# Entropy and Topology of Regular Black Holes

Chen Lan<EMAIL_ADDRESS>Yan-Gang Miao<EMAIL_ADDRESS>corresponding author.  School of Physics, Nankai University, 94 Weijin Road,
Tianjin 300071, China.

###### Abstract

We calculate the entropy of spherically symmetric regular black holes by the
path integral and Noether-charge method. Both methods provide an evidence that
the entropy of regular black holes should be proportional to quarter of area,
and there is no violation of entropy/area law at all.

##  I Introduction

Because it is simpler than the path integral or Noether-charge approach [1,
2], the computing method of entropy

$S_{c}=\int_{r_{\rm ext}}^{r_{\rm H}}(\mathrm{d}M\beta)_{c},\qquad\beta=1/T$
(1)

is widely used for investigating the thermodynamics in regular black holes
(RHBs), e.g. [3, 4, 5, 6, 7, 8, 9]. This method is based on the first law of
thermodynamics and the direct correspondence between internal energy and BH
mass, $\braket{U}=M$. As we pointed in previous work [10], the conditional
entropy $S_{c}$ for RBHs is incorrect because it is not consistent with the
result computed from the first principle. In this work, we discuss the entropy
for more types of BHs by two different ways mentioned above, i.e. the path
integral and Wald’s method.

Let us start with RHBs with a single shape function

$f(r)=1-\frac{2M}{r}\sigma\left(r,M,\alpha_{i}\right),$ (2)

At first, we rewrite $\sigma$ and its derivatives via the curvature invariants

$\displaystyle\frac{\sigma(r)}{r^{3}}$
$\displaystyle=\frac{1}{24M}\left(R+3\sqrt{2E}+2\sqrt{3W}\right),$ (3)
$\displaystyle\frac{\sigma^{\prime}(r)}{r^{2}}$
$\displaystyle=\frac{1}{8M}\left(R+\sqrt{2E}\right),$ (4)
$\displaystyle\frac{\sigma^{\prime\prime}(r)}{r}$
$\displaystyle=\frac{1}{4M}\left(R-\sqrt{2E}\right).$ (5)

The regularity at $r=0$ implies $\lim_{r\to 0}\sigma/r^{3}=\text{const}$ from
the perspective of finite curvatures. For the case, when $\sigma$ has a power
expansion around zero, $\sigma=\sum_{n=3}^{N}a_{n}r^{n}$, where $N$ could be
infinite, we find the extreme case

$R\sim O(1),\qquad E\sim O(r^{2}),\qquad W\sim O(r^{2})$ (6)

and $E\sim W\sim R^{2}$. For generalized case, the above asymptotic behaviors
can be recast in the following

$\lim_{r\to 0}R=\text{const},\qquad\lim_{r\to
0}\frac{E}{r^{2}}=\text{const},\qquad\lim_{r\to
0}\frac{W}{r^{2}}=\text{const}.$ (7)

In other words, the leading terms of $\sigma$ and its derivatives should be

$\sigma\sim\frac{R^{(0)}}{24M}r^{3},\qquad\sigma^{\prime}\sim\frac{R^{(0)}}{8M}r^{2},\qquad\sigma^{\prime\prime}\sim\frac{R^{(0)}}{4M}r,$
(8)

as $r\to 0$, where $R^{(0)}$ is the limit of Ricci curvature at zero. As a
result, the metric of RHBs tends to flat as the radius approaches to zero,
which, in turn, provides the same situation of geodesic completeness as the
zero point of flat spacetime. In other words, even though the null or timelike
geodesic arrives at $r=0$ in finite affine parameter, it does not affect the
completeness of the spacetime, because the zero point is not even singular
point of metric at all.

We end this introductory section by emphasizing an advantage of new
representation obtained by replacing $\sigma$ in Eq. (2) by Eq. (3),

$f(r)=1-\frac{\lambda}{3}r^{2},\qquad 4\lambda=R+3\sqrt{2E}+2\sqrt{3W}.$ (9)

This representation provides a simple way to check whether a BH model is
regular by directly observing the asymptotic behavior of $\lambda r^{2}$
around zero. In particular, when $\lambda r^{2}$ is a rational function of
$r$, the balance of the lowest power of $r$ aprearing in numerator and
denominator for a RHB should not be greater than two.

##  II  The classical action and partition functions

To observe the entropy, we start with the partition function of gravitational
sector in the framework of Einstein’s gravity. After Wick rotation
$t\to-\mathrm{i}\tau$, it can be derived at zero-loop approximation [1]

$\mathcal{Z}=\int\mathcal{D}g\;\mathrm{e}^{-I}\approx\mathrm{e}^{-I_{\rm
cl}},$ (10)

where the Euclidean action $I_{\rm cl}$ consists of three parts [11]

$I_{\rm cl}=I_{\rm EH}+I_{\rm GHY}-I_{0}+\dots.$ (11)

The Einstein-Hilbert action $I_{\rm EH}$, Gibbons-Hawking-York boundary term
$I_{\rm GHY}$

$I_{\rm
EH}=-\frac{1}{16\uppi}\int_{\mathcal{M}}\mathrm{d}^{4}x\sqrt{-g}\,R,\qquad
I_{\rm
GHY}=-\frac{1}{8\uppi}\int_{\partial\mathcal{M}}\mathrm{d}^{3}x\sqrt{-h}\,K,$
(12)

and the counterterm $I_{0}$ for regularizing the IR divergence appearing in
$I_{\rm GHY}$

$I_{0}=-\frac{1}{8\uppi}\int_{\partial\mathcal{M}}\mathrm{d}^{3}x\sqrt{-h}\,K_{0}.$
(13)

where $R$ is Ricci curvature of bulk, $K$ and $K_{0}$ are extrinsic curvature
of surface and background reference separately. The ellipsis omits the matter
sectors, which will not be involved in calculating of internal energy and
entropy of RBHs. This is related to the so-called Christodoulou-Ruffini
irreducible mass [12, 13], and we will call the action without matter sector
as irreducible part. The detailed explanation will be given in the App. A
based on Reissner-Nordström BH.

###  II.1 Regular black holes with single shape function

For the spherically symmetric RBHs

$g_{\mu\nu}=\operatorname{diag}\\!\left\\{-f(r),f^{-1}(r),r^{2},r^{2}\sin^{2}(\theta)\right\\}$
(14)

with single shape function Eq. (2), one can compute the corresponding
curvature invariants

$R=\frac{2M}{r^{2}}\left(2\sigma^{\prime}+r\sigma^{\prime\prime}\right),\qquad
K=\frac{2r-Mr\sigma^{\prime}-3M\sigma}{r^{3/2}\sqrt{r-2M\sigma}}$ (15)

and extrinsic curvature $K_{0}=2/r$ of the surface embedded in flat spacetime.
Here the prime denotes the derivative with respect to $r$. Since
$\sigma(r)\sim O(r^{n})$ with $n\geq 3$ as $r\to 0$, the derivatives of
$\sigma$ can not globally vanish. Thus the EH part will make a contribution,
which is different from that one of traditional singular models, such as
Schwarzschild BH. Considering the measure of bulk
$\sqrt{-g}=r^{2}\sin(\theta)$ and surface
$\sqrt{-h}=r^{3/2}\sin(\theta)\sqrt{r-2M\sigma}$, the integrals Eq. (12) can
be taken off

$I_{\rm
EH}=-\frac{M}{2}\beta\left(r\sigma^{\prime}+\sigma\right)\Big{|}_{r_{\rm
H}}^{\infty},$ (16)

where the integration of Euclidean time $\tau$ runs from $0$ to $\beta=1/T$
and radial $r$ from outer horizon $r_{\rm H}$ to $\infty$. On the other side,
taking into account that the RHBs are asymptotic to Schwarzschild BH at
infinity [10]

$\lim_{r\to\infty}\sigma=1,\qquad\lim_{r\to\infty}r\sigma^{\prime}=0,$ (17)

we obtain

$I_{\rm EH}=\frac{\beta}{2}\left(r_{\rm H}-M\right)-\uppi r_{\rm H}^{2},$ (18)

and the GHY term is

$I_{\rm
GHY}=\frac{M}{2}\beta\left(r\sigma^{\prime}+3\sigma\right)-r\beta\sim\frac{3M}{2}\beta-r\beta,$
(19)

which is divergent as a manner of $r$ as $r\to\infty$ and regularized by the
couterterm

$I_{0}=-r\beta\sqrt{1-\frac{2M\sigma}{r}}\sim-r\beta+M\beta+O\left(r^{-1}\right).$
(20)

Summarizing all above contributions, we arrive at classical Euclidean action

$I_{\rm cl}=-\uppi r_{\rm H}^{2}+\frac{r_{\rm H}}{2}\beta.$ (21)

Then the internal energy and entropy of RBHs are clearly in order

$\braket{U}=\frac{\partial I_{\rm cl}}{\partial\beta}=\frac{r_{\rm
H}}{2},\qquad S=\beta\braket{U}-I_{\rm cl}=\uppi r_{\rm H}^{2},$ (22)

where we have supposed that $\beta$ and $r_{\rm H}$ are two independent
variables, which is equivalent to a _isometric_ process $V_{\rm th}\equiv
4\uppi r_{\rm H}^{3}/3=\text{const}$. Meanwhile, it is worth to note that the
internal energy $\braket{U}$ given here refers to “the matter-independent
energy” equivalent to the irreducible mass [12].

Having the classical action Eq. (21) at hand, one can also calculate the
thermal pressure $P_{\rm q}$ by

$P_{\rm q}=-\frac{\partial\left(\beta^{-1}I_{\rm cl}\right)}{\partial V_{\rm
th}}=-\frac{\beta-4\uppi r_{\rm H}}{8\uppi\beta r_{\rm H}^{2}},$ (23)

and it is reasonable to ask wether this pressure consists with the pressure
$P_{\rm G}\equiv G^{r}_{\;r}/(8\uppi)$ generated by Gliner vacuum (or vacuum-
like medium) at horizon, which is discussed in [10]. After calculating the
Einstein tensor of metric Eq. (14), one can read off the pressure of Gliner
vacuum from $r$-$r$ component,

$P_{\rm G}=-\frac{M\sigma^{\prime}\left(r_{\rm H}\right)}{4\uppi r_{\rm
H}^{2}},$ (24)

while $\sigma^{\prime}\left(r_{\rm H}\right)$ can be represented via $\beta$,
i.e.

$\sigma^{\prime}\left(r_{\rm H}\right)=\frac{\beta-4\uppi r_{\rm H}}{2\beta
M}.$ (25)

Therefore one can verify that the Gliner pressure is exactly the thermal
pressure $P_{\rm q}$. Not only for RHBs, but also for all traditional singular
BHs, such as RN-AdS BH, a thermal pressure can be derived from classical
action, if the thermal volume $V_{\rm th}$ is regarded as BH’s volume, and
this thermal pressure must consist with Gliner pressure.

###  II.2 The isometric process

To derive the entropy/area law Eq. (22), we have supposed that $r_{\rm H}$ is
independent with $\beta$. To understand its meaning, let us consider a typical
example, Hayward BH with shape function [14]

$f_{e}=1-\frac{2M}{r}\frac{r^{3}}{r^{3}+e^{3}}\quad\text{or}\quad
f_{\Lambda}=1-\frac{2M}{r}\frac{r^{3}}{r^{3}+\frac{6M}{\Lambda}}.$ (26)

These two parameterizations are mathematically equivalent, but have different
physical explanations. We are primarily going to use the former one in the
following.

Hayward BH is of two “constraints”, one comes from $f(r_{\rm H})=0$, the other
is $\beta=4\uppi/f^{\prime}(r_{\rm H})$,

$1-\frac{2Mr_{\rm H}^{2}}{e^{3}+r_{\rm
H}^{3}}=0,\qquad\beta=\frac{2\uppi\left(e^{3}+r_{\rm
H}^{3}\right)^{2}}{Mr_{\rm H}\left(r_{\rm H}^{3}-2e^{3}\right)}.$ (27)

The “constraints” imply that the four variables $\beta$, $M$, $e$ and $r_{\rm
H}$ are related to each other. In other words, there are only two of them are
independent in this system. Since the shape function depends on two parameters
$M$ and $e$, this character is obvious. Now, instead of choosing original $M$
and $e$ as independent variables, we prefer to take two others, $\beta$ and
$r_{\rm H}$. This is what we mean by “isometric process” in previous
subsection.

The corresponding sectors are separately

$\begin{split}I_{\rm EH}=-\frac{\beta e^{3}M\left(e^{3}-2r_{\rm
H}^{3}\right)}{2\left(e^{3}+r_{\rm H}^{3}\right)^{2}}&,\qquad I_{\rm
GHY}=-\frac{\beta
r\left[2e^{3}r^{2}(2r-3M)+2e^{6}+r^{5}(2r-3M)\right]}{2\left(e^{3}+r^{3}\right)^{2}},\\\
&I_{0}=-\beta r\sqrt{1-\frac{2Mr^{2}}{e^{3}+r^{3}}}.\end{split}$ (28)

Thus the total contribution is

$I_{\rm cl}=\frac{\beta M\left(4e^{3}r_{\rm H}^{3}+r_{\rm
H}^{6}\right)}{2\left(e^{3}+r_{\rm H}^{3}\right)^{2}}.$ (29)

To reduce the parameters, we apply the constraints Eq. (27) to replace $M$ and
$e$

$M\to\frac{3\beta r_{\rm H}}{4\left(\beta+2\uppi r_{\rm H}\right)},\qquad
e^{3}\to-\frac{r_{\rm H}^{3}\left(4\uppi r_{\rm
H}-\beta\right)}{2\left(\beta+2\uppi r_{\rm H}\right)}$ (30)

and obtain a reexpressed action

$I_{\rm cl}=\frac{\beta r_{\rm H}}{2}-\uppi r_{\rm H}^{2}.$ (31)

Since $\beta$ and $r_{\rm H}$ are independent, we can conclude that

$\braket{U}_{\rm iso}\equiv\frac{\partial I_{\rm
cl}}{\partial\beta}=\frac{r_{\rm H}}{2},\quad S_{\rm
iso}\equiv\beta\braket{U}-I_{\rm H}=\uppi r_{\rm H}^{2}.$ (32)

Furthermore, the choice of independent variables corresponds to the different
processes, and may lead to different results. To see it clearly, let us now
take $\beta$ and $M$ as independent variables, which is equivalent to keep $M$
being constant. Now the internal energy should be calculated by

$\braket{U}_{M}=\frac{\partial I_{\rm cl}}{\partial\beta}+\frac{\partial
I_{\rm cl}}{\partial r_{\rm H}}\left(\frac{\partial r_{\rm
H}}{\partial\beta}\right)_{M}.$ (33)

The results show the difference from Eq. (32)

$\braket{U}_{M}=\frac{r_{\rm H}}{2}+\frac{3e^{3}r_{\rm H}\left(2e^{3}-r_{\rm
H}^{3}\right)}{4\left(e^{3}+r_{\rm H}^{3}\right)^{2}},\qquad S_{M}=\uppi
r_{\rm H}^{2}-\frac{3\uppi e^{3}r_{\rm H}^{2}}{e^{3}+r_{\rm H}^{3}}.$ (34)

Alternatively, if the charge $e$ is supposed to be fixed, one has

$\braket{U}_{e}=\frac{r_{\rm H}}{2}+\frac{3e^{3}r_{\rm H}\left(2e^{3}-r_{\rm
H}^{3}\right)}{20e^{3}r_{\rm H}^{3}+4e^{6}-2r_{\rm H}^{6}},\qquad S_{e}=\uppi
r_{\rm H}^{2}-\frac{6\uppi e^{3}r_{\rm H}^{2}\left(e^{3}+r_{\rm
H}^{3}\right)}{10e^{3}r_{\rm H}^{3}+2e^{6}-r_{\rm H}^{6}}.$ (35)

On the other side, to highlight the difference, let us write down the
conditional entropy

$S_{c}=\int^{r_{\rm H}}_{r_{\rm ext}}\mathrm{d}\tilde{r}_{\rm
H}\beta\left(\frac{\partial M}{\partial\tilde{r}_{\rm H}}\right)_{e}=\uppi
r_{\rm H}^{2}-\frac{2\uppi e^{3}}{r_{\rm H}}.$ (36)

It notes that the conditional entropy is not consistent with any of above
three cases.

###  II.3 The isometric entropy of Gauss-Bonnet theory

To see how the isometric process affects the results of singular BHs, let us
see one more example, the well-known Gauss-Bonnet (GB) BH. We are going to
show its entropy in various processes. The Euclidean action of Einstein-Gauss-
Bonnet (EGB) sector is

$I_{\rm EGB}=-\frac{1}{16\uppi}\int\mathrm{d}^{d}x\sqrt{-g}(R+\alpha B),\qquad
B=K-4R_{2}+R^{2},$ (37)

where $\alpha$ is coupling constant, while $B$ is GB scalar. Moreover there is
an additional boundary term due to GB correction besides GHY term [15, 16,
17],

$I_{M}=-\frac{\alpha}{16\uppi}\int_{\partial\mathcal{M}}\mathrm{d}^{d-1}x\sqrt{-h}\,\mathcal{C},$
(38)

where $\mathcal{C}=J-2G_{ab}K^{ab}$, the definition of $J$ can be found in
Ref. [17]. Then applying the equation of motion

$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\alpha\mathcal{H}_{\mu\nu}=0,$ (39)

one can rewrite the EGB sector by

$I_{\rm EGB}=\frac{1}{8(d-4)\uppi}\int\mathrm{d}^{d}x\sqrt{-g}\,R.$ (40)

The shape function of BH solution can be cast in the following [18]

$f=1+\frac{r^{2}}{2\tilde{\alpha}}\left(1-\sqrt{1+\frac{64\uppi\tilde{\alpha}M}{(d-2)A_{d-2}r^{d-1}}}\right)$
(41)

with redefined coupling constant $\tilde{\alpha}=\alpha(d-3)(d-4)$. The
corresponding contributions are separately

$I_{\rm
EGB}=\frac{2M\beta}{(d-4)(d-2)}-\frac{A_{d-2}\beta}{8(d-4)\uppi}\left[(d-2)r_{\rm
H}^{d-3}-r_{\rm H}^{d-2}f^{\prime}(r_{\rm H})\right],$ (42) $I_{\rm
GHY}=-\frac{(d-2)A_{d-2}\beta}{8\uppi}r_{\infty}^{d-3}+\frac{d-1}{d-2}M\beta,$
(43)

where $r_{\infty}$ denotes the large value of radius, and signifies IR
divergence. The contribution of GB term is

$I_{\rm M}=-\frac{(d-2)\tilde{\alpha}A_{d-2}\beta}{24\uppi}r_{\infty}^{d-5},$
(44)

which is subtracted completely by the counterpart with flat
$\mathcal{C}_{0}=2(d-4)(d-3)(d-2)/(3r^{3})$, i.e. the GB term does not
contribute to the final results. The IR divergence in GHY term is subtracted
by a similar term with $K_{0}=(d-2)/r$,

$I_{0}=-\frac{(d-2)\beta A_{d-2}r^{d-3}_{\infty}}{8\uppi}+M\beta.$ (45)

Summarizing all above formulas, we arrive at

$I_{\rm lc}=I_{\rm EGB}+I_{\rm GHY}-I_{0}=\frac{A_{d-2}r_{\rm
H}^{d-5}}{16\uppi(d-4)}\left[(d-2)\left(\tilde{\alpha}-r_{\rm
H}^{2}\right)\beta+8\uppi r_{\rm H}^{3}\right],$ (46)

which is exactly the same classical action obtained in the seminal work of
Myers and Simon [16]. The internal energy then can be calculated by fixing the
coupling constant

$\braket{U}_{\alpha}=\frac{\partial I_{\rm lc}}{\partial\beta}+\frac{\partial
I_{\rm lc}}{\partial r_{\rm H}}\left(\frac{\partial r_{\rm
H}}{\partial\beta}\right)_{\alpha},$ (47)

which gives a direct correspondence between internal energy and BH mass
$\braket{U}_{\alpha}=M$. This is the reason that why the conditional entropy
$S_{c}$ provides the same result under the fixed coupling constant $\alpha$
[18]. Meanwhile the entropy is calculated by $S_{\alpha}=\beta M-I_{\rm cl}$,

$S_{\alpha}=\frac{A_{d-2}r_{\rm
H}^{d-2}}{4}\left[1+\frac{(d-2)}{(d-4)}\frac{2\tilde{\alpha}}{r_{\rm
H}^{2}}\right].$ (48)

The deviation is shown in second term of square bracket. Although it breaks
the entropy/area law, it should have the same name with the one in
Schwarzschild BH, i.e. Bekenstein-Hawking entropy, because it is derived by
same method originated in Ref. [1]

Choosing $\alpha$ being constant is completely reasonable, but let us see what
the isometric process gives for this singular BH

$\braket{U}_{\rm iso}=\frac{A_{d-2}(d-2)r_{\rm
H}^{d-7}}{8\uppi(d-4)(d-1)}\left[(d-5)\tilde{\alpha}^{2}+2(d-3)\tilde{\alpha}r_{\rm
H}^{2}-r_{\rm H}^{4}\right],$ (49) $S_{\rm iso}=\frac{A_{d-2}r_{\rm
H}^{d-6}}{4(d-4)(d-1)}\left[4(d-2)\tilde{\alpha}^{2}+4(d-2)\tilde{\alpha}r_{\rm
H}^{2}-dr_{\rm H}^{4}\right].$ (50)

The entropy/area law is violated as well.

##  III Conformal regular black holes

Now we turn to consider the RHBs in the framework of conformal invariant
model, whose Euclidean action is [19]

$I_{\rm
conf}=-\frac{1}{2}\int\mathrm{d}^{4}x\sqrt{-g}\phi\left(\frac{1}{6}R\phi-\Box\phi\right).$
(51)

Since the equation of motion is $R=6\Box\phi/\phi$, this sector does not
contribute to classical action in general. The contributions from boundary
term and background reference are [20]

$I_{\rm GHY}=-\int\mathrm{d}^{3}x\sqrt{-h}\frac{\phi^{2}}{6}K,\qquad
I_{0}=-\int\mathrm{d}^{3}x\sqrt{-h}\frac{\phi^{2}}{6}K_{0}.$ (52)

Since under the transformation $\tilde{h}_{ij}=\Omega^{2}h_{ij}$, one has [21]

$\sqrt{-\tilde{h}}=\Omega^{3}\sqrt{-h},\qquad\tilde{K}=\frac{1}{\Omega}\left(K+3\mathcal{L}_{\bm{n}}\Omega\right),$
(53)

where $\mathcal{L}_{\bm{n}}\Omega\equiv n^{\mu}\partial_{\mu}\Omega$ is Lie
derivative of $\Omega$ along the normal vector field $\bm{n}$. Thus the
actions with tildes become

$\displaystyle\tilde{I}_{\rm GHY}$
$\displaystyle=-\int\mathrm{d}^{3}x\sqrt{-\tilde{h}}\frac{\tilde{\phi}^{2}}{6}\left[\tilde{K}-3n^{\mu}\partial_{\mu}\ln(\Omega)\right],$
(54) $\displaystyle\tilde{I}_{0}$
$\displaystyle=-\int\mathrm{d}^{3}x\sqrt{-\tilde{h}}\frac{\tilde{\phi}^{2}}{6}\left[\tilde{K}_{0}-3n_{0}^{\mu}\partial_{\mu}\ln(\Omega)\right].$
(55)

It implies that both $I_{\rm GHY}$ and $I_{0}$ are not conformal invariant
independently, but their difference vanishes

$\Delta
I=\frac{1}{2}\int\mathrm{d}^{3}x\sqrt{-\tilde{h}}\,\tilde{\phi}^{2}(n^{\mu}-n^{\mu}_{0})\partial_{\mu}\ln(\Omega)$
(56)

as $r\to\infty$, because $n^{\mu}$ is asymptotic to $n^{\mu}_{0}$. Thus one
can make a conclusion that a regular black hole, received by a conformal
change to another singular BH, should have the same classical action with that
singular one.

Now let us study a specific regular model. The metric reads
$\tilde{g}_{\mu\nu}=\Omega^{2}g_{\mu\nu}$, where $g_{\mu\nu}$ is metric of
Schwazschild black hole, and the scale factor is $\Omega=(1+L^{2}/r^{2})^{n}$
[22]. The field can be solved from the equation of motion,
$\tilde{\phi}=\Omega^{-1}\phi$ with

$\phi=\frac{c_{1}}{2M}\ln\left(1-\frac{2M}{r}\right)+c_{2}.$ (57)

where $c_{1}$ and $c_{2}$ are integral constants. It notes that $\phi$
diverges at horizon $r_{\rm H}=2M$, except $c_{1}=0$. Thus to avoid
divergency, one could choose constants, such that $\phi=c_{2}$. The
corresponding contributions are

$I_{\rm GHY}=-\frac{2\uppi}{3}\beta(2r-3M)\phi^{2},\qquad
I_{0}=\Omega^{-1}\tilde{I}_{0}=-\frac{4\uppi}{3}\beta
r\phi^{2}\sqrt{1-\frac{2M}{r}},$ (58)

and

$\tilde{I}_{\rm
GHY}=\frac{2\uppi\Omega^{2}\beta\tilde{\phi}^{2}}{3\left(L^{2}+r^{2}\right)}\left[L^{2}(-12Mn+3M+6nr-2r)+r^{2}(3M-2r)\right].$
(59)

Summarizing all above formulas, one obtains the Euclidean action

$I_{\rm cl}=\tilde{I}_{\rm cl}=\frac{2\uppi}{3}\beta
c_{2}^{2}M+O\left(r^{-1}\right),$ (60)

where in both cases $\beta=8\uppi M$. Therefore the internal energy and
entropy can be carried out

$\braket{U}=\frac{4}{3}\uppi c_{2}^{2}M,\qquad
S=\frac{4}{3}\uppi^{2}c_{2}^{2}r_{\rm H}^{2}.$ (61)

Restoring the correct factor by $6\equiv 8\uppi$ and the field by
$\phi^{2}=c_{2}^{2}$, one can arrive at $\braket{U}=\phi^{2}M$ and
$S=\phi^{2}\uppi r_{\rm H}^{2}$. It signifies that the entropy is conformal
invariant. In fact, if one starts with a partition function, conformally
symmetric at the quantum level, i.e., both Lagrangian and measure of
functional integral are invariant under a conformal change, all the
thermodynamical properties, such as temperature and entropy must also be
conformal invariant.

##  IV The Wald entropy

Now we check the entropies of above models by Wald’s Noether-charge approach.
Since the model Eq. (14) is of simple Lagrangian $R$, its entropy density can
be read off directly from the Table I in Ref. [23]; while the conformal RHBs
corresponds to the case $F(\phi,R)$, and the entropy density is
$\partial_{R}F(\phi,R)=\phi^{2}$. The conformal invariance of the entropy can
be shown below. Starting with $2\mathcal{L}=\phi^{2}R/(8\uppi)-\phi\Box\phi$,
one can find [24]

$\begin{split}S_{\rm W}&=-\frac{1}{2}\oint_{r=r_{\rm
H},t=\text{const}}\phi^{2}\left(\frac{\delta R}{\delta
R_{trtr}}\right)^{(0)}r^{2}\mathrm{d}\Sigma^{2}\\\
&=-\frac{1}{4}\oint_{r=r_{\rm
H},t=\text{const}}\phi^{2}g^{rr}g^{tt}r^{2}\mathrm{d}\Sigma^{2}\\\
&=\frac{1}{4}\oint_{r=r_{\rm
H},t=\text{const}}\phi^{2}r^{2}\mathrm{d}\Sigma^{2},\end{split}$ (62)

which is conformal invariant by $\phi\to\Omega^{-1}\phi$ and
$r^{2}\mathrm{d}\Sigma^{2}_{2}\to\Omega^{2}r^{2}\mathrm{d}\Sigma^{2}_{2}$. For
$d$-dimensional case, the scalar field and the area transform via
$\phi\to\Omega^{(2-d)/2}\phi$ and
$r^{d-2}\mathrm{d}\Sigma^{2}_{d-2}\to\Omega^{d-2}r^{d-2}\mathrm{d}\Sigma^{2}_{d-2}$
respectively. Ultimately, we arrive at $S_{W}=\phi^{2}(r_{\rm H})A_{d-2}r_{\rm
H}^{d-2}/4$, where $\phi(r_{\rm H})$ should be understood as reasonably
classical solution of scalar field at horizon. As to the EGB theory, one can
obtain from the original action Eq. (37) that

$\begin{split}S_{\rm W}&=\frac{A_{d-2}r_{\rm
H}^{d-2}}{4}-\frac{\alpha}{2}\oint_{r=r_{\rm
H},t=\text{const}}\left(\frac{\delta B}{\delta
R_{rtrt}}\right)^{(0)}r^{d-2}\mathrm{d}\Sigma_{d-2}\\\ &=\frac{A_{d-2}r_{\rm
H}^{d-2}}{4}\left[1-\frac{2\alpha(d-3)(d-2)(f-1)}{r^{2}}\right]_{r=r_{\rm
H}}.\end{split}$ (63)

Since $f(r_{\rm H})=0$, it will recover the result obtained by path-integral
method, Eq. (48).

##  V Topology of regular black holes

Although the conformal RBHs and RBHs with single shape function are all
regular from the perspective of finiteness of curvature invariants and
completeness of geodesics, they have several different features. To
distinguish these two classes of RBHs, we dub the latter as RBH1, the former
as RBH2.

First of all, by absorbing the conformal factor into radius
$r\to\Omega^{-1}\rho$, one can rewrite the metric of RBH2, whose product of
first two components $g_{tt}g_{\rho\rho}$ is no longer a constant. In other
words, RBH2 are RBHs with two different shape functions.

Second, the total Ricci curvature and GB scalar of RBH1 is finite, i.e., the
integrals of $R$ and $B$ over entire manifold are not divergent constants. As
a matter of fact, given the metric Eq. (14) one can define two functions

$\xi(r)=\frac{1}{4M}\int\mathrm{d}r\int_{0}^{\uppi}\mathrm{d}\theta\sqrt{-g}\;R,\qquad\chi(r)=\frac{1}{32M^{2}}\int\mathrm{d}r\int_{0}^{\uppi}\mathrm{d}\theta\sqrt{-g}\;B.$
(64)

The both integrations can be taken off,

$\xi=r\sigma^{\prime}+\sigma,\qquad\chi=\frac{\sigma}{r^{3}}\left(r\sigma^{\prime}-\sigma\right).$
(65)

Thus the total Ricci curvature and GB scalar are separately

$\xi_{e}=\xi(\infty)-\xi(\varnothing),\qquad\chi_{e}=\chi(\infty)-\chi(\varnothing),$
(66)

where $\varnothing$ denotes the lower boundary. For $\varnothing=0$, taking
into account $\sigma\sim O(r^{n})$ with $n\geq 3$ as $r\to 0$ and $\sigma\sim
O(1)$ as $r\to\infty$, one has $\xi(0)=\chi(0)=\chi(\infty)=0$ and then
concludes that $\xi_{e}=\sigma(\infty)$ and $\chi_{e}=0$. For
$\varnothing=r_{+}$, one gets $\xi_{e}=r_{+}\sigma^{\prime}(r_{+})+2M/r_{+}$
and
$\chi_{e}=-\left[1-2M\sigma^{\prime}\left(r_{+}\right)\right]/(4M^{2}r_{+})$.
While for RBH2, all these topological quantities are divergent, because the
integration measure is divergent at center.

Moreover, the different topologies for RHB1 and RHB2 give a strong suggestion
to distinguish these two types of RHBs. The pathological topology of RHB2 may
shed light on the explanation, why the energy conditions are violated in the
entire parameter space [22]. Therefore finding a conformal RHB without
pathological topology and violation of energy conditions are still an open
problem.

##  VI Conclusions and disscusions

Although the entropy/area law is verified for RHBs by approximated methods or
phenomenological demonstrations, see e.g. Refs. [25, 26, 27, 28, 29, 30, 31],
we provide a solid evidence that entropy/area law still holds from path
integral and Noether-charge approaches. This result is against the conditional
entropy $S_{c}$ widely used before, of which the motivation comes from
absorbing the anomaly of mass term into entropy, rather than the internal
energy. As to the anomalies in first law (or Smarr formula) with respect to
charge or cosmological constant, which are regarded as real extension of phase
space, they can be canceled completely by introducing the pressure of Gliner
vacuum at horizon [10]. The path-integral method also supports this idea,
i.e., if the thermal volume of RBHs is applied, the Gliner pressure can be
derived from partition function.

The RHBs are originally created to conquer the UV incompleteness, and indeed
the classical divergence can be removed, from the perspective of finiteness of
curvature invariants and completeness of geodesics. From quantum level, it is
known that the one-loop correction of entropy for singular BHs has logarithm
divergent at UV limit [32]. Thus it is reasonable to check whether UV
divergence still exist in RHBs. Intuitively, since the essential singularity
is removed, the coordinate singularity can also be removed by choosing a well-
defined coordinates. As a result, UV divergence may be improved in one-loop
correction. We will show this in the near future.

##  Appendix A The isometric process of Reissner-Nordström black hole

First let us list the some facts for RN black hole, i.e. the entropy,
temperature

$S=\uppi r_{\rm H}^{2},\qquad\beta=\frac{2\uppi r_{\rm H}^{3}}{Mr_{\rm
H}-Q^{2}},$ (67)

and the total action [1]

$I_{\rm cl}=\frac{\beta}{2}(M-Q\Phi_{\rm H})=\uppi r_{\rm H}^{2}.$ (68)

where $\Phi_{\rm H}=Q/r_{\rm H}$. Then one can work out the enthalphy

$\mathcal{H}=M-\frac{Q^{2}}{r_{\rm H}}$ (69)

by substituting all above facts into

$S=\beta\mathcal{H}-I_{\rm cl}.$ (70)

The Smarr formula can be recovered by

$\frac{I_{\rm cl}}{\beta}=G=\mathcal{H}-\frac{S}{\beta},$ (71)

where $G$ is Gibbs energy. Alternatively, one can treat Eq.(70) as a
differential equation of $\beta(r_{\rm H})$ because of
$\mathcal{H}\equiv\partial I_{\rm cl}/\partial\beta$. The solution of the
differential equation is then

$\beta=c_{1}r_{\rm H}.$ (72)

Here $c_{1}$ is an integration constant. It implies that when one uses
$\mathcal{H}\equiv\partial I_{\rm cl}/\partial\beta$ to calculate the
enthalphy, $c_{1}$ is supposed to be a constant, i.e.

$\mathcal{H}\equiv\left(\frac{\partial I_{\rm
cl}}{\partial\beta}\right)_{c_{1}}.$ (73)

Equating Eq.(72) and Eq.(67), one arrives at

$1-\frac{4\uppi}{c_{1}}=\Phi_{\rm H}^{2}.$ (74)

The $c_{1}$ be constant signifies that the potential $\Phi_{\rm H}$ is fixed
when one calculates the partial derivative.

Now we apply the isometric process to recalculate the internal energy and
entropy of RN black hole. Removing the contribution of vector field, one has
an irreducible action $I_{\rm ir}=M\beta/2$. To calculate the physical
quantities further, one can make a replacement

$Q\to r_{\rm H}\sqrt{1-\frac{4\uppi r_{\rm H}}{\beta}},\qquad M\to r_{\rm
H}\left(1-\frac{2\uppi r_{\rm H}}{\beta}\right),$ (75)

which is obtained similarly to Eq. (27). Thus the classical action becomes the
same as Eq. (31), in turn, one obtains the internal energy $\braket{U}=M_{\rm
ir}$ with a fixed thermal volume. Here $M_{\rm ir}=r_{\rm H}/2$ refers to
irreducible mass [12], i.e. the charge-independent mass, and connects with
mass $M$ via $M=M_{\rm ir}+Q^{2}/(4M_{\rm ir})$. It’s this quantity that has
one-to-one correspondence with surface area [13]. As a result, one will
recover the entropy/area law in the isometric process by Eq. (70) and Smarr
formula by Eq. (71).

To recover the correct entropy and Smarr formula in the isometric process, we
have removed the matter sector, and calculated all the quantities based on the
irreducible sector. The motivation comes from the RBHs. For a given regular
metric, the action of matter sector is not unique, or sometimes is unknown.
However, the physical quantities of one BH with different parameterization,
such as Eq.(26), should be the same, no matter what kind of matter sector is
introduced. It reflects a fact that the entropy of BH should not depend on the
matter sector.

###### Acknowledgements.

The authors are grateful to Hao Yang (Nankai University) for valuable remarks.
This work was supported in part by the National Natural Science Foundation of
China under Grant No. 11675081.

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