# Seebeck and Nernst coefficients of magnetized hot QCD medium with number
conserving kernel

Salman Ahamad<EMAIL_ADDRESS>and Binoy Krishna
<EMAIL_ADDRESS>
Department of Physics,
Indian Institute of Technology Roorkee, Roorkee 247667, India

###### Abstract

We have studied the thermoelectric response of a hot and magnetized QCD medium
created in the noncentral events at heavy-ion collider experiments. The
collisional aspects of the medium have been embedded in the relativistic
Boltzmann transport equation (RBTE) using Bhatnagar-Gross-Krook (BGK)
collision integral, which insures the particle number conservation, unlike the
commonly used relaxation time approximation (RTA). We have incorporated the
thermal medium effects in the guise of a quasiparticle model, where the
interaction among the partons is embodied in the medium dependent mass of the
quarks, which has been evaluated using the framework of the perturbative
thermal QCD with a magnetic field in the background. In the absence of $B$,
the Seebeck coefficient for individual quark flavors gets slightly reduced in
the BGK term in comparison to naive RTA, while it gets enhanced for the
composite partonic medium. In the presence of strong magnetic field ($B$), the
BGK term enhances the Seebeck coefficient for the individual flavors as well
as that for medium. The medium Seebeck coefficient rises with the strength of
quark chemical potential ($\mu$) in the absence as well as that in the strong
$B$. We observe chirality dependence in the Seebeck and Nernst coefficients in
the weak magnetic field as the degeneracy in the masses of left- and right-
handed chiral modes gets lifted in the weak $B$. Seebeck coefficient is larger
for left-handed modes contrary to the Nernst coefficient which is higher for
the right-handed modes. Seebeck coefficient is almost similar in the weak
field for both the collision term, while Nernst coefficient gets reduced
(enhanced) for $L$ ($R$) modes considerably in the BGK collision term.

##  1 Introduction

A transition from the hadronic matter to a deconfined phase of quark gluon
plasma (QGP) takes place in heavy-ion collision experiments at Relativistic
Heavy Ion Collider (RHIC) and Large Hadron Collider (LHC). In non-central
collisions, a very intense magnetic field of the order of $m_{\pi}^{2}$ at
RHIC [1] and of $15m_{\pi}^{2}$ at LHC [2]) is also produced, which persists
in the medium for a considerable amount of time due to the finite electrical
conductivity of the medium. This magnetic field leads to the modification in
the thermodynamical [3, 4] and transport properties [5, 6, 7, 8, 9, 10, 11] of
the hot and dense quark matter and also induces novel phenomena such as the
chiral magnetic effect[12, 1], magnetic and inverse magnetic catalysis[13, 14,
15, 16], axial magnetic effect[17, 18], chiral vortical effect in rotating QGP
[19, 20], the conformal anomaly and production of soft photons[21, 22]. In
addition to this, the dilepton production rate[23, 24, 25], dispersion
relation in the magnetized thermal QED[26], refractive indices and decay
constants[27, 21] have been explored in the magnetic field background.

Transport coefficients are crucial input parameters needed in the dissipative
hydrodynamics and transport simulation to describe the evolution of the
partonic medium created post-collision at RHIC and LHC. Shear viscosity
quantifies the response of the medium to the transverse momentum gradients
while bulk viscosity to the pressure gradients. Both shear and bulk
viscosities have been studied in the magnetic field extensively in different
models [5, 7, 8, 9, 10]. Electrical and thermal conductivities measure the
response of the system to the electromagnetic fields and thermal gradients in
the medium, respectively. Electrical conductivity plays an important role in
the elongation of the lifetime of the magnetic field created in non-central
collisions, while thermal conductivity controls the sound attenuation in the
medium through the Prandtl number. Both electrical and thermal conductivities
have been extensively studied in phenomenological models as well as using
perturbative methods [6, 28, 29, 30, 31, 32, 33]. On the other hand, the
transport coefficient corresponding to the thermoelectric response is known as
Seebeck coefficient, which measures the ability of any material to convert the
thermal gradient into the electric current. Thermoelectric properties of the
materials have been mainly studied in the context of condensed matter physics
over the years. There have been numerous studies regarding the thermoelectric
properties of the various condensed matter systems such as superconductors
[34, 35, 36, 37], the graphene superconductor junction [38], correlated
quantum dots coupled to superconducting electrode [39], high temperature
cuprates [40], ferromagnet-superconductor hybrid junction [41] and low
dimensional correlated organic metals [42].

We have recently explored the charge, heat, and momentum transport
coefficients in the kinetic theory approach using the BGK type collision
integral [43, 44] and have observed a substantial amount of difference in
comparison to the RT approximation. Motivated by our earlier studies, we are
now interested in thermoelectric effects in the strongly interacting matter
produced in the heavy-ion collisions where a thermal gradient is present
between the central and peripheral regions of the fireball. In addition to the
temperature gradient, a finite baryon chemical potential is also needed to
observe the thermoelectric effect in strongly interacting matter unlike the
condensed matter systems, where only one type of the charge carriers
participate in the transport process. Contrary to that, in strongly
interacting medium, both positive and negative charge carriers take part in
the transport phenomena. In the absence of the quark chemical potential, both
particles and anti-particles have equal numbers, so no net thermoelectric
effect is observed. The Seebeck effect in the absence of a magnetic field has
been studied recently for a hot hadron gas in hadron resonance gas model [45]
and for the QGP phase in the ambit of Nambu–Jona Lasinio (NJL) model [46]. In
the presence of a magnetic field, the thermoelectric response of the hot QCD
medium has been explored earlier [47, 48, 49]. One of us has also studied the
effect of the strong as well as weak magnetic field on the Seebeck and Nernst
coefficients of the hot QCD medium in [50, 51, 52]. In the earlier works, we
have used the relativistic kinetic theory approach, where the collisional
effects of the medium have been incorporated with RT type collision integral.
But, the widely used RTA collision integral has a drawback that it violates
the conservation of the particle number and current. Taking this fact into
consideration, we have used a more realistic BGK-type collision integral,
which ensures the particle number and current conservation in the medium. Such
a type of collision integral has been used earlier in the calculation of the
refractive index [53], dielectric permitivity [54], heavy quark energy loss
[55] and collective modes of the hot QCD medium [56]. We have calculated the
Seebeck and Nernst coefficients using RBTE with BGK type collision integral
and have incorporated the thermal medium effects in the framework of a quasi-
particle model [57]. Quasi-particle models are widely used to study the
thermodynamical and transport properties of the hot QCD medium. The basic idea
behind these models is that a strongly interacting system of massless quarks
and gluons can be described in terms of the quasi-particles having medium
dependent masses generated as a result of the collective effects in the
medium. There are many quasi-particle models in the literature, few of them
are, effective QCD models like Nambu-Jona-Lasinio (NJL) and PNJL model based
quasi-particle model [58, 59, 60, 61], Effective fugacity model [62] and the
recently proposed quasi-particle model based on the Gribov–Zwanziger
quantization [63, 64, 65]. We have used a quasi-particle model [57] wherein
the medium dependence enters through the dispersion relations of the quark and
gluon quasi-particles originated as a result of the collective excitation in
the medium. The medium dependent mass of the quark has been computed from the
pole of the full propagator resummed using the Dyson-Schwinger equation. We
have employed the perturbative thermal QCD in magnetic field background to
calculate the self-energy of the quark.

In current work, our main focus is threefold: (1) to check the sensitivity of
the thermoelectric transport coefficients viz Seebeck and Nernst coefficients
on the collision integrals of different types, (2) how dimensional reduction
in the quark dynamics in the strong magnetic field affects the thermoelectric
properties of the medium, and (3) to study the chirality dependence in the
thermoelectric phenomenon arising in the weak magnetic field due to lifting of
the degeneracy in the masses of the left-and right-hand chiral modes. Keeping
the strength of the magnetic field and the various scales of the medium in
mind, we have explored two regimes of the magnetic field, the strong magnetic
field regime ($|qB|>>T^{2}>>m_{f}^{2}$) and the weak magnetic field regime
($T^{2}>>|qB|>>m_{f}^{2}$). In the magnetic field, the motion of the quarks is
quantized in the transverse direction leading to the discrete energy spectrum
in terms of the Landau levels. When the strength of the magnetic field is
large, the energy separation between the consecutive Landau levels become
large ( of the order $\sqrt{|qB|}$), consequently, the quarks get confined in
the lowest Landau level (LLL) only. Moreover, in the weak magnetic field case,
the magnetic field acts as a perturbation to the thermal medium propagator.
The degeneracy in the mass of the different chiral modes is lifted, as a
result, the mass depends on the handedness of the collective excitations
leading to the chirality dependence in the transport coefficients [11, 52]. We
notice that in the absence of the magnetic field, the magnitude of the Seebeck
coefficient for the individual $u$, $d$ and $s$ quarks gets reduced in the BGK
collision integral while for the composite medium, it gets enhanced. In the
strong magnetic field, magnitude of the Seebeck coefficient gets enhanced in
BGK collision term for individual flavors as well as for composite medium. In
case of the weak magnetic field, the Seebeck coefficient is not much sensitive
to the collision integral and found to be almost similar in both BGK and RTA
collision term. Its magnitude is higher for the $L$ modes. In addition, a hall
type Nernst coefficient also appears, which quantifies the thermoelectric
response in the transverse direction. Nernst coefficient gets reduced
(enhanced) in the BGK term in the case of $L$ ($R$) modes for indivisual
flavors as well as for the composite medium.

The present manuscript has been organized as follows: In section 2, we have
discussed the quasi-particle model and thermal mass of the quarks in the
thermal and magnetic field background obtained using the perturbative thermal
QCD. In subsection 3.1 and 3.2, we have calculated the Seebeck coefficient in
the absence and in the presence of the magnetic field, respectively. We
disccussed the results in section 4 and finally we conclude in section 5.

##  2 Quasi-particle model

The central feature of quasiparticle models is that a strongly interacting
system of massless quarks and gluons can be described in terms of the massive,
weakly interacting quasiparticles originating due to the collective
excitations. There are many quasi-particle models such as NJL and PNJL models
[58, 59, 60, 61], which are based on the respective effective QCD models,
effective fugacity model [62] and the recently proposed quasiparticle model
based on the Gribov–Zwanziger quantization [63, 64, 65]. In this study, we
have used quasiparticle model [57] which has only a single adjustable
parameter and the medium dependence enters through the dispersion relation of
the quark quasi-particles. The temperature and magnetic field-dependent mass
of the quark has been computed from the pole of resummed propagator obtained
from Dyson-Schwinger equation. The respective self-energy is calculated in the
framework of imaginary-time formalism of the thermal QCD. The quasiparticle
mass of $i^{th}$ flavor is written phenomenologically as [57]

$\displaystyle m_{i}^{2}=m_{i,0}^{2}+\sqrt{2}m_{i,0}m_{i,T}+m_{i,T}^{2},$ (1)

where $m_{i,0}$ and $m_{i,T}$ are the current quark mass and medium generated
quark mass. $m_{i,T}$ has been calculated using the HTL perturbation theory as
[66, 67]

$\displaystyle m_{iT}^{2}=\frac{g^{\prime
2}T^{2}}{6}\left(1+\frac{\mu^{2}}{\pi^{2}T^{2}}\right),$ (2)

where $g^{\prime}=\sqrt{4\pi\alpha_{s}}$ refers to the coupling constant which
depends on the temperature as

$\displaystyle\alpha_{s}(T)=\frac{g^{\prime
2}}{4\pi}=\frac{6\pi}{(33-2N_{f})\ln\left(\frac{Q}{\Lambda_{QCD}}\right)},$
(3)

and $Q$ is set at $2\pi\sqrt{T^{2}+\frac{\mu^{2}}{\pi^{2}}}$.

Now, we will include the strong magnetic field in the quasiparticle
description. The quasi particle mass in the presence of strong $B$ can be
generalized as

$\displaystyle m^{2}_{i,s}=m_{i0}^{2}+\sqrt{2}m_{i0}m_{iB,T}+m_{iB,T}^{2},$
(4)

where $m_{iB,T}$ is the medium generated mass, which is obtained from the pole
the resummed quark propagator. The inverse of the resummed propagator can be
written using the Dyson-Schwinger equation as

$\displaystyle
S^{-1}(p_{\parallel})=\gamma^{\mu}p_{\parallel\mu}-\Sigma(p_{\parallel}),$ (5)

where $\Sigma(p_{\parallel})$ is the quark self-energy at finite $T$ and
strong $B$, which has been calculated as [68]

$\displaystyle\Sigma(p_{\parallel})=\frac{g^{2}|q_{i}B|}{3\pi^{2}}\left[\frac{\pi
T}{2m_{i,0}}-\ln(2)+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}-\frac{31\mu^{4}\zeta(5)}{32\pi^{4}T^{4}}\right]\left[\frac{\gamma^{0}p_{0}}{p_{\parallel}^{2}}+\frac{\gamma^{3}p_{z}}{p_{\parallel}^{2}}+\frac{\gamma^{0}\gamma^{5}p_{z}}{p_{\parallel}^{2}}+\frac{\gamma^{3}\gamma^{5}p_{0}}{p_{\parallel}^{2}}\right],$
(6)

where $g=\sqrt{4\pi\alpha_{s}}$ is the running coupling which depends on $T$,
$B$ and $\mu$ as

$\displaystyle\alpha_{s}(\Lambda^{2},eB)=\frac{g^{2}}{4\pi}=\frac{\alpha_{s}(\Lambda^{2})}{1+b_{1}\alpha_{s}(\Lambda^{2})\ln\left(\frac{\Lambda^{2}}{\Lambda^{2}+eB}\right)},$
(7)

with

$\displaystyle\alpha_{s}(\Lambda^{2})=\frac{1}{b_{1}\ln\left(\frac{\Lambda^{2}}{\Lambda_{\overline{MS}}^{2}}\right)},$
(8)

and $\Lambda$ is set at $2\pi\sqrt{T^{2}+\frac{\mu^{2}}{\pi^{2}}}$ for quarks,
$b_{1}=\frac{11N_{c}-2N_{f}}{12\pi}$ and $\Lambda_{\overline{MS}}=0.176$ GeV.

In the presence of heat bath and magnetic field, the Lorentz (boost) and
rotational invariance of the system get broken. In such a non-trivial
background, the covariant form of the quark-self energy
$\Sigma(p_{\parallel})$ can be written as [69, 4]

$\displaystyle\Sigma(p_{\parallel})=A\gamma^{\mu}u_{\mu}+B\gamma^{\mu}b_{\mu}+C\gamma^{5}\gamma^{\mu}u_{\mu}+D\gamma^{5}\gamma^{\mu}b_{\mu},$
(9)

where $u^{\mu}(1,0,0,0)$ and $b^{\mu}(0,0,0,-1)$ correspond to the heat bath
and the magnetic field, respectively. $A,B,C$ and $D$ are the form factors
which are given in the LLL approximation as [68]

$\displaystyle A$ $\displaystyle=$ $\displaystyle\frac{1}{4}{\rm
Tr}[\Sigma\gamma^{\mu}u_{\mu}]=\frac{g^{2}|q_{i}B|}{3\pi^{2}}\left[\frac{\pi
T}{2m_{i,0}}-\ln{(2)}+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}-\frac{31\mu^{4}\zeta(5)}{32\pi^{4}T^{4}}\right]\frac{p_{0}}{p_{\parallel}^{2}},$
(10) $\displaystyle B$ $\displaystyle=$ $\displaystyle-\frac{1}{4}{\rm
Tr}[\Sigma\gamma^{\mu}b_{\mu}]=\frac{g^{2}|q_{i}B|}{3\pi^{2}}\left[\frac{\pi
T}{2m_{i,0}}-\ln{(2)}+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}-\frac{31\mu^{4}\zeta(5)}{32\pi^{4}T^{4}}\right]\frac{p_{z}}{p_{\parallel}^{2}},$
(11) $\displaystyle C$ $\displaystyle=$ $\displaystyle\frac{1}{4}{\rm
Tr}[\gamma^{5}\Sigma\gamma^{\mu}u_{\mu}]=-\frac{g^{2}|q_{i}B|}{3\pi^{2}}\left[\frac{\pi
T}{2m_{i,0}}-\ln{(2)}+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}-\frac{31\mu^{4}\zeta(5)}{32\pi^{4}T^{4}}\right]\frac{p_{z}}{p_{\parallel}^{2}},$
(12) $\displaystyle D$ $\displaystyle=$ $\displaystyle-\frac{1}{4}{\rm
Tr}[\gamma^{5}\Sigma\gamma^{\mu}b_{\mu}]=-\frac{g^{2}|q_{i}B|}{3\pi^{2}}\left[\frac{\pi
T}{2m_{i,0}}-\ln{(2)}+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}-\frac{31\mu^{4}\zeta(5)}{32\pi^{4}T^{4}}\right]\frac{p_{0}}{p_{\parallel}^{2}},$
(13)

where $\zeta(3)$ and $\zeta(5)$ are the Riemann zeta function. We can further
cast the quark self-energy (9) using the chirality projection operators as

$\displaystyle\Sigma(p_{\parallel})=P_{R}[(A-B)\gamma^{\mu}u_{\mu}+(B-A)\gamma^{\mu}b_{\mu}]P_{L}+P_{L}[(A+B)\gamma^{\mu}u_{\mu}+(B+A)\gamma^{\mu}b_{\mu}]P_{R},$
(14)

where $P_{R}$ and $P_{L}$ are the right- and left-handed chiral projection
operators, respectively, which are given by

$\displaystyle P_{R}=\frac{(1+\gamma^{5})}{2}$ (15) $\displaystyle
P_{L}=\frac{(1-\gamma^{5})}{2}.$ (16)

The inverse of the resummed quark propagator can be expressed in terms of the
$P_{R}$ and $P_{L}$ as

$\displaystyle
S^{-1}(p_{\parallel})=P_{R}\gamma^{\mu}X_{\mu}P_{L}+P_{L}\gamma^{\mu}Y_{\mu}P_{R},$
(17)

where

$\displaystyle\gamma^{\mu}X_{\mu}$ $\displaystyle=$
$\displaystyle\gamma^{\mu}p_{\parallel\mu}-(B-A)\gamma^{\mu}b_{\mu}-(A-B)\gamma^{\mu}u_{\mu},$
(18) $\displaystyle\gamma^{\mu}Y_{\mu}$ $\displaystyle=$
$\displaystyle\gamma^{\mu}p_{\parallel\mu}-(B+A)\gamma^{\mu}b_{\mu}-(A+B)\gamma^{\mu}u_{\mu}.$
(19)

On inverting Eq. (25), we get

$\displaystyle
S(p_{\parallel})=\frac{1}{2}\left[P_{L}\frac{\gamma^{\mu}X_{\mu}}{X^{2}/2}P_{R}+P_{R}\frac{\gamma^{\mu}Y_{\mu}}{Y^{2}/2}P_{L}\right],$
(20)

where

$\displaystyle\frac{X^{2}}{2}=X_{1}^{2}=\frac{1}{2}\left[p_{0}-(A-B)\right]^{2}-\frac{1}{2}\left[p_{z}+(B-A)\right]^{2},$
(21)
$\displaystyle\frac{Y^{2}}{2}=Y_{1}^{2}=\frac{1}{2}\left[p_{0}-(A+B)\right]^{2}-\frac{1}{2}\left[p_{z}+(B+A)\right]^{2}~{}.$
(22)

To get the thermal mass, we take static limit ($p_{0}=0,~{}p_{z}\rightarrow
0$) of the poles of the effective propagator (20) (of either $X_{1}^{2}$ or
$Y_{1}^{2}$ )

$\displaystyle m_{i,B}^{2}=\frac{g^{2}|q_{i}B|}{3\pi^{2}}\left[\frac{\pi
T}{2m_{i,0}}-\ln{(2)}+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}-\frac{31\mu^{4}\zeta(5)}{32\pi^{4}T^{4}}\right],$
(23)

which depends on the magnetic field, temperature and quark chemical potential.

The effective quark mass for $i^{th}$ flavor in the case of a weak magnetic
field can be parameterized like the earlier cases as

$\displaystyle m^{2}_{i,w}=m_{i0}^{2}+\sqrt{2}m_{i0}m_{i,L/R}+m_{i,L/R}^{2},$
(24)

where $m_{i,L/R}$ refers to the thermal mass for the left- or right-handed
chiral mode of $i^{th}$ flavor which can be evaluated from the Dyson-Schwinger
equation in weak magnetic field

$\displaystyle S^{*-1}(P)$ $\displaystyle=$ $\displaystyle\not{P}-\Sigma(P).$
(25)

$\Sigma(P)$ is the quark self energy in the weakly magnetized thermal medium.
The general structure of quark self-energy in the covariant form at finite $T$
and $B$ can be written as [70]

$\displaystyle\Sigma(P)=-A^{\prime}\not{P}-B^{\prime}\not{u}-C^{\prime}\gamma_{5}\not{u}-D^{\prime}\gamma_{5}\not{b},$
(26)

where $A^{\prime},B^{\prime},C^{\prime},D^{\prime}$ are the structure
functions, which can be evaluated by taking the appropriate contractions of
Eq. (26) as [70]

$\displaystyle A^{\prime}(p_{0},|{\bf{p}}|)$ $\displaystyle=$
$\displaystyle\frac{m_{th}^{2}}{|{\bf{p}}|^{2}}Q_{1}\Bigg{(}\frac{p_{0}}{|{\bf{p}}|}\Bigg{)},$
(27) $\displaystyle B^{\prime}(p_{0},|{\bf{p}}|)$ $\displaystyle=$
$\displaystyle-\frac{m_{th}^{2}}{|{\bf{p}}|}\Bigg{[}\frac{p_{0}}{|{\bf{p}}|}Q_{1}\Big{(}\frac{p_{0}}{|{\bf{p}}|}\Big{)}-Q_{0}\Bigg{(}\frac{p_{0}}{|{\bf{p}}|}\Bigg{)}\Bigg{]},$
(28) $\displaystyle C^{\prime}(p_{0},|{\bf{p}}|)$ $\displaystyle=$
$\displaystyle-4g^{2}C_{F}M^{2}\frac{p_{z}}{|{\bf{p}}|^{2}}Q_{1}\Bigg{(}\frac{p_{0}}{|{\bf{p}}|}\Bigg{)},$
(29) $\displaystyle D^{\prime}(p_{0},|{\bf{p}}|)$ $\displaystyle=$
$\displaystyle-4g^{2}C_{F}M^{2}\frac{1}{|{\bf{p}}|}Q_{0}\Bigg{(}\frac{p_{0}}{|{\bf{p}}|}\Bigg{)},$
(30)

where

$\displaystyle M^{2}(T,\mu,B)=\frac{|q_{i}B|}{16\pi^{2}}\left(\frac{\pi
T}{2m_{i0}}-\ln{2}+\frac{7\mu^{2}\zeta(3)}{8\pi^{2}T^{2}}\right),$ (31)

where $\zeta$ corresponds to the Riemann zeta function. $Q_{0}$ and $Q_{1}$
are given by

$\displaystyle Q_{0}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right),$ (32) $\displaystyle
Q_{1}(x)$ $\displaystyle=$
$\displaystyle\frac{x}{2}\ln\left(\frac{x+1}{x-1}\right)-1=xQ_{0}(x)-1.$ (33)

Self-energy (26) can be written in the basis of right and left-hand chiral
projection operators as

$\displaystyle\Sigma(P)=-P_{R}(A^{\prime}\not{P}+(B^{\prime}+C^{\prime})\not{u}+D^{\prime}\not{b})P_{L}-P_{L}(A^{\prime}\not{P}+(B^{\prime}-\mathcal{C})\not{u}-D^{\prime}\not{b})P_{R}.$
(34)

We can re-write the inverse fermion propagator (25) using (34) as

$\displaystyle
S^{*-1}(P)=\not{P}+P_{R}\left[A^{\prime}\not{P}+\left(B^{\prime}+C^{\prime}\right)\not{u}+D^{\prime}\not{b}\right]P_{L}+P_{L}\left[A^{\prime}\not{P}+\left(B^{\prime}-C^{\prime}\right)\not{u}-D^{\prime}\not{b}\right]P_{R},$
(35)

which can further be simplified as

$\displaystyle S^{*-1}(P)=P_{R}\not{L}P_{L}+P_{L}\not{R}P_{R}.$ (36)

Since $P_{L,R}\gamma^{\mu}=\gamma^{\mu}P_{R,L}$ and
$P_{L}\not{P}P_{L}=P_{R}\not{P}P_{R}=P_{L}P_{R}\not{P}=0$, $\not{L}$ and
$\not{R}$ are given by

$\displaystyle\not{L}$ $\displaystyle=$
$\displaystyle(1+A^{\prime})\not{P}+(B^{\prime}+C^{\prime})\not{u}+D^{\prime}\not{b},$
(37) $\displaystyle\not{R}$ $\displaystyle=$
$\displaystyle(1+A^{\prime})\not{P}+(B^{\prime}-C^{\prime})\not{u}-D^{\prime}\not{b}.$
(38)

After inverting Eq. (36), we get the effective quark propagator as

$\displaystyle
S^{*}(P)=\frac{1}{2}\left[P_{L}\frac{\not{L}}{L^{2}/2}P_{R}+P_{R}\frac{\not{R}}{R^{2}/2}P_{L}\right],$
(39)

where

$\displaystyle
L^{2}=(1+A^{\prime})^{2}P^{2}+2(1+A^{\prime})(B^{\prime}+C^{\prime})p_{0}-2D^{\prime}(1+A^{\prime})p_{z}+(B^{\prime}+C^{\prime})^{2}-D^{\prime
2},$ (40) $\displaystyle
R^{2}=(1+A^{\prime})^{2}P^{2}+2(1+A^{\prime})(B^{\prime}-C^{\prime})p_{0}+2D^{\prime}(1+A^{\prime})p_{z}+(B^{\prime}-C^{\prime})^{2}-D^{\prime
2}.$ (41)

Now in order to get the quark thermal mass in weakly magnetized thermal QCD
medium, we take the static limit ($p_{0}=0,|{\bf{p}}|\rightarrow 0$) of
$L^{2}/2$ and $R^{2}/2$ modes,333We have expanded the Legendre functions
appearing in the structure functions in power series of
$\frac{|\bf{p}|}{p_{0}}$ and have considered only upto $\mathcal{O}(g^{2})$ we
get

$\displaystyle\frac{L^{2}}{2}\arrowvert_{p_{0}=0,{|\bf{p}|}\rightarrow
0}=m_{th}^{2}+4g^{2}C_{F}M^{2},$ (42)
$\displaystyle\frac{R^{2}}{2}|_{p_{0}=0,{|\bf{p}|}\rightarrow
0}=m_{th}^{2}-4g^{2}C_{F}M^{2}.$ (43)

The masses of the left- and right-handed modes are given by

$\displaystyle m_{L}^{2}=m_{th}^{2}+4g^{2}C_{F}M^{2},$ (44) $\displaystyle
m_{R}^{2}=m_{th}^{2}-4g^{2}C_{F}M^{2},$ (45)

respectively. We will use these thermally generated masses in the dispersion
relation of the quarks to calculate the Seebeck and Nernst coefficients in the
forthcoming sections.

##  3 Theromoelectric response of a thermal QCD medium

In the kinetic theory approach, the evolution of the phase space distribution
function is given by the relativistic Boltzmann transport equation which reads
as

$\displaystyle p^{\mu}\frac{\partial f}{\partial
x^{\mu}}+q~{}F^{\rho\sigma}p_{\sigma}\frac{\partial f}{\partial
p^{\rho}}=C[f],$ (46)

where $f=f_{eq}+\delta f$; $\delta f$ is small deviation from the equilibrium
and $F^{\rho\sigma}$ corresponds to the electromagnetic field strength tensor.
$C[f]$ is the collision term, which provides microscopic input to the RBTE. In
general $C[f]$ is non-linear in $f$, but Anderson and Witting proposed a
simple collision integral exploiting the mean free path arguments, which is
known as relaxation time approximation (RTA)

$\displaystyle
C[f]=-\frac{p^{\mu}u_{\mu}}{\tau}\left(f(\vec{x},\vec{p},t)-f_{\rm
eq}(|\vec{p}|)\right),$ (47)

where $\tau$ is the relaxation time. The problem with the collision term (47)
is that it violates the particle number and current conservation. This
shortcoming is the artifact of the linearization of the collision term
otherwise in principle the full collision term respects all the conservation
laws. Later this shortcoming in collision term (47) was circumvented by
Bhatnagar-Gross-Krook (BGK) by modifying the RTA as [71, 72]

$\displaystyle
C[f]=-\frac{p^{\mu}u_{\mu}}{\tau}\left(f(\vec{x},\vec{p},t)-\frac{n(\vec{x},t)}{n_{\mathrm{e}q}}f_{\rm
eq}(|\vec{p}|)\right),$ (48)

where ${n(\vec{x},t)}$ and ${n_{\mathrm{e}q}}$ are the out of equilibrium and
equilibrium number density, respectively. The collision term (48) respects the
conservation of the particle number instantaneously i.e.

$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}C[f]=0.$ (49)

In what follows, we will apply the framework discussed herewith to study the
thermoelectric response of the hot QCD medium in the absence as well as in the
presence of the magnetic field.

###  3.1 Seebeck coefficient in the absence of the magnetic field

In this subsection, we will evaluate the Seebeck coefficient of the thermal
QCD medium composed of $u$, $d$ and $s$ quarks (and their anti-particles). In
the presence of the thermal gradient, the charge carriers (quarks) will move
from the hotter regions to the colder ones. As a result, a current is induced
in the medium which can be written for single quark flavor as

$\displaystyle
J_{\mu}=g\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p_{\mu}}{\omega}~{}(q\delta
f(x,p)+\bar{q}\delta\bar{f}(x,p)),$ (50)

where $\delta f(x,p)$ ($\delta\bar{f}(x,p)$) refers to the infinitesimal
deviation in the phase space density of quarks (anti-quarks), respectively.
$g$ corresponds to the degeneracy factor.

The Boltzmann transport equation (46) in the presence of the temperature
gradient with BGK collision integral can be written as

$\displaystyle{\bf p}.\frac{\partial f}{\partial{\bf r}}+q{\bf
E.p}\frac{\partial f}{\partial p^{0}}+qp_{0}{\bf E.}\frac{\partial
f}{\partial{\bf
p}}=-p^{\mu}u_{\mu}\nu\left(f-nn_{\mathrm{e}q}^{-1}f_{\mathrm{e}q}\right),$
(51)

where

$\displaystyle n$ $\displaystyle=$ $\displaystyle
g\int\frac{d^{3}p}{(2\pi)^{3}}(f_{\mathrm{e}q}+\delta f),$ (52) $\displaystyle
n_{\mathrm{e}q}$ $\displaystyle=$ $\displaystyle
g\int\frac{d^{3}p}{(2\pi)^{3}}f_{\mathrm{e}q},$ (53)

and $\nu$ is the collision frequency, which is given by the inverse of the
relaxation time ($\tau$) for the quarks as [73]

$\displaystyle\tau(T)=\frac{1}{5.1T\alpha_{s}^{2}\log\left(\frac{1}{\alpha_{s}}\right)[1+0.12(2N_{f}+1)]},$
(54)

where $\alpha_{s}$ is the running coupling constant (3).

The RBTE (51) can be recast after some simplification as (see Appendix A)

$\displaystyle\delta f-gn_{\mathrm{e}q}^{-1}f_{\mathrm{e}q}\int_{p}\delta f$
$\displaystyle=$
$\displaystyle\frac{p}{\omega}(\omega-\mu)\tau\left(-\frac{1}{T^{2}}\right)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)\nabla_{r}T(r)$
(55) $\displaystyle+2q\beta\tau~{}\frac{{\bf E\cdot
p}}{\omega}f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right),$

which can be further solved for $\delta f$ upto first order as

$\displaystyle\delta f=\delta f^{(0)}+gn_{\mathrm{e}q}^{-1}f_{\rm
eq}\int_{p^{\prime}}\delta f^{(0)},$ (56)

where

$\displaystyle\delta
f^{(0)}=\frac{p}{\omega}(\omega-\mu)\tau\left(-\frac{1}{T^{2}}\right)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)\nabla_{r}T(r)+\frac{2\beta
q\tau}{\omega}{\bf E.p}~{}f_{\mathrm{e}q}(1-f_{\mathrm{e}q}).$ (57)

Similarly, the deviation in the distribution function of the anti-quarks can
be calculated as

$\displaystyle\delta\bar{f}=\delta\bar{f}^{(0)}+gn_{\mathrm{e}q}^{-1}\bar{f}_{\rm
eq}\int_{p^{\prime}}\delta\bar{f}^{(0)},$ (58)

where

$\displaystyle\delta\bar{f}^{(0)}=\frac{p}{\omega}(\omega+\mu)\tau\left(-\frac{1}{T^{2}}\right)\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\nabla_{r}T(r)+2\beta\bar{q}\tau\frac{\bf
E.p}{\omega}~{}\bar{f}_{\mathrm{e}q}(1-\bar{f}_{\mathrm{e}q}).$ (59)

Now substituting $\delta f$ and $\delta\bar{f}$ in the Eq. (50) to obtain the
space part of induced current which reads

$\displaystyle J_{k}$ $\displaystyle=$ $\displaystyle
qg\tau\int\frac{d^{3}p}{(2\pi)^{3}}~{}\bigg{[}\bigg{\\{}\frac{p_{k}^{2}}{\omega^{2}}(\omega-\mu)\left(\frac{-1}{T^{2}}\right)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)\nabla_{r}T(r)+2\frac{p_{k}^{2}}{\omega^{2}}qE_{k}\beta
f_{\mathrm{e}q}(1-f_{\mathrm{e}q})\bigg{\\}}$ (60) $\displaystyle+$
$\displaystyle\frac{g}{n_{\mathrm{e}q}}~{}\frac{p_{k}}{\omega}f_{\mathrm{e}q}\int_{p^{\prime}}\bigg{\\{}\frac{p_{k}}{\omega}(\omega-\mu)\left(\frac{-1}{T^{2}}\right)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)\nabla_{r}T(r)+2\frac{p_{k}}{\omega}qE_{k}\beta
f_{\mathrm{e}q}(1-f_{\mathrm{e}q})\bigg{\\}}\bigg{]}$ $\displaystyle+$
$\displaystyle\bar{q}g\tau\int\frac{d^{3}p}{(2\pi)^{3}}~{}\bigg{[}\bigg{\\{}\frac{p_{k}^{2}}{\omega^{2}}(\omega+\mu)\left(\frac{-1}{T^{2}}\right)\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\nabla_{r}T(r)+2\frac{p_{k}^{2}}{\omega^{2}}\bar{q}E_{k}\beta\bar{f}_{\mathrm{e}q}(1-\bar{f}_{\mathrm{e}q})\bigg{\\}}$
$\displaystyle+$
$\displaystyle\frac{g}{n_{\mathrm{e}q}}~{}\frac{p_{k}}{\omega}\bar{f}_{\mathrm{e}q}\int_{p^{\prime}}\bigg{\\{}\frac{p_{k}}{\omega}(\omega+\mu)\left(-\frac{1}{T^{2}}\right)\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\nabla_{r}T(r)+2\frac{p_{k}}{\omega}\bar{q}E_{k}\beta\bar{f}_{\mathrm{e}q}(1-\bar{f}_{\mathrm{e}q})\bigg{\\}}\bigg{]}.$

In the steady state, the net current in the medium beocomes zero i.e. $\bf J=$
0\. Putting the induced current (60) to zero, we get a relation between the
thermal gradient in the medium and electric field as

$\displaystyle{\bf E}$ $\displaystyle=$
$\displaystyle\frac{1}{2Tq}\left(\frac{L_{1}+L_{2}}{L_{3}+L_{4}}\right)~{}\nabla_{r}T(r),$
(61) $\displaystyle\equiv$ $\displaystyle S~{}\nabla_{r}T(r).$

Here $S$ is the Seebeck coefficient, which is given by

$\displaystyle S=\frac{1}{2Tq}\left(\frac{L_{1}+L_{2}}{L_{3}+L_{4}}\right),$
(62)

where

$\displaystyle L_{1}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\bigg{\\{}\frac{p^{2}}{3\omega^{2}}(\omega-\mu)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)+\frac{g}{n_{\mathrm{e}q}}\frac{p}{\omega}f_{\mathrm{e}q}\int_{p^{\prime}}\frac{p^{\prime}}{\omega^{\prime}}(\omega^{\prime}-\mu)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)\bigg{\\}},$
(63) $\displaystyle L_{2}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\bigg{\\{}\frac{p^{2}}{3\omega^{2}}(\omega+\mu)\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)+\frac{g}{\bar{n}_{\mathrm{e}q}}\frac{p}{\omega}\bar{f}_{\mathrm{e}q}\int_{p^{\prime}}\frac{p^{\prime}}{\omega^{\prime}}(\omega^{\prime}+\mu)\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\bigg{\\}},$
(64) $\displaystyle L_{3}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\bigg{\\{}\frac{p^{2}}{3\omega^{2}}f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)+\frac{g}{n_{\mathrm{e}q}}\frac{p}{\omega}f_{\mathrm{e}q}\int_{p^{\prime}}\frac{p^{\prime}}{\omega}f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)\bigg{\\}},$
(65) $\displaystyle L_{4}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\bigg{\\{}\frac{p^{2}}{3\omega_{i}^{2}}\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)+\frac{g}{\bar{n}_{\mathrm{e}q}}\frac{p}{\omega}\bar{f}_{\mathrm{e}q}\int_{p^{\prime}}\frac{p^{\prime}}{\omega^{\prime}}\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\bigg{\\}}.$
(66)

Upto this point, we have only considered a single quark flavor, we will now
focus on the hot QCD medium with multiple quark flavors. In our case, we have
considered three flavour ($u$,$d$ and $s$ quarks and their anti-particles)
quark gluon plasma. The total induced current in the medium can be written as
the vector sum of the currents due to individual flavors as

$\displaystyle\vec{J}$ $\displaystyle=$
$\displaystyle\vec{J_{a}}+\vec{J_{b}}+......$ (67) $\displaystyle=$
$\displaystyle\left(\frac{q^{2}_{1}g_{1}\tau_{1}}{T}(L_{3}+L_{4})_{a}+\frac{q^{2}_{2}g_{2}\tau_{2}}{T}(L_{3}+L_{4})_{b}+....\right)\vec{E}$
$\displaystyle-\left(\frac{q_{1}g_{1}\tau_{1}}{T^{2}}(L_{1}+L_{2})_{a}+\frac{q_{2}g_{2}\tau_{2}}{T^{2}}(L_{1}+L_{2})_{b}+.....\right)\nabla_{r}T(r).$

In the steady state condition, the total induced current vanishes i.e.
$\vec{J}=0$. We get

$\displaystyle\vec{E}=\frac{1}{2T}\frac{\sum_{i}q_{i}g_{i}\tau_{i}(L_{1}+L_{2})_{i}}{\sum_{i}q^{2}_{i}g_{i}\tau_{i}(L_{3}+L_{4})_{i}}\nabla_{r}T(r),$
(68)

which gives the seebeck coefficient for the medium as

$\displaystyle S_{\rm
tot}=\frac{1}{2T}\frac{\sum_{i}q_{i}g_{i}\tau_{i}(L_{1}+L_{2})_{i}}{\sum_{i}q^{2}_{i}g_{i}\tau_{i}(L_{3}+L_{4})_{i}}.$
(69)

Since all the flavors have same relaxation time and degeneracy factor, so the
total Seebeck coefficient for the medium can be expressed in terms of the
Seebeck coefficient of the individual flavor as

$\displaystyle S_{\rm
tot}=\frac{\sum_{i}S_{i}q^{2}_{i}(L_{3}+L_{4})_{i}}{\sum_{i}q^{2}_{i}(L_{3}+L_{4})_{i}}.$
(70)

In the next subsection, we will explore how the presence of the background
magnetic field modulates the thermoelectric response of the hot QCD medium.

###  3.2 Seebeck and Nernst coefficient in the presence of the magnetic field

In this section, we will calculate the Seeebck coefficient of the hot QCD
medium in the presence of the magnetic field. Firstly, we will consider the
strong field regime, where the the energy of the individual quark is quantized
via Landau quantization. Then, we will explore the weak field limit, where the
magnetic field dependence in the transport coefficients enter through the
cyclotron frequency, which menifests a classical description of the charged
particle in the magnetic field.

####  3.2.1 The strong magnetic field case

In the presence of strong magnetic field, the quark energy gets quantized as
[74]

$\displaystyle\omega=\sqrt{p_{3}^{2}+m^{2}+2n|qB|},$ (71)

where $n=0,1,2....$ correspond to the discrete Landau levels and the phase
space factor takes the form [74]

$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}\rightarrow\sum_{n=0}^{\infty}\frac{|qB|}{2\pi}\int\frac{dp_{3}}{2\pi}(2-\delta_{n0}).$
(72)

Since we are interested in the strong magnetic field limit with scale
hierarchy ($|qB|>>T^{2}>>m_{f}{{}^{2}}$), the quarks are confined to the
lowest Landau levels i.e. $n=0$. As a result, the motion of the quarks is
confined in the longitudinal direction and becomes purely along the strong
magnetic field (${\bf B}=B\hat{z}$). A dimensional reduction in the quark
dynamics takes place from $3+1$ to $1+1$ dimensions rendering the induced
current along the $z$ direction as

$\displaystyle J_{3}=\sum
g\frac{|q_{i}B|}{4\pi^{2}}\int{dp_{3}}~{}\frac{p_{3}}{\omega}~{}(q\delta
f^{B}(x,p)+\bar{q}\delta\bar{f}^{B}(x,p)),$ (73)

where $\delta f^{B}(x,p)$ and $\delta\bar{f}^{B}(x,p)$ are the deviations in
the quark and anti-quark distribution functions, respectively. The RBTE (51)
in the presence of the strong B is given by

$\displaystyle p^{0}\frac{\partial f^{B}}{\partial x^{0}}+p^{3}\frac{\partial
f^{B}}{\partial x^{3}}+qF^{03}p_{3}\frac{\partial f^{B}}{\partial
p^{0}}+qF^{30}p_{0}\frac{\partial f^{B}}{\partial
p^{3}}=-p^{\mu}u_{\mu}\nu^{B}\left(f^{B}-n^{B}{n_{\mathrm{e}q}^{B~{}-1}}f_{\mathrm{e}q}^{B}\right),$
(74)

where $f^{B}=f_{\mathrm{e}q}^{B}+\delta f^{B}$; $f_{\mathrm{e}q}^{B}$ is the
equilibrium distribution function

$\displaystyle f_{\mathrm{e}q}^{B}=\frac{1}{e^{\beta\omega}+1}.$ (75)

Here $\omega=\sqrt{p_{3}^{2}+m_{f}^{2}}$; $m_{f}$ is the quasiparticle mass
(4) and $\nu_{B}$ is the collision frequency which is given by the inverse of
the relaxation time [75]

$\displaystyle\tau_{B}(p_{3};T,|qB|)=\frac{\omega(e^{\beta\omega}-1)}{\alpha_{s}C_{F}m_{f}^{2}(e^{\beta\omega}+1)}{\left(\int\frac{dp^{\prime}_{3}}{\omega^{\prime}(e^{\beta\omega^{\prime}}+1)}\right)}^{-1}.$
(76)

In Eq. (74), $n_{\mathrm{e}q}^{B}$ and $n^{B}$ are the equilibrium and non-
equilibrium number densities of quarks in strong $B$, which are given as

$\displaystyle n_{\mathrm{e}q}^{B}$ $\displaystyle=$
$\displaystyle\frac{g|qB|}{4\pi^{2}}\int dp_{3}~{}f_{\rm eq}^{B},$ (77)
$\displaystyle n^{B}$ $\displaystyle=$
$\displaystyle\frac{g|qB|}{4\pi^{2}}\int dp_{3}~{}(f_{\rm eq}^{B}+\delta
f^{B}).$ (78)

In order to obtain $\delta f^{B}$, we simplify Eqn (74) as

$\displaystyle\delta f^{B}-g{n_{\mathrm{e}q}^{B}}^{-1}f_{\rm
eq}^{B}\int_{p_{3}}\delta f^{B}$ $\displaystyle=$
$\displaystyle\tau^{B}\frac{p_{3}}{\omega}(\omega-\mu)\left(-\frac{1}{T^{2}}\right)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)(\nabla T)_{z}$ (79)
$\displaystyle+2q\beta\tau^{B}\frac{p_{3}E_{3}}{\omega}f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right),$

which is further solved for $\delta f^{B}$ upto first order in iteration as

$\displaystyle\delta f^{B}=\delta f^{B(0)}+g{n_{\mathrm{e}q}^{B}}^{-1}f_{\rm
eq}^{B}\int_{p^{\prime}_{3}}\delta f^{B(0)},$ (80)

where

$\displaystyle\delta
f^{B(0)}=\frac{p_{3}}{\omega}\tau^{B}(\omega-\mu)\left(-\frac{1}{T^{2}}\right)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)(\nabla
T)_{z}+\frac{2q\beta\tau_{i}^{B}}{\omega}p_{3}E_{3}~{}f_{\rm eq}^{B}(1-f_{\rm
eq}^{B}).$ (81)

Similarly, we can write for the anti-quarks as

$\displaystyle\delta\bar{f}^{B}=\delta\bar{f}^{B(0)}+g{n_{\mathrm{e}q}^{B}}^{-1}\bar{f}_{\rm
eq}^{B}\int_{p^{\prime}_{3}}\delta\bar{f}^{B(0)},$ (82)

where

$\displaystyle\delta\bar{f}^{B(0)}=\tau^{B}\frac{p_{3}}{\omega}(\omega+\mu)\left(-\frac{1}{T^{2}}\right)\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm eq}^{B}\right)(\nabla
T)_{z}+\frac{2\bar{q}\beta\tau^{B}}{\omega}p_{3}E_{3}~{}\bar{f}_{\rm
eq}^{B}(1-\bar{f}_{\rm eq}^{B}).$ (83)

Now substituting $\delta{f}^{B}$ and $\delta\bar{f}^{B}$ in Eq. (73) to obtain
the induced current in the strong magnetic field

$\displaystyle J_{3}$ $\displaystyle=$
$\displaystyle\frac{\beta}{4\pi^{2}}qg|qB|\int{dp_{3}}\bigg{[}\bigg{\\{}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}(\omega-\mu)\left(\frac{-1}{T^{2}}\right)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)(\nabla
T)_{z}+\frac{p_{3}^{2}}{\omega^{2}}~{}2E_{3}q\tau^{B}(p_{3})~{}f_{\rm
eq}^{B}(1-f_{\rm eq}^{B})\bigg{\\}}$ $\displaystyle+$
$\displaystyle\frac{g}{{n_{\mathrm{e}q}^{B}}}\frac{p_{3}}{\omega_{i}}f_{\rm
eq}^{B}\int_{p^{\prime}_{3}}\bigg{\\{}\frac{p^{\prime}_{3}}{\omega^{\prime
2}}\tau^{B}(\omega-\mu)\left(\frac{-1}{T^{2}}\right)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)(\nabla
T)_{z}+\frac{p^{\prime}_{3}}{\omega^{\prime}}~{}2E_{3}q\tau^{B}(p_{3}^{\prime})~{}f_{\rm
eq}^{B}~{}(1-f_{\rm eq}^{B})\bigg{\\}}\bigg{]}$ $\displaystyle+$
$\displaystyle\frac{\beta}{4\pi^{2}}\bar{q}g|qB|\int{dp_{3}}\bigg{[}\bigg{\\{}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}(\omega+\mu)\left(\frac{-1}{T^{2}}\right)\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm eq}^{B}\right)(\nabla
T)_{z}+\frac{p_{3}^{2}}{\omega^{2}}~{}2E_{3}\bar{q}\tau^{B}(p_{3})~{}\bar{f}_{\rm
eq}^{B}(1-\bar{f}_{\rm eq}^{B})\bigg{\\}}$ $\displaystyle+$
$\displaystyle\frac{g}{{\bar{n}_{\mathrm{e}q}^{B}}}\frac{p_{3}}{\omega}\bar{f}_{\rm
eq}^{B}\int_{p^{\prime}_{3}}\bigg{\\{}\frac{p^{\prime}_{3}}{\omega^{2}}\tau^{B}(\omega^{\prime}+\mu)\left(\frac{-1}{T^{2}}\right)\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm eq}^{B}\right)(\nabla
T)_{z}+\frac{p^{\prime}_{3}}{\omega^{\prime}}~{}2E_{3}\bar{q}\tau^{B}(p_{3}^{\prime})~{}\bar{f}_{\rm
eq}^{B}~{}(1-\bar{f}_{\rm eq}^{B})\bigg{\\}}\bigg{]}.$

In the condition of the state of equilibrium $J_{3}$ becomes zero. The
electric field can be related to the $(\nabla T)_{z}$ as

$\displaystyle{\bf E}$ $\displaystyle=$
$\displaystyle\frac{1}{2qT}\left(\frac{H_{1}+H_{2}}{H_{3}+H_{4}}\right)(\nabla
T)_{z},$ (85) $\displaystyle\equiv$ $\displaystyle S_{B}~{}(\nabla T)_{z}.$
(86)

$S_{B}$ here corresponds to the Seebeck coefficient in the presence of strong
magnetic field, which reads

$\displaystyle
S_{B}=\frac{1}{2qT}\left(\frac{H_{1}+H_{2}}{H_{3}+H_{4}}\right),$ (87)

where the integrals $H_{1}$, $H_{2}$, $H_{3}$ and $H_{4}$ are given by

$\displaystyle H_{1}$ $\displaystyle=$
$\displaystyle\frac{|qB|}{4\pi^{2}}\int{dp_{3}}\bigg{\\{}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}(\omega-\mu)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)+g\frac{p_{3}}{\omega}\frac{f_{\rm
eq}^{B}}{n_{\mathrm{e}q}^{B}}\int_{p^{\prime}_{3}}\frac{p^{\prime}_{3}}{\omega^{\prime}}\tau^{B}(\omega^{\prime}-\mu)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)\bigg{\\}},$ (88) $\displaystyle H_{2}$
$\displaystyle=$
$\displaystyle-\frac{|qB|}{4\pi^{2}}\int{dp_{3}}\bigg{\\{}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}(\omega+\mu)\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm
eq}^{B}\right)+g\frac{p_{3}}{\omega}\frac{\bar{f}_{\rm
eq}^{B}}{\bar{n}_{\mathrm{e}q}^{B}}\int_{p^{\prime}_{3}}\frac{p^{\prime}_{3}}{\omega^{\prime}}\tau^{B}(\omega^{\prime}+\mu)\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm eq}^{B}\right)\bigg{\\}},$ (89) $\displaystyle
H_{3}$ $\displaystyle=$
$\displaystyle\frac{|qB|}{4\pi^{2}}\int{dp_{3}}\bigg{\\{}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)+g\frac{p_{3}}{\omega}\frac{f_{\rm
eq}^{B}}{n_{\mathrm{e}q}^{B}}\int_{p^{\prime}_{3}}\frac{p^{\prime}_{3}}{\omega^{\prime}}\tau^{B}f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)\bigg{\\}},$ (90) $\displaystyle H_{4}$
$\displaystyle=$
$\displaystyle\frac{|qB|}{4\pi^{2}}\int{dp_{3}}\bigg{\\{}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm
eq}^{B}\right)+g\frac{p_{3}}{\omega}\frac{\bar{f}_{\rm
eq}^{B}}{\bar{n}_{\mathrm{e}q}^{B}}\int_{p^{\prime}_{3}}\frac{p^{\prime}_{3}}{\omega^{\prime}}\tau^{B}\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm eq}^{B}\right)\bigg{\\}}.$ (91)

For the hot QCD medium consisting of $u$, $d$ and $s$ quarks, the $3$rd
component of the induced current can be written as the vector sum of the
individual currents as

$\displaystyle J_{3}$ $\displaystyle=$
$\displaystyle(J_{3})_{a}+(J_{3})_{b}+...........,$ (93) $\displaystyle=$
$\displaystyle\left(\frac{q^{2}_{1}g_{1}|q_{1}B|}{T}(H_{3}+H_{4})_{a}+\frac{q^{2}_{2}g_{2}|q_{2}B|}{T}(H_{3}+H_{4})_{b}+.....\right)E_{z}$
$\displaystyle-\left(\frac{q_{1}g_{1}|q_{1}B|}{T^{2}}(H_{1}+H_{2})_{a}+\frac{q_{2}g_{2}|q_{2}B|}{T^{2}}(H_{1}+H_{2})_{b}+....\right)(\nabla
T)_{z},$

which vanishes in the steady state i.e. $J_{3}=0$ and gives

$\displaystyle
E_{z}=\frac{1}{2T}\frac{\sum_{i}q_{i}g_{i}(H_{1}+H_{2})_{i}}{\sum_{i}q^{2}_{i}g_{i}(H_{3}+H_{4})_{i}}(\nabla
T)_{z}.$ (94)

We extract the Seebeck coefficient for the composite medium as

$\displaystyle S^{B}_{\rm
tot}=\frac{1}{2T}\frac{\sum_{i}q_{i}|q_{i}B|(H_{1}+H_{2})_{i}}{\sum_{i}q^{2}_{i}|q_{i}B|(H_{3}+H_{4})_{i}},$
(95)

which can be expressed in terms of the Seebeck coefficient of the single quark
flavor as

$\displaystyle S^{B}_{\rm
tot}=\frac{\sum_{i}S_{i}|q_{i}|^{3}(H_{3}+H_{4})_{i}}{\sum_{i}|q_{i}|^{3}(H_{3}+H_{4})_{i}}.$
(96)

In the strong magnetic field, there is no current in the transverse direction
due to the one-dimensional (LLL) quark dynamics. Hence the Nernst coefficient
which measures the thermoelectric response in the transverse direction,
vanishes. In the next subsection, we will explore the weak magnetic field
regime, where the quark dynamics is not affected by the Landau quantization,
rather the magnetic field dependence enters via the cyclotron frequency, which
manifests the semi-classical description. In this scenario, the Nernst
coefficient would also appear.

####  3.2.2  The weak magnetic field case

In the weak magnetic field, the dispersion relation of the charged particle is
not directly affected by the magnetic field rather $B$ acts as a perturbation.
So the 1+1 dimensional Landau level kinematics is not applicable. The induced
four current due to a single quark flavour is given by

$\displaystyle
J^{\mu}=g\int\frac{d^{3}p}{(2\pi)^{3}}\frac{p^{\mu}}{\epsilon}[q\delta
f+\bar{q}\delta\bar{f}],$ (97)

where $\epsilon=\sqrt{p^{2}+m^{2}}$. $\delta f~{}(\delta\bar{f})$ is the
deviation in the quark (anti-quark) distribution function, which will be
calculated from RBTE in the weak magnetic field.

The RBTE (46) in the presence of the Lorenz force can be written as (see the
Appendix)

$\displaystyle\frac{\partial f}{\partial t}+\vec{v}.\frac{\partial
f}{\partial\vec{r}}+\vec{F}.\frac{\partial f}{\partial\vec{p}}$
$\displaystyle=$
$\displaystyle-\frac{1}{\tau}\left(f-\frac{n}{n_{0}}f_{0}\right),$ (98)

where $\vec{F}=q~{}(\vec{E}+\vec{v}\times\vec{B})$. Since magnetic field is
not a dominat energy scale here, we have used the same relaxation time
calculated in the absence of $B$ (54). The magnetic field and chemical
potential dependence in $\tau$ will enter through the strong coupling contant
(7). $n$ and $n_{0}$ are given by Eqs. (52) and (53), respectively except the
mass in the dispersion relation will be replaced by the thermo-magnetic mass
calculated in weak magnetic field in section 3.

Without the loss of generality, let us consider the electric field in the $xy$
plane i.e. $\vec{E}=E_{x}\hat{x}+E_{y}\hat{y}$ and magnetic field in $z$
direction i.e $\vec{B}=B\hat{z}$. Then for QCD medium, which is homogeneous in
time, Eq. (98) takes the form

$\displaystyle qB\tau\left(v_{x}\frac{\partial f}{\partial
p_{y}}-v_{y}\frac{\partial f}{\partial
p_{x}}\right)-\tau\vec{v}.\frac{\partial f}{\partial\vec{r}}-\tau
q\vec{E}.\frac{\partial f}{\partial\vec{p}}=\left(\delta
f-gn_{0}^{-1}f_{0}\int\delta f\right),$ (99)

which further can be solved for $\delta f$ upto first order as

$\displaystyle\delta f=\delta f^{(a)}+gn_{0}^{-1}f_{0}\int\delta f^{(a)},$
(100)

where

$\displaystyle\delta f^{(a)}=qB\tau\left(v_{x}\frac{\partial f}{\partial
p_{y}}-v_{y}\frac{\partial f}{\partial
p_{x}}\right)-\tau\vec{v}.\frac{\partial f}{\partial\vec{r}}-\tau
q\vec{E}.\frac{\partial f}{\partial\vec{p}}.$ (101)

In order to solve Eq. (100), we take an ansatz [76]

$\displaystyle\delta f=\delta f^{(b)}+gn_{0}^{-1}f_{0}\int_{p^{\prime}}\delta
f^{(b)},$ (102)

where

$\displaystyle\delta f^{b}=f-f_{0}=-\tau q\vec{E}.\frac{\partial
f_{0}}{\partial\vec{p}}-\vec{\lambda}.\frac{\partial f_{0}}{\partial\vec{p}}.$
(103)

Equating Eqs. (100) and (102), we get

$\displaystyle\vec{\lambda}.\frac{\partial
f_{0}}{\partial\vec{p}}-qB\tau\left(v_{y}\frac{\partial f}{\partial
p_{x}}-v_{x}\frac{\partial f}{\partial
p_{y}}\right)-\tau\vec{v}.\frac{\partial
f}{\partial\vec{r}}=gn_{0}^{-1}f_{0}\left(\int_{p^{\prime}}\delta
f^{(a)}+\int_{p^{\prime}}\delta f^{(b)}\right).$ (104)

We calculate $\frac{\partial f}{\partial p_{y}}$ and $\frac{\partial
f}{\partial p_{x}}$ using the ansatz (102) as

$\displaystyle\left(v_{x}\frac{\partial f}{\partial p_{y}}-v_{y}\frac{\partial
f}{\partial p_{x}}\right)=\bigg{(}v_{y}\lambda_{x}+v_{y}\tau
qE_{x}-v_{x}\lambda_{y}-v_{x}\tau qE_{y}\bigg{)}\frac{\partial
f_{0}}{\partial\epsilon}\frac{1}{\epsilon},$ (105)

where we have retained terms which are linear in the velocity only.

Now substituting Eq. (105) in (104) and after doing some re-arrangement, we
get

$\displaystyle v_{x}\left[\frac{\lambda_{x}}{\tau}-\omega_{c}\tau
qE_{y}-\omega_{c}\lambda_{y}+\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial x}\right]+v_{y}\left[\frac{\lambda_{y}}{\tau}+\omega_{c}\tau
qE_{x}+\omega_{c}\lambda_{x}+\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial y}\right]$
$\displaystyle+\frac{gT}{n_{0}\tau}\left(\int_{p^{\prime}}\delta
f^{(a)}+\int_{p^{\prime}}\delta f^{(b)}\right)=0,$ (106)

which on equating the coefficient of $v_{x}$ and $v_{y}$ on both side gives

$\displaystyle\frac{\lambda_{x}}{\tau}-\omega_{c}\tau
qE_{y}-\omega_{c}\lambda_{y}+\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial x}=0,$ (107) $\displaystyle\frac{\lambda_{y}}{\tau}+\omega_{c}\tau
qE_{x}+\omega_{c}\lambda_{x}+\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial y}=0.$ (108)

We solve above Eqs. (107) and (108) to get $\lambda_{x}$ and $\lambda_{y}$
which are given by

$\displaystyle\lambda_{x}=-\frac{\omega_{c}^{2}\tau^{3}}{1+\omega_{c}^{2}\tau^{2}}qE_{x}-\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial
x}+\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}qE_{y}-\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial y},$ (109)
$\displaystyle\lambda_{y}=-\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}qE_{x}+\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial
x}-\frac{\omega_{c}^{2}\tau^{3}}{1+\omega_{c}^{2}\tau^{2}}qE_{y}-\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)\frac{\partial
T}{\partial y}.$ (110)

Now we substitute $\lambda_{x}$ and $\lambda_{y}$ in (102) to obtain $\delta
f$ which reads

$\displaystyle\delta f=\delta f^{(b)}+gn_{0}^{-1}f_{0}\int_{p^{\prime}}\delta
f^{(b)},$ (111)

where

$\displaystyle\delta f^{(b)}$ $\displaystyle=$ $\displaystyle\frac{\partial
f_{0}}{\partial\epsilon}\left[-\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}qv_{x}+\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}qv_{y}\right]E_{x}$
(112) $\displaystyle+\frac{\partial
f_{0}}{\partial\epsilon}\left[-\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}qv_{y}-\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}qv_{x}\right]E_{y}$
$\displaystyle+\frac{\partial
f_{0}}{\partial\epsilon}\left[\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{x}-\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{y}\right]\frac{\partial
T}{\partial x}$ $\displaystyle+\frac{\partial
f_{0}}{\partial\epsilon}\left[\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{y}+\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{x}\right]\frac{\partial
T}{\partial y}.$

Similarly, deviation in the anti-quark distribution function can be evaluated
as (replacing q with -q and $\omega_{c}$ with \- $\omega_{c}$ in Eq. (112))

$\displaystyle\delta\bar{f}=\delta\bar{f}^{(b)}+g\bar{n}_{0}^{-1}\bar{f}_{0}\int_{p^{\prime}}\delta\bar{f}^{(b)}$
(113)

where

$\displaystyle\delta\bar{f}^{(b)}$ $\displaystyle=$
$\displaystyle\frac{\partial\bar{f}_{0}}{\partial\epsilon}\left[\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}qv_{x}+\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}qv_{y}\right]E_{x}$
(114)
$\displaystyle+\frac{\partial\bar{f}_{0}}{\partial\epsilon}\left[\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}qv_{y}-\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}qv_{x}\right]E_{y}$
$\displaystyle+\frac{\partial\bar{f}_{0}}{\partial\epsilon}\left[\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{x}+\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{y}\right]\frac{\partial
T}{\partial x}$
$\displaystyle+\frac{\partial\bar{f}_{0}}{\partial\epsilon}\left[\frac{\tau}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{y}-\frac{\omega_{c}\tau^{2}}{1+\omega_{c}^{2}\tau^{2}}\left(\frac{\epsilon-\mu}{T}\right)v_{x}\right]\frac{\partial
T}{\partial y}.$

We substitute $\delta f$ and $\delta\bar{f}$ in (97) to get the $x$ and $y$
components of the induced current density

$\displaystyle J_{x}$ $\displaystyle=$ $\displaystyle qg\left[(q\beta
I_{1})E_{x}+(q\beta I_{2})E_{y}+(\beta^{2}I_{3})\frac{\partial T}{\partial
x}+(\beta^{2}I_{4})\frac{\partial T}{\partial y}\right],$ (115) $\displaystyle
J_{y}$ $\displaystyle=$ $\displaystyle qg\left[(-q\beta I_{2})E_{x}+(q\beta
I_{1})E_{y}+(-\beta^{2}I_{4})\frac{\partial T}{\partial
x}+(\beta^{2}I_{3})\frac{\partial T}{\partial y}\right].$ (116)

The integrals $I_{1}$, $I_{2}$, $I_{3}$ and $I_{4}$ in the above equations are
given by

$\displaystyle I_{1}=I_{1q}+I_{1\bar{q}},$ (117) $\displaystyle
I_{2}=I_{2q}+I_{2\bar{q}},$ (118) $\displaystyle I_{3}=I_{3q}+I_{3\bar{q}},$
(119) $\displaystyle I_{4}=I_{4q}+I_{4\bar{q}},$ (120)

where

$\displaystyle I_{1q}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}f_{0}(1-f_{0})+\frac{g}{n_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}f_{0}(1-f_{0})\right\\},$
$\displaystyle I_{1\bar{q}}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}\bar{f}_{0}(1-\bar{f}_{0})+\frac{g}{\bar{n}_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}\bar{f}_{0}(1-\bar{f}_{0})\right\\},$
$\displaystyle I_{2q}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}f_{0}(1-f_{0})+\frac{g}{n_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}f_{0}(1-f_{0})\right\\},$
$\displaystyle I_{2\bar{q}}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}\bar{f}_{0}(1-\bar{f}_{0})+\frac{g}{\bar{n}_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}\bar{f}_{0}(1-\bar{f}_{0})\right\\},$
$\displaystyle I_{3q}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon-\mu)f_{0}(1-f_{0})+\frac{g}{n_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon^{\prime}-\mu)f_{0}(1-f_{0})\right\\},$
$\displaystyle I_{3\bar{q}}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon+\mu)\bar{f}_{0}(1-\bar{f}_{0})+\frac{g}{\bar{n}_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon^{\prime}+\mu)\bar{f}_{0}(1-\bar{f}_{0})\right\\},$
$\displaystyle I_{4q}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon-\mu)f_{0}(1-f_{0})+\frac{g}{n_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon^{\prime}-\mu)f_{0}(1-f_{0})\right\\},$
$\displaystyle I_{4\bar{q}}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\left\\{\frac{p^{2}}{3\epsilon^{2}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon+\mu)\bar{f}_{0}(1-\bar{f}_{0})+\frac{g}{\bar{n}_{0}}\frac{p}{\epsilon}f_{0}\int_{p^{\prime}}\frac{p^{\prime}}{\epsilon^{\prime}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}(\epsilon^{\prime}+\mu)\bar{f}_{0}(1-\bar{f}_{0})\right\\}.$

In the state of equilibrium, the components of the induced current density
along $x$ and $y$ direction vanishes i.e. $J_{x}=J_{y}=0$. We can write from
Eqs. (115) and (116)

$\displaystyle C_{1}E_{x}+C_{2}E_{y}+C_{3}\frac{\partial T}{\partial
x}+C_{4}\frac{\partial T}{\partial y}$ $\displaystyle=$ $\displaystyle 0,$
(121) $\displaystyle-C_{2}E_{x}+C_{1}E_{y}-C_{4}\frac{\partial T}{\partial
x}+C_{3}\frac{\partial T}{\partial y}$ $\displaystyle=$ $\displaystyle 0,$
(122)

provided $C_{1}=qI_{1}$, $C_{2}=qI_{2}$, $C_{3}=\beta I_{3}$, $C_{4}=\beta
I_{4}$. The electric field components are related to the temperature
gradients, Seebeck and Nernst coefficients via a matrix equation as

$\displaystyle\begin{pmatrix}E_{x}\\\
E_{y}\end{pmatrix}=\begin{pmatrix}S&N|B|\\\
-N|B|&S\end{pmatrix}\begin{pmatrix}\frac{\partial T}{\partial x}~{}\\\
\frac{\partial T}{\partial y}\end{pmatrix}.$ (123)

We solve Eqs. (121) and (122) for $E_{x}$ and $E_{y}$ as

$\displaystyle
E_{x}=\left(-\frac{C_{1}C_{3}+C_{2}C_{4}}{C_{1}^{2}+C_{2}^{2}}\right)\frac{\partial
T}{\partial
x}+\left(-\frac{C_{2}C_{3}-C_{1}C_{4}}{C_{1}^{2}+C_{2}^{2}}\right)\frac{\partial
T}{\partial y},$ (124) $\displaystyle
E_{y}=\left(-\frac{C_{1}C_{3}+C_{2}C_{4}}{C_{1}^{2}+C_{2}^{2}}\right)\frac{\partial
T}{\partial
y}-\left(-\frac{C_{2}C_{3}-C_{1}C_{4}}{C_{1}^{2}+C_{2}^{2}}\right)\frac{\partial
T}{\partial x},$ (125)

which give the Seebeck and Nernst coefficients

$\displaystyle S$ $\displaystyle=$
$\displaystyle-\frac{(C_{1}C_{3}+C_{2}C_{4})}{C_{1}^{2}+C_{2}^{2}},$ (126)
$\displaystyle N|B|$ $\displaystyle=$
$\displaystyle\frac{(C_{2}C_{3}-C_{1}C_{4})}{C_{1}^{2}+C_{2}^{2}},$ (127)

respectively. The integrals $C_{2}$ and $C_{4}$ vanishes in the absence of the
magnetic field, as a result Nernst coefficient would also vanish.

Now we will compute the Seebeck and Nernst coefficients for the medium. In the
medium composed of $u$ and $d$ light quarks, the $x$ and $y$ components of the
current can be written as the sum of the indivisual contributions as

$\displaystyle J_{x}$ $\displaystyle=$
$\displaystyle\sum_{a=u,d}\left[q_{a}(I_{1})_{a}E_{x}+q_{a}(I_{2})_{a}E_{y}+\beta(I_{3})_{a}\frac{\partial
T}{\partial x}+\beta(I_{4})_{a}\frac{\partial T}{\partial y}\right],$ (128)
$\displaystyle J_{y}$ $\displaystyle=$
$\displaystyle\sum_{a=u,d}\left[q_{a}(I_{2})_{a}E_{x}+q_{a}(I_{1})_{a}E_{y}-\beta(I_{4})_{a}\frac{\partial
T}{\partial x}+\beta(I_{3})_{a}\frac{\partial T}{\partial y}\right].$ (129)

The Seebeck and Nernst coefficients of the medium can be extracted by imposing
the steady state condition (i.e. putting $J_{x}=J_{y}=0$) as

$\displaystyle S^{B^{\prime}}_{tot}$ $\displaystyle=$
$\displaystyle-\frac{(K_{1}K_{3}+K_{2}K_{4})}{K_{1}^{2}+K_{2}^{2}},$ (130)
$\displaystyle N|B|$ $\displaystyle=$
$\displaystyle\frac{(K_{2}K_{3}-K_{1}K_{4})}{K_{1}^{2}+K_{2}^{2}}.$ (131)

where

$\displaystyle K_{1}$ $\displaystyle=$
$\displaystyle\sum_{a=u,d}q_{a}(I_{1})_{a},$ (132) $\displaystyle K_{2}$
$\displaystyle=$ $\displaystyle\sum_{a=u,d}q_{a}(I_{2})_{a},$ (133)
$\displaystyle K_{3}$ $\displaystyle=$
$\displaystyle\sum_{a=u,d}\beta(I_{3})_{a},$ (134) $\displaystyle K_{4}$
$\displaystyle=$ $\displaystyle\sum_{a=u,d}\beta(I_{4})_{a}.$ (135)

##  4 Results and discussion

Now we will discuss the results obtained in the previous sections. In Fig. 1
(a), we have shown the variation of the Seebeck coefficient for u quarks as a
function of temperature. We have noticed that the magnitude of the Seebeck
coefficient decreases with $T$ in BGK as well as RTA collision terms. It is
not much sensitive to the collision term, however, its magnitude gets slightly
reduced (around $3\%$) in the BGK term in the temperature range $160-400$ MeV.
The sign of the Seebeck coefficient in the case of $d$ and $s$ quarks get
reversed due to its negative charge, otherwise, it shows similar trends w.r.t.
$T$ and collision terms. [see Fig. 1 (b) and Fig. 2 (a)]. In the case of the
composite medium, the Seebeck coefficient ($S_{\rm tot}$) [Fig. 2 (b)] has
been found to be positive. We notice a considerable enhancement in the
magnitude of $S_{\rm tot}$ in BGK collision term, which is around $12\%$ at
lower $T$ (160 MeV) and $25\%$ at high $T$ (400 MeV) [In Fig. 3 (b)]. We
further study the effects of the quark chemical potential ($\mu$) on the
Seebeck coefficient of the medium in the BGK collision term in Fig. 3 (a) for
$\mu=40,60,80$ MeV and have found that the magnitude of $S_{\rm tot}$
increases as we raise the strength of $\mu$. We can conclude that both BGK
collision term and baryon asymmetry in the medium enhance its ability to
convert the temperature gradient into the current.

|
---|---
(a) |  (b)
Figure 1: Temperature dependence of the Seebeck coefficient in the absence of magnetic field: (a) for $u$ quarks and (b) for $d$ quarks  |
---|---
(a) |  (b)
Figure 2: Temperature dependence of the Seebeck coefficient in the absence of magnetic field: (a) for $s$ quarks and (b) for composite medium |
---|---
(a) |  (b)
Figure 3: Left panel: Temperature dependence of Seebeck coefficient for the
medium with BGK collision term in absence of magnetic field for different
strengths of $\mu$. Right panel: Ratio of Seebeck coefficients in BGK to that
in RTA collision integral with temperature

We have studied the combined effects of the strong magnetic field and
collision integral on the thermoelectric response of the hot QCD medium in
Figs. 4 and 5. We notice that the Seebeck coefficient for the individual quark
flavours as well as for the combined medium gets enhanced in the BGK collision
integral. We have plotted the ratio of the Seebeck coefficient calculated with
BGK collision term to that with RTA i.e. (${\rm BGK}/{\rm RTA}$) in Fig. 5 (b)
to examine the relative competition between the two collision integrals. We
observe that the ratio is greater than one for the temperature range $160-400$
MeV. The enhancement is around $15\%$ for the $u$ and $d$ quarks in the region
near the critical temperature ($T_{c})$ and further increases with the
temperature. For the $s$ quark the enhancement is around $15\%$ near $T_{c}$
and reaches up to $35\%$ when the temperature of the medium is around $400$
MeV. In the case of the medium, the enhancement is around $10\%$ near $T_{c}$
but decreases as we go towards higher temperatures. $S^{B}_{\rm tot}$
increases with the strength of $\mu$ [Fig. 5 (a)]. The Seebeck coefficient
increases with the strength of $B$ for indivisual flavors as well as for the
combined medium. The strong $B$ enhances the medium Seebeck coefficient in
comparison to $B=0$ case, which indicates that dimensional reduction in the
quark dynamics due to the strong magnetic field enhances the ability of the
hot QCD medium to change the temperature gradient into the current.

|
---|---
(a) |  (b)
Figure 4: Temperature dependence of the Seebeck coefficient in the strong magnetic field: (a) for $u$ quarks and (b) for $d$ quarks  |
---|---
(a) |  (b)
Figure 5: Temperature dependence of the Seebeck coefficient in the strong magnetic field: (a) for $s$ quarks and (b) for composite medium |
---|---
(a) |  (b)
Figure 6: Left panel: Temperature dependence of Seebeck coefficient for the
medium with BGK collision term in strong magnetic field for different
strengths of $\mu$. Right panel: Ratio of Seebeck coefficient in BGK to that
in RTA collision integral with temperature

In Fig. 7, we have displayed the variation of the Seebeck coefficient with
temperature in the weak magnetic field ($eB=0.3~{}m_{\pi}^{2}$) for $u$ (left)
and $d$ (right) quarks. The magnitude of the Seebeck coefficient decreases
with $T$. We notice similar trends for the medium [see Fig. 8 (a)]. It is not
much sensitive to the collision integral, rather it depends on the chirality
of the quark quasiparticles. In both the collision terms, the magnitude of the
Seebeck coefficient is higher for the left-handed modes in comparison to that
of right-handed ones. We have calculated the ratio of the Seebeck coefficient
with the BGK collision term to that calculated using the RTA with $T$ for both
left- and right-handed particles in Fig. 9 (a) and (b), respectively. The
ratio is less than one, which shows that the Seebeck coefficient is reduced in
the BGK term. However, the reduction is very small around 2-3 $\%$ for left-
handed modes, and essentially nonexistent for right-handed modes. We have also
studied the effects of the baryon asymmetry on the thermoelectric phenomenon
in Fig. 8 (b) for $\mu=40$ and $60$ MeV and have noticed that Seebeck
coefficient increases as we increase the strength of $\mu$.

|
---|---
(a) |  (b)
Figure 7: Seebeck coefficient as a function of temperature in the weak magnetic field (a) for u quarks, (b) for d quarks. |
---|---
(a) |  (b)
Figure 8: (a) Seebeck coefficient as a function of temperature for the composite medium, (b) Seebeck coefficient as a function of temperature in the BGK collision term for different strengths of $\mu$.  |
---|---
(a) |  (b)
Figure 9:  (a) Ratio of Seebeck coefficient in BGK to that in RTA as a function of temperature for $L$ modes, (b) Ratio of Seebeck coefficient in BGK to that in RTA as a function of temperature for $R$ modes. |
---|---
(a) |  (b)
Figure 10: Nernst coefficient as a function of temperature in the weak magnetic field (a) for u quarks, (b) for d quarks. |
---|---
(a) |  (b)
Figure 11:  (a) Nernst coefficient as a function of temperature for the composite medium, (b) Nernst coefficient as a function of temperature in the BGK collision term for different strengths of $\mu$.  |
---|---
(a) |  (b)
Figure 12:  (a) Ratio of Nernst coefficient in BGK to that in RTA as a
function of temperature for $L$ modes, (b) Ratio of Nernst coefficient in BGK
to that in RTA as a function of temperature for $R$ modes.

In Fig. 10, we have examined the temperature and collision integral dependence
of the Nernst coefficient for $u$ (left panel) and $d$ (right panel) quarks.
The magnitude of the Nernst coefficient decreases with temperature. We observe
similar trends in the case of combined medium [seen in Fig. 11 (a)]. One
important observation we notice in the case of $u$ and $d$ quarks is that in
the RTA there is no difference in the $L$ and $R$ modes but in BGK approach .
The ratio BGK/RTA for the Nernst coefficient is less than one in the case of
left-handed modes, which shows that BGK collision integral causes a reduction
in the magnitude of the Nernst coefficient [Fig. 12 (a)]. Contrary to it, for
the right-handed modes, the magnitude gets enhanced [see Figs. 12 (b)].
Enhancement is much pronounced in the low-temperature region and further
diminishes as temperature rises. In the case of the left-handed modes, this
ratio is in the range of $0.65-0.68$, while for right-handed modes $1.2-1.9$
for the temperature domain $160-400$ MeV. In Fig 11 (b), we have shown the
medium Nernst coefficient in BGK collision integral as a function of the
temperature for $\mu=40$ and $60$ MeV. We notice that $\mu$ dependence is not
much visible in the case of $L$ modes except near the transition temperature,
where it gets slightly reduced for $\mu=60$ MeV. In the case of $R$ modes, the
Nernst coefficient is larger for $\mu=60$ MeV in comparison to $\mu=40$ MeV at
a fixed value of temperature.

##  5 Conclusion

To conclude, we have studied the thermoelectric response of a hot and
magnetized QCD medium produced in the non-central collisions at RHIC and LHC.
We have employed Boltzmann transport equation linearized by the BGK collision
integral, which conserves the particle number and current instantaneously. We
incorporate the medium effects via dispersion relation wherein the
temperature, chemical potential and magnetic field dependent masses have been
calculated using the imaginary-time formalism of the finite temperature QCD.
In the absence of $B$, the Seebeck coefficient gets reduced in the BGK
collision term for $u$, $d$ and $s$ quarks whereas it gets enhanced for the
composite medium. In the presence of strong magnetic field, the magnitude of
the individual as well as medium Seebeck coefficient get lifted in the BGK
collision term. Seebeck coefficient of the medium gets enhanced with the quark
chemical potential in the absence as well as in the presence of the strong
$B$. In addition to the Seebeck coefficient, Nernst coefficient also appears
in the weak magnetic field. The Seebeck coefficient is not much sensitive to
the collision term but it does depend on the chirality of the quark quasi-
particles. It has greater magnitude for the $L$-modes. On the other hand,
Nernst coefficient gets changed drastically in the BGK collision term and its
magnitude is higher for $R$-modes. Both Seebeck and Nernst coefficients
increase with $\mu$ for both $L$ and $R$ modes.

A nonvanishing Seebeck coefficient will modify the electric as well as heat
current in the medium. The electric current in the presence of Seebeck effect
becomes $J=\sigma_{el}E-\sigma_{el}S\nabla T$, while thermal conductivity gets
modified as $\kappa=\kappa_{0}-T\sigma_{el}S^{2}$. Both electrical and thermal
conductivities should take positive values in accordance with the second law
of thermodynamics. i.e. $T\partial_{\mu}S^{\mu}>0$. Hence a positive Seebeck
coefficient will always reduce the electric current and the thermal
conductivity. It will be also interesting to take the thermoelectric effects
into the account in the calculation of the entropy production, which has been
completely neglected in [77, 78]. Moreover, thermoelectric coefficients could
also be relevant in the context of the spin Hall effect (SHE). In SHE, a
transverse spin current is generated due to the external electric field but
the life-time of such electric field produced in the heavy ion collisions
could be too small to observe the SHE. The electric field produced due to the
temperature gradients in the medium may induce spin Hall effect in a hot and
dense strongly interacting matter produced in heavy-ion collisions [79]. So
the study of the various implications of thermoelectric effects in the hot and
dense medium needs further investigation.

## Acknowledgements

S. A. Khan would like to thank Debarshi Dey and Pushpa Panday for insightful
discussions during the course of this work.

##  Appendix A Derivation of equation (55)

BGK collision term is given by (51) as

$\displaystyle C[f]$ $\displaystyle=$
$\displaystyle-p^{\mu}u_{\mu}\nu_{i}\left(f_{i}-n_{{}_{i}}n_{\mathrm{e}q,i}^{-1}f_{\mathrm{e}q,i}\right)$
(A.136) $\displaystyle=$
$\displaystyle-p^{\mu}u_{\mu}\nu_{i}\left(f_{i}-\frac{g_{i}\int_{p}(f_{\mathrm{e}q,i}+\delta
f_{i})}{n_{\mathrm{e}q,i}}f_{\mathrm{e}q,i}\right)$ $\displaystyle=$
$\displaystyle-p^{\mu}u_{\mu}\nu_{i}\left(f_{i}-\frac{(g_{i}\int_{p}f_{\mathrm{e}q,i}+g_{i}\int_{p}\delta
f_{i})}{n_{\mathrm{e}q,i}}f_{\mathrm{e}q,i}\right)$ $\displaystyle=$
$\displaystyle-p^{\mu}u_{\mu}\nu_{i}\left(\delta
f_{i}-g_{i}n_{\mathrm{e}q,i}^{-1}f_{\mathrm{e}q,i}\int_{p}\delta
f_{i}\right).$

##  Appendix B Boltzmann Equation in the weak magnetic field

The RBTE (46) can be written with the BGK collision integral as

$\displaystyle p^{\mu}\frac{\partial f}{\partial
x^{\mu}}+q~{}F^{\prime\sigma}\frac{\partial f}{\partial
p^{\sigma}}=-p^{\mu}u_{\mu}\nu_{i}\left(f_{i}-n_{{}_{i}}n_{\mathrm{e}q,i}^{-1}f_{\mathrm{e}q,i}\right)$
(B.137)

where
$F^{\prime\sigma}=qF^{\sigma\rho}p_{\rho}=(p^{0}\vec{v}.\vec{F},p^{0}\vec{F})$,
is the relativistic generalisation of the classical electromagnetic force
$F=q(\vec{E}+\vec{v}\times\vec{B})$. We can write Eqn. (B.137) using
$F^{0i}=-E^{i}$ and $2F_{ij}=\epsilon_{ijk}B^{k}$ ($\epsilon_{ijk}$ is the
completely anti-symmetric Levi-Civita tensor) as

$\displaystyle\frac{\partial f}{\partial t}+\vec{v}.\frac{\partial
f}{\partial\vec{r}}+\frac{\vec{F}.\vec{p}}{p^{0}}\frac{\partial f}{\partial
p^{0}}+\vec{F}.\frac{\partial
f}{\partial\vec{p}}=-\nu_{i}\left(f_{i}-n_{{}_{i}}n_{\mathrm{e}q,i}^{-1}f_{\mathrm{e}q,i}\right)$
(B.138)

considering $p^{0}$ as an independent variable

$\displaystyle\frac{\partial}{\partial\vec{p}}\rightarrow\frac{\partial
p^{0}}{\partial\vec{p}}\frac{\partial}{\partial
p^{0}}+\frac{\partial}{\partial\vec{p}}=\frac{\vec{p}}{p^{0}}\frac{\partial}{\partial
p^{0}}+\frac{\partial}{\partial\vec{p}}$ (B.139)

Eqn. (B.138) takes the form

$\displaystyle\vec{v}.\frac{\partial
f}{\partial\vec{r}}+\vec{F}.\frac{\partial
f}{\partial\vec{p}}=-\nu_{i}\left(f_{i}-n_{{}_{i}}n_{\mathrm{e}q,i}^{-1}f_{\mathrm{e}q,i}\right)$
(B.140)

##  Appendix C Seebeck coefficient in relaxation time approximation

The Seebeck coefficient has been calculated earlier in the kinetic theory
framework using the relaxation time approximation in the absence of the
magnetic field as [50]

$\displaystyle S=\frac{1}{2Tq_{i}}\left(\frac{L_{1}}{L_{2}}\right)$ (B.141)
$\displaystyle L_{1}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p^{2}}{3\omega^{2}}\bigg{\\{}(\omega-\mu)f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)+(\omega+\mu)\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\bigg{\\}}$
(B.142) $\displaystyle L_{2}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p^{2}}{3\omega^{2}}\bigg{\\{}f_{\mathrm{e}q}\left(1-f_{\mathrm{e}q}\right)+\bar{f}_{\mathrm{e}q}\left(1-\bar{f}_{\mathrm{e}q}\right)\bigg{\\}}$
(B.143)

and in the presence of strong magnetic field as [50]

$\displaystyle S_{B}=\frac{1}{2qT}\left(\frac{H_{1}}{H_{2}}\right)$ (B.144)

where

$\displaystyle H_{1}$ $\displaystyle=$
$\displaystyle\int{dp_{3}}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}\bigg{\\{}(\omega-\mu)f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)+(\omega+\mu)\bar{f}_{\rm
eq}^{B}\left(1-\bar{f}_{\rm eq}^{B}\right)\bigg{\\}}$ (B.145) $\displaystyle
H_{2}$ $\displaystyle=$
$\displaystyle\int{dp_{3}}\frac{p_{3}^{2}}{\omega^{2}}\tau^{B}\bigg{\\{}f_{\rm
eq}^{B}\left(1-f_{\rm eq}^{B}\right)+\bar{f}_{\rm eq}^{B}\left(1-\bar{f}_{\rm
eq}^{B}\right)\bigg{\\}}$ (B.146)

In other work [51], the thermoelectric response of the hot QCD medium has been
studied in weak magnetic field, where Seebeck and Nernst coefficients are
found to be

$\displaystyle S$ $\displaystyle=$
$\displaystyle-\frac{(C_{1}C_{3}+C_{2}C_{4})}{C_{1}^{2}+C_{2}^{2}},$ (B.147)
$\displaystyle N|B|$ $\displaystyle=$
$\displaystyle\frac{(C_{2}C_{3}-C_{1}C_{4})}{C_{1}^{2}+C_{2}^{2}},$ (B.148)

provided $C_{1}=qI_{1}$, $C_{2}=qI_{2}$, $C_{3}=\beta I_{3}$ and $C_{4}=\beta
I_{4}$. The integrals $I_{1}$, $I_{2}$,$I_{3}$ and $I_{4}$ are given by

$\displaystyle I_{1}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p^{2}}{3\epsilon^{2}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}\bigg{\\{}f_{0}(1-f_{0})+\bar{f}_{0}(1-\bar{f}_{0})\bigg{\\}}$
(B.149) $\displaystyle I_{2}$ $\displaystyle=$
$\displaystyle\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p^{2}}{3\epsilon^{2}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}\bigg{\\{}f_{0}(1-f_{0})-\bar{f}_{0}(1-\bar{f}_{0})\bigg{\\}}$
(B.150) $\displaystyle I_{3}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p^{2}}{3\epsilon^{2}}\frac{\tau}{(1+\omega_{c}^{2}\tau^{2})}\bigg{\\{}(\epsilon-\mu)f_{0}(1-f_{0})-(\epsilon+\mu)\bar{f}_{0}(1-\bar{f}_{0})\bigg{\\}}$
(B.151) $\displaystyle I_{4}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}p}{(2\pi)^{3}}~{}\frac{p^{2}}{3\epsilon^{2}}\frac{\omega_{c}\tau^{2}}{(1+\omega_{c}^{2}\tau^{2})}\bigg{\\{}(\epsilon-\mu)f_{0}(1-f_{0})+(\epsilon+\mu)\bar{f}_{0}(1-\bar{f}_{0})\bigg{\\}}$
(B.152)

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