{1}\Big{)}\bigg{)}$
$\displaystyle+$
$\displaystyle(\xi-1)e^{2\alpha_{0}}\rho_{0}u\Big{(}G(4\xi-1)m_{0}+2\xi
r\left(2\pi G\xi r^{2}\rho_{0}-1\right)\Big{)}$ $\displaystyle\noindent
CO_{r}$ (27) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
re^{2\alpha_{0}}(r-2Gm_{0})\Big{(}p_{1}^{\prime}+(1-\xi)\rho_{0}\alpha_{1}^{\prime}\Big{)}+Ge^{2\alpha_{0}}\left(m_{0}-4\pi\xi
r^{3}\rho_{0}\right)(p_{1}+\rho_{1})$ $\displaystyle+$
$\displaystyle(1-\xi)r\rho_{0}(r-2Gm_{0})\dot{\epsilon}+(1-\xi)\left(e^{\alpha_{0}}\right)r^{2}\rho_{0}\dot{u}$
$\displaystyle\noindent EQ_{tt}$ (28) $\displaystyle 0$ $\displaystyle=$
$\displaystyle 4\pi r^{2}\rho_{1}-m_{1}^{\prime}$ $\displaystyle\noindent
EQ_{tr}$ (29) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\dot{m}_{1}-4\pi(\xi-1)r^{2}e^{\alpha_{0}}\rho_{0}u$
$\displaystyle\noindent EQ_{rr}$ (30) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-4\pi
Gr^{2}(r-2Gm_{0})p_{1}+(r-2Gm_{0})^{2}(\alpha_{1}^{\prime}+e^{-2\alpha_{0}}\dot{\epsilon})+G\left(8\pi
G\xi r^{2}\rho_{0}-1\right)m_{1}$ $\displaystyle\noindent EQ_{\theta\theta}$
(31) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
G\bigg{(}32\pi^{2}G^{2}\xi r^{5}\rho_{0}^{2}-Gm_{0}\Big{(}8\pi
G(4\xi-1)r^{2}\rho_{0}+1\Big{)}+4\pi G(3\xi-2)r^{3}\rho_{0}+r\bigg{)}m_{1}$
$\displaystyle-$ $\displaystyle G(2Gm_{0}-r)\left(Gm_{0}+4\pi G\xi
r^{3}\rho_{0}-r\right)m_{1}^{\prime}+Gr^{2}e^{-2\alpha_{0}}(2Gm_{0}-r)\ddot{m}_{1}$
$\displaystyle-$ $\displaystyle 8\pi
Gr^{2}(r-2Gm_{0})^{2}p_{1}-e^{-2\alpha_{0}}(r-2Gm_{0})^{2}\left(Gm_{0}+4\pi
Gr^{3}\rho_{0}-r\right)\dot{\epsilon}$ $\displaystyle-$
$\displaystyle(r-2Gm_{0})^{2}\Big{(}-Gm_{0}+4\pi
G(2\xi+1)r^{3}\rho_{0}-r\Big{)}\alpha_{1}^{\prime}+re^{-2\alpha_{0}}(r-2Gm_{0})^{3}\dot{\epsilon}^{\prime}$
$\displaystyle+$ $\displaystyle r(r-2Gm_{0})^{3}\alpha_{1}^{\prime\prime}$

As in Sec. II, it is useful to put these equations into a dimensionless form.
With the following definitions,

$\displaystyle r$ $\displaystyle=$ $\displaystyle xR_{0},\quad t=\tau
R_{0},\quad\epsilon(r,t)=\bar{\epsilon}(x,\tau),\quad\alpha_{0}(r)=\bar{\alpha}_{0}(x),\quad\alpha_{1}(r,t)=\bar{\alpha}_{1}(x,\tau)$
$\displaystyle u^{0}(r,t)$ $\displaystyle=$
$\displaystyle\bar{u}^{0}(x,\tau)=e^{-\bar{\alpha}_{0}(x)}\left(1-\bar{\alpha}_{1}(x,\tau)\right),\quad
u(r,t)=\bar{u}(x,\tau),\quad p_{1}(r,t)=\bar{p}_{1}(x,\tau)/(2GR_{0}^{2}),$
$\displaystyle\rho_{0}(r)$ $\displaystyle=$
$\displaystyle\bar{\rho}_{0}(x)(M_{0}/R_{0}^{3})=\bar{\rho}_{0}(x)/(2GR_{0}^{2}),\
\ \rho_{1}(r,t)=\bar{\rho}_{1}(x,\tau)/(2GR_{0}^{2}),$ $\displaystyle
m_{0}(r)$ $\displaystyle=$
$\displaystyle\bar{m}_{0}(x)M_{0}=\bar{m}_{0}(x)(R_{0}/2G),\ \
m_{1}(r,t)=\bar{m}_{1}(x,\tau)(R_{0}/2G),$ (32)

and $\bar{\rho}_{0}(x)$ replaced by $\bar{m}_{0}^{\prime}(x)/4\pi x^{2}$ per
Eq.(11), Eq.(26)-Eq.(31) become

$\displaystyle\noindent CO_{t}$ (33) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\xi-1)\xi x\bar{m}_{0}^{\prime}\dot{m}_{1}-2(\xi-1)\xi
x(x-\bar{m}_{0})e^{\bar{\alpha}_{0}}\bar{m}_{0}^{\prime}\bar{u}^{\prime}$
$\displaystyle+$
$\displaystyle(\xi-1)e^{\bar{\alpha}_{0}}\bar{m}_{0}^{\prime}\Big{(}\xi
x\left(\xi\bar{m}_{0}^{\prime}-4\right)+(4\xi-1)\bar{m}_{0}\Big{)}\bar{u}+8\pi\xi
x^{3}(x-\bar{m}_{0})\dot{\bar{\rho}}_{1}$ $\displaystyle\noindent CO_{r}$ (34)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle 2\pi
xe^{2\bar{\alpha}_{0}}\left(\bar{m}_{0}-\xi
x\bar{m}_{0}^{\prime}\right)(\bar{p}_{1}+\bar{\rho}_{1})-(\xi-1)xe^{\bar{\alpha}_{0}}\bar{m}_{0}^{\prime}\dot{\bar{u}}$
$\displaystyle-$
$\displaystyle(\xi-1)(x-\bar{m}_{0})\bar{m}_{0}^{\prime}(\dot{\bar{\epsilon}}+e^{2\bar{\alpha}_{0}}\bar{\alpha}_{1}^{\prime})+4\pi
x^{2}(x-\bar{m}_{0})e^{2\bar{\alpha}_{0}}\bar{p}_{1}^{\prime}$
$\displaystyle\noindent EQ_{tt}$ (35) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\bar{m}_{1}^{\prime}-4\pi x^{2}\bar{\rho}_{1}$
$\displaystyle\noindent EQ_{tr}$ (36) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\dot{\bar{m}}_{1}-(\xi-1)e^{\bar{\alpha}_{0}}\bar{m}_{0}^{\prime}\bar{u}$
$\displaystyle\noindent EQ_{rr}$ (37) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\left(\xi\bar{m}_{0}^{\prime}-1\right)\bar{m}_{1}-4\pi
x^{2}(x-\bar{m}_{0})\bar{p}_{1}+2(x-\bar{m}_{0})^{2}(e^{-2\bar{\alpha}_{0}}\dot{\bar{\epsilon}}+\bar{\alpha}_{1}^{\prime})$
$\displaystyle\noindent EQ_{\theta\theta}$ (38) $\displaystyle 0$
$\displaystyle=$
$\displaystyle(x-\bar{m}_{0})\Big{(}x\left(\xi\bar{m}_{0}^{\prime}-2\right)+\bar{m}_{0}\Big{)}\bar{m}_{1}^{\prime}-2(x-\bar{m}_{0})^{2}e^{-2\bar{\alpha}_{0}}\Big{(}x\left(\bar{m}_{0}^{\prime}-2\right)+\bar{m}_{0}\Big{)}\dot{\bar{\epsilon}}$
$\displaystyle+$
$\displaystyle\bigg{(}\bar{m}_{0}\Big{(}(1-4\xi)\bar{m}_{0}^{\prime}-1\Big{)}+x\Big{(}\xi\bar{m}_{0}^{\prime
2}+(3\xi-2)\bar{m}_{0}^{\prime}+2\Big{)}\bigg{)}\bar{m}_{1}$ $\displaystyle-$
$\displaystyle
2(x-\bar{m}_{0})^{2}\bigg{(}x\Big{(}(2\xi+1)\bar{m}_{0}^{\prime}-2\Big{)}-\bar{m}_{0}\bigg{)}\bar{\alpha}_{1}^{\prime}+2x^{2}(\bar{m}_{0}-x)e^{-2\bar{\alpha}_{0}}\ddot{\bar{m}}_{1}$
$\displaystyle-$ $\displaystyle 16\pi
x^{2}(x-\bar{m}_{0})^{2}\bar{p}_{1}+4x(x-\bar{m}_{0})^{3}(e^{-2\bar{\alpha}_{0}}\dot{\bar{\epsilon}}^{\prime}+\bar{\alpha}_{1}^{\prime\prime})$

Variables with a bar on top and a subscript 0 are functions of $x$, and
variables with a bar on top and a subscript 1, as well as $\bar{u}$ and
$\bar{\epsilon}$, are functions of $x$ and $\tau$. A prime on a function of
$(x,\tau)$ now represents $\partial/\partial x$, and a dot now represents
$\partial/\partial\tau$.

##  IV Time-independent perturbation and solutions

The first-order quantities
$\bar{p}_{1},\bar{\rho}_{1},\bar{m}_{1},\bar{u},\bar{\epsilon},\bar{\alpha}_{1}$
are to be solved from Eq.(33) to Eq.(38), but these equations still look too
complicated to be tackled directly. Fortunately, the time-independent version
of these equations can be solved, and they in turn provide a stepping stone
for the solution of the general time-dependent equations.

In the time-independent scenario, a small amount of matter is added to a black
hole of mass $M_{0}$, radius $R_{0}$, density $\rho_{0}(r)$, and internal mass
$m_{0}(r)$, to obtain a black hole of mass
$M_{01}=M_{0}+M_{1}:=(1+\sigma)M_{0}$, radius
$R_{01}=R_{0}+R_{1}=(1+\sigma)R_{0}$, density
$\rho_{01}(r)=\rho_{0}(r)+\rho_{1}(r)$, and internal mass
$m_{01}(r)=m_{0}(r)+m_{1}(r)$. We will assume $0<\sigma\ll 1$ to enable
perturbation theory to be applied.

Since the perturbed and the unperturbed solutions are now both static, we can
drop all the time dependences and all the time-derivative terms in Eq.(33) to
Eq.(38), the $\tau$ dependence in all functions, as well as the non-diagonal
quantities $\bar{u}$ and $\bar{\epsilon}$. As a result, Eq.(33) and Eq.(36)
are identically zero, and the remaining four equations become

$\displaystyle\noindent CO_{r}$ (39) $\displaystyle 0$ $\displaystyle=$
$\displaystyle 2\pi x\left(\bar{m}_{0}-\xi
x\bar{m}_{0}^{\prime}\right)(\bar{p}_{1}+\bar{\rho}_{1})-(\xi-1)(x-\bar{m}_{0})\bar{m}_{0}^{\prime}\bar{\alpha}_{1}^{\prime}+4\pi
x^{2}(x-\bar{m}_{0})\bar{p}_{1}^{\prime}$ $\displaystyle\noindent EQ_{tt}$
(40) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\bar{m}_{1}^{\prime}-4\pi x^{2}\bar{\rho}_{1}$
$\displaystyle\noindent EQ_{rr}$ (41) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\left(\xi\bar{m}_{0}^{\prime}-1\right)\bar{m}_{1}-4\pi
x^{2}(x-\bar{m}_{0})\bar{p}_{1}+2(x-\bar{m}_{0})^{2}\bar{\alpha}_{1}^{\prime}$
$\displaystyle\noindent EQ_{\theta\theta}$ (42) $\displaystyle 0$
$\displaystyle=$
$\displaystyle(x-\bar{m}_{0})\Big{(}x\left(\xi\bar{m}_{0}^{\prime}-2\right)+\bar{m}_{0}\Big{)}\bar{m}_{1}^{\prime}$
$\displaystyle+$
$\displaystyle\bigg{(}\bar{m}_{0}\Big{(}(1-4\xi)\bar{m}_{0}^{\prime}-1\Big{)}+x\Big{(}\xi\bar{m}_{0}^{\prime
2}+(3\xi-2)\bar{m}_{0}^{\prime}+2\Big{)}\bigg{)}\bar{m}_{1}$ $\displaystyle-$
$\displaystyle
2(x-\bar{m}_{0})^{2}\bigg{(}x\Big{(}(2\xi+1)\bar{m}_{0}^{\prime}-2\Big{)}-\bar{m}_{0}\bigg{)}\bar{\alpha}_{1}^{\prime}$
$\displaystyle-$ $\displaystyle 16\pi
x^{2}(x-\bar{m}_{0})^{2}\bar{p}_{1}+4x(x-\bar{m}_{0})^{3}\bar{\alpha}_{1}^{\prime\prime}$

These equations still look very formidable, but fortunately their solutions
can be deduced from the scaled solutions of Sec II. Remember in this regard
that a subscript 0 has now been added to the quantities in that section.

According to Sec. II, the scaled density $\bar{\rho}_{0}(x)$ and the scaled
internal mass $\bar{m}_{0}(x)$ obey the differential equations Eq.(10) and
Eq.(11), whose numerical solution for $\xi=0.16$ is given in Fig. 1 as an
illustration. These scaled variables are related to the actual variables by
$r=xR$, $\rho(r)=(M/R^{3})\bar{\rho}(x)=\bar{\rho}(x)/(2GR^{2})$, and
$m(r)=\bar{m}(x)M=\bar{m}(x)(R/2G)$. Thus

$\displaystyle\rho_{0}(r)$ $\displaystyle=$
$\displaystyle\bar{\rho}_{0}(x_{0})/2GR_{0}^{2},\qquad
m_{0}(r)=\bar{m}_{0}(x_{0})R_{0}/2G,\qquad x_{0}=r/R_{0}$
$\displaystyle\rho_{01}(r)$ $\displaystyle=$
$\displaystyle\bar{\rho}_{0}(x_{01})/2GR_{01}^{2},\qquad
m_{01}(r)=\bar{m}_{0}(x_{01})R_{01}/2G,\qquad x_{01}=r/R_{01}.$

Note that $\bar{\rho}_{01}$ has been replaced by $\bar{\rho}_{0}$ and
$\bar{m}_{01}$ has been replaced by $\bar{m}_{0}$. This is allowed because the
zeroth order scaled functions are universal, independent of $R$ and $M$ except
through their arguments $x$.

In order to have a uniform notation later, we shall replace all $x_{0}$ by $x$
from now on. With $R_{01}=(1+\sigma)R_{0}$, and $\sigma\ll 1$, one gets
$x_{01}\simeq x(1-\sigma)$. Hence

$\displaystyle\bar{\rho}(x_{01})$ $\displaystyle\simeq$
$\displaystyle\bar{\rho}_{0}(x)-\sigma x{\bar{\rho}}_{0}^{\prime}(x),$ (43)
$\displaystyle\bar{m}(x_{01})$ $\displaystyle\simeq$
$\displaystyle\bar{m}_{0}(x)-\sigma x\bar{m}_{0}^{\prime}(x),$ (44)
$\displaystyle\rho_{01}(r)$ $\displaystyle\simeq$
$\displaystyle{\bar{\rho}_{0}(x)-\sigma x\bar{\rho}_{0}^{\prime}(x)\over
2GR_{0}^{2}(1+\sigma)^{2}}\simeq\rho_{0}(r)-{\sigma\over
2GR_{0}^{2}}\left(2\bar{\rho}_{0}(x)+x\bar{\rho}_{0}^{\prime}(x)\right),$ (45)
$\displaystyle m_{01}(r)$ $\displaystyle\simeq$
$\displaystyle\left(\bar{m}_{0}(x)-2\sigma
x\bar{m}_{0}^{\prime}(x)\right)(1+\sigma){R_{0}\over 2G}=m_{0}(r)+{\sigma
R_{0}\over 2G}\left(\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\right).$ (46)

The scaled perturbed density and internal mass are then

$\displaystyle\bar{\rho}_{1}(x)$ $\displaystyle\equiv$
$\displaystyle\left(\rho_{01}(r)-\rho(r)\right)2GR_{0}^{2}=-\sigma\left(2\bar{\rho}_{0}(x)+x\bar{\rho}_{0}^{\prime}(x)\right)=\\!-\\!\sigma\bar{m}^{\prime\prime}_{0}(x)/4\pi
x,$ (47) $\displaystyle\bar{m}_{1}(x)$ $\displaystyle\equiv$
$\displaystyle\left(m_{01}(r)-m(r)\right){2G\over
R_{0}}=\sigma\left(\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\right),$ (48)

Since $\bar{m}_{0}(1)=M_{0}$ and $\bar{m}_{0}^{\prime}(1)=0$, the perturbed
mass added to the original black hole of mass $M_{0}$ is
$\bar{m}_{1}(1)=\sigma M_{0}=M_{1}$, which is the expected value.

The equation of state $p(x)=-\xi\rho(x)$ valid for both the unperturbed black
hole and the perturbed black hole implies

$\displaystyle\bar{p}_{1}(x)=-\xi\bar{\rho}_{1}(x).$ (49)

Also, from Eq.(12),

$\displaystyle\bar{\alpha}_{1}^{\prime}(x)$ $\displaystyle=$
$\displaystyle{\xi\over
1-\xi}\left({1\over\bar{\rho}_{0}}{d\bar{\rho}_{1}(x)\over
dx}-{\bar{\rho}_{1}(x)\over\bar{\rho}_{0}(x)^{2}}{d\bar{\rho}_{0}(x)\over
dx}\right)$ (50) $\displaystyle=$
$\displaystyle-{\xi\sigma\bigg{(}\\!-\\!x\bar{m}_{0}^{\prime\prime}(x)^{2}+\bar{m}_{0}^{\prime}(x)\Big{(}\bar{m}_{0}^{\prime\prime}(x)+x\bar{m}_{0}^{\prime\prime\prime}(x)\Big{)}\bigg{)}\over(1-\xi)\bar{m}_{0}^{\prime}(x)^{2}}.$

Another way to deduce Eq.(50) is to start from the scaled version of Eq.(9),
which reads $\bar{\alpha}(x)=\ln\bar{\rho}(x)/2\gamma$. Hence

$\displaystyle\bar{\alpha}_{1}(x)={1\over
2\gamma}\ln\bigg{(}{\bar{\rho}_{0}(x)+\bar{\rho}_{1}(x)\over\bar{\rho}_{0}(x)}\bigg{)}\simeq{\bar{\rho}_{1}(x)\over
2\gamma\bar{\rho}_{0}(x)}=-{\sigma\xi\over
1-\xi}{x\bar{m}_{0}^{\prime\prime}(x)\over\bar{m}_{0}^{\prime}(x)},$ (51)

whose $x$-derivative then gives Eq.(50).

The higher derivatives of $\bar{m}_{0}(x)$ appearing in Eq.(47) to Eq.(51) can
be converted into functions of $\bar{m}_{0}^{\prime}(x)$ and $\bar{m}_{0}(x)$
by using Eq.(10) and Eq.(11). For example,

$\displaystyle\bar{m}_{0}^{\prime\prime}(x)$ $\displaystyle=$
$\displaystyle\frac{\bar{m}_{0}^{\prime}(x)\Big{(}\xi
x\Big{(}(\xi-1)\bar{m}_{0}^{\prime}(x)+4\Big{)}+(1-5\xi)\bar{m}_{0}(x)\Big{)}}{2\xi
x(x-\bar{m}_{0})},$ $\displaystyle\bar{m}_{0}^{\prime\prime\prime}(x)$
$\displaystyle=$
$\displaystyle\frac{\bar{m}^{\prime}_{0}(x)}{4\xi^{2}x^{2}(x-\bar{m}_{0}(x))^{2}}\bigg{(}2\xi
x^{2}\Big{(}(\xi-1)\xi^{2}\bar{m}_{0}^{\prime}(x)^{2}+\left(5\xi^{2}-6\xi+1\right)\bar{m}_{0}^{\prime}(x)+4\xi\Big{)}$
(52) $\displaystyle-$
$\displaystyle\xi(5\xi-1)x\bar{m}_{0}(x)\Big{(}3(\xi-1)\bar{m}_{0}^{\prime}(x)+4\Big{)}+\left(15\xi^{2}-8\xi+1\right)\bar{m}_{0}(x)^{2}\bigg{)}.$

Substituting Eq.(52) into Eq.(50), one gets

$\displaystyle\bar{\alpha}_{1}^{\prime}(x)$ $\displaystyle=$
$\displaystyle\frac{\sigma\bigg{(}\bar{m}_{0}(x)\big{(}(1-7\xi)\bar{m}_{0}^{\prime}(x)+2\big{)}+x\bar{m}_{0}^{\prime}(x)\big{(}(\xi+1)\xi\bar{m}_{0}^{\prime}(x)+4\xi-2\big{)}\bigg{)}}{4(x-\bar{m}_{0}(x))^{2}}.$
(53)

It can be checked, with the help of Eq.(52), that Eq.(47) to Eq.(50) are
indeed the solutions of Eq.(39) to Eq.(42).

Before leaving this section, let us look at the signs of various quantities
and their physical implications. To start with, the injected mass is expected
to increase both $\bar{\rho}(x)$ and $\bar{m}(x)$, causing $\bar{\rho}_{1}(x)$
and $\bar{m}_{1}(x)$ to be positive. This can be seen analytically to be the
case from Eq.(47) and Eq.(48), taking into account the behavior of
$\bar{\rho}_{0}(x)$ and $\bar{m}_{0}(x)$ discussed in Sec. II. The property
$\bar{m}_{0}^{\prime\prime}(x)<0\ (\Rightarrow\bar{\rho}_{1}(x)>0)$ is a
consequence of the decrease of $\bar{m}_{0}^{\prime}(x)$ from $\sim\beta
x^{\beta-1}$ at small $x$, towards 0 at $x=1$. It can be seen from the
convexity of $\bar{m}_{0}(x)$ in Fig. 1, as well as the exact expression of
$\bar{m}_{0}^{\prime\prime}(x)$ in Eq.(52). The property
$\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)>0\ (\Rightarrow\bar{m}_{1}(x)>0)$
results from the increase of $\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\sim
x^{\beta}(1-\beta)>0$ at small $x$ towards 1 at $x=1$, staying positive all
the way through.

Since $\xi<1$ and $\bar{m}_{0}^{\prime}(x)>0$, it also follows that
$\bar{m}_{0}(x)-x\xi\bar{m}_{0}^{\prime}(x)>0$ and
$\bar{m}_{0}(x)-x\xi^{2}\bar{m}_{0}^{\prime}(x)>0$. The former is needed in
Eq.(10) to keep $d\bar{\rho}_{0}(x)/dx$ negative and to ensure that
$\bar{\rho}_{0}(x)$ decreases steadily from infinity to 0. Both of these two
inequalities will be used later to study the behavior of time-dependent
solutions.

##  V Solution of the first-order time-dependent equations

Returning now to the general case, where matter is injected into a black hole
of mass $M_{0}$ and radius $R_{0}=2GM_{0}$ over a period of time. In that case
$M(t):=M_{0}(1+\sigma(\tau))$, with $\sigma(\tau)\ll 1$ at all time so that
perturbation theory can be used. The task now is to find the solution of the
six time-dependent first-order equations Eq.(33) to Eq.(38). Fortunately, with
the help of the solution for the time-independent solutions obtained in the
last section, solutions of the time-dependent equations can also be found.

Since Eq.(47), Eq.(48), and Eq.(51) are true for any $\sigma$, we might expect
their time-dependent solutions to be simply

$\displaystyle\bar{\rho}_{1}(x,\tau)$ $\displaystyle=$
$\displaystyle-\sigma(\tau)\left(2\bar{\rho}_{0}(x)+x\bar{\rho}_{0}^{\prime}(x)\right)=-\sigma(\tau)\bar{m}_{0}^{\prime\prime}(x)/4\pi
x:=\hat{\rho}_{1}(x)\sigma(\tau),$ (54) $\displaystyle\bar{m}_{1}(x,\tau)$
$\displaystyle=$
$\displaystyle\sigma(\tau)\left(\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\right):=\hat{m}_{1}(x)\sigma(\tau),$
(55) $\displaystyle\bar{\alpha}_{1}(x,\tau)$ $\displaystyle=$
$\displaystyle{\bar{\rho}_{1}(x,\tau)\over
2\gamma\bar{\rho}_{0}(x)}=-{\sigma(\tau)\xi\over
1-\xi}{x\bar{m}_{0}^{\prime\prime}(x)\over\bar{m}_{0}^{\prime}(x)}:=\hat{\alpha}_{1}(x)\sigma(\tau).$
(56)

A priori it is possible that they may also depend on
$\dot{\sigma}(\tau),\ddot{\sigma}(\tau)$, etc., which vanish when $\sigma$ is
a constant. However, the following argument shows that this does not happen,
so the response of internal density adjustment to an external mass addition is
instantaneous.

Suppose
$\bar{m}_{1}(x,\tau)=\sigma(\tau)\mu_{1}(x)+\dot{\sigma}(\tau)\mu_{2}(x)$,
where $\bar{m}_{1}(x,\tau)$ is the mass of a spherical ball with a scaled
radius $x\leq 1$ at scaled time $\tau$. As such, $\bar{m}_{1}(x,\tau)$ must be
positive, and be a monotonically increasing function of $x$ and of $\tau$ with
$\bar{m}_{1}(1,\tau)=M_{0}\sigma(\tau)$. Since $\sigma(\tau)$ is arbitrary, at
any time one can choose $\sigma(\tau)\gg\dot{\sigma}(\tau)$, or
$\sigma(\tau)\ll\dot{\sigma}(\tau)$, so both $\mu_{1}(x)$ and $\mu_{2}(x)$
must be non-negative monotonically increasing functions of $x$. The condition
$\bar{m}_{1}(1,\tau)=M_{0}\sigma(\tau)$ shows that $\mu_{1}(1)=M_{0}$ and
$\mu_{2}(1)=0$. The requirement that $\mu_{2}(x)$ be monotonically increasing
then shows that $\mu_{2}(x)=0$ for all $x$. Thus $\bar{m}(x,\tau)$ cannot
depend on $\dot{\sigma}(\tau)$, and similarly it cannot depend on the higher
derivatives of $\sigma(\tau)$. Consequently Eq.(55) should be the time-
dependent solution of $\bar{m}_{1}(x,\tau)$.

By Eq.(35), Eq.(54) should also be the time-dependent solution of
$\bar{\rho}_{1}(x,\tau)$, which shows that Eq.(56) must also be true.

The rest of the variables can be obtained from Eq.(33) to Eq.(38), with the
help of Eq.(52), Eq.(54), Eq.(55), and Eq.(56).

From Eq.(36), one gets

$\displaystyle\bar{u}(x,\tau)=-\frac{e^{-\bar{\alpha}_{0}(x)}\big{(}\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\big{)}\dot{\sigma}(\tau)}{(1-\xi)\bar{m}_{0}^{\prime}(x)}:=\hat{u}(x)\dot{\sigma}(\tau).$
(57)

With Eq.(57), and Eq.(54), Eq.(55), the right-hand side of Eq.(33) can be
computed. The result is zero, so Eq.(33) is satisfied, further confirming the
validity of Eq.(54) and Eq.(55).

Next, from Eq.(37) one gets

$\displaystyle\dot{\bar{\epsilon}}$ $\displaystyle=$
$\displaystyle\frac{e^{2\bar{\alpha}_{0}}}{4(x-\bar{m}_{0})^{2}}\bigg{(}8\pi
x^{2}(x-\bar{m}_{0})\bar{p}_{1}-\sigma(\tau)\Big{(}\xi
x\big{(}(\xi-1)\bar{m}_{0}^{\prime}+4\big{)}+(1-5\xi)\bar{m}_{0}\Big{)}\bar{m}_{0}^{\prime}\bigg{)}.$
(58)

Substituting this into Eq.(34), an ordinary differential equation (in $x$) for
$p_{1}(x,\tau)$ emerges,

$\displaystyle 0$ $\displaystyle=$ $\displaystyle 16\pi\xi
x^{3}(x-\bar{m}_{0})^{2}e^{2\bar{\alpha}_{0}}\bar{p}_{1}^{\prime}-8\pi\xi
x^{2}(x-\bar{m}_{0})e^{2\bar{\alpha}_{0}}\Big{(}(2\xi-1)x\bar{m}_{0}^{\prime}-\bar{m}_{0}\Big{)}\bar{p}_{1}$
(59) $\displaystyle-e^{2\bar{\alpha}_{0}}\bar{m}_{0}^{\prime}\bigg{(}\xi
x^{2}\bar{m}_{0}^{\prime}\Big{(}(\xi\\!-\\!1)\xi\bar{m}_{0}^{\prime}\\!-\\!6\xi+2\Big{)}\\!+\\!2\xi
x\bar{m}_{0}\Big{(}\\!-\\!\left(\xi^{2}\\!-\\!4\xi\\!+\\!1\right)\bar{m}_{0}^{\prime}\\!+\\!\xi\\!+\\!1\Big{)}+(1\\!-\\!5\xi)\bar{m}_{0}^{2}\bigg{)}\sigma(\tau)$
$\displaystyle+4\xi
x^{2}(x-\bar{m}_{0})\left(x\bar{m}_{0}^{\prime}-\bar{m}_{0}\right)\sigma^{\prime\prime}(\tau).$

This equation is considerably simplified if we make use of the equation of
state. In the static case, the equation of state is
$\bar{p}(x)=-\xi\bar{\rho}(x)$, with $\xi$ being a constant. In the time
dependent case, $\xi(x,\tau):=-\bar{p}(x,\tau)/\bar{\rho}(x,\tau)$ may no
longer be a constant. If we expand $\xi(x,\tau)=\xi+\xi_{1}(x,\tau)$
perturbatively, then the first-order equation of state becomes

$\displaystyle\bar{p}_{1}(x,\tau)=-\xi\bar{\rho}_{1}(x,\tau)-\xi_{1}(x,\tau)\bar{\rho}_{0}(x).$
(60)

The differential equation obtained from Eq.(59) and Eq.(60) for
$\xi_{1}(x,\tau)$ is considerably simpler,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\xi_{1}(x,\tau)^{\prime}-f(x)\xi_{1}(x,\tau)-g(x)\ddot{\sigma}(\tau),$
$\displaystyle f(x)$ $\displaystyle=$
$\displaystyle\frac{\bar{m}_{0}(x)-\xi^{2}x\bar{m}_{0}^{\prime}(x)}{2\xi
x(\bar{m}_{0}(x)-x)},$ $\displaystyle g(x)$ $\displaystyle=$
$\displaystyle\frac{xe^{-2\bar{\alpha}_{0}(x)}\left(\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\right)}{(\bar{m}_{0}(x)-x)\bar{m}_{0}^{\prime}(x)}.$
(61)

The use of $\xi_{1}$ not only simplifies Eq.(59), it also reduces Eq.(58) to
the simpler form

$\displaystyle\dot{\bar{\epsilon}}(x,\tau)=\frac{e^{2\bar{\alpha}_{0}(x)}\bar{m}_{0}^{\prime}(x)\xi_{1}(x,\tau)}{2(\bar{m}_{0}(x)-x)}.$
(62)

Substituting Eq.(58) into Eq.(38) and using Eq.(60), one also gets a
differential equation for $\xi_{1}(x,\tau)$, which turns out to be identical
to the one shown in Eq.(61). This shows that Eq.(38) is not independent of the
other five first-order equations.

To summarize, the solution of the six equations Eq.(33) to Eq.(38) is given by
Eq.(47), Eq.(48), Eq.(51), Eq.(60), and Eq.(62), with $\xi_{1}(x,\tau)$
satisfying the differential equation Eq.(61).

To solve Eq.(61), we need to specify boundary conditions. First of all, it
follows from Eq.(57) that $\bar{u}(0,\tau)=0$ (see also the discussion in the
next section). With no flow at the origin, the equation of state there must
remain static, hence $\xi_{1}(0,\tau)=0$. Secondly, for static perturbation
when the driving term $g(x)\ddot{\sigma}(\tau)=0$ in Eq.(61), we must also
have $\xi_{1}(x,\tau)=0$. With these conditions, the solution of Eq.(61) must
be of the form $\xi_{1}(x,\tau)=\hat{\xi}_{1}(x)\ddot{\sigma}(\tau)$, with

$\displaystyle\hat{\xi}_{1}(x)=h(x)\int_{0}^{x}{g(y)\over h(y)}dy,$ (63)

where $h(x)$ is the solution of the homogeneous differential equation
$h^{\prime}(x)-f(x)h(x)=0$, whose solution is

$\displaystyle h(x)=h_{0}\exp\left(\int_{a}^{x}f(z)dz\right).$ (64)

Different lower limit $a$ corresponds to a different amplitude $h_{0}$, but it
can be seen from Eq.(63) that the solution of $\hat{\xi}_{1}(x)$ is completely
independent of the amplitude, so the choice of $h_{0}$ and $a$ is immaterial.

##  VI Summary, discussions, and numerical illustrations

###  VI.1 Summary

When matter enters a black hole, its character is altered. Most of its global
quantum numbers such as isospin and baryonic number are lost. What then is the
property of this new kind of ‘black hole matter’? Being inside a black hole,
that cannot be directly measured, but it must still satisfy the Einstein
equations and the covariant conservation laws, from which some property can be
deduced theoretically.

In the previous paper Lam , we found that matter inside a neutral, static,
spherical, and non-rotational black hole can be modelled by an ideal fluid,
whose pressure $p_{0}(r)$ and density $\rho_{0}(r)$ obey an equation of state
$p_{0}(r)=-\xi\rho_{0}(r)$ for some constant $\xi>0$. The negative pressure
present is needed to counter gravitational attraction to allow the black hole
matter to fill the interior of the black hole. From the Einstein equations and
the covariant conservation laws, we found that a solution exists when the
constant $\xi$ is confined to a narrow range between $1/7=0.1429$ and
$3-2\sqrt{2}=0.1716$. Within this range, the internal density $\rho_{0}(r)$
which vanishes at the event horizon to satisfy continuity, rises steadily and
becomes divergent at the origin as demanded by the singularity theorem.

In the present article, the dynamical response of this matter is probed by
injecting additional matter into the black hole over a period of time. The
time-dependent metric and the energy-momentum tensor for this situation are
given in Eq.(16) and Eq.(17). The resulting Einstein equations and covariant
conservation laws are listed in Apendix A. These equations are so complicated
and so non-linear that there is no hope to solve them exactly, so we resort to
perturbation theory. The unperturbed quantities come from the initial static
black hole with mass $M_{0}$, and the perturbed quantities arise from the
injected mass $M_{0}\sigma(\tau)$. The total black hole mass grows from
$M_{0}$ to $M(t)=M_{0}(1+\sigma(\tau))$, with $0\leq\sigma(\tau)\ll 1$ at all
time to allow perturbation theory to be used.

With the perturbation expansion given in Eq.(22), the resulting six first-
order Einstein equations and conservation laws are shown in Eq.(33) to
Eq.(38), where $x=r/R_{0}$ is the scaled radial distance and $\tau=t/R_{0}$ is
the scaled time. A prime represents $\partial/\partial x$, and a dot indicates
$\partial/\partial\tau$.

The solution of these six equations can be obtained and is shown in Eq.(54),
Eq.(55), Eq.(56), Eq.(57), Eq.(62), and Eq.(63). For convenience, they are
reproduced below.

$\displaystyle\bar{\rho}_{1}(x,\tau)$ $\displaystyle=$
$\displaystyle-\sigma(\tau)\left(2\bar{\rho}_{0}(x)+x\bar{\rho}_{0}^{\prime}(x)\right)=-\sigma(\tau)\bar{m}_{0}^{\prime\prime}(x)/4\pi
x:=\hat{\rho}_{1}(x)\sigma(\tau),$ (65) $\displaystyle\bar{m}_{1}(x,\tau)$
$\displaystyle=$
$\displaystyle\sigma(\tau)\left(\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\right):=\hat{m}_{1}(x)\sigma(\tau),$
(66) $\displaystyle\bar{\alpha}_{1}(x,\tau)$ $\displaystyle=$
$\displaystyle{\bar{\rho}_{1}(x,\tau)\over
2\gamma\bar{\rho}_{0}(x)}=-{\sigma(\tau)\xi\over
1-\xi}{x\bar{m}_{0}^{\prime\prime}(x)\over\bar{m}_{0}^{\prime}(x)}:=\hat{\alpha}_{1}(x)\sigma(\tau),$
(67) $\displaystyle\bar{u}(x,\tau)$ $\displaystyle=$
$\displaystyle-\frac{e^{-\bar{\alpha}_{0}(x)}\big{(}\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\big{)}\dot{\sigma}(\tau)}{(1-\xi)\bar{m}_{0}^{\prime}(x)}:=\hat{u}(x)\dot{\sigma}(\tau),$
(68) $\displaystyle\dot{\bar{\epsilon}}(x,\tau)$ $\displaystyle=$
$\displaystyle\frac{e^{2\bar{\alpha}_{0}(x)}\bar{m}_{0}^{\prime}(x)\hat{\xi}_{1}(x)\sigma^{\prime\prime}(\tau)}{2(\bar{m}_{0}(x)-x)}:=\hat{\epsilon}(x)\ddot{\sigma}(\tau),$
(69) $\displaystyle\xi_{1}(x,\tau)$ $\displaystyle=$
$\displaystyle\hat{\xi}_{1}(x)\ddot{\sigma}(\tau),\qquad\hat{\xi}_{1}(x)=h(x)\int_{0}^{x}{g(y)\over
h(y)}dy,$ (70) $\displaystyle h(x)=h_{0}\exp\left(\int_{a}^{x}f(z)dz\right)$
$\displaystyle f(x)=\frac{\bar{m}_{0}(x)-\xi^{2}x\bar{m}_{0}^{\prime}(x)}{2\xi
x(\bar{m}_{0}(x)-x)},$ $\displaystyle
g(x)=\frac{xe^{-2\bar{\alpha}_{0}(x)}\left(\bar{m}_{0}(x)-x\bar{m}_{0}^{\prime}(x)\right)}{(\bar{m}_{0}(x)-x)\bar{m}_{0}^{\prime}(x)},$

where $\xi_{1}(x,\tau)$ is related to $\bar{p}_{1}(x,\tau)$ through Eq.(60).

###  VI.2 Discussions and limitations

The three quantities $\bar{m}_{1}(x,\tau),\bar{\alpha}_{1}(x,\tau)$, and
$\bar{\epsilon}(x,\tau)$ describe how the spacetime metric is modified. The
other three quantities $\bar{\rho}_{1}(x,\tau),\bar{u}(x,\tau)$, and
$\xi_{1}(x,\tau)$ tell us how the energy-momentum tensor is affected by the
injection of matter, namely, the dynamical response of black hole matter. In
each of the six quantities, the spatial ($x$) and the temporal ($\tau$)
dependences factorize. The temporal function of
$\bar{m}_{1}(x,\tau),\bar{\alpha}_{1}(x,\tau),\bar{\rho}_{1}(x,\tau)$ is
$\sigma(\tau)$, so the response to an external injection of matter is
instantaneous. Both $\bar{u}(x,\tau)$ and $\bar{\epsilon}(x,\tau)$ are
proportional to $\dot{\sigma}(\tau)$, causing the non-diagonal elements of the
metric and the momentum tensor to vanish for time-independent perturbations
when $\dot{\sigma}(\tau)=0$. Modification of the static equation of state is
encoded in $\xi_{1}(x,\tau)$, which is proportional to $\ddot{\sigma}(\tau)$,
so the static equation of state
$\bar{p}_{1}(x,\tau)=-\xi\bar{\rho}_{1}(x,\tau)$ remains valid if external
matter is fed in at a constant rate, when $\ddot{\sigma}(\tau)=0$. How the
static equation of state is modified under an accelerated injection of matter
will be discussed below.

The $x$ dependences of these six quantities are given by functions with a hat
on top, e.g., $\bar{\rho}_{1}(x,\tau)=\hat{\rho}_{1}(x)\sigma(\tau)$. It can
be seen from Eq.(65) to Eq.(70), and the inequalities discussed at the end of
Sec. IV, that each of $\hat{\rho}_{1}(x)$, $\hat{m}_{1}(x)$,
$\hat{\alpha}_{1}(x)$, $-\hat{u}(x)$, $\hat{\epsilon}(x)$, $f(x)$, $g(x)$, and
$\hat{\xi}_{1}(x)$ is positive in the range $x\in(0,1)$.

Physically, the positivity of $\hat{\rho}_{1}(x)$ and $\hat{m}_{1}(x)$ come
about because matter is added to the black hole. This injection causes matter
to flow inward towards, giving rise to $-\hat{u}(x)>0$. The positivity of
$\hat{\xi}_{1}(x)$ is the most interesting consequence, showing an increase in
stiffness of black hole matter to counter an accelerated inward flow, as we
shall explain later.

To better understand these first-order hat functions, we need to know their
behavior near the boundaries $x=0$ and $x=1$. Using the behavior of the
zeroth-order functions near the boundary given in Sec. II, and summarized in
the first few rows of Table 1 and Table 2, the behavior of the first-order
functions can be deduced from Eq.(65) to Eq.(70) and is displayed in the
remaining rows of Table 1 and Table 2. Coefficients are not given except for
the function $f(x)$, because its coefficient affects the $x$-power of $h(x)$
via the homogeneous differential equation $h^{\prime}(x)-f(x)h(x)=0$.

$\begin{array}[]{|c||c|c|}\hline\cr&x\sim 0&x\sim 1\\\
\hline\cr\hline\cr\bar{\rho}_{0}(x)&x^{\beta-3}&(1-x)^{\gamma}\\\
\bar{m}_{0}(x)&x^{\beta}&1\\\
\bar{m}_{0}^{\prime}(x)&x^{\beta-1}&(1-x)^{\gamma}\\\
\bar{m}_{0}^{\prime\prime}(x)&x^{\beta-2}&(1-x)^{\gamma-1}\\\
e^{2\bar{\alpha}_{0}(x)}&x^{-2\zeta}&(1-x)\\\
\hat{\rho}_{1}(x)&x^{\beta-3}&(1-x)^{\gamma-1}\\\
\hat{m}_{1}(x)&x^{\beta}&1\\\ \hat{\alpha}_{1}(x)&1&(1-x)^{-1}\\\
\hat{u}(x)&x^{\zeta+1}&(1-x)^{-{1\over 2}-\gamma}\\\
f(x)&\simeq\omega/x&\simeq 1/2\xi(1-x)\\\ h(x)&x^{\omega}&(1-x)^{1/2\xi}\\\
g(x)&x^{\delta-1}&(1-x)^{-2-\gamma}\\\
\hat{\xi}_{1}(x)&x^{\delta}&(1-x)^{-\gamma-1}\\\
\hat{\epsilon}(x)&x^{2-\beta}&1\\\
\hat{\rho}_{1}(x)/\bar{\rho}_{0}(x)&1&1/(1-x)\\\
\hat{m}_{1}(x)/\bar{m}_{0}(x)&1&1\\\
\hat{\alpha}_{1}(x)/\bar{\alpha}_{0}(x)&1/|\ln(x)|&1/(1-x)\ln|(1-x)|\\\
\hline\cr\end{array}$

Table 1. Boundary behavior of various functions of $x$. Coefficients are not
shown except for $f(x)$. To emphasize that point a $\simeq$ symbol is added to
that row

$\begin{array}[]{|c|c|c|c|}\hline\cr&{\rm definition}&{\rm as\ a\ function\
of}\ \xi&\xi=0.16\\\ \hline\cr\gamma&\gamma&(1-\xi)/2\xi&2.625\\\
\beta&\beta&(-1+7\xi)/\xi(1+\xi)&0.646552\\\
\zeta&(3-\beta)/2\gamma&(1-3\xi)/(1+\xi)&0.448276\\\
\omega&(1-\beta\xi^{2})/2\xi&(1+2\xi-7\xi^{2})/2\xi(1+\xi)&3.07328\\\
\delta&(3-\beta)(1+\gamma^{-1})&-3+1/\xi&2.94143\\\ \hline\cr\end{array}$

Table 2. Various parameters (column 1) as functions of $\xi$ (column 2)

and their values at $\xi=0.16$

Shown in Table 1 are also the ratios $\hat{\rho}_{1}(x)/\bar{\rho}_{0}(x)$,
$\hat{m}_{1}(x)/\bar{m}_{0}(x)$, $\hat{\alpha}_{1}(x)/\bar{\alpha}_{0}(x)$,
because they affect where perturbation theory is valid. For perturbation
theory to make sense, the first-order quantities in Eq.(22) must be much
smaller than the corresponding zeroth order quantities, and that we must also
have $|\bar{u}(x,\tau)|\ll 1$, $|\bar{\epsilon}(x,\tau)|\ll 1$, and
$|\xi_{1}(x,t)|\ll\xi$. If we examine Table 1, we see that perturbation theory
would be valid provided $\sigma(\tau),\dot{\sigma}(\tau)$, and
$|\ddot{\sigma}(\tau)|$ are small enough, and provided we stay far enough away
from $x=1$ because of the divergence of $\hat{\rho}_{1}(x)/\bar{\rho}_{0}(x),\
\hat{\alpha}_{1}(x)/\bar{\alpha}_{0}(x),\ \hat{u}(x)$, and $\hat{\xi}_{1}(x)$
at $x=1$. How close we can get near $x=1$ depend on how small
$\sigma(\tau),\dot{\sigma}(\tau)$, and $|\ddot{\sigma}(\tau)|$ are.

The physical origin of the divergence of $\hat{\rho}_{1}(x)/\bar{\rho}_{0}(x)$
is not hard to understand. Boundary condition specifies that $\rho_{0}(r)=0$
at $r=R$, but $\rho(r)=\rho_{0}(r)+\rho_{1}(r)$ vanishes only at
$r=R_{0}(1+\sigma)$. Hence the ratio of $\rho_{1}(r)/\rho_{0}(r)$ is infinite
at $r=R_{0}$, or equivalently $\hat{\rho}_{1}(x)/\bar{\rho}_{0}(x)$ is
infinite at $x=1$.

The divergence of $\hat{u}(x)$ comes about because the internal density
adjustment to an addition of external matter is instantaneous, both
proportional to $\sigma(\tau)$. By pushing a finite amount of matter across
the event horizon at a very small amount of time (actually 0), the flow rate
at the event horizon has to be infinite.

The divergence of $\hat{\xi}_{1}(x)$ and its implication will be discussed in
the next subsection.

###  VI.3 Static and dynamic property of black hole matter

$\xi_{1}(x,\tau)=0$ if $\ddot{\sigma}(\tau)=0$, so the static equation of
state $\bar{p}(x,\tau)=-\xi\bar{\rho}(x,\tau)$ remains unchanged when
$\dot{\sigma}(\tau)$ is a constant, in which case the flow velocity
$\bar{u}(x,\tau)$ is also a constant at every $x$. More generally,
$\xi_{1}(x,\tau)$ and $\dot{\bar{u}}(x,\tau)$ have the same time dependence
$\ddot{\sigma}(\tau)$, suggesting a relation between the two. Their $x$
dependence both grow steadily from 0 to infinity at the event horizon, so that
the modification $\xi_{1}(x,\tau)$ becomes larger at those $x$ when the
acceleration $\dot{\bar{u}}(x,\tau)$ is bigger. One might therefore understand
the change $\xi_{1}(x,\tau)$ to be caused by the acceleration
$\dot{\bar{u}}(x,\tau)$, to produce more (less) negative pressure to counter
the inward (outward) force brought on by the acceleration (deceleration) of
the injected matter.

That the equation of state can be affected by external conditions is also seen
in ordinary matter. The constant $\xi$ for non-relativistic matter is $\xi=0$,
but if heat is added to cause the molecules to move relativistically, then
$\xi=-\frac{1}{3}$. If the temperature is low near the center of a spherical
system and is high near its boundary at $r=R_{0}$, then $\xi$ would increase
steadily as $r$ increases. This is analogous to the behavior of the dark hole
matter, but with temperature replacing dark hole fluid acceleration.

###  VI.4 Numerical illustration

The zeroth order quantities $\bar{\rho}_{0}(x)$ and $\bar{m}_{0}(x)$ for
$\xi=0.16$ are displayed in Fig. 1. Various first-order quantities for the
same $\xi=0.16$ are displayed in Figs. 2-6 to illustrate the discussions of
the previous sections.

Fig. 2. for $\xi=0.16$

Fig. 3. for $\xi=0.16$

Fig. 4. for $\xi=0.16$

Fig. 5. for $\xi=0.16$

Fig. 6. for $\xi=0.16$

##  Appendix A Einstein equations and the covariant conservation laws

The computed results of the covariant conservation law
$CO_{\nu}:=\nabla^{\mu}T_{\mu\nu}=0$ and the Einstein equation
$EQ_{\mu\nu}=G_{\mu\nu}-8\pi GT_{\mu\nu}=0$ obtained from Eq.(16) and Eq.(17)
are displayed below. Arguments $(r,t)$ of functions are not shown, a dot
represents the time derivative $\partial/\partial t$, and a prime represents
the spatial derivative $\partial/\partial r$. Only non-zero components are
listed. $EQ_{\phi\phi}$ is also not listed because it is the same as
$EQ_{\theta\theta}$.

$\displaystyle\noindent CO_{t}$ (71) $\displaystyle 0$ $\displaystyle=$
$\displaystyle r\rho\bigg{(}G\Big{(}\dot{m}r^{2}-m\epsilon r+(r-2Gm)\epsilon
m^{\prime}r+2Gm^{2}\epsilon\Big{)}-r(r-2Gm)^{2}\epsilon^{\prime}\bigg{)}u^{2}$
$\displaystyle+$
$\displaystyle(r-2Gm)\Bigg{(}r(r-2Gm)\big{(}(r-2Gm)\epsilon^{2}+e^{2\alpha}r\big{)}\rho(u^{0})^{\prime}$
$\displaystyle+$ $\displaystyle
u^{0}\bigg{(}e^{2\alpha}r\Big{(}\rho\big{(}Gm^{\prime}r+3(r-2Gm)\alpha^{\prime}r+2r-5Gm\big{)}+r(r-2Gm)\left(p^{\prime}+\rho^{\prime}\right)\Big{)}$
$\displaystyle+$
$\displaystyle\epsilon\Big{(}\epsilon\left(2\rho+r\left(p^{\prime}+\rho^{\prime}\right)\right)(r-2Gm)^{2}+r\rho\left(\epsilon^{\prime}(r-2Gm)^{2}+2Gr\dot{m}\right)\Big{)}\bigg{)}\Bigg{)}u$
$\displaystyle+$ $\displaystyle
r(r-2Gm)\Bigg{(}\bigg{(}\epsilon\Big{(}2\rho\dot{\epsilon}+\epsilon\left(\dot{p}+\dot{\rho}\right)\Big{)}(r-2Gm)^{2}+e^{2\alpha}\Big{(}r(r-2Gm)\left(\dot{p}+\dot{\rho}\right)$
$\displaystyle+$
$\displaystyle\rho\big{(}\epsilon\alpha^{\prime}(r-2Gm)^{2}+2r\dot{\alpha}(r-2Gm)+Gr\dot{m}\big{)}\Big{)}\bigg{)}(u^{0})^{2}$
$\displaystyle+$
$\displaystyle(r-2Gm)\left((r-2Gm)\epsilon^{2}+e^{2\alpha}r\right)\rho\left(2\dot{(u^{0})}+u^{\prime}\right)u^{0}+(r-2Gm)\left((r-2Gm)\epsilon
p^{\prime}-r\dot{p}\right)\Bigg{)}$ $\displaystyle+$ $\displaystyle
p\Bigg{(}r\bigg{(}G\Big{(}\dot{m}r^{2}-m\epsilon r+(r-2Gm)\epsilon
m^{\prime}r+2Gm^{2}\epsilon\Big{)}-r(r-2Gm)^{2}\epsilon^{\prime}\bigg{)}u^{2}$
$\displaystyle+$
$\displaystyle(r-2Gm)\bigg{(}r(r-2Gm)\Big{(}(r-2Gm)\epsilon^{2}+e^{2\alpha}r\Big{)}(u^{0})^{\prime}$
$\displaystyle+$ $\displaystyle
u^{0}\Big{(}e^{2\alpha}r\big{(}Gm^{\prime}r+3(r-2Gm)\alpha^{\prime}r+2r-5Gm\big{)}$
$\displaystyle+$
$\displaystyle\epsilon\big{(}2G\dot{m}r^{2}+(r-2Gm)^{2}\epsilon^{\prime}r+2(r-2Gm)^{2}\epsilon\big{)}\Big{)}\bigg{)}u$
$\displaystyle+$ $\displaystyle
r(r-2Gm)u^{0}\Big{(}(r-2Gm)\left((r-2Gm)\epsilon^{2}+e^{2\alpha}r\right)\left(2\dot{(u^{0})}+u^{\prime}\right)$
$\displaystyle+$ $\displaystyle
u^{0}\left(2\epsilon\dot{\epsilon}(r-2Gm)^{2}+e^{2\alpha}\left(\epsilon\alpha^{\prime}(r-2Gm)^{2}+2r\dot{\alpha}(r-2Gm)+Gr\dot{m}\right)\right)\Big{)}\Bigg{)}$
$\displaystyle\noindent CO_{r}$ (72) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\Bigg{(}e^{2\alpha}r\bigg{(}\rho\Big{(}2\left(Gm^{\prime}r+r-3Gm\right)+r(r-2Gm)\alpha^{\prime}\Big{)}+r(r-2Gm)\left(p^{\prime}+\rho^{\prime}\right)\bigg{)}$
$\displaystyle+$
$\displaystyle\epsilon\left(\epsilon\left(2\rho+r\left(p^{\prime}+\rho^{\prime}\right)\right)(r-2Gm)^{2}+r\rho\left(2(r-2Gm)^{2}\epsilon^{\prime}-Gr\dot{m}\right)\right)\Bigg{)}u^{2}$
$\displaystyle+$ $\displaystyle
r\Bigg{(}(r-2Gm)\Big{(}(r-2Gm)\epsilon^{2}+e^{2\alpha}r\Big{)}\rho\left(\dot{(u^{0})}+2u^{\prime}\right)$
$\displaystyle+$ $\displaystyle
u^{0}\bigg{(}\epsilon\Big{(}\rho\dot{\epsilon}+\epsilon\left(\dot{p}+\dot{\rho}\right)\Big{)}(r-2Gm)^{2}+e^{2\alpha}\Big{(}r(r-2Gm)\left(\dot{p}+\dot{\rho}\right)$
$\displaystyle+$
$\displaystyle\rho\left(-2\epsilon\alpha^{\prime}(r-2Gm)^{2}+r\dot{\alpha}(r-2Gm)+3Grm^{(0,1)}\right)\Big{)}\bigg{)}\Bigg{)}u$
$\displaystyle+$ $\displaystyle
r(r-2Gm)\Bigg{(}e^{4\alpha}(r-2Gm)\rho\alpha^{\prime}(u^{0})^{2}+(r-2Gm)\epsilon\left(\dot{p}+u^{0}\epsilon\rho\dot{u}\right)$
$\displaystyle+$ $\displaystyle
e^{2\alpha}\left(u^{0}\rho\left(r\dot{u}-(r-2Gm)u^{0}\left(\epsilon\dot{\alpha}-\dot{\epsilon}\right)\right)+(r-2Gm)p^{\prime}\right)\Bigg{)}$
$\displaystyle+$ $\displaystyle
p\Bigg{(}\bigg{(}e^{2\alpha}r\Big{(}2\left(Gm^{\prime}r+r-3Gm\right)+r(r-2Gm)\alpha^{\prime}\Big{)}$
$\displaystyle+$
$\displaystyle\epsilon\Big{(}2\epsilon(r-2Gm)^{2}+r\big{(}2(r-2Gm)^{2}\epsilon^{\prime}-Gr\dot{m}\big{)}\Big{)}\bigg{)}u^{2}$
$\displaystyle+$ $\displaystyle
r\Bigg{(}(r-2Gm)\Big{(}(r-2Gm)\epsilon^{2}+e^{2\alpha}r\Big{)}\left(\dot{(u^{0})}+2u^{\prime}\right)$
$\displaystyle+$ $\displaystyle
u^{0}\bigg{(}3e^{2\alpha}Gr\dot{m}+(r-2Gm)\Big{(}e^{2\alpha}r\dot{\alpha}+(r-2Gm)\epsilon\left(\dot{\epsilon}-2e^{2\alpha}\alpha^{(1,0)}\right)\Big{)}\bigg{)}\Bigg{)}u$
$\displaystyle+$ $\displaystyle
r(r-2Gm)u^{0}\left(\left((r-2Gm)\epsilon^{2}+e^{2\alpha}r\right)\dot{u}+e^{2\alpha}(r-2Gm)u^{0}\left(-\epsilon\dot{\alpha}+\dot{\epsilon}+e^{2\alpha}\alpha^{\prime}\right)\right)\Bigg{)}$
$\displaystyle\noindent EQ_{tt}$ (73) $\displaystyle 0$ $\displaystyle=$
$\displaystyle
e^{4\alpha}\epsilon\bigg{(}\epsilon\Big{(}4G^{2}m^{2}+2r\alpha^{\prime}(r-2Gm)^{2}-r^{2}\Big{)}-2Gr^{2}\dot{m}-2r\epsilon^{\prime}(r-2Gm)^{2}\bigg{)}$
$\displaystyle-$ $\displaystyle
2Gr^{2}m^{\prime}e^{6\alpha}-e^{2\alpha}\epsilon^{2}(r-2Gm)^{2}\left(2r\dot{\alpha}\epsilon-2r\dot{\epsilon}-\epsilon^{2}\right)$
$\displaystyle-$ $\displaystyle 8\pi
Gr^{2}\Big{(}\epsilon^{2}(r-2Gm)+re^{2\alpha}\Big{)}^{2}\left((p+\rho)\left(e^{2\alpha}u^{0}-u\epsilon\right)^{2}-pe^{2\alpha}\right)$
$\displaystyle\noindent EQ_{tr}$ (74) $\displaystyle 0$ $\displaystyle=$
$\displaystyle
2Gr^{2}e^{4\alpha}\Big{(}r\dot{m}-m^{\prime}\epsilon(r-2Gm)\Big{)}-\epsilon^{5}(r-2Gm)^{3}$
$\displaystyle-$ $\displaystyle 8\pi
Gr^{2}(r-2Gm)\Big{(}\epsilon^{2}(r-2Gm)+re^{2\alpha}\Big{)}^{2}\left((p+\rho)\left(u\epsilon-e^{2\alpha}u^{0}\right)\left(\frac{ru}{r-2Gm}+u^{0}\epsilon\right)+p\epsilon\right)$
$\displaystyle+$ $\displaystyle
e^{2\alpha}\epsilon(r-2Gm)^{2}\bigg{(}-2r\epsilon\Big{(}\epsilon^{\prime}(r-2Gm)+r\dot{\alpha}\Big{)}+\epsilon^{2}\Big{(}2r\alpha^{\prime}(r-2Gm)-2Gm-r\Big{)}$
$\displaystyle+$ $\displaystyle 2r^{2}\dot{\epsilon}\bigg{)}$
$\displaystyle\noindent EQ_{rr}$ (75) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-8\pi
Gr\Big{(}\epsilon^{2}(r-2Gm)+re^{2\alpha}\Big{)}^{2}\bigg{(}(p+\rho)\Big{(}ru+(r-2Gm)u^{0}\epsilon\Big{)}^{2}+rp\bigg{)}$
$\displaystyle+$ $\displaystyle
e^{2\alpha}\bigg{(}-2r^{2}\epsilon\Big{(}2G\dot{m}+\dot{\alpha}(r-2Gm)\Big{)}$
$\displaystyle+$
$\displaystyle\epsilon^{2}(r-2Gm)\Big{(}2Grm^{\prime}+4r\alpha^{\prime}(r-2Gm)-4Gm-r\Big{)}+2r^{2}\dot{\epsilon}(r-2Gm)\bigg{)}$
$\displaystyle+$
$\displaystyle\epsilon^{3}(r-2Gm)\Big{(}-2Gr\dot{m}+2\epsilon^{\prime}(r-2Gm)^{2}-\epsilon(r-2Gm)\Big{)}+2re^{4\alpha}\Big{(}r\alpha^{\prime}(r-2Gm)-Gm\Big{)}$
$\displaystyle\noindent EQ_{\theta\theta}$ (76) $\displaystyle 0$
$\displaystyle=$
$\displaystyle-\epsilon\dot{\epsilon}\epsilon^{\prime}r^{5}+\epsilon^{2}\dot{\epsilon}^{\prime}r^{5}+G\epsilon\dot{m}\dot{\epsilon}r^{4}-G\epsilon^{2}\ddot{m}r^{4}+8Gm\epsilon\dot{\epsilon}\epsilon^{\prime}r^{4}$
$\displaystyle-$ $\displaystyle
8Gm\epsilon^{2}\dot{\epsilon}^{\prime}r^{4}-4G^{2}\epsilon^{2}\dot{m}^{2}r^{3}-4G^{2}m\epsilon\dot{m}\dot{\epsilon}r^{3}+4G^{2}m\epsilon^{2}\ddot{m}r^{3}-24G^{2}m^{2}\epsilon\dot{\epsilon}\epsilon^{\prime}r^{3}$
$\displaystyle+$ $\displaystyle
24G^{2}m^{2}\epsilon^{2}\dot{\epsilon}^{\prime}r^{3}+8G^{3}m\epsilon^{2}\dot{m}^{2}r^{2}+4G^{3}m^{2}\epsilon\dot{m}\dot{\epsilon}r^{2}-4G^{3}m^{2}\epsilon^{2}\ddot{m}r^{2}$
$\displaystyle+$ $\displaystyle
32G^{3}m^{3}\epsilon\dot{\epsilon}\epsilon^{\prime}r^{2}-32G^{3}m^{3}\epsilon^{2}\dot{\epsilon}^{\prime}r^{2}-16G^{4}m^{4}\epsilon\dot{\epsilon}\epsilon^{\prime}r+16G^{4}m^{4}\epsilon^{2}\dot{\epsilon}^{\prime}r$
$\displaystyle+$ $\displaystyle
e^{4\alpha}(r-2Gm)^{2}\bigg{(}(r-2Gm)\alpha^{\prime\prime}r^{2}+\left(r\alpha^{\prime}+1\right)\Big{(}G\left(m-rm^{\prime}\right)+r(r-2Gm)\alpha^{\prime}\Big{)}\bigg{)}$
$\displaystyle+$ $\displaystyle
e^{2\alpha}\bigg{(}\epsilon^{2}\Big{(}2\alpha^{\prime}\left(r\alpha^{\prime}+1\right)+r\alpha^{\prime\prime}\Big{)}(r-2Gm)^{4}-\epsilon\Big{(}\left(r\alpha^{\prime}+1\right)\epsilon^{\prime}(r-2Gm)^{2}$
$\displaystyle+$ $\displaystyle
r\dot{\alpha}\left(Gm^{\prime}r-r+Gm\right)-Gr\dot{m}\left(r\alpha^{\prime}+1\right)\Big{)}(r-2Gm)^{2}$
$\displaystyle+$ $\displaystyle
Gr^{3}\dot{m}\dot{\alpha}(r-2Gm)+r\Big{(}r\left(-\left(\dot{\alpha}\epsilon^{\prime}-\dot{\epsilon}^{\prime}\right)(r-2Gm)^{2}-Gr\ddot{m}\right)$
$\displaystyle-$
$\displaystyle(r-2Gm)\dot{\epsilon}\big{(}Gm+r\left(Gm^{\prime}-1\right)\big{)}\Big{)}(r-2Gm)-3G^{2}r^{3}\dot{m}^{2}\bigg{)}$
$\displaystyle-$ $\displaystyle 8G\pi
r(r-2Gm)^{2}p\Big{(}(r-2Gm)\epsilon^{2}+e^{2\alpha}r\Big{)}^{2}$

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