# AdS black brane coupled to non-abelian logarithmic gauge theory and color DC
conductivity

Mehdi Sadeghi
Department of Physics, School of Sciences,
Ayatollah Boroujerdi University, Boroujerd, Iran Corresponding author: Email:
<EMAIL_ADDRESS>
###### Abstract

In this paper, we consider Einstein-Hilbert gravity in the presence of
cosmological constant and non-abelian nonlinear electromagnetic field of
logarithmic type which is minimally coupled to gravity. First, black brane
solution of this model is introduced and then color non-abelian DC
conductivity is calculated for this solution by using AdS/CFT duality. Our
result recovers the Yang-Mills model in $\lambda\to 0$ limit.

PACS numbers: 11.10.Jj, 11.10.Wx, 11.15.Pg, 11.25.Tq

Keywords: Non-abelian color DC Conductivity, Black brane , AdS/CFT duality

##  1 Introduction

Maxwell’s electromagnetic field theory that is a linear theory, suffers from
some problems [1, 2, 3], singularity in the origin of particle, infinite self-
energy, description of the self-interaction of virtual electron-positron pairs
[4], the radiation propagation inside specific materials [5]. Heisenberg and
Euler have shown that quantum electrodynamics (QED) due to loop corrections
gives nonlinear electrodynamics (NED) [4]. NED’s coupled to general relativity
(GR) can explain inflation and current acceleration of the universe [6]-[9].
Logarithmic gauge theory [10],[11],[12],[13],[14], arcsin-electrodynamics
[15], Born-Infeld theory [16, 17] have been introduced as nonlinear
electrodynamics (NED) theories to solve some of problems in Maxwell’s theory.
The NED field gives more information in higher magnetized neutron stars and
pulsars [18][19] and this NED theories remove both of the big bang and black
hole singularities [20]-[21] by modifying space-time geometry.
There are three kinds of Born-Infeld (BI) actions, Born-Infeld nonlinear
electromagnetic (BINEF), logarithmic form of nonlinear electromagnetic field
(LNEF) and exponential form of nonlinear electromagnetic field (ENEF). The
color DC conductivity bound of Einstein–Born–Infeld AdS black brane was
studied in [22] and the same calculation for logarithmic form of Born-Infeld
(BI) action is investigated in this paper. We also will consider the
exponential form of nonlinear electromagnetic field in our future work.
We want to study the field theory dual of the logarithmic gauge theory by
AdS/CFT duality [23]. Therefore, we should calculate the transport
coefficients like conductivity and shear viscosity to entropy density.
Hydrodynamics is an effective theory for describing near-equilibrium phenomena
at late time and large distance [24, 25, 26]. Conservation equations together
with constitutive relations in a gradient expansion around equilibrium is
constructed hydrodynamics equations. These equations up to first order of
derivative expansion are as follows,

$\displaystyle\nabla_{\mu}J^{\mu}=0\,\,\,\,\,,\,\,\nabla_{\mu}T^{\mu\nu}=0,$
(1) $\displaystyle J^{\mu}=n\,u^{\mu}-\sigma
TP^{\mu\nu}\partial_{\nu}(\frac{\mu}{T}),$ $\displaystyle
T^{\mu\nu}=(\rho+p)u^{\mu}u^{\nu}+pg^{\mu\nu}-\sigma^{\mu\nu},$
$\displaystyle\sigma^{\mu\nu}={P^{\mu\alpha}P^{\nu\beta}}[\eta(\nabla_{\alpha}u_{\beta}+\nabla_{\beta}u_{\alpha})+(\zeta-\frac{2}{3}\eta)g_{\alpha\beta}\nabla.u],$
$\displaystyle P^{\mu\nu}=g^{\mu\nu}+u^{\mu}u^{\nu},$

where $n$, $\eta$, $\zeta$, $\sigma^{\mu\nu}$ , $\sigma$ and $P^{\mu\nu}$ are
charge density, shear viscosity, bulk viscosity, shear tensor, conductivity
and projection operator, respectively [27]. $\eta$, $\zeta$ and $\sigma$ are
known as transport coefficients.

The fluid-gravity correspondence as a subclass of gauge-gravity duality states
that black holes in AdS space-time are dual to stationary solutions of the
equations of relativistic hydrodynamics [25]-[30]. Conductivity is calculated
by Green-Kubo formula and Witten prescription.

$\sigma^{ij}_{ab}(k_{\mu})=-\mathop{\lim}\limits_{\omega\to
0}\frac{1}{\omega}\Im G^{ij}_{ab}(k_{\mu}),$ (2)

where $a,b$ and $i,j$ indices refer to $SU(2)$ group symmetry and
translational symmetry in spatial directions of $x,y$, respectively. There is
no difference between upper and lower indices on the conformal field theory
(CFT) side because the metric of the boundary theory is flat spacetime with
Minkowski metric. There is a lower bound for DC conductivity as
$\sigma\geq\frac{1}{e^{2}}=1$ where $e$ is the charge of the gauge field but
$e$ is not the unit of charge in the boundary theory [31, 32]. This bound is
also saturated in graphene[33]. This bound is violated for massive gravity
[34], models with abelian case [35] and non-abelian Born-Infeld theory[22]. In
this paper, we want to investigate the effects of non-abelian logarithmic
gauge theory as a nonlinear electrodynamics gauge field on the conductivity
bound of strongly coupled dual theory.

##  2  AdS black brane coupled to non-abelian logarithmic gauge theory
Solution

The 4-dimensional action of non-abelian logarithmic gauge theory with negative
cosmological constant is [36],[37],

$\displaystyle S=\frac{1}{16\pi G}\int
d^{4}x\sqrt{-g}\bigg{[}R-2\Lambda-\frac{1}{\lambda}\log\big{(}1+\lambda\mathcal{F}\big{)}\bigg{]},$
(3)

where $R$ is the Ricci scalar, $l$ the AdS radius,
$\mathcal{F}={\bf{Tr}}(F_{\mu\nu}^{(a)}F^{(a)\,\,\mu\nu})$ is Yang-Mills
invariant and $\lambda$ is the value of the non linear coupling constant
parameter. The trace element stands for ${\bf{Tr}}(.)=\sum_{i=1}^{3}(.).$
$F_{\mu\nu}^{(a)}$ is the $SU(2)$ Yang-Mills field strength tensor,

$\displaystyle
F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}-i[A_{\mu},A_{\nu}],$
(4)

in which the gauge coupling constant is 1, $A_{\nu}$’s are the $SU(2)$ gauge
group Yang-Mills potentials. When $\lambda\to 0$ the logarithmic term gets
transformed into standard linear non-abelian Yang-Mills term [38].
We get to the equations of motion by varying the action 3 with respect to
$g_{\mu\nu}$ and $A^{(a)}_{\nu}$,

$R_{\mu\nu}-\tfrac{1}{2}g_{\mu\nu}R+\Lambda
g_{\mu\nu}+\tfrac{1}{2\lambda}g_{\mu\nu}\log(1+\lambda\mathcal{F})-\frac{2F^{(a)}_{\mu}\,{}^{\alpha}F^{(a)}_{\nu\alpha}}{1+\
\lambda\mathcal{F}}=0,$ (5)
$\nabla_{\mu}\bigg{(}\frac{F^{{(a)}\mu\nu}}{1+\lambda\mathcal{F}}\bigg{)}=0.$
(6)

Our goal is to find asymptotically AdS black brane solution in four
dimensional flat symmetric spacetime, so we consider the following ansatz,

$ds^{2}=-f(r)dt^{2}+\frac{dr^{2}}{f(r)}+\frac{r^{2}}{l^{2}}(dx^{2}+dy^{2}),$
(7)

where $f(r)$ is the metric function that should be determined and $l$ is a
constant relating to the cosmological constant.
We write the gauge field in terms of matrix $\sigma_{3}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}$, which is the generators of $SU(2)$ [38].

${\bf{A}}^{(a)}=h(r)dt\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$ (8)

Eq.6 can be written as below by considering 6 and 7,

$2h^{\prime}\Big{(}\lambda
h^{\prime}\left(rh^{\prime\prime}-2h^{\prime}\right)+1\Big{)}+rh^{\prime\prime}=0.$
(9)

One could solve the above differential equation to obtain the function $h(r)$
as following,

$h(r)=\mu-\frac{4q^{2}}{3r}\,_{2}F_{1}\left[\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{8\lambda
q^{2}}{r^{4}}\right]+\frac{r^{3}}{12q\lambda}\left(\sqrt{1+\frac{8\lambda
q^{2}}{r^{4}}}-1\right),$ (10)

$\mu$ is a constant of integration and called chemical potential of quantum
field that locates on the boundary of AdS spacetime. It is found by applying
the regularity condition on the horizon i.e. $A_{t}(r_{h})=0$ [39].

$\mu=\frac{4q^{2}}{3r_{h}}\,_{2}F_{1}\left[\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{8\lambda
q^{2}}{r_{h}^{4}}\right]-\frac{r_{h}^{3}}{12q\lambda}\left(\sqrt{1+\frac{8\lambda
q^{2}}{r_{h}^{4}}}-1\right),$ (11)

for the $\lambda=0$, $h(r)$ will be as,

$\displaystyle h(r)=\frac{q}{r},$ (12)

where $q$ is integration constant which is related to the electric charge. The
invariant scalar
$\mathcal{F}_{YM}={\bf{Tr}}(F_{\mu\nu}^{(a)}F^{(a)\,\,\mu\nu})$ for fields is,

$F_{tr}^{(a)}=-F_{rt}^{(a)}=h^{\prime}(r)\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}=\frac{2qr^{-2}}{1+\sqrt{1+\frac{8\lambda
q^{2}}{r^{4}}}}\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},$ (13)

by setting $\lambda=0$, our results decrease to non-abelian Yang-Mills
solution.
The Bianchi identity is satisfied,

$\nabla_{\alpha}F^{(a)}_{\mu\nu}+\nabla_{\nu}F^{(a)}_{\alpha\mu}+\nabla_{\mu}F^{(a)}_{\nu\alpha}=0.$
(14)

By considering $tt$ component of Eq. (5),

$rf^{\prime}(r)+f(r)+\Lambda r^{2}+\frac{r^{2}}{2\lambda}\log\left(1-4\lambda
h^{\prime 2}\right)+\frac{2r^{2}h^{\prime 2}}{1-4\lambda h^{\prime 2}}=0,$
(15)

where $1-4\lambda h^{\prime 2}>0$.
$f(r)$ is obtained by solving Eq.(15)

$\displaystyle f(r)=\frac{m}{r}-\frac{\Lambda
r^{2}}{3}+\frac{16q^{2}}{9r^{2}}\,_{2}F_{1}\left[\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{8\lambda
q^{2}}{r^{4}}\right]$
$\displaystyle+\frac{r^{2}}{6\lambda}\Bigg{[}\log\Big{(}\frac{1+\sqrt{1+\frac{8\lambda
q^{2}}{r^{4}}}}{2}\Big{)}+\frac{5}{3}\Bigg{(}1-\sqrt{1+\frac{8\lambda
q^{2}}{r^{4}}}\Bigg{)}\Bigg{]},$ (16)

where $m$ is an integration constant and ${}_{2}F_{1}$ is the hypergeometric
function. Notice $r$ is the radial coordinate that put us from bulk to the
boundary.
We notice that the topology of event horizon in our model is flat so the
extrinsic curvature is zero, $\kappa=0$. But, the solution of Einstein’s
equations in [40] was considered black hole so the extrinsic curvature is 1,
$\kappa=1$.
Hawking temperature for this black brane is,

$\displaystyle T$
$\displaystyle=\frac{f^{\prime}(r_{h})}{4\pi}=\frac{-1}{4\pi}\Bigg{(}\frac{m}{r_{h}^{2}}+\frac{2\Lambda
r_{h}}{3}+\frac{16q^{2}}{9\pi
r_{h}^{3}}\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{8q^{2}\lambda}{r_{h}^{4}}\right)+\frac{16q^{2}}{9r_{h}^{3}\sqrt{1-\frac{8\lambda
q^{2}}{r_{h}^{4}}}}$ $\displaystyle+\frac{\bigg{(}2\sqrt{\frac{8\lambda
q^{2}}{r_{h}^{4}}+1}-2+\log 8\bigg{)}r_{h}^{8}+8\lambda q^{2}r_{h}^{4}(\log
8-2)}{9\lambda r_{h}^{3}\left(8\lambda
q^{2}+r_{h}^{4}\right)}+\frac{3r_{h}^{4}}{\lambda}\log\left(1+\sqrt{1+\frac{8\lambda
q^{2}}{r_{h}^{4}}}\right)\Bigg{)}.$ (17)

Event horizon is located on where $f(r_{h})=0$ and $m$ is identified by this
condition.

##  3 Color non-abelian DC conductivity

For calculating retarded Green’s function $<J^{\mu}(\omega)J^{\nu}(-\omega)>$,
we perturb the gauge field as $A_{\mu}\to A_{\mu}+\tilde{A}_{\mu}$ and expand
the action up to second order of perturbed part, $\tilde{A}_{\mu}$ [35]\-
[47]. By reading $J_{a}^{\mu}$ from the perurbed part of the action as,

$S\to S+\int d^{4}x\,\,A^{a}_{\mu}J_{a}^{\mu}.$ (18)

Retarded Green’s function read from AdS/CFT correspondence as follows,

$\sigma^{\mu\nu}_{ab}(\omega)=\frac{1}{i\omega}<J^{\mu}_{a}(\omega)J^{\nu}_{b}(-\omega)>=\frac{\delta^{2}S}{\delta\tilde{A}^{0\,(a)}_{\mu}\delta\tilde{A}^{0(b)}_{\nu}},$
(19)

where $\tilde{A}^{0}_{\nu}$ is the value of gauge field on the boundary. We
can calculate non-abelian color DC conductivity by using Green-Kubo formula,

$\sigma^{ij}_{ab}(k_{\mu})=-\mathop{\lim}\limits_{\omega\to
0}\frac{1}{\omega}\Im G^{ij}_{ab}(k_{\mu}),$ (20)

So DC conductivities are related to the retarded Green’s function of the
boundary current $\mathop{J}^{i}$. The conductivities along $i$ and $j$
directions can be expressed as $\sigma^{ij}_{ab}(x,y)=\frac{\delta
J^{i}_{a}(x)}{\delta E^{b}_{j}(y)}$ respectively. There is a rotation symmetry
in $xy$ plane so color DC conductivity is a scalar quantity,

$\sigma^{ij}_{ab}=\sigma_{ab}\delta^{ij}.$ (21)

Now, we consider the perturbation part of gauge field as
$\tilde{A}_{x}=\tilde{A}_{x}(r)e^{-i\omega t}$ in which $\omega$ is small
since hydrodynamics regime is valid. By inserting the perturbed part of gauge
field into action Eq.(3) and keeping terms up to second order of $\tilde{A}$
we have,

$\displaystyle S^{(2)}$ $\displaystyle=\int d^{4}x\frac{4}{f\left(1-4\lambda
h^{\prime
2}\right)}\left[f^{2}\left((\partial_{r}\tilde{A}_{x}^{(1)})^{2}+(\partial_{r}\tilde{A}_{x}^{(2)})^{2}+(\partial_{r}\tilde{A}_{x}^{(3)})^{2}\right)\right.$
$\displaystyle\left.-\Big{(}(\tilde{A}_{x}^{(1)})^{2}+(\tilde{A}_{x}^{(2)})^{2}+\tilde{A}_{x}^{(3)})^{2}\Big{)}\omega^{2}+4\Big{(}\tilde{A}_{x}^{(1)})^{2}+\tilde{A}_{x}^{(2)}\Big{)}h^{2}\right].$
(22)

By variation of the action $S^{(2)}$ with respect to $\tilde{A}_{x}^{a}$ we
have,

$f\left(f\tilde{A}_{x}^{(1)^{\prime}}\right)^{\prime}+(\omega^{2}-4h^{2})\tilde{A}_{x}^{(1)}+\frac{8\lambda
f^{2}\tilde{A}_{x}^{(1)^{\prime}}h^{\prime}h^{\prime\prime}}{1-4\lambda
h^{\prime 2}}=0,$ (23)
$f\left(f\tilde{A}_{x}^{(2)^{\prime}}\right)^{\prime}+(\omega^{2}-4h^{2})\tilde{A}_{x}^{(2)}+\frac{8\lambda
f^{2}\tilde{A}_{x}^{(2)^{\prime}}h^{\prime}h^{\prime\prime}}{1-4\lambda
h^{\prime 2}}=0,$ (24)

$f\left(f\tilde{A}_{x}^{(3)^{\prime}}\right)^{\prime}+\omega^{2}\tilde{A}_{x}^{(3)}+\frac{8\lambda
f^{2}\tilde{A}_{x}^{(3)^{\prime}}h^{\prime}h^{\prime\prime}}{1-4\lambda
h^{\prime 2}}=0.$ (25)

Firstly, we solve Eq.(23), Eq.(24) and Eq.(25) near the horizon. Therefore, by
expanding theses equations near event horizon and considering the relation
$f\sim 4\pi T(r-r_{h})$ from definition of Hawking temperature we find the
solution of $\tilde{A}_{x}^{(a)}$ as follows,

$\displaystyle\tilde{A}_{x}^{(a)}\sim(r-r_{h})^{z_{a}}\,,\qquad
a=1,2,3\,\,\,\,,$ (26)

where,

$\displaystyle z_{1}$ $\displaystyle=z_{2}=\pm
i\frac{\sqrt{h(r_{h})^{2}+\omega^{2}}}{4\pi T},$ (27) $\displaystyle z_{3}$
$\displaystyle=\pm i\frac{\omega}{4\pi T}.$ (28)

To solve Eq.(23), Eq.(24) and Eq.(25), we apply these following ansatzs,

$\displaystyle\tilde{A}_{x}^{(1)}=\tilde{A}^{(1)}_{\infty}\Big{(}\frac{21\lambda}{r^{2}(2-7\lambda\Lambda)}f\Big{)}^{z_{1}}\Big{(}1+i\omega
h_{1}(r)+\cdots\Big{)},$ (29)
$\displaystyle\tilde{A}_{x}^{(2)}=\tilde{A}^{(2)}_{\infty}\Big{(}\frac{21\lambda}{r^{2}(2-7\lambda\Lambda)}f\Big{)}^{z_{2}}\Big{(}1+i\omega
h_{2}(r)+\cdots\Big{)},$ (30)
$\displaystyle\tilde{A}_{x}^{(3)}=\tilde{A}^{(3)}_{\infty}\Big{(}\frac{21\lambda}{r^{2}(2-7\lambda\Lambda)}f\Big{)}^{z_{3}}\Big{(}1+i\omega
h_{3}(r)+\cdots\Big{)},$ (31)

where $\tilde{A}^{(a)}_{\infty}$ is the value of fields in the boundary and
$z_{i}$’s are the minus sign of 27 and 28.
By inserting Eq.29 into Eq.(23) and considering to the first order of
$\omega$, we obtain,

$\displaystyle-4r^{2}h^{2}\left(4\mathit{k}h^{\prime 2}-1\right)\left(4\pi
Th_{1}-\log\left(-\frac{21\mathit{k}f}{r^{2}(7\mathit{k}\Lambda-2)}\right)\right)$
$\displaystyle+rf\Bigg{(}\left(4\mathit{k}h^{\prime
2}-1\right)\bigg{(}f^{\prime}\left(4\pi
rTh_{1}^{\prime}+2\right)-rf^{\prime\prime}\bigg{)}+8\mathit{k}rf^{\prime}h^{\prime}h^{\prime\prime}\Bigg{)}$
$\displaystyle+2f^{2}\Bigg{(}2\pi
r^{2}Th_{1}^{\prime\prime}\left(4\mathit{k}h^{\prime
2}-1\right)-4\mathit{k}h^{\prime}\bigg{(}2rh^{\prime\prime}\left(2\pi
rTh_{1}^{\prime}+1\right)+h^{\prime}\bigg{)}+1\Bigg{)}=0.$ (32)

The equation for $h_{2}(r)$ is the same as $h_{1}(r)$.
By inserting Eq.31 into Eq.(25) and considering to the first order of
$\omega$, we obtain,

$\displaystyle\frac{1}{r}\Bigg{(}\left(4\lambda h^{\prime
2}-1\right)\bigg{(}f^{\prime}\left(4\pi
Trh_{3}^{\prime}+2\right)-rf^{\prime\prime}\bigg{)}+8\lambda
rf^{\prime}h^{\prime}h^{\prime\prime}\Bigg{)}$
$\displaystyle+\frac{2}{r^{2}}f\Bigg{(}1-4\lambda
h^{\prime}\bigg{(}2rh^{\prime\prime}\left(2\pi
rTh_{3}^{\prime}+1\right)+h^{\prime}\bigg{)}+2\pi
r^{2}Th_{3}^{\prime\prime}\left(1-4\lambda h^{\prime 2}\right)\Bigg{)}=0.$
(33)

The solution of $h_{3}(r)$ is as,

$\displaystyle
h_{3}(r)=C_{3}+\int_{1}^{r}\Bigg{(}\frac{uf^{\prime}(u)-2f(u)}{4\pi
Tuf(u)}+\frac{C_{4}\left(1-4\lambda
h^{\prime}(u)^{2}\right)}{f(u)}\Bigg{)}du,$ (34)

where $C_{3}$ and $C_{4}$ are integration constants. The behavior of
$h_{3}(r)$ near horizon is,

$h_{3}\approx\bigg{(}\frac{C_{4}(1-4\lambda h^{\prime}(r_{h})^{2})+1}{4\pi
T}\bigg{)}\log(r-r_{h})+\text{finite terms}.$ (35)

$C_{4}$ is determined by applying regularity of $h_{3}(r)$ at horizon as,

$C_{4}=\frac{1}{4\lambda h^{\prime}(r_{h})^{2}-1}.$ (36)

By plugging the solution of $\tilde{A}_{x}^{(3)}$ in Eq.(3) and variation with
respect to $\tilde{A}^{(3)}_{\infty}$ , Green’s function can be read as,

$G_{xx}^{(33)}(\omega,\vec{0})=\frac{(\tilde{A}_{x}^{(3)})^{*}f(r)\partial_{r}\tilde{A}_{x}^{(3)}}{4\lambda
h^{\prime 2}-1}\bigg{|}_{r\to\infty}=i\omega C_{4},$ (37)

by substituting $f(r)$ , $h_{3}(r)$ , $h^{\prime}(r)$ in above equation and
using Eq.(20) we have,

$\displaystyle\sigma_{xx}^{(33)}=-\mathop{\lim}\limits_{\omega\to
0}\frac{1}{\omega}\Im G^{ij}(k_{\mu})=\frac{1}{1-4\lambda h^{\prime
2}(r_{h})},$ (38)

The conductivity bound is violated for $1-4\lambda h^{\prime 2}(r_{h})>1$ and
it is preserved for $0<1-4\lambda h^{\prime 2}(r_{h})\leq 1$ .
In the limit of $\lambda\to 0$ we have non-abelian Yang-Mills theory and the
conductivity bound is saturated.

$\displaystyle\sigma_{xx}^{(33)}=1.$ (39)

The value of $\sigma_{xx}^{(11)}$ and $\sigma_{xx}^{(22)}$ is calculated by
the same procedure of $\sigma_{xx}^{(33)}$ as following,

$\displaystyle\sigma_{xx}^{(11)}=\sigma_{xx}^{(22)}=0.$ (40)

It means that the color non-abelian DC conductivity in terms of color indices
is diagonal and also Ohm’s law is diagonal in this model. If we consider the
gauge field 8 in $\sigma_{1}$ or $\sigma_{2}$ directions, conductivity will be
non-zero in these directions.

##  4 Conclusion

We introduced Einstein non-abelian logarithmic gauge theory AdS black brane
solution and calculated the non-abelian color DC conductivity for this model.
There is a conjecture that conductivity is bounded by the universal value
$\sigma\geq 1$ 111 We consider $\frac{1}{e^{2}}=1$. Our outcome shows that the
conductivity bound is violated for $1-4\lambda h^{\prime 2}(r_{h})>1$ and
preserved for $0<1-4\lambda h^{\prime 2}(r_{h})\leq 1$ in non-abelian
logarithmic gauge theory but this bound is saturated for Yang-Mills
theory.[48]. The conductivity bound can be violated in nonlinear model,
depending on the form of the nonlinear theory, or coupling of $F^{2}$ to the
scalar field. Our interpretation is that nonlinearities are somehow similar to
electron-electron couplings and therefore the violation related to Mott
insulators. It can be concluded that the conductivity is the derivative of the
Lagrangian with respect to $F^{2}$ computed at the event horizon and it is
saturated for the Yang-Mills theory but it can be violated for the nonlinear
model.
The ratio of shear viscosity to entropy density for this model is
$\frac{\eta}{s}=\frac{1}{4\pi}$. Therefore, the Kovtun-Son-Starinet (KSS)
bound [49] that it stated this value for Einstein-Hilbert gravity is
$\frac{\eta}{s}=\frac{1}{4\pi}$ is preserved for this model. $\frac{\eta}{s}$
is proportional to inverse squared coupling of field theory,
$\frac{\eta}{s}\sim\frac{1}{\lambda^{2}}$. It means that the coupling of the
field theory dual to our model is the same as the coupling of the field theory
dual to the Einstein AdS black brane solution, but the color conductivity is
different.

Acknowledgment We would like to thank Shahrokh Parvizi, Matteo Baggioli,
Mojtaba Shahbazi and Komeil Babaei for useful comments and suggestions. We
also thank the referees of CJP for the valuable comments which helped us to
improve the manuscript.

Data Availability  Data generated or analyzed during this study are provided
in full within the published article.

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