# Irreversible Boltzmann samplers in dense liquids: weak-coupling
approximation and mode-coupling theory

Federico Ghimenti  Laboratoire Matière et Systèmes Complexes (MSC), Université
Paris Cité & CNRS (UMR 7057), 75013 Paris, France  Ludovic Berthier
Laboratoire Charles Coulomb (L2C), Université de Montpellier & CNRS (UMR
5221), 34095 Montpellier, France  Gulliver, UMR CNRS 7083, ESPCI Paris, PSL
Research University, 75005 Paris, France  Grzegorz Szamel  Department of
Chemistry, Colorado State University, Fort Collins, Colorado 80523, United
States of America  Frédéric van Wijland  Laboratoire Matière et Systèmes
Complexes (MSC), Université Paris Cité & CNRS (UMR 7057), 75013 Paris, France

###### Abstract

Exerting a nonequilibrium drive on an otherwise equilibrium Langevin process
brings the dynamics out of equilibrium but can also speedup the approach to
the Boltzmann steady-state. Transverse forces are a minimal framework to
achieve dynamical acceleration of the Boltzmann sampling. We consider a simple
liquid in three space dimensions subjected to additional transverse pairwise
forces, and quantify the extent to which transverse forces accelerate the
dynamics. We first explore the dynamics of a tracer in a weak coupling regime
describing high temperatures. The resulting acceleration is correlated with a
monotonous increase of the magnitude of odd transport coefficients (mobility
and diffusivity) with the amplitude of the transverse drive. We then develop a
nonequilibrium version of the mode-coupling theory able to capture the effect
of transverse forces, and more generally of forces created by additional
degrees of freedom. Based on an analysis of transport coefficients, both odd
and longitudinal, both for the collective modes and for a tracer particle, we
find a systematic acceleration of the dynamics. Quantitatively, the gain,
which is guaranteed throughout the ergodic phase, turns out to be a decreasing
function of temperature beyond a temperature crossover, in particular as the
glass transition is approached. Our theoretical results are in good agreement
with available numerical results.

##  I Introduction

When interested in the dynamics of dense liquids [1], the mode-coupling
approximation [2, 3] is a method of choice. It is a versatile tool designed to
analyze the dynamics of simple liquids and of other soft-matter systems in
realistic space dimensions ($d=2$ or $3$). It has famously been applied to
critical dynamics [4], to glass-formers [3], to polymers [5] and colloidal
assemblies [6, 7] up to hard-condensed matter systems [8]. These systems share
the common trait that they are in thermal equilibrium. In recent years, the
mode-coupling approximation has been extended to some nonequilibrium settings,
such as sheared liquids [9], granular fluids [10, 11], or, even more recently,
to systems of self-propelled particles [12, 13, 14, 15]. Here, our goal is to
use these works as an inspiration to develop a mode-coupling approximation for
systems of particles driven by nonequilibrium forces that nevertheless sample
the Boltzmann distribution in their steady-state. Such systems are of special
interest because, for specifically designed nonequilibrium forces, they can
sample the Boltzmann distribution faster than their reversible equilibrium
counterparts. This is of course an interesting property when dynamical
evolution is intrinsically sluggish, as in dense and cold liquids [16].

When confronted with the problem of sampling the Boltzmann distribution for a
system of interacting particles, a practitioner of simulations can resort to
several standard techniques [17, 18] such as Metropolis-Hastings Monte Carlo,
Brownian and Newtonian Dynamics. These techniques endow the system with
reversible dynamics, ensuring that at long times the equilibrium Boltzmann
distribution is sampled. It is a practical challenge to reduce as much as
possible the time required to achieve convergence to the equilibrium
distribution. This convergence time proves overwhelmingly long for many
systems of interest, from proteins to supercooled liquids and neural networks.
Recently, several techniques were proposed to accelerate the dynamics [16],
including the very efficient swap Monte Carlo algorithm that achieves dramatic
speedup for size polydisperse liquids [19, 20].

An alternative approach, which has recently garnered attention, is to exploit
an irreversible dynamics, namely one that does not respect detailed balance,
but that at the same time ensures that the Boltzmann distribution is properly
sampled. The idea is that by going out-of-equilibrium a faster exploration of
the phase space is sometimes possible. To achieve this goal one could add, for
example, a drift perpendicular to the gradient flow of the system, allowing
for an out of equilibrium dynamics with the aforementioned property. We will
refer to such an orthogonal drift as transverse forces. The possibility of
using transverse forces was first envisaged in [21], where it was also shown
that the presence of a nonequilibrium drive increases the spectral gap of the
associated dynamical evolution operator, allowing for shorter convergence
time. A similar proof was later discussed in [22] in the context of Markov
Chains. More detailed analysis of the exact amount of the acceleration were
carried out in the case of a particle in a harmonic well [23, 24]. Generic
inequalities concerning asymptotic variance (a projection of the convergence
time) [25, 26] and large deviation functions [27, 28, 29] were also obtained,
all demonstrating a better performance of the transverse force dynamics with
respect to its equilibrium counterpart, in agreement with existing
mathematical theorems.

The acceleration cannot however increase indefinitely. In [30] it was shown
that the spectral gap saturates in the limit of a transverse force with
infinite amplitude, although little is known about this actual limiting value.
As a matter of fact, despite the abundance of mathematical inequalities and
rigorous bounds, there exist very few quantitative results regarding the
speedup actually obtained in systems of theoretical and practical interest.
Efforts in this direction consist in some studies on benchmark neural network
problems [31, 32, 33, 34], in numerical results on double well and $XY$ models
[35], and the zero-range process [28]. Results on the barrier crossing problem
in the presence of transverse forces were also obtained [36]. Previously, some
of us studied the effect of transverse forces on the hard sampling problem of
a mean-field spin glass belonging to the RFOT universality class [37]. In a
recent Letter [38], we addressed the dynamics of dense liquids by a
combination of numerical methods and of analytical approaches, based on the
dynamical-mean-field theory and on the mode-coupling approximation. The former
infinite-dimensional approach has been further detailed in [39]. The present
work addresses the acceleration witnessed in finite-dimensional simulations
[38] by exploring the corresponding phenomenology through the lens of the
mode-coupling approximation. We wish to quantify theoretically the
acceleration provided by irreversible dynamics in a generic system of
interacting particles in finite dimensions.

After a presentation of our model in Sec. II, we devote Sec. III to a study of
the dynamics in the simpler weak coupling approximation, achieved for instance
in a high-temperature regime. We use this approach to qualitatively explain
the acceleration that transverse forces exert and the emergence of odd
transport coefficients that help characterize these driven dynamics. In Sec.
IV we present the core of our mode-coupling approximation for transverse
forces. The standard schematic approximation is shown to be missing the gist
of transverse forces which leads us to present an improved version of the
schematic approach. The emergence of odd transport is also discussed. In Sec.
V, we complement the analysis of collective modes by an in-depth description
of the dynamics of a tracer particle. In particular, the mismatch between the
diffusivity tensor and the mobility tensor helps us shape a more intuitive
picture of how acceleration operates. In [38] it was argued that the basic
physics at work in the acceleration provided by transverse forces was of a
similar nature as that induced by what is known in the mathematical community
as lifting [40]. These two families of nonequilibrium drives are not usually
treated on an equal footing. By constructing a mode-coupling approach for a
lifted process, and by comparing our results with those of Sec. IV, we are
able to better substantiate our claim. Conclusions and outlook are finally
gathered in Sec. VIII, with pointers in the direction of applications to
chiral active matter.

##  II Dynamics with transverse forces

Our starting point is the following dynamics for a fluid of interacting
particles with positions $\text{\bf r}_{i}$ in three dimensions [38]

$\mathbf{\dot{r}}_{i}=\mu_{0}\left(\mathbf{1}+\gamma\text{\bf
A}\right)\mathbf{F}_{i}+\sqrt{2\mu_{0}T}\bm{\xi}_{i},$ (1)

with
$\mathbf{F}_{i}\equiv-\sum_{i}\bm{\nabla}_{i}V(\lvert\mathbf{r}_{i}-\mathbf{r}_{j}\rvert)$
a conservative force arising from a pairwise, isotropic interaction potential,
and the $\bm{\xi}_{i}$’s are independent Gaussian white noises with
independent components. The bare mobility $\mu_{0}$ and the temperature $T$
are related to the diffusion constant of a free particle $D_{0}$ through
Einstein’s relation $D_{0}=\mu_{0}T$, setting the Boltzmann constant to unity.
This dynamics differs from its overdamped equilibrium counterpart due to the
presence of an antisymmetric matrix $\text{\bf A}=-\text{\bf A}^{T}$. The
resulting additional transverse force $\text{\bf A}\text{\bf
F}_{i}=A_{\alpha\beta}F_{i,\beta}$ injects into the system a nonequilibrium
current that nevertheless preserves the Boltzmann distribution in the
nonequilibrium steady-state. This means in particular that in the presence of
transverse forces, the dressed mobility and diffusivity will pick up an odd
component (as most other transport coefficients, such as viscosity), and we do
not expect the Einstein relation to extend to the full dressed diffusivity and
mobility tensors.

We would now like to take a few lines to discuss how we choose the matrix A.
For concreteness, we control the strength of the nonequilibrium drive through
the parameter $\gamma$ while imposing the condition that the Frobenius norm of
A is unity, namely

$\sqrt{\sum_{i,j}A_{ij}^{2}}=1.$ (2)

Under this condition, the matrix A reads

$\text{\bf A}=\begin{bmatrix}0&A_{12}&A_{13}\\\ -A_{12}&0&A_{23}\\\
-A_{13}&-A_{23}&0\end{bmatrix},$ (3)

with $\sqrt{A_{12}^{2}+A_{13}^{2}+A_{23}^{2}}=1$. We now show that upon a
suitable set of rotations of coordinates the matrix A can be recast in a
simpler form. Indeed, if R is an orthogonal matrix representing a rotation in
three dimensions, i.e. $\det(\text{\bf R})=1$ and $\text{\bf R}^{T}\text{\bf
R}=\mathbf{1}$, then applying R to both sides of Eq. (1) yields

$\mathbf{\dot{r}^{\prime}}_{i}=\mu_{0}\left(\mathbf{1}+\gamma\text{\bf
A}^{\prime}\right)\mathbf{F}^{\prime}_{i}+\sqrt{2\mu_{0}T}\bm{\xi}^{\prime}_{i},$
(4)

where we have defined $\text{\bf r}_{i}^{\prime}\equiv\text{\bf R}\text{\bf
r}_{i}$, $\text{\bf F}_{i}^{\prime}\equiv\text{\bf R}\text{\bf F}_{i}$,
$\bm{\xi}_{i}^{\prime}\equiv\text{\bf R}\bm{\xi}_{i}$ and $\text{\bf
A}^{\prime}\equiv\text{\bf R}\text{\bf A}\text{\bf R}^{T}$. The dynamics in
Eq. (4) is the same as in Eq. (1), in a rotated reference frame. Of course the
force $\text{\bf F}^{\prime}_{i}$ is still a central force in the rotated
reference frame,

$\displaystyle F^{\prime}_{i,\alpha}$
$\displaystyle=\sum_{j}R_{\alpha\beta}\partial_{r_{i,\beta}}V(\lvert\text{\bf
r}_{i}-\text{\bf r}_{j}\rvert)$ (5)
$\displaystyle=\sum_{j}R_{\alpha\beta}\frac{\partial r_{i,\beta}}{\partial
r^{\prime}_{i,\gamma}}\partial_{r^{\prime}_{i,\gamma}}V(\lvert\text{\bf
r}_{i}-\text{\bf r}_{j}\rvert)$
$\displaystyle=\sum_{j}R_{\alpha\beta}R^{T}_{\beta\gamma}\partial_{r^{\prime}_{i,\gamma}}V(\lvert\text{\bf
r}^{\prime}_{i}-\text{\bf r}^{\prime}_{j}\rvert)$
$\displaystyle=\sum_{j}\partial_{r^{\prime}_{i,\alpha}}V(\lvert\text{\bf
r}^{\prime}_{i}-\text{\bf r}^{\prime}_{j}\rvert),$

and the noises $\bm{\xi}_{i}^{\prime}$ have the same statistics as the noises
$\bm{\xi}_{i}$:

$\displaystyle\left\langle\xi^{\prime}_{i,\alpha}(t)\xi^{\prime}_{j,\beta}(t^{\prime})\right\rangle$
$\displaystyle=R_{\alpha\alpha^{\prime}}R_{\beta\beta^{\prime}}\left\langle\xi_{i,\alpha^{\prime}}(t)\xi_{j,\beta^{\prime}}(t^{\prime})\right\rangle$
(6)
$\displaystyle=R_{\alpha\alpha^{\prime}}R_{\beta\beta^{\prime}}\delta_{ij}\delta_{\alpha^{\prime}\beta^{\prime}}\delta(t-t^{\prime})$
$\displaystyle=\delta_{ij}\delta_{\alpha\beta}\delta(t-t^{\prime}).$

The matrix $\text{\bf A}^{\prime}$ is still an antisymmetric matrix satisfying
Eq. (2). Without loss of generality, we can take for A the following explicit
form:

$\text{\bf A}\equiv\begin{bmatrix}0&-1&0\\\ 1&0&0\\\ 0&0&0\end{bmatrix},$ (7)

which is mathematically similar to A in Eq. (3) up to a rotation $\text{\bf
R}=\text{\bf R}_{\text{\bf x}}(\theta)\text{\bf R}_{\text{\bf z}}(\phi)$, with
$\text{\bf R}_{\text{\bf x}}(\theta)$, $\text{\bf R}_{\text{\bf z}}(\phi)$
rotation around the $x$ and $z$ axis respectively, satisfying
$\tan(\phi)=-\frac{A_{23}}{A_{13}}$,
$\tan(\theta)=-\frac{1}{A_{12}}\sqrt{A^{2}_{13}+A_{23}^{2}}$. Note that for a
nonzero matrix A the dynamics is no longer isotropic. It is invariant only by
rotations generated by the matrix A in Eq. (7). We identify the plane left
invariant by this rotations as the $(xy)$ plane.

We have now set the stage for our analysis of transverse forces. We begin with
a weak-coupling approximation, which follows an approach often termed the
random-phase approximation [41].

##  III Weak coupling approximation

Before entering the details of the mode-coupling theory derivation, it is
useful to consider the problem of a tracer diffusing in a system interacting
with transverse forces in the special case in which fluctuations in the
surrounding fluid can be considered as weak. Following [42], we assume that
the tracer produces a small perturbation in the density field of its
environment, a procedure known as the random phase approximation, and derive
an expression for the transport coefficients of the tracer, namely the
longitudinal and the odd diffusion constants. This gives a first grasp on the
speedup generated by transverse forces. In spite of throwing an interesting
physical light, this approach does not easily extend to low temperatures where
the weak coupling hypothesis is not fulfilled.

###  III.1 Equation of motion for a tracer

We consider a tagged particle with position $\text{\bf r}_{0}$ diffusing in a
fluid under the action of transverse forces. We include the possibility for
the tagged particle to be subjected to a constant external force $\text{\bf
F}_{\text{ext}}$, which will be used to probe the mobility of the tagged
particle. The equation of motion for the tracer is

$\begin{split}\dot{\text{\bf r}}_{0}&=\mu_{0}\text{\bf
F}_{\text{ext}}-\mu_{0}(\mathbf{1}+\gamma\text{\bf
A})\sum_{i>0}\bm{\nabla}_{0}V(\text{\bf r}_{0}(t)-\text{\bf r}_{i}(t))\\\
&+\sqrt{2\mu_{0}T}\bm{\xi}_{0}.\end{split}$ (8)

This is Eq. (1), with the role of particle $0$ being singled out. All other
particles in the fluid also interact according to Eq. (1). We introduce the
density field of the fluid without the tracer, $n(\text{\bf
r},t)\equiv\sum_{i>0}\delta(\text{\bf r}-\text{\bf r}_{i}(t))$ and we can then
rewrite Eq. (8) as

$\begin{split}\dot{\text{\bf r}}_{0}&=\mu_{0}\text{\bf
F}_{\text{ext}}-(\mathbf{1}+\gamma\text{\bf A})(\bm{\nabla}V*n)(\text{\bf
r}_{0}(t),t)\\\ &+\sqrt{2\mu_{0}T}\bm{\xi}_{0}(t).\end{split}$ (9)

with the convolution between two functions denoted by $*$: $(f*g)(\text{\bf
r})\equiv\int d\text{\bf r}f(\text{\bf r}-\text{\bf r}^{\prime})g(\text{\bf
r}^{\prime})$. Equation (9) needs to be complemented with the Dean-Kawasaki
[43] equation for $n(\text{\bf r},t)$:

$\begin{split}\partial_{t}n(\text{\bf
r},t)&=\mu_{0}T\bm{\nabla}^{2}n(\text{\bf r},t)\\\
&+\mu_{0}\bm{\nabla}\cdot\left[\left(\mathbf{1}+\gamma\text{\bf
A}\right)n(\text{\bf r},t)(\bm{\nabla}V*n)(\text{\bf r},t)\right]\\\
&+\mu_{0}\bm{\nabla}\cdot\left[(\mathbf{1}+\gamma\text{\bf A})\cdot
n(\text{\bf r},t)\bm{\nabla}V(\text{\bf r}-\text{\bf r}_{0}(t))\right]\\\
&+\bm{\nabla}\cdot\sqrt{2T\mu_{0}n(\text{\bf r},t)}\bm{\chi}(\text{\bf
r},t),\end{split}$ (10)

where the Gaussian noise $\bm{\xi}(\text{\bf r},t)$ has correlations
$\left\langle\bm{\xi}(\text{\bf r},t)\otimes\bm{\xi}(\text{\bf
r}^{\prime},t^{\prime})\right\rangle=\mathbf{1}\delta(\text{\bf r}-\text{\bf
r}^{\prime})\delta(t-t^{\prime})$. In order to solve Eqs. (9, 10) we resort to
the random phase approximation.

###  III.2 Random phase approximation

In the language of the collective coordinate $n$, the random phase
approximation is a simple linearization of the Dean-Kawasaki equation (10) for
the local and fluctuating density field $n(\text{\bf
r},t)=\sum_{i}\delta(\text{\bf r}-\text{\bf r}_{i}(t))$ [42, 44, 45, 46].
Assuming that the homogeneous density of the system
$\rho_{0}\equiv\frac{N}{V}$ is large enough so that density fluctuations
remain small, we split $n(\text{\bf r},t)$ into

$n(\text{\bf r},t)=\rho_{0}\left(1+\frac{1}{\sqrt{\rho_{0}}}\phi(\text{\bf
r},t)\right)$ (11)

with $\phi(\text{\bf r},t)$ a fluctuating field. To linear order in
$\phi/\sqrt{\rho_{0}}$, which we assume to be small, Eq. (10) becomes

$\begin{split}\partial_{t}\phi=&\mu_{0}T\bm{\nabla}^{2}\phi+\mu_{0}\bm{\nabla}\cdot\left(\bm{\nabla}\rho_{0}V*\phi\right)(\text{\bf
r},t)\\\ &+\mu_{0}\bm{\nabla}^{2}\sqrt{\rho_{0}}V(\text{\bf r}-\text{\bf
r}_{0}(t))+\bm{\nabla}\cdot\sqrt{2T\mu_{0}}\bm{\chi}(\text{\bf
r},t).\end{split}$ (12)

In deriving Eq. (12) from Eq. (10) we used the fact that, to linear order in
$\phi/\sqrt{\rho_{0}}$, $\bm{\nabla}\cdot\gamma\text{\bf A}n(\text{\bf
r},t)(\bm{\nabla}V*n)(\text{\bf r},t)=0$ because A is anti-symmetric, while
the nonzero contribution from $\bm{\nabla}\cdot\gamma\text{\bf A}\cdot
n(\text{\bf r},t)\bm{\nabla}V(\text{\bf r}-\text{\bf r}_{0}(t))$ is of order
$\rho_{0}^{-1}$. The linearized equation for the density field is therefore
not directly influenced by the presence of the transverse forces. Instead the
latter manifestly appear in the equation of motion of the tracer. To proceed
toward the weak coupling approximation, we consider the case where the
interaction potential between the particle is soft, with $V(r)$ finite as
$r\to 0$. Upon introducing the Fourier transform of a function $f$ by the
convention $f(\text{\bf k})\equiv\int\differential\text{\bf
r}\text{e}^{-i\text{\bf k}\cdot\text{\bf r}}f(\text{\bf r})$, we obtain from
Eq. (9):

$\begin{split}\dot{\text{\bf r}}_{0}(t)&=\mu_{0}\text{\bf F}_{\text{ext}}\\\
&-i\mu_{0}(\mathbf{1}+\gamma\text{\bf A})\int\frac{\differential\text{\bf
k}}{(2\pi)^{d}}\text{e}^{i\text{\bf k}\cdot\text{\bf r}_{0}(t)}\text{\bf
k}\sqrt{\rho_{0}}V(\text{\bf k})\phi(\text{\bf k},t)\\\
&+\sqrt{2T\mu_{0}}\bm{\xi}_{0}(t),\end{split}$ (13)

where $\gamma$ appears explicitly in the right-hand side. It is formally
possible to express $\phi(\text{\bf k},t)$ as a functional of the tracer
position $\text{\bf r}_{0}$ upon integrating Eq. (12), and then Eq. (13)
becomes a self-contained equation for $\text{\bf r}_{0}$. We assume that the
bath and the tracer have evolved for very long times before we start observing
them so that the initial condition for the density fluctuation $\phi(\text{\bf
k},t)$ can be overlooked, namely

$\begin{split}\phi(\text{\bf k},t)&=\int_{-\infty}^{t}\differential
s\Biggl{[}-\mu_{0}k^{2}\sqrt{\rho_{0}}V(\text{\bf k})e^{-i\text{\bf
k}\cdot\text{\bf r}_{0}(s)}\\\ &+i\sqrt{2\mu_{0}T}\text{\bf
k}\cdot\bm{\chi}(\text{\bf
k},s)\Biggr{]}e^{-\mu_{0}Tk^{2}\left[1+\beta\rho_{0}V(\text{\bf
k})\right](t-s)},\end{split}$ (14)

the correlations of the noise in Fourier space being
$\langle\bm{\chi}(\text{\bf k},t)\otimes\bm{\chi}(\text{\bf
k}^{\prime},t^{\prime})\rangle=\mathbf{1}(2\pi)^{d}\delta(t-t^{\prime})\delta(\text{\bf
k}+\text{\bf k}^{\prime})$. Substitution of Eq. (14) into Eq. (13) leads to

$\begin{split}\dot{\text{\bf r}}_{0}(t)&=\mu_{0}\text{\bf
F}_{\text{ext}}+\mu_{0}\int_{-\infty}^{t}\differential s\text{\bf F}(\text{\bf
r}_{0}(t)-\text{\bf r}_{0}(s),t-s)\\\ &+\bm{\Xi}(\text{\bf
r}(t),t)+\sqrt{2\mu_{0}T}\bm{\xi}_{0}(t).\end{split}$ (15)

The interaction between the tracer and the bath splits into a deterministic
force

$\begin{split}\text{\bf F}(\text{\bf r}_{0}(t),t)&\equiv
i\mu_{0}\int\frac{\differential\text{\bf
k}}{(2\pi)^{3}}(\mathbf{1}+\gamma\text{\bf A})\cdot\text{\bf k}\\\ &\times
k^{2}\rho_{0}V^{2}(\text{\bf k})\text{e}^{i\text{\bf k}\cdot\text{\bf
r}_{0}(t)-\mu_{0}Tk^{2}\left[1+\beta\rho_{0}V(\text{\bf
k})\right]t},\end{split}$ (16)

and a Gaussian colored noise $\bm{\Xi}$, independent of $\bm{\xi}_{0}$. The
memory kernel $\langle\bm{\Xi}(\text{\bf r}(t),t)\otimes\bm{\Xi}(\text{\bf
r}_{0}(t^{\prime}),t^{\prime})\rangle=\text{\bf G}(\text{\bf
r}_{0}(t)-\text{\bf r}_{0}(t^{\prime}),t-t^{\prime})$ has the following
expression:

$\begin{split}\text{\bf G}(\text{\bf
r}_{0}(t),t)&\equiv\frac{(T\mu_{0})^{2}}{\rho_{0}}\int_{\text{\bf
k}}(\mathbf{1}+\gamma\text{\bf
A})\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}(\mathbf{1}-\gamma\text{\bf
A})\\\ &\times\frac{\left[\beta\rho_{0}V(\text{\bf
k})\right]^{2}}{1+\beta\rho_{0}V(\text{\bf k})}\text{e}^{i\text{\bf
k}\cdot\text{\bf r}_{0}(t)-\mu_{0}Tk^{2}\left[1+\beta\rho_{0}V(\text{\bf
k})\right]t}.\end{split}$ (17)

We are now in a position to define, and then to determine, the various
transport coefficients quantifying the dynamics of the tracer. Our computation
rests on a path-integral formalism, which we briefly outline in the next
subsection.

###  III.3 Averaging over the tracer’s trajectories

We are interested in the computation of the diffusivity tensor, obtained from
the velocity-velocity autocorrelation function [47]

$\begin{split}\text{\bf D}&\equiv\int_{0}^{+\infty}\differential
t\langle\dot{\text{\bf r}}_{0}(t)\otimes\dot{\text{\bf r}}_{0}(0)\rangle\\\
&=\lim_{t_{f}\to\infty}\langle[\text{\bf r}_{0}(t_{f})-\text{\bf
r}_{0}(0)]\otimes\dot{\text{\bf r}}_{0}(0)\rangle,\end{split}$ (18)

with $\text{\bf F}_{\text{ext}}=\mathbf{0}$, along with that of the mobility
tensor $\bm{\mu}$, defined by the response to an external force as

$\langle\text{\bf r}_{0}(t_{f})-\text{\bf
r}_{0}(0)\rangle\equiv\bm{\mu}\text{\bf F}_{\text{ext}}t_{f}$ (19)

as $t_{f}\to\infty$ and $\text{\bf F}_{\text{ext}}\to 0$. The angular brackets
denote averages over the realization of the noise. Using a path-integral
formalism, these averages over the noise realization can be rewritten as
weighted averages over the tracer’s trajectories. If $f(\text{\bf
r}_{0}(t_{f}))$ is a generic function of the tracer position at the final time
$t_{f}$ then within the path-integral formalism we have

$\langle f(\text{\bf r}_{0}(t_{f}))\rangle=\int\mathcal{D}\text{\bf
p}_{0}(t)\mathcal{D}\text{\bf r}_{0}(t)f(\text{\bf
r}_{0}(t_{f}))\text{e}^{-S[\text{\bf r}_{0}(t),\text{\bf p}_{0}(t)]},$ (20)

where the Janssen-De Dominicis functional integral $\int\mathcal{D}\text{\bf
p}_{0}(t)\mathcal{D}\text{\bf r}_{0}(t)$ is over all the trajectories
$\text{\bf r}_{0}(t),\text{\bf p}_{0}(t)$. The response field $\text{\bf
p}_{0}(t)$ is an auxiliary field encoding stochastic fluctuations. The
trajectories are weighted by a trajectory-dependent exponential factor
$\text{e}^{-S[\text{\bf r}_{0}(t),\text{\bf p}_{0}(t)]}$. The corresponding
action features two contributions, expressing respectively the motion of a
free tracer and its interactions with the bath:

$S[\text{\bf r}_{0}(t),\text{\bf p}_{0}(t)]=S_{\text{free}}[\text{\bf
r}_{0}(t),\text{\bf p}_{0}(t)]+S_{\text{int}}[\text{\bf r}_{0}(t),\text{\bf
p}_{0}(t)],$ (21)

with

$\begin{split}S_{\text{free}}[\text{\bf r}_{0}(t),\text{\bf
p}_{0}(t)]&\equiv-i\int\differential t\text{\bf
p}_{0}(t)\cdot\left[\dot{\text{\bf r}}_{0}(t)-\mu_{0}\text{\bf
F}_{\text{ext}}\right]\\\ &+\mu_{0}T\int\differential t\text{\bf
p}_{0}(t)^{2},\end{split}$ (22)

and

$\begin{split}S_{\text{int}}&[\text{\bf r}_{0}(t),\text{\bf p}_{0}(t)]\equiv
i\int\differential t\int_{-\infty}^{t}\differential s\text{\bf
p}_{0}(t)\cdot\text{\bf F}(\text{\bf r}_{0}(t)-\text{\bf r}_{0}(s),t-s)\\\
&+\int\differential t\int_{-\infty}^{t}\differential s\text{\bf
p}_{0}(t)\cdot\text{\bf G}(\text{\bf r}_{0}(t)-\text{\bf
r}_{0}(s),t-s)\cdot\text{\bf p}_{0}(s).\end{split}$ (23)

To evaluate the dynamical averages in Eqs. (18, 19), we resort to a small
coupling approximation between the bath and the tracer and expand in powers of
the small coupling parameter. This is carried out explicitly in the next
subsection.

###  III.4 Weak-coupling approximation

We now assume that the coupling between the bath and the tracer is weak with
respect to the thermal forces. This is expressed by requiring that

$h\equiv\frac{\sqrt{\rho_{0}}V}{T}\ll 1.$ (24)

From Eqs. (16, 17) we see that both F and G are of order $h^{2}$. From Eq.
(23), this implies that $S_{\text{int}}$ is of order $h^{2}$ as well. To
second-order in $h$, dynamical averages can thus be computed by expanding the
exponential $\text{e}^{-S}$:

$\langle f(\text{\bf r}_{0}(t_{f}))\rangle=\frac{\langle f(\text{\bf
r}_{0}(t_{f}))\rangle_{\text{free}}-\langle f(\text{\bf
r}_{0}(t_{f}))S_{\text{int}}\rangle_{\text{free}}}{1-\langle
S_{\text{int}}\rangle_{\text{free}}}+O(h^{4}),$ (25)

where $\langle\ldots\rangle_{\text{free}}\equiv\int\mathcal{D}\text{\bf
r}_{0}(t)\mathcal{D}\text{\bf p}_{0}(t)\ldots\text{e}^{-S_{\text{free}}}$
refers to an average over the dynamics of a free tracer. The action
$S_{\text{free}}$ is quadratic, which ensures that linear and quadratic
functionals of the trajectories $\text{\bf r}_{0}(t)$, $\text{\bf p}_{0}(t)$
can be computed exactly. Since we are interested in the tracer’s displacement,
we define the one-point quantity $\Delta\text{\bf r}_{0}(t,s)\equiv\text{\bf
r}_{0}(t)-\text{\bf r}_{0}(s)$ with $t\geq s$. We then have

$\begin{split}\langle\Delta\text{\bf
r}_{0}(t,s)\rangle_{\text{free}}&=\mu_{0}\text{\bf F}_{\text{ext}}(t-s),\\\
\langle\text{\bf p}_{0}(t)\rangle_{\text{free}}&=0,\end{split}$ (26)

from which we recover the result that the mobility tensor for the free tracer
is $\bm{\mu}_{\text{free}}=\mu_{0}\mathbf{1}$. The two-point averages read

$\begin{split}\left\langle\text{\bf p}_{0}(t)\otimes\text{\bf
p}_{0}(s)\right\rangle_{\text{free}}&=\mathbf{0}\\\
\left\langle\Delta\text{\bf r}_{0}(t,s)\otimes\text{\bf
p}_{0}(s^{\prime})\right\rangle_{\text{free}}&=-i\mathbf{1}\chi_{[s,t)}(s^{\prime})\\\
\langle\left[\Delta\text{\bf r}_{0}(t,0)-\mu_{0}\text{\bf
F}_{\text{ext}}t\right]\\\ \otimes\left[\Delta\text{\bf
r}_{0}(t^{\prime},s^{\prime})-\mu_{0}\text{\bf
F}_{\text{ext}}(t^{\prime}-s^{\prime})\right]\rangle_{\text{free}}&=2\mathbf{1}\mu_{0}T\text{L}([s,t]\cap[s^{\prime},t^{\prime}]),\end{split}$
(27)

where $\chi_{[s,t)}(s^{\prime})$ is the characteristic function of the
interval $[s,t)$ (it is $1$ if $s^{\prime}\in[s,t)$ and $0$ otherwise). The
quantity $\text{L}[a,b]$ returns the length of the interval $[a,b]$. Note that
for $\text{\bf F}_{\text{ext}}=0$ the last equality in Eq. (27) yields the
mean-squared displacement of a free particle, $\langle\Delta\text{\bf
r}_{0}(t_{f},0)^{2}\rangle_{\text{free}}=2d\mu_{0}Tt_{f}$.

We now evaluate $\langle S_{\text{int}}\rangle_{\text{free}}$. This average
requires the evaluation over the dynamical action for a free particle of two
contributions. The first one is

$\begin{split}\langle\text{\bf p}_{0}(t)\text{e}^{i\text{\bf
k}\cdot\Delta\text{\bf
r}_{0}(t,s)}\rangle_{\text{free}}&=i\text{e}^{-\mu_{0}Tk^{2}(t-s)+i\text{\bf
k}\cdot\text{\bf F}_{\text{ext}}(t-s)}\\\ &\times\langle\text{\bf
p}_{0}(t)\Delta\text{\bf r}_{0}(t,s)\rangle_{\text{free}}\cdot\text{\bf k}\\\
&=\mathbf{0},\end{split}$ (28)

where we made use of Wick’s theorem and of the second equality of Eq. (27).
Similarly one can show that the other term vanishes, namely $\langle\text{\bf
p}_{0}(t)\otimes\text{\bf p}_{0}(s)\text{e}^{i\text{\bf k}\cdot\Delta\text{\bf
r}_{0}(t,s)}\rangle=\mathbf{0}$. We thus conclude that

$\left\langle S_{\text{int}}\right\rangle_{\text{free}}=0.$ (29)

In the next two subsections we will compute the diffusion and the mobility
tensors to leading order in $h$.

###  III.5 Diffusion tensor

We specialize the calculation to the case where no external force is applied
to the tracer particle, $\text{\bf F}_{\text{ext}}=0$. The calculation of the
diffusion tensor to second order in the weak coupling $h$ requires computing
two free-particle dynamical averages, stemming respectively from the drift
term F of Eq. (16) and from the memory term G of Eq. (17). They are given by
(with $s\leq t$):

$\begin{split}\langle\Delta
r_{0,a}(t_{f},0)\dot{r}_{0,b}(0)p_{0,c}(t)\text{e}^{i\text{\bf
k}\cdot\Delta\text{\bf
r}(t,s)}\rangle_{\text{free}}&=-2i\delta_{ac}\mu_{0}Te^{-\mu_{0}Tk^{2}[1+\beta\rho_{0}V(\text{\bf
k})](t-s)}k_{b}\chi_{[s,t)]}(0)\chi_{[0,t_{f})}(t)\\\ \left\langle\Delta
r_{0,a}(t_{f},0)\dot{r}_{0,b}(0)p_{0,c}(t)p_{0,d}(s)\text{e}^{i\text{\bf
k}\cdot\Delta\text{\bf
r}_{0}(t,s)}\right\rangle_{\text{free}}&=\left[\delta_{bd}\delta_{ac}\delta(s-0)-2\mu_{0}T\delta_{ac}k_{d}k_{b}\chi_{[s,t)}(0)\right]\chi_{[0,t_{f}]}(t)\text{e}^{-\mu_{0}Tk^{2}(t-s)}.\end{split}$
(30)

Using these results, one can directly compute the diffusivity matrix

$\begin{split}\text{\bf
D}(\gamma,T)&\approx\mu_{0}T\mathbf{1}-\lim_{t_{f}\to\infty}\langle\Delta\text{\bf
r}_{0}(t_{f},0)\otimes\text{\bf
r}_{0}(0)S_{\text{int}}\rangle_{\text{free}}\\\
&=\mu_{0}T\mathbf{1}-\lim_{t_{f}\to+\infty}i\int\differential
t\int_{-\infty}^{t}\differential s\langle\Delta\text{\bf
r}_{0}(t_{f},0)\otimes\text{\bf r}_{0}(0)\left[\text{\bf
p}_{0}(t)\cdot\text{\bf F}(\Delta\text{\bf
r}_{0}(t,s),t-s))\right]\rangle_{\text{free}}\\\
&-\lim_{t_{f}\to+\infty}\int\differential t\int_{-\infty}^{t}\differential
s\langle\Delta\text{\bf r}_{0}(t_{f},0)\otimes\text{\bf
r}_{0}(0)\left[\text{\bf p}_{0}(t)\cdot\text{\bf G}(\Delta\text{\bf
r}_{0}(t,s),t-s))\cdot\text{\bf p}_{0}(s)\right]\rangle_{\text{free}}\\\
&=\mu_{0}T\Biggl{[}\mathbf{1}-\frac{1}{2\rho_{0}}\int_{\text{\bf
k}}(\mathbf{1}+\gamma\text{\bf
A})\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}\frac{[\beta\rho_{0}V(\text{\bf
k})]^{2}}{\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf k})\right]^{2}}\\\
&+\frac{1}{4\rho_{0}}\int_{\text{\bf k}}(\mathbf{1}+\gamma\text{\bf
A})\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}(\mathbf{1}-\gamma\text{\bf
A})\frac{[\beta\rho_{0}V(\text{\bf
k})]^{3}}{\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf
k})\right]^{2}\left[1+\beta\rho_{0}V(\text{\bf
k})\right]}\Biggr{]}.\end{split}$ (31)

For $\gamma=0$, we obtain

$\begin{split}\text{\bf
D}(0,T)&=\mu_{0}T\left[\mathbf{1}-\frac{1}{2\rho_{0}}\int_{\text{\bf
k}}\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}\frac{[\beta\rho_{0}V(\text{\bf
k})]^{2}}{\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\beta\rho_{0}V(\text{\bf k})\right]}\right]\\\
&=\mu_{0}T\mathbf{1}\left[1-\frac{1}{2d\rho_{0}}\int_{\text{\bf
k}}\frac{[\beta\rho_{0}V(\text{\bf
k})]^{2}}{\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\beta\rho_{0}V(\text{\bf k})\right]}\right],\end{split}$
(32)

where we used, in the second line, the fact that upon integration over all the
wavevectors $\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}$ can be
replaced, using the isotropy of the integrand in Eq. (32), by
$d^{-1}\mathbf{1}$. We thus recover the equilibrium result for the
longitudinal diffusion constant obtained in [44, 42]. The equilibrium
diffusion tensor of Eq. (32) is diagonal, and the interaction of the tracer
with the bath reduces its ability to diffuse. When $\gamma\neq 0$, the
diffusion tensor acquires an antisymmetric contribution proportional to
$\gamma$, and the diagonal part picks up a contribution proportional to
$\gamma^{2}$. These additional terms lead respectively to odd diffusivity and
enhanced diffusion. To see this explicitly, we can write the tensor products
of Eq. (31) in the $(xyz)$ basis. In this basis, the matrix A displayed in Eq.
(7) reads $\text{\bf A}=\text{\bf e}_{y}\otimes\text{\bf e}_{x}-\text{\bf
e}_{x}\otimes\text{\bf e}_{y}$. The second tensor product inside the brackets
of Eq. (31) is, neglecting the term that vanishes upon integration over all
the wavevectors,

$\begin{split}(\mathbf{1}+\gamma\text{\bf
A})\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}&=\sum_{i=x,y,z}\hat{k}_{i}^{2}\text{\bf
e}_{i}\otimes\text{\bf e}_{i}\\\ &+\gamma\left[\text{\bf
e}_{y}\otimes\text{\bf e}_{x}\hat{k}_{x}^{2}-\text{\bf e}_{x}\otimes\text{\bf
e}_{y}\hat{k}_{y}^{2}\right].\end{split}$ (33)

The term proportional to $\gamma$ is antisymmetric upon integration over the
wavevector k, and it contributes to the odd diffusivity.

The third tensor in Eq. (31) is

$\begin{split}(\mathbf{1}+\gamma\text{\bf
A})&\bm{\hat{\textbf{k}}}\otimes\bm{\hat{\textbf{k}}}(\mathbf{1}-\gamma\text{\bf
A})=\sum_{i=x,y,z}\hat{k}_{i}^{2}\text{\bf e}_{i}\otimes\text{\bf e}_{i}\\\
&+\gamma\left[\text{\bf e}_{x}\otimes\text{\bf e}_{y}\hat{k}_{x}^{2}-\text{\bf
e}_{y}\otimes\text{\bf e}_{x}\hat{k}_{y}^{2}\right]\\\ &+\gamma\left[\text{\bf
e}_{y}\otimes\text{\bf e}_{x}\hat{k}_{x}^{2}-\text{\bf e}_{x}\otimes\text{\bf
e}_{y}\hat{k}_{y}^{2}\right]\\\ &+\gamma^{2}\left[\text{\bf
e}_{x}\otimes\text{\bf e}_{x}\hat{k}_{x}^{2}+\text{\bf e}_{y}\otimes\text{\bf
e}_{y}\hat{k}_{y}^{2}\right].\end{split}$ (34)

Upon integration over k the terms proportional to $\gamma$ cancel out, leaving
only a contribution proportional to $\gamma^{2}$ to the longitudinal
diffusion.

The diffusion tensor in the $(xyz)$ basis is therefore given by

$\begin{split}\text{\bf D}&=D_{xx}(\gamma)\left[\text{\bf
e}_{x}\otimes\text{\bf e}_{x}+\text{\bf e}_{y}\otimes\text{\bf
e}_{y}\right]+D_{zz}\text{\bf e}_{z}\otimes\text{\bf e}_{z}\\\
&+D_{\perp}\left(\text{\bf e}_{y}\otimes\text{\bf e}_{x}-\text{\bf
e}_{x}\otimes\text{\bf e}_{y}\right),\end{split}$ (35)

with $D_{\parallel}(\gamma)$ given by

$\begin{split}D_{\parallel}(\gamma)&=\mu_{0}T\Biggl{(}1-\frac{1}{2d\rho_{0}}\int_{\text{\bf
k}}\frac{\left[\beta\rho_{0}V(\text{\bf
k})\right]^{2}}{\left[1+\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf k})\right]}\\\
&+\frac{\gamma^{2}}{4d\rho_{0}}\int_{\text{\bf
k}}\frac{\left[\beta\rho_{0}V(\text{\bf
k})\right]^{3}}{\left[1+\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf
k})\right]^{2}}\Biggr{)}.\end{split}$ (36)

The odd diffusion constant reads

$D_{\perp}(\gamma)=-\gamma\frac{\mu_{0}T}{2d\rho_{0}}\int_{\text{\bf
k}}\frac{\left[\beta\rho_{0}V(\text{\bf
k})\right]^{2}}{\left[1+\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf k})\right]}.$ (37)

For systems with equilibrium dynamics (that is, when $\gamma=0$), Onsager
reciprocity relations impose this quantity to vanish. However, when departing
from equilibrium, as is the case in the presence of transverse forces, odd
transport coefficients need not vanish as they are a priori not prohibited by
time-reversal invariance. In our case, a nonzero odd diffusivity reveals the
presence of a directed swirling motion for the tagged particle. Of course, for
$\gamma=0$, the odd diffusivity vanishes, as expected in equilibrium.

We see from Eq. (36) that the diffusivity of the tracer is enhanced by
transverse forces, as the contribution proportional to $\gamma^{2}$ is
positive. It is remarkable that this diffusion enhancement is captured even
for a system where the linearized bath relaxation is not affected by the
transverse forces. An enhancement of diffusions in odd system was also found
in the case where the oddity is induced by external magnetic fields [48, 49],
and is not, as in our case, a consequence of the nonequilibrium dynamics of
the system.

We now take advantage of these explicit expressions to discuss the temperature
dependence of the speedup. First, we observe that both terms in the integrals
of Eq. (36) are decreasing functions of the temperature. This implies that the
quantity $D_{\parallel}(0)/\mu_{0}T$ is an increasing function of the
temperature. Thus the efficiency of the transverse forces, defined as the
ratio of the diffusion coefficient with and without transverse forces, is

$\frac{D_{\parallel}(\gamma)}{D_{\parallel}(0)}=1+\gamma^{2}\frac{I}{D_{\parallel}(0)/\mu_{0}T},$
(38)

with $I\equiv\frac{1}{2d\rho_{0}}\int_{\text{\bf
k}}\frac{\left[\beta\rho_{0}V(\text{\bf
k})\right]^{3}}{\left[1+\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf k})\right]^{2}}$. We
conclude that the efficiency of transverse forces increases upon cooling down
the system. This is consistent with the increase of the odd diffusivity of the
tracer. An analogous trend was recently found for binary mixture with
nonreciprocal interactions between particles of different species [46].

It would be remarkable that the monotonous trend of the efficiency extends to
low temperatures or high densities. In the following sections, using the mode
coupling theory, we will show that this trend does not hold for very cool or
very dense systems, where the small coupling approximation in Eq. (24) breaks
down. Before moving to the mode-coupling formalism, we briefly examine the
mobility of the tracer which contains additional information, since the
Einstein relation between the diffusivity and mobility tensors does not
necessarily hold.

###  III.6 Mobility tensor

The mobility tensor of the tracer can be obtained from the weak coupling
expansion of Eq. (19):

$\langle\Delta\text{\bf r}_{0}(t_{f},0)\rangle=\mu_{0}T\text{\bf
F}_{\text{ext}}t_{f}-\langle\Delta\text{\bf
r}_{0}(t_{f},0)S_{\text{int}}\rangle_{\text{free}}+o(t_{f}).$ (39)

The second term involves the following averages, which can be obtained using
Wick’s theorem and the dynamical averages of a free tracer displayed in Eq.
(27):

$\begin{split}\langle\ \Delta r_{0,a}(t_{f},0)\text{\bf
p}_{0,b}(t)\text{e}^{i\text{\bf k}\cdot\Delta\text{\bf
r}_{0}(t,s)}\rangle_{\text{free}}&=i\delta_{ab}\chi_{[0,t_{f})}\text{e}^{-\mu_{0}Tk^{2}(t-s)+i\mu_{0}\text{\bf
k}\cdot\text{\bf F}_{\text{ext}}}\\\ \langle\Delta
r_{0,a}(t_{f},0)p_{0,b}(t)p_{0,c}(s)\text{e}^{i\text{\bf
k}\cdot\Delta\text{\bf
r}_{0}(t,s)}\rangle_{\text{free}}&=-i\delta_{ab}k_{c}\chi_{[0,t_{f})}\text{e}^{-\mu_{0}Tk^{2}(t-s)+i\mu_{0}\text{\bf
k}\cdot\text{\bf F}_{\text{ext}}}.\end{split}$ (40)

Plugging these results into Eq. (39) we obtain

$\begin{split}\langle\Delta\text{\bf
r}(t_{f},0)\rangle&\approx\mu_{0}\text{\bf
F}_{\text{ext}}t_{f}-(\mathbf{1}+\gamma\text{\bf A})\text{\bf
F}_{\text{ext}}\frac{\mu_{0}}{2d\rho_{0}}\int_{\text{\bf
k}}\frac{\left[\rho_{0}\beta V(\text{\bf
k})\right]^{2}}{\left[1+\beta\rho_{0}V(\text{\bf
k})\right]\left[1+\frac{1}{2}\beta\rho_{0}V(\text{\bf k})\right]}t_{f}\\\
&\equiv\left(\mathbf{1}\mu_{\parallel}+\text{\bf A}\mu_{\perp}\right)\text{\bf
F}_{\text{ext}}t_{f}.\end{split}$ (41)

When $\gamma=0$, the Einstein relation
$D_{\parallel}(0,T)=T\mu_{\parallel}(0,T)$ is recovered. When $\gamma\neq 0,$
the presence of a transverse component to the force on the tracer does not
affect the longitudinal mobility of the tracer, and its expression is the same
as in equilibrium [42]. However, an odd mobility coefficient [50] proportional
to $\gamma$ appears. If the tracer is pulled by a force $\text{\bf
F}_{\text{ext}}$ along a given direction in the $(xy)$ plane, it will also
move along the transverse direction parallel to $\text{\bf A}\text{\bf
F}_{\text{ext}}$: this is the physical meaning of the emergent odd mobility.
As the temperature is lowered, the longitudinal mobility decreases while the
odd mobility increases.

The main take-home message of this section is the increase of the sampling
efficiency upon decreasing the temperature. In what follows we analyse what
happens when the small coupling approximation breaks down. When the strength
of transverse forces increases, we ask about the asymptotic behavior of the
efficiency.

##  IV Mode-coupling theory in the presence of transverse forces

The dynamical evolution of the system of $N$ particles with positions
$\text{\bf r}_{i}$ is governed by the operator $\Omega_{\gamma}$

$\Omega_{\gamma}\equiv
D_{0}\sum_{i}\bm{\nabla}_{i}\cdot\left[\bm{\nabla}_{i}-\left(\mathbf{1}+\gamma\text{\bf
A}\right)\beta\mathbf{F}_{i}\right].$ (42)

When $\gamma=0$, $\Omega_{\gamma}$ is the usual Smoluchowski evolution
operator of the equilibrium dynamics. The evolution of the probability
distribution $\rho(\mathbf{r}^{N},t)$ of the system reads

$\partial_{t}\rho(\mathbf{r}^{N},t)=\Omega_{\gamma}\rho(\mathbf{r}^{N},t).$
(43)

We can thus write the formal expression for $\rho(\text{\bf r}^{N},t)$ given
its initial condition $\rho(\text{\bf r}^{N})$:

$\rho(\text{\bf r}^{N},t)=\text{e}^{\Omega_{\gamma}t}\rho(\text{\bf
r}^{N},0).$ (44)

We denote the average value of any function $f(\text{\bf r}^{N},t)$ by
$\langle f(\text{\bf r}^{N},t)\rangle$, and it is fully determined by the
knowledge of $\rho(\text{\bf r}^{N},t)$:

$\displaystyle\left\langle f(\text{\bf r}^{N},t)\right\rangle$
$\displaystyle\equiv\int d\text{\bf r}^{N}f(\text{\bf
r}^{N})\text{e}^{\Omega_{\gamma}t}\rho(\text{\bf r}^{N})$ (45)
$\displaystyle=\int d\text{\bf
r}^{N}\left[\text{e}^{\Omega^{\dagger}_{\gamma}t}f(\text{\bf
r}^{N})\right]\rho(\text{\bf r}^{N})$ $\displaystyle=\int d\text{\bf
r}^{N}\text{e}^{\Omega_{-\gamma}t}f(\text{\bf r}^{N})\rho(\text{\bf r}^{N})$
$\displaystyle=\left\langle\text{e}^{\Omega_{-\gamma}t}f(\text{\bf
r}^{N})\right\rangle,$

where we have used the adjoint operator $\Omega^{\dagger}_{\gamma}$, defined
as

$\Omega^{\dagger}_{\gamma}\equiv
D_{0}\sum_{i}\left[\bm{\nabla}_{i}+\left(\mathbf{1}+\gamma\text{\bf
A}\right)\beta\mathbf{F}_{i}\right]\cdot\bm{\nabla}_{i},$ (46)

and the following identity, which is a consequence of the breaking of detailed
balance:

$\cdots\Omega_{\gamma}f(\text{\bf
r}^{N})\Biggr{\rangle}=\cdots\left(\Omega^{\dagger}_{-\gamma}f(\text{\bf
r}^{N})\right)\Biggr{\rangle}.$ (47)

The steady state solution of Eq. (44) is the Boltzmann distribution
$\rho_{\text{\tiny{B}}}(\mathbf{r}^{N})=\frac{\text{e}^{-\beta\mathcal{H}(\mathbf{r}^{N})}}{\int
d\mathbf{r}^{N}\text{e}^{-\beta\mathcal{H}(\mathbf{r}^{N})}}$ with
$\mathcal{H}(\mathbf{r}^{N})\equiv\frac{1}{2}\sum_{i\neq
j}V(\lvert\mathbf{r}_{i}-\mathbf{r}_{j}\rvert)$. Since we are interested in
the steady state dynamics of the system, we assume that the initial condition
is also sampled from the Boltzmann distribution, i.e. $\rho(\text{\bf
r}^{N})=\rho_{\text{\tiny{B}}}(\text{\bf r}^{N})$.

We are interested in the fluctuating density mode $n(\text{\bf q},t)$, defined
as the Fourier transform of the fluctuating density field $n(\text{\bf
r},t)=\sum_{i}\delta(\text{\bf r}-\text{\bf r}_{i}(t))-\rho_{0}$:

$n(\text{\bf q},t)\equiv\sum_{i}\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{i}(t)},$ (48)

evaluated at the wavevector q, and the dynamical structure factor $S(\text{\bf
q},t)$:

$\begin{split}S(\text{\bf q},t)&\equiv\frac{1}{N}\left\langle n^{*}(\text{\bf
q})\left(\text{e}^{\Omega^{\dagger}_{\gamma}t}n(\text{\bf
q})\right)\right\rangle\\\ &=\frac{1}{N}\left\langle n^{*}(\text{\bf
q})\text{e}^{\Omega_{-\gamma}t}n(\text{\bf q})\right\rangle.\end{split}$ (49)

The initial condition $S(\text{\bf q})$ is the equilibrium structure factor of
the system. Another quantity of interest is the self-part of the intermediate
scattering function, $F_{s}(q,t)$:

$F_{s}(\text{\bf q},t)\equiv\frac{1}{N}\sum_{i}\left\langle
n_{i}^{*}(\text{\bf q})\text{e}^{\Omega_{-\gamma}t}n_{i}(\text{\bf
q})\right\rangle,$ (50)

with $n_{i}(\text{\bf q})\equiv\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{i}}$.

In the next section, we implement the Mori-Zwanzig projection operator
formalism to obtain an equation of motion for the dynamical density
correlations.

###  IV.1 Projection operator formalism

We start our calculation by introducing the Laplace transform

$S(\text{\bf q},z)\equiv\int_{0}^{+\infty}S(\text{\bf q},t)\text{e}^{-zt},$
(51)

which, once applied to the time derivative of Eq. (49) yields

$zS(\text{\bf q},z)-S(\text{\bf q})=\frac{1}{N}\left\langle n^{*}(\text{\bf
q})\Omega_{-\gamma}\frac{1}{z-\Omega_{-\gamma}}n(\text{\bf q})\right\rangle.$
(52)

We introduce an operator that projects along the density mode $n(\text{\bf
q})$:

$\mathcal{P}\equiv\frac{1}{NS(\text{\bf q})}n(\text{\bf
q})\Bigr{\rangle}\Bigl{\langle}n^{*}(\text{\bf q}),$ (53)

and its orthogonal counterpart $\mathcal{Q}\equiv\mathcal{I}-\mathcal{P}$.
Using the resolvent identity [3]

$\begin{split}\mathcal{Q}\frac{1}{z-\Omega_{-\gamma}}&=\mathcal{Q}\frac{1}{z-\Omega_{-\gamma}\mathcal{Q}}\\\
&+\mathcal{Q}\frac{1}{z-\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\Omega_{-\gamma}\mathcal{P}\frac{1}{z-\Omega_{-\gamma}},\end{split}$
(54)

we obtain

$\begin{split}zS(\text{\bf q},z)-S(\text{\bf q})&=\frac{1}{N}\left\langle
n^{*}(\text{\bf
q})\Omega_{-\gamma}\frac{1}{z-\Omega_{-\gamma}}(\mathcal{P}+\mathcal{Q})n(\text{\bf
q})\right\rangle\\\ &=\Biggl{[}\frac{1}{NS(q)}\left\langle n^{*}(\text{\bf
q})\Omega_{-\gamma}n(\text{\bf q})\right\rangle+\frac{1}{NS(q)}\left\langle
n^{*}(\text{\bf
q})\Omega_{-\gamma}\mathcal{Q}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\Omega_{-\gamma}n(\text{\bf
q})\right\rangle\Biggr{]}S(\text{\bf q},z).\\\ \end{split}$ (55)

Using the projection operators, we have separated the contributions to the
evolution of the structure factor into an exponentially decaying part with
frequency $-N^{-1}\left\langle n^{*}(\text{\bf q})\Omega_{-\gamma}n(\text{\bf
q})\right\rangle$ and a memory kernel. The frequency term reads

$\begin{split}-\frac{1}{N}\langle n(\text{\bf
q})^{*}\Omega_{-\gamma}n(\text{\bf q})\rangle&=i\frac{D_{0}}{N}\text{\bf
q}\cdot\sum_{i}\langle\text{e}^{i\text{\bf q}\cdot\text{\bf
r}_{i}}\cdot\left[\left(\bm{\nabla}_{i}-\gamma\text{\bf A}\beta\text{\bf
F}_{i}\right)n(\text{\bf q})\right]\rangle\\\ &=-D_{0}q^{2}.\end{split}$ (56)

This term, which encodes the diffusive decay of the structure factor in the
absence of interactions, is left unaffected by the transverse forces.

Using an integration by parts, together with the fact that the Boltzmann
distribution describes the steady state of the system, the memory kernel
$\mathbf{\widetilde{\text{ \bf M}}}$ can be written as

$\begin{split}\text{\bf
q}\cdot\frac{(D_{0}\beta)^{2}}{V\rho_{0}}\left\langle\sum_{i}e^{i\text{\bf
q}\cdot\text{\bf r}_{i}}((\mathbf{1}-\gamma\text{\bf A})\text{\bf
F}_{i}+iT\text{\bf
q})\mathcal{Q}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\sum_{j}e^{i\text{\bf
q}\cdot\text{\bf r}_{j}}(\text{\bf F}_{i}(\mathbf{1}-\gamma\text{\bf
A})-iT\text{\bf q})\right\rangle\cdot\text{\bf q}\equiv D_{0}\text{\bf
q}\cdot\mathbf{\widetilde{\text{ \bf M}}}(\text{\bf q},z)\cdot\text{\bf
q}.\end{split}$ (57)

Due to the presence of projection operators $\mathcal{Q}$, the formula for
$\mathbf{\widetilde{\text{ \bf M}}}$ can be simplified,

$\begin{split}\mathbf{\widetilde{\text{ \bf M}}}(\text{\bf
q},z)=\frac{(D_{0}\beta)^{2}}{V\rho_{0}}\left\langle\sum_{i}e^{i\text{\bf
q}\cdot\text{\bf r}_{i}}((\mathbf{1}-\gamma\text{\bf A})\text{\bf
F}_{i}+iT\text{\bf
q})\mathcal{Q}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\sum_{j}e^{i\text{\bf
q}\cdot\text{\bf r}_{j}}(\text{\bf F}_{i}(\mathbf{1}-\gamma\text{\bf
A})-iT\text{\bf q})\right\rangle\\\
=\frac{(D_{0}\beta)^{2}}{V\rho_{0}}\left\langle\sum_{i}e^{i\text{\bf
q}\cdot\text{\bf r}_{i}}((\mathbf{1}-\gamma\text{\bf A})\text{\bf
F}_{i})\mathcal{Q}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\sum_{j}e^{i\text{\bf
q}\cdot\text{\bf r}_{j}}\text{\bf F}_{i}(\mathbf{1}-\gamma\text{\bf
A})\right\rangle.\end{split}$ (58)

The memory kernel can be expressed in terms of the correlations between
projected force density Fourier modes,

$\begin{split}\mathcal{Q}{\bf j}(\text{\bf
q})&\equiv\mathcal{Q}\sum_{i}\text{\bf F}_{i}\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{i}}.\end{split}$ (59)

Note that due to the projection operator we have
$\ldots\mathcal{Q}\sum_{i}\text{\bf F}_{i}\text{e}^{-i\text{\bf
q}\cdot\text{\bf
r}_{i}}\rangle=\ldots\mathcal{Q}T\sum_{i}\bm{\nabla}_{i}\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{i}}\rangle$ and it is the latter form that is used in the
derivation of the mode-coupling approximation, Eq. (70). The memory kernel
$\mathbf{\widetilde{\text{ \bf M}}}$ thus reads

$\mathbf{\widetilde{\text{ \bf M}}}(\text{\bf q},z)=(1-\gamma\text{\bf
A})\mathbf{\widetilde{\text{ \bf K}}}(\text{\bf q},z)(1-\gamma\text{\bf A}),$
(60)

with

$\mathbf{\widetilde{\text{ \bf K}}}(\text{\bf
q},z)\equiv\frac{D_{0}\beta^{2}}{V\rho_{0}}\left\langle{\bf j}(\text{\bf
q})^{*}\mathcal{Q}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}{\bf
j}(\text{\bf q})\right\rangle.$ (61)

Equation (55) becomes

$zS(\text{\bf q},z)-S(q)=\frac{D_{0}}{S(q)}\text{\bf
q}\cdot\left[-\mathbf{1}+\mathbf{\widetilde{\text{ \bf M}}}(\text{\bf
q},z)\right]\cdot\text{\bf q}S(\text{\bf q},z).$ (62)

When $\gamma=0$, the memory kernel $\mathbf{\widetilde{\text{ \bf M}}}$ is
diagonal, and only the correlations among longitudinal particle currents
contribute. We thus fall back onto the equilibrium case. In the presence of
transverse forces, correlations arising from transverse currents contribute to
the dynamics of the structure factor. Cross-correlations among longitudinal
and transverse currents might as well influence the dynamics. This is the new
feature entering the design of our mode-coupling approximation. In the next
section, we expand on the irreducible representation of the memory kernel.

###  IV.2 The irreducible memory kernel

We now introduce the irreducible memory kernel. Following Kawasaki [51] and
Vogel and Fuchs [52] we define an irreducible operator

$\Omega^{\text{irr}}_{-\gamma}\equiv
D_{0}\mathcal{Q}\sum_{j}\bm{\nabla}_{j}\mathcal{Q}_{j}\cdot\left(\mathbf{1}-\gamma\text{\bf
A}\right)\left[-\beta\text{\bf F}_{j}+\bm{\nabla}_{j}\right]\mathcal{Q},$ (63)

with $\mathcal{Q}_{j}\equiv 1-\mathcal{P}_{j}$ and
$\mathcal{P}_{j}\equiv\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{j}}\rangle\langle\text{e}^{i\text{\bf q}\cdot\text{\bf r}_{j}}$ a single-
particle projection operator. The definition of the irreducible operator comes
from a formal extension of the equilibrium case. Using the fact that particles
are statistically equivalent we obtain
$\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}=\Omega^{\text{\tiny{irr}}}_{-\gamma}-\delta\Omega_{-\gamma}$,
with

$\delta\Omega_{-\gamma}=\frac{D_{0}\beta^{2}}{N}\mathcal{Q}{\bf j}(\text{\bf
q})\rangle\cdot\left(\mathbf{1}-\gamma\text{\bf A}\right)\langle{\bf
j}(\text{\bf q})^{*}\mathcal{Q}.$ (64)

Using Eq. (64) and the resolvent identity

$\begin{split}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}&=\frac{1}{z-\Omega^{\text{\tiny{irr}}}_{-\gamma}}\\\
&+\frac{1}{z-\Omega^{\text{\tiny{irr}}}_{-\gamma}}\delta\Omega_{-\gamma}\frac{1}{z-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}},\end{split}$
(65)

we can express the reducible memory kernel $\mathbf{\widetilde{\text{ \bf
K}}}$ in Eq. (61) as a function of an irreducible memory kernel $\text{\bf
K}(\text{\bf q},z)$, where the evolution of the system is governed by
$\Omega^{\text{\tiny{irr}}}_{-\gamma}$:

$\text{\bf K}(\text{\bf
q},z)\equiv\frac{D_{0}\beta^{2}}{\rho_{0}V}\left\langle{\bf j}(\text{\bf
q})^{*}\mathcal{Q}\frac{1}{z-\Omega^{\text{\tiny{irr}}}_{-\gamma}}\mathcal{Q}{\bf
j}(\text{\bf q})\right\rangle.$ (66)

The relationship between the reducible and the irreducible memory kernels is

$\mathbf{\widetilde{\text{ \bf K}}}(\text{\bf q},z)=\left[\mathbf{1}+\text{\bf
K}(\text{\bf q},z)(\mathbf{1}-\gamma\text{\bf A})\right]^{-1}\text{\bf
K}(\text{\bf q},z).$ (67)

The introduction of the irreducible memory kernel is a necessary step in order
to avoid an unphysical negative viscosity in the system, which may appear
within an approximate evaluation of the reducible kernel [53].

Approximations are now needed to compute $\text{\bf K}(\text{\bf q},z)$. In
the next section we apply the mode-coupling approximation to the memory
kernel.

###  IV.3 Mode-coupling expansion of the memory matrix

To perform the mode-coupling expansion, we follow Szamel and Löwen [54]. The
basic idea behind the mode-coupling approximation is to decompose the current
field into a sum of products of density modes:

$\mathcal{Q}{\bf j}(\text{\bf q})\approx\frac{1}{2}\sum_{\text{\bf
k}}\frac{n(\text{\bf k})n(\text{\bf q}-\text{\bf k})}{N^{2}S(\text{\bf
k})S(\text{\bf q}-\text{\bf k})}\left\langle n^{*}(\text{\bf
k})n^{*}(\text{\bf q}-\text{\bf k})\mathcal{Q}{\bf j}(\text{\bf
q})\right\rangle,$ (68)

where the factor $\frac{1}{2}$ comes from a Gaussian factorization of a static
multi-point density correlator. The central mode-coupling approximation is the
factorization of the time-dependent multi-point density correlator, combined
to the approximation
$\Omega^{\text{\tiny{irr}}}_{-\gamma}\approx\Omega_{-\gamma}$:

$\displaystyle\left\langle n(\text{\bf k}^{\prime})^{*}n(\text{\bf
q}-\text{\bf
k}^{\prime})^{*}\text{e}^{\Omega^{\text{\tiny{irr}}}_{-\gamma}t}n(\text{\bf
k})n(\text{\bf q}-\text{\bf k})\right\rangle$ (69)
$\displaystyle\approx\left\langle n^{*}(\text{\bf
k}^{\prime})e^{\Omega_{-\gamma}t}n(\text{\bf k})\right\rangle\left\langle
n^{*}(\text{\bf q}-\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n(\text{\bf q}-\text{\bf
k})\right\rangle$ $\displaystyle+\left\langle n^{*}(\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n(\text{\bf q}-\text{\bf
k})\right\rangle\left\langle n^{*}(\text{\bf q}-\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n(\text{\bf k})\right\rangle$
$\displaystyle=N^{2}S(\text{\bf k},t)S(\text{\bf q}-\text{\bf
k},t)\left[\delta_{\text{\bf k},\text{\bf k}^{\prime}}+\delta_{\text{\bf
k}^{\prime}\text{\bf q}-\text{\bf k}}\right].$

We now compute the expectation value in Eq. (68) with the aid of a convolution
approximation

$\displaystyle\left\langle n(\text{\bf k})^{*}n(\text{\bf q}-\text{\bf
k})^{*}\mathcal{Q}{\bf j}(\text{\bf q})\right\rangle=\left\langle n(\text{\bf
k})^{*}n(\text{\bf q}-\text{\bf k})^{*}{\bf j}(\text{\bf q})\right\rangle$
(70) $\displaystyle-\frac{1}{NS(\text{\bf q})}\left\langle n(\text{\bf
k})^{*}n(\text{\bf q}-\text{\bf k})^{*}n(\text{\bf
q})\right\rangle\left\langle n(\text{\bf q}){\bf j}(\text{\bf
q})\right\rangle$ $\displaystyle\approx T\left\langle n(\text{\bf
k})^{*}n(\text{\bf q}-\text{\bf
k})\sum_{i}\bm{\nabla}_{i}\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{i}}\right\rangle$ $\displaystyle-\frac{T}{NS(\text{\bf q})}NS(\text{\bf
k})S(\text{\bf q}-\text{\bf k})S(\text{\bf q})\left\langle n(\text{\bf
q})^{*}\sum_{i}\bm{\nabla}_{i}\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{i}}\right\rangle$
$\displaystyle=-T\left\langle\sum_{i}\bm{\nabla}_{i}\left[n(\text{\bf
k})^{*}n(\text{\bf q}-\text{\bf k})^{*}\right]\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{i}}\right\rangle$ $\displaystyle+\frac{T}{NS(\text{\bf
q})}NS(\text{\bf k})S(\text{\bf q}-\text{\bf k})S(\text{\bf
q})\left\langle\sum_{i}\bm{\nabla}_{i}\left[n(\text{\bf
q})^{*}\right]\text{e}^{-i\text{\bf q}\cdot\text{\bf r}_{i}}\right\rangle$
$\displaystyle=-iNT\left[\text{\bf k}S(\text{\bf q}-\text{\bf
k})+\left(\text{\bf q}-\text{\bf k}\right)S(\text{\bf k})-\text{\bf
q}S(\text{\bf k})S(\text{\bf q}-\text{\bf k})\right]$
$\displaystyle=iNT\rho_{0}S(\text{\bf k})S(\text{\bf q}-\text{\bf
k})\left[\text{\bf k}c(\text{\bf k})+\left(\text{\bf q}-\text{\bf
k}\right)c(\text{\bf q}-\text{\bf k})\right],$

where in the last step we have introduced the direct correlation function
$c(\text{\bf k})$, related to $S(\text{\bf k})$ via the Ornstein-Zernike
relation $\rho_{0}c(\text{\bf k})=1-\frac{1}{S(\text{\bf k})}$. We can now
inject the expansion (68) into the memory kernel K of Eq. (66). We then use
Eq. (69) and Eq. (70). After replacing the sum over the wavevectors with an
integral, $\sum_{\text{\bf k}}\to V\int_{\text{\bf k}}$, we obtain

$\text{\bf K}(\text{\bf q},t)\approx\frac{D_{0}\rho_{0}}{2}\int_{\text{\bf
k}}\text{\bf V}_{\text{\bf k},\text{\bf q}}\otimes\text{\bf V}_{\text{\bf
k},\text{\bf q}}S(\text{\bf k},t)S(\text{\bf q}-\text{\bf k},t),$ (71)

where the vertex

$\text{\bf V}_{\text{\bf k},\text{\bf q}}\equiv\text{\bf k}c(k)+(\text{\bf
q}-\text{\bf k})c(\lvert\text{\bf q}-\text{\bf k}\rvert)$ (72)

is the mode-coupling vertex of the equilibrium dynamics.

To proceed further, we decompose the kernel K over an orthonormal basis that
depends on the wavevector q and on the matrix A. The axes of this basis are
defined by

$\begin{split}\text{\bf e}_{1}\equiv\text{\bf e}_{\text{\bf
q}1}&\equiv\frac{\text{\bf A}^{T}\text{\bf A}\text{\bf q}}{\lvert\text{\bf
A}\text{\bf q}\rvert}\\\ \text{\bf e}_{2}\equiv\text{\bf e}_{\text{\bf
q}2}&\equiv\frac{\text{\bf A}\text{\bf q}}{\lvert\text{\bf A}\text{\bf
q}\rvert}\\\ \text{\bf e}_{3}\equiv\text{\bf e}_{\text{\bf
q}3}&\equiv\frac{(\mathbf{1}-\text{\bf A}^{T}\text{\bf A})\text{\bf
q}}{\lvert(\mathbf{1}-\text{\bf A}^{T}\text{\bf A})\text{\bf q}\rvert}.\\\
\end{split}$ (73)

Looking at the expression for A in Eq. (7), we see that $\text{\bf e}_{1}$ is
the normalized projection of q on the $(xy)$ plane, $\text{\bf e}_{2}$ is the
normalized vector orthogonal to $\text{\bf e}_{1}$ in the $(xy)$ plane, as
selected by the matrix A, and $\text{\bf e}_{3}$ is the normalized projection
of q along the $z$-direction. We refer to this basis as the $\text{\bf
A}-\text{\bf q}$ basis. The matrix $\text{\bf K}(\text{\bf q},t)$ is diagonal
in the basis defined in Eq. (73), within the mode-coupling approximation. To
prove this, we consider the decomposition of $\text{\bf K}(\text{\bf q},t)$ in
the $\text{\bf A}-\text{\bf q}$ basis, namely

$\text{\bf K}(\text{\bf q},t)=\sum_{i,j=1}^{3}K_{ij}(\text{\bf q},t)\text{\bf
e}_{i}\otimes\text{\bf e}_{j}.$ (74)

The matrix element $K_{ij}$ reads

$\begin{split}K_{ij}(\text{\bf q},t)&=\frac{D_{0}\rho_{0}}{2}\int_{\text{\bf
k}}[k_{i}c(k)+(q_{i}-k_{i})c(\lvert\text{\bf q}-\text{\bf k}\rvert)]\\\
&\times[k_{j}c(k)+(q_{j}-k_{j})c(\lvert\text{\bf q}-\text{\bf
k}\rvert)]S(\text{\bf k},t)S(\text{\bf q}-\text{\bf k},t).\end{split}$ (75)

We now consider the symmetries of the dynamics with transverse forces. The
first symmetry transformation we exploit is a reflection of direction $3$ (the
axis $z$). This symmetry imposes

$\text{\bf K}(\mathcal{R}_{3}\text{\bf q},t)=\text{\bf K}(\text{\bf q},t),$
(76)

with $\mathcal{R}_{3}\text{\bf q}\equiv\text{\bf q}-2\text{\bf
e}_{3}(\text{\bf q}\cdot\text{\bf e}_{3})$ the operator that reflects a vector
with respect to direction $3$. By construction, we have $\text{\bf
e}_{\text{\bf q}i}=\text{\bf e}_{\mathcal{R}_{3}\text{\bf q}i}$ for $i=1,2$,
and $\text{\bf e}_{\text{\bf q}3}=-\text{\bf e}_{\mathcal{R}_{3}\text{\bf
q}3}$. This implies that $K_{3i}(\text{\bf
q},t)=K_{3i}(\mathcal{R}_{3}\text{\bf q},t)$ for $i=1,2$. On the other hand, a
change of variable $k_{3}\to-k_{3}$ in the momentum integration of
$K_{3i}(\mathcal{R}_{3}\text{\bf q},t)$ shows that

$K_{3i}(\mathcal{R}_{3}\text{\bf q},t)=-K_{3i}(\text{\bf q},t),$ (77)

where we used the fact that $S(\mathcal{R}_{3}\text{\bf k},t)=S(\text{\bf
k},t)$ due to the symmetries of the dynamics with transverse forces. Combining
Eqs. (76, 77) we see that $K_{3i}(\text{\bf q},t)=-K_{3i}(\text{\bf q},t)$ for
$i=1,2$, thus implying that $K_{3i}(\text{\bf q},t)=0$ for $i=1,2$.

The second symmetry we exploit is obtained by considering a rotation within
the $(xy)$ plane by an angle $\pi/2$. The resulting invariance imposes

$\text{\bf K}(\mathcal{R}_{\frac{\pi}{2}}\text{\bf q},t)=\text{\bf
K}(\text{\bf q},t),$ (78)

with $\mathcal{R}_{\frac{\pi}{2}}\text{\bf q}\equiv\text{\bf A}\text{\bf
q}+\text{\bf e}_{3}q_{3}$ the operator that rotates a vector by an angle
$\pi/2$ in the $(xy)$ plane. By construction, $\text{\bf e}_{\text{\bf
q}3}=\text{\bf e}_{\mathcal{R}_{\frac{\pi}{2}}\text{\bf q}3}$, $\text{\bf
e}_{\text{\bf q}1}=-\text{\bf e}_{\mathcal{R}_{\frac{\pi}{2}}\text{\bf q}2}$
and $\text{\bf e}_{\text{\bf q}2}=\text{\bf
e}_{\mathcal{R}_{\frac{\pi}{2}}\text{\bf q}1}$. This implies $K_{12}(\text{\bf
q},t)=-K_{21}(\mathcal{R}_{\frac{\pi}{2}}\text{\bf q},t)$. On the other hand,
a change of variable $\text{\bf k}\to\mathcal{R}_{\frac{\pi}{2}}\text{\bf k}$
in the momentum integral of $K_{21}(\mathcal{R}_{\frac{\pi}{2}}\text{\bf
q},t)$ shows that

$K_{21}(\mathcal{R}_{\frac{\pi}{2}}\text{\bf q},t)=-K_{12}(\text{\bf q},t),$
(79)

where we used the fact that $S(\mathcal{R}_{\frac{\pi}{2}}\text{\bf
k},t)=S(\text{\bf k},t)$ due to the symmetries of the dynamics with transverse
forces. Combining Eqs. (78, 79) we see that $K_{12}(\text{\bf
q},t)=-K_{12}(\text{\bf q},t)$ for $i=1,2$, thus implying that
$K_{12}(\text{\bf q},t)=K_{21}(\text{\bf q},t)=0$. This concludes our proof
that the kernel $\text{\bf K}(\text{\bf q},t)$ is diagonal in the $\text{\bf
A}-\text{\bf q}$ basis, i.e.

$\text{\bf K}(\text{\bf q},t)=\sum_{i=1}^{3}K_{ii}(\text{\bf q},t)\text{\bf
e}_{i}\otimes\text{\bf e}_{i}.$ (80)

Using this diagonal approximation, we can give an expression of
$\mathbf{\widetilde{\text{ \bf K}}}(\text{\bf q},z)$. Noting that the matrix A
reads, in the $\text{\bf A}-\text{\bf q}$ basis,

$\text{\bf A}=\text{\bf e}_{2}\otimes\text{\bf e}_{1}-\text{\bf
e}_{1}\otimes\text{\bf e}_{2},$ (81)

we can invert the matrix in Eq. (67), substitute the result in Eq. (60) and
Eq. (62) to finally obtain

$\begin{split}&zS(\text{\bf q},z)-S(\text{\bf q})=-D_{0}\text{\bf
q}\cdot\Biggl{[}\frac{1+(1+\gamma^{2})K_{22}}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}\text{\bf
e}_{1}\otimes\text{\bf e}_{1}+\frac{1}{1+K_{33}}\text{\bf
e}_{3}\otimes\text{\bf e}_{3}\\\
&+\gamma\frac{K_{11}+K_{22}+(1+\gamma^{2})K_{11}K_{22}}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}\left[\text{\bf
e}_{1}\otimes\text{\bf e}_{2}-\text{\bf e}_{2}\otimes\text{\bf
e}_{1}\right]+\frac{(1+K_{11})K_{22}+K_{11}(K_{22}-1)\gamma^{2}}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}\text{\bf
e}_{2}\otimes\text{\bf e}_{2}\Biggr{]}\cdot\text{\bf q}S(\text{\bf
q},z),\end{split}$ (82)

where we have omitted the dependence on $(\text{\bf q},z)$ of the memory
kernel $K_{ii}$, for simplicity. Note that only the first line of the term in
the square bracket contributes to the decay of $S(\text{\bf q},z)$, due to the
projection along the mode q. The off-diagonal, anti-symmetric term on the
second line hints at the odd transport properties of the dynamics, which will
be investigated below.

Before turning to the discussion of longitudinal and odd transport, we study
the decay of the dynamical structure factor to a (possibly nonzero) plateau.
To do so, we introduce the normalized dynamical density correlations
$\phi(\text{\bf q},t)\equiv\frac{S(\text{\bf q},t)}{S(\text{\bf q})}$. Its
dynamics is readily obtained from Eq. (82):

$z\phi(\text{\bf q},z)-1=-\frac{D_{0}}{S(q)}\text{\bf
q}\cdot\Biggl{[}\frac{1+(1+\gamma^{2})K_{22}}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}\text{\bf
e}_{1}\otimes\text{\bf e}_{1}+\frac{1}{1+K_{33}}\text{\bf
e}_{3}\otimes\text{\bf e}_{3}\Biggr{]}\cdot\text{\bf q}\phi(\text{\bf q},z).$
(83)

For $\gamma=0$, we are back to the known equilibrium situation. The isotropy
of the dynamics is restored, $K_{11}=K_{33}$ and we consistently find:

$z\phi(\text{\bf q},z)-1=-\frac{D_{0}}{S(q)}q^{2}\frac{\phi(\text{\bf
q},z)}{1+K_{\parallel}(\text{\bf q},z)}$ (84)

where $K_{\parallel}$ is the longitudinal mode-coupling kernel

$\begin{split}K_{\parallel}(\text{\bf
q},t)&=\frac{D_{0}\rho_{0}}{2q^{2}}\int_{\text{\bf k}}\left[\text{\bf
q}\cdot\text{\bf k}c(k)+\text{\bf q}\cdot(\text{\bf q}-\text{\bf
k})c(\lvert\text{\bf q}-\text{\bf k}\rvert)\right]^{2}\\\ &\times S(\text{\bf
k},t)S(\text{\bf q}-\text{\bf k},t).\end{split}$ (85)

For the general situation with $\gamma\neq 0$, it is instructive to consider
the case where $q_{3}=0$ and rewrite Eq. (83) in the time domain:

$\begin{split}\partial_{t}\phi(\text{\bf q},t)+D_{0}q^{2}\phi(\text{\bf
q},t)+D_{0}q^{2}(1+\gamma^{2})K_{22}*\phi(\text{\bf q},t)\\\
=-\left[\left(K_{11}+K_{22}\right)+(1+\gamma^{2})K_{11}*K_{22}\right]*\partial_{t}\phi(\text{\bf
q},t),\end{split}$ (86)

where $*$ denotes a convolution in time, $f*g(t)=\int_{0}^{t}\differential
tf(t-\tau)g(\tau)$. With this expression, it is clear that when $\gamma\neq 0$
the transverse currents affect both the relaxation rate and the friction
kernel of the system.

In order to extract a physical picture from these equations, we inject an
approximate form for the static structure factor in which the structure is
sharply localized at a given wavevector. As a result, the next subsection
explores the resulting so-called schematic approximation.

###  IV.4 Schematic mode-coupling theory

The first schematic approximation we consider follows the historical one [55]
where the structure factor is strongly peaked for a wavevector of modulus
$q_{0}$, $S(\text{\bf q})\approx 1+S_{0}\delta(\lvert\text{\bf q}\rvert-
q_{0})$, which implies in turn that $\rho_{0}c(\text{\bf q})=1$ if
$\lvert\text{\bf q}\rvert=q_{0}$ and $0$ otherwise. Within this approximation,
$K_{22}=0$. In this case, Eq. (83) has the following property: if
$\phi(\text{\bf k},t)$ is rotationally invariant at $t=0$, then
$\phi(\text{\bf k},t)$ remains rotationally invariant at all subsequent times
$t$. The memory kernel thus reads

$\begin{split}K_{11}(q_{0},t)&=K_{33}(q_{0},t)\\\
&=\frac{D_{0}S_{0}^{2}}{2\rho_{0}}\int\displaylimits_{\lvert\text{\bf
k}\rvert=\lvert\text{\bf q}-\text{\bf
k}\rvert=q_{0}}\frac{d\mathbf{k}}{(2\pi)^{3}}\phi(\text{\bf
k},t)\phi(\text{\bf q}-\text{\bf k},t)\\\
&=\lambda_{\text{eq}}\phi^{2}(q_{0},t).\end{split}$ (87)

where
$\lambda_{\text{eq}}\equiv\frac{\sqrt{3}}{16\pi^{2}}\frac{D_{0}S_{0}^{2}q_{0}}{\rho_{0}}\phi^{2}(q_{0},t)$.
Upon substitution into Eq. (83) we obtain, denoting by $\phi_{2}(q_{0},z)$ the
Laplace transform of $\phi^{2}(q_{0},t)$,

$z\phi(q_{0},z)-1=-D_{0}q_{0}^{2}\frac{1}{1+\lambda_{\text{eq}}\phi_{2}(q_{0},z)}\phi(q_{0},z).$
(88)

This equation is identical to the one obtained in the equilibrium case for
$\gamma=0$.

We briefly review here how an ergodicity breaking scenario is predicted from
this equation [56]. If we denote by $f_{0}$ the nonergodicity parameter
$f_{0}\equiv\lim\displaylimits_{t\rightarrow+\infty}\phi(q_{0},t)$, we have

$\displaystyle\lim\displaylimits_{z\rightarrow 0}\phi(q_{0},z)$
$\displaystyle=\frac{f_{0}}{z},$ (89)
$\displaystyle\lim\displaylimits_{z\rightarrow 0}\phi_{2}(q_{0},z)$
$\displaystyle=\frac{f^{2}_{0}}{z}.$

Taking the limit $z\rightarrow 0$ on both sides of Eq. (88) and keeping only
the diverging part yields

$\frac{f_{0}}{1-f_{0}}=\frac{\lambda_{\text{eq}}}{D_{0}q_{0}^{2}}f^{2}_{0}.$
(90)

Equation (90) admits a nonzero solution if and only if
$\frac{\lambda_{\text{eq}}}{D_{0}q_{0}^{2}}\geq 4$. This is the ergodicity
breaking predicted by the mode-coupling theory of the colloidal glass
transition in equilibrium [54].

We now give an expression for the high-temperature relaxation time in the
equilibrium case. Let us assume that the system is ergodic,
$\lambda_{\text{eq}}<4D_{0}q^{2}$, and that
$\phi(q_{0},t)=e^{-\frac{t}{\tau_{0}}}$, from which it follows that
$\phi(q_{0},z=0)=\tau_{0}$ and $\phi_{2}(q_{0},z=0)=\frac{\tau_{0}}{2}$.
Equation (88) evaluated at $z=0$ gives an expression for the relaxation time
$\tau_{0}$

$\tau_{0}=\frac{1}{q_{0}^{2}D_{0}-\frac{\lambda_{\text{eq}}}{2}}.$ (91)

This result is physical only for $\lambda_{\text{eq}}<2q_{0}^{2}D_{0}$, which
corresponds to the ergodic phase. At higher values of $\lambda_{\text{eq}}$
another functional form for the decay of $\phi$ must be assumed [55, 56].

Implementing the conventional schematic approximation does not allow us to
capture any qualitative change in the evolution equation of the dynamical
structure factor. However, the fact that dynamics is accelerated is a
mathematical statement, and so we should blame the nature of the schematic
approximation. There is also the possibility that the mode-coupling
approximation itself does not capture the acceleration provided by transverse
forces, but our analysis, developed in the remainder of the paper, shows that
this is not the case.

The historical schematic approximation spuriously transforms a set of
anisotropic equations for time-correlations into an isotropic one. We thus
need to take a step back and consider an alternative schematic approximation
that does not overlook the anisotropy, which we believe is an essential
ingredient. This is what we implement in the next subsection.

###  IV.5 New ansatz for the ergodic phase: relaxation speedup

We now resort to an approximate form of the static structure factor that
preserves the chiral character of the dynamics, and which, as will appear,
implies a faster relaxation in the ergodic phase.

We keep the main ingredient of the schematic approximation, namely the idea
that the structure factor is sharply localized for wavevectors around a given
modulus $q_{0}$. Nevertheless, we replace the sharp delta-function with a
rounded peak of nonzero width $\epsilon$, which is assumed to be small with
respect to $q_{0}$, but nonzero. This will be enough to produce an interesting
interplay of modes in various directions. In practice, we resort to the
following approximate forms for $S(k)$ and for the product
$S(\mathbf{k})S(\mathbf{q}-\mathbf{k})$:

$\displaystyle S(k)$ $\displaystyle\approx
1+S_{0}\frac{1}{\epsilon^{d}}\eta\left(\frac{k-q_{0}}{\epsilon}\right)$ (92)
$\displaystyle S(\lvert\mathbf{k}\rvert)S(\mathbf{q}-\mathbf{k})$
$\displaystyle\approx\frac{S^{2}_{0}}{\epsilon^{d}}\eta\left(\frac{k-q_{0}}{\epsilon}\right)\eta\left(\frac{\lvert\text{\bf
q}-\text{\bf k}\rvert-q_{0}}{\epsilon}\right)$

with $\epsilon^{-1}\eta(k)$ a function normalized to unity with an absolute
maximum at $k=0$. An example of such $\eta(k)$ could be a Gaussian function

$\epsilon^{-d}\eta(k)\equiv\frac{1}{\sqrt{2\pi\epsilon^{2d}}}e^{-\frac{k^{2}}{2\epsilon^{2}}},$
(93)

but the specific shape of $\eta$ is not relevant. Such a function will serve
as an approximation of the Dirac distribution. Using the Ornstein-Zernike
relation, the direct correlation function is given by

$\rho_{0}c(k)\approx\frac{\frac{S_{0}}{\epsilon^{d}}\eta\left(\frac{k-q_{0}}{\epsilon}\right)}{1+\frac{S_{0}}{\epsilon^{d}}\eta\left(\frac{k-q_{0}}{\epsilon}\right)}.$
(94)

When computing $K_{22}(\text{\bf q},t)$, we focus on a wavevector q such that
$\lvert\text{\bf q}\rvert=q_{0}$ and expand around wavevectors $\text{\bf
k}_{0}$ such that $k_{0}=q_{0}$ and $\lvert\text{\bf q}-\text{\bf
k}_{0}\rvert=q_{0}$. We then rewrite k as

$\text{\bf k}=\text{\bf k}_{0}+\epsilon\text{\bf p}$ (95)

in the integrals of Eq. (75). Note that $\epsilon$ has the dimension of a
wavevector, and hence p is dimensionless. The approximate values of $k$ and
$\lvert\text{\bf q}-\text{\bf k}\rvert$ up to order $\epsilon^{2}$ are then

$\displaystyle k$ $\displaystyle=\lvert\text{\bf k}_{0}+\epsilon\text{\bf
p}\rvert=q_{0}+\epsilon\frac{\text{\bf p}\cdot\text{\bf
k}_{0}}{q_{0}}+\frac{\epsilon^{2}}{2q_{0}}\left[p^{2}-\frac{\left(\text{\bf
p}\cdot\text{\bf k}_{0}\right)^{2}}{q_{0}^{2}}\right]$ (96)
$\displaystyle\lvert\text{\bf q}-\text{\bf k}\rvert$ $\displaystyle\approx
q_{0}+\epsilon\frac{\text{\bf p}\cdot\left(\text{\bf q}-\text{\bf
k}_{0}\right)}{q_{0}}+\frac{\epsilon^{2}}{2q_{0}}\left[p^{2}-\frac{\left(\text{\bf
p}\cdot\left(\text{\bf q}-\text{\bf
k}_{0}\right)\right)^{2}}{q_{0}^{2}}\right].$

We now expand the direct correlation function in powers of $\epsilon$. We
assume the function $\eta$ to be even, so that $\rho_{0}c^{\prime}(q_{0})=0$.
We thus obtain

$\begin{split}\rho_{0}c(k)&\approx\rho_{0}c(q_{0})+\epsilon^{2}\frac{(\text{\bf
p}\cdot\text{\bf k}_{0})^{2}}{2q_{0}^{2}}\rho_{0}c^{\prime\prime}(q_{0})\\\
&=1+\epsilon^{2}\frac{(\text{\bf p}\cdot\text{\bf
k}_{0})^{2}}{2S_{0}q_{0}^{2}}\frac{\eta^{\prime\prime}_{0}}{\eta_{0}^{2}},\end{split}$
(97)

where we used the notation $\eta_{0}\equiv\eta(0)$,
$\eta^{\prime\prime}_{0}\equiv\eta^{\prime\prime}(0)$. Substituting Eq. (97)
into the expression for the memory kernel $K_{22}(\text{\bf q},t)$, given in
Eq. (75), we find to lowest order in $\epsilon$,

$\begin{split}K_{22}(\text{\bf q},t)&=\frac{D_{0}\rho_{0}}{2}\int_{\text{\bf
k}}\left(\frac{\text{\bf A}\text{\bf q}\cdot\text{\bf k}}{|\text{\bf
A}\text{\bf q}|}\right)^{2}\left[c(k)-c(|\text{\bf q}-\text{\bf
k}|)\right]^{2}\\\ &\times S(k)S(|\text{\bf q}-\text{\bf k}|)\phi(\text{\bf
k},t)\phi(\text{\bf q}-\text{\bf k},t)\\\
&\approx\epsilon^{4}\frac{D_{0}(\eta^{\prime\prime}_{0})^{2}}{8\rho_{0}q_{0}^{4}\eta_{0}^{4}}\int_{\text{\bf
p}}\left(\frac{\text{\bf A}\text{\bf q}\cdot\text{\bf k}_{0}}{|\text{\bf
A}\text{\bf q}|}\right)^{2}(\text{\bf p}\cdot\text{\bf q})^{2}\\\
&\times\left[2\text{\bf p}\cdot\text{\bf k}_{0}-\text{\bf p}\cdot\text{\bf
q}\right]^{2}\eta\left(\frac{\text{\bf p}\cdot\text{\bf
k}_{0}}{q_{0}}\right)\eta\left(\frac{\text{\bf p}\cdot(\text{\bf q}-\text{\bf
k}_{0})}{q_{0}}\right)\\\ &\times\phi(\text{\bf k}_{0},t)\phi(\text{\bf
q}-\text{\bf k}_{0},t).\end{split}$ (98)

The integrand is nonzero as long as $\text{\bf A}\text{\bf q}\neq 0$. Allowing
for a small momentum shell of thickness $\epsilon$ was enough for anisotropy
to play a role, in contrast to the standard schematic approximation.

The previous result justifies the following approximation for the memory
kernel $\text{\bf K}(\text{\bf q},z)$, when $|\text{\bf q}|=q_{0}$ and
$\text{\bf A}\text{\bf q}\neq 0$:

$\text{\bf K}(\text{\bf q},z)=\lambda\phi_{2}(q_{0},z)(\text{\bf
e}_{1}\otimes\text{\bf e}_{1}+\text{\bf e}_{3}\otimes\text{\bf
e}_{3})+\nu\phi_{2}(q_{0},z)\text{\bf e}_{2}\otimes\text{\bf e}_{2},$ (99)

with $\lambda$ and $\nu$ two parameters that depend continuously on density
and temperature. Moreover, as a further simplification, we assume that all the
modes relax as the ones in the $(xy)$ plane. We speculate that the anisotropy
of the system would simply lead to a renormalization of $\gamma$ for our
theory, thus weakly affecting the results discussed in the following.

Within the above approximation, Eq. (83) reads

$z\phi-1=-q_{0}^{2}D_{0}\phi\frac{1+\nu\phi_{2}(1+\gamma^{2})}{1+\phi_{2}(\lambda+\nu)+\phi_{2}^{2}\lambda\nu\left(1+\gamma^{2}\right)}.$
(100)

For $\gamma=0$, this expression must match its equilibrium counterpart.
Therefore we must have $\lambda=\lambda_{\text{eq}}$. In the next paragraph we
investigate the glass transition and the dynamic acceleration in the ergodic
phase.

####  IV.5.1 Ergodicity Breaking

We substitute Eq. (89) into Eq. (100) and keep only the leading diverging
terms of order $1/z^{2}$:

$\lambda_{\text{eq}}\nu\left(1+\gamma^{2}\right)\left(1-f_{0}\right)f_{0}=q_{0}^{2}D_{0}\nu\left(1+\gamma^{2}\right).$
(101)

This expression simplifies into Eq. (90), namely
$\frac{1}{1-f_{0}}=\frac{\lambda_{\text{eq}}}{q_{0}^{2}D_{0}}f_{0}$, which
establishes that dynamic ergodicity breaking occurs at exactly the same
location as in equilibrium.

####  IV.5.2 Speedup of the relaxation time

Using $\phi(q_{0},z=0)=\tau_{0}$, $\phi_{2}(q_{0},0)=\frac{\tau_{0}}{2}$ and
substituting into Eq. (100) with $z=0$ we obtain an equation for $\tau_{0}$:

$\tau_{0}^{2}\left[\left(1+\gamma^{2}\right)\left(\frac{\lambda}{4}-q_{0}^{2}D_{0}\right)\nu\right]+\tau_{0}\left[\frac{1}{2}\left(\lambda+\nu\right)-q_{0}^{2}D_{0}\right]+1=0.$
(102)

The positive solution for any value of $\gamma$ is

$\tau_{0}^{-1}(\gamma)=\frac{4}{-\lambda-\nu+2q_{0}^{2}D_{0}+\sqrt{\lambda^{2}+\nu^{2}+4\nu
q_{0}^{2}D_{0}(1+2\gamma^{2})+4q_{0}^{2}D_{0}^{2}-2\lambda(\nu+2\gamma^{2}\nu+2q_{0}^{2}D_{0})}}.$
(103)

This describes the relaxation time for exponential relaxation in the ergodic
phase. The relaxation is exponential as long as
$\lambda_{\text{eq}}<2q_{0}^{2}D_{0}$, as in equilibrium. As a consistency
check, $\tau_{0}(0)$ matches the equilibrium solution, while
$\tau_{0}(\gamma)<\tau_{0}(0)$ for all values of $\gamma$. This proves that
transverse forces accelerate the relaxation of the system.

In the limit $\gamma\to\infty$, the relaxation time reads:

$\tau_{0}(\gamma)=\frac{1}{\gamma\nu}\sqrt{\frac{2}{\left(q_{0}^{2}D_{0}-\frac{\lambda}{2}\right)}}.$
(104)

In this limit, the relaxation time goes to $0$ linearly in $1/\gamma$. If
correct, this implies a relaxation time that goes to zero for a large
amplitude of the transverse forces. However, in numerical applications this
method would be hindered by errors in the discretized equation of motion as
increasing $\gamma$ would in practice demand smaller and smaller timesteps.
Note also that the $\gamma^{1}$-dependence of the acceleration at large
$\gamma$ differs from the naive $\gamma^{2}$ acceleration one might anticipate
based on a simpler dimensional analysis as argued in [38]. This modified
scaling results from a many-body correlation effect.

We now move to lower temperatures close to the dynamical glass transition,
where the exponential form for the relaxation is no longer valid.

####  IV.5.3 Early and late $\beta$-relaxations close to the transition

We now study the time-dependence of $\phi$ near its intermediate-time plateau
within the ergodic phase, but very close to the mode-coupling critical
temperature. We study whether the known power-law regimes in this region, as
found in [55], are affected by the transverse forces. Our starting point is
Eq. (100), which we recall here:

$\frac{q_{0}^{2}D_{0}\phi(z)}{1-z\phi(z)}=\frac{1+\left(\lambda+\nu\right)\phi_{2}(z)+\left(1+\gamma^{2}\right)\lambda\nu\phi_{2}(z)^{2}}{1+\left(1+\gamma^{2}\right)\nu\phi_{2}(z)}.$
(105)

The system loses ergodicity for $\lambda\geq 4q_{0}^{2}D_{0}$. In the glass
state close to the ergodicity breaking we set
$\lambda=4q_{0}^{2}D_{0}(1+\varepsilon)$ with $\varepsilon\ll 1$, and we
obtain

$\lim_{t\to+\infty}\phi(t)=\frac{1}{2}\left(1+\varepsilon^{1/2}\right).$ (106)

If instead we set

$\lambda=4q_{0}^{2}D_{0}\left(1-\varepsilon\right)$ (107)

we are in the ergodic phase close to the glass transition. To describe the
approach of $\phi$ to the plateau we assume, in the limit $\varepsilon\to 0$
the following scaling form, taken from Eq. (106):

$\phi(t)=\frac{1}{2}+\varepsilon^{1/2}g(\tau),$ (108)

with $\tau\equiv\varepsilon^{\alpha}t$ a rescaled time. We begin by assuming
that $\alpha>1$ and we shall later check that this is self-consistently
correct. When considering the Laplace transform, we use a rescaled variable
$z\equiv\varepsilon^{\alpha}\zeta$. The Laplace transforms of $\phi(t)$ and
$\phi^{2}(t)$ are therefore

$\displaystyle\phi(\zeta)$
$\displaystyle=\varepsilon^{-\alpha}\left[\frac{1}{2\zeta}+\varepsilon^{1/2}g(\zeta)\right],$
(109) $\displaystyle\phi_{2}(\zeta)$
$\displaystyle=\varepsilon^{-\alpha}\left[\frac{1}{4\zeta}+\varepsilon^{1/2}g(\zeta)+\varepsilon
g_{2}(\zeta)\right].$

We thus substitute Eq. (109) into Eq. (105) and search for the leading order
in $\varepsilon$. To do so, we assume that we are in a regime where
$\lvert\varepsilon^{1/2}g(\tau)\rvert\ll 1$. The orders $\varepsilon^{0}$ and
$\varepsilon^{1/2}$ are self-consistently satisfied, while at order
$\varepsilon$ we obtain

$8\zeta g^{2}(\zeta)-4g_{2}(\zeta)=-\frac{1}{\zeta}.$ (110)

This is the same equation as that obtained in equilibrium [55]. The exponents
controlling the approach to and the departure from the plateau are left
unchanged by the transverse dynamics.

For completeness, we also recall how to compute these exponents below. For the
early $\beta$-relaxation we take

$g(\tau)\equiv a_{0}\tau^{-a}$ (111)

for $\tau\ll 1$, which implies that

$\displaystyle g(\zeta)$ $\displaystyle=a_{0}\zeta^{a-1}\Gamma(1-a),$ (112)
$\displaystyle g_{2}(\zeta)$
$\displaystyle=a_{0}^{2}\zeta^{2a-1}\Gamma(1-2a),$

for $\zeta\gg 1$, with $\Gamma(x)\equiv\int_{0}^{+\infty}t^{x-1}e^{-t}dt$ the
gamma function. Substitution into Eq. (110) yields to leading order in $\zeta$

$4a_{0}^{2}\zeta^{2a-1}\left[2\Gamma(1-a)^{2}-\Gamma(1-2a)\right]=0.$ (113)

This equation has two solutions $a=-1$ and $a=0.395$. Among these, we retain
only the physical one with $a=0.395$. This solution is consistent with the
earlier assumption that $\alpha>1$. In fact, if we want

$\phi(t)=\frac{1}{2}+a_{0}t^{-a}$ (114)

to hold, then we must have $\alpha=1/2a>1$. We also assess in which range of
times this solution is valid. Since we want
$\lvert\varepsilon^{1/2}g(\tau)\rvert\ll 1$ and $\tau\ll 1$ we must have

$\varepsilon^{\frac{1}{2a}}a_{0}^{1/a}\ll\tau\ll 1,$ (115)

which implies

$a_{0}^{1/a}\ll t\ll\varepsilon^{-1/2a}.$ (116)

For the late $\beta$-relaxation we assume

$g(\tau)\equiv-b_{0}\tau^{b}$ (117)

for $\tau\gg 1$, which implies that

$\displaystyle g(\zeta)$ $\displaystyle=-b_{0}\zeta^{-b-1}\Gamma(1+b),$ (118)
$\displaystyle g_{2}(\zeta)$
$\displaystyle=b_{0}^{2}\zeta^{-2b-1}\Gamma(1+2b),$

for $\zeta\ll 1$. Substitution in Eq. (110) yields to leading order in $\zeta$

$4a_{0}^{2}\zeta^{-2b-1}\left[2\Gamma(1+b)^{2}-\Gamma(1+2b)\right]=0.$ (119)

This equation has the solution $b=1$. We also assess in which range of times
this solution is valid. Since we want $\lvert\varepsilon^{1/2}g(\tau)\rvert\ll
1$ and $\tau\gg 1$ we must have

$1\ll\tau\ll\varepsilon^{\frac{1}{2b}}b_{0}^{-1/b},$ (120)

which implies

$\varepsilon^{-1/2a}\ll
t\ll\varepsilon^{-\frac{1}{2a}-\frac{1}{2b}}b_{0}^{-1/b}.$ (121)

This approach does not allow us to compute the values of the two constants
$a_{0}$ and $b_{0}$. These will however exhibit a dependence on $\gamma$,
which, we expect, should result in a marginal shortening of the time required
to approach and leave the plateau in the temperature regime close to the
ergodicity breaking transition.

####  IV.5.4 Divergence of the relaxation time

In this subsection we show that the exponent characterizing the divergence of
the relaxation time close to criticality is also unchanged by the transverse
forces.

We start by assuming the following form for $\phi(t)$ for $t\gg
a_{0}^{1/\alpha}$:

$\phi(t)\equiv\frac{1}{2}e^{2\varepsilon^{1/2}g(\varepsilon^{\alpha}t)}.$
(122)

This expression naturally gives Eq. (108) whenever
$\lvert\varepsilon^{1/2}g(\varepsilon^{\alpha}t)\rvert\ll 1$, which
corresponds to the plateau regime investigated earlier. When
$t\gg\varepsilon^{-\frac{1}{2a}-\frac{1}{2b}}b_{0}^{-1/b}$ we have

$\phi(t)=\frac{1}{2}e^{-2b_{0}\varepsilon^{\alpha+1/2}t}.$ (123)

The Laplace transforms read

$\begin{split}\phi(z)&=\frac{1}{2(z+2b_{0}\varepsilon^{\alpha+1/2})},\\\
\phi_{2}(z)&=\frac{1}{4(z+4b_{0}\varepsilon^{\alpha+1/2})}.\end{split}$ (124)

Using the scaling $z=\varepsilon^{\alpha+1/2}\zeta$ we see that, to leading
order in $\varepsilon$, Eq. (124) satisfies Eq. (105), which confirms that the
ansatz of Eq. (123) is correct. Our schematic mode-coupling theory therefore
predicts that, close to the critical temperature $T_{\text{MCT}}$, the
relaxation time $\tau_{\alpha}$ diverges as a power law,

$\tau_{\alpha}\sim(T-T_{\text{MCT}})^{-1/2a+1/2b},$ (125)

with $\frac{1}{2a}+\frac{1}{2b}\approx 1.7658$, which is again the same
exponent as in equilibrium [55].

###  IV.6 Beyond the schematic approximation

Using insights gained from the schematic approximation, we now prove some
general properties of the dynamics with transverse forces, as derived in Eq.
(83), in particular regarding the location of the glass transition and the
acceleration of the dynamics.

####  IV.6.1 Location of the glass transition

To locate the glass transition, we assume that when ergodicity breaking
occurs, the plateau is the same for all wavevectors with the same modulus:

$\begin{split}\phi(\text{\bf q},z\to 0)&\approx\frac{\phi_{\infty}(\text{\bf
q})}{z},\\\ K_{11}(\text{\bf q},z\to 0)&\approx K_{33}(\text{\bf q},z\to
0)\approx\frac{K_{\parallel,\infty}(\text{\bf q})}{z},\\\ K_{22}(\text{\bf
q},z\to 0)&\approx\frac{K_{\perp,\infty}(\text{\bf q},z\to 0)}{z}.\end{split}$
(126)

Taking the limit $z\to 0$ in Eq. (83), and using Eq. (126), the contribution
from transverse forces cancels, leading to

$q^{2}D_{0}\frac{\phi_{\infty}(\text{\bf q})}{1-\phi_{\infty}(\text{\bf
q})}=K_{\parallel,\infty}(\text{\bf q}).$ (127)

This equation is the same as the one governing the equilibrium case. This
establishes that the glass transition takes place for the same value of the
parameters (density, temperature) as in equilibrium when $\gamma=0$.

####  IV.6.2 Acceleration in the ergodic phase

In the ergodic phase, however, some acceleration can be achieved even if the
location of the glass transition is the same. To support this assertion, we
begin by self-consistently assuming that transverse forces accelerate the
decay of the dynamical structure factor. In practice, we start from the
postulate that

$\phi_{\text{\tiny{f}}}(\text{\bf q})\leq\phi_{\text{\tiny{eq, f}}}(\text{\bf
q}),$ (128)

where $\phi_{\text{\tiny{f}}}\equiv\phi(\text{\bf
q},z=0)\equiv\int_{0}^{+\infty}\differential\tau\phi(\text{\bf q},\tau)$ is
the time-integrated normalized dynamical structure factor, and
$\phi_{\text{\tiny{eq,f}}}$ is the same quantity for equilibrium dynamics,
obtained from Eq. (83) when $\gamma=0$. Due to the mode-coupling expression of
the memory kernel, Eq. (75), it follows that $K_{ij,\text{\tiny{f}}}(\text{\bf
q})\equiv K_{ij}(\text{\bf q},z=0)\leq K_{ij,\text{\tiny{eq,f}}}(\text{\bf
q})\equiv K_{ij,\text{\tiny{eq}}}(\text{\bf q},z=0)$. This means that

$\frac{1+(1+\gamma^{2})K_{22,\text{\tiny{f}}}}{(1+K_{11,\text{\tiny{f}}})(1+K_{22,\text{\tiny{f}}})+\gamma^{2}K_{11,\text{\tiny{f}}}K_{22,\text{\tiny{f}}}}\geq\frac{1}{1+K_{11,\text{\tiny{eq,f}}}},$
(129)

where we omitted the dependence on the q argument for clarity. This inequality
self-consistently proves that the transverse force dynamics accelerates the
decay rate of $\phi$ in the ergodic phase.

####  IV.6.3 Limiting behavior for large $\gamma$

We explore the asymptotic behavior of the relaxation in the ergodic in the
presence of very strong transverse forces. In the ergodic phase, the $z\to 0$
limit of Eq. (83) reads

$\begin{split}1&=-D_{0}\text{\bf
q}\cdot\Biggl{[}\frac{1+(1+\gamma^{2})K_{22,\text{f}}}{(1+K_{11,\text{f}})(1+K_{22,\text{f}})+\gamma^{2}K_{11,\text{f}}K_{22,\text{f}}}\text{\bf
e}_{1}\otimes\text{\bf e}_{1}\\\ &+\frac{1}{1+K_{33,\text{f}}}\text{\bf
e}_{3}\otimes\text{\bf e}_{3}\Biggr{]}\cdot\text{\bf
q}\phi_{{\text{f}}}.\end{split}$ (130)

The analysis of the exponential relaxation in the schematic approximation
suggests that, in the limit $\gamma\to\infty$, one should expect

$\lim_{\gamma\to\infty}\phi_{\text{f}}=\frac{\overline{\phi_{\text{f}}}}{\gamma},$
(131)

where $\overline{\phi_{\text{\tiny{f}}}}$ is a constant that does not depend
on $\gamma$. Our starting point is the assumption that a similar scaling holds
for the memory kernels, namely

$\lim_{\gamma\to\infty}K_{ii,\text{f}}=\frac{\overline{K}_{ii,\text{f}}}{\gamma},$
(132)

with $\overline{K}_{ii,\text{f}}$ a quantity independent from $\gamma$. This
scaling can be explicitly checked in the case of exponential relaxation. If we
substitute the asymptotic behaviors of Eqs. (131, 132) into Eq. (130), the
dependence on $\gamma$ disappears. This result consistently demonstrates that
the relaxation time decreases linearly in $\gamma^{-1}$ in the limit of strong
transverse forces. Note that, by contrast to the discussion in Sec. IV.5.2
which was limited to exponential decays, this result is valid over the whole
ergodic phase.

In this section, we have investigated the relaxation properties of the
collective density modes in the presence of transverse forces, demonstrating
the existence of a speedup, and establishing its asymptotic behavior for
strong drift. We have also realized that transverse forces do not affect the
location of the transition to the nonergodic regime. This, in turn, triggers a
number of questions related to the microscopic mechanisms and dynamical
pathways opened by transverse forces that support these observed collective
behaviors. To answer this question, we turn to the analysis of the motion of
an individual particle. The transport coefficients of this tracer, along with
the diffusivity and mobility tensors, will help us shape a qualitative
picture.

##  V Dynamics of a tracer with transverse forces in the mode-coupling
approach

We now analyze the motion of a tracer within the mode-coupling framework. We
first introduce the general setting and the relevant dynamical quantities.

Following Ernst and Dorfman [57], we consider the motion of a tagged tracer
particle with label $0$ in a liquid of $N$ particles with transverse forces.
The tracer initial position at time $t=0$, $\text{\bf r}_{0}$, is fixed at the
origin, and it influences the distribution of the surrounding bath. The
initial condition for the probability distribution of the total system thus
reads

$P_{N+1}(\text{\bf r}^{N+1},t=0)=V\delta(\text{\bf r}_{0})\rho_{\text{\tiny
B}}(\text{\bf r}^{N+1}).$ (133)

Averages over the initial distribution $P_{N+1}(\text{\bf r}^{N+1},0)$ will be
denoted by $\langle\ldots\rangle_{0}$. Averages over the Boltzmann
distribution are denoted by $\langle\ldots\rangle$, as usual. Starting from
$t=0$, a constant external force $\text{\bf F}_{\text{ext}}$ is applied on the
tracer. The evolution operator becomes $\Omega^{\text{ext}}_{\gamma}$, defined
as

$\begin{split}\Omega_{\gamma,\text{ext}}=&\Omega_{\gamma}+\delta\Omega^{\text{ext}}\\\
=&D_{0}\sum_{i}\bm{\nabla}_{i}\cdot\left[\bm{\nabla}_{i}-\left(\mathbf{1}+\gamma\bm{A}\right)\beta\mathbf{F}_{i}\right]\\\
&-D_{0}\bm{\nabla}_{0}\cdot\beta\text{\bf F}_{\text{ext}}.\end{split}$ (134)

We denote the Fourier transform of the tracer density as

$n_{0}(\text{\bf q})\equiv\text{e}^{-i\text{\bf q}\cdot\text{\bf r}_{0}}.$
(135)

The large-time, long-wavelength limit of this Fourier transform can be used to
obtain relevant transport coefficients. A small q expansion of the time
derivative of Eq. (135) yields constitutive equations of motion for the tracer

$\begin{split}\langle\partial_{t}n_{0}(\text{\bf q},t)\rangle_{0}&=-i\text{\bf
q}\cdot\langle\dot{\text{\bf r}}_{0}(t)\rangle_{0}\\\ &-\text{\bf
q}\cdot\int_{0}^{t}\differential\tau\langle\dot{\text{\bf
r}}_{0}(\tau)\otimes\dot{\text{\bf r}}_{0}(0)\rangle_{0}\cdot\text{\bf q}\\\
&-\text{\bf q}\cdot\langle\dot{\text{\bf r}}_{0}(t)\otimes\text{\bf
r}_{0}\rangle_{0}\cdot\text{\bf q}\\\ &+O(q^{3})\\\ &\approx-i\text{\bf
q}\cdot\bm{\mu}\cdot\text{\bf F}_{\text{ext}}-\text{\bf q}\cdot\text{\bf
D}\cdot\text{\bf q}.\end{split}$ (136)

In the second line, we have taken a large time limit, $t\to\infty$, and used
the definition of the diffusivity tensor D, given by Eq. (18), and the
mobility tensor $\bm{\mu}$, given by Eq. (19). Note that the mobility and the
diffusivity are computed within a linear response formalism, where the
intensity of the external force is small, $\lvert\text{\bf
F}_{\text{ext}}\rvert\to 0$. Moreover, as for the weak coupling analysis of
the tracer, we are interested in the expression of the diffusivity tensor
obtained for $\text{\bf F}_{\text{ext}}=0$, and neglect the dressing of the
diffusivity that comes from the presence of the dragging external force.

###  V.1 Equation of motion of the tracer

We introduce a projection operator tailored to the space of the tracer density
fluctuations

$\mathcal{P}_{0}\equiv\sum_{\text{\bf q}}\ldots n_{0}^{*}(\text{\bf
q})\rangle\langle n_{0}(\text{\bf q})\ldots$ (137)

and the associated orthogonal projector
$\mathcal{Q}_{0}\equiv\mathcal{I}-\mathcal{P}_{0}$. Note that, in contrast
with the other projection operators used in this work, $\mathcal{P}_{0}$
contains a summation over all wavevectors, since these are all included inside
the $\delta$-function of the initial condition in Eq. (133). With this
definition, we have

$\begin{split}\mathcal{P}_{0}P_{N+1}(\text{\bf
r}^{N+1};t=0)&=\frac{1}{V}\sum_{\text{\bf q}}n_{0}(\text{\bf q})\rangle\langle
n_{0}^{*}(\text{\bf q})V\delta(\text{\bf r}_{0})\rangle\\\ &=\sum_{\text{\bf
q}}n_{0}(\text{\bf q})\rangle\\\ &=V\delta(\text{\bf
r}_{0})\rho_{\text{\tiny{B}}}(\text{\bf r}^{N+1})\\\ &=P_{N+1}(\text{\bf
r}^{N};t=0),\end{split}$ (138)

and thus

$\mathcal{Q}_{0}P_{N+1}(\text{\bf r}^{N},t=0)=0.$ (139)

Our goal is to derive an equation of motion for the evolution of the average
tracer density, $\langle n_{0}(\text{\bf
q})\text{e}^{\Omega_{\gamma,\text{ext}}t}\rangle_{0}$. In this notation, the
probability distribution $P_{N+1}(\text{\bf r}^{N+1},0)$ stands to the right
of the evolution operator $\text{e}^{\Omega_{\gamma,\text{ext}}t}$, which acts
on the said distribution. The time derivative of this quantity reads, after a
Fourier transformation

$\begin{split}\left\langle n_{0}(\text{\bf
q})\Omega_{\gamma,\text{ext}}\frac{1}{z-\Omega_{\gamma,\text{ext}}}\right\rangle_{0}&=\left\langle
n_{0}(\text{\bf
q})\Omega_{\gamma,\text{ext}}\mathcal{P}_{0}\frac{1}{z-\Omega_{\gamma,\text{ext}}}\right\rangle_{0}+\left\langle
n_{0}(\text{\bf
q})\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}\frac{1}{z-\Omega_{\gamma,\text{ext}}}\right\rangle_{0}\\\
&=\Biggl{[}\langle n_{0}(\text{\bf
q})\Omega_{\gamma,\text{ext}}n_{0}^{*}(\text{\bf q})\rangle+\left\langle
n_{0}(\text{\bf
q})\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}\frac{1}{z-\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}}\mathcal{Q}_{0}\Omega_{\gamma,\text{ext}}n_{0}^{*}(\text{\bf
q})\right\rangle\Biggr{]}n_{0}(\text{\bf q},z),\end{split}$ (140)

with $n_{0}(\text{\bf q},z)=\left\langle n_{0}(\text{\bf
q})\frac{1}{z-\Omega_{\gamma,\text{ext}}}\right\rangle_{0}$. We used the same
resolvent identity as in Eq. (65) (with $\mathcal{P}_{0}$ and
$\mathcal{Q}_{0}$ instead of $\mathcal{P}$ and $\mathcal{Q}$), namely

$\frac{1}{z-\Omega_{\gamma,\text{ext}}}=\frac{1}{z-\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}}+\frac{1}{z-\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}}\mathcal{Q}_{0}\Omega_{\gamma,\text{ext}}\mathcal{P}_{0}\frac{1}{z-\Omega_{\gamma,\text{ext}}},$
(141)

together with the fact that
$\left\langle\ldots\mathcal{Q}_{0}\frac{1}{z-\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}}\right\rangle_{0}=0$.
The frequency matrix reads

$\langle n_{0}(\text{\bf q})\Omega_{\gamma,\text{ext}}n_{0}^{*}(\text{\bf
q})\rangle=-D_{0}q^{2}-iD_{0}\beta\text{\bf q}\cdot\text{\bf F}_{\text{ext}},$
(142)

and it contains the mobility of a free tracer, $D_{0}\beta$.

The memory kernel reads instead

$\begin{split}&\left\langle n_{0}(\text{\bf
q})\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}\frac{1}{z-\mathcal{Q}_{0}\Omega_{\gamma,\text{ext}}\mathcal{Q}_{0}}\mathcal{Q}_{0}\Omega_{\gamma,\text{ext}}n_{0}^{*}(\text{\bf
q})\right\rangle=\left\langle n_{0}^{*}(\text{\bf
q})\Omega_{-\gamma,\text{\tiny{ext}}}\mathcal{Q}_{0}\frac{1}{z-\mathcal{Q}_{0}\Omega_{-\gamma,\text{\tiny{ext}}}\mathcal{Q}_{0}}\mathcal{Q}_{0}\Omega_{-\gamma,\text{\tiny{ext}}}n_{0}(\text{\bf
q})\right\rangle\\\ &=-i\text{\bf q}\cdot
D_{0}^{2}\Biggl{\langle}\text{e}^{i\text{\bf q}\cdot\text{\bf
r}_{0}}\left[(\mathbf{1}-\gamma\text{\bf A})\cdot\beta\text{\bf
F}_{0}-i\text{\bf q}+\beta\text{\bf
F}_{\text{ext}}\right]\mathcal{Q}_{0}\frac{1}{z-\mathcal{Q}_{0}\Omega_{-\gamma,\text{\tiny{ext}}}\mathcal{Q}_{0}}\mathcal{Q}_{0}\bm{\nabla}_{0}\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{0}}\Biggr{\rangle}\cdot\left[(1-\gamma\text{\bf A})\cdot
i\text{\bf q}-\beta\text{\bf F}_{\text{ext}}\right].\end{split}$ (143)

The first term $\text{\bf F}_{\text{ext}}$ on the right hand side does not
contribute, since it belongs to the space of the tracer’s density modes. It
thus vanishes when the projector $\mathcal{Q}_{0}$ acts on its right. To
extract the mobility of the tracer, we are interested in terms linear in
$\text{\bf F}_{\text{ext}}$. One of these terms comes from the last $\text{\bf
F}_{\text{ext}}$ appearing on the right-hand side of Eq. (143). Another term
comes in principle from the expansion of the operator
$\Omega_{\gamma,\text{ext}}$. However, the physical meaning of this term is a
dressing of the diffusion matrix by means of the external force, as it yields
a contribution proportional to $q^{2}$. The tracer’s equation of motion can
thus be cast in the form

$\begin{split}z\langle&n_{0}(q,z)\rangle_{0}-1=[-i\text{\bf
q}\cdot\bm{\mu}(\text{\bf q},z)\cdot\text{\bf F}_{\text{ext}}\\\ &-\text{\bf
q}\cdot\left[\text{\bf D}(\text{\bf q},z)+|\text{\bf
F}_{\text{ext}}|\delta\text{\bf D}(\text{\bf q},z)\right]\cdot\text{\bf
q}]\text{\bf n}(\text{\bf q},z),\end{split}$ (144)

where $\delta\text{\bf D}(\text{\bf q},z)$ is the correction to the
diffusivity tensor due to the applied external force. We are not interested in
this term, and we focus on the mobility and diffusivity tensors, that read
respectively

$\bm{\mu}(\text{\bf
q},z)=D_{0}\beta\left[\mathbf{1}-(\mathbf{1}-\gamma\text{\bf
A})\mathbf{\widetilde{\text{ \bf K}}}_{0}(\text{\bf q},z)\right],$ (145)

and

$\text{\bf D}(\text{\bf
q},z)=D_{0}\left[\mathbf{1}-(\mathbf{1}-\gamma\text{\bf
A})\mathbf{\widetilde{\text{ \bf K}}}_{0}(\text{\bf
q},z)(\mathbf{1}-\gamma\text{\bf A})\right].$ (146)

The tracer-memory kernel can be expressed in terms of the projected
longitudinal tracer force density Fourier modes

$\mathbf{\widetilde{\text{ \bf K}}}_{0}(\text{\bf q},z)\equiv
D_{0}\beta^{2}\left\langle{\bf j}_{0}^{*}(\text{\bf
q})\mathcal{Q}_{0}\frac{1}{z-\mathcal{Q}_{0}\Omega_{-\gamma}\mathcal{Q}_{0}}\mathcal{Q}_{0}{\bf
j}_{0}(\text{\bf q})\right\rangle,$ (147)

with $\mathcal{Q}_{0}{\bf j}_{0}(\text{\bf q})\equiv\mathcal{Q}_{0}\text{\bf
F}_{0}\text{e}^{-i\text{\bf q}\cdot\text{\bf r}_{0}}$. Note that due to the
projection operator we have $\ldots\mathcal{Q}_{0}\text{\bf
F}_{0}\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{0}}\rangle=\ldots\mathcal{Q}_{0}T\bm{\nabla}_{0}\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{0}}\rangle$ and it is the latter form that is used in the
definition of the irreducible evolution operator, Eq. (151) and the derivation
of the mode-coupling vertex, Eq. (154).

Equations (145, 146) express the transport coefficients of the tracer as a
function of the current-current, memory tensor $\mathbf{\widetilde{\text{ \bf
K}}}_{0}$. As usual in a mode-coupling approach, we first give an irreducible
representation of this memory kernel and then expand it within the mode-
coupling approximation.

The reduction of $\mathbf{\widetilde{\text{ \bf K}}}$ is achieved by
introducing an irreducible evolution operator

$\Omega_{{-\gamma}}^{\text{irrs}}=\Omega_{-\gamma}+D_{0}\beta^{2}\sum_{\text{\bf
k}}\mathcal{Q}_{0}{\bf j}_{0}(\text{\bf
k})\rangle\cdot(\mathbf{1}-\gamma\text{\bf A})\cdot\langle{\bf
j}_{0}^{*}(\text{\bf k})\mathcal{Q}_{0}$ (148)

and exploiting the identity

$\begin{split}\frac{1}{z-\mathcal{Q}_{0}\Omega_{-\gamma}\mathcal{Q}_{0}}&=\frac{1}{z-\Omega_{{-\gamma}}^{\text{irrs}}}\Biggl{[}1-\frac{D_{0}\beta^{2}}{V}\sum_{\text{\bf
k}}\mathcal{Q}_{0}{\bf j}_{0}(\text{\bf k})\rangle\\\
&\cdot(\mathbf{1}-\gamma\text{\bf A})\cdot\langle{\bf j}_{0}^{*}(\text{\bf
k})\mathcal{Q}_{0}\frac{1}{z-\mathcal{Q}_{0}\Omega_{-\gamma}\mathcal{Q}_{0}}\Biggr{]}.\end{split}$
(149)

We obtain from Eq. (147)

$\mathbf{\widetilde{\text{ \bf K}}}_{0}(\text{\bf
q},z)=\left[\mathbf{1}+\text{\bf K}_{0}(\text{\bf
q},z)(\mathbf{1}-\gamma\text{\bf A})\right]^{-1}\cdot\text{\bf
K}_{0}(\text{\bf q},z),$ (150)

where we have introduced the irreducible memory kernel for the tracer

$\text{\bf K}_{0}(\text{\bf q},z)\equiv\beta^{2}D_{0}\left\langle{\bf
j}^{*}_{0}(\text{\bf
q})\mathcal{Q}_{0}\frac{1}{z-\Omega_{{-\gamma}}^{\text{irrs}}}\mathcal{Q}_{0}{\bf
j}_{0}(\text{\bf q})\right\rangle.$ (151)

As a consequence of the linear response treatment in the definition of the
$\text{\bf K}_{0}(\text{\bf q},z)$ the evolution operator involves the
dynamics with transverse forces in the absence of the external perturbation.
This allows us to simplify the tracer’s memory kernel, as was done previously
for the memory kernel related to the dynamics structure factor, $\text{\bf
K}(\text{\bf q},z)$. Before discussing further this point, we apply the mode-
coupling approximation scheme to $\text{\bf K}_{0}$.

###  V.2 Mode-coupling expansion of the tracer’s memory kernel

The mode-coupling expansion can be carried out in an analogous way to what was
done in Sec. IV.3. The tracer’s current is expanded along the following
density product:

$\mathcal{Q}_{0}{\bf j}_{0}(\text{\bf q})\approx\sum_{\text{\bf
k}}\frac{n(\text{\bf k})n_{0}(\text{\bf q}-\text{\bf k})}{NS(\text{\bf
k})}\left\langle n^{*}(\text{\bf k})n_{0}^{*}(\text{\bf q}-\text{\bf
k})\mathcal{Q}_{0}{\bf j}_{0}(\text{\bf q})\right\rangle.$ (152)

The Gaussian approximation now becomes

$\displaystyle\left\langle n^{*}(\text{\bf k}^{\prime})n_{0}^{*}(\text{\bf
q}-\text{\bf
k}^{\prime})\text{e}^{\Omega_{{-\gamma}}^{\text{irrs}}t}n(\text{\bf
k})n_{0}(\text{\bf q}-\text{\bf k})\right\rangle$ (153)
$\displaystyle\approx\left\langle n^{*}(\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n(\text{\bf
k})\right\rangle\left\langle n_{0}^{*}(\text{\bf q}-\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n_{0}(\text{\bf q}-\text{\bf
k})\right\rangle$ $\displaystyle+\left\langle n^{*}(\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n_{0}(\text{\bf q}-\text{\bf
k})\right\rangle\left\langle n_{0}^{*}(\text{\bf q}-\text{\bf
k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n(\text{\bf k})\right\rangle$
$\displaystyle=NS(\text{\bf k},t)F_{s}(\text{\bf q}-\text{\bf
k},t)\delta_{\text{\bf k},\text{\bf k}^{\prime}}+S(\text{\bf k},t)S(\text{\bf
q}-\text{\bf k},t)\delta_{\text{\bf k}^{\prime},\text{\bf q}-\text{\bf k}}$
$\displaystyle\approx NS(\text{\bf k},t)F_{s}(\text{\bf q}-\text{\bf
k},t)\delta_{\text{\bf k},\text{\bf k}^{\prime}}.$

In the last passage we have neglected the term of order $1$ compared to the
term of order $N^{2}$. In addition, we noted that since the tracer is
equivalent to any other particle, we have $\left\langle n_{0}^{*}(\text{\bf
q}-\text{\bf k}^{\prime})\text{e}^{\Omega_{-\gamma}t}n_{0}(\text{\bf
q}-\text{\bf k})\right\rangle=F_{s}(\text{\bf q}-\text{\bf k})$, with the
self-intermediate correlation function $F_{s}$ defined in Eq. (50).

We now compute the average in Eq. (152) with the help of a convolution
approximation

$\displaystyle\left\langle n(\text{\bf k})^{*}n_{0}^{*}(\text{\bf q}-\text{\bf
k})\mathcal{Q}_{0}{\bf j}_{0}(\text{\bf q})\right\rangle$ (154)
$\displaystyle=\left\langle n(\text{\bf k})^{*}n_{0}^{*}(\text{\bf
q}-\text{\bf k}){\bf j}_{0}(\text{\bf q})\right\rangle$
$\displaystyle-\left\langle n(\text{\bf k})^{*}n_{0}^{*}(\text{\bf
q}-\text{\bf k})n_{0}(\text{\bf q})\right\rangle\left\langle
n^{*}_{0}(\text{\bf q}){\bf j}_{0}(\text{\bf q})\right\rangle$
$\displaystyle=T\left\langle n(\text{\bf k})^{*}n_{0}^{*}(\text{\bf
q}-\text{\bf k})\bm{\nabla}_{0}\text{e}^{-i\text{\bf q}\cdot\text{\bf
r}_{0}}\right\rangle$ $\displaystyle-TS(k)\left\langle n^{*}_{0}(\text{\bf
q})\bm{\nabla}_{0}\text{e}^{-i\text{\bf q}\cdot\text{\bf r}_{0}}\right\rangle$
$\displaystyle=-T\left\langle\bm{\nabla}_{0}\left[n(\text{\bf
k})^{*}n_{0}^{*}(\text{\bf q}-\text{\bf k})\right]\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{0}}\right\rangle$
$\displaystyle+TS(k)\left\langle\bm{\nabla}_{0}\left[n^{*}_{0}(\text{\bf
q})\right]\text{e}^{-i\text{\bf q}\cdot\text{\bf r}_{0}}\right\rangle$
$\displaystyle=-iT\left[\text{\bf k}+\left(\text{\bf q}-\text{\bf
k}\right)S(k)-\text{\bf q}S(k)\right]$
$\displaystyle=-iT\rho_{0}S(k)c(k)\text{\bf k},$

with $\rho_{0}c(k)=1-\frac{1}{S(k)}$ the direct correlation function. With
these results, substitution of Eq. (152) in the irreducible memory kernel of
Eq. (151) yields

$\text{\bf K}_{0}(\text{\bf q},t)\approx D_{0}\rho_{0}\int_{\text{\bf
k}}\text{\bf k}\otimes\text{\bf k}c(k)^{2}F_{s}(\text{\bf q}-\text{\bf
k},t)S(\text{\bf k},t).$ (155)

This is the mode-coupling expression of the tracer’s memory kernel. Using the
symmetries of the dynamics, as done in Sec. IV.3, one can show that $\text{\bf
K}_{0}$ is diagonal in the $\text{\bf A}-\text{\bf q}$ basis given by Eq.
(73):

$\text{\bf K}_{0}(\text{\bf q},t)=\sum_{i}K_{0,ii}(\text{\bf q},t)\text{\bf
e}_{i}\otimes\text{\bf e}_{i}.$ (156)

Now that we have a diagonal decomposition of the tracer’s memory kernel, we
can focus on physical quantities of interest for the tracer dynamics.

###  V.3 Self-intermediate scattering function

The dynamics of the self-intermediate scattering function $F_{s}$ can be read
from the tracer’s dynamics. It is given by

$zF_{s}(\text{\bf q})-1=-D_{0}\text{\bf
q}\cdot\Biggl{[}\frac{1+(1+\gamma^{2})K_{0,22}}{(1+K_{0,11})(1+K_{0,22})+\gamma^{2}K_{0,11}K_{0,22}}\text{\bf
e}_{1}\otimes\text{\bf e}_{1}+\frac{1}{1+K_{0,33}}\text{\bf
e}_{3}\otimes\text{\bf e}_{3}\Biggr{]}\cdot\text{\bf q}F_{s}(\text{\bf q},z).$
(157)

This equation of motion is formally similar to Eq. (83), with the collective
memory kernel being replaced by the tracer’s memory kernel. For a wavevector q
lying in the $(xy)$ plane, we get, using an inverse Laplace transform

$\partial_{t}F_{s}+D_{0}q^{2}F_{s}+D_{0}q^{2}(1+\gamma^{2})K_{0,22}*F_{s}=-\left[\left(K_{0,11}+K_{0,22}\right)+(1+\gamma^{2})K_{0,11}*K_{0,22}\right]*\partial_{t}F_{s},$
(158)

where the dependence on $(\text{\bf q},t)$ has been omitted for clarity. This
equation was announced in an earlier work (Eq. (11) in [38]). The properties
of this dynamics can be analyzed in the same way illustrated for the dynamical
structure factor in Sec. IV.5.

###  V.4 Diffusion tensor

The diffusion tensor at any finite wavevector and frequency is obtained by
substituting the mode-coupling expression for $\text{\bf K}_{0}$, given in Eq.
(156), into Eq. (146). The result reads, in the $\text{\bf A}-\text{\bf q}$
basis introduced in Eq. (73)

$\begin{split}&\text{\bf D}(\text{\bf
q},z)=D_{0}\Biggl{[}\frac{1+(1+\gamma^{2})K_{0,22}}{(1+K_{0,11})(1+K_{0,22})+\gamma^{2}K_{0,11}K_{0,22}}\text{\bf
e}_{1}\otimes\text{\bf e}_{1}+\frac{1}{1+K_{0,33}}\text{\bf
e}_{3}\otimes\text{\bf e}_{3}\\\
&+\gamma\frac{K_{0,11}+K_{0,22}+(1+\gamma^{2})K_{0,11}K_{0,22}}{(1+K_{0,11})(1+K_{0,22})+\gamma^{2}K_{0,11}K_{0,22}}\left[\text{\bf
e}_{1}\otimes\text{\bf e}_{2}-\text{\bf e}_{2}\otimes\text{\bf
e}_{1}\right]+\frac{(1+K_{0,11})K_{0,22}+K_{0,11}(K_{0,22}-1)\gamma^{2}}{(1+K_{0,11})(1+K_{0,22})+\gamma^{2}K_{0,11}K_{0,22}}\text{\bf
e}_{2}\otimes\text{\bf e}_{2}\Biggr{]}.\end{split}$ (159)

The expression of the diffusion tensor is one of the main results of this
work. From the large wavelength, small frequency limit of Eq. (159) we obtain
the diffusion matrix D. In Cartesian coordinates, it reads

$\begin{split}\text{\bf D}&=D_{\parallel,x}\left[\text{\bf
e}_{x}\otimes\text{\bf e}_{x}+\text{\bf e}_{y}\otimes\text{\bf
e}_{y}\right]\\\ &+D_{\parallel,z}\text{\bf e}_{z}\otimes\text{\bf e}_{z}\\\
&+D_{\perp}\left[\text{\bf e}_{y}\otimes\text{\bf e}_{x}-\text{\bf
e}_{z}\otimes\text{\bf e}_{y}\right].\end{split}$ (160)

We now discuss the expression and the behavior of the different diffusion
constants that appear in the diffusion tensor D.

####  V.4.1 Longitudinal diffusion

The dynamics with transverse forces is anisotropic. As a consequence, there
are two types of longitudinal diffusion constants, one related to the
diffusion in the $(xy)$ plane (within which rotational invariance is
preserved), $D_{\parallel,x}$ and one related to the diffusion along the $z$
direction, $D_{\parallel},z$. Their expressions are

$\begin{split}D_{\parallel,x}(\gamma)&=D_{0}\frac{1+(1+\gamma^{2})K_{0,11}^{\infty}}{(1+K_{0,11}^{\infty})(1+K_{0,22}^{\infty})+\gamma^{2}K_{0,11}^{\infty}K_{0,22}^{\infty}}\\\
D_{\parallel,z}(\gamma)&=D_{0}\frac{1}{1+K_{0,33}^{\infty}},\end{split}$ (161)

where we introduced the notation $K_{0,ii}^{\infty}\equiv K_{0,ii}(\text{\bf
q}\to 0,z\to 0)$ to denote the long wavelength, zero frequency of the tracer’s
memory kernel. This coefficient reads explicitly

$K_{0,ii}^{\infty}=\frac{D_{0}}{\rho_{0}}\int_{0}^{+\infty}\differential
t\int_{\text{\bf k}}[k_{i}\rho_{0}c(k)]^{2}F_{s}(-\text{\bf k},t)S(\text{\bf
k},t).$ (162)

For $\gamma=0$, the equilibrium result is duly recovered:

$D_{aa,\text{\tiny{eq}}}=D_{0}\frac{1}{1+K_{0,\parallel,\text{\tiny{eq}}}^{\infty}},$
(163)

for $a=x,y,z$. When $\gamma\neq 0$ the anisotropy of the dynamics is revealed
through the different expressions of the diffusion constants.

The dynamics is indeed accelerated. If we assume that in the ergodic phase
$K_{0,ii}^{\infty}\leq K_{0,ii,\text{\tiny{eq}}}^{\infty}$, we self-
consistently obtain $D_{aa}(\gamma)\geq D_{aa}(0)$.

In the ergodic phase in the limit of strong drive, the memory kernel scales as
$\gamma^{-1}$. This implies that the efficiency of transverse forces with
respect to the equilibrium dynamics, computed as the ratio
$D(\gamma)\equiv\sum_{a=x,y,z}D_{aa}(\gamma)$ over $D(0)$, grows as

$\frac{D(\gamma)}{D(0)}\sim\gamma.$ (164)

We now investigate two limiting cases: in the high temperature regime, the
memory kernels vanish because the fluid enters an effectively noninteracting
limit and therefore $D(\gamma)=D_{0}$, while at $T=T_{\text{MCT}}$ the memory
kernels diverge as predicted by the schematic approach of Sec. IV.4, implying
that $D(\gamma)=0$ and dynamical arrest occurs. We have therefore

$\displaystyle\lim_{T\to\infty}\frac{D(\gamma)}{D(0)}$
$\displaystyle=c(\gamma),$ (165) $\displaystyle\lim_{T\to
T_{\text{MCT}}}\frac{D(\gamma)}{D(0)}$ $\displaystyle=1,$

With $c(\gamma)>1$ a constant that depends on the intensity of the driving
force. These conditions, together with the acceleration in the ergodic phase,
demonstrate that the ratio of the diffusion constants
$\frac{D(\gamma)}{D_{0}}$ exhibits a maximum as a function of the temperature
in the ergodic phase. This maximum is observed in numerical simulations [38].

####  V.4.2 Odd diffusivity

The odd diffusion constant is encoded in the antisymmetric part of the
diffusivity tensor:

$D_{\perp}(T,\gamma)=-D_{0}\gamma\frac{K^{\infty}_{0,11}+K^{\infty}_{0,22}+(1+\gamma^{2})K^{\infty}_{0,11}K^{\infty}_{0,22}}{(1+K^{\infty}_{0,11})(1+K^{\infty}_{0,22})+\gamma^{2}K^{\infty}_{0,11}K^{\infty}_{0,22}}.$
(166)

For an asymptotically large driving, $\gamma\to\infty$, the odd diffusion
constant grows linearly in $\gamma$, $D_{\perp}(\gamma,T)\sim\gamma$. This
suggests that strong transverse forces manifest themselves with an
increasingly swirling motion.

Close to ergodicity breaking where the memory kernel diverges, we obtain

$\lim_{T\to T_{\text{MCT}}}D_{\perp}=-\gamma D_{0}.$ (167)

Therefore, even if the longitudinal diffusion constant associated to particle
transport goes to $0$ at the critical temperature, the odd diffusion constant
remains nonzero. Together with the vanishing of the diffusion constant, this
suggests a physical picture where particles perform a swirling motion inside
the permanent local cage made by their neighbors. An analogous situation can
be shown to arise in the much simpler model of a particle in an harmonic well
under the action of transverse forces [38].

###  V.5 Mobility tensor

The mobility tensor is an alternative quantity characterizing the nature of
the dynamics with a physical content distinct from that of the diffusivity
tensor due to the breaking of the Einstein relation. The mobility tensor is
obtained from Eq. (145), using the mode-coupling expansion of the tracer’s
memory kernel given in Eq. (156). The result reads

$\begin{split}\bm{\mu}(\text{\bf
q},z)&=D_{0}\beta\Biggl{[}\frac{1+K_{22}}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}\text{\bf
e}_{1}\otimes\text{\bf e}_{1}+\frac{1}{1+K_{33}}\text{\bf
e}_{3}\otimes\text{\bf e}_{3}\\\ &+\frac{\gamma\left(K_{0,11}\text{\bf
e}_{2}\otimes\text{\bf e}_{1}-K_{0,22}\text{\bf e}_{1}\otimes\text{\bf
e}_{2}\right)}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}+\frac{1+K_{11}}{(1+K_{11})(1+K_{22})+\gamma^{2}K_{11}K_{22}}\text{\bf
e}_{2}\otimes\text{\bf e}_{2}\Biggr{]}.\end{split}$ (168)

In the small frequency and large wavelength limit of these expression, we
obtain the mobility of the tracer,

$\begin{split}\bm{\mu}&=\mu_{\parallel,x}\left[\text{\bf
e}_{x}\otimes\text{\bf e}_{x}+\text{\bf e}_{y}\otimes\text{\bf
e}_{y}\right]+\mu_{\parallel,z}\text{\bf e}_{z}\otimes\text{\bf e}_{z}\\\
&+\mu_{\perp}\left[\text{\bf e}_{x}\otimes\text{\bf e}_{y}-\text{\bf
e}_{y}\otimes\text{\bf e}_{x}\right],\end{split}$ (169)

with the longitudinal mobilities $\mu_{\parallel,x}$,$\mu_{\parallel,z}$, and
odd mobility $\mu_{\perp}$ given by

$\begin{split}\mu_{\parallel,x}&=D_{0}\beta\frac{1+K_{0,11}^{\infty}}{(1+K_{0,11}^{\infty})^{2}+\gamma^{2}K_{0,11}^{\infty}},\\\
\mu_{\parallel,z}&=D_{0}\beta\frac{1}{1+K_{0,33}^{\infty}},\\\
\mu_{\perp}&=-\gamma
D_{0}\beta\frac{K_{0,11}^{\infty}}{(1+K_{0,11}^{\infty})^{2}+\gamma^{2}K_{0,11}^{\infty}}.\end{split}$
(170)

The anisotropic character of transverse forces in three dimensions manifests
itself through the fact that $\mu_{\parallel,z}\neq\mu_{\parallel,x}$. In the
large $\gamma$-limit, all nonzero entries of the mobility tensor converge (in
modulus) to the mobility of a free tracer, $D_{0}\beta$. This is in contrast
with the behavior of the diffusivity tensor, whose nonzero entries grow
linearly in $\gamma$. This is a consequence of the breakdown of the
fluctuation-dissipation theorem, arising from the nonequilibrium nature of the
dynamics. A variant of the fluctuation-dissipation theorem survives for the
longitudinal components of the diffusion and mobility tensor, as
$[\mu(\mathbf{1}+\gamma\text{\bf A})]_{aa}=\beta[\text{\bf D}]_{aa}$ with
$a=x,y,z$. This variant of the fluctuation-dissipation theorem was also
derived in the dynamical mean-field treatment of transverse forces [39].

As the glass transition is approached, all components of the mobility tensor
vanish. Physically, this means that in the dynamically arrested glass, a weak
external force cannot set the particles in motion, neither in the longitudinal
nor in the transverse direction with respect to the external force.

The vanishing of longitudinal and odd mobilities is one of the facets of the
glass transition in the mode-coupling theory of transverse forces. Another
facet is the divergence of the viscosity tensor, which will be explored in the
next section.

##  VI Odd viscosity

This section is devoted to the study of the emerging odd viscosity [58, 59,
60, 61, 62]. Following [52, 60], we compute the viscosity tensor in the long
wavelength and long times limits by looking at the response of the stress to
an external perturbation. To do so, we define a microscopic stress tensor in
the presence of transverse forces and develop its linear response theory,
leading to the definition of the viscosity tensor, and in particular to the
identification of its odd component.

###  VI.1 Stress tensor and transverse forces

Following [52], we start by defining the microscopic stress tensor in the
presence of transverse forces from the equation

$\partial_{t}n(\text{\bf q})=\Omega^{\dagger}_{\gamma}n(\text{\bf q})\equiv-
D_{0}\beta q_{a}q_{b}\sigma_{ab}(\text{\bf q}).$ (171)

This identity, which only holds when inserted into the average
$\langle\ldots\rangle$, yields

$\sigma_{ab}(\text{\bf q})\equiv\sum_{i}\left[i(\delta_{bc}+\gamma
A_{bc})\frac{q_{a}F_{i,c}}{q^{2}}+T\delta_{ab}\right]\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{i}}.$ (172)

For $\gamma=0$, we recover the expression of the microscopic equilibrium
stress tensor, $\sigma^{\text{eq}}(\text{\bf q})$, given by

$\sigma^{\text{eq}}_{ab}(\text{\bf
q})\equiv\sum_{i}\left[i\frac{q_{a}F_{i,b}}{q^{2}}+T\delta_{ab}\right]\text{e}^{-i\text{\bf
q}\cdot\text{\bf r}_{i}}.$ (173)

For equilibrium dynamics, the viscosity is defined as the response function of
$\mathcal{Q}\sigma^{\text{eq}}$ to particle current fluctuations, as discussed
in [63] for underdamped Langevin dynamics and extended in [52] for the
overdamped case. In the presence of transverse forces one must substitute
$\sigma^{\text{eq}}$ with $\sigma$ given by Eq. (172). This gives rise to
coupling between parallel and longitudinal components of the stress tensor,
which eventually results in odd viscosity.

Note that the projection of the stress tensor in the space orthogonal to the
density fluctuations, $\mathcal{Q}\sigma_{ab}(\text{\bf q})$, is related with
its equilibrium counterpart by the following relation

$\mathcal{Q}\sigma_{ab}(\text{\bf
q})=\mathcal{Q}\sigma_{ac}^{\text{eq}}(\delta_{cb}-\gamma A_{cb}),$ (174)

where the Einstein summation convention is being used. We also have

$\mathcal{Q}j_{a}(\text{\bf
q})=-iq_{b}\mathcal{Q}\sigma_{ba}^{\text{eq}}(\text{\bf q}),$ (175)

with $\mathcal{Q}j_{a}(\text{\bf q})$ the projected force density Fourier
modes defined in Eq. (59).

In the next section, we pursue our program by developing a linear response
theory of the microscopic stress.

###  VI.2 Linear response theory

We consider the perturbation produced by a weak external velocity field made
by a single Fourier mode $\text{\bf v}(\text{\bf
r},t)\equiv\frac{1}{V}\text{\bf v}(t)\text{e}^{i\text{\bf q}\cdot\text{\bf
r}}$. The evolution operator associated to the system is now
$\Omega_{\gamma}+\delta\Omega$, with

$\displaystyle\delta\Omega$
$\displaystyle\equiv-\frac{1}{V}\sum_{i}\bm{\nabla}_{i}\cdot\text{\bf
v}(t)\text{e}^{i\text{\bf q}\cdot\text{\bf r}_{i}}.$ (176)

As a consequence of the perturbation, the probability distribution associated
to the system becomes $\rho_{\text{\tiny{B}}}(\text{\bf
r}^{N})+\delta\rho(\text{\bf r}^{N},t)$. The equation of motion for the
perturbation $\delta\rho(\text{\bf r}^{N},t)$ reads, to linear order,

$\partial_{t}\delta\rho(\text{\bf
r}^{N},t)=\Omega_{\gamma}\delta\rho(\text{\bf
r}^{N},t)+\delta\Omega(t)\rho_{\text{\tiny{B}}}(\text{\bf r}^{N}).$ (177)

The solution to this equation is given by

$\displaystyle\delta\rho(\text{\bf
r}^{N},t)=\int_{-\infty}^{t}d\tau\text{e}^{\Omega_{\gamma}(t-\tau)}\delta\Omega(\tau)\rho_{\text{\tiny{B}}}(\text{\bf
r}^{N})$ (178)
$\displaystyle=-\frac{\beta}{V}\int_{-\infty}^{t}\differential\tau\text{e}^{\Omega_{\gamma}(t-\tau)}v_{b}(\tau)iq_{a}\sigma^{\text{eq}*}_{ab}(\text{\bf
q})\rho_{\text{\tiny{B}}}(\text{\bf r}^{N}).$

The average of the stress tensor in linear response reads

$\displaystyle\left\langle\mathcal{Q}\sigma_{ab}(\text{\bf
q},t)\right\rangle^{\text{lr}}$
$\displaystyle=-\frac{\beta}{V}\int_{-\infty}^{t}\differential\tau
iq_{c}v_{d}(\tau)$ (179)
$\displaystyle\times\left\langle\sigma^{\text{eq}*}_{cd}(\text{\bf
q})\text{e}^{\Omega_{-\gamma}(t-\tau)}\mathcal{Q}\sigma_{\alpha\beta}(\text{\bf
q})\right\rangle$
$\displaystyle=-\frac{\beta}{V}\int_{-\infty}^{t}\differential\tau
iq_{c}v_{d}(\tau)$
$\displaystyle\times\left\langle\sigma^{\text{eq}*}_{cd}(\text{\bf
q})\text{e}^{\Omega_{-\gamma}(t-\tau)}\mathcal{Q}\sigma_{ab}(\text{\bf
q})\right\rangle\theta(t-\tau).$

The average $\langle\ldots\rangle^{\text{lr}}$ is an average over the dynamics
described by the operator $\Omega_{\gamma}+\delta\Omega$, neglecting terms of
order higher than linear in $\text{\bf v}(t)$. We are interested in computing
the variation of the stress with respect to the small perturbation. To do
this, it is more suitable to work in Fourier space, by denoting the Fourier
transform of a function $f(t)$ as
$f(\omega)\equiv\int_{-\infty}^{+\infty}f(t)\text{e}^{-i\omega t}$. Within the
linear response framework, we can take the perturbation v to be a
monochromatic plane wave, i.e. $\text{\bf v}(t)=\text{\bf
v}(\omega)\text{e}^{-i\omega t+0_{+}t}$. The symbol $0_{+}$ has to be
understood as a small positive infinitesimal quantity used to ensure that the
perturbation goes to $0$ at $t=-\infty$. Substitution into Eq. (179) yields

$\frac{\partial\left\langle\mathcal{Q}\sigma_{ab}(\text{\bf
q},\omega)\right\rangle^{\text{lr}}}{\partial
iq_{c}v_{d}}=\frac{\beta}{V}\left\langle\sigma^{\text{eq}*}_{cd}(\text{\bf
q})\frac{1}{-i\omega-\Omega_{-\gamma}}\mathcal{Q}\sigma_{ab}(\text{\bf
q})\right\rangle.$ (180)

This is the response of $\mathcal{Q}\sigma_{ab}(\text{\bf q})$ to an external
velocity gradient. To obtain the viscosity, we need to work out the response
of $\mathcal{Q}\sigma_{ab}(\text{\bf q})$ to a change in the gradients of the
particle current. However, within linear response, we can work out a relation
between the particle currents and the applied velocity field, thus making the
calculation of the viscosity tensor possible. This is done next.

###  VI.3 The viscosity tensor

Following [63, 53, 52], we introduce the average particle current $\text{\bf
J}(\text{\bf q},t)$, governing the evolution of the density mode $n(\text{\bf
q})$

$\left\langle\partial_{t}n(\text{\bf q},t)\right\rangle^{\text{lr}}\equiv
i\rho_{0}\text{\bf q}\cdot\text{\bf J}(\text{\bf q},t).$ (181)

The left hand side of Eq. (181) splits into

$\left\langle\partial_{t}n(\text{\bf
q},t)\right\rangle^{\text{lr}}=\left\langle n(\text{\bf
q},t)\delta\Omega(t)\right\rangle+\left\langle n(\text{\bf
q},t)\Omega_{\gamma}\right\rangle^{\text{lr}},$ (182)

which implies

$J_{a}(\text{\bf q},t)=v_{a}(\text{\bf
q},t)-\frac{D_{0}\beta}{\rho_{0}}q_{b}\left\langle\sigma_{ab}(\text{\bf
q},t)\right\rangle^{\text{lr}}.$ (183)

Thus, the particle current corresponds to sum of the imposed solvent velocity
v and of a term that originates from the change in the stress tensor.

The viscosity tensor is defined as the response function describing the change
of the projected stress tensor due to the gradient of the average particle
current [63]:

$\eta_{abcd}(\text{\bf
q},\omega)\equiv\frac{\partial\left\langle\mathcal{Q}\sigma_{ab}(\text{\bf
q},\omega)\right\rangle^{\text{lr}}}{\partial iq_{c}J_{d}(\text{\bf
q},\omega)}.$ (184)

Our aim is now to show that the viscosity tensor is related to the following
correlation function,

$C^{\text{irr}}_{abcd}(\text{\bf
q},\omega)\equiv\frac{\beta}{V}\left\langle\sigma^{\text{eq}*}_{ab}(\text{\bf
q})\mathcal{Q}\frac{1}{-i\omega-\Omega^{\text{irr}}_{-\gamma}}\mathcal{Q}\sigma_{cd}(\text{\bf
q})\right\rangle,$ (185)

which encodes the correlation between the equilibrium stress tensor and the
stress tensor in presence of transverse forces, evolving with the irreducible
operator $\Omega^{\text{irr}}_{-\gamma}$ defined in Eq. (63).

We first introduce an auxiliary correlation function $C^{\mathcal{Q}}_{abcd}$,

$C^{\mathcal{Q}}_{abcd}\equiv\frac{\beta}{V}\left\langle\sigma_{ab}^{\text{eq}*}(\text{\bf
q})\mathcal{Q}\frac{1}{-i\omega-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\sigma_{cd}(\text{\bf
q})\right\rangle,$ (186)

which is the analogue of $C^{\text{irr}}_{abcd}$ but evolves with the
projected evolution operator $\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}$. Using
the operator identity

$\begin{split}\frac{1}{-i\omega-\mathcal{Q}\Omega_{-\gamma}}&=\frac{1}{-i\omega-\Omega_{-\gamma}}\\\
&+\frac{1}{-i\omega-\Omega_{-\gamma}}\mathcal{P}\Omega_{-\gamma}\frac{1}{-i\omega-\mathcal{Q}\Omega_{-\gamma}},\end{split}$
(187)

we obtain

$\begin{split}\frac{\partial\langle\mathcal{Q}\sigma_{ab}(\text{\bf
q})\rangle^{\text{lr}}}{\partial
iq_{c}v_{d}}&=\Biggl{[}\delta_{ce}\delta_{dg}\\\
&-\frac{D_{0}\beta}{S(q)}q_{f}\frac{\partial\langle n(\text{\bf
q},t)\rangle^{\text{lr}}}{\partial iq_{c}v_{d}}(\delta_{fg}+\gamma
A_{fg})q_{e}\Biggr{]}C^{\mathcal{Q}}_{egab}.\end{split}$ (188)

On the other hand, the correlators $C^{\mathcal{Q}}_{abcd}$ and
$C^{\text{irr}}_{abcd}$ are related to each other. We can exploit the
resolvent identity given by Eq. (65) and the fact that the operator
$\delta\Omega_{-\gamma}$ defined in Eq. (64) can be expressed in terms of the
stress tensor through Eq. (175),

$\delta\Omega_{-\gamma}=\frac{D_{0}\beta^{2}}{N}q_{a}q_{d}\mathcal{Q}\sigma^{\text{eq}}_{ab}(\text{\bf
q})\rangle[\delta_{bc}-\gamma
A_{bc}]\langle\sigma_{dc}^{\text{eq}}\mathcal{Q}.$ (189)

This yields

$\begin{split}C^{\text{irr}}_{abcd}&=C^{\mathcal{Q}}_{abcd}-\frac{D_{0}\beta^{2}}{\rho_{0}V}q_{e}q_{f}\\\
&\times\langle\sigma^{\text{eq}*}_{ab}(\text{\bf
q})\frac{1}{-i\omega-\mathcal{Q}\Omega_{-\gamma}\mathcal{Q}}\mathcal{Q}\sigma^{\text{eq}}_{ef}(\text{\bf
q})\rangle(\delta_{fg}-\gamma A_{fg})C^{\text{irr}}_{fgcd}\\\
&=C^{\mathcal{Q}}_{abcd}-\frac{D_{0}\beta}{\rho_{0}}q_{e}q_{f}C^{\mathcal{Q}}_{abeg}C^{\text{irr}}_{fgcd}.\end{split}$
(190)

This equation is analogous to the relation between the memory kernels K and
$\mathbf{\widetilde{\text{ \bf K}}}$ given by Eq. (67). Contracting both sides
of Eq. (190) with the quantity inside the brackets in Eq. (188) and
rearranging the terms gives

$\begin{split}\frac{\partial\langle\mathcal{Q}\sigma_{ab}(\text{\bf
q})\rangle^{\text{lr}}}{\partial
iq_{c}v_{d}}&=\Biggl{[}\delta_{ce}\delta_{df}+\frac{D_{0}\beta}{\rho_{0}}q_{e}q_{g}\frac{\partial\langle\mathcal{Q}\sigma_{gf}(\text{\bf
q})\rangle^{\text{lr}}}{\partial iq_{c}v_{d}}\\\
&-q_{e}q_{g}\frac{D_{0}}{\rho_{0}S(q)}\frac{\partial\langle n(\text{\bf
q})\rangle^{\text{lr}}}{\partial iq_{c}v_{d}}(\delta_{gf}+\gamma
A_{gf})\Biggr{]}C^{\text{irr}}_{efab}.\end{split}$ (191)

On the other hand, from Eq. (183) we have

$\begin{split}\frac{\partial iq_{e}J_{f}}{\partial
iq_{c}v_{d}}&=\delta_{ce}\delta_{df}+\frac{D_{0}\beta}{\rho_{0}}q_{e}q_{g}\frac{\partial\langle\mathcal{Q}\sigma_{hf}(\text{\bf
q})\rangle^{\text{lr}}}{\partial iq_{c}v_{d}}\\\
&-q_{e}q_{g}\frac{D_{0}}{\rho_{0}S(q)}\frac{\partial\langle n(\text{\bf
q})\rangle^{\text{lr}}}{\partial iq_{c}v_{d}}\delta_{gf}.\end{split}$ (192)

Combining Eqs. (191, 192), using the chain rule and the definition of the
viscosity tensor given by Eq. (184) we obtain

$\begin{split}\frac{\partial iq_{e}J_{f}}{\partial
iq_{c}v_{d}}\eta_{abef}(\text{\bf q},\omega)&=\Biggl{[}\frac{\partial
iq_{e}J_{f}}{\partial iq_{c}v_{d}}\\\ &-\gamma
q_{e}q_{g}\frac{D_{0}}{\rho_{0}S(q)}\frac{\partial\langle n(\text{\bf
q})\rangle^{\text{lr}}}{\partial
iq_{c}v_{d}}A_{gf}\Biggr{]}C^{\text{irr}}_{efab}(\text{\bf
q},\omega).\end{split}$ (193)

This equation relates the irreducible stress-stress correlator with the
viscosity. For $\gamma=0$, we recover the relation derived in [52],

$\eta_{abcd}(\text{\bf
q},\omega)=\frac{\beta}{V}\langle\sigma^{\text{eq}*}_{cd}(\text{\bf
q})\frac{1}{-i\omega-\Omega^{\text{irr}}_{0}}\mathcal{Q}\sigma^{\text{eq}}_{ab}(\text{\bf
q})\rangle.$ (194)

When $\gamma\neq 0$, and additional term stemming from the influence of the
transverse forces and proportional to the response of the density field to
external currents appears. Moreover, the stress-stress correlator
$C^{\text{irr}}$ depends on $\gamma$ both through the definition of the stress
$\sigma_{ab}$ and the dynamical evolution operator
$\Omega_{-\gamma}^{\text{irr}}$.

In the hydrodynamic limit, when $q\to 0$ and $\omega\to 0$, the second term on
the right hand side of Eq. (193) is subleading compared to the first one,
leading to

$\eta_{abcd}(\text{\bf q}\to 0,\omega\to 0)=C^{\text{irr}}_{cdab}(\text{\bf
q}\to 0,\omega\to 0).$ (195)

This equation relates the hydrodynamic viscosity with the stress-stress
irreducible correlator. In the next two sections, we use this formula to
compute, within the mode-coupling approximation, the shear and odd
viscosities.

###  VI.4 Shear viscosity

We start with the shear viscosity $\eta_{xyxy}$. Due to the rotational
invariance of the dynamics in the $(xy)$ plane, we evaluate the $\text{\bf
q}\to 0$ limit with $\text{\bf q}=q\text{\bf e}_{x}$. From Eq. (195) we obtain

$\begin{split}\eta_{xyxy}&=\lim_{q\to 0}\lim_{\omega\to
0}\frac{\beta}{V}\left\langle\sigma^{\text{eq}*}_{xy}(\text{\bf
q})\mathcal{Q}\frac{1}{-i\omega-\Omega_{-\gamma}^{\text{irr}}}\mathcal{Q}\sigma_{xy}^{\text{eq}}(\text{\bf
q})\right\rangle\\\
&-\gamma\frac{\beta}{V}\left\langle\sigma^{\text{eq}*}_{xy}(\text{\bf
q})\mathcal{Q}\frac{1}{-i\omega-\Omega_{-\gamma}^{\text{irr}}}\mathcal{Q}\sigma_{xx}^{\text{eq}}(\text{\bf
q})\right\rangle.\end{split}$ (196)

We now replace $\mathcal{Q}\sigma_{ab}^{\text{eq}}(\text{\bf q})$ with