The canonical definition of energy-momentum tensor is basically obtained by
differentiating the matter Lagrangian with respect to the metric field. This
definition was assumed to satisfy the null energy condition $(T_{vv}\geq 0)$
in Bhattacharyya:2021jhr , which immediately implies that $T_{vv}$ is of the
order ${\cal O}(\epsilon^{2})$. We have, however, seen in this paper that
there exist terms that are $\mathcal{O}(\epsilon)$, and hence, no longer
satisfy the null energy condition. This is well known in examples of non-
minimally coupled Lagrangians Barcelo:2000zf ; Flanagan_1996 ; Wall:2018ydq ;
Chatterjee:2015uya . It will be interesting to see if we can come up with a
prescription for a covariant definition of energy-momentum tensor that
satisfies the null energy condition. This definition would apply to arbitrary
matter coupling with gravity but desirably would not pick up the
$\mathcal{O}(\epsilon)$ contributions in $T_{vv}$.

###### Acknowledgements.

We would like to thank Sayantani Bhattacharyya for collaboration at the
initial stage and for comments on the draft. We are also thankful to Jyotirmoy
Bhattacharya, Sayantani Bhattacharyya, Diptarka Das, Anirban Dinda, Alok
Laddha, R. Loganayagam, Milan Patra, Shuvayu Roy, Amitabh Virmani for several
discussions and important inputs. We would also like to thank Áron Kovács and
Harvey Reall for comments on the draft. P.B. would like to acknowledge the
support provided by the grant SB/SJF/2019-20/08. P.D. would like to duly
acknowledge the Council of Scientific and Industrial Research (CSIR), New
Delhi for financial assistance through the Senior Research Fellowship (SRF)
scheme. We finally would like to thank the people of India for their steady
support for research in basic sciences.

##  Appendix A A short review of the gauge choice and boost weight analysis

This appendix serves as a quick recapitulation of the gauge choice and boost
weight analysis of Bhattacharyya:2021jhr . Thus, we will be brief here and we
refer the reader to Bhattacharyya:2021jhr  for more details.

The most general dynamical black hole metric can always be expressed by the
following metric 444444This fact has been explicitly argued for in the
appendix A of Bhattacharyya:2016xfs .

$ds^{2}=2~{}dv~{}dr-r^{2}X(r,v,x^{i})~{}dv^{2}+2~{}r~{}\omega_{i}(r,v,x^{i})~{}dv~{}dx^{i}+h_{ij}(r,v,x^{i})dx^{i}dx^{j}\,.$
(126)

As noted in Bhattacharya:2019qal ; Bhattacharyya:2021jhr , the above metric
choice does not completely fix the gauge freedom. We are allowed to do two
types of coordinate redefinition:

$\displaystyle v$ $\displaystyle\rightarrow$
$\displaystyle\tilde{v}=f_{1}(x^{i})\,v+f_{2}(x^{i}),\quad\text{with an
appropriate re-definition of}\,r\,,$ (127) $\displaystyle x^{i}$
$\displaystyle\rightarrow$ $\displaystyle\tilde{x}^{i}=g^{i}(x^{j})\,.$ (128)

The second coordinate freedom eq.(128) is just a relabelling of the spatial
tangents and it represents the residual diffeomorphism invariance of the
spatial metric $h_{ij}$. This freedom can be implemented by lifting the
partial derivatives $\partial_{i}$ to covariant derivatives $\nabla_{i}$
compatible with $h_{ij}$. The first coordinate freedom eq.(127) is non-trivial
and the transformation under this coordinate change constrains the structure
of the relevant components of covariant tensors. We will only be interested in
a sub-class of the full residual symmetry of the gauge given by,

$r\rightarrow\tilde{r}=\lambda~{}r,~{}~{}v\rightarrow\tilde{v}={v\over\lambda}\,.$
(129)

This will be called the boost transformation 454545The nomenclature follows
that of Wall:2015raa ; Bhattacharya:2019qal ; Bhattacharyya:2021jhr .. The
infinitesimal version of this transformation is generated by the vector field

$\xi=\xi^{\mu}\partial_{\mu}=v\partial_{v}-r\partial_{r}\,.$ (130)

Now the metric in our gauge eq.(126) that remains stationary under this vector
field eq.(130) is given by

$ds^{2}=2\,dv\,dr-r^{2}\,X(rv,x^{i})\,dv^{2}+2\,r\,\omega_{i}(rv,x^{i})\,dv\,dx^{i}+h_{ij}(rv,x^{i})\,dx^{i}\,dx^{j}\,.$
(131)

The vector field $\xi^{\mu}$ is a Killing vector which generates the Killing
horizon of this stationary space-time. This form of the stationary metric
constrains the dynamics away from it.

To describe the dynamics, we will employ a perturbative expansion in the
amplitude of the dynamics and we will be working up to linearized order in the
dynamics. Mathematically, this means that we can decompose the metric
$g_{\mu\nu}$ given in eq.(126) as follows 464646Here $\epsilon$ is very small.

$g_{\mu\nu}=g^{eq}_{\mu\nu}+\epsilon\,\delta g_{\mu\nu}\,,$ (132)

where $g^{eq}_{\mu\nu}$ is the equilibrium metric given by eq.(131). Notice
that here the metric components arising in $g^{eq}_{\mu\nu}$ namely,
$X,\omega_{i},h_{ij}$ have a specific functional dependence on the coordinates
(i.e) $X(rv,x^{i}),\omega_{i}(rv,x^{i}),h_{ij}(rv,x^{i})$. The components in
$\delta g_{\mu\nu}$ have arbitrary dependence. This is summarized below:

$\begin{split}X&=X(rv,x^{i})+\epsilon\,\delta X(r,v,x^{i})\,,\\\
\omega_{i}&=\omega_{i}(rv,x^{i})+\epsilon\,\delta\omega_{i}(r,v,x^{i})\,,\\\
h_{ij}&=h_{ij}(rv,x^{i})+\epsilon\,\delta h_{ij}(r,v,x^{i})\,.\end{split}$
(133)

Such a Background plus fluctuation split enforces structures on covariant
tensors. Suppose one considers a generic covariant tensor of the form

${\cal C}\sim(\partial_{r})^{q_{r}}(\partial_{v})^{q_{v}}{\cal D}\,,$ (134)

constructed out of the metric components, where $\mathcal{D}$
474747$\mathcal{D}$ is typically constructed out of the metric components
$X,\omega_{i},h_{ij}$ and $\nabla_{i}$. doesn’t have explicit $\partial_{v}$
and $\partial_{r}$ derivatives, and evaluate it on the full metric eq.(132),
$\mathcal{D}$ will have the following structure

$\mathcal{D}=\mathcal{D}(rv,x^{i})+\epsilon\,\mathcal{D}(r,v,x^{i})\,.$ (135)

Thus, we can use the following criterion to distinguish between equilibrium
configuration and non-equilibrium configuration

$\begin{split}\text{Equilibrium
configuration}\rightarrow~{}&(\partial_{r})^{q_{r}}(\partial_{v})^{q_{v}}{\cal
D}(rv,x)|_{r=0}=0~{}\text{(for $q_{v}>q_{r}$)}\,,\\\ \text{Non-equilibrium
configuration}\rightarrow~{}&(\partial_{r})^{q_{r}}(\partial_{v})^{q_{v}}{\cal
D}(r,v,x)|_{r=0}\neq 0~{}\text{(for $q_{v}>q_{r}$)}\,.\end{split}$ (136)

The non-equilibrium configuration essentially evaluates to
$\mathcal{O}(\epsilon)$. Thus, whenever extra $\partial_{v}$ derivatives are
distributed between product of two or more different factors of $\mathcal{D}$,
the resulting term is non-linear in dynamics, i.e.,
$\mathcal{O}(\epsilon^{2})$. This type of analysis carries over
straightforwardly to fields propagating on this space-time and this is
discussed in detail in section §3.

To study these dynamics of small fluctuations away from stationary solutions,
it is very useful to describe how covariant quantities transform under the
boost transformation given in eq.(129). We define a quantity $w$ called boost
weight as the power of $\lambda$ to which a covariant tensor transforms under
the boost transformation eq.(129):

${\cal A}\rightarrow\widetilde{\cal A}=\lambda^{w}{\cal
A}\,,~{}~{}\text{under}~{}~{}\\{r\rightarrow\tilde{r}=\lambda~{}r,~{}v\rightarrow\tilde{v}=\lambda^{-1}v\\}~{}~{}\Rightarrow~{}~{}\text{boost
weight of ${\cal A}$ is $w$}\,.$ (137)

Now under this transformation, the operators $\partial_{v}$ and $\partial_{r}$
transform non-trivially as follows:

$\partial_{r}\rightarrow\partial_{\tilde{r}}=\lambda^{-1}\,\partial_{r}\,,~{}~{}\partial_{v}\rightarrow\partial_{\tilde{v}}=\lambda\,\partial_{v}\,.$
(138)

Thus, if a covariant tensor has positive boost weight, the structure has more
number of $\partial_{v}$ derivatives than $\partial_{r}$ derivatives. This
implies that, when the tensor is evaluated on the horizon using the full
metric eq.(132), the equilibrium contribution vanishes by the argument of
eq.(136). Hence, it is a non-equilibrium configuration, i.e.,
$\mathcal{O}(\epsilon)$. This helps in characterizing whether terms are
$\mathcal{O}(1)$ (equilibrium) or $\mathcal{O}(\epsilon)$ (non-equilibrium).

##  Appendix B A derivation of the generalized zeroth law

In this appendix, our aim is to explicitly show how to derive eq.(28) from
eq.(27). Working in our gauge for the metric eq.(2), the derivation goes as
follows,

$\begin{split}\mathcal{L}_{\xi}A^{eq}_{\mu}+D_{\mu}\Lambda&=\xi^{\alpha}D_{\alpha}A^{eq}_{\mu}+A^{eq}_{\alpha}D_{\mu}\xi^{\alpha}+D_{\mu}\Lambda\\\
&=\xi^{\alpha}F^{eq}_{\alpha\mu}+D_{\mu}(A^{eq}_{\alpha}\xi^{\alpha}+\Lambda)\,.\end{split}$
(139)

Further, contracting with $\xi^{\mu}$ on both sides and setting the RHS to
zero by demanding eq.(27), we get

$\begin{split}\xi^{\mu}D_{\mu}(A^{eq}_{\alpha}\xi^{\alpha}+\Lambda)=0\,,\end{split}$
(140)

such that one can solve it using $\xi$ from eq.(4) and obtain

$\begin{split}A^{eq}_{\alpha}\xi^{\alpha}+\Lambda=f(rv,x^{i})\,.\end{split}$
(141)

We now assume that the gauge potential is smooth at $r=0$. Thus,

$f(rv,x^{i})|_{r=0}=f^{(0)}(x^{i})\,,$ (142)

and $f^{(0)}(x^{i})$ is non-singular 484848An example involving the Reissner-
Nordstorm black hole, where the gauge transformation is used to remove the
singular part of the gauge field at the horizon, is given in Section I of
Gao:2003ys . Now, we use the $x^{i}$ component of the stationarity condition
given by eq.(27) to set this to a constant as follows

$\begin{split}\mathcal{L}_{\xi}A^{eq}_{i}+D_{i}\Lambda^{eq}=&\xi^{\alpha}F^{eq}_{\alpha
i}+D_{i}(A^{eq}_{\alpha}\xi^{\alpha}+\Lambda)=0\,\implies
D_{i}(A^{eq}_{\alpha}\xi^{\alpha}+\Lambda)|_{r=0}=0\,,\\\ &\implies
f^{(0)}(x^{i})=c\,.\end{split}$ (143)

In the second step, we have evaluated the expression of the RHS on $r=0$:
$\xi^{\alpha}F^{eq}_{\alpha i}|_{r=0}=vF^{eq}_{vi}|_{r=0}=0$. This follows
because $F_{vi}$ is a component of a covariant tensor that has positive boost
weight. Thus, an equilibrium configuration evaluates to zero. In the third
step, $c$ is a constant and it can be set to zero by using the gauge freedom
in $\Lambda$. Putting the results in eq.(141) and eq.(143) together with
setting $c$ to zero,

$(A^{eq}_{\alpha}\xi^{\alpha}+\Lambda)|_{r=0}=0\,.$ (144)

This is a simplified version of the argument presented in Proposition 3.1 of
Prabhu:2015vua . This is also explained in section III of Gao:2001ut . This
completes the proof of eq.(28).

##  Appendix C Modified generic structure of any covariant tensor with
positive boost weight

In this appendix, we will show that “Result:1” of appendix-E of
Bhattacharyya:2021jhr  will be generically modified once we consider gauge
fields coupled to gravity. Our aim will be to give a detailed analysis
justifying eq.(79) once we have additional building blocks in the form of
gauge field components $A_{v},A_{r},A_{i}$. In the scalar case, the explicit
$v$ dependence of $E_{vv}$ was through boost invariant quantities like the
metric functions $(X,\omega_{i},h_{ij})$ and the scalar field $\phi$. The
structure of $E_{vv}$ was constrained by how $\partial_{v}$ and $\partial_{r}$
derivatives act on them. Without the gauge fields in Bhattacharyya:2021jhr
and in the scalar case of §5.2, $\partial_{v}$ and $\partial_{r}$ were the
only objects that had non-trivial boost weights. Thus, from Result:1 of
appendix E of Bhattacharyya:2021jhr , we have that, any covariant tensor of
boost weight $a+1>0$ that doesn’t explicitly contain $\xi^{\mu}$ can always be
re-expressed as follows

$\begin{split}t&{}^{(k)}_{(a+1)}|_{r=0}=\tilde{T}_{(-k)}\partial^{(k+a+1)}_{v}T_{(0)}|_{r=0}+\mathcal{O}(\epsilon^{2})\\\
&=\partial^{a+1}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}~{}(^{m+a}C_{m})~{}\tilde{T}_{(-k+m)}\partial^{k-m}_{v}T_{(0)}\right]+(-1)^{k}~{}^{k+a}C_{a}~{}\tilde{T}_{(0)}\partial^{a+1}_{v}T_{(0)}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(145)

where $\tilde{T}_{(-k+m)}=\partial^{m}_{v}\tilde{T}_{(-k)}$ and we are not
being very careful with the indices and the numbers in the $()$ of subscripts
denote the boost weights. But once we introduce a gauge field $A_{\mu}$, we
have $A_{v}$ and $A_{r}$ as additional components that transform non-trivially
under boost transformation. Thus, Result:1 must be modified by accounting for
these extra fields. However, $A_{v}$ and $A_{r}$ cannot arbitrarily appear in
the covariant tensors constructed out of a gauge-invariant Lagrangian of the
form considered in eq.(64). The only place where $A_{\mu}$ explicitly appears
is in the term $U^{r\mu}_{0}\delta A_{\mu}$ of $\Theta^{r}$ and we have seen
in §6.3 that this term doesn’t contribute to eq.(76). Thus, for the
modification of Result:1, we can completely ignore this term. Since the
Lagrangian is gauge invariant, in all other terms of $\Theta^{\mu}$ and
$Q^{\mu\nu}$, $A_{v}$ and $A_{r}$ always arise in the form of $F_{vr}$,
$F_{vi}$, $F_{ri}$ (where $F_{\mu\nu}$ is the field strength tensor). Of these
$F_{vr}$ is boost invariant but $F_{vi}$ and $F_{ri}$ have non-trivial boost
weights.

From the details of appendix-A, we know that
$F_{vi}|_{r=0}\sim\mathcal{O}(\epsilon)$ as it is a covariant tensor of boost
weight $+1$. From the structure of $\Theta^{\mu}$ and $Q^{\mu\nu}$ worked out
in section §6.1, it is clear that, Result:1 will be generically modified by
incorporating new non zero boost weight elements in $F_{vi}$ and $F_{ri}$ as
494949The ’$i$’ index can appear in $\tilde{T}_{(-k)}$ as well but we consider
this form without loss of generality. Also, we are not taking into account the
$U^{\mu\nu}_{0}\delta A_{\nu}$ term.

$\begin{split}t^{(k)}_{(a+1)}|_{r=0}=&\tilde{T}_{(-k)}\,\partial^{k+a+1}_{v}\,T_{(0)}|_{r=0}+\tilde{T}_{1(-k)}\,(F_{vi})^{m_{1}}\partial^{k-m_{1}+a+1}_{v}\,T^{i}_{1(0)}|_{r=0}\\\
&+\tilde{T}_{2(-k)}\,\partial^{k-m_{2}+a+1}_{v}\,((F_{vi})^{m_{2}}\,T^{i}_{2(0)})|_{r=0}\\\
&+\tilde{T}_{3(-k)}\,(F_{ri})^{m_{3}}\,\partial^{k+m_{3}+a+1}_{v}\,T^{i}_{3(0)}|_{r=0}\\\
&+\tilde{T}_{4(-k)}\,\partial^{k+m_{4}+a+1}_{v}\,((F_{ri})^{m_{4}}\,T^{i}_{4(0)})|_{r=0}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(146)

where 505050We have separated out all the occurrences of $F_{vi}$ and
$F_{ri}$, i.e., $\tilde{T}_{n(-k)}$ and $T_{n(0)}$ can only contain gauge
field terms in the form $F_{vr}$ and $F_{ij}$. we assume $k+a+1>m_{1},m_{2}$.
Now, since $F_{vi}|_{r=0}\sim\mathcal{O}(\epsilon)$,

$\tilde{T}_{1(-k)}\,(F_{vi})^{m_{1}}\,\partial^{k-m_{1}+a+1}_{v}\,T^{i}_{1(0)}|_{r=0}\sim\mathcal{O}(\epsilon^{2})\,,$
(147)

and 515151The only term of type $T_{2}$ that survives is $m_{2}=1$.

$\begin{split}\tilde{T}_{2(-k)}\,\partial^{k-m_{2}+a+1}_{v}((F_{vi})^{m_{2}}\,T^{i}_{2(0)})|_{r=0}&=\tilde{T}_{2(-k)}\,\partial^{k-1+a+1}_{v}(F_{vi}\,T^{i}_{2(0)})|_{r=0}+\mathcal{O}(\epsilon^{2})\\\
&=\tilde{T}_{2(-k)}\,\partial^{k+a}_{v}(F_{vi}\,T^{i}_{2(0)})|_{r=0}\,.\end{split}$
(148)

Now,
$\tilde{T}_{3(-k)}\,(F_{ri})^{m_{3}}\,\partial^{k+m_{3}+a+1}_{v}\,T^{i}_{3(0)}|_{r=0}$
is already of the form accounted for by Result:1. So, there is no new analysis
that is needed for this term. Similarly, by using the $m_{4}$ $\partial_{v}$
derivatives, we can easily see that

$\tilde{T}_{4(-k)}\,\partial^{k+m_{4}+a+1}_{v}\,((F_{ri})^{m_{4}}\,T^{i}_{4(0)})|_{r=0}=\tilde{T}_{4(-k)}\,\partial^{k+a+1}_{v}(\mathcal{C}_{(0)})+\mathcal{O}(\epsilon^{2})\,.$
(149)

Thus, this term also doesn’t require new analysis. As a result, any covariant
tensor with boost weight $a+1$ takes the following form

$t^{(k)}_{(a+1)}|_{r=0}=\tilde{T}_{5(-k)}\,\partial^{k+a+1}_{v}\,T_{5(0)}|_{r=0}+\tilde{T}_{2(-k)}\,\partial^{k+a}_{v}(F_{vi}\,T^{i}_{2(0)})+\mathcal{O}(\epsilon^{2})\,.$
(150)

Now, the term with $F_{vi}$ can be re-organized in the following manner (if
$a>0$) using eq.(145) where we substitute $a-1$ for $a$ and
$F_{vi}T^{i}_{2(0)}$ for $T_{(0)}$:

$\begin{split}\tilde{T}_{2(-k)}\,\partial^{k+a}_{v}\,(F_{vi}\,T^{i}_{2(0)})|_{r=0}=&\partial^{a}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}~{}^{m+a-1}C_{m}\,\tilde{T}_{2(-k+m)}\partial^{k-m}_{v}(F_{vi}\,T^{i}_{2(0)})\right]\\\
&~{}+(-1)^{k}~\,{}^{k+a-1}C_{a-1}\,\tilde{T}_{2(0)}\,\partial^{a}_{v}(F_{vi}\,T^{i}_{2(0)})+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(151)

Thus, Result:1 is modified as

$\begin{split}t^{(k)}_{(a+1)}&=\partial^{a+1}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}~{}(^{m+a}C_{m})~{}\tilde{T}_{5(-k+m)}\partial^{k-m}_{v}T_{5(0)}\right]\\\
&~{}~{}+(-1)^{k}~{}^{k+a}C_{a}~{}\tilde{T}_{5(0)}\partial^{a+1}_{v}T_{5(0)}\\\
&~{}~{}+\partial^{a}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}~{}^{m+a-1}C_{m}\,\tilde{T}_{2(-k+m)}\partial^{k-m}_{v}(F_{vi}\,T^{i}_{2(0)})\right]\\\
&~{}~{}+(-1)^{k}~\,{}^{k+a-1}C_{a-1}\,\tilde{T}_{2(0)}\,\partial^{a}_{v}(F_{vi}\,T^{i}_{2(0)})+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(152)

What it essentially states is that, ignoring the co-efficients, we have

${\mathbb{T}}_{(k+1)}=\partial_{v}^{k+1}C_{(0)}+A_{(0)}~{}\partial_{v}^{(k+1)}B_{(0)}+\partial^{k}_{v}(F_{vi}\,D^{i}_{(0)})+E_{(0)}\,\partial^{k}_{v}(F_{vi}\,G^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\,.$
(153)

This is a generalization of eq.(3.33) of Bhattacharyya:2021jhr  with gauge
fields. This completes the proof of eq.(79).

##  Appendix D Details regarding the structure of $\Theta^{r}$ for gauge
theories

###  D.1 Gauge invariant terms in $\Theta^{r}$

In this appendix, we show that $\Theta^{\mu}$ inherits some additional
structure from the gauge-invariant Lagrangian given by eq.(64). In this
regard, we will explicitly prove eq.(81) and this can be argued for by working
out $\Theta^{\mu}$ for the Lagrangian given by eq.(64). For example, if the
Lagrangian is of the form

$L=L(F_{\mu\nu},D_{\alpha_{1}}F_{\mu\nu},D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu},...)\,,$
(154)

then we consider the following term in the standard variation,

$\begin{split}\dfrac{\partial L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\delta
D_{\alpha_{1}}F_{\mu\nu}&=\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\left[D_{\alpha_{1}}\delta
F_{\mu\nu}-\delta\Gamma^{\beta}_{\alpha_{1}\mu}F_{\beta\nu}-\delta\Gamma^{\beta}_{\alpha_{1}\nu}F_{\mu\beta}\right]\\\
&=\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\left[D_{\alpha_{1}}\delta
F_{\mu\nu}+\text{Terms proportional to $D_{\alpha_{1}}\delta
g_{\mu\gamma}$}\right]\,.\end{split}$ (155)

The Terms proportional to $D_{\alpha_{1}}\delta g_{\mu\gamma}$ clearly
contribute to Terms proportional to $\delta g_{\mu\gamma}$ in $\Theta^{\rho}$.
These terms are gauge invariant and they have no role to play in the proof of
eq.(81). For example, consider the following term in eq.(155):

$\begin{split}-\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\delta\Gamma^{\beta}_{\alpha_{1}\mu}F_{\beta\nu}&=-\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}F_{\beta\nu}\left[\dfrac{1}{2}g^{\beta\gamma}D_{\alpha_{1}}\delta
g_{\mu\gamma}+\dots\right]\\\
&=-D_{\alpha_{1}}\left[\dfrac{1}{2}\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}F^{\gamma}_{~{}\nu}\delta
g_{\mu\gamma}\right]+\dfrac{1}{2}D_{\alpha_{1}}\left[\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}F^{\gamma}_{~{}\nu}\right]\delta
g_{\mu\gamma}+\dots\,.\end{split}$ (156)

(i.e.) such terms contribute a term of the form $S^{\nu\alpha\beta}\delta
g_{\alpha\beta}$ in $\Theta^{\nu}$ where $S^{\nu\alpha\beta}$ is gauge
invariant. Thus, we only need to analyze

$\begin{split}\dfrac{\partial
L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}D_{\alpha_{1}}\delta
F_{\mu\nu}&=2\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}D_{\alpha_{1}}\delta
D_{\mu}A_{\nu}\\\
&=2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\delta
D_{\mu}A_{\nu}\right]-2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\right]\delta
D_{[\mu}A_{\nu]}\\\
&=2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\delta
D_{\mu}A_{\nu}\right]-2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\right]D_{[\mu}\delta
A_{\nu]}\\\
&=2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\delta
D_{\mu}A_{\nu}\right]-2D_{\mu}\left[D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\right]\delta
A_{\nu}\right]\\\
&~{}~{}+2D_{\mu}D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\mu\nu})}\right]\delta
A_{\nu}\,.\end{split}$ (157)

In the third step, we have crucially used the fact that the indices $\mu$ and
$\nu$ are antisymmetric to interchange $\delta$ and $D_{\mu}$. We can do this
because

$\begin{split}\delta D_{\mu}A_{\nu}&=D_{\mu}\delta
A_{\nu}-\delta\Gamma^{\rho}_{\mu\nu}\delta A_{\rho}\\\ \implies\delta
D_{[\mu}A_{\nu]}&=D_{[\mu}\delta A_{\nu]}\,.\end{split}$ (158)

Thus, the contribution of the first term in eq.(155) to $\Theta^{\rho}$ is

$\begin{split}\Theta^{\rho}&=2\dfrac{\partial\,L}{\partial(D_{\rho}F_{\mu\nu})}\delta
D_{\mu}A_{\nu}-D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\rho\nu})}\right]\delta
A_{\nu}~{}+S^{\rho\alpha\beta}_{1}\delta g_{\alpha\beta}\\\
&=\dfrac{\partial\,L}{\partial(D_{\rho}F_{\mu\nu})}\delta
F_{\mu\nu}-D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{\alpha_{1}}F_{\rho\nu})}\right]\delta
A_{\nu}+S^{\rho\alpha\beta}_{1}\delta g_{\alpha\beta}\,.\end{split}$ (159)

Notice the anti-symmetry predicted by eq.(81) is present in the first term of
the above equation. The lessons learned from the structures arising from the
simplest term can be straightforwardly carried over to more complicated
examples. In what follows, we won’t carefully work out the exact form of the
terms that contribute to $S^{\nu\alpha\beta}\delta g_{\alpha\beta}$ in
$\Theta^{\nu}$. We will indicate those terms which will eventually contribute
to $S^{\mu\alpha\beta}$ by “$\dots$”. For example, another possible term in
the standard variation gives

$\begin{split}&\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}\delta
D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu}=2\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}D_{(\alpha_{1}}\delta
D_{\alpha_{2})}D_{\mu}A_{\nu}+\dots\\\
&=2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}\delta
D_{\alpha_{2}}D_{\mu}A_{\nu}\right]-2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}\right]\delta
D_{\alpha_{2}}D_{\mu}A_{\nu}\\\
&=2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}\delta
D_{\alpha_{2}}D_{\mu}A_{\nu}\right]-2D_{\alpha_{2}}\left[D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}\right]\delta
D_{\mu}A_{\nu}\right]\\\
&~{}~{}+2D_{\alpha_{2}}D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\mu\nu})}\right]\delta
D_{\mu}A_{\nu}+\dots\,.\end{split}$ (160)

Thus from a similar analysis of the term of eq.(155), the contribution of this
term to $\Theta^{\rho}$ can be worked out to be

$\begin{split}\Theta^{\rho}&=2\dfrac{\partial\,L}{\partial(D_{(\rho}D_{\alpha_{2})}F_{\mu\nu})}\delta
D_{\alpha_{2}}D_{\mu}A_{\nu}-2D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\rho)}F_{\mu\nu})}\right]\delta
D_{\mu}A_{\nu}\\\
&~{}~{}+2D_{\alpha_{2}}D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\rho\nu})}\right]\delta
A_{\nu}+~{}S^{\rho\alpha\beta}_{2}\delta g_{\alpha\beta}\\\
&=\dfrac{\partial\,L}{\partial(D_{(\rho}D_{\alpha_{2})}F_{\mu\nu})}\delta
D_{\alpha_{2}}F_{\mu\nu}-D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\rho)}F_{\mu\nu})}\right]\delta
F_{\mu\nu}\\\
&~{}~{}+2D_{\alpha_{2}}D_{\alpha_{1}}\left[\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2})}F_{\rho\nu})}\right]\delta
A_{\nu}+S^{\rho\alpha\beta}_{2}\delta g_{\alpha\beta}\,.\end{split}$ (161)

Here $S^{\rho\alpha\beta}_{2}$ is again gauge invariant. The anti-symmetry of
eq.(81) is again seen in the first 2 terms of the above equation. These
observations can be generalized straightforwardly to a term containing any
number of derivatives acting on the gauge field given by

$\dfrac{\partial\,L}{\partial(D_{(\alpha_{1}}D_{\alpha_{2}}\dots
D_{\alpha_{n})}F_{\mu\nu})}\delta D_{(\alpha_{1}}D_{\alpha_{2}}\dots
D_{\alpha_{n})}F_{\mu\nu}\,.$ (162)

As we repeatedly interchange $\delta$ and $D_{\alpha_{i}}$ to extract
$\Theta^{\mu}$, not only do we get terms of the form $S^{\mu\alpha\beta}\delta
g_{\alpha\beta}$ in $\Theta^{\mu}$, but also we get terms of the form given in
eq.(81) by integration by parts manipulation. In the end, there will be one
term of the form $\delta D_{\mu}A_{\nu}$. But since we vary the Lagrangian
with respect to gauge invariant quantities like $D_{(\alpha_{1}}\dots
D_{\alpha_{i})}F_{\mu\nu}$, there will always be an antisymmetry in the
indices of the gauge field and the covariant derivative acting on it. This is
a direct consequence of the gauge invariant structure of the Lagrangian given
by eq.(64).

###  D.2 Non gauge invariant terms in $\Theta^{r}$

In this appendix, we will detail the structure of non gauge invariant terms in
$\Theta^{r}$. First we will prove eq.(83). From the arguments of appendix D.1,
we note an additional structure in the $U^{\mu\nu}_{0}\delta A_{\nu}$ term of
$\Theta^{\mu}$. From eq.(159) and eq.(161), we see that

$U^{\mu\nu}_{0}=-U^{\nu\mu}_{0}\,.$ (163)

This is because the term with $\delta A_{\nu}$ is contracted with a term which
contains $F_{\rho\nu}$. Thus, the only non-gauge invariant term in
$\Theta^{\mu}$ still has additional structure due to gauge invariance of the
Lagrangian.

We will now detail the calculations of §6.3 and show that this non-gauge
invariant term drops out from our main equation eq.(76). In
$\Theta^{r}(\delta_{\xi}A_{\mu})|_{r=0}$ (here $\delta_{\xi}A_{\mu}$ is given
by eq.(66), i.e.,
$\delta_{\xi}A_{\mu}=\mathcal{L}_{\xi}A_{\mu}+D_{\mu}\Lambda$), we have

$U^{r\mu}_{0}\delta_{\xi}A_{\mu}|_{r=0}=U^{rv}_{0}\delta_{\xi}A_{v}|_{r=0}+U^{ri}_{0}\delta_{\xi}A_{i}|_{r=0}\,.$
(164)

The RHS can be straightforwardly evaluated using eq.(2.20) of
Bhattacharyya:2021jhr  with the additional gauge transformation contribution
$D_{\mu}\Lambda$ given according to eq.(66). As a result, we have the
following

$\begin{split}U^{rv}_{0}\delta_{\xi}A_{v}|_{r=0}&=U^{rv}_{0}(1+v\partial_{v})A_{v}+U^{rv}_{0}\partial_{v}\Lambda=U^{rv}_{0}\partial_{v}(vA_{v}+\Lambda)\\\
&=\partial_{v}\left[U^{rv}_{0}(vA_{v}+\Lambda)\right]+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(165)

where we have used the eq.(31), i.e.,
$(vA_{v}+\Lambda)|_{r=0}\sim\mathcal{O}(\epsilon)$. Similarly, for the other
component we have,

$\begin{split}U^{ri}_{0}\delta_{\xi}A_{i}|_{r=0}&=U^{ri}_{0}\left(vF_{vi}+\partial_{i}[vA_{v}+\Lambda]\right)=\mathcal{O}(\epsilon^{2})\,,\end{split}$
(166)

where in the first step, we have used eq.(66) and in the second step, we have
used $F_{vi}|_{r=0}\sim\mathcal{O}(\epsilon)$ and
$(vA_{v}+\Lambda)|_{r=0}\sim\mathcal{O}(\epsilon)$. Thus, the $\Theta^{r}$
contribution of $U^{r\nu}_{0}\delta A_{\nu}$ is

$\partial_{v}\left[U^{rv}_{0}(vA_{v}+\Lambda)\right]+\mathcal{O}(\epsilon^{2})\,.$
(167)

In $Q^{\mu\nu}$ analogously, we only need to consider
$Q^{\mu\nu}=U^{\mu\nu}_{0}(A^{\alpha}\xi_{\alpha}+\Lambda)$,

$Q^{rv}|_{r=0}=U^{rv}_{0}(v\,A_{v}+\Lambda)\,,\hskip
56.9055ptQ^{ri}|_{r=0}=U^{ri}_{0}(v\,A_{v}+\Lambda)\sim\mathcal{O}(\epsilon^{2})\,.$
(168)
$\begin{split}D_{\mu}Q^{r\mu}|_{r=0}&=\dfrac{1}{\sqrt{h}}\partial_{v}(\sqrt{h}\,Q^{rv})+\nabla_{i}Q^{ri}=\dfrac{1}{\sqrt{h}}\partial_{v}\left[\sqrt{h}\,(U^{rv}_{0}[v\,A_{v}+\Lambda])\right]\\\
&=\partial_{v}(U^{rv}_{0}[v\,A_{v}+\Lambda])+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(169)

From eq.(167), we know that the $\Theta^{r}$ contribution of
$U^{r\nu}_{0}\delta A_{\nu}$ is

$\partial_{v}\left[U^{rv}_{0}(vA_{v}+\Lambda)\right]+\mathcal{O}(\epsilon^{2})\,.$
(170)

Thus,

$-\Theta^{r}+D_{\mu}Q^{r\mu}=\mathcal{O}(\epsilon^{2})\,,$ (171)

for $U^{\mu\nu}_{0}\delta A_{\nu}$. The only non-gauge invariant term drops
out within our approximation from eq.(76).

###  D.3 Final form of $\Theta^{r}$

In this appendix, we show that the additional terms that arise due to the
modification of Result:1 with gauge fields in eq.(93) will not affect the
arguments following eqn.(3.33) up to eqn.(3.41) of Bhattacharyya:2021jhr
(arguments of Section 3.2.2 of Bhattacharyya:2021jhr ), i.e., we must show
that

$\begin{split}\sum_{k}\tilde{\mathbb{T}}_{(-k)}(k+1+v\partial_{v})\mathbb{T}_{(k+1)}=(1+v\partial_{v})\mathcal{M}_{(1)}+v\partial^{2}_{v}\mathcal{N}_{(0)}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(172)

where $\mathbb{T}_{(k+1)}$ is given by the extra terms present in eq.(79):

$\mathbb{T}_{k+1}=\partial^{k}_{v}(F_{vi}\,D^{i}_{(0)})+E_{(0)}\,\partial^{k}_{v}(F_{vi}\,G^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\,.$
(173)

Now we collect a couple of results that we need to prove eq.(172): 525252Note
that these results are not contradicting eq.(145) since here we are pulling
out an extra $\partial_{v}$ derivative when compared to eq.(145).

$\begin{split}\tilde{T}_{(-n)}\partial^{n}_{v}(F_{vi}T^{i}_{(0)})&=\partial_{v}\left[\sum^{n-1}_{m=0}(-1)^{m}\tilde{T}_{(-n+m)}\partial^{n-(m+1)}_{v}(F_{vi}T^{i}_{(0)})\right]\\\
&~{}~{}+(-1)^{n}\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(174)
$\begin{split}\tilde{T}_{(-n)}\partial^{n+1}_{v}(F_{vi}T^{i}_{(0)})&=\partial^{2}_{v}\left[\sum^{n-1}_{m=0}(-1)^{m}(m+1)\tilde{T}_{(-n+m)}\partial^{n-(m+1)}_{v}(F_{vi}T^{i}_{(0)})\right]\\\
&~{}~{}+(-1)^{n}(n+1)\partial_{v}(\tilde{T}_{(0)}F_{vi}T^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(175)

(where $\tilde{T}_{(-n+m)}=\partial^{m}_{v}\tilde{T}_{(-n)}$) These can be
straightforwardly proved by method of induction (see the last part of this
appendix D.3 for the proofs of eq.(174) and eq.(175))

Now,

$\begin{split}\tilde{\mathbb{T}}_{(-k)}(k+1+v\partial_{v})\mathbb{T}_{(k+1)}&=\tilde{\mathbb{T}}_{(-k)}(k+1+v\partial_{v})\left[\partial^{k}_{v}(F_{vi}\,D^{i}_{(0)})+E_{(0)}\,\partial^{k}_{v}(F_{vi}\,G^{i}_{(0)})\right]\\\
&=(k+1)\tilde{\mathbb{T}}_{(-k)}\partial^{k}_{v}(F_{vi}\,D^{i}_{(0)})+v\tilde{\mathbb{T}}_{(-k)}\partial^{k+1}_{v}(F_{vi}\,D^{i}_{(0)})\\\
&\,+(k+1)\tilde{\mathbb{T}}_{(-k)}E_{(0)}\partial^{k}_{v}(F_{vi}\,G^{i}_{(0)})+v\tilde{\mathbb{T}}_{(-k)}E_{(0)}\partial^{k+1}_{v}(F_{vi}\,G^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\\\
&\sim(k+1)X_{(-k)}\partial^{k}_{v}(F_{vi}\,Y^{i}_{(0)})+vX_{(-k)}\partial^{k+1}_{v}(F_{vi}\,Y^{i}_{(0)})\,.\end{split}$
(176)

Using the results, eq.(174) and eq.(175),

$\begin{split}X_{(-k)}\partial^{k}_{v}(F_{vi}Y^{i}_{(0)})&=\partial_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}\,Y^{i}_{(0)})\right]\\\
&~{}~{}+(-1)^{k}X_{(0)}F_{vi}Y^{i}_{(0)}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(177)
$\begin{split}X_{(-k)}\partial^{k+1}_{v}(F_{vi}Y^{i}_{(0)})&=\partial^{2}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}(m+1)X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}\,Y^{i}_{(0)})\right]\\\
&~{}~{}+(-1)^{k}(k+1)\,\partial_{v}(X_{(0)}F_{vi}Y^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(178)

Using these, one can straightforwardly show that the terms in $\Theta^{r}$ are
of the desired form:

$\begin{split}(k+1)&X_{(-k)}\partial^{k}_{v}(F_{vi}Y^{i}_{(0)})+vX_{(-k)}\partial^{k+1}_{v}(F_{vi}Y^{i}_{(0)})\\\
&=(k+1)\partial_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\\\
&~{}+(-1)^{k}(k+1)X_{(0)}F_{vi}Y^{i}_{(0)}\\\
&~{}+v\partial^{2}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}(m+1)X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\\\
&~{}+(-1)^{k}(k+1)\partial_{v}\left[X_{(0)}F_{vi}Y^{i}_{(0)}\right]\,,\\\
&=(1+v\partial_{v})\left[(-1)^{k}(k+1)X_{(0)}F_{vi}Y^{i}_{(0)}\right]\\\
&~{}+(k+1)\partial_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\\\
&~{}+v\partial^{2}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}(k+1+m-k)X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\,,\\\
&=(1+v\partial_{v})\left[\partial_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\right.\\\
&~{}\left.+(-1)^{k}(k+1)X_{(0)}F_{vi}Y^{i}_{(0)}\right]\\\
&-v\partial^{2}_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}(k-m)X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\,,\\\
&\sim(1+v\partial_{v})\mathcal{M}_{(1)}+v\partial^{2}_{v}\mathcal{N}_{(0)}\,.\end{split}$
(179)

where

$\begin{split}\mathcal{M}_{(1)}&=\partial_{v}\left[\sum^{k-1}_{m=0}(-1)^{m}X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\right]\\\
&~{}+(-1)^{k}(k+1)X_{(0)}F_{vi}Y^{i}_{(0)}\,,\\\
\mathcal{N}_{(0)}&=\sum^{k-1}_{m=0}(-1)^{m}(k-m)X_{(-k+m)}\partial^{k-(m+1)}_{v}(F_{vi}Y^{i}_{(0)})\,.\end{split}$
(180)

This completes the proof of eq.(172).

We have already seen how $U^{r\nu}\delta A_{\nu}$ term drops out from the
eq.(76) in §6.3. Hence, even if we have non-zero $A_{v}$ and $A_{r}$,
$\Theta^{r}$ given by eq.(84) takes the required form needed for our proof,
i.e., $\Theta^{r}$ takes the form

$\Theta^{r}=(1+v\partial_{v})\mathcal{A}_{(1)}+v\partial^{2}_{v}\mathcal{B}_{(0)}+\mathcal{O}(\epsilon^{2})\,.$
(181)

Most importantly note that $\mathcal{N}_{(0)}$ given by eq.(95), which
contributes $\mathcal{J}^{v}$ (according to eq.(101)) is gauge-invariant.

It is interesting to note that the anti-symmetry in $U^{\mu\nu}_{0}$ was
crucial for ensuring the gauge invariance of entropy current. Suppose, we
didn’t have anti-symmetry, then we would have a term of the form $U^{rr}_{0}$.
Then, we have

$\begin{split}U^{rr}_{0}\mathcal{L}_{\xi}A_{r}&=U^{rr}_{0}(-1+v\partial_{v})A_{r}\\\
&=(1+v\partial_{v})\mathcal{C}_{(1)}+v\partial^{2}_{v}\mathcal{D}_{(0)}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(182)

where we used eqn(3.51) of Bhattacharyya:2021jhr  in the second step. Note
that $\mathcal{D}_{(0)}$ will explicitly contain $A_{r}$ type terms and it
will contribute to $\mathcal{J}^{v}$. This will spoil the gauge invariance of
$\mathcal{J}^{v}$.

### Proof of intermediary result eq.(174)

We will prove eq.(174) using the method of induction. First we check for
$n=1$:

$\tilde{T}_{(-1)}\partial_{v}(F_{vi}T^{i}_{(0)})=\partial_{v}\left[\tilde{T}_{(-1)}F_{vi}T^{i}_{(0)}\right]-\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}\,.$
(183)

This is trivially of the form predicted in eq.(174). Now we assume that the
result is valid for $n$:

$\begin{split}\tilde{T}_{(-n)}\partial^{n}_{v}\left[F_{vi}T^{i}_{(0)}\right]=\partial_{v}\left[\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})-(\partial_{v}\tilde{T}_{(-n)})\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})\right.\\\
+\left.(\partial^{2}_{v}\tilde{T}_{(-n)})\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})-\dots\right]+(-1)^{n}\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(184)

Consider the expression for $n+1$:

$\begin{split}\tilde{T}_{(-n-1)}\partial^{n+1}_{v}(F_{vi}T^{i}_{(0)})&=\partial_{v}\left[\tilde{T}_{(-n-1)}\partial^{n}_{v}(F_{vi}T^{i}_{0})\right]-\tilde{T}_{(-n)}\partial^{n}_{v}(F_{vi}T^{i}_{(0)})\,,\\\
&=\partial_{v}\left[\tilde{T}_{(-n-1)}\partial^{n}_{v}(F_{vi}T^{i}_{0})\right]-\partial_{v}\left[\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})\right.\\\
&~{}\left.-\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})+\tilde{T}_{(-n+2)}\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})-\dots\right]\\\
&~{}-(-1)^{n}\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}+\mathcal{O}(\epsilon^{2})\,,\\\
&=\partial_{v}\left[\tilde{T}_{(-n-1)}\partial^{n}_{v}(F_{vi}T^{i}_{(0)})-\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})\right.\\\
&~{}+\left.\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})-\tilde{T}_{(-n+2)}\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})+\dots\right]\\\
&~{}+(-1)^{n+1}\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(185)

where in the second step, we have used the result for $n$ in eq.(184). This is
of the required form in eq.(174) we tried to prove.

### Proof of intermediary result eq.(175)

We will prove eq.(175) by using method of induction. For $n=1$, we have

$\begin{split}\tilde{T}_{(-1)}\partial^{2}_{v}(F_{vi}T^{i}_{(0)})&=\partial_{v}\left[\tilde{T}_{(-1)}\partial_{v}(F_{vi}T^{i}_{(0)})\right]-\partial_{v}\tilde{T}_{(-1)}\partial_{v}(F_{vi}T^{i}_{(0)})\\\
&=\partial^{2}_{v}(\tilde{T}_{(-1)}F_{vi}T^{i}_{(0)})-2\partial_{v}(\tilde{T}_{(0)}F_{vi}T^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(186)

This is trivially of the form given by eq.(175). Now we assume that the result
is valid for $n$:

$\begin{split}\tilde{T}_{(-n)}\partial^{n+1}_{v}(F_{vi}T^{i}_{(0)})&=\partial^{2}_{v}\left[\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})-2\,\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})\right.\\\
&+\left.3\,\tilde{T}_{(-n+2)}\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})+\dots\right]+(-1)^{n}(n+1)\partial_{v}(\tilde{T}_{(0)}F_{vi}T^{i}_{(0)})+\mathcal{O}(\epsilon^{2})\,.\end{split}$
(187)

For $n+1$, we have

$\begin{split}\tilde{T}_{(-n-1)}\partial^{n+2}_{v}(F_{vi}T^{i}_{(0)})&=\partial_{v}\left[\tilde{T}_{(-n-1)}\partial^{n+1}_{v}(F_{vi}T^{i}_{(0)})\right]-\tilde{T}_{(-n)}\partial^{n+1}_{v}(F_{vi}T_{(0)})\,,\\\
&=\partial_{v}\left[\tilde{T}_{(-n-1)}\partial^{n+1}_{v}(F_{vi}T^{i}_{(0)})\right]-\partial^{2}_{v}\left[\tilde{T}_{(-n)}\partial_{v}(F_{vi}T^{i}_{(0)})\right.\\\
&\left.-2\,\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})+3\,\tilde{T}_{(-n+2)}\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})-\dots\right]\\\
&-(-1)^{n}(n+1)\partial_{v}\left[\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}\right]+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(188)

where in the second equality, we have used the previous equation. Now, we will
use the result of eq.(174) for $n+1$ in the above equation since we already
proved it for all $n$ in appendix D. RHS of the above equation thus becomes

$\begin{split}&=\partial^{2}_{v}\left[\tilde{T}_{(-n-1)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})-\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})+\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})-\dots\right]\\\
&~{}~{}+(-1)^{n+1}\partial_{v}\left[\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}\right]\\\
&+\partial^{2}_{v}\left[-\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})+2\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})-3\tilde{T}_{(-n+2)}\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})-\dots\right]\\\
&~{}~{}-(-1)^{n}(n+1)\partial_{v}\left[\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}\right]+\mathcal{O}(\epsilon^{2})\,,\\\
&=\partial^{2}_{v}\left[\tilde{T}_{(-n-1)}\partial^{n}_{v}(F_{vi}T^{i}_{(0)})-2\tilde{T}_{(-n)}\partial^{n-1}_{v}(F_{vi}T^{i}_{(0)})+3\tilde{T}_{(-n+1)}\partial^{n-2}_{v}(F_{vi}T^{i}_{(0)})\right.\\\
&\left.-4\tilde{T}_{(-n+2)}\partial^{n-3}_{v}(F_{vi}T^{i}_{(0)})+\dots\right]+(-1)^{n+1}(n+2)\partial_{v}\left[\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}\right]+\mathcal{O}(\epsilon^{2})\,,\\\
&=\partial^{2}_{v}\left[\sum^{n}_{m=0}(-1)^{m}(m+1)\tilde{T}_{(-n-1+m)}\partial^{n+1-(m+1)}_{v}(F_{vi}T^{i}_{(0)})\right]\\\
&~{}~{}+(-1)^{n+1}(n+2)\,\partial_{v}\left[\tilde{T}_{(0)}F_{vi}T^{i}_{(0)}\right]+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(189)

This is of the required form in eq.(175) we tried to prove.

##  Appendix E Details of the proof of the physical process first law

### Conservation of $\alpha^{\mu}$:

Here, we will quickly show why $\alpha^{\mu}$ given in eq.(116) is conserved.
For this, we will need the following expression derived in eq.(71):

$\Theta^{\mu}-\zeta^{\mu}\mathcal{L}_{(A_{\mu}g_{\mu\nu})}=-2\mathcal{E}^{\mu\nu}\zeta_{\nu}-\,G^{\mu}(A^{\nu}\zeta_{\nu}+\Lambda)+D_{\nu}Q^{\mu\nu}\,.$
(190)

The LHS is the usual Noether current
$N^{\mu}=\Theta^{\mu}-\zeta^{\mu}\mathcal{L}_{(A_{\mu}g_{\mu\nu})}$ in the
language of doi:10.1063/1.528801 . The variation of the LHS of eq.(190)
results in the following identity derived in Iyer:1994ys :

$\delta
N^{\mu}=\omega^{\mu}(\psi,\delta\psi,\mathcal{L}_{\zeta}\psi)+\dfrac{1}{2}D_{\nu}(\zeta^{\mu}\Theta^{\nu}-\zeta^{\nu}\Theta^{\mu})\,.$
(191)

Here $\psi$ collectively denotes the fields $A_{\mu}$ and $g_{\mu\nu}$.
$\omega^{\mu}$ is defined by the anti-symmetrized variation of $\Theta^{\mu}$
doi:10.1063/1.528801  535353Here we substitute $\delta_{2}=\mathcal{L}_{\xi}$
to obtain eq.(191).:

$\omega^{\mu}(\psi,\delta_{1}\psi,\delta_{2}\psi)=\delta_{1}\Theta^{\mu}(\psi,\delta_{2}\psi)-\delta_{2}\Theta^{\mu}(\psi,\delta_{1}\psi)\,.$
(192)

If we now restrict to the case $\zeta^{\mu}=\xi^{\mu}$ of eq.(4), which is a
Killing symmetry of the background spacetime and the fields $\psi$, then we
have $\mathcal{L}_{\xi}\psi=0$, and this implies

$\omega^{\mu}(\psi,\delta\psi,\mathcal{L}_{\xi}\psi)=0\,.$ (193)

Substituting eq.(193) in eq.(191), we have

$\delta
N^{\mu}=\dfrac{1}{2}D_{\nu}(\xi^{\mu}\Theta^{\nu}-\xi^{\nu}\Theta^{\mu})\implies
D_{\mu}\delta N^{\mu}=0\,.$ (194)

The variation of the RHS of eq.(191) results in

$\delta N^{\mu}=-\alpha^{\mu}+D_{\nu}\delta Q^{\mu\nu}\,,$ (195)

where $\alpha^{\mu}=2\,\delta\mathcal{E}^{\mu\nu}\zeta_{\nu}+\,\delta
G^{\mu}(A^{\nu}\zeta_{\nu}+\Lambda)$. Using eq.(194) in eq.(195), we
straightforwardly have

$D_{\mu}\alpha^{\mu}=0\,.$ (196)

### Alternate perspective on the physical process first law:

In this section, we wish to emphaisze the point of view taken in
Chakraborty:2017kob  to prove the physical process version of the first law.
This serves as an alternative perspective of the same. We start with the
following generic structure of the entropy

$S=4\pi\int_{\mathcal{H}}d^{d-2}x\,\sqrt{h}\,(1+s_{w})\,,$ (197)

where $s_{w}=s^{\text{HD}}_{w}+s_{cor}$. $s^{\text{HD}}_{w}$ denotes the Wald
entropy contribution from the higher derivative terms and $s_{cor}$ denotes
the out-of-equilibrium extension of the entropy in dynamical situations
Wall:2015raa ; Bhattacharya:2019qal ; Bhattacharyya:2021jhr . For
concreteness, we consider a theory of the form

$\mathcal{L}_{(A_{\mu}g_{\mu\nu})}=\sqrt{-g}\left(R-\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}+L_{\text{non-
min}}\right)\,.$ (198)

Thus, it is clear why we chose to explicitly mention the area term
(contributed by the Einstein-Maxwell term) in eq.(197); the $L_{\text{non-
min}}$ is of the form given in eq.(111) and it will contribute to $s_{w}$. The
equations of motion are thus

$\mathcal{E}_{\mu\nu}=-R_{\mu\nu}+\dfrac{1}{2}g_{\mu\nu}R+T^{\text{EM}}_{\mu\nu}+\widetilde{T}_{\mu\nu}=0\,,$
(199)

where $T^{\text{EM}}_{\mu\nu}$ denotes the contribution from the Maxwell
Lagrangian and $\widetilde{T}_{\mu\nu}$ denotes the contribution from the
higher derivative non-minimal couplings. We have to relate the change in
entropy of eq.(197) to the source term given in eq.(119). The variation of the
entropy given by eq.(197) due to some infalling matter is given by 545454Here
we are using our setup described in the gauge eq.(2).

$\delta
S=4\pi\int_{\mathcal{H}}d^{d-2}x\,\dfrac{d}{dv}[\sqrt{h}\,(1+s_{w})]\,dv=4\pi\int_{\mathcal{H}}\sqrt{h}\,d^{d-2}x\int
dv\,\vartheta_{k}\,,$ (200)

where $\vartheta_{k}$ is the generalized expansion given by

$\vartheta_{k}=\theta_{k}+s_{w}\,\theta_{k}+\dfrac{d\,s_{w}}{dv}\,.$ (201)

Here $\theta_{k}=\dfrac{1}{\sqrt{h}}\partial_{v}\sqrt{h}$ is the expansion of
null generators $k^{\mu}=(\partial_{v})^{\mu}$ of the horizon within General
Relativity. The generalized expansion $\vartheta_{k}$ reduces to $\theta_{k}$
when $s_{w}=0$ as for GR. Using integration by parts in eq.(200) to find the
change in entropy between two different stationary slices $v=v_{1}$ and
$v=v_{2}$ of the horizon, we have

$\delta
S=4\pi\left(\int_{\mathcal{H}}\sqrt{h}\,d^{d-2}x\,v\,\vartheta_{k}\right)^{v_{2}}_{v_{1}}-4\pi\int_{\mathcal{H}}dA\,dv\,v\dfrac{d\,\vartheta_{k}}{dv}\,.$
(202)

Now,

$\begin{split}\dfrac{d\,\vartheta_{k}}{dv}|_{\mathcal{H}}&=(1+s_{w})\dfrac{d\,\theta_{k}}{dv}+\theta_{k}\dfrac{d\,s_{w}}{dv}+\dfrac{d^{2}\,s_{w}}{dv^{2}}\\\
&=(1+s_{w})\left(-\dfrac{(d-3)}{d-2}\theta^{2}_{k}-\sigma^{2}_{k}-R_{\mu\nu}k^{\mu}k^{\nu}\right)+\theta_{k}\dfrac{d\,s_{w}}{dv}+\dfrac{d^{2}\,s_{w}}{dv^{2}}\\\
&\approx-
R_{\mu\nu}k^{\mu}k^{\nu}+\dfrac{d^{2}\,s_{w}}{dv^{2}}-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}+\mathcal{O}(\epsilon^{2})\\\
&=\dfrac{d^{2}\,s_{w}}{dv^{2}}-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}-\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}-T^{\text{EM}}_{\mu\nu}k^{\mu}k^{\nu}-\dfrac{1}{2}\delta
T_{\mu\nu}k^{\mu}k^{\nu}\\\
&\approx\dfrac{d^{2}\,s_{w}}{dv^{2}}-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}-\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}-\dfrac{1}{2}\delta
T_{\mu\nu}k^{\mu}k^{\nu}+\mathcal{O}(\epsilon^{2})\,.\end{split}$ (203)

Here, in the second step we have used the $d$ dimensional Raychaudhuri
equation for the congruence of null generators $k^{\mu}$. In the third step,
we have used the amplitude approximation of our gauge in eq.(6) to cast away
all the terms that are quadratic in the amplitude. In the fourth step, we have
basically used the EOM eq.(199) with a source term $\dfrac{1}{2}\delta
T_{\mu\nu}$. In the last step, we have used the fact that
$T^{\text{EM}}_{\mu\nu}k^{\mu}k^{\nu}\approx\mathcal{O}(\epsilon^{2})$ in our
gauge 555555This essentially amounts to the fact that the Maxwell stress
tensor satisfies the null energy condition..

Thus, using the final result of eq.(203) in eq.(202),

$\begin{split}\delta
S&=4\pi\left(\int_{\mathcal{H}}\sqrt{h}\,d^{d-2}x\,v\,\vartheta_{k}\right)^{v_{2}}_{v_{1}}+2\pi\int_{\mathcal{H}}dA\,dv\,v\,\delta
T_{\mu\nu}k^{\mu}k^{\nu}\\\
&~{}~{}-4\pi\int_{\mathcal{H}}dA\,dv\,v\,\left(\dfrac{d^{2}\,s_{w}}{dv^{2}}-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}-\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}\right)\,.\end{split}$
(204)

Since we have assumed that the black hole was stationary before the matter
fell in and it finally settles down to a stationary state after the
perturbation has ended, we can set the boundary term (first term in the above
equation) to zero. Now we use the main result of our paper eq.(100) which
proves that even when we consider a non-minimally coupled gauge theory,
$\widetilde{T}_{vv}$ still has the entropy current structure of the following
form when evaluated on the horizon:

$\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}|_{\mathcal{H}}=\widetilde{T}_{vv}|_{r=0}=\partial_{v}\left[\frac{1}{\sqrt{h}}\partial_{v}\left(\sqrt{h}\,\widetilde{\cal
J}^{v}\right)+\nabla_{i}\widetilde{\cal J}^{i}\right]+{\cal
O}(\epsilon^{2})\,.$ (205)

We can thus use the results of section 5 of Bhattacharyya:2021jhr  to
basically state that

$\int_{\mathcal{H}}dA\,dv\,v\,\left(\dfrac{d^{2}\,s_{w}}{dv^{2}}-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}-\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}\right)=\mathcal{O}(\epsilon^{2})\,.$
(206)

This has been argued for in eq.(5.7) to eq.(5.10) of Bhattacharyya:2021jhr
and the main requirement was the fact that $\widetilde{T}_{vv}$ had the
entropy current structure given by eq.(205). Finally, eq.(204) becomes

$\delta S=2\pi\int_{\mathcal{H}}dA\,dv\,v\,\delta
T_{\mu\nu}k^{\mu}k^{\nu}+\mathcal{O}(\epsilon^{2})\,.$ (207)

The Killing field $\xi^{\mu}$ and the null generators of the horizon $k^{\mu}$
are related as $\xi^{\mu}=\kappa\,v\,k^{\mu}$, where $\kappa$ is the surface
gravity. Thus, finally we get

$\delta S=\dfrac{2\pi}{\kappa}\int_{\mathcal{H}}\delta
T_{\mu\nu}k^{\mu}\xi^{\nu}\,.$ (208)

Combining eq.(119) and eq.(208), we get the desired first law relation that we
originally set out to prove

$\delta M-\Omega_{H}\delta I-\Phi_{bh}\delta q=\dfrac{\kappa}{2\pi}\delta
S\,.$ (209)

The changes in the mass $M$, angular momentum $I$, charge $q$ and the entropy
$S$ are induced by some physical matter falling into the black hole. This
completes the proof of the physical process first law for arbitrary non-
minimally coupled gauge theories. We emphasize that the non-trivial part of
the proof was basically the structure of the entropy current eq.(205) leading
to the result of eq.(206). The result of eq.(206) was first established for
theories of gravity in Bhattacharyya:2021jhr  and we have extended it to non-
minimally coupled gauge theories here.

### Verification of eq.(206) through an example:

In the proof given above, eq.(206) was the crucial step. We argued that this
relation is true even for arbitrary non-minimal coupling of gauge fields with
gravity. To illustrate this through an explicit example, let us consider a
gauge invariant Lagrangian of the form given in eq.(198)

$\mathcal{L}=\sqrt{-g}\left(R-\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}+\alpha\,R_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}\right)\,.$
(210)

The standard variation gives 565656The symmetrization of indices $\alpha$ and
$\beta$ has a factor of $\dfrac{1}{2}$.

$\begin{split}\delta\mathcal{L}&=\sqrt{-g}\left[-R^{\alpha\beta}+\dfrac{1}{2}g^{\alpha\beta}R+\dfrac{1}{2}\left(F^{\alpha\nu}F^{\beta}_{~{}\nu}-\dfrac{1}{4}g^{\alpha\beta}F_{\mu\nu}F^{\mu\nu}\right)\right]\delta
g_{\alpha\beta}\\\
&+\alpha\,\sqrt{-g}\Bigg{[}\dfrac{1}{2}g^{\alpha\beta}R_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}-3R^{(\alpha}_{~{}~{}\nu\rho\sigma}F^{\beta)\nu}F^{\rho\sigma}+2D_{\nu}D_{\rho}(F^{\alpha\nu}F^{\rho\beta})\Bigg{]}\delta
g_{\alpha\beta}\\\ &+~{}\sqrt{-g}\left[D_{\mu}F^{\mu\sigma}-4\alpha
D_{\rho}\left(R^{\rho\sigma}_{\hskip
11.38092pt\mu\nu}F^{\mu\nu}\right)\right]\delta A_{\sigma}\\\
&+\sqrt{-g}D_{\rho}\left[g^{\rho\alpha}D^{\sigma}\delta
g_{\alpha\sigma}-g^{\alpha\sigma}D^{\mu}\delta
g_{\alpha\sigma}-F^{\rho\nu}\delta A_{\nu}\right]\\\
&+\alpha\,\sqrt{-g}D_{\rho}\Bigg{[}2F^{\mu\nu}F^{\rho\sigma}D_{\nu}\delta
g_{\sigma\mu}-2D_{\nu}(F^{\mu\rho}F^{\nu\sigma})\delta
g_{\sigma\mu}+4R^{\rho\sigma}_{\hskip 11.38092pt\mu\nu}F^{\mu\nu}\delta
A_{\sigma}\Bigg{]}\,.\end{split}$ (211)

The total derivative term, $\theta^{\rho}$ is given by

$\begin{split}\theta^{\rho}&=g^{\rho\alpha}D^{\sigma}\delta
g_{\alpha\sigma}-g^{\alpha\sigma}D^{\mu}\delta
g_{\alpha\sigma}-F^{\rho\nu}\delta A_{\nu}\\\
&~{}~{}+\alpha\left[2F^{\mu\nu}F^{\rho\sigma}D_{\nu}\delta
g_{\sigma\mu}-2D_{\nu}(F^{\mu\rho}F^{\nu\sigma})\delta
g_{\sigma\mu}+4R^{\rho\sigma}_{\hskip 11.38092pt\mu\nu}F^{\mu\nu}\delta
A_{\sigma}\right]\,.\end{split}$ (212)

The Noether charge $Q^{\rho\nu}$ is given by

$\begin{split}Q^{\rho\nu}&=D^{\nu}\xi^{\rho}-D^{\rho}\xi^{\nu}-F^{\rho\nu}(A_{\alpha}\xi^{\alpha}+\Lambda)\\\
&~{}~{}+\alpha\,\left[2F^{\rho\nu}F^{\mu\sigma}D_{\sigma}\xi_{\mu}+2D_{\sigma}(F^{\rho\mu}F^{\sigma\nu}-F^{\rho\sigma}F^{\mu\nu})\xi_{\mu}\right.\\\
&~{}~{}+\left.2D_{\mu}(F^{\rho\nu}F^{\mu\sigma})\xi_{\sigma}+4R^{\rho\nu}_{\hskip
11.38092pt\alpha\beta}F^{\alpha\beta}(A_{\gamma}\xi^{\gamma}+\Lambda)\right]\,.\end{split}$
(213)

From these expressions, one can readily obtain the following

$\mathcal{E}_{vv}|_{r=0}=\partial_{v}\left[\dfrac{1}{\sqrt{h}}\partial_{v}(\sqrt{h}\,\left[1-2\alpha\,F^{rv}F^{rv}\right])-\nabla_{i}(4\alpha\,F^{rv}F^{ri})\right]\,.$
(214)

Thus, we have the components of the Entropy current

$\mathcal{J}^{v}=1-2\alpha\,F^{rv}F^{rv},\hskip
85.35826pt\mathcal{J}^{i}=-4\alpha\,F^{ri}F^{rv}\,.$ (215)

If we filter out the non-minimal component of the EOM,

$\widetilde{T}_{vv}|_{r=0}=\partial_{v}\left[\dfrac{1}{\sqrt{h}}\partial_{v}(\sqrt{h}\,\left[-2\alpha\,F^{rv}F^{rv}\right])-\nabla_{i}(4\alpha\,F^{rv}F^{ri})\right]+\mathcal{O}(\epsilon^{2})\,.$
(216)

From $\mathcal{J}^{v}$ of eq.(214), we can read off the expression for entropy
from eq.(197) (or from eq.(5.6) of Bhattacharyya:2021jhr ) as

$S=4\pi\int_{\mathcal{H}}\sqrt{h}\,d^{d-2}x\,\left(1-2\alpha\,F^{rv}F^{rv}\right)\,,$
(217)

and $s_{w}=-2\alpha\,F^{rv}F^{rv}$. Thus, we have

$\begin{split}\dfrac{d^{2}\,s_{w}}{dv^{2}}-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}-\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}&=\partial_{v}\left[\dfrac{1}{\sqrt{h}}\partial_{v}(\sqrt{h}\,s_{w})\right]-\widetilde{T}_{vv}\\\
&=\partial_{v}\left[\nabla_{i}(4\alpha\,F^{rv}F^{ri})\right]+\mathcal{O}(\epsilon^{2})\,,\end{split}$
(218)

where we have used eq.(216) in the second step. Finally, we have eq.(206):

$\begin{split}\int_{\mathcal{H}}dA\,dv\,v\,\left(\dfrac{d^{2}\,s_{w}}{dv^{2}}\right.&\left.-s_{w}\,R_{\mu\nu}k^{\mu}k^{\nu}-\widetilde{T}_{\mu\nu}k^{\mu}k^{\nu}\right)\\\
&=\int_{\mathcal{H}}dA\,dv\,v\,\partial_{v}\left[\nabla_{i}(4\alpha\,F^{rv}F^{ri})\right]+\mathcal{O}(\epsilon^{2})=\mathcal{O}(\epsilon^{2})\,,\end{split}$
(219)

where we have used the assumption of compact horizons in the final step to set
the total derivative term to zero. This illustrates the proof of the physical
process first law from the perspective of Chakraborty:2017kob  for a
particular example.

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