Magneto-vortical effect in strongly coupled plasma

Abstract

Based on a holographic model incorporating both the chiral anomaly and the gravitational anomaly, we study the effect of magneto-vortical coupling on the transport properties of a strongly coupled plasma. The focus of present work is on the generation of a vector charge density and an axial current, as response to vorticity in a magnetized plasma. The transport coefficients parameterizing the vector charge density and axial current are calculated both analytically (in the weak magnetic field limit) and also numerically (for general values of the magnetic field). We find the generation of vector charge receives both non-anomalous and anomalous contributions, with the non-anomalous contribution dominating in the limit of a strong magnetic field and the anomalous contribution sensitive to both chiral anomaly and gravitational anomaly. On the contrary, we find the axial current is induced entirely due to the gravitational anomaly, thus we interpret the axial current generation as chiral vortical effect. The corresponding chiral vortical conductivity is found to be suppressed by the magnetic field. By the Onsager relation, these transport coefficients are responsible for the generation of a thermal current due to a transverse electric field or a transverse axial magnetic field, which we call the thermal Hall effect and the thermal axial magnetic effect, respectively.

1 Introduction

The effect of magnetic field and vorticity on QCD matter has attracted much attention over the past few years. At very high temperature, when quarks become asymptotically free, a charged-neutral QCD matter can be either magnetized in the magnetic field or polarized in the vorticity field. Close to the chiral phase transition, when the interaction among quarks becomes strong, more interesting phenomena such as inverse magnetic catalysis [1,2,3,4] and vector meson condensation [5, 6] can emerge. Similarly, the vorticity field may suppress the chiral condensation [7, 8].

When the QCD matter carries net vector charge or axial charge densities, the chiral anomaly and gravitational anomaly can induce a variety of anomalous transport phenomena such as the chiral magnetic effect (CME) [9,10,11], the chiral vortical effect (CVE) [12,13,14] and the chiral separation effect (CSE) [15, 16].

Recently, the interplay of a strong magnetic field and a vorticity was found to lead to new transport phenomena such as dynamical generation of a vector charge [17]; see also [18,19,20,21]. Under the lowest Landau level (LLL) approximation, Hattori and Yin found the generation of a vector charge from spin–vorticity coupling as [17]¹

$$\begin{aligned} J^t= q_f \frac{C_A}{2} (\vec {B}\cdot \vec {{\Omega }}), \end{aligned}$$

(1)

with $C_A=\frac{1}{2\pi ^2}$ the chiral anomaly coefficient. In fact, such a contribution should be viewed as a large B, free limit of a QED plasma. More generally, one would expect from the viewpoint of polarizable matter [22] that

$$\begin{aligned} J^t= \xi (B,T) (\vec {B}\cdot \vec {{\Omega }}). \end{aligned}$$

(2)

Moreover, if we could associate an effective chemical potential for the generated vector charge, this can further give rise to the generation of an axial current by the chiral anomaly and gravitational anomaly. Note that the vector charge susceptibility is ${\chi }=C_A|q_f|B$ in the LLL approximation, and thus (1) corresponds to the effective chemical potential ${\mu }_{\mathrm{eff}}=\frac{\mathrm{sgn}(q_f)}{2}{\vec {\Omega }}\cdot {{\hat{B}}}$ with ${\hat{B}}={\vec {B}}/|B|$. The vector charge imbalance would result in an axial current through the chiral separation effect [15, 16]

$$\begin{aligned} \vec {J}_5= |q_f| \frac{C_A}{2} (\vec {B}\cdot \vec {{\Omega }}) {\hat{B}}. \end{aligned}$$

(3)

Again, this is a large B, free limit of a QED plasma. More generally, one would expect an extra contribution from the gravitational anomaly, which always induces a temperature-dependent contribution to the axial current, even in the absence of the chiral imbalance [14, 23, 24]. Therefore, we expect the more general axial current

$$\begin{aligned} \vec {J}_5 = \sigma (B,T) \vec {{\Omega }}. \end{aligned}$$

(4)

It is worth noting that the physical picture behind (1) and (3) is the spectral flow: a shift in the background vector gauge field leads to an opposite energy shift for right- and left-handed fermions, generating a net axial charge. In order for the spectral flow picture to generate vector charge, we would need an axial gauge field, whose couplings to right- and left-handed fermions differ in sign, thus leading to the same energy shift for them. In the analysis of Hattori and Yin [17], the role of the axial gauge field is played by the vorticity. Indeed, in free theory, we have ${\vec {S}}={\vec {J}}_5$ so that we can identify ${\vec {A}}_5$ with ${\vec {{\Omega }}}$ by comparing the coupling ${\vec {\Omega }}\cdot {\vec {S}}$ with ${\vec {J}}_5\cdot {\vec {A}}_5$. However, in an interacting theory, the “equivalence” of ${\vec {A}}_5$ with ${\vec {\Omega }}$ is far from obvious. First of all, even in a free theory the presence of an axial gauge field as a source poses an ambiguity in the definition of currents: consistent current and covariant current could differ by terms proportional to the axial gauge field [25]. A similar ambiguity does not exist in the case with the vorticity as a source. Secondly, in an interacting theory, the vorticity couples to the angular momentum as a whole. The separation of the spin from the total angular momentum is often ambiguous. Therefore, it is desirable to go away from the free theory limit to test the robustness of the mechanism. In this paper, we go to the opposite limit, where the theory is strongly coupled. Specifically, we will study the response of a strongly coupled magnetized plasma to the vorticity field by a holographic model.

The rest of the paper is organized as follows: In Sect. 2, we present the setup of the holographic model. In Sect. 3, we turn on a metric perturbation as a proxy for the vorticity in the magnetized plasma. We will study the response of the vector charge density and axial current to the vorticity. In Sect. 4, we will present both analytic results in the small B regime and numerical results for general B. In Sect. 5, we use the Onsager relation to obtain the thermal Hall effect and the thermal axial magnetic effect. We conclude and discuss implications of our results in Sect. 6. Details of the computations are collected in Appendices A, B and C.

2 Holographic setup: magnetic brane in $AdS_5$

2.1 Gravity action and dictionary

We extend the holographic model initially considered in [26, 27] by including both vector and axial gauge fields. The full action is

$$\begin{aligned} \begin{aligned} S&=\frac{1}{2\kappa ^2}\int d^5x \sqrt{-g}\left\{ R[g]+12-\frac{1}{4}(F^V)^2\right. \\&\quad \left. - \frac{1}{4} (F^a)^2+ \epsilon ^{MNPQR} A_M \times \left[ \frac{1}{3}\alpha (F^a)_{NP} (F^a)_{QR}\right. \right. \\&\quad \left. \left. + \alpha (F^V)_{NP}(F^V)_{QR}+ \lambda R^Y_{~XNP} R^X_{~~YQR} \right] \right\} \\&\quad +\frac{1}{\kappa ^2}\int d^4x \sqrt{-\gamma } K[\gamma ] + \underline{S_{\mathrm{{CSK}}}}+ S_{\mathrm{c.t.}}, \end{aligned} \end{aligned}$$

(5)

where $F^V=dV$ and $F^a=dA$. The last line of (5) corresponds to boundary terms defined on the hypersurface $\Sigma$ of constant r. The notation $\gamma$ denotes the determinant of the induced metric $\gamma _{\mu \nu }$ on $\Sigma$:

$$\begin{aligned} \mathrm{d}s^2|_{\Sigma }=g_{MN}\mathrm{d}x^M \mathrm{d}x^N|_{\Sigma }=\gamma _{\mu \nu } \mathrm{d}x^\mu \mathrm{d}x^\nu . \end{aligned}$$

(6)

We also need the out-pointing unit normal vector of the surface $\Sigma$:

$$\begin{aligned} n_M= \frac{\partial _M r}{\sqrt{g^{AB} \partial _A r \partial _B r}}. \end{aligned}$$

(7)

Moreover, $K=\gamma ^{\mu \nu } K_{\mu \nu }$ whereas $K_{\mu \nu }$ is the extrinsic curvature tensor

$$\begin{aligned} K_{\mu \nu }= & {} \frac{1}{2} {\mathfrak {L}}_n \gamma _{\mu \nu }=\frac{1}{2}\left( n^M \partial _M \gamma _{\mu \nu }\right. \nonumber \\&\left. +\gamma _{\mu N} \partial _\nu n^N+ \gamma _{\nu N} \partial _\mu n^N\right) . \end{aligned}$$

(8)

The Levi-Civita tensor is $\epsilon ^{MNPQR}= \epsilon (MNPQR)/\sqrt{-g}$ whereas $\epsilon (MNPQR)$ is the Levi–Civita symbol under the convention $\epsilon (rtxyz)=+1$. The purely gauge Chern–Simons action ($\alpha$-terms) mimics the chiral anomaly, while the mixed gauge-gravitational Chern–Simons term ($\lambda$-term) is to model the gravitational anomaly of the boundary field theory.

As explained in [27], in order to get a correct form of the gravitational anomaly (i.e. guarantee the gauge variation of the bulk action to be a total derivative), one needs to add the term

$$\begin{aligned} S_{\mathrm{{CSK}}}=-\frac{4}{\kappa ^2}\lambda \int d^4x \sqrt{-\gamma } n_M \epsilon ^{MNPQR} A_N K_{PL} {\tilde{\nabla }}_Q K_R^L,\nonumber \\ \end{aligned}$$

(9)

where ${\tilde{\nabla }}$ is compatible with the induced metric $\gamma _{AB}$. The counter-term action is

$$\begin{aligned} S_{\mathrm{c.t.}}=-\frac{1}{2\kappa ^2} \int d^4x \sqrt{-\gamma }\left( 6+ \frac{1}{2} R[\gamma ] -{\mathcal {C}}_t\right) , \end{aligned}$$

(10)

where ${\mathcal {C}}_t$ cancels the logarithmic divergences [28, 29]

$$\begin{aligned} \begin{aligned} {\mathcal {C}}_t&= \frac{1}{4} \log r \left[ \left( F^V\right) _{\mu \nu } \left( F^V\right) ^{\mu \nu } + \left( F^a\right) _{\mu \nu } \left( F^a\right) ^{\mu \nu } \right] \\&\quad + \log \frac{1}{r^2} \left( \frac{1}{8} R^{\mu \nu }[\gamma ]R_{\mu \nu }[\gamma ] -\frac{1}{24} R^2[\gamma ]\right) . \end{aligned} \end{aligned}$$

(11)

Note that ${\mathcal {C}}_t$ non-vanishes only when non-trivial sources (either external gauge fields or non-flat boundary metric) are turned on for the boundary theory. In addition, in ${\mathcal {C}}_t$ we employ the minimal subtraction scheme so that it will not generate finite contribution to the boundary currents and stress tensor.

According to the holographic dictionary, the expectation values of the stress tensor and currents of the boundary theory are defined as

$$\begin{aligned} T_{\mu \nu }\equiv & {} \lim _{r\rightarrow \infty } \frac{-2r^2}{\sqrt{-\gamma }} \frac{\delta S}{\delta \gamma ^{\mu \nu }},\quad J^\mu \equiv \lim _{r\rightarrow \infty } \frac{\delta S}{\delta V_\mu } ,\quad \nonumber \\&J^\mu _5\equiv \lim _{r\rightarrow \infty } \frac{\delta S}{\delta A_\mu }. \end{aligned}$$

(12)

Explicitly, the vector current is (from now on, we set $2{\kappa }^2=1$ for convenience)

$$\begin{aligned} J^\mu= & {} \lim _{r\rightarrow \infty } \sqrt{-\gamma } \left\{ n_M\left( F^V\right) ^{\mu M}+ 4\alpha n_M \epsilon ^{M\mu N QR} A_N \right. \nonumber \\&\left. \left( F^V\right) _{QR} -{\tilde{\nabla }}_\nu \left( F^V\right) ^{\nu \mu } \log r \right\} . \end{aligned}$$

(13)

However, the axial current and stress tensor are somewhat subtle and complicated:

$$\begin{aligned} J_5^\mu= & {} \lim _{r\rightarrow \infty } \sqrt{-\gamma } \left\{ n_M\left( F^a\right) ^{\mu M}+ \frac{4}{3}\alpha n_M \epsilon ^{M\mu N QR} A_N \right. \nonumber \\&\left. \left( F^a\right) _{QR} -{\tilde{\nabla }}_\nu \left( F^a\right) ^{\nu \mu } \log r +J_{\mathrm{CSK}}^\mu \right\} ,\end{aligned}$$

(14)

$$\begin{aligned} T_{\mu \nu }= & {} -2\lim _{r\rightarrow \infty }r^2\left( K_{\mu \nu }-K\gamma _{\mu \nu } +3\gamma _{\mu \nu } \right. \nonumber \\&\left. - \frac{1}{2} {\mathcal {G}}_{\mu \nu }[\gamma ] - T_{\mu \nu }^{\mathrm{Gra}}\right) + T_{\mu \nu }^{{\mathcal {C}}}, \end{aligned}$$

(15)

where $T_{\mu \nu }^{{\mathcal {C}}}$ arises from the functional derivative of ${\mathcal {C}}_t$, $J_{\mathrm{CSK}}^\mu$ is due to the added action $S_{\mathrm{CSK}}$, and $T_{\mu \nu }^{\mathrm{Gra}}$ comes from the gravitational Chern–Simons term. The expressions for all of them are [27, 30]

$$\begin{aligned} J_{\mathrm{CSK}}^\mu&=-8\lambda n_M \epsilon ^{M\mu PQR}K_{PL} {{\tilde{\nabla }}}_Q K^L_R, \nonumber \\ T_{\mu \nu }^{{\mathcal {C}}}&= T_{\mu \nu }^{{\mathcal {C}}_1}+ \lim _{r\rightarrow \infty } \frac{1}{4} r^6 \log r \left[ \gamma _{\mu \nu } (F^V)_{\alpha \beta } (F^V)^{\alpha \beta } \right. \nonumber \\&\quad \left. - 4 (F^V)_{\mu \alpha } (F^V)_\nu ^{~\alpha }\right] + (V\rightarrow a), \nonumber \\ T^{\mu \nu }_{\mathrm{Gra}}&= 4\lambda \epsilon ^{(\mu \alpha \beta \rho }\left[ \frac{1}{2} (F^a)_{\alpha \beta }R^{\nu )}_{~\rho }[\gamma ] +{{\tilde{\nabla }}}_\delta \left( A_\alpha R^{\delta \nu )}_{~~\beta \rho }[\gamma ] \right) \right] , \end{aligned}$$

(16)

where $T_{\mu \nu }^{{\mathcal {C}}_1}$ vanishes for a flat boundary. $T^{\mu \nu }_{\mathrm{Gra}}$ was first derived in [30] based on the ADM decomposition approach. Above, we stick to the consistent current formalism. Indeed, in the absence of a background for the axial gauge field, there will be no difference between the consistent current and covariant current [25]. The authors of [26] presented a thorough analysis for the holographic renormalization of the model, but did not get the term $S_{\mathrm{CSK}}$. Additionally, the authors of [26] addressed the fact that the gravitational Chern–Simons term will make a contribution to the boundary stress tensor. See also [30, 31] for more recently updated formulas for stress tensor and axial current of the boundary theory. The holographic model does correctly describe the chiral/gravitational anomalies for the boundary field theory [27]:

$$\begin{aligned}&{\hat{\nabla }}_\mu J^\mu =0,\quad {\hat{\nabla }}_\mu J_5^\mu = 8\alpha {\hat{\epsilon }}^{\alpha \beta \rho \delta } {{\hat{F}}}_{\alpha \beta } {{\hat{F}}}_{\rho \delta }\nonumber \\&\qquad + \lambda {\hat{\epsilon }}^{\alpha \beta \rho \delta } {{\hat{R}}}^\tau _{~\kappa \alpha \beta } {{\hat{R}}}^\kappa _{~ \tau \rho \delta }, \end{aligned}$$

(17)

where a hat means that the corresponding quantity is defined on the boundary.

Under the variation

$$\begin{aligned}&g_{MN}\rightarrow g_{MN}+\delta g_{MN},\qquad V_M\rightarrow V_M+\delta V_M,\nonumber \\&A_M\rightarrow A_M+ \delta A_M, \end{aligned}$$

(18)

one obtains the Einstein equation

$$\begin{aligned} 0=E_{MN}\equiv R_{MN}-\frac{1}{2}R g_{MN}-6 g_{MN}- T_{MN}^{\mathrm{bulk}}, \end{aligned}$$

(19)

and anomalous Maxwell equations

$$\begin{aligned} 0= & {} EV^M \equiv \nabla _N\left( F^V\right) ^{NM}\nonumber \\&+ 2\alpha \epsilon ^{MNPQR} \left( F^a \right) _{NP} \left( F^V\right) _{QR}, \end{aligned}$$

(20)

$$\begin{aligned} 0= & {} EA^{M}\equiv \nabla _N \left( F^a\right) ^{NM} \nonumber \\&+ \alpha \epsilon ^{MNPQR} \left[ \left( F^V\right) _{NP} \left( F^V\right) _{QR} +\left( F^a\right) _{NP} \left( F^a\right) _{QR}\right] \nonumber \\&+ \lambda \epsilon ^{MNPQR} R^Y_{~XNP} R^X_{~~YQR}. \end{aligned}$$

(21)

The bulk stress tensor $T_{MN}^{\mathrm{bulk}}$ could be split into two parts,

$$\begin{aligned} T_{MN}^{\mathrm{bulk}}= T_{MN}^{\mathrm{Maxwell}}- \nabla _X \Theta ^X_{MN}, \end{aligned}$$

(22)

where

$$\begin{aligned} T_{MN}^{\mathrm{Maxwell}}= & {} \frac{1}{2} \left( F^V\right) _{RM} \left( F^V\right) ^R_{~N} -\frac{1}{8} g_{MN} \left( F^V\right) ^2 \nonumber \\&+ \frac{1}{2} \left( F^a\right) _{RM} \left( F^a\right) ^R_{~N} -\frac{1}{8} g_{MN} \left( F^a\right) ^2 \end{aligned}$$

(23)

$$\begin{aligned} \Theta ^X_{MN}= & {} \lambda \epsilon ^{QRSTU}\left( g_{QM} R^X_{~NRS}+ g_{QN} R^X_{~MRS}\right) \left( F^a\right) _{TU}.\nonumber \\ \end{aligned}$$

(24)

Alternatively, the Einstein equation could be rewritten as

$$\begin{aligned} 0=E_{MN}=R_{MN}+4g_{MN}-{\tilde{T}}^{\mathrm{bulk}}_{MN} \end{aligned}$$

(25)

where

$$\begin{aligned} {\tilde{T}}^{\mathrm{bulk}}_{MN}&=\frac{1}{2}(F^V)_{RM}(F^V)^R_{~N} -\frac{1}{12}g_{MN}(F^V)^2 \nonumber \\&+ \frac{1}{2}(F^a)_{RM} (F^a)^R_{~N}-\frac{1}{12}g_{MN}(F^a)^2 \nonumber \\&- \nabla _X \Theta ^X_{MN}+ \frac{1}{3}g_{MN}g^{AB}\nabla _X \Theta ^X_{AB}. \end{aligned}$$

(26)

2.2 Neutral magnetic brane background

To proceed, we consider the background solution of the holographic model (5). For simplicity, we focus on a neutral magnetized plasma. To this end, we turn on a constant magnetic field along the z-direction,

$$\begin{aligned} V=Bx \mathrm{d}y \Rightarrow \vec {B}=B {\hat{z}}, \end{aligned}$$

(27)

which obviously breaks the SO(3) rotational symmetry to $SO(2)_\perp$ on the xy-plane. As a result, the background metric takes the form

$$\begin{aligned} \mathrm{d}s^2= & {} 2\mathrm{d}r\mathrm{d}t-f(r)\mathrm{d}t^2+ e^{2W_T(r)}(\mathrm{d}x^2\nonumber \\&+\mathrm{d}y^2)+ e^{2 W_L(r)} \mathrm{d}z^2. \end{aligned}$$

(28)

Note that in writing down (28), the ingoing Eddington–Finkelstein coordinate has been employed in order to avoid a coordinate singularity. The background metric (28) has an event horizon at $r=r_h$ so that

$$\begin{aligned} f(r\simeq r_h)=0+f^\prime (r_h)(r-r_h)+\cdots , \end{aligned}$$

(29)

while $W_T,W_L$ are regular at $r=r_h$. The Hawking temperature, identified as the temperature of the dual gauge theory, is

$$\begin{aligned} T= \frac{\partial _r(f(r))}{4\pi }\Bigg |_{r=r_h}. \end{aligned}$$

(30)

Generically, both $W_T(r)$ and $W_L(r)$ will depend on $r_h$ non-trivially.

It is a simple exercise to check that, given the above ansatz (27) and (28), both the gauge and the gravitational Chern–Simons terms do not affect the bulk equations of motion. Therefore, the background geometry is simply the “magnetic brane” solution initially studied in [32].

The ordinary differential equations (ODEs) for the metric functions in (28) are

$$\begin{aligned} E_{rr}&=0:\quad 0=W_L^{\prime \;2}+2W_T^{\prime \;2}+W_L^{\prime \prime }+2W_T^{\prime \prime }, \end{aligned}$$

(31)

$$\begin{aligned} E_{rt}&=0=E_{tt}:\quad 24+B^2e^{-4W_T(r)}\nonumber \\&=3f^\prime (W_L^\prime +2W_T^\prime )+ 3f^{\prime \prime }, \end{aligned}$$

(32)

$$\begin{aligned} E_{xx}&=0=E_{yy}:\quad 12-B^2e^{-4W_T(r)}\nonumber \\&=3f^\prime W_T^\prime +3f(W_L^\prime W_T^\prime +2W_T^{\prime \;2}+ W_T^{\prime \prime }), \end{aligned}$$

(33)

$$\begin{aligned} E_{zz}&=0:\quad 24+B^2e^{-4W_T(r)}\nonumber \\&=6f^\prime W_L^\prime +6f(W_L^{\prime \;2}+ 2W_L^\prime W_T^{\prime } + W_L^{\prime \prime }), \end{aligned}$$

(34)

where the prime denotes a derivative with respect to r. Equations (33) and (34) look different from those of [32]. However, suitable combinations of the above equations give rise to the results of [32]:

$$\begin{aligned}&2\times (33)-(34) \Rightarrow 2f(W_T^{\prime \prime }-W_L^{\prime \prime }) + 2\left[ f^\prime +f(W_L^\prime +2 W_T^\prime )\right] \nonumber \\&\quad (W_T^\prime - W_L^{\prime })=-B^2 e^{-4W_T(r)}, \nonumber \\&4\times (33)+(34) \Rightarrow 2f^\prime (2W_T^\prime + W_L^\prime ) +4f W_T^\prime (W_T^\prime + 2 W_L^\prime )\nonumber \\&\quad =24-B^2 e^{-4W_T(r)}. \end{aligned}$$

(35)

Obviously, not all the equations in (31)–(34) are independent: we will take (31) as the constraint and solve the remaining ones to determine the metric functions $f(r),W_T(r),W_L(r)$.

In order to fully determine $f(r),W_T(r),W_L(r)$, we have to impose two boundary conditions for each of them. For f(r), we impose

$$\begin{aligned} f(r=r_h)=0;\quad \underline{f(r\rightarrow \infty ) \rightarrow r^2}. \end{aligned}$$

(36)

However, it is found that the second boundary condition (underlined above) is automatically satisfied by the bulk EOMs. This requires one to impose another condition for f(r), which is explained in Appendix 1. For $W_T(r)$ and $W_L(r)$, the boundary conditions are

$$\begin{aligned}&W_T(r),\,\, W_L(r) \rightarrow \log r, \quad \mathrm{as} \quad r\rightarrow \infty , \end{aligned}$$

(37)

$$\begin{aligned}&12-B^2 e^{-4 W_T(r)}= 3 f^\prime W_T^\prime , \quad \mathrm{at}\quad r=r_h, \end{aligned}$$

(38)

$$\begin{aligned}&24+ B^2 e^{-4 W_T(r)}= 6 f^\prime W_L^\prime , \quad \mathrm{at}\quad r=r_h, \end{aligned}$$

(39)

where the last two equations are read off from (33) and (34) by requiring regularity of $W_T(r), W_L(r)$ at the horizon $r=r_h$.

We solve the bulk EOMs (31)–(34) analytically when the magnetic field is weak (i.e., $B/T^2\ll 1$) and numerically when B is general. The calculational details as well as the main results are deferred to Appendix A.

3 Fluctuation in the bulk theory: general consideration

3.1 Bulk perturbations

In this section, we study the linear response of the magnetized plasma to a fluid vorticity. A weak fluid vorticity $\vec {\Omega }$ would be mimicked by a gravito-magnetic field [27, 31]. More precisely, one perturbs the boundary Minkowski spacetime (where the fluid flows) as

$$\begin{aligned} \mathrm{d}s^2_{\mathrm{M}}&= -\mathrm{d}t^2 +\mathrm{d}{\vec {x}}^2 + 2 h_{ti}(t,\vec x)\mathrm{d}t \mathrm{d}x^i, \quad \nonumber \\&\qquad \mathrm{with}\,\, h_{ti}(t,\vec x)=u_i(t,\vec x). \end{aligned}$$

(40)

Then the vorticity is generated at linear order in $h_{ti}$ as

$$\begin{aligned} {\Omega }^i=\frac{1}{2}{\epsilon }^{ijk}\nabla _ju_k=\frac{1}{2}{\epsilon }^{ijk}{\partial }_jh_{tk}, \end{aligned}$$

(41)

with the unperturbed fluid velocity $u^{\mu }=(1,0,0,0)$. Thus, the curl of $h_{ti}$ could be thought of as a fluid vorticity. We take

$$\begin{aligned} h_{ti}(\vec x)=e^{iqx} h_{ty}(q) \delta _{yi}, \end{aligned}$$

(42)

which gives rise to a stationary vorticity along the z-direction, i.e. parallel to the magnetic field. Then we can obtain the Kubo formulas for the transport coefficients defined in (2) and (4) as

$$\begin{aligned} \xi = \frac{2}{B}\lim _{q\rightarrow 0}\frac{\langle J^t T^{ty}\rangle }{iq}, \quad \sigma = 2 \lim _{q\rightarrow 0} \frac{\langle J^z_5 T^{ty}\rangle }{iq}. \end{aligned}$$

(43)

These will be used in holographic calculations.

To turn on a gravito-magnetic field in the bulk, it is convenient to use the Poincaré coordinate system so that the bulk metric takes a diagonal form,

$$\begin{aligned} \mathrm{d}s^2= & {} -f(r)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{f(r)}+e^{2W_T(r)}\nonumber \\&\times \left( \mathrm{d}x^2+\mathrm{d}y^2\right) + e^{2W_L(r)} \mathrm{d}z^2,\nonumber \\ \end{aligned}$$

(44)

where we still denote the time of the bulk theory by t. On top of the background (44) and (27), it is consistent to turn on the following fluctuation modes:

$$\begin{aligned} \delta (\mathrm{d}s^2)&= 2 e^{W_T(r)}\left[ \delta g_{ty}(r,t,x)\mathrm{d}t\mathrm{d}y + \delta g_{xy}(r,t,x) \mathrm{d}x\mathrm{d}y \right] , \nonumber \\ \delta V&= \delta V_t(r,t,x) \mathrm{d}t + \delta V_x(r,t,x)\mathrm{d}x , \nonumber \\ \delta A&= \delta A_z(r,t,x) \mathrm{d}z, \end{aligned}$$

(45)

while setting all the rest corrections to zero. Here, we have assumed (t, x)-dependent fluctuations for a reason to be discussed in the next subsection and we consider a plane wave ansatz:

$$\begin{aligned}&\delta g_{ty}(r,t,x)\sim e^{-i\omega t+iq x} \delta g_{ty}(r),\nonumber \\&\delta g_{xy}(r,t,x) \sim e^{-i\omega t+iq x} \delta g_{xy}(r), \nonumber \\&\delta V_t(r,t,x) \sim e^{-i\omega t+i qx} \delta V_t(r), \nonumber \\&\delta V_x(r,t,x) \sim e^{-i\omega t+i qx} \delta V_x(r), \nonumber \\&\delta A_z(r,t,x) \sim e^{-i\omega t+i qx} \delta A_z(r). \end{aligned}$$

(46)

In what follows we record the bulk equations of motion for the fluctuation modes. First, we consider the constraint equations. The constraint $E_{ry}=0$ is

$$\begin{aligned} 0&= \lambda \omega q e^{-W_L} \delta A_z\left( -\frac{2B^2 e^{-4W_T}}{f} + \frac{2f^\prime W_L^\prime }{f} \right. \nonumber \\&\quad \left. + \frac{4f^\prime W_T^\prime }{f}-4W_L^\prime W_T^\prime - 8 W_T^{\prime 2} \right) \nonumber \\&\quad +\frac{1}{2}iq \partial _r\delta g_{xy}+ \frac{i\omega e^{2W_T}}{2f} \partial _r\delta g_{ty} + \frac{1}{2} Be^{-2W_T} \partial _r \delta V_x. \end{aligned}$$

(47)

The constraint $EV^r=0$ gives

$$\begin{aligned} 0=8\omega \alpha B \delta A_z + q e^{W_L} f \partial _r \delta V_x + \omega e^{W_L +2 W_T} \partial _r \delta V_t. \end{aligned}$$

(48)

Next, we turn to the dynamical components of the bulk EOMs. The Einstein equation $E_{ty}=0$ reads

$$\begin{aligned} 0&= \partial _r\left( e^{W_L+4W_T} \partial _r \delta g_{ty}\right) - \frac{B^2 e^{W_L}}{f(r)} \delta g_{ty}- \frac{q^2 e^{W_L+2W_T}}{f(r)} \delta g_{ty} \nonumber \\&\quad -\frac{\omega q e^{W_L+2W_T}}{f} \delta g_{xy} + \frac{i\omega B e^{W_L}}{f} \delta V_x\nonumber \\&\quad + \frac{i q B e^{W_L}}{f(r)} \delta V_t + iq \lambda e^{2W_T} \partial _r \delta A_z \nonumber \\&\quad \times \left( 4B^2 e^{-4W_T} -4 f^\prime W_L^\prime -8 f^\prime W_T^\prime \right. \nonumber \\&\quad \left. + 8 f W_T^\prime W_L^\prime +16 f W_T^{\prime 2}\right) + iq \lambda e^{2W_T} \delta A_z \nonumber \\&\quad \times \left( 2B^2 e^{-4W_T} f^{-1} f^\prime -48 f^{-1}f^\prime - 4B^2 e^{-4W_T} W_L^\prime \right. \nonumber \\&\quad \left. + 4f^{-1} f^{\prime 2}W_L^\prime + 8f^\prime W_L^{\prime 2} -20 B^2 e^{-4W_T} W_T^\prime \right. \nonumber \\&\quad \left. +96 W_T^\prime + 8 f^{-1} f^{\prime 2} W_T^\prime +16 f^\prime W_T^\prime W_L^\prime \right. \nonumber \\&\quad \left. -16 f W_T^\prime W_L^{\prime 2}- 48 f W_L^\prime W_T^{\prime 2} -32 f W_T^{\prime 3} \right) . \end{aligned}$$

(49)

The Einstein equation $E_{xy}=0$ is

$$\begin{aligned} 0&= \partial _r\left( e^{W_L+2W_T} f\partial _r \delta g_{xy}\right) + \frac{\omega ^2 e^{W_L+ 2W_T}}{f} \delta g_{xy} \nonumber \\&\quad + \frac{\omega q e^{W_L+2W_T}}{f} \delta g_{ty}. \end{aligned}$$

(50)

The Maxwell equation $EV^t=0$ is

$$\begin{aligned} 0&= \partial _r\left( e^{W_L+2W_T}\partial _r \delta V_t\right) + 8\alpha B \partial _r \delta A_z - \frac{\omega q e^{W_L}}{f} \delta V_x\nonumber \\&\quad - \frac{q^2 e^{W_L}}{f}\delta V_t -\frac{iq B e^{W_L}}{f}\delta g_{ty}. \end{aligned}$$

(51)

The Maxwell equation $EV^x=0$ is

$$\begin{aligned} 0&= \partial _r\left( e^{W_L}f \partial _r \delta V_x\right) + \frac{\omega ^2 e^{W_L}}{f}\delta V_x \nonumber \\&\quad + \frac{\omega q e^{W_L}}{f} \delta V_t + \frac{i\omega B e^{W_L}}{f} \delta g_{ty}. \end{aligned}$$

(52)

Finally, $EA^z=0$ yields

$$\begin{aligned} 0&= \partial _r\left( e^{2W_T-W_L}f \partial _r \delta A_z\right) + \frac{\omega ^2 e^{2W_T-W_L}}{f} \delta A_z \nonumber \\&\quad - q^2 e^{-W_L}\delta A_z+ iq \lambda e^{2W_T} \left( -4 B^2 e^{-4W_T} + 4 f^\prime W_L^\prime \right. \nonumber \\&\quad \left. + 8 f^\prime W_T^\prime - 8f W_T^\prime W_L^\prime -16 f W_T^{\prime 2}\right) \partial _r \delta g_{ty}+ 8\alpha B \partial _r \delta V_t. \end{aligned}$$

(53)

3.2 Adiabatic limit and boundary conditions

From the Kubo formulas (43), it seems as if we could set ${\omega }=0$ from the beginning. However, the boundary condition at the horizon cannot be uniquely determined in this case. This ambiguity is related to the ambiguity in the Kubo formula itself. It can be evaluated in any equilibrium state, be it charged or neutral. For our purpose, it should be evaluated in the unperturbed neutral plasma state. The boundary condition to use should correspond to the neutral state. In practice, we specify the state as follows: the state is realized by turning on the vorticity field adiabatically to the original neutral magnetized plasma.

We will seek solutions to (47) through (53) in the adiabatic limit ${\omega }\rightarrow 0$. To this end, we expand the bulk perturbations in powers of ${\omega }$:

$$\begin{aligned} X=X^{(0)}+{\omega }X^{(1)}+\cdots , \end{aligned}$$

(54)

with $X={\delta }g_{xy},\;{\delta }V_x,\;{\delta }g_{ty},\;{\delta }V_t$ and ${\delta }A_z$. In fact, we only need the leading order solution $X^{(0)}$, for which we suppress the superscript (0). The fields decouple into two sets $\{{\delta }g_{xy},\;{\delta }V_x \}$ and $\{{\delta }g_{ty},\;{\delta }V_t,\;{\delta }A_z\}$. The set $\{{\delta }g_{ty},\;{\delta }V_t,\;{\delta }A_z\}$ satisfies the following equations:

$$\begin{aligned}&\partial _r\left( e^{W_L+4W_T} \partial _r \delta g_{ty}\right) - \frac{B^2 e^{W_L}}{f(r)} \delta g_{ty}- \frac{q^2 e^{W_L+2W_T}}{f(r)} \delta g_{ty}\nonumber \\&\quad + \frac{i q B e^{W_L}}{f(r)} \delta V_t + iq \lambda e^{2W_T} \partial _r \delta A_z \nonumber \\&\quad \times \left( 4B^2 e^{-4W_T} -4 f^\prime W_L^\prime -8 f^\prime W_T^\prime \right. \nonumber \\&\quad \left. + 8 f W_T^\prime W_L^\prime +16 f W_T^{\prime 2}\right) \nonumber \\&\quad + iq \lambda e^{2W_T} \delta A_z \left( 2B^2 e^{-4W_T} f^{-1} f^\prime -48 f^{-1} f^\prime \right. \nonumber \\&\quad \left. - 4B^2 e^{-4W_T} W_L^\prime + 4f^{-1} f^{\prime 2}W_L^\prime \right. \nonumber \\&\quad \left. + 8f^\prime W_L^{\prime 2} -20 B^2 e^{-4W_T} +96 W_T^\prime + 8 f^{-1} f^{\prime 2} W_T^\prime \right. \nonumber \\&\quad \left. +16 f^\prime W_T^\prime W_L^\prime W_T^\prime -16 f W_T^\prime W_L^{\prime 2} \right. \nonumber \\&\quad \left. - 48 f W_L^\prime W_T^{\prime 2} -32 f W_T^{\prime 3} \right) =0, \end{aligned}$$

(55)

$$\begin{aligned}&\partial _r\left( e^{W_L+2W_T}\partial _r \delta V_t\right) + 8\alpha B \partial _r \delta A_z \nonumber \\&\quad - \frac{q^2 e^{W_L}}{f}\delta V_t -\frac{iq B e^{W_L}}{f}\delta g_{ty}=0, \end{aligned}$$

(56)

$$\begin{aligned}&\partial _r\left( e^{2W_T-W_L}f \partial _r \delta A_z\right) - q^2 e^{-W_L}\delta A_z + iq \lambda e^{2W_T} \nonumber \\&\quad \left( -4 B^2 e^{-4W_T} + 4 f^\prime W_L^\prime + 8 f^\prime W_T^\prime \right. \nonumber \\&\left. - 8f W_T^\prime W_L^\prime -16 f W_T^{\prime 2}\right) \partial _r \delta g_{ty}+ 8\alpha B \partial _r \delta V_t=0. \end{aligned}$$

(57)

The boundary conditions on the horizon need to be derived by matching with the horizon solutions in the limit ${\omega }\rightarrow 0$. We elaborate on the derivation in Appendix B. The resultant boundary conditions on the horizon are given by

$$\begin{aligned}&{\delta }g_{ty}(r=r_h)=0, \qquad {\delta }V_t(r=r_h)=0, \quad \nonumber \\&\quad {\delta }A_z(r=r_h)=\text {constant}. \end{aligned}$$

(58)

The free parameters for the three fields can be chosen as horizon derivatives of ${\delta }g_{ty}$, ${\delta }V_t$ and horizon value of ${\delta }A_z$. The three parameters on the horizon can be mapped to boundary values of the three fields. We can further simplify Eqs. (55)–(57) by considering the limit $q\rightarrow 0$. Note that ${\delta }g_{ty}$ has an opposite parity to those of ${\delta }V_t$ and ${\delta }A_z$, and ${\delta }g_{ty}$ is the only field sourced on the AdS boundary. Therefore, the AdS boundary conditions are

$$\begin{aligned}&\delta g_{ty} \xrightarrow []{r\rightarrow \infty } h_{ty}, \quad \mathrm{others} \xrightarrow []{r\rightarrow \infty } 0. \end{aligned}$$

(59)

We expect the scaling behaviors ${\delta }g_{ty}\sim {\mathcal {O}}(q^0)$, and ${\delta }V_t,\;{\delta }A_z\sim {\mathcal {O}}(q)$. Defining ${\delta }V_t=iq{\delta }{\tilde{V}}_t$ and ${\delta }A_z=iq{\delta }{\tilde{A}}_z$, we can further simplify (55) through (57) by keeping the leading terms in the q-expansion

$$\begin{aligned}&\partial _r\left( e^{W_L+4W_T} \partial _r \delta g_{ty}\right) - \frac{B^2 e^{W_L}}{f(r)} \delta g_{ty}=0, \nonumber \\&\partial _r\left( e^{W_L+2W_T}\partial _r \delta {\tilde{V}}_t\right) + 8\alpha B \partial _r \delta {\tilde{A}}_z -\frac{ B e^{W_L}}{f}\delta g_{ty}=0, \nonumber \\&\partial _r\left( e^{2W_T-W_L}f \partial _r \delta {\tilde{A}}_z\right) + \lambda e^{2W_T} \left( -4 B^2 e^{-4W_T} + 4 f^\prime W_L^\prime \right. \nonumber \\&\left. + 8 f^\prime W_T^\prime - 8f W_T^\prime W_L^\prime -16 f W_T^{\prime 2}\right) \partial _r \delta g_{ty}+ 8\alpha B \partial _r \delta {\tilde{V}}_t=0. \end{aligned}$$

(60)

4 The correlators $\langle J^t T^{ty} \rangle$ and $\langle J_5^z T^{ty}\rangle$

In this section, we calculate the generation of $J^t$ and $J_5^z$ as linear responses to the external source $h_{ty}$. Note that $h_{ty}$ couples to $T^{ty}$ in the boundary theory. We can express the responses in terms of retarded correlators $\langle J^t T^{ty} \rangle$ and $\langle J_5^z T^{ty}\rangle$ as

$$\begin{aligned} J^t({\omega },q)&=\langle J^t({\omega },q) T^{ty}(-{\omega },-q) \rangle h_{ty}({\omega },q),\nonumber \\ J_5^z({\omega },q)&=\langle J_5^z({\omega },q) T^{ty}(-{\omega },-q)\rangle h_{ty}({\omega },q), \end{aligned}$$

(61)

with the retarded correlators defined as

$$\begin{aligned}&\langle J^t({\omega },q) T^{ty}(-{\omega },-q) \rangle \nonumber \\&\quad =\int \frac{\mathrm{d}t\mathrm{d}^3x}{(2\pi )^4}(-i)e^{-i{\omega }t+i\vec {q}\cdot \vec {x}}\theta (t)\langle [J^t(t,x),T^{ty}(0)]\rangle , \nonumber \\&\langle J^z_5({\omega },q) T^{ty}(-{\omega },-q) \rangle \nonumber \\&\quad =\int \frac{\mathrm{d}t\mathrm{d}^3x}{(2\pi )^4}(-i)e^{-i{\omega }t+i\vec {q}\cdot \vec {x}}\theta (t)\langle [J^z_5(t,x),T^{ty}(0)]\rangle . \end{aligned}$$

(62)

Equation (61) relates the above retarded correlators in equilibrium to the responses of $J^t$ and $J_5^z$ to $h_{ty}$, which are exactly the quantities we will calculate. So, instead of directly computing the retarded correlators (62), we will read off them by computing the one-point functions of $J^t$ and $J_5^z$ (with the only source $h_{ty}$ turned on).

The bulk EOMs (60) will be solved under the boundary conditions (58) and (59). This section will be further split into two parts: an analytical study when the magnetic field is weak versus a numerical study when the value of the magnetic field is generic. In these two complementary studies, we will utilize the results of the background metric functions summarized in Appendix A.

In the limit ${\omega }\rightarrow 0$ and $q\rightarrow 0$, the vector charge density and the axial current are (in terms of the bulk fields)

$$\begin{aligned}&J^t=\lim _{r\rightarrow \infty } \left\{ e^{W_L+2W_T} \partial _r \delta V_t - \frac{e^{W_L}}{\sqrt{f(r)}} \left( B \partial _x \delta g_{ty} \right) \log r \right\} , \nonumber \\&J_5^z=\lim _{r\rightarrow \infty } \left\{ -f(r)e^{2W_T-W_L} \partial _r \delta A_z \right\} . \end{aligned}$$

(63)

Near the AdS boundary, the bulk fluctuations behave as

$$\begin{aligned}&\delta g_{ty} \xrightarrow []{r\rightarrow \infty } h_{ty}+ \frac{B^2 h_{ty}}{4r^4} \log \frac{r_h}{r} + \frac{t_{ty}}{r^4}+ {\mathcal {O}}(r^{-5}), \nonumber \\&\delta {{\tilde{V}}}_t \xrightarrow []{r\rightarrow \infty } \frac{B h_{ty}}{2r^2} \log \frac{r_h}{r} + \frac{v_t^2}{r^2} + {\mathcal {O}}(r^{-3}\log r), \nonumber \\&\delta {{\tilde{A}}}_z \xrightarrow []{r\rightarrow \infty } \frac{a_z^2}{r^2}+ {\mathcal {O}}(r^{-3}). \end{aligned}$$

(64)

So, the vector charge density and axial current for the boundary theory are

$$\begin{aligned} J^t= -2 v_t^2 iq - B iq h_{ty} \left( \log r_h+ \frac{1}{2}\right) , \quad J_5^z= 2 iq a_z^2. \end{aligned}$$

(65)

Below we solve for $J^t$ and $J_5^z$ perturbatively in B and also numerically for generic B.

4.1 Weak magnetic field: a perturbative study

When the magnetic field B is weak, the bulk fluctuations are expandable,

$$\begin{aligned}&\delta g_{ty}= \delta g_{ty}^{[0]}+ \epsilon ^2\delta g_{ty}^{[2]}+ \cdots ,\nonumber \\&\delta {{\tilde{V}}}_t= \epsilon \delta {{\tilde{V}}}_t^{[1]}+ \epsilon ^3 \delta {{\tilde{V}}}_t^{[3]}+ \cdots , \nonumber \\&\delta {{\tilde{A}}}_z= \delta {{\tilde{A}}}_z^{[0]} + \epsilon ^2 \delta {{\tilde{A}}}_z^{[2]}+ \cdots , \end{aligned}$$

(66)

where $\epsilon \sim B$. At the lowest order ${\mathcal {O}}(\epsilon ^0)$, first we have

$$\begin{aligned} 0=\partial _r \left( r^5 \partial _r \delta g_{ty}^{[0]}\right) \Longrightarrow \delta g_{ty}^{[0]}= \left( 1-\frac{r_h^4}{r^4}\right) h_{ty}. \end{aligned}$$

(67)

Then we have

$$\begin{aligned} 0=\partial _r\left( (r^3-r_h^4/r)\partial _r \delta {{\tilde{A}}}_z^{[0]}\right) + \frac{48 \lambda r_h^4}{r^2} \partial _r \delta g_{ty}^{[0]}, \end{aligned}$$

(68)

whose solution is

$$\begin{aligned} \delta {{\tilde{A}}}_z^{[0]}= 8\lambda \left( \frac{r_h^4}{r^4}+ 2\log \frac{r^2+r_h^2}{r^2} \right) h_{ty}. \end{aligned}$$

(69)

At the first order ${\mathcal {O}}(\epsilon ^1)$,

$$\begin{aligned} 0= \partial _r\left( r^3 \partial _r \delta {{\tilde{V}}}_t^{[1]}\right) +8\alpha B \partial _r \delta {{\tilde{A}}}_z^{[0]}- \frac{B}{r}h_{ty}, \end{aligned}$$

(70)

which is solved by

$$\begin{aligned} \delta {{\tilde{V}}}_t^{[1]}&= B h_{ty} \left\{ \frac{\log (r_h/r)}{2r^2} - \frac{128(\log 2)\alpha \lambda }{r^2}+ \frac{32\alpha \lambda }{3r^6}(r_h^4-r^4) \right. \nonumber \\&\quad \left. + 64 \alpha \lambda \left( \frac{1}{r^2}+ \frac{1}{r_h^2} \right) \log \frac{r^2+r_h^2}{r^2} \right\} \nonumber \\&\quad \xrightarrow []{r\rightarrow \infty } \left\{ - \frac{\log r}{2r^2} +\frac{1}{r^2} \left( \frac{1}{2}\log r_h + \frac{160}{3} \alpha \lambda \right. \right. \nonumber \\&\quad \left. \left. -128 \log 2 \; \alpha \lambda \right) \right\} B h_{ty} + {\mathcal {O}}(r^{-3}). \end{aligned}$$

(71)

At the second order ${\mathcal {O}}(\epsilon ^2)$:

$$\begin{aligned} 0= \partial _r\left( r^5 \partial _r \delta g_{ty}^{[2]}\right) - \frac{B^2}{r}h_{ty}+ 8r_h^4 W_T^{(2)\prime }(r) h_{ty}. \end{aligned}$$

(72)

Here, we would like to recall that $W_T^{(2)}$ is obtained in (114). The solution for $\delta g_{ty}^{[2]}$ would be

$$\begin{aligned} \delta g_{ty}^{[2]}(r)&=- B^2 h_{ty}\int _r^\infty \frac{\log x}{x^5}\mathrm{d}x \nonumber \\&\qquad + 8r_h^4 h_{ty} \int _r^\infty \frac{W_T^{(2)}(x)}{x^5}\mathrm{d}x + \frac{C}{r^4}, \end{aligned}$$

(73)

where the integration constant C is fixed as

$$\begin{aligned} \delta g_{ty}^{[2]}(r=r_h)&=0 \Longrightarrow C= B^2r_h^4 h_{ty}\int _{r_h}^\infty \frac{\log x}{x^5} \mathrm{d}x \nonumber \\&\quad - 8r_h^8 h_{ty}\int _{r_h}^\infty \frac{W_T^{(2)}(x)}{x^5}\mathrm{d}x \nonumber \\&= \frac{1}{24}B^2 h_{ty} (1+6\log r_h). \end{aligned}$$

(74)

So,

$$\begin{aligned} \partial _r \delta g_{ty}^{[2]}(r)&= h_{ty} \left\{ \frac{B^2\log r}{r^5} - \frac{8r_h^4}{r^5} W_T^{(2)}(r)\right. \nonumber \\&\quad \left. - \frac{1}{6r^5}B^2(1+6\log r_h)\right\} . \end{aligned}$$

(75)

The equation for $\delta {{\tilde{A}}}_z^{[2]}$ is

$$\begin{aligned} 0&=\partial _r\left[ (r^3-r_h^4/r)\partial _r \delta {{\tilde{A}}}_z^{[2]}\right] + \alpha H_1(r)\nonumber \\&\quad + \alpha ^2 \lambda H_2(r)+ \lambda H_3(r), \end{aligned}$$

(76)

where

$$\begin{aligned} H_1(r)&= h_{ty} \left( -\frac{4B^2}{r^3}- \frac{8B^2\log (r_h/r)}{r^3}\right) , \nonumber \\ H_2(r)&=h_{ty}\left[ -\frac{512 B^2 r_h^4}{r^7}+ \frac{512B^2(12\log 2-5)}{3r^3}\right. \nonumber \\&\quad \left. -\frac{1024 B^2 }{r^3}\log \left( 1+\frac{r_h^2}{r^2}\right) \right] , \nonumber \\ H_3(r)&= h_{ty} \left\{ - \frac{80 B^2 r_h^4}{3r^7} + \frac{16B^2 r_h^2}{3r^3(r^2+r_h^2)} \right. \nonumber \\&\quad \left. + \frac{64r_h^2(r_h^4-r_p^4)}{r(r^2+r_h^2)}\left( \frac{1}{r^2+r_h^2}+ \frac{1}{r^2} \right) \right. \nonumber \\&\quad \left. + \left[ \frac{32B^2 r_h^2}{3r(r^2+r_h^2)^2} + \frac{32B^2 r_h^2}{3r^3 (r^2+r_h^2)} - \frac{48 B^2 r_h^4}{r^7}\right] \right. \nonumber \\&\quad \left. \log \frac{r_h}{r} - \frac{768r_h^8 W_T^{(2)}}{r^7} \right. \nonumber \\&\quad \left. +32 r_h^2 W_T^{(2)\prime } \left[ \frac{7r_h^2(r_h^6-r^6)}{r^6} + \frac{4(r_h^4-r^4)}{r^2(r^2+r_h^2)}\right] \right\} . \end{aligned}$$

(77)

The solution and the near-boundary expansion for $\delta {{\tilde{A}}}_z^{[2]}$ are

$$\begin{aligned} \delta {{\tilde{A}}}_z^{[2]}(r)&=\int _r^\infty \frac{\mathrm{d}x}{x^3-r_h^4/x} \nonumber \\&\quad \times \int _{r_h}^x \left[ \alpha H_1(y) + \alpha ^2 \lambda H_2(y) + \lambda H_3(y)\right] \mathrm{d}y \nonumber \\&\quad \xrightarrow []{r\rightarrow \infty } -\frac{4\lambda h_{ty}(B^2-6r_h^4+6r_p^4)}{3r_h^2r^2}+ {\mathcal {O}}(r^{-3}). \end{aligned}$$

(78)

Up to ${\mathcal {O}}(B^2)$, the vector charge density and axial current on the boundary are

$$\begin{aligned}&J^t= \left[ \frac{64}{3}(12\log 2-5)\alpha \lambda - \log r_h - \frac{1}{2}\right] B iq h_{ty}+ \cdots , \nonumber \\&J_5^z=32\lambda r_h^2 iq h_{ty} - \frac{8\lambda }{3r_h^2}\left( B^2-6r_h^4+6r_p^4\right) iq h_{ty} + \cdots . \end{aligned}$$

(79)

The transport coefficients $\xi$ and $\sigma$ are

$$\begin{aligned} \xi&= \frac{128}{3}(12\log 2-5)\alpha \lambda - 2\log r_h -1+ {\mathcal {O}}(B/T^2), \nonumber \\ \sigma&= r_h^2 \left[ 64{\lambda }- \frac{32\lambda B}{3r_h^2}+ {\mathcal {O}}(B^2/T^4) \right] , \end{aligned}$$

(80)

where we have substituted the perturbative expression (116) for $r_p$.

The transport coefficient ${\xi }$ contains both non-anomalous contribution and anomalous contribution proportional to ${\alpha }{\lambda }$. The non-anomalous contribution is consistent with the prediction of the non-anomalous magnetohydrodynamics (MHD) [33], which for a neutral plasma has²

$$\begin{aligned} \Delta J^t=2 \left( M_{{\Omega },{\mu }}{\vec {B}}\cdot {\vec {{\Omega }}}-2p_{,B^2}{\vec {B}}\cdot {\vec {{\Omega }}} \right) , \end{aligned}$$

(81)

with $M_{\Omega }\equiv \frac{{\partial }{\mathcal {F}}}{{\partial }(B\cdot {\Omega })}$ being the magneto-vortical susceptibility and $2p_{,B^2}\equiv 2\frac{{\partial }p}{{\partial }(B^2)}$ being the magnetic susceptibility. Here p is the pressure and ${\mathcal {F}}$ is the free energy density [33]. Note that in (81) we ignored terms nonlinear in ${\Omega }$. Both susceptibilities $M_{\Omega }$ and $2p_{,B^2}$ can be calculated independently. In the weak B field limit, the magnetic susceptibility $2p_{,B^2}$ can be calculated from the perturbative background we already obtain in Appendix A. While the magneto-vortical susceptibility $M_{\Omega }$ vanishes for a neutral plasma by charge conjugation symmetry, $M_{{\Omega },{\mu }}\equiv \partial M_{\Omega }/\partial \mu$ does not. The quantity $M_{{\Omega },{\mu }}$ can even be calculated in a charged plasma at $B=0$. In Appendix C, we calculate both susceptibilities with the following results:

$$\begin{aligned} 2p_{,B^2}=\log r_h,\quad M_{{\Omega },{\mu }}=-\frac{1}{2}. \end{aligned}$$

(82)

The negative value of $M_{{\Omega },{\mu }}$ is consistent with the fact that spin–vorticity coupling lowers/raises energy of particle/anti-particle. Clearly, (81) and (82) are in perfect agreement with (79).

The dependence $\log r_h$ may look odd at the first sight. To restore units, we should use the replacement $\log r_h\rightarrow \log (r_h L)$. In fact, this transport coefficient is scheme dependent. The appearance of the AdS radius L comes from the fact that we use 1/L as our renormalization scale [34]. Other physically significant renormalization scales could be used, which could alter this term [34]. It is also interesting to note that the scheme dependence is related conformal anomaly. In fact, a different scheme would correspond to adding a finite counter-term as

$$\begin{aligned} {\Delta }S_{\mathrm{c.t.}}=-\frac{1}{2\kappa ^2} \int \mathrm{d}^4x \sqrt{-\gamma }\left( \frac{a}{4}\left( F^V\right) _{\mu \nu } \left( F^V\right) ^{\mu \nu }\right) . \end{aligned}$$

(83)

Such a counter-term would give the following contribution to the vector current:

$$\begin{aligned} {\Delta }J^{\mu }=-\frac{a}{2\kappa ^2}\sqrt{-{\gamma }}\nabla _{\nu }F^{{\mu }{\nu }}. \end{aligned}$$

(84)

In the presence of $h_{ty}(x)$, we can easily obtain ${\Delta }J^t\sim a {\vec {B}}\cdot {\vec {{\Omega }}}$. Therefore the combination $M_{{\Omega },{\mu }}-2p_{,B^2}$ can be shifted by a constant. Note that the scheme dependence of the vector charge density is absent in a free theory.

The anomalous contribution is proportional to ${\lambda }{\alpha }B$. The EOM (60) suggests the following chain of responses: ${\delta }A_z\sim {\mathcal {O}}({\lambda })$ is induced in response to vorticity and then backreaction of ${\delta }A_z$ to ${\delta }V^t$ gives $J^t\sim {\mathcal {O}}({\lambda }{\alpha }B)$. This corresponds to the backreaction of $J_5^z$ generated by CVE to $J^t$ on the field theory side. A possible $\alpha ^2B^2$-term would emerge at the next order ${\mathcal {O}}(B^2)$. In this case, (60) suggests the following chain of responses: a non-anomalous contribution to $J^t$ is generated by magneto-vortical coupling. Then $J_5^z$ is induced by CSE. The backreaction of $J_5^z$ to $J^t$ would give the $O(\alpha ^2B^2)$ contribution. However, the above reasoning is not quite accurate. As we show below, in fact CSE is not generated in the presence of $J^t$. Nevertheless the bulk profile of ${\delta }A_z$ does backreact to ${\delta }V_t$ to give $\alpha ^2B^2$ correction to $J^t$.

In contrast to ${\xi }$, the transport coefficient ${\sigma }$ is scheme independent. We can add an analogous counter-term as (83). It would not contribute to $J_5^z$ in the absence of a background axial gauge field. The structure of ${\sigma }$ is relatively simple. Aside from the $T^2$-correction to the CVE (i.e., the first piece in ${\sigma }$), ${\sigma }$ encodes the correction to chiral vortical coefficient from B. However, there is no contribution proportional to ${\alpha }$. In other words, no CSE is seen despite the generation of $J^t$. This is in contrast to the naive expectation from CSE

$$\begin{aligned} J_5^z=8{\alpha }B\frac{J^t}{\chi }, \end{aligned}$$

(85)

with $\chi$ being the vector charge susceptibility. In fact, from the holographic model, the absence of CSE holds more generally: if we integrate (57) from the horizon to an arbitrary r, we obtain (with ${\lambda }=0$)

$$\begin{aligned} e^{2W_T-W_L}f\partial _r \delta {{\tilde{A}}}_z+ 8 \alpha B \delta {{\tilde{V}}}_t(r)=0, \end{aligned}$$

(86)

where the horizon boundary conditions (58) have been utilized to fix the integration constant. Taking $r\rightarrow \infty$ and noting ${\tilde{V}}_t(r\rightarrow \infty )=0$, we have

$$\begin{aligned} J_5^z=0,\quad \mathrm{when} \quad \lambda =0. \end{aligned}$$

(87)

Following [35], we should identify the difference of ${\delta }V_t$ on the boundary and on the horizon as the vector chemical potential

$$\begin{aligned} {\mu }={\delta }V_t(r=\infty )-{\delta }V_t(r=r_h). \end{aligned}$$

(88)

It then follows naturally from (86) that $J_5^z=8{\alpha }B{\mu }$. In our case, we have ${\mu }=0$ but $J^t\ne 0$.

To understand the physical difference between $J^t$ and ${\mu }$, we note ${\mu }$ ($-{\mu }$) is the extra energy cost to create one unit of particle (anti-particle), but $J^t$ depends on actual distribution of particles and anti-particles. In our case ${\mu }=0$ implies that it costs the same energy to create both particle and anti-particle. Indeed, we can view ${\delta }V_t$ as a zero mode as it vanishes both on the horizon and on the boundary, which supports the picture of vanishing energy cost for creating a particle. Since the state is obtained by the adiabatic limit, it means $J^t$ is generated dynamically³. The CSE seems to be only sensitive to the energy difference ${\mu }$, not the charge density. The absence of CSE can also been understood from the scheme independence $J^z_5$: unlike $J^t$, $J^z_5$ is unaffected by the choice of scheme from (63). This makes natural for $J_5^z$ to depend on ${\mu }$, rather than on $J^t$.

Fig. 1

The correlator $\langle J^t T^{ty}\rangle /(iq)=J^t/\partial _x h_{ty}$ in units of $r_h^2$ (left) and the transport coefficient $\xi$ (right) as a function of $B/T^2$ when $\alpha =\lambda =0$

4.2 Generic magnetic field: a numerical study

For generic values of B, we will solve the fluctuation EOMs (60) through the shooting technique. First, we find the near-horizon solution:

$$\begin{aligned}&\delta g_{ty}= \delta g_{ty}^1 (r-r_h)+ \delta g_{ty}^2 (r-r_h)^2+ \cdots , \nonumber \\&\delta {{\tilde{V}}}_t= \delta {{\tilde{V}}}_t^1 (r-r_h) + \delta {{\tilde{V}}}_t^2 (r-r_h)^2+ \cdots , \nonumber \\&\delta {{\tilde{A}}}_z= \delta {{\tilde{A}}}_z^0+ \delta {{\tilde{A}}}_z^1 (r-r_h) + \delta {{\tilde{A}}}_z^2 (r-r_h)^2 + \cdots , \end{aligned}$$

(89)

where, thanks to the horizon condition (58), only $\delta g_{ty}^1$, $\delta {{\tilde{V}}}_t^1$ and $\delta {{\tilde{A}}}_z^0$ are undetermined. Then we will choose a reasonable value of $\delta g_{ty}^1$ (this corresponds to turning on a specific source $h_{ty}$) and fine-tune $\delta {{\tilde{V}}}_t^1, \delta {{\tilde{A}}}_z^0$ until $\delta {{\tilde{V}}}_t(r=\infty )= \delta {{\tilde{A}}}_z(r=\infty )=0$ are satisfied. From the numerical solution, we can read off the expectation values of $J^t$ and $J_5^z$, as responses to the source $h_{ty}$ only. In practical numerics, we set the horizon data $\delta g_{ty}^1=-1$.

However, there is one problem in the procedure mentioned above. Since we intend to solve the background EOMs (32)–(34) using the initial conditions (122) and (124), we should be careful in solving the fluctuation EOMs (60). More precisely, the correct solutions are

$$\begin{aligned} \delta g_{ty}= \frac{\delta g_{ty}^*}{\sqrt{v(b)}}, \quad \delta V_t= \delta V_t^*, \quad \delta A_z= \frac{\delta A_z^*}{\sqrt{w(b)}}, \end{aligned}$$

(90)

where the starred functions $\delta g_{ty}^*$, $\delta V_t^*$ and $\delta A_z^*$ are solved from (60) using the “incorrect” numerical background metric functions, as discussed in Appendix A. Adapted to the tilde variables, we have

$$\begin{aligned} \frac{\delta {{\tilde{V}}}_t}{B}&= \frac{\delta V_t}{iq B}=v^{3/2}(b) \delta {{\tilde{V}}}_t^*, \nonumber \\ {{\tilde{A}}}_z&\equiv \frac{\delta A_z}{iq}= \sqrt{\frac{v(b)}{w(b)}} \delta {{\tilde{A}}}_z^*. \end{aligned}$$

(91)

Here, for the sake of numerical calculation, we have further re-scaled the term $\delta {{\tilde{V}}}_t$ of (60) by a factor of B.

For convenience, we set $r_h=1$ in our numerical calculations. So, the dimensionful quantities ($J^t, J_5^z$ etc.) to be plotted in Sects. 4.2.1 and 4.2.2 should be understood as in units of proper powers of $r_h$.

4.2.1 Non-anomalous effects: $\alpha =\lambda =0$

When the chiral anomaly and gravitational anomaly are turned off (i.e., $\alpha =\lambda =0$), the transport properties of the magnetized plasma get non-anomalous contributions from the medium only. The medium effects are not covered by the study of [17] since the calculations therein are essentially based on the vacuum state. In this situation, as we discussed in the previous section we only see a dynamically generated vector charge density $J^t$, whereas the correlator $\langle J^z_5 T^{ty}\rangle /(iq)$ (and thus $J_5^z$) vanishes identically as seen from (87). In Fig. 1, we show the correlator $\langle J^t T^{ty}\rangle /(iq)$ and the transport coefficient $\xi$ as a function of $B/T^2$. For the purpose of probing the strong magnetic field limit, we have improved our numerical calculations and generate plots up to $B/T^2\sim 3000$.

Fig. 2

In the strong B limit, the numerical result of $\xi$ (dots) is fitted by the function (92) when $\alpha =\lambda =0$

Fig. 3

Left: the CVE conductivity ${\sigma }$ in units of $r_h^2$ (with anomaly coefficient ${\lambda }$ factorized out) as a function of $B/T^2$. Right: $B{\sigma }/{\lambda }$ in unit of $r_h^4$, demonstrating the asymptotic behavior for ${\sigma }$ at strong magnetic field. Here, in each plot the three curves (dashed orange, solid blue, dotted green) corresponding to the different choice in (93) perfectly collapse into a single one

Fig. 4

The anomalous correction $\langle \delta J^t T^{ty}\rangle /(iq)$ in unit of $r_h^2$ as a function of $B/T^2$: $\alpha =1/20, \lambda =1/50$ for dashed curve and $\alpha =1/20, \lambda =1/20$ for solid one

From the left panel of Fig. 1, it seemingly implies a quasi-linear growth for $\langle J^t T^{ty}\rangle /(iq)$ as $B/T^2$ is increased, which as we will show is inaccurate. The right panel of Fig. 1 reveals more information: $-{\xi }$ approaches 1 from above. In Fig. 2, we fit our numerical result for $\xi$ in the strong magnetic field limit by the following function:

$$\begin{aligned} -{\xi }\simeq 1.0001779 + 32.692107 \frac{\log (B/T^2)}{B/T^2} + \frac{124.89109}{B/T^2}. \end{aligned}$$

(92)

It is tempting to conclude that $-\xi \rightarrow 1$ asymptotically. The correction in (92) can be understood as the $v_t^2$ term in the general expression (65) by noting that $r_h=1$ in our numerical results.

Fig. 5

The anomalous correction $\delta \xi$ as a function of $B/T^2$: $\alpha =1/20, \lambda =1/50$ for the dashed curve and $\alpha =1/20, \lambda =1/20$ for the solid one

4.2.2 Anomalous effects: $\lambda \ne 0$

We now turn attention to the anomalous contributions to the transport properties of the magnetized plasma. While the two anomaly coefficients, ${\alpha }, {\lambda }$, are fixed for a specific QFT on the boundary, we here take a phenomenological viewpoint and think of ${\alpha },{\lambda }$ as free parameters. First of all, taking ${\lambda }=0$ will kill $J_5^z$ completely, as seen from the bulk EOMs (60). Thus, our representative choices for the anomaly coefficients are

$$\begin{aligned} ({\alpha },{\lambda })=(0,1/50), \; (1/20, 1/50), \; (1/20,1/20). \end{aligned}$$

(93)

We begin with the fate of the CVE conductivity ${\sigma }$. First, the last equation of (60) could be formally integrated from the horizon to the AdS boundary, yielding

$$\begin{aligned} J_5^z= \sigma {\Omega }+ 8{\alpha }{\mu }B, \end{aligned}$$

(94)

where ${\mu }$ is defined in (88). Here, we stress that the CVE conductivity ${\sigma }$ depends on ${\lambda }$ linearly and is independent of ${\alpha }$. In the left panel of Fig. 3, we plot the CVE conductivity $\sigma$ as a function of $B/T^2$, taking all choices for ${\alpha },{\lambda }$ from (93). From the plot, we obviously see perfect overlapping of different curves, confirming our claim that ${\sigma }$ linearly depends on ${\lambda }$ only. Intriguingly, the magneto-vortical coupling effect tends to suppress the CVE conductivity and eventually renders it vanishing at large magnetic field. Asymptotically, $\sigma \sim B^{-1}$ as demonstrated in the right panel of Fig. 3. Similar suppression effects due to the quark mass [38, 39] and spacetime curvature [40] are seen. Our findings are in contrast to the proposal of [17] that the magneto-vortical coupling (through the chiral anomaly) generates a term to ${\sigma }$ linear in B when $B/T^2$ becomes very large.

Next, we consider anomalous contributions to the generation of the vector charge density $J^t$ and the transport coefficient $\xi$. Given that $J^t$ always gets a non-anomalous contribution, we find it more transparent to consider

$$\begin{aligned} \delta J^t \equiv J^t- J^t|_{{\alpha }={\lambda }=0}, \quad \delta \xi \equiv \xi - \xi |_{{\alpha }={\lambda }=0}. \end{aligned}$$

(95)

In accord with the different choices for ${\alpha },{\lambda }$ as made in (93), we show anomalous corrections $\langle \delta J^t T^{ty}\rangle /(iq)$ and $\delta \xi$ in Figs. 4 and 5, respectively. From the second equation of (60), it is clear that taking ${\alpha }=0$ makes anomalous contributions $\delta J^t$ and $\delta \xi$ to be zero. So, for each panel of Figs. 4 and 5, we have only two non-trivial curves.

Fig. 6

The anomalous correction $\langle \delta J^t T^{ty}\rangle /(iq)$ in unit of $r_h^2$ as a function of $\log (B/T^2)$: $\alpha =1/20, \lambda =1/50$ for the squared points and $\alpha =1/20, \lambda =1/20$ for the circled points

From Fig. 4, we read that the anomalous contribution to the vector charge ${\delta }J^t$ has opposite sign to their non-anomalous counterpart for a weak magnetic field. As the magnetic field becomes stronger, ${\delta }J^t$ changes sign and continues to grow mildly at large $B/T^2$. More precisely, the numerical results at large B imply the following asymptotic behaviors for the anomalous corrections:

$$\begin{aligned} \delta J^t \simeq \log B, \quad \delta \xi \simeq \frac{\log B}{B}, \end{aligned}$$

(96)

which are clearly confirmed by the plots of Fig. 6. It is worth noting that from (92) and (96) the large B limit of ${\xi }$ is dominated by the non-anomalous medium contribution.

5 Thermal Hall effect and thermal axial magnetic effect

In the previous section, we have obtained the following transport coefficients:

$$\begin{aligned}&J^t= \xi (B,T)({\vec {B}}\cdot {\vec \Omega }),\nonumber \\&J_5^z={\sigma }(B,T){\Omega }, \end{aligned}$$

(97)

which can be viewed as responses of $J^t$ and $J_5^z$ to $h_{ty}$ (cf. (61)):

$$\begin{aligned} J^t(q)&=\langle J^t(q)T^{ty}(-q)\rangle _B h_{ty}(q)=\frac{iq\xi (B,T) B}{2}h_{ty}(q),\nonumber \\ J_5^z(q)&=\langle J_5^z(q)T^{ty}(-q)\rangle _B h_{ty}(q)=\frac{iq {\sigma }(B,T)}{2}h_{ty}(q), \end{aligned}$$

(98)

where we have substituted the expression for $\Omega$ in (41). The correlators in (98) give the following transposed correlators by the Onsager relation:

$$\begin{aligned}&\langle T^{ty}(-q)J^t(q)\rangle _{-B}=\frac{1}{2}iq{\xi }(B,T)B,\nonumber \\&\langle T^{ty}(-q)J_5^z(q)\rangle _{-B}=\frac{1}{2}iq{\sigma }(B,T). \end{aligned}$$

(99)

The subscript indicates the expectation values are taken with $-B$. We can then obtain the response of $T^{ty}$ to both $A_t$ and $A_z^5$ using (99) with the signs of both q and B flipped

$$\begin{aligned}&T^{ty}(q)=\langle T^{ty}(q)J^t(-q)\rangle _BA_t(q)=\frac{1}{2}{\xi }(-B,T)E_x(q)B,\nonumber \\&T^{ty}(q)=\langle T^{ty}(q)J_5^z(-q)\rangle _BA_z^5(q)=\frac{1}{2}{\sigma }(-B,T)B_{5y}(q), \end{aligned}$$

(100)

where we have used $E_x(q)=-iqA_t(q)$ and $B_{5y}(q)=-iqA_z^5(q)$. In a neutral plasma, (100) corresponds to the generation of thermal current by transverse electric field and transverse axial magnetic field, which we coin the thermal Hall effect and the thermal axial magnetic effect, respectively. Below we will derive (100) more rigorously using the time-reversal symmetry. We start with the following correlators for the responses to $h_{ty}$:

$$\begin{aligned}&\langle J^t(q)T^{ty}(-q)\rangle =\frac{iq\xi B}{2}, \nonumber \\&\langle J^z_5(q)T^{ty}(-q)\rangle =\frac{iq {\sigma }}{2}, \end{aligned}$$

(101)

with the correlators in (101) being the limit ${\omega }\rightarrow 0$ of the following retarded correlator:

$$\begin{aligned}&\langle O_a({\omega },q)O_b(-{\omega },-q)\rangle =\int \frac{\mathrm{d}t\mathrm{d}^3x}{(2\pi )^4}e^{-i{\omega }t+i\vec {q}\cdot \vec {x}}G^R_{ab}(t,x,B)\nonumber \\&\quad =\int \frac{\mathrm{d}t\mathrm{d}^3x}{(2\pi )^4}(-i)e^{-i{\omega }t+i\vec {q}\cdot \vec {x}}\theta (t)\langle [O_a(t,x),O_b(0)]\rangle . \end{aligned}$$

(102)

By time-reversal symmetry, we can obtain the transposed correlators by [41]

$$\begin{aligned}&G^R_{ab}(t,x,B)=\gamma _a\gamma _bG^R_{ba}(t,-x,-B) \nonumber \\&\quad \Rightarrow \langle O_a({\omega },q)O_b(-{\omega },-q)\rangle _B\nonumber \\&\qquad =\gamma _a\gamma _b\langle O_b({\omega },-q)O_a(-{\omega },q)\rangle _{-B} \end{aligned}$$

(103)

with $\gamma _{a}=\pm 1$ corresponding to eigenvalues of operator $O_a$ under time-reversal. Note that B flips sign under time-reversal, which we indicate in the subscript. For the operators of our interest ${T^{ty},J^t,J_5^z}$, we have $\gamma ={-,+,+,}$ respectively. Therefore, (101) and (103) give the following transposed correlators (with the direction of B reversed and ${\omega }=0$):

$$\begin{aligned}&\langle T^{ty}(-q)J^t(q)\rangle _{-B}=-\frac{1}{2}iq\xi (B) B, \nonumber \\&\langle T^{ty}(-q)J^z_5(q)\rangle _{-B}=-\frac{1}{2}iq{\sigma }(B). \end{aligned}$$

(104)

Note that $J^t(q)$ and $J^z_5(q)$ couple to sources $V_t(-q)$ and $A_z(-q)$, respectively. We can rewrite (104) in a more intuitive way:

$$\begin{aligned}&\langle T^{ty}(-q)\rangle _{-B}=\frac{1}{2}\xi (B)BE_x(-q), \nonumber \\&\langle T^{ty}(-q)\rangle _{-B}=\frac{1}{2}{\sigma }(B) B_{5y}(-q). \end{aligned}$$

(105)

The thermal Hall effect contains both non-anomalous and anomalous contributions, where the non-anomalous contribution can be understood from non-anomalous MHD [33]. Naively, turning on $E_x$ necessarily induces a steady flow along $v_y$ due to the Lorentz force acting on positive and negative charge carriers. However, this is not true for a stationary state. The stationary state can be obtained simply by setting ${\omega }=0$ in (47) through (53). In this case, the dynamics of the fields ${\delta }g_{ty},\;{\delta }V_t,\;{\delta }A_z$ decouple from ${\delta }V_x,\; {\delta }g_{xy}$. We can thus consistently set ${\delta }V_x$ and ${\delta }g_{xy}$ to zero, leading to vanishing $J^x$ and $T^{xy}$. This strongly constrains the hydrodynamic analysis. Note that both $J^x$ and $T^{xy}$ contain the dissipative terms as follows:

$$\begin{aligned} J^x&={\sigma }\left( E_x-{\partial }_x{\mu }+v_yB\right) ,\\ T^{xy}&=\eta {\partial }_xv_y+\cdots . \end{aligned}$$

The shear contribution in $T^{xy}$ cannot be canceled by other terms. The only possibility for $T^{xy}$ to vanish is to have $v_y=0$. This implies that an inhomogeneous vector charge density is needed for the stationary state: $E_x-{\partial }_x{\mu }=0$. Indeed, this is consistent with the holographic analysis if we identify ${\mu }=V_t(r=\infty )-V_t(r=r_h)$. With $v_y=0$, the non-anomalous contribution to the thermal current is given by [33, 42,43,44])

$$\begin{aligned} T^{ty}=-E_xB\left( 2p_{,B^2}-M_{{\Omega },{\mu }}\right) . \end{aligned}$$

(106)

The first term can be identified as $-E_xM$ by noting $2p_{,B^2}B=M$. The second term can be interpreted as $-P_xB$ if we identify $E_xM_{{\Omega },{\mu }}$ as an effective polarization $P_x$. Apart from the non-anomalous contribution, we also obtain an anomalous contribution that requires at least the chiral anomaly to exist. This can be seen from the middle equation in (60). When ${\alpha }=0$, the dynamics of ${\delta }A_z$ decouples from that of ${\delta }V_t$, leaving only a non-anomalous contribution. On the contrary, the thermal axial magnetic effect contains only an anomalous effect. Its existence relies on the gravitational anomaly.

6 Conclusion

In this work, based on a holographic model, we considered the effects of the magneto-vortical coupling on the transport properties of a strongly coupled plasma. First of all, even when the chiral and gravitational anomalies are turned off, the coupling of a magnetic field and a weak fluid’s vorticity dynamically generate a contribution to the vector charge density $J^t$, which we refer to as a non-anomalous medium contribution. The non-anomalous medium contribution in $J^t$ grows linearly in B and the relevant transport $\xi$ approaches a constant in the strong magnetic field limit. However, a similar non-anomalous contribution is not observed for the axial current.

Secondly, the magneto-vortical coupling also generates anomalous contributions to the vector charge density $J^t$ and axial current $J_5^z$. Thanks to the absence of a background for the vector chemical potential, the anomalous contribution to $J_5^z$ is completely induced by the gravitational anomaly (i.e., insensitive to the chiral anomaly). In this sense, it would be more natural to interpret the anomalous contribution to $J_5^z$ as a CVE contribution [27], rather than a CSE contribution. The corresponding chiral vortical conductivity receives a correction at finite B. In particular, in the strong magnetic field limit, the magneto-vortical coupling renders the CVE conductivity vanishing asymptotically. This is quantitatively different from the conclusion of [17]. In contrast to that of $J_5^z$, the anomalous contribution to $J^t$ requires the chiral anomaly to exist. The presence of a gravitational anomaly can also affect the generation of $J^t$. Thus, the anomalous contribution to $J^t$ would contain more fruitful physics. Particularly, in the strong magnetic field limit, the anomalous part of $J^t$ seemingly grows logarithmically as a function of B. This is to be compared with the results of [17] where a term linear in B was generated in $J^t$ by the chiral anomaly.

Our findings summarized above would necessitate the formulating of a consistent/complete anomalous magnetohydrodynamics. This requires one to consistently add novel transport phenomena induced by anomalies into the non-anomalous magnetohydrodynamics [33, 42]. While our study treated the magentic field as external, the case with a dynamical electromagnetic field is also interesting. The corresponding anomalous MHD has been initially considered in [45] by assuming a small chiral anomaly coefficient. In fact, holographic models corresponding to a dynamical electromagnetic field have been proposed in [46, 47]. Including anomalies in the model would allow us to study anomalous MHD without further assumptions on the anomaly coefficients. We leave this for future work.

Last but not least, the transport phenomena we discussed are dissipationless. It is possible to derive them by including anomalies in the partition function approach [48]. This would allow us to obtain more complete dissipationless transport phenomena. We hope to address this in the future.

Appendices

Appendix A: Details of solving the background metric functions

In this appendix, we collect calculational details of solving the background metric functions.

When the magnetic field is weak (i.e. $B/T^2\ll 1$), we construct the bulk metric functions perturbatively [49]:

$$\begin{aligned} f(r)&=r^2-\frac{r_h^4}{r^2}+\epsilon ^2 f^{(2)}(r) +\epsilon ^4 f^{(4)}(r)+\cdots , \nonumber \\ W_T(r)&=\log r+ \epsilon ^2 W_T^{(2)}(r)+ \epsilon ^4 W_T^{(4)}(r) +\cdots , \nonumber \\ W_L(r)&=\log r+ \epsilon ^2 W_L^{(2)}(r)+ \epsilon ^4 W_L^{(4)}(r) +\cdots , \end{aligned}$$

(107)

where a formal parameter $\epsilon \sim B/T^2$ is introduced to mark the perturbative expansion. $f^{(n)}(r)$, $W_T^{(n)}(r)$ and $W_L^{(n)}(r)$ are solved from the bulk equations. We are interested in $n=2$.

At ${\mathcal {O}}(\epsilon ^2)$, the constraint equation (31) yields

$$\begin{aligned} W_L^{(2)}+ 2 W_T^{(2)}= c_1+ \frac{c_2}{r}, \end{aligned}$$

(108)

where the integration constant $c_1$ should be set to zero due to asymptotic AdS requirement (37). By redefinition of the radial coordinate r, we could also set $c_2=0$. Thus,

$$\begin{aligned} W_L^{(2)}+ 2 W_T^{(2)}=0. \end{aligned}$$

(109)

Then the dynamical equation (32) is solved as

$$\begin{aligned}&r^3\partial _r^2f^{(2)}+3r^2 \partial _r f^{(2)} \nonumber \\&\quad = \frac{B^2}{3r} \Rightarrow f^{(2)}=c_f^1+ \frac{c_f^2}{r^2} - \frac{B^2\log r}{6r^2}. \end{aligned}$$

(110)

Substituting (109) and (110) into the dynamical equations (33) and (34), we obtain

$$\begin{aligned}&0=\left( r^5-r r_h^4\right) W_T^{(2)\prime \prime }(r) +\left( 5 r^4-r_h^4\right) W_T^{(2)\prime }(r) +\frac{B^2}{6 r}+2 c_f^1 r, \nonumber \\&0=\left( r^5-r r_h^4\right) W_L^{(2)\prime \prime }(r)+\left( 5 r^4-r_h^4\right) W_L^{(2)\prime }(r) -\frac{B^2}{3 r}+2 c_f^1 r. \end{aligned}$$

(111)

Obviously, in order to be consistent with (109), one has to set $c_f^1=0$. The integration constant $c_f^2$ is fixed by the location of the horizon, which will be presumably shifted due to the presence of a magnetic field,

$$\begin{aligned} \left( r^2-\frac{r_h^4}{r^2}+ f^{(2)}(r)\right) \Bigg |_{r=r_p^{(2)}}&=0 \Rightarrow c_f^2\nonumber \\&= -\left( (r_p^{(2)})^4-r_h^4\right) + \frac{B^2\log r_p^{(2)}}{6(r_p^{(2)})^2}, \end{aligned}$$

(112)

where $r_p^{(2)}$ represents the location of the event horizon at ${\mathcal {O}}(\epsilon ^2)$. Then

$$\begin{aligned} f^{(2)}= - \frac{(r_p^{(2)})^4-r_h^4}{r^2}+ \frac{B^2}{6r^2} \log \frac{r_p^{(2)}}{r}. \end{aligned}$$

(113)

From (33), $W_T^{(2)}$ is solved as

$$\begin{aligned}&\partial _r\left[ r(r^4-r_h^4)\partial _r W_T^{(2)}(r)\right] +\frac{B^2}{6 r}=0 \nonumber \\&\quad \Rightarrow W_T^{(2)}=\frac{B^2}{6}\int _r^\infty \frac{\log (x/r_h)}{x(x^4-r_h^4)}\mathrm{d}x\nonumber \\&\quad = \frac{B^2}{6}\int _r^\infty \frac{\log (x/r_p^{(2)})}{x(x^4-(r_p^{(2)})^4)}\mathrm{d}x \nonumber \\&\quad =\frac{B^2}{288 r_h^4} \left\{ \pi ^2 -24\log ^2\frac{r_h}{r}\log \left( 1-\frac{r_h^4}{r^4}\right) \right. \nonumber \\&\qquad \left. + 12i\pi \log \frac{r}{r_h} -3 \text {Li}_2\left( \frac{r^4}{r_h^4}\right) \right\} , \end{aligned}$$

(114)

where in the last equality of the second line we have made use of the fact that the difference between $r_p^{(2)}$ and $r_h$ is of ${\mathcal {O}}(B^2)$. At ${\mathcal {O}}(\epsilon ^2)$, the relation between the location of the event horizon and the Hawking temperature becomes

$$\begin{aligned} r_p^{(2)}-\frac{B^2}{24(r_p^{(2)})^3}+ {\mathcal {O}}(B^4)= \pi T, \end{aligned}$$

(115)

which is solved as

$$\begin{aligned} r_p^{(2)}=r_h+ \frac{B^2}{24 r_h^3} + {\mathcal {O}}(B^4),\quad \mathrm{with} \quad r_h=\pi T. \end{aligned}$$

(116)

When the value of B is generic, we have to solve the metric functions numerically. We find it more convenient to make the change of variables

$$\begin{aligned}&f(r) \rightarrow r^2 U(r),\quad W_T(r)\rightarrow \log r + \frac{1}{2} \log V(r),\nonumber \\&\quad W_L(r)\rightarrow \log r + \frac{1}{2} \log W(r), \end{aligned}$$

(117)

followed by

$$\begin{aligned} u=\frac{r_h}{r}\in [0,1]. \end{aligned}$$

(118)

Then the dynamical bulk equations (32)–(34) turn into

$$\begin{aligned} 0&=U''(u) +U'(u) \left( \frac{V'(u)}{V(u)}+\frac{W'(u)}{2 W(u)}-\frac{5}{u}\right) \nonumber \\&+U(u) \left( \frac{8}{u^2}-\frac{2 V'(u)}{u V(u)}-\frac{W'(u)}{u W(u)}\right) \nonumber \\&-\frac{B^2 u^2}{3 V(u)^2} -\frac{8}{u^2}, \nonumber \\ 0=&V''(u)+ V'(u) \left( \frac{U'(u)}{U(u)}+\frac{W'(u)}{2 W(u)}-\frac{5}{u}\right) \nonumber \\&+V(u) \left( -\frac{8}{u^2 U(u)}+\frac{8}{u^2} -\frac{2 U'(u)}{u U(u)} \right. \nonumber \\&\left. -\frac{W'(u)}{u W(u)}\right) + \frac{2 B^2 u^2}{3 U(u) V(u)} , \nonumber \\ 0=&W''(u)+ W'(u) \left( \frac{U'(u)}{U(u)}+\frac{V'(u)}{V(u)}-\frac{4}{u}\right) \nonumber \\&-\frac{W'(u)^2}{2 W(u)}+ W(u) \left( \frac{-\frac{B^2 u^2}{3 V(u)^2} -\frac{8}{u^2}}{U(u)} \right. \nonumber \\&\left. + \frac{8}{u^2}-\frac{2 U'(u)}{u U(u)}-\frac{2 V'(u)}{u V(u)}\right) , \end{aligned}$$

(119)

where, since we have set $r_h=1$ above, B should be understood as $B/r_h^2$.

Near the AdS boundary $u=0$, the metric functions U, V, W are expanded as

$$\begin{aligned} U(u\rightarrow 0)&= 1+ U_b^1u+ \frac{1}{4}(U_b^1)^2 u^2\nonumber \\&\quad + \frac{B^2}{6(V_b^0)^2} u^4\log u + U_b^4 u^4 + \cdots , \nonumber \\ V(u\rightarrow 0)&= V_b^0 + V_b^0 U_b^1u+ \frac{1}{4}V_b^0(U_b^1)^2 u^2\nonumber \\&\quad - \frac{B^2}{12 V_b^0} u^4\log u + V_b^4 u^4 + \cdots , \nonumber \\ W(u\rightarrow 0)&= W_b^0 + W_b^0 U_b^1 u + \frac{1}{4}W_b^0 (U_b^1)^2 u^2\nonumber \\&\quad + \frac{W_b^0 B^2}{6 (V_b^0)^2} u^4\log u + W_b^4 u^4 \cdots , \end{aligned}$$

(120)

where we have made use of the constraint equation (31). Obviously, the asymptotic boundary conditions only give rise to “two” effective requirements! The regularity requirements will yield another three conditions. Just as in the fixing of $c_2$, we can utilize the freedom of redefining the radial coordinate u and set $U_b^1=0$.

To summarize, the boundary conditions at $u=0$ (the AdS boundary) are

$$\begin{aligned} U'(u=0)=0,\quad V(u=0)=W(u=0) = 1, \end{aligned}$$

(121)

while at the event horizon $u=1$

$$\begin{aligned}&U(u=1)=0, \nonumber \\&U'(1)V'(1)-8V(1)-2U'(1)V(1)+\frac{2B^2}{3V(1)}=0, \nonumber \\&U'(1) W'(1)-2 W(1) U'(1)-8 W(1) -\frac{B^2 W(1)}{3 V(1)^2}=0. \end{aligned}$$

(122)

To find the numeric solutions, one can proceed in two different ways. The first approach would be to directly solve (119) under the boundary conditions (121) and (122). A second approach would be to replace the boundary conditions (121) by the following conditions at the horizon:

$$\begin{aligned}&U(u\rightarrow 1)= 0+ U^1_h(u-1)+ U_h^2(u-1)^2+\cdots , \nonumber \\&V(u\rightarrow 1)=V_h^0 + V_h^1(u-1)+\cdots , \nonumber \\&W(u\rightarrow 1)= W_h^0+ W_h^1(u-1)+ \cdots , \end{aligned}$$

(123)

where

$$\begin{aligned} U_h^1=-4, \quad V_h^0=1, \quad W_h^0=1. \end{aligned}$$

(124)

Note that the choice of $U_h^1$ will set $\pi T=1$. However, solving (119) under the initial conditions (122) and (124), near the boundary $u=0$ the solution will behave as

$$\begin{aligned}&U(u\rightarrow 0) \rightarrow 1,\quad V(u\rightarrow 0) \rightarrow v(b), \quad \nonumber \\&\quad W(u\rightarrow 0) \rightarrow w(b). \end{aligned}$$

(125)

Then the correct solution would be obtained by a further rescaling of the boundary coordinates,

$$\begin{aligned} x\rightarrow \sqrt{v(b)}x, \quad y\rightarrow \sqrt{v(b)}y,\quad z\rightarrow \sqrt{w(b)}z. \end{aligned}$$

(126)

Due to the “incorrect” asymptotic boundary behavior (125), we have relabeled the magnetic field by b in (125) and (126). When solving the EOMs (119) under the initial conditions (122) and (124), the same relabeling should be made. Recalling the definition of the magnetic field $F^V= b\mathrm{d}x \wedge dy$, the physical magnetic field B (in units of $r_h^2$) should be

$$\begin{aligned} B= \frac{b}{v(b)}. \end{aligned}$$

(127)

Finally, we would like to point out that the background solution obtained with conditions (122) and (124) does not necessarily satisfy $U_b^1=0$ (cf. (120)).

Appendix B: Horizon boundary conditions from matching

We first seek solutions to (47) through (53) near the horizon with ingoing boundary conditions. We obtain the following series solutions:

$$\begin{aligned}&{\delta }A_z=a_0(r-r_h)^{\beta }+a_1(r-r_h)^{{\beta }+1}+\cdots ,\nonumber \\&{\delta }V_{x}=b_0(r-r_h)^{\beta }+b_1(r-r_h)^{{\beta }+1}+\cdots ,\nonumber \\&{\delta }g_{ty}=c_1(r-r_h)^{{\beta }+1}+c_2(r-r_h)^{{\beta }+2}+\cdots ,\nonumber \\&{\delta }V_t=d_1(r-r_h)^{{\beta }+1}+d_2(r-r_h)^{{\beta }+2}+\cdots ,\nonumber \\&{\delta }g_{xy}=e_0(r-r_h)^{\beta }+e_1(r-r_h)^{{\beta }+1}+\cdots , \end{aligned}$$

(128)

with ${\beta }=-\frac{i{\omega }}{f'(r_h)}$. Here, $a_0$, $b_0$ and $e_0$ are free parameters, while all the rest coefficients are completely determined by them. For instance, $c_1$ and $d_1$ are

$$\begin{aligned} c_1&=\frac{if'(r_h)}{f'(r_h)-i{\omega }}\nonumber \\&\quad \times \left\{ -iBe^{-4W_T(r_h)} b_0 -4{\lambda }B^2 e^{-W_L(r_h)-6W_T(r_h)} q a_0 \right. \nonumber \\&\left. \quad +e^{-W_L(r_h)-2W_T(r_h)}q\right. \nonumber \\&\quad \times \left. \left[ e^{W_L(r_h)}e_0 +4{\lambda }f'(r_h) \left( W_L'(r_h) +2W_T'(r_h)\right) a_0 \right] \right\} ,\nonumber \\ d_1&=\frac{if'(r_h)}{f'(r_h)-i{\omega }} e^{-W_L(r_h)-2W_T(r_h)} \left[ e^{W_L(r_h)} q b_0 + 8iB{\alpha }a_0\right] . \end{aligned}$$

(129)

The three parameters $a_0$, $b_0$ and $e_0$ do not match the five sources to the fields ${\delta }A_z$, ${\delta }V_x$, ${\delta }g_{ty}$, ${\delta }V_t$ and ${\delta }g_{xy}$. The remaining two parameters come from pure gauge solutions, which are by gauge transformation of the trivial solution

$$\begin{aligned} {\delta }V_x&=-BC_1,\quad {\delta }g_{ty}=-i{\omega }C_1,\quad {\delta }g_{xy}=iqC_1,\quad \nonumber \\&\qquad \text {others}=0;\nonumber \\ {\delta }V_t&=-i{\omega }C_2,\quad {\delta }V_x=iqC_2,\quad \text {others}=0. \end{aligned}$$

(130)

The horizon solutions are to be matched with the lowest order solutions in (54) near the horizon region. Since the horizon solutions also contain ${\delta }V_x$ and ${\delta }g_{xy}$, we also need the lowest order solutions to them. To the lowest order in ${\omega }$, the EOMs of ${\delta }V_x$ and ${\delta }g_{xy}$ decouple:

$$\begin{aligned}&\frac{1}{2}iq{\partial }_r{\delta }g_{xy}+\frac{1}{2}Be^{-2W_T}{\partial }_r{\delta }V_x=0,\nonumber \\&{\partial }_r\left( e^{W_L+2W_T}f{\partial }_r{\delta }g_{xy}\right) =0,\nonumber \\&{\partial }_r\left( e^{W_L}f{\partial }_r{\delta }V_{x}\right) =0. \end{aligned}$$

(131)

They are clearly solved by constant solutions. Matching with the horizon solutions, we simply have

$$\begin{aligned} {\delta }V_x=b_0,\quad {\delta }g_{xy}=e_0. \end{aligned}$$

(132)

Note that we can set the above two solutions to zero by adding pure gauge solutions. In the limit ${\omega }\rightarrow 0$, the pure gauge solutions do not change the horizon values of ${\delta }g_{ty}$, ${\delta }V_t$ and ${\delta }A_z$:

$$\begin{aligned}&{\delta }g_{ty}(r=r_h)=0,\quad {\delta }V_t(r=r_h)=0,\quad \nonumber \\&\quad {\delta }A_z(r=r_h)=\mathrm{constant}. \end{aligned}$$

(133)

The ${\omega }\rightarrow 0$ limit of (129) determines the horizon derivative of ${\delta }g_{ty}$ and ${\delta }V_t$. For the decoupled EOMs (55) through (57), we can take the horizon derivatives of ${\delta }g_{ty}$ and ${\delta }V_t$, and horizon value of ${\delta }A_z$ as free parameters.

Appendix C: Magnetic and magneto-vortical susceptibilities

In this appendix, we calculate the magnetic susceptibility $2p_{,B^2}$ and magneto-vortical susceptibility $M_{{\Omega }}$ independently as a confirmation to our claim in (82).

Let us begin with the magnetic susceptibility $2p_{,B^2}$ and the magnetization M. For the equilibrium state (corresponding to the magnetic brane background), the stress tensor for the boundary theory is computed as

$$\begin{aligned}&T^{tt}=\lim _{r\rightarrow \infty } \left\{ -2r^6 \left[ -\frac{3}{f(r)} +\frac{W_L^\prime + 2 W_T^\prime }{\sqrt{f(r)}} - \frac{B^2 e^{-4W_T}r^6 \log r}{2f(r)} \right] \right\} , \nonumber \\&T^{xx}=T^{yy}=\lim _{r\rightarrow \infty }\left\{ -2r^6 \left[ 3e^{-2W_T} - \frac{e^{-2W_T}f^\prime (r)}{2\sqrt{f(r)}} \right. \right. \nonumber \\&\left. \left. - \sqrt{f(r)} e^{-2 W_T} (W_L^\prime + W_T^\prime ) \right] - \frac{1}{2}B^2 e^{-6 W_T} r^6 \log r \right\} , \nonumber \\&T^{zz}=\lim _{r\rightarrow \infty } \left\{ - 2r^6 \left[ 3e^{-2W_L}- \frac{e^{-2W_L}f^\prime (r)}{2\sqrt{f(r)}} -2 e^{-2W_L} \sqrt{f(r)} W_T^\prime \right] \right. \nonumber \\&\left. + \frac{1}{2} B^2 e^{-2W_L- 4W_T} r^6\log r\right\} . \end{aligned}$$

(134)

With the analytical solution presented in Appendix A, it is straightforward to compute the various components of $T^{\mu \nu }$:

$$\begin{aligned}&T^{tt}=3r_h^4+3(r_p^4-r_h^4)- \frac{1}{2}B^2\log r_h + {\mathcal {O}}(B^3), \nonumber \\&T^{xx}=T^{yy}= r_h^4+(r_p^4-r_h^4) -\frac{1}{6}B^2 - \frac{1}{2}B^2 \log r_h + {\mathcal {O}}(B^3), \nonumber \\&T^{zz}= r_h^4+(r_p^4-r_h^4) - \frac{1}{6}B^2+ \frac{1}{2}B^2 \log r_h + {\mathcal {O}}(B^3). \end{aligned}$$

(135)

To extract the energy density, pressure and magnetization, we compare (135) with the MHD formalism [43] (see Eq. 14 there). Here, we would like to point out that the AdS/CFT computations give rise to the medium contributions (denoted as $T^{\mu \nu }_{\mathrm{F0}}$ in [43]). Consequently,

$$\begin{aligned} \varepsilon = T^{tt}, \quad p_\perp = T^{xx}= T^{yy}, \quad p_\parallel = T^{zz}. \end{aligned}$$

(136)

The pressure p is identified as $p_\parallel$

$$\begin{aligned} p=p_\parallel = \frac{1}{2}B^2 \log r_h + {\mathcal {O}}(B^3)\Longrightarrow 2p_{,B^2}= \log r_h, \end{aligned}$$

(137)

where we used the perturbative expression for $r_p$ in (116). The magnetization M could be computed as

$$\begin{aligned} p_\perp = p_\parallel -MB \Longrightarrow M= B \log r_h+ {\mathcal {O}}(B^2). \end{aligned}$$

(138)

Now we move on to the calculation of the magneto-vortical susceptibility $M_{\Omega }$. In the zero magnetic field situation, we calculate $M_{\Omega }$ based on the following Kubo formula [33]:

$$\begin{aligned} M_\Omega = -\lim _{q_x, q_y\rightarrow 0} \frac{\langle T^{ty} J^x \rangle }{q_y q_x}. \end{aligned}$$

(139)

Since $M_{\Omega }$ is C-odd, we need to consider the finite density RN-$AdS_5$ background:

$$\begin{aligned}&\mathrm{d}s^2= -f(r) \mathrm{d}t^2+ \frac{\mathrm{d}r^2}{f(r)} + r^2(\mathrm{d}x^2+ \mathrm{d}y^2+ \mathrm{d}z^2), \nonumber \\&V= \left( \frac{Q}{r_h^2}- \frac{Q}{r^2}\right) \mathrm{d}t, \quad A=0, \end{aligned}$$

(140)

where

$$\begin{aligned} f(r)= r^2\left( 1-\frac{r_h^4}{r^4} + \frac{Q^2}{3r^6}- \frac{Q^2}{3r_h^2 r^4}\right) . \end{aligned}$$

(141)

For consistency, we turn on the following fluctuations on top of (140):

$$\begin{aligned} \delta (\mathrm{d}s^2)&= 2r^2 \left[ \delta g_{tx}(r,x,y)\mathrm{d}t\mathrm{d}x+ \delta g_{ty}(r,x,y)\mathrm{d}t\mathrm{d}y \right] , \nonumber \\ \delta V&= \delta V_x(r,x,y)\mathrm{d}x+ \delta V_y(r,x,y)\mathrm{d}y. \end{aligned}$$

(142)

We turn attention to the Fourier space by assuming a plane wave ansatz for the fluctuations,

$$\begin{aligned}&\delta g_{tx}(r,x,y)\sim e^{i q_x x+ iq_y y} \delta g_{tx}(r),\nonumber \\&\delta g_{ty}(r,x,y)\sim e^{i q_x x+ iq_y y} \delta g_{ty}(r),\nonumber \\&\delta V_x(r,x,y)\sim e^{i q_x x+ iq_y y} \delta V_x(r),\nonumber \\&\delta V_y(r,x,y)\sim e^{i q_x x+ iq_y y} \delta V_y(r). \end{aligned}$$

(143)

To proceed, we collect the EOMs for the fluctuations in (142). The constraint equations $E_{rt}=0$ and $EV^r=0$ give rise to

$$\begin{aligned}&r^{-2}f(r)\partial _r\left( q_x \delta g_{tx} + q_y \delta g_{ty}\right) \nonumber \\&\qquad - \left( r^{-2}f(r) \right) ^\prime \left( q_x \delta g_{tx} + q_y \delta g_{ty}\right) =0, \end{aligned}$$

(144)

$$\begin{aligned}&rf(r) \partial _r(q_x \delta V_x+ q_y \delta V_y) + 2Q (q_x \delta g_{tx} + q_y \delta g_{ty})=0. \end{aligned}$$

(145)

The dynamical components of the Einstein equations $E_{tx}=E_{ty}=0$ are

$$\begin{aligned}&\partial _r(r^5 \partial _r \delta g_{tx}) + 2Q \partial _r \delta V_x +\frac{r^3}{f(r)} \nonumber \\&\qquad \times \left( q_x q_y \delta g_{ty}- q_y^2 \delta g_{tx}\right) =0, \end{aligned}$$

(146)

$$\begin{aligned}&\partial _r(r^5 \partial _r \delta g_{ty}) + 2Q \partial _r \delta V_y +\frac{r^3}{f(r)} \left( q_x q_y \delta g_{tx}- q_x^2 \delta g_{ty}\right) =0. \end{aligned}$$

(147)

The dynamical components of the vector Maxwell equations $EV^x=EV^y=0$ are

$$\begin{aligned}&\partial _r\left[ rf(r)\partial _r \delta V_x\right] + 2Q \partial _r \delta g_{tx} + \frac{1}{r} \left( q_x q_y \delta V_y- q_y^2 \delta V_x\right) =0, \end{aligned}$$

(148)

$$\begin{aligned}&\partial _r\left[ rf(r)\partial _r \delta V_y\right] + 2Q \partial _r \delta g_{ty} + \frac{1}{r} \left( q_x q_y \delta V_x- q_x^2 \delta V_y\right) =0. \end{aligned}$$

(149)

Near the AdS boundary, we impose

$$\begin{aligned} \delta V_x \xrightarrow []{r\rightarrow \infty } v_x, \quad \mathrm{others} \xrightarrow []{r\rightarrow \infty } 0, \end{aligned}$$

(150)

while at the horizon we have

$$\begin{aligned} \delta g_{tx}, \delta g_{ty} \xrightarrow []{r\rightarrow r_h} 0, \quad \delta V_x, \delta V_y ~\mathrm{are~regualr~at}~r=r_h. \end{aligned}$$

(151)

Given (139), we solve (146)–(149) in the small momenta limit. The fluctuation modes could be expanded as

$$\begin{aligned}&\delta g_{tx}= \delta g_{tx}^{(0)} + \epsilon \,\delta g_{tx}^{(1)} + \epsilon ^2 \delta g_{tx}^{(2)}, \nonumber \\&\delta g_{ty}= \delta g_{ty}^{(0)} + \epsilon \,\delta g_{ty}^{(1)} + \epsilon ^2 \delta g_{ty}^{(2)}, \nonumber \\&\delta V_x= \delta V_x^{(0)} + \epsilon \,\delta V_x^{(1)} + \epsilon ^2 \delta V_x^{(2)}, \nonumber \\&\delta V_y= \delta V_y^{(0)} + \epsilon \, \delta V_y^{(1)} + \epsilon ^2 \delta V_y^{(2)}, \end{aligned}$$

(152)

where $\epsilon \sim q_x, q_y$.

At the lowest order ${\mathcal {O}}(\epsilon ^0)$, the solutions are simply given by

$$\begin{aligned} \delta V_x^{(0)}= v_x, \quad \delta g_{tx}^{(0)}= \delta g_{ty}^{(0)}= \delta V_y^{(0)}=0. \end{aligned}$$

(153)

At the first order ${\mathcal {O}}(\epsilon ^1)$, there are no non-trivial solutions,

$$\begin{aligned} \delta g_{tx}^{(1)}= \delta g_{tx}^{(1)}= \delta V_y^{(1)}= \delta V_y^{(1)}=0. \end{aligned}$$

(154)

At the second order ${\mathcal {O}}(\epsilon ^2)$, the equations we need are

$$\begin{aligned}&\partial _r(r^5 \partial _r \delta g_{ty}^{(2)}) +2 Q \partial _r \delta V_y^{(2)}=0, \end{aligned}$$

(155)

$$\begin{aligned}&\partial _r\left[ rf(r)\partial _r \delta V_y^{(2)}\right] + 2Q \partial _r \delta g_{ty}^{(2)} + \frac{q_x q_y}{r} \delta V_x^{(0)}=0. \end{aligned}$$

(156)

In $\delta g_{ty}^{(2)}$, we will track the term linear in Q ($\sim \mu$) only. Therefore, the Q-term in (156) could be discarded. Furthermore, we can simply take $f(r) \rightarrow r^2(1-r_h^4/r^4)$. Equation (156) is solved as

$$\begin{aligned} \delta V_y^{(2)}= -q_x q_y v_x \int _{r}^{\infty } \frac{\log (r_h/x)}{x^3(1-r_h^4/x^4)} \mathrm{d}x. \end{aligned}$$

(157)

Finally, Eq. (155) is solved as

$$\begin{aligned} \delta g_{ty}^{(2)}= 2Q q_x q_y v_x \int _r^\infty \frac{\mathrm{d}x}{x^5} \int _{r_h}^x \frac{\log (r_h/y)}{y^3(1-r_h^4/y^4)}\mathrm{d}y + \frac{C_0}{r^4}, \end{aligned}$$

(158)

where the integration constant $C_0$ is fixed as

$$\begin{aligned}&2Q q_x q_y v_x \int _{r_h}^\infty \frac{\mathrm{d}x}{x^5} \int _{r_h}^x \frac{\log (r_h/y)}{y^3(1-r_h^4/y^4)}\mathrm{d}y + \frac{C_0}{r_h^4}\nonumber \\&\quad =0 \Rightarrow C_0= \frac{\pi ^2-8}{64r_h^2} Q q_x q_y v_x. \end{aligned}$$

(159)

Near the AdS boundary,

$$\begin{aligned} \delta g_{ty}^{(2)} \xrightarrow []{r\rightarrow \infty } - \frac{Qq_x q_y v_x}{8 r_h^2 r^4}+ {\mathcal {O}}(r^{-5}), \end{aligned}$$

(160)

which is translated to

$$\begin{aligned} T^{ty}= \frac{Q}{2r_h^2} q_x q_y v_x \Rightarrow M_\Omega = -\frac{Q}{2r_h^2} = - \frac{1}{2} \mu . \end{aligned}$$

(161)