# A note on the physical process first law of black hole mechanics

Antoine Rignon-Bret
Aix Marseille Univ., Univ. de Toulon, CNRS, CPT, UMR 7332, 13288 Marseille,
France

###### Abstract

I give a simple proof of the physical process first law of black hole
thermodynamics including charged black holes, in which all perturbations are
computed on the horizon.

##  1 Introduction

One of the most mysterious and promising topics in modern physics is the deep
connection between the laws of black hole mechanics and the fundamental
principles of thermodynamics. Originally stated in the seminal paper of
Bardeen, Carter and Hawking [1], the analogy between the laws of black hole
mechanics and thermodynamics has been taken more seriously after the deep
insights of Bekenstein about black hole entropy [2], the introduction of the
generalized second law [3], and the discovery by Hawking that black holes
radiate [4, 5]. The first law of black hole thermodynamics is very general and
robust. It can be extended to the case of charged black holes using the
Hamiltonian formalism [6], to any covariant theory of gravity using the
Lagrangian formalism [7, 8] including cases where arbitrary fields with
internal degrees of freedom are present [9]. In general, it relates the
variations of asymptotic charges, as the ADM mass $M$ and angular momentum $J$
and the electric charge $Q$ to the variation of the black hole area $A$. This
is a relation between phase space variables, on the same footing as the usual
first law of black hole thermodynamics.

However, while usual thermodynamic relations deal with ideal transitions
between equilibrium states characterized by perfectly stationary solutions, it
is sometimes instructive to consider realistic transitions bringing one
stationary state into another one. Indeed, it is a priori non trivial that a
new equilibrium state is reached after introducing some perturbations into the
system. Equivalently, in black holes mechanics, we are interested in the
evolution of the black hole if some piece of matter is thrown into it. If we
assume that the black hole settles down to a new stationary state after some
small energy, angular momentum and electric charge fluxes have crossed the
horizon, it would be consistent to recover an identity similar to the
equilibrium state version of the first law. This version of the first law of
black hole thermodynamics is called the Physical Process First Law (PPFL) and
its relation to the equilibrium state version of the first law is subtle [10,
11]. The problem of energy and angular momentum perturbations to the Kerr
black hole has been studied first by Hawking and Hartle [12]. The usual PPFL
has been derived by Wald in [13] for uncharged black holes and extended to the
charged case by Gao and Wald in [14].

However, while the mass and angular momentum variations are calculated
directly on the horizon for the perturbed Kerr black hole [13], it is not the
case for the derivation in the charged case in [14]. Indeed, this derivation
has the inconvenience of relying on the covariant phase space techniques. In
particular, the black hole mass and angular momentum variations are computed
at spatial infinity, as the ADM mass and angular momentum. It might be a bit
unexpected for a physical process first law to involve variations of
asymptotic quantities rather than the mass or angular momentum of the matter
fields crossing the black hole horizon. In fact, Gao and Wald wrote 111see
footnote 5 of their paper that the ADM charges $M$ and $J$ should hold as the
definition for the final mass and angular momentum of the black hole even if
not all the matter fell into it [14], but they did not explicitly show that
this is indeed the case. As they work with the ADM masses and angular
momentum, they cannot distinguish the mass and angular momentum of the black
hole from the one of the matter outside. In contrast, the derivation presented
below is similar to the original derivation of the PPFL from Wald [13] for the
uncharged black hole, where the mass and angular momentum variations are given
locally and are not the perturbed ADM Hamiltonians at infinity. We compute
everything on the horizon and do not need to introduce the covariant phase
space formalism.

##  2 Derivation of the PPFL

Figure 1: Penrose diagramm sketching a piece of matter falling into a black
hole (blue arrow). The null event horizon is represented in red, and is
defined as the boundary of the past of future null infinity, as usual. The
apparent horizon is a dashed black line.

Let us consider a stationary black hole, with an event horizon ${\cal N}$
parameterized by some null affine parameter $v$ vanishing on the black hole
bifurcation surface $v=0$. This eternal black hole has mass $M$, angular
momentum $J$ and charge $Q$. The Killing vector $\xi$ generating the null
geodesics of the event horizon is defined as

$\xi^{\mu}=(\frac{\partial}{\partial
t})^{\mu}+\Omega_{H}(\frac{\partial}{\partial\phi})^{\mu}\stackrel{{\scriptstyle{\cal
N}}}{{=}}\kappa vn^{\mu}$ (2.1)

where $\frac{\partial}{\partial t}$ is a Killing vector which is timelike and
normalized to one at infinity, $\frac{\partial}{\partial\phi}$ is an
axisymmetric Killing field generating closed orbits of length $2\pi$,
$n^{\mu}=(\frac{\partial}{\partial v})^{\mu}$ the null normal with vanishing
unaffinity, $\Omega_{H}$ the horizon’s angular velocity, and $\kappa$ being
the unaffinity of $\xi$ on ${\cal N}$, which turns out to be the surface
gravity of the stationary black hole. As the black hole is charged, there is
some electromagnetic potential $A_{\mu}$ alongside the metric $g_{\mu\nu}$,
which are the Kerr-Newman solutions of the Einstein-Maxwell equations. We use
the same gauge for the Kerr-Newman solution as in [15]. This is the stationary
gauge for which the electromagnetic potential satisfies the stationarity
requirement $\pounds_{\xi}A_{\mu}=0$, in addition to
$\pounds_{\xi}g_{\mu\nu}=0$. Furthermore, in this gauge, $A_{\mu}$ vanishes at
spatial infinity. It is also the same gauge as the one used in [14, 16].

Now, let us assume that the stationary black hole is perturbed by some small
amount of matter propagating on spacetime and crossing the black hole horizon
at some point, such that at very late times ($v$ going to infinity), the black
hole has settled down into a stationary state (see Figure.1). Hence, the event
horizon ${\cal N}$ becomes a Killing horizon a large amount of time after the
incoming matter crossed ${\cal N}$, and during the whole process, the
spacetime is a slightly perturbed Kerr-Newman spacetime. If we consider some
(charged) perturbed matter field of order $\epsilon$, its corresponding stress
energy tensor is of order $\epsilon^{2}$ and so the perturbation to the
stationary background metric is of order $\epsilon^{2}$ at most. Furthermore,
the charge current $J^{\mu}$ is also of order $\epsilon^{2}$ 222For instance,
in Dirac fields electrodynamics, a Dirac field would be of order $\epsilon$
and the associated current will be
$J^{\mu}=i\bar{\psi}\gamma^{\mu}\psi=O(\epsilon^{2})$. Same for scalar
electrodynamics, if the scalar field $\phi$ is of order $\epsilon$, the
current is given by
$J^{\mu}=i(\bar{\phi}\partial^{\mu}\phi-\partial^{\mu}\bar{\phi}\phi)=O(\epsilon^{2}).$.
Hence, we have

$\displaystyle\phi$ $\displaystyle=O(\epsilon)$ $\displaystyle
T_{\mu\nu}^{matter}$ $\displaystyle=O(\epsilon^{2})$ $\displaystyle J^{\mu}$
$\displaystyle=O(\epsilon^{2})$ (2.2)

from which it implies by using the equations of motion

$\displaystyle\pounds_{\xi}g_{\mu\nu}$ $\displaystyle=O(\epsilon^{2})$
$\displaystyle\pounds_{\xi}A_{\mu}$ $\displaystyle=O(\epsilon^{2})$ (2.3)

as $\xi$ being of course the background Killing, and so that
$\xi^{\mu}\stackrel{{\scriptstyle{\cal N}}}{{=}}\kappa vn^{\mu}$ is still null
everywhere on the (dynamical) event horizon. The Raychaudhuri equation on the
perturbed event horizon ${\cal N}$ for the affine normal vector $n$ reads

$\frac{d\theta_{n}}{dv}=-\theta_{n}^{2}-\sigma_{n,\mu\nu}\sigma_{n}^{\mu\nu}-T_{\mu\nu}n^{\mu}n^{\nu}$
(2.4)

where $\theta_{n}$ is the expansion of $n^{\mu}$, $\sigma_{n,\mu\nu}$ the
shear of $n^{\mu}$, and there is no twist $\omega_{\mu\nu}$ because the null
vector $n$ is hypersurface orthogonal. The expansion and the shear are of
order $\epsilon^{2}$ from (2.3), so if we keep only the leading order terms in
(2.4) we get

$\frac{d\theta_{n}}{dv}=-T_{\mu\nu}n^{\mu}n^{\nu}+O(\epsilon^{4})$ (2.5)

Now, we can multiply both sides of the equation (2.4) by $\kappa v$ and
integrate by part the left hand side. The boundary terms vanish as we
integrate between the bifurcation surface of the background spacetime located
at $v=0$ and infinity, where
$v\theta_{n}\underset{v\rightarrow+\infty}{\rightarrow}0$. Hence we get from
(2.5)

$\frac{\kappa}{8\pi}\Delta A=\int_{\cal
N}T_{\mu\nu}\xi^{\mu}n^{\nu}\epsilon_{N}+O(\epsilon^{4})$ (2.6)

Terms on both sides of this equation are of order $\epsilon^{2}$. We can
notice that the null energy conditions are satisfied, the RHS is positive, and
so is the LHS, in agreement with Hawking’s area theorem [17]. Furthermore, we
would like to write the RHS of 2.6 as a linear combination of the mass,
angular momentum and charge of the matter perturbation crossing the null
horizon. First, consider the case where the black hole is not charged.
Therefore, $T_{\mu\nu}=T_{\mu\nu}^{matter}$ and if $\xi$ is a background
Killing vector 2.1, we have from 2.2, 2.3

$\nabla_{\mu}(T_{\nu}^{\mu}\xi^{\nu})=O(\epsilon^{4})$ (2.7)

as $T_{\mu\nu}$ is symmetric and divergence free if the equations of motion
are imposed. Furthermore, as $\frac{\partial}{\partial t}$ and
$\frac{\partial}{\partial\phi}$ are also Killing vectors of the background
metric, we also have
$\nabla_{\mu}\big{(}T_{\nu}^{\mu}(\frac{\partial}{\partial
t})^{\nu}\big{)}=O(\epsilon^{4})$ and
$\nabla_{\mu}\big{(}T_{\nu}^{\mu}(\frac{\partial}{\partial\phi})^{\nu}\big{)}=O(\epsilon^{4})$.
Hence, the currents

$j_{t}^{\mu}=T_{\nu}^{\mu}(\frac{\partial}{\partial t})^{\nu}$ (2.8)

and

$j_{\phi}^{\mu}=-T_{\nu}^{\mu}(\frac{\partial}{\partial\phi})^{\nu}$ (2.9)

are conserved up to $\epsilon^{4}$ terms, so they are conserved at first order
in perturbation of the metric and the stress energy tensor, i.e at order
$\epsilon^{2}$. The physical meaning of these quantities is clear when we look
at these currents at infinity, and so we identify $j_{t}$ as the energy
current and $j_{\phi}$ as the angular momentum current. Hence, to make sense
of 2.6, we can apply Gauss’s theorem between Cauchy surfaces $\Sigma_{1}$ and
$\Sigma_{2}$ intersecting the null event horizon in two cross sections $S_{1}$
and $S_{2}$ respectively. In order to relate these considerations to (2.6), we
suppose that $S_{1}$ is the bifurcation surface located at $v=0$ and that
$S_{2}$ is in the very far future ($v\longrightarrow+\infty$) such that the
perturbations on the horizon vanish well before $S_{2}$. Hence, 2.6 becomes

$\frac{\kappa}{8\pi}\Delta A=\Delta M-\Omega_{H}\Delta J+O(\epsilon^{4})$
(2.10)

However, if the black hole is charged, the relation (2.7) is not satisfied.
Indeed, we have

$T_{\mu\nu}=T_{\mu\nu}^{EM}+T_{\mu\nu}^{matter}$ (2.11)

with [18]

$T^{EM}_{\mu\nu}=\frac{1}{4}(F_{\mu\rho}F_{\nu}^{\rho}-\frac{1}{4}g_{\mu\nu}F^{\rho\sigma}F_{\rho\sigma})$
(2.12)

which is of order $1$. Hence

$\nabla_{\mu}(T_{\nu}^{\mu}\xi^{\nu})=T_{\mu\nu}^{EM}\nabla^{(\mu}\xi^{\nu)}+O(\epsilon^{4})=O(\epsilon^{2})$
(2.13)

is of order $\epsilon^{2}$ and not of order $\epsilon^{4}$ as in a non-charged
case (2.7). Therefore, we cannot apply Gauss theorem, because the current is
not divergence free at order $\epsilon^{2}$. Thus we have to proceed a bit
differently. It can be proven from (2.12) and using Bianchi identity that

$\nabla^{\mu}T^{EM}_{\mu\nu}=J^{\mu}F_{\mu\nu}$ (2.14)

which is not conserved of course if the electromagnetic field is coupled to
matter. However, on-shell, the full stress energy tensor $T_{\mu\nu}$ is still
divergence free on-shell and then we get from (2.14)

$-\nabla^{\mu}T^{matter}_{\mu\nu}=J^{\mu}F_{\mu\nu}$ (2.15)

Then we contract both sides of the equation by $\xi^{\nu}$, and using
$F_{\mu\nu}=\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}$, the charge conservation
$\nabla_{\mu}J^{\mu}=0$ and the identity
$\pounds_{\xi}A_{\mu}=\xi^{\nu}\nabla_{\nu}A_{\mu}+A_{\nu}\nabla_{\mu}\xi^{n}$,
we get from (2.15)

$\nabla^{\mu}(T^{matter}_{\mu\nu}\xi^{\nu}+J_{\mu}A_{\nu}\xi^{\nu})=T_{\mu\nu}^{matter}\nabla^{\mu}\xi^{\nu}+J^{\mu}\pounds_{\xi}A_{\mu}$
(2.16)

Furthermore, from (2.2) and (2.3), we can conclude that

$\nabla^{\mu}(T^{matter}_{\mu\nu}\xi^{\nu}+J_{\mu}A_{\nu}\xi^{\nu})=O(\epsilon^{4})$
(2.17)

Hence, the current

$j^{\mu}=(T^{matter})_{\nu}^{\mu}\xi^{\nu}+J^{\mu}A_{\nu}\xi^{\nu}$ (2.18)

is conserved at order $\epsilon^{4}$, and so are the energy current

$j_{t}^{\mu}=(T^{matter})_{\nu}^{\mu}(\frac{\partial}{\partial
t})^{\nu}+J^{\mu}A_{\nu}(\frac{\partial}{\partial t})^{\nu}$ (2.19)

and angular momentum current 333It can be seen that the current $j_{t}$ and
$j_{\phi}$ correspond to the energy and angular momentum currents respectively
when evaluated at infinity, because $\lim_{r\rightarrow\infty}A_{\mu}=0$ in
the stationary gauge.

$j_{\phi}^{\mu}=-\big{[}(T^{matter})_{\nu}^{\mu}(\frac{\partial}{\partial\phi})^{\nu}+J^{\mu}A_{\nu}(\frac{\partial}{\partial\phi})^{\nu}\big{]}$
(2.20)

Again, the physical meaning of these currents is clear if we look them on the
asymptotically flat spacetime. Indeed, at infinity, the background
electromagnetic potential $A_{\mu}$ vanishes in the stationary gauge we chose,
and so at infinity the energy and angular momentum are given only by the
stress energy tensor of the charged matter, as expected in
Minkowski’spacetime. Therefore, we should make appear these conserved currents
in (2.6), in order to apply Gauss theorem between the Cauchy slices
$\Sigma_{1}$ and $\Sigma_{2}$ the piece of the event horizon ${\cal N}$
comprised between $S_{1}$ and $S_{2}$. Hence, we write (2.6) as

$\displaystyle\frac{\kappa}{8\pi}\Delta A$ $\displaystyle=\int_{\cal
N}(T^{matter}_{\mu\nu}\xi^{\mu}n^{\nu}+T^{EM}_{\mu\nu}\xi^{\mu}n^{\nu})\epsilon_{N}+O(\epsilon^{4})$
(2.21) $\displaystyle=\int_{\cal
N}[(T^{matter}_{\mu\nu}+J_{\nu}A_{\mu})\xi^{\mu}n^{\nu}-J^{\nu}A_{\mu}\xi^{\mu}n_{\nu}\big{)}\epsilon_{\cal
N}+\int_{\cal N}T^{EM}_{\mu\nu}\xi^{\mu}n^{\nu}\epsilon_{\cal N}]$ (2.22)

Let us start by looking at the first integral. The first term in the integral
is our conserved current. Furthermore, we can see that
$J^{\nu}n_{\nu}\epsilon_{\cal N}=-dQ$ is the infinitesimal electric charge
crossing the horizon, and $\Phi_{H}=-\xi^{\mu}A_{\mu}$ is the electrostatic
potential of the horizon, constant on a non-expanding horizon. Indeed, we have

$\pounds_{\xi}A=di_{\xi}A+i_{\xi}F$ (2.23)

as $F=dA$. If we take the pullback of (2.23) on the non-expanding null horizon
${\cal N}$, we get
$\underset{\leftarrow}{di_{\xi}A}=d\underset{\leftarrow}{i_{\xi}A}=0$ as $\xi$
is tangent to ${\cal N}$. Hence, $d\Phi_{H}=O(\epsilon^{2})$ on the perturbed
horizon, and from (2.17) the first integral of (2.22) can be written as

$\int_{\cal
N}[(T^{matter}_{\mu\nu}+J_{\nu}A_{\mu})\xi^{\mu}n^{\nu}-J^{\nu}A_{\mu}\xi^{\mu}n_{\nu}]\epsilon_{\cal
N}=\Delta M-\Omega_{H}\Delta J-\Phi_{H}\Delta Q+O(\epsilon^{4})$ (2.24)

Now, we should look at the second integral of the RHS in (2.22). On a non-
expanding horizon, we have $T_{\mu\nu}^{EM}\xi^{\mu}n^{\nu}=0$ from the
Raychaudhuri equation (2.4), ie. $T_{\mu\nu}^{EM}\xi^{\mu}n^{\nu}=\kappa
vF_{\mu\nu}\xi^{\nu}F_{\rho}n^{\rho}=0$. It means that the vector
$F^{\mu\nu}n_{\nu}$ is null. Furthermore, $F_{\mu\nu}n^{\mu}n^{\nu}=0$ by
antisymmetry of $F$, so the vector $F^{\mu\nu}n_{\nu}$ is tangent to ${\cal
N}$. As it is tangent to ${\cal N}$ and null, it means that it is proportional
to $n^{\mu}$, and we can write $F^{\mu}_{\nu}n^{\nu}=\alpha n^{\mu}$. From
this, we deduce that on the dynamical event horizon we have

$F^{\mu\nu}n_{\nu}=\alpha n^{\mu}+\epsilon^{2}\gamma^{\mu}$ (2.25)

as the first order of perturbation in $F$ is $\epsilon^{2}$ (as we use the
electromagnetic equations of motion and we know that $J^{\mu}$ is of order
$\epsilon^{2}$) for some vector $\gamma^{\mu}$. Then, by antisymmetry of $F$

$\displaystyle 0$ $\displaystyle=F^{\mu\nu}n_{\mu}n_{\nu}=(\alpha
n^{\mu}+\epsilon^{2}\gamma^{\mu})n_{\mu}=\epsilon^{2}\gamma^{\mu}n_{\mu}$
(2.26)

so $\gamma^{\mu}n_{\mu}=0$ and $\gamma^{\mu}$ is tangent to ${\cal N}$.
Therefore, on the dynamical null horizon, we get

$\displaystyle T^{EM}_{\mu\nu}\xi^{\mu}n^{\nu}$ $\displaystyle=\kappa v(\alpha
n^{\mu}+\epsilon^{2}\gamma^{\mu})(\alpha n_{\mu}+\epsilon^{2}\gamma_{\mu})$
$\displaystyle=\kappa v\gamma^{\mu}\gamma_{\mu}\epsilon^{4}=O(\epsilon^{4})$
(2.27)

Hence, from (2.6), (2.24) and (2.27), we deduce the PPFL for charged black
holes

$\frac{\kappa}{8\pi}\Delta A=\Delta M-\Omega_{H}\Delta J-\Phi_{H}\Delta
Q+O(\epsilon^{4})$ (2.28)

where here, $\Delta M$, $\Delta J$ and $\Delta Q$ correspond to the energy,
angular momentum and charge of the piece of matter flowing through the event
horizon. We should notice that the term $T_{\mu\nu}\xi^{\mu}n^{\nu}$, usually
associated to an energy flow, is more related to the entropy variation than to
the energy variation. Furthermore, though we restricted ourselves to
electromagnetism, there should not be any difficulties to extend the
derivation above to arbitrary Yang-Mills fields and charged matter.

## Acknowledgements

I would like to thank Alejandro Perez, Simone Speziale and Robert Wald for
helpful discussions.

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