Comments on Entropy Calculations in Gravitational Systems

S. P. de Alwis†

Physics Department, University of Colorado,

Boulder, CO 80309 USA

Abstract

We discuss the logic of, and some puzzles in, the various approaches to
thermodynamics of gravitational systems. In particular the blackhole, deSitter
(dS), black hole in dS (SdS) and in Anti-deSitter SAdS backgrounds are
considered. After reviewing the original calculations of Hawking and Gibbons
we discuss an alternative Hamiltonian method. This justifies the lowest order
Euclidean calculation but is free of the problems associated with the latter
when going to higher orders. To conclude we address the sign issue in dS
thermodynamics.

†<EMAIL_ADDRESS>
##  1 Summary

We begin by summarizing the classic calculations of Gibbons and Hawking [1]
(GH1) and [2] (GH2).

  * •

The original (Euclidean path integral) calculation in GH1 for the black hole
action, got (with $\hbar=c=G=1$), $I=\beta F=\beta r_{{\rm h}}/4=\beta
M-\frac{\beta}{\beta_{{\rm h}}}A_{{\rm h}}/4$ leading to the identification
$S=A_{{\rm h}}/4=4\pi M^{2}$ if we identify the average energy in a canonical
ensemble as $<E>=M$. Here the subscript ${\rm h}$ denotes horizon and
$\beta_{{\rm h}}=\frac{2\pi}{\kappa_{{\rm h}}}=4\pi r_{{\rm h}}$ with
$\kappa_{{\rm h}}$ being the surface gravity at the horizon (as measured at
infinity!), and $r_{h}=2M$ is the horizon radius . For general beta the
Euclidean metric has a conical singularity which is removed precisely at
$\beta=\beta_{h}$ i.e. the Hawking temperature. The original derivation of the
temperature was however based on using the behavior of quantum fields in a
black hole (of dS) space-time [3][2] and did not depend on Euclidean path
integral methods.

  * •

However this result for the entropy forces us to conclude that the canonical
ensemble makes no sense - it is divergent since the density of states rises as
$\exp\left(8\pi E^{2}\right)$, assuming the entropy of a black hole of energy
(mass) $E$ is given by the usual formula above, overwhelming the Boltzmann
factor $\exp\left(-\beta E\right)$. One can as advocated by York[4, 5] insert
a wall at a finite distance from the horizon, that is coupled to a heat bath
at a fixed temperature to define equilibrium thermodynamics \- but that still
does not get rid of the problem with the density of states. However as pointed
out in [6] the microcanonical ensemble is well-defined. Furthermore in
Schwarzchild Anti-deSitter (SAdS) space the entropy grows only as $E^{2/3}$ so
that the canonical ensemble makes sense [7].

  * •

In GH1 the BH entropy comes entirely from the GHY boundary term (at infinity)
since the bulk term vanishes for a (Euclidean) vacuum solution. In the dS case
(in the static patch) on the other hand the entire classical action comes from
the bulk term since there is no boundary at $r=0$.

  * •

For the extended (Kruskal-Szekeres) diagram for the black hole as well as for
dS space in global coordinates the entropy (at least to leading order) is
zero. This is consistent with identifying these space times with quantum
mechanical pure states [8].

  * •

It is unclear how to define the thermodynamics of SdS space since the two
horizons are at different temperatures $\kappa_{{\rm
b}}/2\pi\neq\kappa_{c}/2\pi$. Gibbons and Hawking [2] advocate putting a
perfectly reflecting wall in-between the two horizons. But this is artificial
and surely not the same as the original system. The system itself is
thermodynamically unstable [9, 10].

##  2 Review of original calculations and comments thereof

We will first quote some basic formulae111See for example the excellent
account given by Eric Poisson [11]. for the gravitational action in both
Lagrangian and Hamiltonian form. Note that we work in units $\hbar=c=G=1$ and
the mostly plus metric convention, with
$R_{\mu\nu\,\,\,\sigma}^{\,\,\,\,\,\lambda}V^{\sigma}\equiv\left[\nabla_{\mu},\nabla_{\nu}\right]V^{\lambda}$
and $R_{\nu\sigma}=R_{\,\,\,\,\nu\mu\sigma}^{\mu}$. The Einstein-Hilbert
action for gravity (without a cosmological constant) is

$I_{H}\left[g_{\alpha\beta}\right]=\frac{1}{16\pi}\int_{{\cal
V}}R\sqrt{-g}d^{4}x.$ (1)

Here ${\cal V}$ is a space-time volume with a boundary which will be denoted
by $\partial{\cal V}$. Under a metric variation we have

$16\pi\delta I_{H}=\int_{{\cal
V}}\left(R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R\right)\delta
g^{\alpha\beta}\sqrt{-g}d^{4}x-\int_{\partial{\cal V}}\epsilon
h^{\alpha\beta}\delta g_{\alpha\beta,\mu}n^{\mu}|h|^{1/2}d^{3}y.$ (2)

The first term contains the Einstein tensor which is the LHS of the Einstein
field equation. In the boundary term the $y$ are coordinates adapted to
$\partial{\cal V}$, $n^{\mu}$ is the unit normal to it with $n^{2}=\epsilon$,
where $\epsilon=\pm 1$ depending on whether the boundary component is time-
like or space-like, and $h$ is the metric on $\partial{\cal V}$ with
$h^{\alpha\beta}=g^{\alpha\beta}-\epsilon n^{\alpha}n^{\beta}$. If we impose
Neumann (N) boundary conditions on $\partial{\cal V}$ the second term vanishes
and the variational principle will lead to the field equations.

However if we wish to impose Dirichlet (D) boundary conditions on
$\partial{\cal V}$, we need to add a boundary term whose variation (after
fixing the metric on the boundary i.e. $\delta g_{\alpha\beta}|_{\partial{\cal
V}}=0$ will cancel the boundary term in (2), thus giving the desired equation
of motion. This term [1] is

$I_{B}\left[g_{\alpha\beta}\right]=\frac{1}{8\pi}\int_{{\cal\partial
V}}\epsilon K|h|^{1/2}d^{3}y.$ (3)

Here $K=\nabla_{\mu}n^{\mu}$ is the trace of the so-called second fundamental
form (i.e. the extrinsic curvature of $\partial{\cal V}$,
$K_{\mu\nu}=\nabla_{\mu}n_{\nu}$ as embedded in ${\cal V}$ where $n_{\mu}$ is
the unit normal to $\partial{\cal V}$). Thus the appropriate action that
should be used in the path integral is

$I=I_{H}+I_{B}-I_{B_{0}},$ (4)

where the last term is given by (3) with $K\rightarrow K_{0}$ i.e. the
extrinsic curvature evaluated in flat space. This is required in order to have
zero action for flat space. Clearly it will not contribute to the equation of
motion. It is important to note that this implies (according to the entropy
calculation of Gibbons and Hawking) empty Minkowski space (the vacuum) will be
assigned zero entropy consistent with standard thermodynamics in flat space
where the ground state is assigned zero entropy.

One might ask why one would need to impose D boundary conditions rather than
N. The reason is that (as explained for instance in [12]) if one wishes to
compute the quantum amplitude for transitions from one metric $h_{1}$ on a
surface (1) to another $h_{3}$ on another surface (3), then in order that the
sum over states with intermediate metrics $h_{2}$ on a surface (2) gives the
correct relation in the classical limit, we need
$I(g_{1}+g_{2})=I(g_{1})+I(g_{2})$, where $g_{1}$ is the metric between the
surfaces 1 and 2 and $g_{2}$ being the metric between 2 and 3. This relation
can only be satisfied with D boundary conditions.

Given this action for gravity, namely eqn. (4), the thermodynamic
interpretation given in [1, 2] follows.

The discussion in [1] (GH1) actually begins by evaluating the Lorentzian
action (4). GH1 first evaluate it for the Schwarzschild black hole with the
metric

$ds^{2}=-\left(1-\frac{r_{s}}{r}\right)dt^{2}+\left(1-\frac{r_{s}}{r}\right)^{-1}+r^{2}d\Omega_{2}^{2},$
(5)

where the last term is the metric on a two sphere with radius $r$ and the
Schwarzschild (horizon) radius $r_{s}$ may be identified as $2M$ where $M$ is
the mass of the black hole.

This has a curvature singularity at $r=0$ as well as a coordinate singularity
at $r=r_{s}$. In view of these the question is how does one evaluate the bulk
action $S_{H}$? GH1 prescription is to evaluate the action in the so-called
Euclidean section where $t\rightarrow-i\tau$ with $\tau$ taken to be real. The
metric is then everywhere smooth provided that $\tau$ is taken to be an
angular coordinate with $\tau$ and $\tau+4\pi r_{s}$ identified. In this case
the bulk term $S_{H}$ is zero since $R=0$ everywhere, and the classical action
comes entirely from the boundary term which now has the topology $S^{1}\times
S^{2}$ and is compact.

To evaluate this we need the unit normal to the boundary $r={\rm constant}$
which is $n_{\mu}=\delta_{\mu}^{r}/\sqrt{1-\frac{r_{s}}{r}}$,
$n^{\mu}=g^{\mu\nu}n_{\nu}=\sqrt{1-\frac{r_{s}}{r}}\delta_{r}^{\mu}$, so we
get
$K=\nabla.n=\frac{1}{\sqrt{g}}\partial_{\mu}\left(\sqrt{g}n^{\mu}\right)=\frac{1}{r^{2}}\partial_{r}\left(r^{2}\sqrt{1-\frac{r_{s}}{r}}\right)$
and $K_{0}=\frac{1}{r^{2}}\partial_{r}\left(r^{2}\right)=\frac{2}{r}$. Also we
have $|h|^{1/2}=\sqrt{1-\frac{r_{s}}{r}}r^{2}\sin\theta$. Hence (defining the
boundary at $r=r_{b}\rightarrow\infty)$

$\displaystyle I_{{\rm bh}}$
$\displaystyle=\frac{1}{8\pi}\int_{0}^{-i\beta}dt\int_{S_{2}}d^{2}x|h|^{1/2}\left(K-K_{0}\right)|_{r=r_{{\rm
b}}}=-i\frac{\beta}{2}\left(2r_{{\rm b}}-\frac{3}{2}r_{s}-2r_{{\rm
b}}\sqrt{1-\frac{r_{s}}{r_{{\rm b}}}}\right)$
$\displaystyle=i\frac{\beta}{2}\left(\frac{r_{s}}{2}+O\left(\frac{r_{s}^{2}}{r_{{\rm
b}}}\right)\right)\overset{\beta=4\pi r_{s},\,r_{{\rm
b}}\rightarrow\infty}{=}i\pi r_{s}^{2}$ (6)

Note that the the final result is $i$ times a quarter the area of the horizon.

GH1 also calculate the classical action for deSitter space. In this case (1)
needs to be replaced by

$I_{H}\left[g_{\alpha\beta}\right]=\frac{1}{16\pi}\int_{{\cal
V}}\left(R-\frac{6}{r_{{\rm ds}}^{2}}\right)\sqrt{-g}d^{4}x,$ (7)

where $r_{{\rm ds}}^{2}=3/\Lambda$ and $\Lambda$ is the (positive)
cosmological constant. Also the space-time has no boundary so there is no
boundary term and no singularity. From the Einstein equation we have
$R=12/r_{{\rm ds}}^{2}$. However if we substitute this into the action we get
infinity from the infinite space time volume.

GH1 argues that one has to again consider the Euclidean section. First let us
consider the so-called static patch representation of (part of) dS space given
by a presentation like that of the black hole in (5):

$ds^{2}=-\left(1-\frac{r^{2}}{r_{{\rm
ds}}^{2}}\right)dt^{2}+\left(1-\frac{r^{2}}{r_{{\rm
ds}}^{2}}\right)^{-1}dr^{2}+r^{2}d\Omega_{2}^{2}.$ (8)

As with the black hole metric presentation in (5), the above is also singular
at the horizon $r=r_{{\rm ds}}$ (where the time-like Killing vector
$\partial/\partial t$ becomes null). Again following the prescription used in
the BH case GH1 propose evaluating the action in the Euclidean section with
$t\rightarrow-i\tau$ (with $\tau$ real). Also as in the previous case the
relation between these coordinates and the global coordinates implies that the
Euclidean variable $\tau$ is periodic with period $2\pi r_{{\rm ds}}$,. This
also avoids the conical singularity at $r=r_{{\rm ds}}$ in the Euclideanized
version of (8). The topology of the Euclidean section is now an $S_{4}$ with
radius $r_{{\rm ds}}$, so the action becomes,

$\displaystyle iI_{{\rm ds}}$ $\displaystyle\rightarrow-I_{{\rm ds}}^{{\rm
Euclidean}}=-\frac{1}{16\pi}\int\sqrt{|g|}d^{4}x_{{\rm
Euclidean}}\left(\frac{6}{r_{{\rm ds}}^{2}}-R\right)=$
$\displaystyle=\frac{1}{16\pi}\int\sqrt{|g|}d^{4}x_{{\rm
Euclidean}}\frac{6}{r_{{\rm ds}}^{2}}=\frac{1}{16\pi}\frac{8\pi^{2}r_{{\rm
ds}}^{4}}{3}\frac{6}{r_{{\rm ds}}^{2}}=\pi r_{{\rm ds}}^{2}.$ (9)

What do these calculations have to do with thermodynamics? This follows
according to GH1 from the representation of the partition function for the
canonical ensemble by a path integral, i.e.

$Z\left(\beta\right)={\rm Tr}e^{-\beta
H}=\int_{\phi\left(t\right)=\phi(t+i\beta)}\left[d\phi\right]e^{iI[\phi]}$
(10)

Here $\phi$ stands for all fields in the system including the metric field.
One then evaluates the RHS in the saddle point approximation by writing
$\phi=\phi_{0}+\bar{\phi}$ in which case we have,

$-\beta F\equiv\ln Z\simeq
iI\left[\phi_{0}\right]+\ln\int_{\bar{\phi}\left(t\right)=\bar{\phi}(t+i\beta)}\left[d\bar{\phi}\right]e^{iI[\phi_{0}\bar{,\phi}]}+\ldots.$
(11)

Here the second term is the one loop correction and the ellipses represent
higher loop terms.

Let us evaluate this for the two systems discussed above, i.e. the
Schwarzschild black hole and deSitter space. In both cases $\phi$ is just the
metric field. For the black hole (implicitly assuming that $\phi_{0}$ just
represents a unique black hole of mass $M$) and ignoring the one-loop
correction we have,

$\beta F_{0}=-iI_{{\rm bh}}[g_{0}]=\pi r_{s}^{2}=\beta<E>-S$ (12)

In the second equality we used (6) and the last one is the usual thermodynamic
relation between free energy $F$ the average energy of the ensemble $<E>$ and
the entropy $S$. GH1 then proceeds to identify $<E>$ with the mass $M$, and
with $\beta=4\pi r_{s},\,M=r_{s}/2$ this gives the famous formula $S=\pi
r_{s}^{2}=\frac{1}{4}A_{{\rm h}}$. A similar argument for dS space gives
using(9)

$\beta F_{0}=I_{{\rm ds}}^{{\rm Euclidean}}=-\pi r_{{\rm ds}}^{2}=\beta<E>-S$
(13)

In this case GH1 identify $<E>$ with zero since space is closed (so there is
no boundary energy) giving again $S=\frac{1}{4}A_{{\rm h}}.$

At this point it is worthwhile noting that the two calculations apparently
give contradictory results for the entropy of empty Minkowski space. The
$M\rightarrow 0$ limit of black hole space time gives Minkowski space. Given
that the calculation of GH1 gave $S_{{\rm bh}}=\pi
r_{s}^{2}=\pi\left(2M\right)^{2}\rightarrow 0$ in the limit, this implies that
the entropy of Minkowski space is zero. This is clearly consistent with the
expectation that quantum mechanically this geometry must represent a vacuum
state. On the other hand one might also argue that Minkowski space is obtained
from deSitter space by taking the limit $r_{{\rm ds}}\rightarrow\infty$ so
that, since GH1 gave $S_{{\rm ds}}=\pi r_{{\rm ds}}^{2}$, one has infinite
entropy in the limit. In other words one is apparently led to assign infinite
entropy to (empty) Minkowski space, which seems rather strange! However It has
been argued by many authors (perhaps the earliest were [13] and [14]), that
the horizon entropy of dS space is the maximum entropy that this space can
hold. With this interpretation then, in the limit of the horizon radius going
infinity, the ensuing infinite entropy of Minkowski space, should also be
interpreted as the maximum entropy that this space can contain, which of
course is reasonable. It is not therefore the entropy of the Minkowski vacuum,
which should indeed be zero.

If we look at the calculations in GH1 we see that the two entropies arise from
quite different terms in the (Euclidean) action. In the black hole case the
space-time volume integral (the Einstein-Hilbert term) gives zero contribution
since the space is Ricci flat. The entropy comes from the boundary term. As
discussed above, GH1 subtracts from the boundary term a term integrated over
the same surface but with $\nabla.n$ evaluated in flat space. It is the
cancellation between these two for Minkowski space which results in zero
entropy for this space. On the other hand in the corresponding calculation for
dS space it is the volume integral which gives the entropy and there is no GHY
boundary term since there is no boundary. Now the integrand in this term (see
eqn. (9)) goes to zero like $\frac{1}{r_{{\rm ds}}^{2}}$ in the Minkowski
limit but the volume integral gives a factor of $r_{{\rm ds}}^{4}$ so the
action and hence the entropy is proportional to $r_{{\rm ds}}^{2}$ ! However
the infinite radius limit of dS space should not be identified with Minkowski
space. The topologies are different and hence the need for a (spatial)
boundary term in the latter case which is absent in the former. Indeed if one
worked in the strict Minkowski limit (even though the volume is still
infinite) the curvature is strictly zero, so that the E-H term is exactly
zero.

###  2.1 Canonical and Microcanonical Ensemble for Black Holes

The canonical ensemble refers to a collection of systems in thermal
equilibrium at some externally fixed temperature $T=\beta^{-1}$. Instead of a
single energy one has a (Boltzmann) distribution of energies. However as is
well known the canonical partition function for Schwarzchild black holes is
ill-defined even with Brown-York boundary conditions [6], (which impose a
physical boundary at a finite distance held at a fixed temperature).

The point is simply that the density of states $D(E)$ (the exponential of the
entropy) increases as $\exp\left(4\pi E^{2}\right)$ (if indeed the entropy is
given by the Bekenstein-Hawking formula). In the Brown-York set up each black
hole is enclosed in a spherical wall at some radius $R>r_{H}=2GM$ held at a
fixed temperature $T=\beta^{-1}$. Nevertheless clearly the density of states
overwhelms the Boltzmann factor and hence the partition function is divergent.

It is more appropriate to consider the micro-canonical ensemble (as advocated
in [6]). Typically this is an ensemble with a small (or vanishingly small)
energy range $E-dE,E+dE$. Entropy is then defined to be $S(E)=\ln D(E)$ where
$D(E)$ is the density of states at energy $E$. Instead of (11) one should
start with the functional integral representation of $D(E)$. This is

$D(E)={\rm Tr}\delta\left(E-\hat{H}\right)=\frac{1}{2\pi}{\rm Tr}\int
dte^{i\left(E-\hat{H}\right)t}=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dte^{iEt}{\cal
N}\int_{\phi\left(0\right)=\phi(t)}\left[d\phi\right]\int\left[d\pi\right]e^{i\int_{0}^{t}dt^{\prime}\left[\int_{\Sigma}\pi\dot{\phi}-H\right]}$
(14)

Here we’ve used the path integral representation for the (trace of the)
unitary evolution operator,

${\rm Tr}e^{i\hat{H}t}={\cal
N}\int_{\phi\left(0\right)=\phi(t)}\left[d\phi\right]\int\left[d\pi\right]e^{i\int_{0}^{t}dt^{\prime}\int_{\Sigma}\pi\dot{\phi}-H]}.$

The normalization constant ${\cal N}$ is defined so that the trace of the unit
operator i.e. ${\rm Tr}\hat{I}={\cal
N}\int_{\phi\left(0\right)=\phi(t=0)}\left[d\phi\right]\int\left[d\pi\right]$
is unity, which implies that we have regularized and renormalized our
definition of the trace such that, as we’ll see below, the extended black hole
or deSitter space time (or in general any stationary asymptotically flat or
deSitter space time), has zero entropy corresponding to a pure state. The
entropy associated with a black hole, or that of de Sitter in static
coordinates must then be identified as entanglement entropy. Actually it makes
more sense to use the original Hamiltonian representation for the path
integral. In this case, we replace $\phi$ by the set $N,N_{a},h_{ab},\psi$,
where we’ve used a $3+1$ split of space time and defined fields on the space-
like surface $\Sigma_{t}$ as discussed earlier. Then the expression for the
density of states becomes

$D(E)=\frac{1}{2\pi}\int dte^{iEt}{\cal
N}\int_{\phi\left(0\right)=\phi(t)}\left[dN\right]\left[dN_{a}\right]\left[dh_{ab}\right]\left[d\pi^{ab}\right]\left[d\psi\right]\left[d\pi_{\psi}\right]e^{i\int_{0}^{t}dt^{\prime}\left[\int_{\Sigma}\pi^{ab}\dot{h}_{ab}+\pi_{\psi}\dot{\psi}-H\right]}.$
(15)

Here $N,N_{a}$ become Lagrange parameters imposing the secondary constraints
(the primary are $\pi_{N}=\pi_{N_{a}}=0$). The Hamiltonian is given by222See
for example Poisson [11] eqn. 4.61. $k$ is defined in eqn.(23) below.  with
matter Hamiltonian ${\cal H}_{m}$ and momentum density $\nabla_{b}{\cal
P}_{m}^{ab}$)

$\displaystyle H$
$\displaystyle=\frac{1}{16\pi}\int_{\Sigma_{t}}\left[N\left((K^{ab}K_{ab}-K^{2}-^{3}R)+{\cal
H}_{{\rm m}}\right)-2N_{a}\nabla_{b}\left(K^{ab}-Kh^{ab}+{\cal P}_{{\rm
m}}^{ab}\right)\right]\sqrt{h}d^{3}y+H_{\partial},$ (16)
$\displaystyle\equiv\int_{\Sigma_{t}}\left[N{\cal H}+N_{a}{\cal
H}^{a}\right]+H_{\partial},$ (17) $\displaystyle H_{\partial}$
$\displaystyle=\frac{1}{8\pi}\ointop_{S_{t}}\left[N\left(k-k_{0}\right)-N_{a}\left(K^{ab}-Kh^{ab}\right)r_{b}\right]\sqrt{\sigma}d^{2}\theta.$
(18)

Here $S_{t}$ is the $S_{2}$ boundary of $\Sigma_{t}$. Doing the functional
integrals over the $N^{\prime}s$ gives

$D(E)=\frac{1}{2\pi}\int dte^{iEt}{\cal
N}\int_{\phi\left(0\right)=\phi(t)}\left[dh_{ab}\right]\left[d\pi^{ab}\right]\left[d\psi\right]\left[d\pi_{\psi}\right]\delta\left[{\cal
H}\right]\delta\left[{\cal
H}^{a}\right]e^{i\int_{0}^{t}dt^{\prime}\left[\int_{\Sigma_{t^{\prime}}}\left(\pi^{ab}\dot{h}_{ab}+\pi_{\psi}\dot{\psi}\right)-H_{\partial}\right]}$
(19)

If we assume that the path integral is saturated by a single stationary
configuration with boundary Hamiltonian $H_{\partial}$ (as assumed in[1]) then
the $t$ integral also yields a delta function and we have

$D(E)=\delta\left(E-H_{\partial}\right){\cal
N}\int\left[dh_{ab}\right]\left[d\pi^{ab}\right]\left[d\psi\right]\left[d\pi_{\psi}\right]\delta\left[{\cal
H}\right]\delta\left[{\cal H}^{a}\right]|_{{\rm
classical}}=\delta\left(E-H_{\partial}\right)e^{S(E)}.$ (20)

so that $\int_{H_{\partial}-\Delta E}^{H_{\partial}+\Delta
E}D(E)dE=e^{S(H_{\partial})}.$

Thus the log of the coefficient of the delta function gives the entropy of
this configuration. i.e. the general expression for the exponential of the
entropy for a stationary configuration is

$e^{S(H_{\partial})}={\cal
N}\int\left[dh_{ab}\right]\left[d\pi^{ab}\right]\left[d\psi\right]\left[d\pi_{\psi}\right]\delta\left[{\cal
H}\right]\delta\left[{\cal H}^{a}\right]|_{H_{\partial}}.$ (21)

Note that this formula is exact for stationary fluctuations around the
original stationary background. In general of course we need to keep the time
derivative terms in the exponent of (19). In the above we’ve assumed that
there is no inner boundary so that the calculation is appropriate for the
extended (two sided) black hole. If one ignored the delta functions in the
measure (they are effectively a quantum correction), then we have in this
approximation $S(M)=0$ which is consistent with the picture of a two sided
black hole being a pure state as observed in [8]333In this reference the
normalization factor is effectively set equal to one. However it is clear from
the above that it has been implicitly defined such that the trace of the unit
operator is one.. Note that the fields in the functional integral here are
defined as functions of global coordinates so that $\Sigma_{t}$ in the
integrand of (19) is taken over both causal wedges (eg. region I and III of
the Kruskal diagram in the black hole case). The same would be true of both dS
and SAdS spaces. In other words the extended field space for all these
configurations should have entropy zero. It is unclear how this generalizes to
spaces with more than one horizon such as SdS Kerr black holes etc..

####  2.1.1 Schwarzchild black hole

If on the other hand we wish to compute the entropy of one sided black hole
(here identified as say region $I$ of the diagram for an eternal black hole)
we need to compute in Schwarzschild coordinates in which case (since they
diverge at the horizon) we need to insert a boundary there. Let us recall
briefly444For details see for example [11] section 4.2. how the first term of
(18) arose.

The boundary $\partial{\cal V}$ of the region of integration in (1) can be
split up into two space-like boundaries $\Sigma_{t_{1}},\Sigma_{t_{2}}$ and a
time-like boundary ${\cal B}$. In the 1+3 split (ADM) formulation the action
for pure gravity becomes (with $I_{m}=\int\sqrt{|g|}{\cal L}_{m}$ being the
matter action - including the cosmological constant),555$n$ is the time-like
normal to $\Sigma_{t}$ and $r$ is the space-like normal to ${\cal B}$, $y,z$
are respectively coordinates on $\Sigma_{t},{\cal B}$ with metrics
$h_{ab},\gamma_{ij}$. Also ${\cal K}=\gamma^{ij}{\cal
K}_{ij}=\gamma^{ij}\left(e_{i}^{\alpha}e_{j}^{\beta}\nabla_{\beta}r_{\alpha}\right)$.
One also has the completeness relation for the metric
$g^{\alpha\beta}=r^{\alpha}r^{\beta}+\gamma^{\alpha\beta}=r^{\alpha}r^{\beta}-n^{\alpha}n^{\beta}+\sigma^{AB}e_{A}^{\alpha}e_{B}^{\beta}$.
Also note that note that ${\cal B}$ is foliated by $S_{t}$.

$\displaystyle 16\pi I\left[g_{\alpha\beta}\right]$ $\displaystyle=\int_{{\cal
V}}R\sqrt{-g}d^{4}x+2\int_{{\cal\partial V}}\epsilon K|h|^{1/2}d^{3}y+I_{{\rm
m}}=\int_{t_{1}}^{t_{2}}dt\int_{\Sigma_{t}}\left({}^{3}R+K^{ab}K_{ab}-K^{2}+{\cal
L}_{m}\right)N\sqrt{h}d^{3}y$
$\displaystyle+2\int_{t_{1}}^{t_{2}}dt\ointop_{S_{t}}\left({\cal
K}+\nabla_{\beta}r_{\alpha}n^{\alpha}n^{\beta}\right)N\sqrt{\sigma}.$ (22)

Note that the first term in the second line comes from the time-like part of
the GH boundary term (2). Using the completeness relation for the metric (see
footnote (5)) we then have

${\cal K}+\nabla_{\beta}r_{\alpha}n^{\alpha}n^{\beta}=\sigma^{AB}k_{AB}=k.$
(23)

This boundary term in the action then gives us the first term of the boundary
Hamiltonian in (18).

On the inner boundary (at the black hole horizon) however there is no GH
boundary term thus the corresponding contribution to the boundary integral
there is $\frac{1}{8\pi}\ointop_{{\cal
H}_{t}}\left[\nabla_{\beta}r_{\alpha}n^{\alpha}n^{\beta}\right]N\sqrt{\sigma}d^{2}\theta$
(there is a minus sign from the fact that the outward normal to ${\cal H}_{t}$
is $-r_{\alpha}$ and another one from going from the action to the
Hamiltonian). This is evaluated by first moving slightly away from the horizon
(to the stretched horizon) where the relevant vectors ($\hat{n},\hat{r}$ are
respectively still time-like and space-like). We have
$n^{\mu}=N^{-1}\frac{dx^{\mu}}{dt}$ (normalized to $n^{2}=-1$) and
$r^{\mu}=N\frac{dx^{\mu}}{dr}$ (normalized to $r^{2}=1$) and $r.n=0$. Then
666All but the last two steps below are the same as in footnote 13 of [15]
except that it’s in Euclidean form.

$\displaystyle Nn^{\mu}n^{\nu}\nabla_{\mu}r_{\nu}$
$\displaystyle=-Nn^{.u}r^{\nu}\nabla_{\mu}n_{\nu}=-Nn^{.u}r^{\nu}\nabla_{\mu}\left(-N\nabla_{\nu}t\right)=N^{2}n^{\mu}r^{\nu}\nabla_{\mu}\nabla_{\nu}t=N^{2}n^{\mu}r^{\nu}\nabla_{\nu}\left(-N^{-1}n_{\mu}\right)$
$\displaystyle=-N^{2}n^{\mu}r^{\nu}\nabla_{\nu}\left(N^{-1}\right)n_{\mu}=-N^{2}n^{\mu}r^{\nu}\left(-N^{-2}\nabla_{\nu}N\right)n_{\mu}=-r^{\nu}\nabla_{\nu}N\rightarrow-N\nabla_{r}N|_{r\rightarrow
r_{{\cal H}_{t}}}=-\kappa,$

where we used $r^{\nu}\nabla_{\nu}t=0$,
$n^{2}=-1,\,n^{\mu}\nabla_{\nu}n_{\mu}=0$, and $\kappa$ is the surface gravity
at the relevant horizon777The surface gravity $\kappa$ is defined by the
acceleration at the horizon,
$\xi^{\beta}\nabla_{\beta}\xi^{\alpha}=\kappa\xi^{\alpha}$ and can be shown to
be constant at the horizon. For static space times it can be shown that
$\kappa=N\frac{dN}{dr}$ where $r$is the radial coordinate and $N=N(r)$. .

$\Delta H_{\partial}=\frac{1}{8\pi}\ointop_{{\cal
H}_{t}}\left[\nabla_{\beta}r_{\alpha}n^{\alpha}n^{\beta}\right]N\sqrt{\sigma}d^{2}\theta=-\frac{1}{8\pi}\kappa{\cal
A_{{\cal H}}}_{{}_{t}}.$ (24)

Thus what one actually gets from the Lorentzian calculation is (in the
classical approximation as above with the path integral saturated by a single
black hole of mass
$M=\frac{1}{8\pi}\ointop_{S_{t}^{\infty}}\left(k-k_{0}\right)N\sqrt{\sigma}$)

$D\left(E\right)=\delta\left(E-\left(M-\frac{1}{8\pi}\kappa{\cal
A}\right)\right).$

How are we to interpret this result? The formula tells us that only states
satisfying the relation $M=E+\frac{1}{8\pi}\kappa{\cal A}$ contribute.
Alternatively if one computed the partition function in this
approximation888Note that here we just have one fixed energy - since we are
focused on just one black hole. So the issue of the divergence of the partion
function does not arise. In other words we are simply identifying what the
free energy is corresponding to the one black hole configuration. we have

$Z=e^{-\beta F}=\int dED(E)e^{-\beta
E}=e^{-\beta\left(M-\frac{\kappa}{2\pi}\frac{1}{4}{\cal A}\right)}.$ (25)

IF we identify $\kappa/2\pi$ as the temperature and $M$ as the internal energy
$U$ then comparing this with the thermodynamic relation $F=U-TS$ we would
obtain the relation $S={\cal A}/4$ . However this classical approximation to
the Lorentzian path integral cannot give an independent derivation of the
Hawking temperature. This should not be surprising since if we restored
Planck’s constant Hawking’s expression for the temperature is
$\kappa\hbar/2\pi$. In other words the temperature is a quantum effect and was
derived by Hawking by computing the Green’s functions of quantum fields in a
black hole background. One should not expect to derive it in the classical
approximation.

Of course once we have the temperature and the identification of the internal
energy $U=<E>$ as the black hole mass $M$ one could have integrated the first
law $dU=TdS$ to get the entropy up to an arbitrary constant i.e.
$S=\frac{1}{4}{\cal A}/\hbar+{\rm constant}$. The above argument fixes the
constant to be zero.

####  2.1.2 AdS black hole (SAdS)

This has the metric

$ds^{2}=-\left(1-\frac{2M}{r}+\frac{r^{2}}{b^{2}}\right)dt^{2}+\left(1-\frac{2M}{r}+\frac{r^{2}}{b^{2}}\right)^{-1}dr^{2}+r^{2}d\Omega_{2}^{2}$
(26)

Here $b\equiv\sqrt{3/|\Lambda|}$ is the AdS radius. For $M=0$ we have (global)
AdS while for $b\rightarrow\infty$ we have the Schwarzchild black hole. There
is only one horizon in this space time which can be identified with the
horizon of the black hole, i.e. the solution of
$\left(1-\frac{2M}{r}+\frac{r^{2}}{b^{2}}\right)=0$. Calling this $r_{+}$ we
have the relation

$2M=r_{+}\left(1+\frac{r_{+}^{2}}{b^{2}}\right).$ (27)

The thermodynamics of this space time using Euclidean methods was worked out
in [7]. The Hawking temperature calculation gives999As is well-known the quick
and easy way of getting the temperature is to Euclideanize the metric and then
require the absence of a conical singularity at $r=r_{+}$ giving
$\beta=\frac{4\pi}{\left(dN^{2}/dr\right)_{r=r_{+}}}$. However this is
equivalent to the original Lorentzian calculation of Hawking in terms of QFT
in a curved background so is independent of Euclidean methods.

$\beta=\frac{4\pi r_{+}b^{2}}{b^{2}+3r_{+}^{2}}.$ (28)

Note that $\beta<\beta_{0}\equiv\frac{2\pi b}{\sqrt{3}}$ (i.e.
$T>T_{0}\frac{\sqrt{3}}{2\pi b}$). Also $T\rightarrow\infty$ for $r_{+}=0$ and
$r_{+}\rightarrow\infty$ and the temperature is minimum at
$r_{+}=r_{0}\equiv\frac{b}{\sqrt{3}}$. Unlike in flat space in AdS a black
hole can be in thermodynamic equilibrium with radiation without artificial
reflecting walls. Note however that for a given temperature there are two
possible solutions for the horizon radius $r_{+}.$ The larger value
corresponds to a black hole with positive specific heat (unlike its flat space
analog) and hence is stable, while the smaller value has negative specific
heat and is unstable to decay into a bath of radiation. This is easily seen
from the graph of temperature vs. radius. So for $r_{+}>r_{0}$ we have a
stable black hole whilst for $r_{+}<r_{0}$ the black hole is unstable and will
decay into radiation. This is the so-called Hawking-Page phase transition.

As for the entropy calculation, in our Hamiltonian framework it proceeds as in
the flat space case except that for $k_{0}$ in (18) one now has $k$ evaluated
in the background AdS space rather than in flat space and the horizon radius
and inverse temperatures are given by (27) and (28). Thus the formula for the
free energy is again given by (25) so that the entropy (after identifying
$\beta=\frac{2\pi}{\kappa}$) is again

$S=\frac{1}{4}{\cal A}=\pi r_{+}^{2},$ (29)

with the horizon radius given in terms of the black hole mass by (27). The
latter formula also tells us that for large $r_{+},M$, the entropy goes as
$M^{2/3}$. So unlike in the flat space case, the density of states does not
swamp the Boltzmann term $-\beta M$ and the canonical ensemble is well-
defined.

####  2.1.3 deSitter space (dS)

Now consider dS. Again as in the black hole case if we identify the spatial
slice $\Sigma_{t}$ as extending over the whole spatial region of global dS we
would get zero entropy (at least classically as in the black hole case). On
the other hand if we take $\Sigma_{t}$ as extending only over the static patch
(i.e. from the dS horizon to $r=0$), we get (since the only boundary is at the
dS horizon ($r=0$ is the origin of coordinates!),

$D\left(E\right)=\delta\left(E+\frac{1}{8\pi}\kappa_{c}{\cal A}_{c}\right).$

Again to interpret this we need to go back to the partition function so,

$Z=e^{-\beta F}=\int dED(E)e^{-\beta
E}=e^{-\beta\left(-\frac{\kappa_{c}}{2\pi}\frac{1}{4}{\cal A}_{c}\right)}.$
(30)

As before identifying $\kappa_{c}/2\pi$ as the temperature of the horizon [2]
and comparing with the formula $F=U-TS$ but now with the internal
energy101010There is no boundary energy associated with dS, so Gibbons and
Hawking take it to be zero [1] $U=0$ we get the Gibbons Hawking result
$S=\frac{1}{4}{\cal A}_{c}$. In this case there is no direct way of comparing
with the thermodynamic identity $dU=TdS$ since the energy is fixed at zero
(actually it is not defined). The Gibbons-Hawking argument proceeds by adding
external energy to the system and identifying it as that which goes into the
thermodynamic identity $dU$. We will discuss the implications of this below
after rederiving the geometric version of the thermodynamic identity.

In computing the surface gravity at the dS horizon from the formula
$\kappa=\frac{1}{2}\nabla_{r}N^{2}$, it should be noted that this gives the
right sign for a black hole where $N^{2}=1-r_{s}/r$, $\kappa=1/2r_{s}$,
whereas for dS space, where $N^{2}=1-r^{2}/r_{{\rm dS}}^{2}$, since the
direction of increasing $r$ is the opposite of the black hole case,
$\kappa=-\frac{1}{2}\nabla_{r}N^{2}$, giving $\kappa=1/r_{{\rm dS}}$.

####  2.1.4 deSitter black hole (SdS)

Now let us consider Schwarzschild deSitter (SdS) space. The metric is

$ds^{2}=-\left(1-\frac{2M}{r}-\frac{\Lambda}{3}r^{2}\right)dt^{2}+\left(1-\frac{2M}{r}-\frac{\Lambda}{3}r^{2}\right)^{-1}dr^{2}+r^{2}d\Omega_{2}^{2}.$
(31)

Now there is a black hole horizon at $r=r_{{\rm b}}$ as well as a cosmological
horizon at $r=r_{{\rm c}}$ (when $\Lambda>0,\,9\Lambda M^{2}<1$)111111If the
last inequality is not satisfied the two real positive roots become complex
and there are no horizons., whose positions are given by the two positive
roots of $r-2M-\frac{\Lambda}{3}r^{3}=0$. The three roots are such that
$r_{-}<0<2M<r_{{\rm b}}<3M<r_{{\rm c}}$ and
$N^{2}=-\frac{1}{r}\frac{\Lambda}{3}\left(r-r_{-}\right)(r-r_{{\rm
b}})(r-r_{{\rm c}})$. The maximum value of the black hole mass is
$M=\Lambda^{-1/2}/3$ where the two horizons coincide - this is the so-called
Narai space. From the above formula we then have, with the upper(lower) sign
for blackhole(cosmological) horizons, because of the different directions in
which the horizon normals point with respect to the direction of increasing
$r$.

$T_{{\rm b,c}}=\frac{1}{2\pi}\kappa_{{\rm
b,c}}=\pm\frac{1}{2\pi}\frac{1}{2}\left(\frac{2M}{r_{{\rm
b,c}}^{2}}-\frac{\Lambda}{3}2r_{{\rm b,c}}\right).$ (32)

There are now two temperatures one associated with the black hole horizon and
the other with the cosmological horizon. The thermodynamics of such are
situation is unclear. In fact the system is not in thermal equilibrium and is
unstable [9][10].

###  2.2 Alternative Lorentzian arguments

####  2.2.1 Lorentzian conical singularity argument

Marolf [16] has given a Lorentzian argument for the (Schwarzchild) black hole
entropy formula. Let us first note that in the Hamiltonian framework, after
imposing the constraints it is clear that there is no singularity in the
integrand for the action, since (for a static or stationary metric) the
Hamiltonian version of the action consists of the boundary terms identified in
the previous subsection.

Alternatively let us take the Lagrangian in 3+1 split form given in eqn. (22).
On a classical solution one must have ${\cal H}K^{ab}K_{ab}-K^{2}-^{3}R)=0$.
Thus we have

$\displaystyle 16\pi I_{{\rm cl}}$
$\displaystyle=2\int_{t_{1}}^{t_{2}}dt\int_{\Sigma_{t}}\left(K^{ab}K_{ab}-K^{2}\right)N\sqrt{h}d^{3}y$
$\displaystyle-2\int_{t_{1}}^{t_{2}}dt\ointop_{S_{t}}\left({\cal
K}+\nabla_{\beta}r_{\alpha}n^{\alpha}n^{\beta}\right)N\sqrt{\sigma}.$

Again it is clear that the Lorentzian action is well defined on a solution.

Nevertheless it has been argued in [16] that the functional integral is better
approximated by slices that are not solutions but contain conical
singularities. By contour deformation into the complex plane one can still
pick up a saddle point and it turns out to be the Euclidean black hole
solution. It would be interesting to see whether this procedure can be applied
to dS, SdS and SAdS spaces as well.

####  2.2.2 Algebraic QFT argument

In a recent work Chandrasekaran Pennington and Witten[17] have discussed black
hole thermodynamics from an Algebraic QFT perspective. Also these authors and
Longo [18], have discussed the thermodynamics of dS spaces from the same point
of view. Apparently the Von Neumann algebras in the two cases are different,
which seems rather curious. The arguments are rather abstract and it is not
clear to us how it relates to our Hamiltonian argument or to the original
Euclidean calculations. Nevertheless it is important to understand the
connection. We leave further discussion of this to future work.

###  2.3 Euclidean calculation

At this point it behoves us to revisit the Euclidean argument. Here one argues
that the partition function can be represented as a functional integral of a
Euclidean action. So we write

$\displaystyle Z$ $\displaystyle={\rm Tr}e^{-\beta H}={\cal
N}\int_{\phi\left(\tau=0\right)=\phi(\tau=\beta)}\left[d\phi\right]\int\left[d\pi\right]e^{\int_{0}^{\beta}d\tau^{\prime}\left[\int_{\Sigma}\pi\dot{\phi}-H\right]}$
$\displaystyle={\cal
N}\int_{\phi\left(\tau=0\right)=\phi(\tau=\beta)}\left[d\phi\right]\int\left[d\pi\right]e^{\int_{0}^{\beta}d\tau^{\prime}\left[\int_{\Sigma}\left(\pi\dot{\phi}-N{\cal
H}-N^{a}{\cal P}_{a}\right)-H_{\partial}\right]}$

Assuming again that the path integral is well-approximated by a single
stationary classical solution (which must necessarily satisfy ${\cal H={\cal
P}}_{a}=0$), we have

$Z=={\rm Tr}e^{-\beta H}\simeq{\cal
N}\int_{\phi\left(\tau=0\right)=\phi(\tau=\beta)}\left[d\phi\right]\int\left[d\pi\right]e^{\int_{0}^{\beta}d\tau^{\prime}\left[-H_{\partial}\right]}$
(33)

The Euclidean metric only describes region I of the black hole or just the
static patch of deSitter space. The horizon, as is well-known, has a conical
singularity which needs to be resolved. In addition to the boundary
Hamiltonian coming from infinity , we have also the term corresponding to
(24). This then gives for the exponent in (33)
$-\frac{1}{8\pi}\int_{0}^{\beta}\kappa{\cal A_{{\cal H}}}d\tau_{{}_{t}}$.
Putting $\kappa d\tau=d\theta$ the integral becomes
$-\frac{1}{8\pi}\int_{0}^{\kappa\beta}{\cal A_{{\cal H}}}d\theta_{{}_{t}}$ and
resolving the conical singularity amounts to making the range of $\theta$
cover the full circle of polar coordinates so that $\kappa\beta=2\pi$ implying
that121212As mentioned before the temperature calculation necessarily involves
quantum mechanics. This is seen here from the fact that $\tau$ has dimensions
of a length (with $c=1$ but $\hbar\neq 1$), whereas $\beta$ which is defined
as an inverse temperature in the definition of $Z$ has dimensions of inverse
energy and $\kappa$ has inverse length dimensions. So restoring the factor of
$\hbar^{-1}$ and $G$ in the functional integral we have
$\hbar\beta\kappa=2\pi$ giving $T=\kappa\hbar/2\pi$ and $S={\cal A}/4G\hbar$.
$T=\kappa/2\pi$ and the integral becomes $-\frac{1}{4}A_{{\cal H}}$ which from
(33) may be identified with the entropy associated with the corresponding
horizon.

All this is well-known of course - all we’ve done in the previous subsection,
is to reformulate the calculation in Hamiltonian framework. However let us now
consider the SdS case. There are two horizons with two different
temperatures131313except in the degenerate Narai case where $r_{{\rm
b}}=r_{{\rm c}}=\Lambda^{-1/2}$ and $T_{{\rm b}}=T_{{\rm c}}=0$ $T_{{\rm
b}}=\kappa_{{\rm b}}/2\pi$ and $T_{{\rm c}}=\kappa_{{\rm c}}/2\pi$. In this
case there is no way to resolve both conical singularities. This means that
this Euclidean configuration will not satisfy the Euclidean equation of
motion, breaking down at one or other conical sigularity depending on which
one was resolved[10]. Indeed it is hard to envisage such a system as being in
thermal equilibrium unless as advocated by Gibbons and Hawking [2] one inserts
a perfectly reflecting wall between the two horizons.

The above calculation is completely equivalent to that using the Gauss-Bonnet
(GB) theorem141414The Gauss Bonnet theorem is valid for compact two
dimensional manifolds. The argument of the above reference relies on the fact
that in the near horizon region the space time factorizes into a two
dimensional cone times an $S_{2}$. presented in [19]. That argument although
specifically aimed at black holes can clearly be used in the dS case as well.
However it cannot be applied to an SdS space. As we’ve discussed above once
one has introduced the range of the Euclidean time variable to be $\beta$ then
the corresponding angular range is $\beta\kappa$. In the application of the GB
theorem one is using it for the complete disc. But in the SdS case one can
complete the disc by taking $\beta\kappa=2\pi$ at one horizon or the other but
clearly not both, since there is only parameter $\beta$ defining the
ensemble151515Here we disagree with [20]. This paper claims to derive the
(admittedly compelling) formula $S_{{\rm SdS}}=\frac{1}{4}\left({\cal A}_{{\rm
b}}+{\cal A}_{{\rm c}}\right)$ by using (in effect) the GB theorem. However as
argued above one cannot use it simultaneously at both horizon. In [21] the
entropy formula for SdS in 3 dimensions is derived. But this space has only
one horizon and hence only one temperature, so the above problem does not
arise..

####  2.3.1 The problem with Euclidean arguments

While the Euclidean calculations in GH1 seem to give physically meaningful
results as long as we just compute the leading (classical) contribution),
going beyond this with the functional integral in the Euclidean section is
problematic. This is due to the well known conformal factor problem. Computing
quantum corrections to the leading order expressions for the entropy will
inevitably involve confronting the wrong sign of the kinetic terms for the
conformal fluctuations. This is of course well-known and indeed was pointed
out in the original papers by Hawking and collaborators (see for instance
Hawking’s essay in [12]). Nevertheless at the leading order the Euclidean
argument gives the correct result - in so far as it agrees with the
thermodynamic identity $dE=TdS$ when the calculation of the temperature in
Hawking’s original paper which is the Lorentzian calculation is substituted
for $T$. Of course this determines $S$ only up to a constant.

In our discussion (subsection (2.1)) and in particular eqn. (15) we gave a
Lorentzian formula which in the leading order justifies the Euclidean
calculation - including the constant. Furthermore in principle this formula is
well-defined (at least perturbatively and subject to the treatment quantum
gravity as an effective theory below the cutoff (at Planck or string scale or
whatever theory UV completes Einstein’s theory). In other words the formula
avoids the problems of the Euclidean path integral precisely because of the
delta functions in the measure which impose the constraints in the quantum
theory.

##  3 Smarr relations and the first law of Gravitational Thermodynamics \-
identifying cosmological entropy

In this section we first review both the Smarr relation and the related first
law of gravitational thermodynamics [22] with the aim of clarifying a sign
issue in its application to dS spacetime.

The Smarr relation for (Kerr) black holes is,

$M-2\Omega_{H}J_{H}-\frac{1}{4\pi}\kappa_{H}{\cal
A}_{H}=\frac{1}{4\pi}\int_{\Sigma}R_{\,\,\,\nu}^{\mu}t^{\nu}d\Sigma_{\mu}$
(34)

Here $M$ is the total mass defined as an integral over the bounding surface at
infinity of the black hole space-time (taken to be asymptotically flat) and
$\Sigma$ extends from the event horizon to the sphere at infinity.
$\Omega_{H},\kappa_{H}$ are the angular velocity and the surface gravity of
the black hole horizon $H$. The corresponding first law of black
thermodynamics is

$\displaystyle\delta M-\Omega_{H}\delta
J_{H}-\frac{\kappa_{H}}{2\pi}\delta\left(\frac{1}{4}{\cal A}_{H}\right)$
$\displaystyle=\frac{1}{8\pi}\delta\int
G_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}-\frac{1}{16\pi}\int G^{\mu\nu}\delta
g_{\mu\nu}t^{\lambda}d\Sigma_{\lambda}$ $\displaystyle=\delta\int
T{}_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}-\frac{1}{2}\int T^{\mu\nu}\delta
g_{\mu\nu}t^{\lambda}d\Sigma_{\lambda}.$ (35)

Here $t^{\mu}$ is the time like Killing vector of the space-time and the
integrals are over the space-like surface between the black hole horizon and
the bounding surface at infinity, with $d\Sigma_{\mu}=n_{\mu}d\Sigma$ being
the oriented surface element. $T_{\mu\nu}$ is the energy momentum tensor
(including a cosmological constant term161616So for a pure cosmological
constant term without dynamical matter $T_{\mu\nu}=-\frac{1}{8\pi}\Lambda
g_{\mu\nu}$. or a possibly non-zero minimum of a scalar potential, of an
external (to the black hole horizon) matter distribution. Consider now SdS
space. Here the surface $\Sigma$ may be taken to be stretched between the
black hole horizon and the cosmological horizon. Denoting the quantities
evaluated at the latter by the subscript $c$ the above two relations take the
form171717In the original derivation [22] the second term in eqn. (37) was not
given but was included in a subsequent essay by B. Carter in [12].

$-\frac{1}{4\pi}\kappa_{c}{\cal
A}_{c}-2\Omega_{H}J_{H}-\frac{1}{4\pi}\kappa_{H}{\cal
A}_{H}=\frac{1}{4\pi}\int R_{\,\,\,\nu}^{\mu}t^{\nu}d\Sigma_{\mu}$ (36)

and

$\displaystyle-\frac{\kappa_{c}}{2\pi}\delta\left(\frac{1}{4}{\cal
A}_{c}\right)-\Omega_{H}\delta
J_{H}-\frac{\kappa_{H}}{2\pi}\delta\left(\frac{1}{4}{\cal A}_{H}\right)$
$\displaystyle=\frac{1}{8\pi}\delta\int
G_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}-\frac{1}{16\pi}\int G^{\mu\nu}\delta
g_{\mu\nu}t^{\lambda}d\Sigma_{\lambda}$ $\displaystyle=\delta\int
T{}_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}-\frac{1}{2}\int T^{\mu\nu}\delta
g_{\mu\nu}t^{\lambda}d\Sigma_{\lambda}.$ (37)

Here $G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}R$ and we’ve used the Einstein equation
in the last line. The terms involving the angular velocity correspond to
chemical potential terms (such as $\mu dN$) in the thermodynamic identity. Let
us ignore these (or consider non-rotating black holes). As discussed above,
Hawking’s QFT calculation established that a black hole appears to an external
observer at infinity as a black body at a temperature
$T_{H}=\kappa_{H}\hbar/2\pi$. A similar argument by Gibbons and Hawking
established that to a observer at (say) the origin of static patch coordinates
in dS would associate a temperature $\kappa_{c}\hbar/2\pi$ to the cosmological
horizon. From (35) after identifying $M$ as the total energy of the system and
setting for the moment the external matter-energy to zero, we would have in
the black hole case $\delta M=T_{H}\delta\left(\frac{1}{4\hbar}{\cal
A}_{H}\right)$, giving the well-known identification of the entropy of a black
hole to be $S=\frac{1}{4\hbar}{\cal A}_{H}$, after fixing the integration
constant to be zero.

On the other hand the corresponding formula for the dS or SdS case is (37).
Taking pure dS with some external matter the formula becomes

$-\frac{\kappa_{c}}{2\pi}\delta\left(\frac{1}{4}{\cal A}_{c}\right)=\delta\int
T{}_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}-\frac{1}{2}\int T^{\mu\nu}\delta
g_{\mu\nu}t^{\lambda}d\Sigma_{\lambda}.$ (38)

Let us take $T^{\mu\nu}=-\Lambda g^{\mu\nu}$, so that initially we have empty
dS space. The second term is then $\Lambda\int g^{\mu\nu}\delta
g_{\mu\nu}t^{\lambda}d\Sigma_{\lambda}=-\Lambda\int\frac{\delta(N\sqrt{h}}{N}d^{3}y$
which even for $\Lambda>0$, as is the case here, may be positive or negative.
However if the result of adding matter is the creation of a small black hole
then this additional term is zero. To see this consider the SdS metric Taking
$\delta g_{\mu\nu}$ to be the result of changing $M\rightarrow M+\delta M$
(and then setting $M=0$) we get

$g^{\mu\nu}\delta g_{\mu\nu}=0$

(true even for $M\neq 0$) since $\sqrt{g}$ is independent of $M$. In this case
from (38) we have, after identifying $\frac{\kappa_{c}}{2\pi}=T_{c}$ as the
temperature of the dS horizon, and $\delta\int
T{}_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}=\delta E$ as the variation of
Killing energy, and then comparing with the thermodynamic identity $\delta
E=T\delta S$, the variation of the entropy associated with the horizon to be
$\delta S_{c}=-\delta\left(\frac{1}{4}{\cal A}_{c}\right)$!

This is a long standing puzzle and various resolutions have been proposed. A
summary of these proposals has been given in [23]. The original paper [2] had
one explanation for it. This was clarified further in [21]181818In [21] the
entropy and the thermodynamic relation for $dS_{3}$ was investigated.
Following arguments of Gibbons and Hawking the authors considered $SdS_{3}$
which only has a horizon at $r_{{\rm b}}=\sqrt{1-8GM}$, and so correspondingly
only one temperature $T=\frac{1}{2\pi}\sqrt{1-8GM}$. They then integrate
$dM=TdS$ and fixing the integration constant so that the maximal mass black
hole (with $M=1/8G$) has zero entropy, which gives them $S=-{\cal A}_{{\rm
b}}/4G$! However as in the above discussion they argue that the correct
interpretaion of the thermodynamic identity is $d(-M)=TdS$ which then gives
positive entropy. See the discussion around figure 8 of that paper.. The
argument is based on the fact that in the space-like slice of the global dS
space is closed so that the total energy (which is a boundary term because of
the constraints) is zero. Suppose that the observer is at the south pole of
the $S_{3}$. Any added energy there must be balanced by negative energy in the
causal patch of the north pole. The thermodynamic relation must reflect what
from the observer’s point of view is happening beyond the horizon of the
observer, i.e. in the north pole patch. Thus the observer would conclude that
in that patch the added energy is negative and hence the relation(38) (after
dropping the last term) should be read as

$\frac{\kappa_{c}}{2\pi}\delta\left(\frac{1}{4}{\cal
A}_{c}\right)=TdS=-\delta\int T{}_{\,\,\,\nu}^{\,u}t^{\nu}d\Sigma_{\mu}=\delta
E$ (39)

where $\delta E$ now refers to the observer’s interpretation of what goes on
behind the horizon.

This argument has been criticized in [23] on the grounds that the time-like
Killing vector in the north pole patch is pointing downwards - meaning that in
that patch the Killing energy (which is the negative of the Killing energy in
the south pole patch) is actually positive. However this is not relevant to
the thermodynamics of the cosmological horizon of the south pole observer who
must interpret what is going on behind the horizon. In other words from the
south pole it appears as if the observed system, i.e. the one beyond the south
pole observer’s horizon, has acquired negative energy and the relevant
relation is (39). Obviously observers on the North pole will infer a similar
relation for the thermodynamics of what is beyond their horizon. There is of
course an inevitable subjectivity in these interpretations since the horizon
itself is observer dependent. The paper [23] also argues that if, as is
expected, the state of global dS is a pure state (as also argued in the
comments after eqn. (19) in this note), then the entropy inside the horizon
must equal the entropy outside so both must decrease. However this is in
contradiction with the argument in [21] since the energy inside is obviously
increasing. But here again the issue is what does the entropy inside mean?
What [23] refers to as entropy inside is actually the horizon entropy as seen
by the observer at the north pole. As argued above this also decreases since
there is complete symmetry between the two observers. In other words once one
interprets entropy as representing a measure of the information inaccessible
to each observer, then there is no conflict191919The work of [23] advocates an
alternate interpretation. It is based on a somewhat different interpretaion of
horizon entropy than what is advocated in [21] and the present work..

Now if we use (39) to solve for the entropy we get

$S=\frac{1}{4}{\cal A}_{c}+{\rm constant}.$

How are we to fix this constant? In the black hole case the constant was fixed
by arguing that when the horizon area goes to zero (i.e. the black hole mass
is zero) we should get flat space, which should have zero entropy, thus fixing
the constant to be zero. In the dS case however the horizon going to zero is
the infinite cosmological constant202020${\cal A}_{c}=4\pi r_{{\rm ds}}^{2},$
where $r_{s}=\sqrt{3/\Lambda}$ so flat space is obtained in the limit
$r_{s}\rightarrow\infty$. case! This seems bizarre to say the least. On the
other hand if we had set the constant to be zero, as is usually done then as
discussed earlier, we would get the entropy of empty flat space (i.e. when
$r_{{\rm ds}}\rightarrow\infty$) to be infinite in sharp contradiction to the
argument in the black hole case where the entropy of flat space was set to
zero! But as we saw above this entropy is the maximum possible in dS space so
its Minkowski limit is the maximum entropy that that space can hold which is
plausibly infinite. However the point is that there is no plausible argument
for fixing this constant unlike in the black hole case. Our Hamiltonian
perspective however did fix both constants.

##  4 Conclusion

In these notes we’ve reviewed the entropy calculations for static space-times
with horizons, based on Euclidean arguments, and presented alternative
Lorentzian arguments. At leading order these serve to justify the leading
order Euclidean calculations. However unlike the Euclidean calculations which
suffer from the well-known problem of the wrong sign kinetic term for
conformal fluctuations, the Lozentzian method is well-defined modulo the usual
problems of perturbative quantum field theories with gravity. In particular
the formulae for the entropy (19) to (21) are free of the conformal factor
issue since the constraints eliminate integration over these unphysical modes.
Approximating the relevant path integrals with a single black hole etc. leads
to the familiar expressions derived by Euclidean methods.

It should be pointed out that we do not have an independent derivation of the
Hawking temperature - for that we need to appeal to Hawking’s original
Lorentzian calculations. Of course once one has the Hawking temperature
formulae one may use the thermodynamic identity to evaluate changes in the
entropy. Our calculation shows how the constant may be fixed. This is
particularly relevant in the dS case since (unlike in the black hole case) the
constant cannot be fixed by taking the flat space limit. We also discussed the
AdS black hole case and noted some puzzles associated with SdS space.

Finally we revisited the geometric version of the first law and argued in
favor of the analysis of [21] (a modified version of that in [2]) for
interpreting the sign of this relation in the dS case.

##  5 Acknowledgements

I wish to thank Sebastian Cespedes, Francesco Muia, and Fernando Quevedo for
discussions. I also wish to thank Don Marolf for clarifying the arguments of
[16] and to Ted Jacobson for email correspondence on [23].

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