# Checking volume-law entropy with Hubeny-Rangamani-Takayanagi surfaces

Tyler Guglielmo  Department of Physics, Theory Group, The University of Texas
at Austin, Austin, Texas 78712, USA  Phuc Nguyen  Department of Physics and
Astronomy, Lehman College, City University of New York, 250 Bedford Park
Boulevard West, Bronx, New York 10468, USA  Department of Mathematics and
Haifa Research Center for Theoretical Physics and Astrophysics, University of
Haifa, Haifa 3498838, Israel

###### Abstract

We check formally that the Hubeny-Rangamani-Takayanagi prescription for
holographic entanglement entropy – when applied to a static black brane
spacetime and to a wide class of subregions that do not lie on a constant time
slice – gives rise to volume-law entropy in the limit of large subregion. By
volume-law entropy, we mean that the entanglement entropy scales with the
volume of the projection of the boundary subregion onto a static time slice
with respect to the boundary thermal state (with the same coefficient as in
the volume law on the static time slice). Our result applies to subregions
that have reflection symmetry as well as strips, and we also present field-
theoretic arguments in support of our holographic findings.

††preprint: APS/123-QED

UTTG 29-2022

###### Contents

  1. I Introduction
  2. II Expectation from the field theory side
    1. II.1 Free massless scalar field
    2. II.2 Conformal perfect fluid
    3. II.3 Projected volume versus other volumes
    4. II.4 How universal is the projected volume scaling?
    5. II.5 Other comments
  3. III A simple holographic example: strips on a boosted time slice
    1. III.1 Boosted Black Brane
    2. III.2 The special case of the BTZ black hole
  4. IV A more general argument
    1. IV.1 The $d=3$ case
      1. IV.1.1 The near-boundary expansions
      2. IV.1.2 The large-$R$ expansion
      3. IV.1.3 The volume law
  5. V Conclusion
  6. A More on the large-$R$ expansion
  7. B Volume-law scaling in arbitrary dimensions

##  I Introduction

The entanglement entropy of a subregion of a field theory in a thermal state
has an extensive component that scales with the volume of the subregion, when
the subregion is large. In holography, previous work by Liu-Mezei [1, 2] has
shown that the Hubeny-Rangamani-Takayanagi (HRT) surface [7, 8, 9] is
consistent with this volume scaling in static situations. See also [3, 4, 5,
6] for related techniques. In time-dependent situations, it is widely expected
that the HRT surface continues to be consistent with the volume law, at least
for an appropriate class of nonthermal states. In this paper, we take the
first steps toward generalizing the Liu-Mezei technique to a particular kind
of nonstatic situation: we will consider entangling subregions that break
staticity in thermal states (and not states with general time dependence).

The rest of the paper is organized as follows. In Sec. II, we describe our
intuition from the field theory side. In Sec. III, we check the volume law for
strips lying on a boosted time slice on the boundary of a black brane
spacetime. In Sec. IV, we present a more general argument for any subregion
that has reflection symmetry. Finally, in Sec. V, we conclude. Some of the
more technical results are relegated to the Appendixes.

##  II Expectation from the field theory side

Intuitively, we expect that the entanglement entropy of the boosted thermal
state is proportional to the volume of the projection of the subregion onto a
“static time slice” (i.e., a slice on which the stress-energy tensor of the
field theory is diagonal). To see this, we can consider a system of qubits at
equal separation in space, each of which is in a maximally mixed state (so, in
particular, the qubits are not entangled among them). Furthermore, let us
suppose that the qubits are stationary, so their worldline is perpendicular to
the $t=0$ slice, as depicted in Fig. 1.

Figure 1: The worldlines of the qubits are in blue. The unboosted and boosted
subregions are in red.

Now consider two subregions (depicted in red in Fig. 1), one of which is on
the unboosted time slice and the other one is on a boosted time slice, such
that the projection of the boosted subregion onto the unboosted time slice is
the unboosted subregion. Simply by counting the number of qubits contained in
the subregions, we see that the entanglement entropy of the two subregions are
equal to each other. In other words, the entanglement entropy of a boosted
subregion scales with the volume of the projection onto a static time slice,
with the same coefficient as in volume-law entropy on the static time slice.

Next, we back up this intuition with a few simple computations. Consider a
field theory in the thermal state $\rho=\frac{e^{-\beta H}}{Z}$ with
$Z=\mathrm{Tr}{(e^{-\beta H})}$. This is the state on the $t=0$ slice. We now
boost the state in the $z$ direction by conjugating $\rho$ with the unitary
operator implement the boost. Using the Poincaré algebra, we find the new
state on the $t=0$ slice to be $\rho_{boosted}=\frac{e^{-\beta
H_{boosted}}}{Z}$ with $H_{boosted}=\gamma H-v\gamma P_{z}$, where $v$ is the
boost velocity, $\gamma$ is the dilation factor $\gamma=1/\sqrt{1-v^{2}}$, and
$P_{z}$ is the translation generator along the $z$ direction.

It will be convenient to rewrite the boosted state in the Gibbs form,

$\rho_{boosted}=\frac{e^{-\beta_{boosted}(H-vP_{z})}}{Z}$ (1)

with $\beta_{boosted}=\gamma\beta$ the “boosted temperature,” and with
$vP_{z}$ playing the role of a work term similar to the product of the
chemical potential and the number of particles. Thus, the state on the $t=0$
slice is a grand canonical ensemble with the unconventional feature that the
temperature is $v$ dependent.

Computing the von Neumann entropy $S_{vN}\equiv-\mathrm{Tr}{(\rho\ln\rho)}$ of
the state, we find

$S_{vN}=\beta_{boosted}\langle H-vP_{z}\rangle_{boosted}+\ln Z$ (2)

and recognize the formula for the Gibbs entropy.

###  II.1 Free massless scalar field

We now specialize to the free massless scalar field in $d+1$ dimensions, and
compute the first term of (2). The expectation value $\langle
H-vP_{z}\rangle_{boosted}$ is

$\langle
H-vP_{z}\rangle_{boosted}=V\int\frac{d^{d}k}{(2\pi)^{d}}\frac{|\textbf{k}|-vk_{z}}{e^{\beta_{boosted}(|\textbf{k}|-vk_{z})}-1},$
(3)

where $V$ is the proper volume on the $t=0$ slice. Using spherical coordinates
in momentum space,

$\langle H-vP_{z}\rangle_{boosted}=\frac{V}{(2\pi)^{d}}\int d\Omega_{d-1}\int
dkk^{d-1}\frac{k(1-v\cos{\theta})}{e^{\beta_{boosted}k(1-v\cos{\theta})}-1},$
(4)

where $\theta$ is the polar angle. By doing the change of variable
$q=\beta_{boosted}k(1-v\cos{\theta})$, one can show that the integral above is
$v$ independent. So we find

$\langle H-vP_{z}\rangle_{boosted}=\langle H\rangle_{unboosted}.$ (5)

The second term of (2) can be computed in a similar manner, and yields very
similar results. So, the entropy is $S_{thermal}\sim\beta_{boosted}\langle
H-vP_{z}\rangle_{boosted}$, which is also $\beta_{boosted}\langle
H\rangle_{unboosted}$. So the $v$ dependence in the thermal entropy is through
a Lorentz boost factor contained in the boosted temperature. That boost factor
can be combined with the proper volume $V$ on the $t=0$ slice to give the
projected volume onto the time slice on which the stress tensor is diagonal.
Thus, we have checked the intuition described in Fig. 1 in the setting of a
free scalar field.

###  II.2 Conformal perfect fluid

Next, instead of a massless free scalar field, let us specialize Eq. (2) to
the case of a conformal fluid in $d+1$ dimensions. The stress-energy tensor of
such a fluid takes the form

$\langle T^{\mu\nu}\rangle=\alpha T^{d+1}(\eta^{\mu\nu}+(d+1)u^{\mu}u^{\nu})$
(6)

for some constant $\alpha$, and where $u^{\mu}$ is the four velocity of the
fluid. This stress tensor is that of a conformal field theory (CFT) dual to a
boosted Schwarzschild-anti-de Sitter (AdS) black hole, so we would expect that
lessons learned here might carry over to other holographic scenarios. In
addition, the Hamiltonian and translation generator are given in terms of the
stress-energy tensor by the standard formulas

$\displaystyle H$ $\displaystyle=\int T_{tt}~{}d^{d}x,$ (7) $\displaystyle
P_{z}$ $\displaystyle=\int T_{zt}~{}d^{d}x.$ (8)

We now compute the $\langle H-vP_{z}\rangle_{boosted}$ in (2). We have

$\langle H-vP_{z}\rangle_{boosted}=\int\left(\langle
T_{tt}\rangle_{boosted}-v\langle T_{zt}\rangle_{boosted}\right)d^{d}x,$ (9)

where $\langle T_{\mu\nu}\rangle_{boosted}$ can be obtained from (6) with the
four velocity taken to be $u^{t}=\gamma$, $u^{z}=-\gamma v$ (note the negative
sign) and all other spatial components vanishing,

$\displaystyle\langle T_{tt}\rangle_{boosted}$ $\displaystyle=\alpha
T^{d+1}(-1+(d+1)\gamma^{2}),$ (10) $\displaystyle\langle
T_{zt}\rangle_{boosted}$ $\displaystyle=\alpha T^{d+1}(d+1)\gamma^{2}v.$ (11)

We then find

$\langle H-vP_{z}\rangle_{boosted}=\alpha T^{d+1}d\int d^{d}x.$ (12)

Now let us compare the answer above with $\langle H\rangle_{unboosted}$. We
have

$\langle H\rangle_{unboosted}=\int\langle T_{tt}\rangle_{unboosted}~{}d^{d}x,$
(13)

where $\langle T_{tt}\rangle_{unboosted}$ is given by (6) with $u^{t}=1$ (and
all other components vanishing),

$\langle T_{tt}\rangle_{unboosted}=\alpha T^{d+1}d.$ (14)

We therefore find

$\langle H-vP_{z}\rangle_{boosted}=\langle H\rangle_{unboosted},$ (15)

just like in the free scalar field case. Beyond this point, the steps are the
same as for the free scalar field case: the von Neumann entropy is
$\sim\beta_{boosted}\langle H-vP_{z}\rangle_{boosted}=\beta_{boosted}\langle
H\rangle_{unboosted}$. The $\gamma$ factor from the boosted inverse
temperature can be combined with the volume factor, yielding the projected
volume.

###  II.3 Projected volume versus other volumes

_A priori_ , one might naively think that the entropy of a subregion lying on
a boosted time slice (in the thermal state) should scale with the proper
volume of the subregion on that time slice. To see why, let us consider a
subregion $A$ lying on an unboosted time slice, and the unitary operator $U$
which implements the boost. $U$ is the exponential of a local integral of the
stress-energy tensor (i.e., the boost generator), so it might seem that $U$ is
the tensor product of local unitaries,

$U\stackrel{{\scriptstyle?}}{{=}}U_{A}\otimes U_{\bar{A}},$ (16)

where $U_{A}$ is the exponential of the integral over the subregion $A$, and
$U_{\bar{A}}$ is the exponential of the integral over the complement
$\bar{A}$. If the equation above is true, then we would conclude that $U$ has
no effect on the entanglement entropy, and thus the volume law is still with
respect to the proper volume on the (now boosted) slice.

However, we know that $U$ does not have the local unitary form as written
above (see [10, 11] for recent studies of the effects on modular Hamiltonians
of the failure of such unitaries to be local). An intuitive way to understand
why $U$ does not factorize is by noting that the stress-energy tensor moves
local operators around, and as such the “factorization” is a mixed combination
of $A$ and $\bar{A}$. Thus, we think that the scaling of the entropy with the
projected volume rather than the proper volume is a consequence of a nonlocal
boost unitary.

###  II.4 How universal is the projected volume scaling?

It is easy enough to find counterexamples to the projected volume statement if
we consider states other than a thermal state. For example, a quenched state –
where entropy evolves in time – is easily seen to be a counterexample. Hence,
we believe the projected volume law should only hold in a thermal state while
holding true for subregions of arbitrary shape. However, this projected volume
law should receive corrections after moving away from a thermal state.

It might be surprising that such a universal behavior can be true, given that
entanglement entropy for a CFT on the torus is nonuniversal and depends on the
operator contents of the theory [12]. However, we note that the projected
volume law is a statement only for very large subregions (larger than the
thermal scale). So, effectively, we are in a high-temperature regime where it
is conceivable that universal statements can be made.

###  II.5 Other comments

We also note the argument in [13], where the authors used strong subadditivity
to argue in a special case that the entanglement entropy on a null strip is
equal to the entanglement entropy of its projection.

##  III A simple holographic example: strips on a boosted time slice

We now move on to discuss holography. In this setting, we can also expect the
scaling of the entropy with the projected volume to come out, basically
because of the coordinate singularity at the horizon of a black hole
spacetime.

We note that what we are after is a finite contribution to the entropy, as
illustrated from the field theory examples of the previous section. As such,
we do not need to use an explicit renormalization scheme to deal with the UV-
divergent part (i.e. the vacuum part) of the entanglement entropy.
Holographically, we impose a fixed near-boundary cutoff at some small value of
the Fefferman-Graham coordinate $z=\epsilon$, and send the size of the
boundary subregion to infinity while keeping $\epsilon$ fixed.

###  III.1 Boosted Black Brane

We first consider a simple example: a strip boundary subregion in a
$3+1$-dimensional boosted black brane (i.e., the strip lies on a boosted time
slice on the boundary). The metric of the Schwarzschild black brane is

$ds^{2}=\frac{L^{2}}{z^{2}}\left[-f(z)dt^{2}+\frac{dz^{2}}{f(z)}+dx^{2}+dy^{2}\right],$
(17)

with $f(z)=1-\frac{z^{3}}{z_{h}^{3}}$, and $z_{h}$ is the $z$ coordinate of
the horizon. Throughout this paper, we will set $z_{h}=1$ for convenience. We
could work with boosted coordinates, but it will be simpler to work with the
static coordinates above. The strip is delimited by the two lines at $x=\pm
R$, $t=\pm t_{0}$ for some half-width $R$ and time $t_{0}$. We parametrize the
HRT surface as $t(z)$, $x(z)$. The induced metric on a codimension-2 spacelike
surface anchored at the strip is

$ds^{2}=\frac{L^{2}}{z^{2}}\left[\left(\frac{1}{f}-ft^{\prime 2}+x^{\prime
2}\right)dz^{2}+dy^{2}\right],$ (18)

and the area functional $\mathcal{A}$ of the surface (evaluated over a segment
of length $\Delta y$ along the $y$ direction) is

$\displaystyle\mathcal{A}$ $\displaystyle=2\Delta
y\int_{\delta}^{z_{m}}dz\frac{L^{2}}{z^{2}}\sqrt{Q},$ (19) $\displaystyle Q$
$\displaystyle=\frac{1}{f}-ft^{\prime 2}+x^{\prime 2},$ (20)

where $\delta$ is a near-boundary cutoff, and $z_{m}$ is the $z$ coordinate of
the “tip” of the HRT surface. The Lagrangian above is independent of $t$ and
of $x$, hence there are two conserved momenta:

$\displaystyle p_{t}$ $\displaystyle=\frac{\partial\mathcal{L}}{\partial
t^{\prime}}=-\left(\frac{L}{z}\right)^{2}\frac{ft^{\prime}}{\sqrt{Q}},$ (21)
$\displaystyle p_{x}$ $\displaystyle=\frac{\partial\mathcal{L}}{\partial
x^{\prime}}=\left(\frac{L^{2}}{z^{2}}\right)\frac{x^{\prime}}{\sqrt{Q}}.$ (22)

We can solve algebraically for $x^{\prime}$ and $t^{\prime}$ as functions of
$z$,

$\displaystyle x^{\prime}$
$\displaystyle=\frac{p_{x}}{\sqrt{p_{t}^{2}+f\left(\frac{L^{4}}{z^{4}}-p_{x}^{2}\right)}},$
(23) $\displaystyle t^{\prime}$
$\displaystyle=-\frac{p_{t}}{f}\frac{1}{\sqrt{p_{t}^{2}+f\left(\frac{L^{4}}{z^{4}}-p_{x}^{2}\right)}},$
(24)

where we have assumed $x^{\prime}<0$ and $t^{\prime}<0$. These assumptions
should hold true over half the HRT surface, and on the other half we have
$x^{\prime}>0$ and $t^{\prime}>0$. Note that these assumptions imply $p_{x}<0$
and $p_{t}>0$. We can then integrate the two equations above to get the HRT
surface,

$\displaystyle x(z)$
$\displaystyle=R+\int_{0}^{z}dz^{\prime}\frac{p_{x}}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}},$ (25) $\displaystyle t(z)$
$\displaystyle=t_{0}-\int_{0}^{z}dz^{\prime}\frac{p_{t}}{f(z^{\prime})}\frac{1}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}}.$ (26)

The two constants $R$ and $t_{0}$ are determined by the momenta $p_{x}$ and
$p_{t}$. To see that, we evaluate the two functions above at $z=z_{m}$ and use
the fact that $x(z_{m})=t(z_{m})=0$. To see that the $x$ and $t$ coordinates
of the tip of the HRT surface are zero, we note that both the boundary
subregion and the bulk metric are symmetric under the double reflection
$x\rightarrow-x$ and $t\rightarrow-t$. Therefore, the tip of the HRT surface
must be preserved under that reflection, hence it is located at $x=t=0$. We
then obtain

$\displaystyle R$
$\displaystyle=-\int_{0}^{z_{m}}dz^{\prime}\frac{p_{x}}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}},$ (27) $\displaystyle t_{0}$
$\displaystyle=\int_{0}^{z_{m}}dz^{\prime}\frac{p_{t}}{f(z^{\prime})}\frac{1}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}}.$ (28)

In the two equations above, $z_{m}$ itself is of course determined by $p_{t}$
and $p_{x}$, via the condition that $x^{\prime}$ blows up at $z=z_{m}$.
Explicitly, this condition is

$p_{t}^{2}+f(z_{m})\left(\frac{L^{4}}{z_{m}^{4}}-p_{x}^{2}\right)=0.$ (29)

Let us solve for $p_{t}$ as a function of $z_{m}$ and $p_{x}$ from the above,
and plug into the two previous equations. Note that the above forces a lower
bound on $|p_{x}|$,

$|p_{x}|\geq\frac{L^{2}}{z_{m}^{2}}.$ (30)

We will call the right-hand side above the negative of the critical value of
$p_{x}$ and denote it by $|p_{x,crit}|$. We then obtain

$R=-\int_{0}^{z_{m}}dz^{\prime}\frac{p_{x}}{\sqrt{H(z^{\prime})}},$ (31)
$t_{0}=\int_{0}^{z_{m}}\frac{dz^{\prime}}{f(z^{\prime})}\frac{\sqrt{-f(z_{m})\left(\frac{L^{4}}{z_{m}^{4}}-p_{x}^{2}\right)}}{\sqrt{H(z^{\prime})}},$
(32)

with

$H(z)\equiv
f(z)\left(\frac{L^{4}}{z^{4}}-p_{x}^{2}\right)-f(z_{m})\left(\frac{L^{4}}{z_{m}^{4}}-p_{x}^{2}\right).$
(33)

The two equations above tell us how $R$ and $t_{0}$ are related to $p_{x}$ and
$z_{m}$. The area functional becomes:

$\mathcal{A}=2\Delta y\int_{\delta}^{z_{m}}dz^{\prime}\frac{L^{4}}{z^{\prime
4}}\frac{1}{\sqrt{H(z^{\prime})}}.$ (34)

Next, we take the large-$R$ limit, with the ratio $R/t_{0}$ kept fixed. We
want to keep this latter ratio fixed so that the strip remains on the same
boosted time slice as we make the half-width larger and larger. We expect that
$z_{m}\approx z_{h}=1$ in the large-$R$ limit (indeed it is well known that
the horizon acts as a barrier to extremal surfaces [14, 15]). So, for the rest
of this section, $z_{m}$ will be taken to be close to the horizon. We also
need to find the approximate value of $p_{x}$. To do so, we need to think
about how the function $H(z)$ looks like.

For generic values of $z_{m}$, $z_{h}$, and $p_{x}$, the function $H(z)$ has
two zeros, one of which is at $z=z_{m}$, and a minimum between the two zeros.
Let $z=z_{M}$ denote the location of this minimum. In addition to the lower
bound (30), $p_{x}$ is also constrained by the fact that we should only
consider values of $p_{x}$ for which $z_{m}$ is the smaller of the two zeros.
Indeed, in the other case, we have that $H(z)$ is negative when $z$ approaches
$z_{m}$. But $H(z)$ occurs under a square root in the $R$ integral (31), so
the integrand is not real near the upper limit of integration (and the
integral has to be real). We plot $H(z)$ for a few representative values of
the parameters in Fig. 2.

Figure 2: Plot of $H(z)$ versus $z$ for three different choices of parameters.
For all three curves, we set $L=1$, $z_{m}=0.9$, and $z_{h}=1$. The values of
$p_{x}$ differ from curve to curve: $p_{x}=-1.7$ (blue), $p_{x}=-1.3$
(yellow), and $p_{x}=(L/z_{m})^{2}\approx-1.234$ (green). The blue curve is in
the unphysical parameter region, since $z_{m}$ is the larger of the two zeros.
The yellow and green curves are in the physical parameter region. At the
critical value of $p_{x}$, the two roots of $H(z)$ are $z_{m}$ and $z_{h}$, as
can be seen by inspecting the green curve.

From the plot of $H(z)$ and Eq. (31), we see that the large-$R$ regime
corresponds to the regime where $z_{M}$ is close to $z_{m}$. In this regime,
we can check this divergence of $R$ by replacing the $H(z)$ in (31) with its
expansion to second order around the minimum,

$H(z)=H(z_{M})+H_{2}(z-z_{M})^{2}+\dots,$ (35)

where

$H_{2}=\frac{1}{2}f^{\prime\prime}(z_{M})\left(\frac{L^{4}}{z_{M}^{4}}-p_{x}^{2}\right)-\frac{4L^{4}}{z_{M}^{5}}f^{\prime}(z_{M})+\frac{10L^{4}}{z_{M}^{6}}f(z_{M}).$
(36)

The precise value of $H_{2}$ will not be important for our purposes. The $R$
integral then takes the form

$R\approx-\int_{z_{r}-\delta}^{z_{r}}dz^{\prime}\frac{p_{x}}{\sqrt{H(z_{M})+H_{2}(z^{\prime}-z_{M})^{2}}},$
(37)

where $z_{r}$ is the zero of the quadratic expression (35), which is smaller
than $z_{M}$, and $\delta$ is some small number.111We note that although
$\delta$ is small – it is not an infinitesimal quantity. We are expanding in
the regime where $z_{M}-z_{r}$ is small. Evaluating the integral, we find

$R\approx\frac{p_{x}}{2\sqrt{H_{2}}}\log{|H(z_{M})|}.$ (38)

So, $R$ diverges when $H(z_{M})=0$, which is to say that the zero at $z=z_{m}$
is also a minimum of $H(z)$. To find out which value of $p_{x}$ corresponds to
that, we solve for $p_{x}$ from the equation $H^{\prime}(z_{m})=0$, and find

$p_{x}=p_{x,crit}.$ (39)

So, $p_{x}$ has to be the critical value for $R$ to diverge.

We now replace the $p_{x}$ in the $R$ integral (31) by the critical value

$R\approx\int_{0}^{z_{m}}dz^{\prime}\frac{|p_{x,crit}|}{\sqrt{H(z^{\prime})}}.$
(40)

Also, we can approximate the area functional (34) by setting some of the
$z^{\prime}$ dependence in the integrand to $z_{m}$ (which is near the
horizon),

$\mathcal{A}\approx 2\Delta
y\int_{\delta}^{z_{m}}dz^{\prime}\frac{L^{2}}{z_{m}^{2}}\frac{|p_{x,crit}|}{\sqrt{H(z^{\prime})}}.$
(41)

Comparing the two previous equations, and dividing by $4G_{N}$, we see that
the entanglement entropy $S$ is

$S\approx s2R\Delta y,$ (42)

with the entropy density $s$ given by

$s=\frac{L^{2}}{4G_{N}z_{h}^{2}}$ (43)

(where we restored $z_{h}$ for clarity). Thus, we have obtained the statement
of volume-law entropy, with the volume given by $2R\Delta y$. We note that
this is the volume of the projection of the strip onto a static time slice,
which is independent of the boost angle (or the ratio $R/t_{0}$) and not the
proper volume of the strip. We also note that, in the analysis above, we did
not need to find out how the ratio $R/t_{0}$ depends on $z_{m}$ and $p_{x}$:
for all values of the boost angle, we have that $z_{m}\approx z_{h}$ and
$p_{x}\approx p_{x,crit}$ in the large-$R$ limit. The exact boost angle of the
strip will depend on how close $z_{m}$ is to $z_{h}$, in comparison with how
close $p_{x}$ is to $p_{x,crit}$.

###  III.2 The special case of the BTZ black hole

Finally, we discuss the special case of the Bañados-Teitelboim-Zanelli (BTZ)
black hole, where boundary subregions are simply intervals. In this case, a
more explicit treatment can be given when compared to the higher-dimensional
cases, since geodesics anchored at the two end points of a boosted boundary
interval can be found explicitly.

Our strategy to find these geodesics will be to map from the Poincaré patch,
where geodesics are easy to write down, to BTZ by a coordinate transformation.
The three-dimensional Poincaré patch metric is

$ds^{2}=\frac{L^{2}}{z^{2}}(-dt^{2}+dx^{2}+dz^{2}).$ (44)

The geodesic connecting the point $(t=0,x=0)$ on the boundary of AdS to the
point $(t=t_{0},x=x_{0})$ on the boundary of AdS can be given in parametric
form by

$z(u)=\sqrt{u(1-u)(-t_{0}^{2}+x_{0}^{2})},$ (45) $t(u)=ut_{0},$ (46)
$x(u)=ux_{0},$ (47)

where $u$ is a parameter ranging from $0$ to $1$. To obtain the expressions
above, we can start by a geodesic on the $t=0$ time slice in the Poincaré
patch and boost it (along the boundary direction $x$). The result is
guaranteed to be a geodesic because boosting along $x$ is an isometry of the
spacetime. By eliminating the parameter $u$, we can also describe the
geodesics by the two functions $t(x)$ and $z(x)$.

In the remainder of this subsection, we set $L=1$ for convenience (but we will
restore $L$ in the next section). We map to AdS-Rindler (otherwise known as
the planar BTZ black hole), by the coordinate transformation

$x^{\pm}=\pm\sqrt{1-r^{-2}}e^{\pm\sigma^{\pm}},$ (48)
$z=\frac{e^{\sigma}}{r},$ (49)

where $x^{\pm}\equiv t\pm x$ and $\sigma^{\pm}\equiv\tau\pm\sigma$. The AdS-
Rindler metric is

$ds^{2}=-(r^{2}-1)d\tau^{2}+\frac{dr^{2}}{r^{2}-1}+r^{2}d\sigma^{2},$ (50)

and the image of the geodesic (45)–(47) is found to be

$\displaystyle\tau(\sigma)$ $\displaystyle=$
$\displaystyle\mathrm{arctanh}\bigg{[}$
$\displaystyle\frac{\sinh{\tau_{R}}\sinh{(\sigma-\sigma_{L})}+\sinh{\tau_{L}}\sinh{(\sigma_{R}-\sigma)}}{\cosh{\tau_{R}}\sinh{(\sigma-\sigma_{L})}+\cosh{\tau_{L}}\sinh{(\sigma_{R}-\sigma)}}\bigg{]},$
$\displaystyle r(\sigma)$ $\displaystyle=$
$\displaystyle\sinh{(\sigma_{R}-\sigma_{L})}\sqrt{\frac{\mathrm{csch}(\sigma-\sigma_{L})\mathrm{csch}(\sigma_{R}-\sigma)}{2[\cosh{(\sigma_{R}-\sigma_{L})}-\cosh{(\tau_{R}-\tau_{L})}}}.$

In the expression above, $(\tau_{L},\sigma_{L})$ and $(\tau_{R},\sigma_{R})$
are the boundary coordinates of the two end points of the geodesic,
respectively. Beyond this point, we will use translation symmetry to set
$-\tau_{L}=\tau_{R}\equiv\tau_{0}$ and $-\sigma_{L}=\sigma_{R}\equiv R$. Using
the explicit formulas for the geodesics above, we can write down the area of
the HRT surface (i.e., the length of the geodesic). After some straightforward
but tedious maths, we obtain

$\displaystyle\mathrm{Area}$ $\displaystyle=$ $\displaystyle
2\sinh{(2R)}\int_{r_{tip}}^{r_{cutoff}}\frac{dr}{r}$
$\displaystyle\bigg{[}\bigg{(}\cosh{2R}-\frac{\sinh^{2}{2R}}{r^{2}(\cosh{2R}-\cosh{2\tau_{0}})}\bigg{)}^{2}-1\bigg{]}^{-1/2},$

with $r_{tip}$ the $r$ coordinate of the tip of the HRT surface, given in
terms of $R$ and $\tau_{0}$ by

$r_{tip}=\frac{\sqrt{2}\cosh{R}}{\sqrt{\cosh{2R}-\cosh{2\tau_{0}}}}$ (54)

and $r_{cutoff}$ some large near-boundary cutoff radial coordinate. The two
expressions above specify the area as a function of $R$ and $\tau_{0}$ or,
equivalently as a function of $R$ and the ratio $\frac{\tau_{0}}{R}$. To check
the projected volume law, we can plot numerically the area as a function of
$R$, at various fixed values of the ratio $\frac{\tau_{0}}{R}$. We show the
plots in Fig. 3.

Figure 3: Plot of the area of the HRT surface versus $R$, for three fixed
values of $\frac{\tau_{0}}{R}$ (namely, $\frac{\tau_{0}}{R}=0.1$ for the blue
curve, $\frac{\tau_{0}}{R}=0.9$ for the orange curve, and
$\frac{\tau_{0}}{R}=0.99$ for the green curve). The $r_{cutoff}$ has been set
to $100$.

As can be seen Fig. 3, the area scales linearly with $R$, with a slope that is
independent of the ratio $\frac{\tau_{0}}{R}$. In this way, we have checked
numerically the projected volume law.

##  IV A more general argument

In this section, we consider more general subregions. We will consider the
$d=3$ case to keep the notation simple and generalize to arbitrary dimensions
in Appendix B.

###  IV.1 The $d=3$ case

We still work with static coordinates in the bulk, but with polar coordinates
instead of Cartesian ones,

$ds^{2}=\frac{L^{2}}{z^{2}}\left[-fdt^{2}+\frac{dz^{2}}{f}+d\rho^{2}+\rho^{2}d\phi^{2}\right],$
(55)

and we consider a boundary subregion described by $\rho=R\tilde{\rho}(\phi)$
and $t=R\tilde{t}(\phi)$, where $R$ is a scaling parameter, which we will take
to be large. The boundary subregion needs not lie on a static time slice or
even a boosted time slice, but we will require that it is symmetric under
$(t,x,y)\rightarrow(-t,-x,-y)$.

The HRT surface is described by $\rho(z,\phi)$ and $t(z,\phi)$. The area
functional is

$\mathcal{A}=\int_{\delta}^{z_{m}}dz\int_{0}^{2\pi}d\phi\mathcal{L}=\int_{\delta}^{z_{m}}dz\int_{0}^{2\pi}d\phi\frac{L^{2}}{z^{2}}\sqrt{Q}$
(56)

with

$Q=-f(t^{\prime}\rho_{,\phi}-\rho^{\prime}t_{,\phi})^{2}+\rho^{2}(-ft^{\prime
2}+\frac{1}{f}+\rho^{\prime 2})-t_{,\phi}^{2}+\frac{\rho_{,\phi}^{2}}{f}.$
(57)

The trick is to study the variation of $A$ with respect to $R$. By a standard
Hamilton-Jacobi argument, we have that this variation is

$\frac{d\mathcal{A}}{dR}=\int
d\phi\left(\Pi_{\rho}^{z}\frac{d\rho}{dR}\bigg{|}^{z_{m}}_{\delta}+\Pi_{t}^{z}\frac{dt}{dR}\bigg{|}_{\delta}^{z_{m}}-\mathcal{H}(z_{m})\frac{dz_{m}}{dR}\right),$
(58)

where $\Pi_{\rho}^{z}\equiv\frac{\partial\mathcal{L}}{\partial\rho^{\prime}}$
and $\Pi_{t}^{z}\equiv\frac{\partial\mathcal{L}}{\partial t^{\prime}}$ are the
conjugate momenta. The prime denotes differentiation with respect to $z$.
Explicitly,

$\displaystyle\Pi^{z}_{\rho}$
$\displaystyle=\frac{L^{2}}{z^{2}\sqrt{Q}}\left[f(t^{\prime}\rho_{,\phi}-\rho^{\prime}t_{,\phi})t_{,\phi}+\rho^{2}\rho^{\prime}\right],$
(59) $\displaystyle\Pi^{z}_{t}$
$\displaystyle=-\frac{L^{2}}{z^{2}\sqrt{Q}}\bigg{[}f(t^{\prime}\rho_{,\phi}-\rho^{\prime}t_{,\phi})\rho_{,\phi}+\rho^{2}ft^{\prime}\bigg{]},$
(60)

and
$\mathcal{H}\equiv\Pi^{z}_{\rho}\rho^{\prime}+\Pi^{z}_{t}t^{\prime}-\mathcal{L}$
is the Hamiltonian density. Explicitly,

$\mathcal{H}=\frac{L^{2}}{z^{2}\sqrt{Q}}\left(-\frac{\rho^{2}}{f}-\frac{\rho_{,\phi}^{2}}{f}+t_{,\phi}^{2}\right).$
(61)

In Eq. (58), the first two terms on the right-hand side come from the change
in the shape of the HRT surface as $R$ is varied, and the last term comes from
the change in $z_{m}$ as $R$ is varied. Technically, the right-hand side of
(58) includes also the term $\int dz\int
d\phi\partial_{\phi}\bigg{(}\Pi^{\phi}_{\rho}\frac{d\rho}{dR}+\Pi^{\phi}_{t}\frac{dt}{dR}\bigg{)}$,
but this latter term vanishes by periodicity in $\phi$.

From the explicit expressions (73) and (74), we see that those conjugate
momenta vanish when evaluated at $z=z_{m}$. This is because partial
derivatives of $\rho$ or $t$ with respect to $\phi$ vanish at the tip of the
HRT surface, by regularity, and also because $\rho(z=z_{m})$ vanishes. For the
same reasons, we see that the Hamiltonian density $\mathcal{H}$ vanishes when
evaluated at $z=z_{m}$. So Eq. (58) simplifies to

$\frac{d\mathcal{A}}{dR}=-\int
d\phi\left(\Pi_{\rho}^{z}\frac{d\rho}{dR}\bigg{|}_{\delta}+\Pi_{t}^{z}\frac{dt}{dR}\bigg{|}_{\delta}\right).$
(62)

####  IV.1.1 The near-boundary expansions

Because the right-hand side above is evaluated at the near-boundary cutoff
$z=\delta$, we see that it is enough to know how the HRT surface looks like
near the boundary to know $\frac{dA}{dR}$. Let us expand $\rho(z,\phi)$ and
$t(z,\phi)$ near the boundary (i.e., as series in $z$),

$\displaystyle\rho$
$\displaystyle=R\tilde{\rho}(\phi)+c_{1}(R,\phi)z+c_{2}{(R,\phi)}z^{2}+\dots,$
(63) $\displaystyle t$
$\displaystyle=R\tilde{t}(\phi)+d_{1}(R,\phi)z+d_{2}{(R,\phi)}z^{2}+\dots,$
(64)

where we will refer to the coefficients $c_{i}$ and $d_{i}$ as the Fefferman-
Graham (FG) coefficients. We claim that the first coefficients $c_{1}$ and
$d_{1}$ vanish. To see this, we will need to solve the equations of motion
perturbatively near the boundary. From the area functional (56) in the $d=3$
case, we find the equations of motion,

$\displaystyle-\frac{2}{z}Q\frac{\partial
Q}{\partial\rho^{\prime}}-\frac{1}{2}\frac{\partial Q}{\partial
z}\frac{\partial Q}{\partial\rho^{\prime}}+Q\frac{\partial}{\partial
z}\frac{\partial Q}{\partial\rho^{\prime}}$ (65) $\displaystyle-$
$\displaystyle\frac{1}{2}\frac{\partial Q}{\partial\phi}\frac{\partial
Q}{\partial\rho_{,\phi}}+Q\frac{\partial}{\partial\phi}\frac{\partial
Q}{\partial\rho_{,\phi}}=0,$ $\displaystyle-\frac{2}{z}Q\frac{\partial
Q}{\partial t^{\prime}}-\frac{1}{2}\frac{\partial Q}{\partial z}\frac{\partial
Q}{\partial t^{\prime}}+Q\frac{\partial}{\partial z}\frac{\partial Q}{\partial
t^{\prime}}$ (66) $\displaystyle-$ $\displaystyle\frac{1}{2}\frac{\partial
Q}{\partial\phi}\frac{\partial Q}{\partial
t_{,\phi}}+Q\frac{\partial}{\partial\phi}\frac{\partial Q}{\partial
t_{,\phi}}=0.$

If we plug the near-boundary expansions (63) and (64) into the equations of
motion above and demand that the equations of motion are satisfied at each
order in $z$, we find that the order $z^{-1}$ yields

$-\frac{2}{z}Q^{(0)}\left(\frac{\partial
Q}{\partial\rho^{\prime}}\right)^{(0)}=0,$ (67)
$-\frac{2}{z}Q^{(0)}\left(\frac{\partial Q}{\partial
t^{\prime}}\right)^{(0)}=0,$ (68)

where the superscript $(0)$ means the term is zeroth order in $z$. We have

$Q^{(0)}=-(d_{1}R\tilde{\rho}_{,\phi}-c_{1}R\tilde{t}_{,\phi})^{2}+R^{2}\tilde{\rho}^{2}(1+c_{1}^{2}-d_{1}^{2})-R^{2}\tilde{t}_{,\phi}^{2}+R^{2}\tilde{\rho}_{,\phi}^{2},$
(69) $\left(\frac{\partial
Q}{\partial\rho^{\prime}}\right)^{(0)}=2R^{2}[c_{1}(\tilde{\rho}^{2}-\tilde{t}_{,\phi}^{2})+d_{1}\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}],$
(70) $\left(\frac{\partial Q}{\partial
t^{\prime}}\right)^{(0)}=2R^{2}[c_{1}\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}-d_{1}(\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2})].$
(71)

The solution to (67) and (68) is, as claimed,

$c_{1}=d_{1}=0.$ (72)

We now plug the near-boundary expansions for $\rho$ and $t$ above into
$\Pi^{z}_{\rho}$ and $\Pi^{z}_{t}$ and use the fact that the first-order
coefficients vanish. We then find, in the $\delta\rightarrow 0$ limit

$\Pi^{z}_{\rho}=\frac{3L^{2}R(d_{3}\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}+c_{3}\tilde{\rho}^{2}-c_{3}\tilde{t}_{,\phi}^{2})}{\sqrt{\tilde{\rho}^{2}-\tilde{t}_{,\phi}^{2}+\tilde{\rho}_{,\phi}^{2}}}+\dots,$
(73)
$\Pi^{z}_{t}=\frac{3L^{2}R(c_{3}\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}-d_{3}\tilde{\rho}^{2}-d_{3}\tilde{\rho}_{,\phi}^{2})}{\sqrt{\tilde{\rho}^{2}-\tilde{t}_{,\phi}^{2}+\tilde{\rho}_{,\phi}^{2}}}+\dots,$
(74)

where the $\dots$ stands for terms that either diverge or go to zero as
$\delta\rightarrow 0$. In other words, we have only written down the term that
remains finite in the $\delta\rightarrow 0$ limit, because intuitively we
expect this finite contribution to give rise to volume-law entanglement
entropy. To further stress the point that we are interested only in the finite
contributions to the entanglement entropy, we are not including divergent
terms associated with the vacuum entropy that should be regularized. We note
also that the quantity
$\tilde{\rho}^{2}-\tilde{t}_{,\phi}^{2}+\tilde{\rho}_{,\phi}^{2}$ is positive
since the subregion is spacelike.

####  IV.1.2 The large-$R$ expansion

In the large-$R$ limit, we guess that $\rho$ and $t$ can be expanded in $1/R$
as

$\displaystyle\rho$
$\displaystyle=R\tilde{\rho}{(\phi)}-\rho_{0}{(z,\phi)}+\frac{\rho_{1}(z,\phi)}{R}+\frac{\rho_{2}(z,\phi)}{R^{2}}+\dots,$
(75) $\displaystyle t$
$\displaystyle=R\tilde{t}{(\phi)}-t_{0}{(z,\phi)}+\frac{t_{1}(z,\phi)}{R}+\frac{t_{2}(z,\phi)}{R^{2}}+\dots$
(76)

(without noninteger powers of $R$). Then $Q$ can be expanded in a series in
$1/R$ without any noninteger powers,

$Q=R^{2}Q^{(-2)}+RQ^{(-1)}+Q^{(0)}+\frac{1}{R}Q^{(1)}+\frac{1}{R^{2}}Q^{(2)}+\dots$
(77)

and similarly for the area functional

$\mathcal{A}=R\mathcal{A}^{(-1)}+\mathcal{A}^{(0)}+\frac{1}{R}\mathcal{A}^{(1)}+\frac{1}{R^{2}}\mathcal{A}^{(2)}+\dots~{}.$
(78)

We treat the expansions above as an ansatz. For the purpose of the volume law,
we will need only the coefficients $\rho_{0}$ and $t_{0}$. In Appendix A, we
take a closer look at the general structure of the higher-order perturbation
theory in $1/R$, as well as check explicitly that the ansatz above matches
with the exact HRT surface in the strip case. The area functional to leading
order in $1/R$ takes the form

$R\mathcal{A}^{(-1)}=R\int_{\delta}^{z_{m}}dz\int_{0}^{2\pi}d\phi\frac{L^{2}}{z^{2}}\sqrt{Q^{(-2)}},$
(79)

with

$\displaystyle Q^{(-2)}$ $\displaystyle=$
$\displaystyle-f(-t_{0}^{\prime}\tilde{\rho}_{,\phi}+\rho_{0}^{\prime}\tilde{t}_{,\phi})^{2}+\tilde{\rho}^{2}\left[-f(t_{0}^{\prime})^{2}+\frac{1}{f}+(\rho_{0}^{\prime})^{2}\right]$
(80) $\displaystyle-$
$\displaystyle\tilde{t}_{,\phi}^{2}+\frac{\tilde{\rho}_{,\phi}^{2}}{f}.$

The equations of motion take the form

$\frac{\partial\mathcal{L}}{\partial
t_{0}^{\prime}}=\frac{L^{2}}{z^{2}}\frac{f}{\sqrt{Q^{(-2)}}}\left[\rho_{0}^{\prime}\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}-t_{0}^{\prime}(\tilde{\rho}_{,\phi}^{2}+\tilde{\rho}^{2})\right]=a(\phi),$
(81)
$\frac{\partial\mathcal{L}}{\partial\rho_{0}^{\prime}}=\frac{L^{2}}{z^{2}}\frac{1}{\sqrt{Q^{(-2)}}}\left[f\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}t_{0}^{\prime}+\rho_{0}^{\prime}(\tilde{\rho}^{2}-f\tilde{t}_{,\phi}^{2})\right]=b(\phi),$
(82)

where $a(\phi)$ and $b(\phi)$ are some functions of $\phi$ only, and not of
$z$. Taking the ratio of the two equations above, and solving for
$t_{0}^{\prime}$, we find

$t_{0}^{\prime}=F(\phi)\rho_{0}^{\prime},$ (83)

with

$F(\phi)=\frac{\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}b(\phi)-\left(\frac{\tilde{\rho}^{2}}{f}-\tilde{t}_{,\phi}^{2}\right)a(\phi)}{a(\phi)\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}+b(\phi)(\tilde{\rho}_{,\phi}^{2}+\tilde{\rho}^{2})}.$
(84)

Evaluating the conjugate momenta at the tip of the HRT surface, and using the
equation above together with regularity at the tip
$(\rho_{0}^{\prime}\rightarrow\infty)$ and the fact that $f(z_{m})$ is close
to zero for large $R$, we then conclude that

$\displaystyle a(\phi)$ $\displaystyle=0,$ (85) $\displaystyle b(\phi)$
$\displaystyle=\frac{L^{2}}{z_{m}^{2}}\tilde{\rho},$ (86)

and we find that $\rho_{0}$ satisfies the differential equation

$\frac{A(\phi,z)\rho_{0}^{\prime}}{\sqrt{B(\phi,z)\rho_{0}^{\prime
2}+C(\phi,z)}}=\frac{L^{2}}{z_{m}^{2}}\tilde{\rho},$ (87)

with

$\displaystyle A(\phi,z)$
$\displaystyle=\frac{L^{2}}{z^{2}}\tilde{\rho}^{2}(1-\xi),$ (88)
$\displaystyle B(\phi,z)$ $\displaystyle=\tilde{\rho}^{2}(1-\xi),$ (89)
$\displaystyle C(\phi,z)$
$\displaystyle=\frac{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}}{f}(1-\xi),$
(90)

where we defined

$\xi\equiv\frac{f\tilde{t}_{,\phi}^{2}}{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}}.$
(91)

Solving for $\rho_{0}^{\prime}$, we find

$\rho_{0}^{\prime}=\sqrt{1+\frac{\tilde{\rho}_{,\phi}^{2}}{\tilde{\rho}^{2}}}\frac{1}{\sqrt{f}}\frac{z^{2}}{\sqrt{z_{m}^{4}(1-\xi)-z^{4}}}.$
(92)

At this point, a few remarks are in order about the nature of our large-$R$
expansion. We note that, in principle, $z_{m}$ in the expression above is
itself a function of $R$ (the size of the boundary subregion determines how
deep the surface penetrates into the bulk), so the $\rho_{0}$ above is not
exactly a quantity zeroth order in $1/R$. It is tempting to resolve this
subtlety by replacing $z_{m}$ by its zeroth-order part, which is the horizon
location $z_{h}=1$. Doing so, however, is unsatisfactory because the size of
the subregion is infinite when $z_{m}$ is at the horizon (as can be seen by
integrating $\rho_{0}^{\prime}$ and checking that there is a divergence near
the horizon). It would be preferable to deal with large but finite subregions.

We prefer to think about the large-$R$ expansion in this way: we hypothesize
that there is a way to split the $R$ dependence of $\rho(z,\phi)$ and
$t(z,\phi)$ into a dependence through $z_{m}$ and a dependence not through
$z_{m}$. Then the large-$R$ expansion consists of expanding the $R$ dependence
not through $z_{m}$, while leaving $z_{m}$ unexpanded. In Appendix A, we
illustrate more concretely how such $z_{m}$ arises in the case of strips.

For the purpose of the volume law, however, it seems harmless to replace
$z_{m}$ by the horizon value $1$, so we set $z_{m}$ to $1$ now.

Expanding in $z$ around $z=0$, we find the leading-order term to be

$\rho_{0}^{\prime}=\frac{1}{\tilde{\rho}}\frac{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}}{\sqrt{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}-\tilde{t}_{,\phi}^{2}}}z^{2}+\dots~{}.$
(93)

Integrating, we find the coefficient $c_{3}$ in the large-$R$ limit to be

$c_{3}=-\frac{1}{3\tilde{\rho}}\frac{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}}{\sqrt{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}-\tilde{t}_{,\phi}^{2}}},$
(94)

and the coefficient $d_{3}$ is found to be

$d_{3}=-\frac{1}{3\tilde{\rho}}\frac{\tilde{\rho}_{,\phi}\tilde{t}_{,\phi}}{\sqrt{\tilde{\rho}^{2}+\tilde{\rho}_{,\phi}^{2}-\tilde{t}_{,\phi}^{2}}}.$
(95)

####  IV.1.3 The volume law

Plugging back the answers for $c_{3}$ and $d_{3}$ in (94) and (95) into the
conjugate momenta $\Pi^{z}_{\rho}$ and $\Pi^{z}_{t}$ in (73), (74), we obtain:

$\displaystyle\Pi^{z}_{\rho}$ $\displaystyle=-L^{2}R\tilde{\rho},$ (96)
$\displaystyle\Pi^{z}_{t}$ $\displaystyle=0.$ (97)

Plugging the two results above back into $\frac{dA}{dR}$, we then find

$\frac{d\mathcal{A}}{dR}=\int_{0}^{2\pi}d\phi L^{2}R\tilde{\rho}(\phi)^{2}.$
(98)

Finally, consider the volume of the projection of the subregion onto a static
time slice. The boundary of that projected subregion is described by
$\rho(\phi)=R\tilde{\rho}(\phi)$ (with $t=0$). The volume of that projected
subregion is found to be:

$\mathrm{Volume}=\int_{0}^{2\pi}d\phi\int_{0}^{R\tilde{\rho}(\phi)}\rho
d\rho=\frac{1}{2}R^{2}\int_{0}^{2\pi}d\phi\tilde{\rho}{(\phi)}^{2}.$ (99)

Differentiating with respect to $R$,

$\frac{d}{dR}\mathrm{Volume}=R\int_{0}^{2\pi}d\phi\tilde{\rho}{(\phi)}^{2}.$
(100)

Comparing the above with (98) for $\frac{d\mathcal{A}}{dR}$, we then see that
$\frac{dA}{dR}$ is proportional to $\frac{d\mathrm{Volume}}{R}$, up to a
factor of $L^{2}$, which is an entropy density. In other words, the area
$\mathcal{A}$ of the HRT surface indeed scales with the volume of the
projected subregion.

##  V Conclusion

In this paper, we have taken the first steps toward establishing a version of
volume-law entanglement entropy, by using HRT surfaces in the eternal black
brane spacetime which are anchored at boundary subregions which break
staticity of the thermal states. We consider strips as well as general
subregions which are reflection symmetric. We emphasize that this work is not
a rigorous proof, but the findings agree with intuition from field theory as
well as more rigorous computations in field theory.

A special case of our result of particular physical interest is that of the
boosted black brane, which is the gravity dual to nonequilibrium steady states
(or NESSs). In particular, the entanglement structure of NESSs was studied in
[16] (without using holography) or in [19] (using the boosted black brane).

A natural direction for future work is to generalize to subregions without
reflection symmetry. Then the techniques used here do not apply since
reflection symmetry guarantees that the tip of the HRT surface stays at the
same boundary coordinates as $R$ is varied. This in turn allows for the
Hamilton-Jacobi argument at the beginning of Sec. IV that relates
$\frac{dA}{dR}$ to the near-boundary behavior of the HRT surface. In the
absence of reflection symmetry, we need a more general technique, perhaps a
“matching technique” similar to the one used in [1, 2]. A different approach
would be to use Gauss’ identity for the first variation of the area as in
Appendix A of [21]. The primary difficulty in Gauss’s identity approach is
computing the vector field $\eta^{a}$ of [21] explicitly in the large-$R$
limit which seems nontrivial at this time.

A different direction for future work is a generalization to other bulk
spacetimes. In particular, it would be interesting to study the volume law in
the context of the hydrodynamic black hole solution [17, 18]. Here again, a
more general technique than the one employed in this paper will be necessary.
We also expect that there will be corrections to the projected volume law in
this case due to the fact that entropy can be generated in these scenarios.
This can be seen from the derivative expansion of a hydrodynamic black hole
where the zeroth-order term is a boosted black hole. The projected volume law
should be true to zeroth order in this case with subsequent corrections.

###### Acknowledgements.

The work of T.G. is supported by the National Science Foundation under Grant
No. PHY-1914679. P.N. acknowledges support from Israel Science Foundation
Grant No. 447/17 for the work done in Sec. II and from U.S. National Science
Foundation Grant No. PHY-1820734 at CUNY for the work done in Sec. III. The
authors are very thankful for Brian Swingle, Mark Mezei, and Matthew Headrick
for useful discussions, and especially for help with the Sec. II. We thank
Josiah Couch for collaboration during the early stage of the project.

##  Appendix A More on the large-$R$ expansion

Next, we perform a consistency check of the large-$R$ expansion by showing
that it is consistent with the exact HRT surface for a strip in $3+1$ bulk
dimensions and also quantify the error of the large-$R$ approximation in this
case. The exact shape of the HRT surface in that case is given by Eqs. (25)
and (26), reproduced here for convenience,

$\displaystyle x(z)$
$\displaystyle=R+\int_{0}^{z}dz^{\prime}\frac{p_{x}}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}},$ (101) $\displaystyle t(z)$
$\displaystyle=t_{0}-\int_{0}^{z}dz^{\prime}\frac{p_{t}}{f(z^{\prime})}\frac{1}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}}.$ (102)

To compare with the large-$R$ approximation, we first need to convert from
Cartesian coordinates to polar coordinates using the well-known relation
$x=\rho\cos{\phi}$ and $y=\rho\sin{\phi}$. Equation (101) becomes

$\rho(z,\phi)=R\tilde{\rho}(\phi)+\int_{0}^{z}dz^{\prime}\frac{p_{x}\tilde{\rho}(\phi)}{\sqrt{p_{t}^{2}+f(z^{\prime})\left(\frac{L^{4}}{z^{\prime
4}}-p_{x}^{2}\right)}},$ (103)

with $\tilde{\rho}(\phi)=\sec{\phi}$. Let us take the derivative with respect
to $z^{\prime}$,

$\rho^{\prime}(z,\phi)=\frac{p_{x}\tilde{\rho}(\phi)}{\sqrt{p_{t}^{2}+f(z)\left(\frac{L^{4}}{z^{4}}-p_{x}^{2}\right)}}.$
(104)

The expression above is exact. In the large-$R$ limit, as is argued in Sec.
III, we have $p_{t}\approx 0$ and $p_{x}\approx-\frac{L^{2}}{z_{m}^{2}}$, and
the above becomes

$\rho^{\prime}(z,\phi)\approx-\tilde{\rho}{(\phi)}\frac{1}{\sqrt{f}}\frac{z^{2}}{\sqrt{z_{m}^{4}-z^{4}}}.$
(105)

Note that we do not expand $z_{m}$ into a series in $1/R$, which is in
accordance with the rule described in Sec. IV. The point now is to check that
the above is the same as $-\rho_{0}^{\prime}$, with $\rho_{0}^{\prime}$ given
by Eq. (92), when we specialize that equation to a strip. We note that the
function $\tilde{t}(\phi)$ is independent of $\phi$ for a strip, so the
quantity $\xi$ in (92) vanishes. Furthermore, the factor
$\sqrt{1+\frac{\tilde{\rho}_{,\phi}^{2}}{\tilde{\rho}^{2}}}$ in that equation
simplifies to $\tilde{\rho}$ when we take $\tilde{\rho}(\phi)=\sec{\phi}$ as
is appropriate for a strip. Then we see that the equation above is indeed the
same as $-\rho_{0}^{\prime}$.

We can similarly check that $t^{\prime}$ as obtained from (102) matches with
$-t_{0}^{\prime}$ as obtained from (138). Both expressions are easily seen to
vanish.

Generically, these calculations show that we are not missing any
“intermediate” terms in the large-$R$ expansion such as $\sqrt{R}$, $R^{1/4}$,
etc., at least for strips. Perhaps the absence of noninteger powers of $1/R$
can be argued more carefully from results such as [20].

Next, we take a closer look at the general structure of the higher-order
perturbation theory in $1/R$. Consider first the term $\mathcal{A}^{(0)}$ in
the action. We find that it is given in terms of $Q$ by

$\mathcal{A}^{(0)}=\frac{1}{2}\int dz\int
d\phi\frac{L^{2}}{z^{2}}\frac{Q^{(-1)}}{\sqrt{Q^{(-2)}}},$ (106)

with

$Q^{(-1)}=F_{1}{(z,\phi)}\rho_{1}^{\prime}+F_{2}{(z,\phi)}t_{1}^{\prime}+F_{3}{(z,\phi)}$
(107)

and

$F_{1}{(z,\phi)}=-2\tilde{\rho}^{2}\rho_{0}^{\prime}+2f\tilde{t}_{,\phi}(\tilde{t}_{,\phi}\rho_{0}^{\prime}-\tilde{\rho}_{,\phi}t_{0}^{\prime}),$
(108)
$F_{2}{(z,\phi)}=2f\tilde{\rho}^{2}t_{0}^{\prime}-2f\tilde{\rho}_{,\phi}(\tilde{t}_{,\phi}\rho_{0}^{\prime}-\tilde{\rho}_{,\phi}t_{0}^{\prime}),$
(109) $\displaystyle F_{3}{(z,\phi)}$ $\displaystyle=$ $\displaystyle
2t_{0,\phi}\tilde{t}_{,\phi}-2\rho_{0}\tilde{\rho}\left(\frac{1}{f}-ft_{0}^{\prime
2}+\rho_{0}^{\prime 2}\right)-\frac{2}{f}\rho_{0,\phi}\tilde{\rho}_{,\phi}$
(110) $\displaystyle+$ $\displaystyle
2f(\tilde{t}_{,\phi}\rho_{0}^{\prime}-\tilde{\rho}_{,\phi}t_{0}^{\prime})(-t_{0}^{\prime}\rho_{0,\phi}+t_{0,\phi}\rho_{0}^{\prime}).$

Note that, in the action $\mathcal{A}^{(0)}$, the $F_{1}$, $F_{2}$, $F_{3}$,
and $Q^{(-2)}$ are known functions. Because the action above is linear in the
first derivatives $\rho_{1}^{\prime}$ and $t_{1}^{\prime}$, we might worry
that there is an inconsistency, unless the quantities

$\frac{L^{2}}{z^{2}}\frac{F_{1}(z,\phi)}{\sqrt{Q^{(-2)}}}$ (111)

and

$\frac{L^{2}}{z^{2}}\frac{F_{2}(z,\phi)}{\sqrt{Q^{(-2)}}}$ (112)

turn out to be independent of $z$ (which is what the equations of motion
predict). And indeed they are, because they are equal to $-2b(\phi)$ and
$-2a(\phi)$ with $a(\phi)$ and $b(\phi)$ as defined in Eqs. (81) and (82). So
the perturbation theory is self-consistent, but we have learned that we need
to go to the next order in the action to find $\rho_{1}$ and $t_{1}$.

So then, consider the term $\mathcal{A}^{(1)}$ in the action. It is given in
terms of $Q$ by

$\mathcal{A}^{(1)}=\int dz\int
d\phi\frac{L^{2}}{z^{2}}\frac{1}{Q^{(-2)}{}^{3/2}}\left(\frac{1}{2}Q^{(0)}Q^{(-2)}-\frac{1}{8}Q^{(-1)}{}^{2}\right),$
(113)

where $Q^{(-2)}$ and $Q^{(-1)}$ have previously been written down, and
$Q^{(0)}$ is found to be

$\displaystyle Q^{(0)}$ $\displaystyle=$ $\displaystyle-
t_{0,\phi}^{2}-2t_{1,\phi}\tilde{t}_{,\phi}-4\rho_{0}\tilde{\rho}(ft_{0}^{\prime}t_{1}^{\prime}-\rho_{0}^{\prime}\rho_{1}^{\prime})$
(114) $\displaystyle+$ $\displaystyle\tilde{\rho}^{2}(-ft_{1}^{\prime
2}+2ft_{0}^{\prime}t_{2}^{\prime}+\rho_{1}^{\prime
2}-2\rho_{0}^{\prime}\rho_{2}^{\prime})$ $\displaystyle+$
$\displaystyle\left(\frac{1}{f}-ft_{0}^{\prime 2}+\rho_{0}^{\prime
2}\right)(\rho_{0}^{2}+2\rho_{1}\tilde{\rho})$ $\displaystyle-$ $\displaystyle
f(t_{0}^{\prime}\rho_{0,\phi}-t_{0,\phi}\rho_{0}^{\prime}-\tilde{t}_{,\phi}\rho_{1}^{\prime}+t_{1}^{\prime}\tilde{\rho}_{,\phi})^{2}$
$\displaystyle+$ $\displaystyle
2f(\tilde{t}_{,\phi}\rho_{0}^{\prime}-t_{0}^{\prime}\tilde{\rho}_{,\phi})(t_{1}^{\prime}\rho_{0,\phi}-t_{1,\phi}\rho_{0}^{\prime}$
$\displaystyle+$ $\displaystyle
t_{0}^{\prime}\rho_{1,\phi}-t_{0,\phi}\rho_{1}^{\prime}+\tilde{t}_{,\phi}\rho_{2}^{\prime}-t_{2}^{\prime}\tilde{\rho}_{,\phi})$
$\displaystyle+$
$\displaystyle\frac{\rho_{0,\phi}^{2}+2\rho_{1,\phi}\tilde{\rho}_{,\phi}}{f}.$

By inspection, the equations of motion for $\rho_{1}$ and $t_{1}$ do not
involve $\rho_{2}$ and $t_{2}$, so we can solve for $\rho_{1}$ and $t_{1}$
from those equations of motion. This is another check that the perturbation
theory is consistent. The differential equations that determine $\rho_{1}$ and
$t_{1}$ have the form of a coupled system of second-order ordinary
differential equations, the solution of which can be written down analytically
in integral form but we will not do so because it is not illuminating. Also,
the equations of motion from the action $\mathcal{A}^{1}$ for $\rho_{2}$ and
$t_{2}$ turn out to be the same consistency check as the one we have seen
coming from $\mathcal{A}^{(0)}$. This is another piece of evidence that the
perturbation is consistent.

##  Appendix B Volume-law scaling in arbitrary dimensions

In this appendix, we generalize the computation in Sec. IV to arbitrary $d$.
The metric is now

$ds^{2}=\frac{L^{2}}{z^{2}}\left[-f(z)dt^{2}+\frac{dz^{2}}{f(z)}+d\rho^{2}+\rho^{2}d\Omega_{d-2}^{2}\right],$
(115)

with

$f(z)=1-z^{d},$ (116)

and we consider a boundary subregion described by $\rho=R\tilde{\rho}(\phi)$
and $t=R\tilde{t}(\phi)$, where $R$ is a scaling parameter that we will take
to be large. The boundary subregion needs not lie on a static time slice or
even a boosted time slice, but – as in the case of strips in Sec. IV – we will
require that it is symmetric under the reflection
$(t,x,y)\rightarrow(-t,-x,-y)$. We write the metric on the $(d-2)$-sphere as

$d\Omega_{d-2}^{2}=\sum_{i=1}^{d-2}g_{i}d\theta_{i}^{2},$ (117)

with $g_{1}=1$, $g_{2}=\sin^{2}{\theta_{1}}$,
$g_{3}=\sin^{2}{\theta_{1}}\sin^{2}{\theta_{2}}$, etc.

The HRT surface is described by two functions $\rho(z,\theta_{i})$ and
$t(z,\theta_{i})$. The induced metric on the surface is

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle\frac{L^{2}}{z^{2}}\bigg{[}Adz^{2}+2B_{i}dzd\theta^{i}+D_{ij}d\theta^{i}d\theta^{j}\bigg{]},$
(118)

where

$\displaystyle A$ $\displaystyle=-f(z)t^{\prime 2}+\frac{1}{f(z)}+\rho^{\prime
2},$ (119) $\displaystyle B_{i}$
$\displaystyle=-ft^{\prime}t_{,i}+\rho^{\prime}\rho_{,i},$ (120)
$\displaystyle D_{ij}$
$\displaystyle=-ft_{,i}t_{,j}+\rho_{,i}\rho_{,j}+\rho^{2}g_{i}\delta_{ij},$
(121)

and $t_{,i}\equiv\frac{\partial t}{\partial\theta^{i}}$. The area functional
is

$\mathcal{A}=\int_{\delta}^{z_{m}}\int
d\Omega_{d-2}\left(\frac{L}{z}\right)^{d-1}\sqrt{Q},$ (122)

with

$\displaystyle Q$ $\displaystyle=$
$\displaystyle\rho^{2(d-2)}\det(d\Omega_{d-2}^{2})\bigg{\\{}-ft^{\prime
2}+\frac{1}{f}+\rho^{\prime 2}$ (123) $\displaystyle+$
$\displaystyle\frac{1}{\rho^{2}}\sum_{i}\frac{1}{g_{i}}\bigg{[}\frac{\rho_{,i}^{2}}{f}-t_{,i}^{2}-f(\rho^{\prime}t_{,i}-t^{\prime}\rho_{,i})^{2}\bigg{]}$
$\displaystyle-$
$\displaystyle\frac{1}{\rho^{4}}\sum_{i<j}\frac{(\rho_{,i}t_{,j}-t_{,i}\rho_{,j})^{2}}{g_{i}g_{j}}\bigg{\\}}.$

The $\frac{d\mathcal{A}}{dR}$ is now given by (again by using the Hamilton-
Jacobi argument)

$\frac{d\mathcal{A}}{dR}=-\int
d\Omega_{d-2}\left(\Pi^{z}_{\rho}\frac{d\rho}{dR}\bigg{|}_{\delta}+\Pi^{z}_{t}\frac{dt}{dR}\bigg{|}_{\delta}\right),$
(124)

where the conjugate momenta are

$\displaystyle\Pi^{z}_{\rho}$ $\displaystyle=$
$\displaystyle\left(\frac{L}{z}\right)^{d-1}\frac{1}{\sqrt{Q}}\rho^{2(d-2)}\mathrm{det}{(d\Omega_{d-2}^{2})}$
(125)
$\displaystyle\bigg{[}\rho^{\prime}-\frac{1}{\rho^{2}}\sum_{i}\frac{ft_{,i}}{g_{i}}(\rho^{\prime}t_{,i}-t^{\prime}\rho_{,i})\bigg{]},$
$\displaystyle\Pi^{z}_{t}$ $\displaystyle=$
$\displaystyle\left(\frac{L}{z}\right)^{d-1}\frac{1}{\sqrt{Q}}\rho^{2(d-2)}\mathrm{det}{(d\Omega_{d-2}^{2})}$
(126)
$\displaystyle\bigg{[}-ft^{\prime}+\frac{1}{\rho^{2}}\sum_{i}\frac{f\rho_{,i}}{g_{i}}(\rho^{\prime}t_{,i}-t^{\prime}\rho_{,i})\bigg{]}.$

We note that the right-hand side of (124) also includes the term $\int dz\int
d\Omega_{d-2}\partial_{i}\left(\Pi^{i}_{\rho}\frac{d\rho}{dR}+\Pi^{i}_{t}\frac{dt}{dR}\right)$.
But we can easily see that this term vanishes by virtue of $\Pi^{i}_{\rho}$
vanishing when $\theta^{i}$ is evaluated at $0$ or $\pi$.

Next, we plug the near-boundary expansions (63) and (64) into the above. Let
us assume that, in the large-$R$ limit, the leading terms in the expansion for
$\rho^{\prime}$ and $t^{\prime}$ are the $d$th FG coefficients,

$\lim_{R\rightarrow\infty}\rho^{\prime}=c_{d}{(R,\phi)}dz^{d-1}+\dots,$ (127)
$\lim_{R\rightarrow\infty}t^{\prime}=d_{d}{(R,\phi)}dz^{d-1}+\dots,$ (128)

and that the FG coefficients of lower order than the $d$th ones are $1/R$
suppressed. We will justify this assumption _a posteriori_ when we look at the
large-$R$ equation of motion later. The conjugate momenta can then be written
in terms of the FG coefficients as

$\displaystyle\Pi^{z}_{\rho}$ $\displaystyle=$ $\displaystyle
L^{d-1}R^{d-2}\tilde{\rho}^{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}}\frac{d}{\sqrt{\tilde{Q}}}$
(129) $\displaystyle\times$
$\displaystyle\bigg{[}c_{d}\tilde{\rho}^{2}-\sum_{i}\frac{\tilde{t}_{,i}}{g_{i}}(c_{d}\tilde{t}_{,i}-d_{d}\tilde{\rho}_{,i})\bigg{]}$
$\displaystyle+$ $\displaystyle\dots,$ $\displaystyle\Pi^{z}_{t}$
$\displaystyle=$ $\displaystyle
L^{d-1}R^{d-2}\tilde{\rho}^{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}}\frac{d}{\sqrt{\tilde{Q}}}$
(130) $\displaystyle\times$
$\displaystyle\bigg{[}-d_{d}\tilde{\rho}^{2}+\sum_{i}\frac{\tilde{\rho}_{,i}}{g_{i}}(c_{d}\tilde{t}_{,i}-d_{d}\tilde{\rho}_{,i})\bigg{]}$
$\displaystyle+$ $\displaystyle\dots,$

where

$\tilde{Q}=\tilde{\rho}^{4}+\tilde{\rho}^{2}\sum_{i}\frac{(\tilde{\rho}_{,i}^{2}-\tilde{t}_{,i}^{2})}{g_{i}}-\sum_{i<j}\frac{(\tilde{\rho}_{,i}\tilde{t}_{,j}-\tilde{t}_{,i}\tilde{\rho}_{,j})^{2}}{g_{i}g_{j}}.$
(131)

We note that $\tilde{Q}$ is a positive-definite quantity due to spacelikeness
of the boundary subregion. [Indeed, if we compute the metric induced by
$d$-dimensional Minkowski space onto the surface
$\rho=\tilde{\rho}{(\theta_{i})}$, $t=\tilde{t}{(\theta_{i})}$, we find that
the determinant of the induced metric is proportional to $\tilde{Q}$.]

Next, in order to obtain the coefficients $c_{d}$ and $d_{d}$ that appear
above, we have to solve the equations of motion in the large-$R$ limit. We
guess that the expansions of $\rho$ and $t$ in $1/R$ are

$\displaystyle\rho$
$\displaystyle=R\tilde{\rho}{(\theta_{i})}-\rho_{0}{(z,\theta_{i})}+\mathcal{O}{(1/R)},$
(132) $\displaystyle t$
$\displaystyle=R\tilde{t}{(\theta_{i})}-t_{0}{(z,\theta_{i})}+\mathcal{O}{(1/R)}.$
(133)

The area functional at large $R$ is

$\mathcal{A}=R^{d-2}\int_{\delta}^{z_{m}}dz\int
d\Omega_{d-2}\left(\frac{L}{z}\right)^{d-1}\sqrt{Q^{\prime\prime}},$ (134)

with

$\displaystyle Q^{\prime\prime}$ $\displaystyle=$
$\displaystyle\tilde{\rho}^{2(d-2)}\mathrm{det}{(d\Omega_{d-2}^{2})}\bigg{\\{}-ft_{0}^{\prime
2}+\frac{1}{f}+\rho_{0}^{\prime 2}$ (135) $\displaystyle+$
$\displaystyle\frac{1}{\tilde{\rho}^{2}}\sum_{i}\frac{1}{g_{i}}\bigg{[}\frac{\tilde{\rho}_{,i}^{2}}{f}-\tilde{t}_{,i}^{2}-f(t_{0}^{\prime}\tilde{\rho}_{,i}-\rho_{0}^{\prime}\tilde{t}_{,i})^{2}\bigg{]}$
$\displaystyle-$
$\displaystyle\frac{1}{\tilde{\rho}^{4}}\sum_{i<j}\frac{(\tilde{\rho}_{,i}\tilde{t}_{,j}-\tilde{t}_{,i}\tilde{\rho}_{,j})^{2}}{g_{i}g_{j}}\bigg{\\}}.$

The equations of motion are

$\displaystyle-\left(\frac{L}{z}\right)^{d-1}\frac{f}{\sqrt{Q^{\prime\prime}}}\tilde{\rho}^{2(d-2)}\mathrm{det}{(d\Omega_{d-2})}$
(136) $\displaystyle\times$
$\displaystyle\bigg{[}t_{0}^{\prime}+\frac{1}{\tilde{\rho}^{2}}\sum_{i}\frac{\tilde{\rho}_{,i}}{g_{i}}(t_{0}^{\prime}\tilde{\rho}_{,i}-\rho_{0}^{\prime}\tilde{t}_{,i})\bigg{]}=a{(\theta^{i})},$
$\displaystyle\left(\frac{L}{z}\right)^{d-1}\frac{1}{\sqrt{Q^{\prime\prime}}}\tilde{\rho}^{2(d-2)}\mathrm{det}{(d\Omega_{d-2})}$
(137) $\displaystyle\times$
$\displaystyle\bigg{[}\rho_{0}^{\prime}+\frac{f}{\tilde{\rho}^{2}}\sum_{i}\frac{\tilde{t}_{,i}}{g_{i}}(t_{0}^{\prime}\tilde{\rho}_{,i}-\rho_{0}^{\prime}\tilde{t}_{,i})\bigg{]}=b{(\theta^{i})},$

for some functions $a$ and $b$ of the $(d-2)$ angles. Taking the ratio of the
two equations above, and solving for $t_{0}^{\prime}$, we find

$t_{0}^{\prime}=G(\theta^{i})\rho_{0}^{\prime},$ (138)

with

$G=\frac{b(\theta^{i})\sum_{i}\frac{\tilde{t}_{,i}\tilde{\rho}_{,i}}{g_{i}}-a(\theta^{i})\left(\frac{\tilde{\rho}^{2}}{f}-\sum_{i}\frac{\tilde{t}_{,i}^{2}}{g_{i}}\right)}{a(\theta^{i})\sum_{i}\frac{\tilde{t}_{,i}\tilde{\rho}_{,i}}{g_{i}}+b(\theta^{i})\left(\tilde{\rho}^{2}+\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}\right)}.$
(139)

By invoking regularity of the HRT surface at the tip, we find $a$ and $b$ to
be

$\displaystyle a{(\theta^{i})}$ $\displaystyle=0,$ (140) $\displaystyle
b{(\theta^{i})}$
$\displaystyle=\left(\frac{L}{z_{m}}\right)^{d-1}\tilde{\rho}^{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}},$
(141)

and we find that $\rho_{0}^{\prime}$ satisfies the algebraic equation

$\frac{\gamma\rho_{0}^{\prime}}{\sqrt{\alpha(\rho_{0}^{\prime})^{2}+\beta}}=\left(\frac{z}{z_{m}}\right)^{d-1},$
(142)

with

$\alpha=1-\frac{f}{\tilde{\rho}^{2}}\frac{\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}\sum_{j}\frac{\tilde{t}_{,j}^{2}}{g_{j}}-\left(\sum_{i}\frac{\tilde{\rho}_{,i}\tilde{t}_{,i}}{g_{i}}\right)^{2}+\tilde{\rho}^{2}\sum_{i}\frac{\tilde{t}_{,i}^{2}}{g_{i}}}{\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}+\tilde{\rho}^{2}},$
(143)
$\beta=\frac{1}{\tilde{\rho}^{2}}\frac{\tilde{\rho}^{2}+\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}}{f}-\frac{\sum_{i<j}\frac{(\tilde{\rho}_{,i}\tilde{t}_{,j}-\tilde{t}_{,i}\tilde{\rho}_{,j})^{2}}{g_{i}g_{j}}+\tilde{\rho}^{2}\sum_{i}\frac{\tilde{t}_{,i}^{2}}{g_{i}}}{\tilde{\rho}^{4}},$
(144)
$\gamma=1-\frac{f}{\tilde{\rho}^{2}}\sum_{i}\frac{\tilde{t}_{,i}^{2}}{g_{i}}+\frac{\left(\sum_{i}\frac{\tilde{\rho}_{,i}\tilde{t}_{,i}}{g_{i}}\right)^{2}f}{\tilde{\rho}^{2}(\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}+\tilde{\rho}^{2})}.$
(145)

Solving for $\rho_{0}^{\prime}$ yields

$\rho_{0}^{\prime}=\frac{\sqrt{\beta}z^{d-1}}{\sqrt{z_{m}^{2d-2}\gamma^{2}-\alpha
z^{2d-2}}}.$ (146)

Expanding the above around $z=0$,

$\rho_{0}^{\prime}=\frac{\sqrt{1+\frac{1}{\tilde{\rho}^{2}}\sum_{i}\frac{(\tilde{\rho}_{,i}^{2}-\tilde{t}_{,i}^{2})}{g_{i}}-\frac{1}{\tilde{\rho}^{4}}\sum_{i<j}\frac{(\tilde{\rho}_{,i}\tilde{t}_{,j}-\tilde{t}_{,i}\tilde{\rho}_{,j})^{2}}{g_{i}g_{j}}}}{\left[1-\frac{1}{\tilde{\rho}^{2}}\sum_{i}\frac{\tilde{t}_{,i}^{2}}{g_{i}}+\frac{1}{\tilde{\rho}^{2}(\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}+\tilde{\rho}^{2})}\left(\sum_{i}\frac{\tilde{\rho}_{,i}\tilde{t}_{,i}}{g_{i}}\right)^{2}\right]}\left(\frac{z}{z_{m}}\right)^{d-1}+\dots~{}.$
(147)

From the above, we see that the near-boundary expansion for
$\rho_{0}^{\prime}$ indeed starts with the $(d-1)$th power, as previously
claimed. $z_{m}$ is near the horizon, which we will set to $1$ again.
Integrating over $z$, we obtain the coefficient $c_{d}$,

$c_{d}=-\frac{1}{d}\frac{\sqrt{1+\frac{1}{\tilde{\rho}^{2}}\sum_{i}\frac{(\tilde{\rho}_{,i}^{2}-\tilde{t}_{,i}^{2})}{g_{i}}-\frac{1}{\tilde{\rho}^{4}}\sum_{i<j}\frac{(\tilde{\rho}_{,i}\tilde{t}_{,j}-\tilde{t}_{,i}\tilde{\rho}_{,j})^{2}}{g_{i}g_{j}}}}{\left[1-\frac{1}{\tilde{\rho}^{2}}\sum_{i}\frac{\tilde{t}_{,i}^{2}}{g_{i}}+\frac{1}{\tilde{\rho}^{2}(\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}+\tilde{\rho}^{2})}\left(\sum_{i}\frac{\tilde{\rho}_{,i}\tilde{t}_{,i}}{g_{i}}\right)^{2}\right]}$
(148)

and also the coefficient $d_{d}$,

$d_{d}=\frac{\sum_{i}\frac{\tilde{\rho}_{,i}\tilde{t}_{,i}}{g_{i}}}{\tilde{\rho}^{2}+\sum_{i}\frac{\tilde{\rho}_{,i}^{2}}{g_{i}}}c_{d}.$
(149)

Plugging the coefficients above into the conjugate momenta $\Pi^{z}_{\rho}$
and $\Pi^{z}_{t}$, we find that they simplify to

$\displaystyle\Pi^{z}_{\rho}$
$\displaystyle=-L^{d-1}R^{d-2}\tilde{\rho}^{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}},$
(150) $\displaystyle\Pi^{z}_{t}$ $\displaystyle=0.$ (151)

Plugging the above into $\frac{dA}{dR}$, we find

$\frac{dA}{dR}=L^{d-1}R^{d-2}\int
d\Omega_{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}}\tilde{\rho}{(\theta^{i})}^{d-1}.$
(152)

Let us now compare with the projected volume. This latter is

$\displaystyle\mathrm{Volume}$ $\displaystyle=$ $\displaystyle\int
d\Omega_{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}}\int_{0}^{R\tilde{\rho}}\rho^{d-2}d\rho$
(153) $\displaystyle=$ $\displaystyle\int
d\Omega_{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}}\frac{(R\tilde{\rho})^{d-1}}{d-1}.$

Differentiating with respect to $R$ yields

$\frac{d\mathrm{Volume}}{dR}=R^{d-2}\int
d\Omega_{d-2}\sqrt{\mathrm{det}{(d\Omega_{d-2}^{2})}}\tilde{\rho}^{d-1}$ (154)

which is the same as $\frac{d\mathcal{A}}{dR}$ up to the factor of entropy
density $L^{d-1}$. Thus, we have managed to verify the volume-law entropy in
arbitrary dimensions.

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