PHYSICAL REVIEW E 103, 042610 (2021)

Nonreciprocal response of a two-dimensional fluid with odd viscosity
Yuto Hosaka

and Shigeyuki Komura*

Department of Chemistry, Graduate School of Science, Tokyo Metropolitan University, Tokyo 192-0397, JapanDavid Andelman†

Raymond and Beverly Sackler School of Physics and Astronomy, Tel Aviv University, Ramat Aviv, Tel Aviv 69978, Israel(Received 30 December 2020; accepted 16 March 2021; published 27 April 2021)
We discuss the linear hydrodynamic response of a two-dimensional active chiral compressible fluid with odd
viscosity. The viscosity coefficient represents broken time-reversal and parity symmetries in the 2D fluid andcharacterizes the deviation of the system from a passive fluid. Taking into account the hydrodynamic coupling
to the underlying bulk fluid, we obtain the odd viscosity-dependent mobility tensor, which is responsible forthe nonreciprocal hydrodynamic response to a point force. Furthermore, we consider a finite-size disk moving
laterally in the 2D fluid and demonstrate that the disk experiences a nondissipative lift force in addition to thedissipative drag one.
DOI: 10.1103/PhysRevE.103.042610

I. INTRODUCTION

Two-dimensional (2D) active chiral fluids have been
predicted to have a new rheological property called odd
viscosity [1]. Over two decades ago, Avron has shown thatwhen time-reversal and parity symmetries are broken, theviscosity tensor of a 2D isotropic fluid could have an antisymmetric (odd) part that does not result in dissipation [1,2].
The origin of the odd viscosity is explained by coarse-grained
theories, such as Onsager’s reciprocal relations [1] and GreenKubo relations for viscosity coefficients [3–5]. Furthermore,in microscopic approaches, it was shown that active chiralfluids composed of self-spinning objects also exhibit odd viscosity [2,6].
To observe odd viscosity in physical systems, several protocols have been proposed. For incompressible fluids, it waspredicted that odd viscosity can emerge at a dynamical boundary that is subjected to no-stress boundary conditions [7,8].
Experimentally, Soni et al. measured odd viscosity in active chiral fluids by observing the boundary dynamics of a
fluid [9]. Odd viscosity has also been measured by using
molecular dynamics simulations and its Green-Kubo representation [4,5].
In addition, more fundamental phenomena in active chiralsystems, such as responses to point forces or finite-size bodies, have been studied theoretically [7,10–14]. Recently, thenonreciprocity of the point force response was investigated ina 3D anisotropic fluid with odd viscosity [10] as well as inan active solid material [11,12]. As for the finite-size body
response, lift force was observed in active chiral granular
media [13] and in a 2D fluid with inertia [14], whereas nosuch force was found in the incompressible limit [7]. Despite*†

komura@tmu.ac.jp
andelman@tauex.tau.ac.il

2470-0045/2021/103(4)/042610(11)these intensive findings, very little is known about the linearhydrodynamic response of a 2D active chiral fluid.
In this paper, we discuss the hydrodynamic response ofa 2D isotropic compressible fluid with odd viscosity, which
can be regarded as a 2D active chiral fluid [1–3]. Taking
into account the 3D bulk fluid coupled to the 2D fluid layer
and employing the lubrication approximation for the 3D
fluid [15–17], we analytically obtain the asymmetric mobility tensor of the 2D fluid in the presence of odd viscosity.
Because of such a nonreciprocal hydrodynamic response, aperpendicular fluid flow develops and breaks the axial symmetry of the flow with respect to the driving force. Extending
the point force response, we derive viscous forces on a rigid
disk that moves laterally in the 2D active chiral fluid. As aconsequence of the nonreciprocal hydrodynamic response due
to odd viscosity, we find that lift force acts on the driven disk.
There are two reasons that motivated us to consider a 2D
compressible fluid with odd viscosity in contact with an underlying 3D fluid. As pointed out in Ref. [7], the velocity field
is independent of the odd viscosity in a 2D incompressible
fluid, when nonslip boundary conditions are imposed on amoving object. This means that the effects of odd viscosity
cannot be directly seen in a 2D incompressible fluid. Moreover, for a pure 2D fluid at low Reynolds number, a linear
relation between the velocity and the viscous drag force acting
on a translating disk cannot be obtained [18,19]. This is known
as the Stokes paradox that originates from the constraint of
momentum conservation in a pure 2D system. This paradox
can be resolved, e.g., by considering momentum decay to anunderlying 3D fluid [20–25].
In the next section, we review the concept of the odd viscosity [1–3]. Then, we introduce in Sec. III the hydrodynamic
equations for the 2D active chiral fluid layer by taking into
account the coupling to the underlying 3D bulk fluid [15–17].
In Sec. IV, we derive the corresponding mobility tensor andcalculate the velocity fields induced by a point force or a force

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HOSAKA, KOMURA, AND ANDELMANPHYSICAL REVIEW E 103, 042610 (2021)

TABLE I. The components associated with the three viscosity coefficients ηd (dilatation), ηs (shear), and ηo (odd) in the ηi jk tensor underindex permutations [see Eq. (2)]. The + (−) sign denotes that the components are symmetric (antisymmetric) under a given index permutation.
Components that include even (odd) number of i j are symmetric (antisymmetric) under the parity transformation in a 2D system (x → −x,y → y) [3].
Components

i↔ j

k↔

i j ↔ k

Parity

δi j δkδik δ j + δi δ jk − δi j δkik δ j +  j δik + i δ jk +  jk δi

+
+
+

+
+
+

+
+
−

+
+
−

Viscosity coefficients
ηd
ηs
ηo

dipole. In Sec. V, we discuss the force and torque acting on amoving disk of a finite size. Finally, a summary of our workand some further comments are given in Sec. VI.

symmetries are broken [2]. In Table I, the above permutationsof the viscosity-tensor components are summarized.
Substituting Eq. (2) into Eq. (1), we obtain the stress tensor
of a 2D compressible fluid with odd viscosity as

II. ODD VISCOSITY

σi j = (ηd − ηs )δi j ∂k vk + ηs (∂ j vi + ∂i v j )Here we briefly review the concept of odd viscosity
and its contribution to the fluid stress tensor in two dimensions [1–3]. First, the strain rate tensor is defined asvk = (∂k v + ∂ vk )/2, where vi is the 2D velocity component, ∂i = ∂/∂ri is the 2D differential operator component,
and r = (x, y). The general linear relation between vk andthe fluid stress tensor σi j is given by1+ ηo (∂ j vi∗ + ∂i v ∗j + ∂ ∗j vi + ∂i∗ v j ),(3)2
where vi∗ ≡ ik vk is the velocity vector rotated clockwise by
π /2 and ∂i∗ ≡ ik ∂k . Since the odd viscosity does not contribute to the energy dissipation, (∂i v j )σi j , the sign of ηo canbe either positive or negative.

σi j = ηi jk vk ,III. ACTIVE CHIRAL FLUID(1)

where ηi jk is the fourth-rank viscosity tensor. Throughout ourwork, we assume summation over repeated indices. Noticethat the choice of the Cartesian coordinates is just done forconvenience and is not a requirement.
In an isotropic fluid, rotational invariance of the system
requires the symmetry of the stress tensor under the exchange
of indices i ↔ j, i.e., σi j = σ ji . This enforces the symmetry
relation of the viscosity tensor ηi jk = η jik , as inferred from
Eq. (1). Since vk is a symmetric tensor by definition, thesymmetry under the exchange k ↔  always holds, leading
to the symmetry relation ηi jk = ηi jk .
Extending the above symmetry argument, Avron introduced a new type of index exchange i j ↔ k, which implies
time-reversal transformation [1–3]. In light of such a pairexchange, the viscosity tensor can generally be split into symmetric (even) and antisymmetric (odd) parts ηi jk = ηiSjk +SAA
ηiAjk , where ηiSjk = ηkij and ηi jk = −ηki j . The antisymmetric term ηiAjk exists as a consequence of broken time-reversal
symmetry in a 2D fluid. Under the assumption that σi j isisotropic, the general viscosity tensor can be written as [2,3]We consider a 2D layer of an active chiral compressible
fluid particularly having odd viscosity, which is flat, thin,
infinitely large and overlays a 3D bulk fluid (e.g., water). Oneof the realizations of such a system is schematically depicted
in Fig. 1. The bulk fluid has a 3D shear viscosity η and is incontact with an impermeable flat wall located at z = 0, where
the fluid velocity vanishes. To clearly see the odd viscosity
effect [7], we suppose that the 2D fluid layer is compressible,
so that it has both 2D dilatational and shear viscosities, ηdand ηs , respectively. In physical systems, such a fluid canbe realized by a monolayer of amphiphiles that are loosely
packed on the interface at z = h [15–17]. More details on thephysical realization of our model will be discussed in Sec. VI.

ηi jk = ηd δi j δk + ηs (δik δ j + δi δ jk − δi j δk )1(2)+ ηo (ik δ j +  j δik + i δ jk +  jk δi ),2where ηd , ηs , and ηo are 2D dilatational, shear, and odd viscosities, respectively, δi j is the Kronecker delta, and i j is the
2D Levi-Civita tensor with xx = yy = 0 and xy = −yx = 1.
The above viscosity tensor ηi jk is symmetric under theparity transformation in a 2D system (x → −x, y → y) [3],and hence it is parity-even. This is because both σi j and vk
are parity-even in Eq. (1). However, terms that include odd
number of i j are parity-odd. Hence, one concludes from
Eq. (2) that ηo exists only if both time-reversal and parity

FIG. 1. Schematic sketch of an active chiral fluid characterized
by odd viscosity. An infinitely large, flat, and thin 2D fluid layer(green) (e.g., a monolayer formed by amphiphiles) is located at z = hhaving 2D dilatational, shear, and odd viscosities, ηd , ηs , and ηo ,respectively. The fluid interface is in contact with air (z > h) and a
3D fluid underneath (0 < z < h) having a 3D shear viscosity η. The
3D fluid is bounded by an impermeable flat wall located at z = 0,
and its velocity is assumed to vanish at z = 0. On the 2D fluid layer
at z = h, both time-reversal and parity symmetries are broken (e.g.,
due to the self-spinning objects injecting energy into the fluid), giving
rise to the possibility of an odd viscosity, ηo , in the 2D layer.

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PHYSICAL REVIEW E 103, 042610 (2021)

In addition to the above viscosity coefficients, we assume
that the 2D layer has odd viscosity, ηo , that is an importantmeasure of how far the fluid departs from passive fluids. We
will not specifically focus on the origin of the odd viscosity,
but it can be attributed, for example, to self-spinning objects
immersed in the 2D fluid layer that break both time-reversal
and parity symmetries [2,26]. In this case, one has to assume that the active rotors are homogeneously distributed in
the 2D fluid layer and their concentration is small enough.
Under this condition, the 2D layer can be regarded as alayer of a continuum active chiral fluid with a constant odd
viscosity, ηo .
For the 2D fluid layer introduced above, the momentum
balance equation at low Reynolds number can be written as

Substituting the obtained fb into Eq. (4), one can show that thehydrodynamic equation for the active chiral layer is−∇+ ∇ · σ + fb + F = 0,(4)

where ∇ = (∂x , ∂y ) stands for the 2D gradient operator,
is the 2D hydrostatic pressure, σ is the stress tensor given
in Eq. (3) that includes the odd viscosity ηo , fb is the force
exerted on the 2D fluid layer by the underlying 3D bulk fluid,
and F is any other force density acting on the 2D fluid.
The bulk underneath the 2D fluid layer is a pure 3D fluid.
We denote its velocity field by the vector u(r, z) and the3D hydrostatic pressure by p(r, z). The corresponding Stokes
equation is˜ p = 0,(5)
η∇˜ 2 u − ∇˜ bewith η being the 3D shear viscosity of the bulk fluid and ∇ing the 3D gradient operator. The incompressibility condition
for u reads
˜ · u = 0.
∇
(6)
To obtain fb , we solve the hydrodynamic equations for thebulk fluid in Eqs. (5) and (6) [15–17]. The boundary conditions on the 3D bulk velocity u(r, z) are the stick (nonslip)
conditions at the bottom surface of the bulk fluid (z = 0) andat its top surface (z = h):u(r, 0) = 0,u(r, h) = v(r),(7)

with v(r) being the in-plane velocity vector of the fluid layer
at z = h. We now use the lubrication approximation where thevertical component of the velocity, uz , is neglected comparedto the in-plane components and the vertical pressure gradientvanishes, i.e., ∂ p/∂z = 0. This assumption is justified as longas h is smaller than any horizontal characteristic length scaleof the bulk fluid velocity. Under these assumptions, the 3D
Stokes equation (5) reduces to ∂ 2 ux− ∇ p = 0.
(8)
η 2
∂z uy
Taking into account the boundary conditions in Eq. (7), theabove equation can be integrated to giveu(r, z) =z2 − zhz∇ p(r) + v(r).
2η
h

(9)

Then, the force exerted on the 2D fluid layer by the bulk fluid
beneath is calculated as [15–17]

η
∂u 
h
(10)
= − ∇ p − v.
fb = −η ∂z z=h2h−∇+ ηd ∇(∇ · v) + ηs ∇ 2 v + ηo ∇ 2 v∗

ηh− ∇ p − v + F = 0,
2
h

(11)

where the divergence of the in-plane velocity is given by thefollowing relation [15,17]:
∇·v =

h2 2
∇ p.
6η

(12)

This relation can be derived by taking the divergence ofEq. (9) and integrating over the lubrication layer (0  z  h)under the incompressibility condition of Eq. (6) and theboundary conditions of Eq. (7). The fourth term on the lefthand side of Eq. (11) indicates that odd viscosity contributes
to the fluid flow that is perpendicular to the one generated byshear viscosity [2].
IV. HYDRODYNAMIC RESPONSE OF A POINT FORCE
A. Mobility tensor

We derive the mobility tensor for the hydrodynamic
response of the 2D fluid layer with odd viscosity. The secondrank mobility tensor G(r) connects the force density F acting
on the fluid layer at position r with its induced velocity at
position r:
vi (r) = d 2 r  Gi j (r − r )Fj (r ).
(13)
To derive Gi j (r), we solve the hydrodynamic equations (11)and (12) in Fourier space and obtain Gi j [k], where k =(kx , ky ), and the square brackets indicate a function in Fourier
space. We introduce two orthogonal unit vectorsk̂ = (kx /k, ky /k),k̄ = (−ky /k, kx /k),(14)

with k = |k|.
In Appendix A, we show that Gi j [k] has the following
expression:
Gi j [k] =ηs (k 2 + κ 2 )k̂i k̂ j + (ηs + ηd )(k 2 + λ2 )k̄i k̄ j −ηo k 2 i j,
ηs (ηs + ηd )(k 2 + κ 2 )(k 2 + λ2 ) + ηo2 k 4
(15)

where
κ2 =

η
,
hηs

λ2 =

4η
.
h(ηs + ηd )(16)

Note that the ratio, η/ηs , gives the inverse length scale because
the 2D viscosity, ηs , has the dimension of Pa s m, while that ofη is Pa s. The lengths scales, κ −1 and λ−1 , correspond to thehydrodynamic screening lengths beyond which the 2D layer
exchanges momentum with the underlying bulk fluid. Importantly, the numerator of Eq. (15) includes an antisymmetrictensor i j .
For simplicity sake, we hereafter assume κ = λ (or equivalently ηd = 3ηs ) in Eq. (15) and consider the following

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FIG. 2. Streamlines of the velocity v(x, y) rescaled by F/(2π ηs ) and generated by a point force, F = F êx δ(r), as a function of κx and κywhile keeping ηd = 3ηs . The force along the x axis is applied at the origin (the black horizontal arrow) for (a) μ = ηo /ηs = 0, (b) μ = 3, and(c) μ = −3 [see Eqs. (13) and (19)]. The blue arrows indicate the flow direction.

simplified mobility tensor
Gi j [k] =(k + κ )(4δi j − 3k̂i k̂ j ) − μk i j,
ηs [4(k 2 + κ 2 )2 + μ2 k 4 ]
2

2

where
μ=

2

ηo
.
ηs

(17)

(18)

The above dimensionless parameter, μ, is a measure of how
far the 2D active chiral fluid departs from the passive fluid,e.g., due to the self-spinning active objects.
As shown in Appendix B, the real space representation of
the mobility tensor can be obtained by the inverse Fouriertransform of Eq. (17) [27]Gi j (r) = C1 (r)δi j + C2 (r)r̂i r̂ j + C3 (r)i j ,the mobility tensor of passive fluids satisfies the symmetry
property G0i j = G0ji , whereas for active chiral fluids, such asymmetry does not hold, i.e., Gi j = G ji as shown in Eq. (19).
This asymmetry gives rise to the fluid velocity perpendicular
to an applied force that results from the nonreciprocal hydrodynamic response.

(19)

where r̂ = r/r is a unit vector (r = |r|), and the three coefficients are given by ∞1
k(k 2 + κ 2 )C1 (r) =dk
2
2π ηs 04(k + κ 2 )2 + μ2 k 4
3J1 (kr),(20)
× 4J0 (kr) −kr ∞1
3k(k 2 + κ 2 )C2 (r) =dk
2
2π ηs 04(k + κ 2 )2 + μ2 k 4
2J1 (kr)× −J0 (kr) +
,
(21)
kr
 ∞
μ
k 3 J0 (kr)
dk
.
(22)
C3 (r) = −22π ηs 04(k + κ 2 )2 + μ2 k 4
In the above, Jn (x) is the Bessel function of the first kind [28].
When ηo = 0 (or μ = 0), C3 vanishes and Gi j (r) reduces tothat of a 2D passive compressible fluid, G0i j (r), which weanalytically derive in Appendix B [see Eq. (B5)]. Hereafter,
the superscript “0” denotes quantities when μ = 0 (vanishingodd viscosity). Under the exchange ηo ↔ −ηo , C1 and C2 of
Eqs. (20) and (21) remain unchanged, whereas C3 of Eq. (22)changes its sign.
We briefly discuss the symmetry property of the mobility
tensor obtained in Eq. (19). From G0i j in Eq. (B5), one sees that

B. Velocity field

With the obtained mobility tensor, we first investigate thevelocity field induced by a point force acting on a 2D fluidlayer with odd viscosity. Substituting Eq. (19) into Eq. (13),
we calculate the velocity field induced by a point force at theorigin, F = F êx δ(r), with êx being a unit vector in the x direction. The obtained velocity field is plotted in Fig. 2 for μ = 0,3, and −3. When μ = 0, we see axisymmetric streamlines thatpass through the applied force (the black horizontal arrow), as
in Fig. 2(a). There are two vortices whose center is located at(κx, κy) ≈ (0, ±2.0). They result from the nature of the 2D
fluid layer with the hydrodynamic screening length, κ −1 .
When μ is finite, however, a perpendicular flow in the y direction starts to develop and accordingly, the axial symmetry
breaks down, as shown in Figs. 2(b) and 2(c). This behavior
results from the nonreciprocal hydrodynamic response in thepresence of the odd viscosity. When μ = ±3, the vortices
approach to the positions (0, ±3.0), meaning that finite values of μ causes an effective increase in the hydrodynamic
length, κ −1 .
We next calculate the flow field generated by a hydrodynamic force dipole that is composed of two point forces
directed oppositely to each other. When a force dipole at theorigin is directed along a given unit vector d̂, its induced
velocity field is given by [29]vi (r) = −Fd̂k ∂k Gi j (r)d̂ j .

(23)Here, F is the force magnitude,  is the distance between thetwo point forces, and the limit r is assumed. In Fig. 3, weplot the force dipole along the x axis (i.e., d̂ = êx ) for μ = 0,3, and −3. When μ = 0, the streamlines have an axial symmetry along the x direction as well as the y direction, as shownin Fig. 3(a). For μ = ±3, however, flows perpendicular to the042610-4

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PHYSICAL REVIEW E 103, 042610 (2021)

FIG. 3. Streamlines of the velocity v(x, y) rescaled by F κ/(2π ηs ) and generated by a force dipole as a function of κx and κy while keeping
ηd = 3ηs . The force dipole along the x axis (d̂ = êx ) is centered at the origin (the black double arrow) for (a) μ = ηo /ηs = 0, (b) μ = 3, and(c) μ = −3 [see Eq. (23)]. The blue arrows indicate the flow direction.

applied forces become dominant, and both mirror symmetries
are broken, as shown in Figs. 3(b) and 3(c).

C. Mobility coefficients

Next, we investigate each component of the mobility tensor
in Eq. (19). If we choose the x axis, without loss of generality,
along the r direction, i.e., r = r êx , the longitudinal, transverse,
and antisymmetric mobility coefficients are given by Gxx =C1 + C2 , Gyy = C1 , and Gxy = −Gyx = C3 , respectively. The
nonzero mobility coefficient Gxy is characteristic of the active
chiral fluid with finite odd viscosity, while Gxx and Gyy remainin the limit of ηo → 0. Note that both Gxx and Gyy also dependon ηo , as seen in Eqs. (20) and (21).
In Fig. 4, we plot Gxx , Gyy , and Gxy as a function of κrfor various values of μ, while keeping ηd = 3ηs . We see thatGxx decreases monotonically with κr for all the μ values, as
shown in Fig. 4(a). The decrease of Gxx is more enhanced forlager μ, whereas Gxx weakly depends on μ for larger κr. This
reflects the fact that, for κr
1, the 3D hydrodynamic effectbecomes more important, and Gxx is almost independent of
the odd viscosity ηo . In Fig. 4(b), however, Gyy takes negativevalues because the 2D fluid layer can flow in the direction

opposite to the applied force [23]. Such a behavior resultsfrom the two vortices shown in Fig. 2.
The mobility coefficient Gxy describes the nonreciprocalhydrodynamic response because it gives the relation between
the applied force Fy and induced velocity vx . When μ = 0, Gxyis always zero as it should be, whereas for μ = ±3, it exhibits
nonmonotonic behavior, as shown in Fig. 4(c). This meansthat when the system is active, an applied force in the 2D fluidlayer generates a perpendicular flow, giving rise to the broken
mirror symmetry with respect to the force direction, evenwhen the fluid layer is isotropic. When ηo is positive, Gxy takes
negative (κr < 2) and positive values (κr > 2), corresponding
to the attractive and repulsive flows, respectively, and viceversa for negative ηo . These flow patterns can lead to either
convergence or dispersion of surrounding inclusions, and thespecific behavior depends on the hydrodynamic screening
length, κ −1 .
V. HYDRODYNAMIC RESPONSE OF A RIGID DISK
A. Boundary integral equation

So far, we have discussed the hydrodynamic response induced by a point force and a force dipole in a 2D fluid

FIG. 4. Plots of the mobility coefficients rescaled by 2π ηs as a function of κr for various values of μ. (a) The longitudinal mobility
coefficient Gxx for μ = 0, 3, and 5 (black, red, and blue lines). (b) The transverse mobility coefficient Gyy for μ = 0, 3, and 5 (black, red, andblue lines). (c) The antisymmetric mobility coefficient Gxy for μ = −3, 0, and 3 (red dashed, black solid, and red solid lines).
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HOSAKA, KOMURA, AND ANDELMANPHYSICAL REVIEW E 103, 042610 (2021)

layer with odd viscosity using the mobility tensor in Eq. (19).
Here we generalize the discussion to the situation where theresponse is induced by a finite-size body moving in the 2D
fluid layer. For a passive fluid, the force acting on such abody can be calculated by using a boundary integral equationthat is based on the Lorentz reciprocal theorem [30,31]. In
Appendix C, we first generalize this theorem for a 2D compressible fluid with finite ηo . Then, in Appendix D, we derive
the corresponding boundary integral equation that is used in
the following analysis.
Consider a circular rigid disk of radius R, which translates
and rotates in the 2D fluid. We assume that a nonslip boundary
condition holds at the disk perimeter and further consider thelimit of κR
1. As detailed in Appendix D, the velocity at
any point on the disk perimeter (R = Rr̂) can be expressed interms of the following boundary integral equation:ds(R ) f j (R )G ji (R − R ), (24)Ui + i jk j Rk = −Cuwhere U and  are the lateral and angular velocities of therigid disk, respectively, and i jk is the 3D Levi-Civita tensor.
The right-hand side of Eq. (24) is a line integral over anunspecified closed curve Cu , and ds(R ) indicates that R is
the integration variable. The boundary integral equation (24)relates the velocities of the disk moving in the 2D fluid layer
with the accompanying unknown force distribution f.
B. Translational and rotational frictions

Since the governing hydrodynamic equation (11) is linear
in v, the translational motion is decoupled from the rotational
one. Hence, the following linear relations hold:(25)
Fd = − · U, Td = − ,where Fid = Cu ds(R ) fi (R ) is the force acting on the disk
and Ti d = i jk Cu ds(R ) Rj fk (R ) is the torque, while  is thetranslational friction tensor andis the rotational friction
coefficient. The minus signs in equation (25) take into accountthat the force and torque act opposite to the velocities. Notethat is a scalar because both Td and  must point to the z
direction in a 2D system.
Using the assumption |R ||R| in Eq. (24) [22], we
obtain the expressions for  and as=

C1 + C2 /2
1
−C3
2
2
(C1 + C2 /2) + C3=

C3
C1 + C2 /2 , (26)

2R2.
C2 − R(∂C1 /∂R)(27)

See Appendix E for the derivation. In the above, the arguments of the three coefficients are omitted, Cn ≡ Cn (R) (n =1, 2, 3), to keep the notation compact. For passive fluids,
the translational friction tensor must be symmetric and positive definite according to the requirement that the dissipatedenergy is positive [32]. For the considered fluid with ηo ,however, the translational friction tensor is allowed to beasymmetric when C3 is nonzero. Notice that the energy
dissipation calculated from Eq. (26) is Ui i j U j ∼ (C1 +C2 /2)U 2 and the antisymmetric part does not contribute to thedissipation.

FIG. 5. Plots of the drag ( ) and lift (⊥ ) coefficients rescaledby 2π ηs as a function of the rescaled disk radius κR. (a) The dragcoefficient  for μ = 0, 3, and 5 (black, red, and blue solid lines).
The dotted line represents the full expression 0 for the drag coefficient when μ = 0 reported in Ref. [15] (see the text for the specificexpression). (b) The lift coefficient ⊥ for μ = −5, −3, 0, 3, and 5(blue dashed, red dashed, black solid, red solid, and blue solid lines).

For a disk translating with the velocity U = (U, 0), we
have the viscous drag and lift coefficients as  = xx and⊥ = yx , respectively. In Fig. 5, we plot  and ⊥ as afunction of the dimensionless radius κR for various valuesof μ. We see that  increases monotonically with increasing the disk size for all μ values, as seen in Fig. 5(a).
For a fixed disk size,  is larger for larger μ values.
The dotted line in Fig. 5(a) is the full analytical result for
μ = 0 (2D passive compressible fluid) [15], 0 /(2π ηs ) =4(κR)2 K2 (κR)/K0 (κR), where Kn (x) is the modified Bessel
5function of the second kind [28]. Our result obtained by using
the boundary integral equation (24) coincides with the analytical one when κR
1.
In Fig. 5(b), we see that ⊥ shows both increasing anddecreasing dependencies on κR for positive and negative values of μ, respectively. However, when μ = 0, ⊥ is always
zero, as it should be. Finite values of ⊥ mean that the disk

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NONRECIPROCAL RESPONSE OF A TWO-DIMENSIONAL …

FIG. 6. Plot of the rotational friction coefficient ( ) rescaled by
4π ηs /κ 2 as a function of the rescaled disk radius κR for μ = 0, 3,
and 5 (black, red, and blue solid lines). The dotted line represents thefull expression 0 for the rotational coefficient when μ = 0 reported
in Ref. [20] (see the text for the specific expression).

translated along the x axis presents a lift motion along they direction. For a fixed disk size, the absolute value of ⊥
increases and the lift force is more enhanced as the absolutevalue of μ increases.
Assuming that the disk is rotating with velocity  =(0, 0, ), we obtain the rotational friction coefficient that isplotted in Fig. 6 as a function of κR. We see that shows anincreasing dependency on κR for all the μ values. The dottedline in Fig. 6 represents the full analytical expression for μ =0 (2D passive incompressible fluid) [20], κ 2 0 /(4π ηs ) =(κR)2 + 21 (κR)3 K0 (κR)/K1 (κR). The solid black line and thedotted line coincide in the limit of κR
1, because the disk
rotation contributes neither to the compression nor to the
expansion of fluids.

VI. DISCUSSION AND CONCLUSION

We have investigated the linear hydrodynamic response ofa 2D fluid layer with broken time-reversal and parity symmetries. Such a 2D active fluid presents a special rheologicalproperty called odd viscosity, characterizing the deviation of
the system from a passive fluid. In our approach, we combine
the concept of the odd viscosity [1–3] and the hydrodynamic
model of a 2D compressible fluid derived by using the lubrication approximation [15–17]. In contrast to well-studied
2D fluids characterized by a shear viscosity [20–25], the additional odd viscosity ηo leads to anomalous flow behavior, i.e.,
nonreciprocal hydrodynamic response.
In the case of a point force and a force dipole, the symmetryof the velocity field in terms of the force direction is broken,
generating flow perpendicular to the applied force (see Figs. 2
and 3). We also analyzed the effects of the odd viscosity
on the mobility tensor, as derived in Eq. (19). In particular,
we investigated the behavior of the antisymmetric mobility
coefficient Gxy that exists only for nonvanishing ηo , as shown

PHYSICAL REVIEW E 103, 042610 (2021)

in Eq. (22). As for the hydrodynamic response of finite-sizebodies, we have investigated the forces acting on a translating
and rotating disk in a 2D fluid layer, using the boundary
integral equation (24). We found that small disks (κR1)
not only undergo a drag force, but also a lift force, whichcannot be seen in an isotropic passive fluid (see Fig. 5) [1].
As a possible biological application, we can relate the2D fluid layer to a monolayer with a low concentration ofactive motor proteins, such as ion pumps [33]. Rheologicalproperties of these system can be investigated by surface microrheology techniques [34]. For typical values such as η ≈10−3 Pa s, ηs ≈ 10−6 Pa s m, and h ≈ 1 nm, we find that theobtained drag and lift forces could be observed in experiments
using a submicrometer probe, i.e., R < 0.1 μm.
In this paper, we have considered a 2D compressible
fluid, which is in contact with a 3D bulk fluid. Such acompressible 2D system can be realized by a dilute Gibbs
monolayer, which is composed of soluble amphiphiles thatcan dissolve into the underlying bulk fluid [15,16]. When theadsorption and desorption processes of soluble amphiphiles
are instantaneous, surface concentration gradients can beeliminated rapidly. This is the situation that we consider inthe present work. When the amphiphile is insoluble, however, the concentration gradient is sustained, giving rise to a
Marangoni flow [17,35]. Although the Marangoni convection
is outside the scope of this paper, it would be interesting toinvestigate the effects of odd viscosity on such convective
phenomena.
In general, odd viscosity can depend on the density ofself-spinning objects, although such an effect was not considered in the present work. By using the 2D Faxén laws,
effective shear viscosity of a fluid membrane with finite-size
suspensions was derived [22,23]. Hence, it is of interest to see
how the nonreciprocal flow field of the 2D fluid with an odd
viscosity ηo changes with the rotor concentration. Moreover,
for a passive fluid at equilibrium, the disk drag coefficient isconnected to its diffusion constant through Einstein’s relation.
However, such an evident relation does not exist in active
chiral fluids and one has to extend the fluctuation dissipation
theorem [36–39]. These interesting questions are left for future investigations.

ACKNOWLEDGMENTS

We thank J. E. Avron, H. Diamant, M. Doi, R. Seto, M. S.
Turner, and K. Yasuda for useful discussions. Y.H. acknowledges support by a Grant-in-Aid for JSPS Fellows (Grant
No. 19J20271) from the Japan Society for the Promotionof Science (JSPS). Y.H. also thanks the hospitality of Tel
Aviv University, where part of this research was conductedunder the TMU-TAU co-tutorial program. S.K. acknowledges
support by a Grant-in-Aid for Scientific Research (C) (Grants
No. 18K03567 and No. 19K03765) from the JSPS, and support by a Grant-in-Aid for Scientific Research on Innovative
Areas “Information physics of living matters” (Grant No.
20H05538) from the Ministry of Education, Culture, Sports,
Science and Technology of Japan. D.A. acknowledges support
from the Israel Science Foundation (ISF) under Grant No.
213/19.

042610-7

HOSAKA, KOMURA, AND ANDELMANPHYSICAL REVIEW E 103, 042610 (2021)
APPENDIX A: DERIVATION OF EQ. (15)

We derive the mobility tensor in Fourier space G[k] as given by Eq. (15). The Fourier transform of v(r) is defined by
d 2kv[k] exp(ik · r),v(r) =(2π )2(A1)

with k = (kx , ky ), and similarly for the 3D pressure p(r) and the force density F(r). In the Fourier space, Eq. (11) becomes−ηs k 2 v[k] − ηd k 2 k̂k̂ · v[k] − ηo k 2 (k̂k̄ · v[k] − k̄k̂ · v[k]) −ηihk p[k]k̂ − v[k] + F[k] = 0,
2
h

(A2)

or equivalently,−ηs k 2 v[k] − ηd k 2 v [k]k̂ − ηo k 2 (v⊥ [k]k̂ − v [k]k̄) −ihηk p[k]k̂ − v[k] + F[k] = 0,
2
h

(A3)

where v [k] = k̂ · v[k] and v⊥ [k] = k̄ · v[k]. For Eq. (12), we haveik k̂ · v[k] = ikv [k] = −h2 2
k p[k].
6η

(A4)

In the derivation of Eq. (A2), we have assumed that the 2D fluid layer is quickly equilibrated with the 3D bulk, and hence the2D pressure is homogeneous in space, i.e., ∇ = 0 [15,17].
Substituting Eq. (A4) into Eq. (A2) to eliminate p[k], we obtain−ηs k 2 v[k] − ηd k 2 v [k]k̂ − ηo k 2 (v⊥ [k]k̂ − v [k]k̄) −Hence, F[k] can be written as F [k](ηs + ηd )k 2 + 4η/h=F⊥ [k]−ηo k 23ηηv [k]k̂ − v[k] + F[k] = 0.
hhηo k 2
ηs k 2 + η/h

(A5)




v [k].
v⊥ [k](A6)

Since the mobility tensor in the Fourier space satisfies the relation v[k] = G[k] · F[k], we obtain Eq. (15).
APPENDIX B: DERIVATION OF EQ. (19) AND G0 (r)Here we perform the inverse Fourier transform of G[k] in Eq. (15) to obtain G(r). By calculating Gii , Gi j r̂i r̂ j , and Gi j i j , weobtain [27]ηs (k 2 + κ 2 ) + (ηs + ηd )(k 2 + λ2 )d 2k
2C1 + C2 =exp(ik · r)(2π )2 ηs (ηs + ηd )(k 2 + κ 2 )(k 2 + λ2 ) + ηo2 k 4
 ∞1
ηs (k 2 + κ 2 ) + (ηs + ηd )(k 2 + λ2 )=dk k
J0 (kr),(B1)
2π ηs 0(ηs + ηd )(k 2 + κ 2 )(k 2 + λ2 ) + ηo2 k 4

d 2 k ηs (k 2 + κ 2 ) cos2 θ + (ηs + ηd )(k 2 + λ2 )(1 − cos2 θ )exp(ik · r)(2π )2ηs (ηs + ηd )(k 2 + κ 2 )(k 2 + λ2 ) + ηo2 k 4
 ∞1
ηs (k 2 + κ 2 )[J0 (kr) − J1 (kr)/(kr)] + (ηs + ηd )(k 2 + λ2 )J1 (kr)/(kr)=dk k,
2π ηs 0(ηs + ηd )(k 2 + κ 2 )(k 2 + λ2 ) + ηo2 k 4

C1 + C2 =(B2)ηo k 2d 2k
exp(ik · r)(2π )2 ηs (ηs + ηd )(k 2 + κ 2 )(k 2 + λ2 ) + ηo2 k 4 ∞ηo
k 3 J0 (kr)
=−
dk
,
2
2π ηs 0(ηs + ηd )(k + κ 2 )(k 2 + λ2 ) + ηo2 k 4

C3 = −(B3)

respectively, where θ is the angle between the vectors k and r. Solving Eqs. (B1) and (B2) when κ = λ, we obtain Eqs. (20),(21), and (22).
Next, we analytically derive the mobility tensor G0 (r) in the absence of the odd viscosity, i.e., μ = 0. In this case, we havefrom Eq. (15)G0i j [k] =k̂i k̂ j
δi j − k̂i k̂ j+.
22
ηs (k + κ ) (ηs + ηd )(k 2 + λ2 )042610-8(B4)

NONRECIPROCAL RESPONSE OF A TWO-DIMENSIONAL …

PHYSICAL REVIEW E 103, 042610 (2021)

The real-space mobility tensor G0 (r) can be obtained by assuming
G0i j (r) = B1 (r)δi j + B2 (r)r̂i r̂ j ,with two coefficients B1 and B2 .
By calculating G0ii and G0i j r̂i r̂ j , we have1
2B1 + B2 =
2π

 ∞
0

(B5)


k
k
+
J0 (kr)dk
ηs (k 2 + κ 2 ) (ηs + ηd )(k 2 + λ2 )

1
1=
K0 (κr) +K0 (λr),2π ηs2π (ηs + ηd )

(B6)



k
J1 (kr)J1 (kr)
+
J
(kr)
−
0
ηs r(k 2 + κ 2 ) (ηs + ηd )(k 2 + λ2 )
kr
0




K1 (κr)1
1
1K1 (λr)1−+.
+
K0 (λr) +
−
=
2π ηs
κr
(κr)2
2π (ηs + ηd )
λr
(λr)2

B1 + B2 =

1
2π



 ∞

dk

Solving Eqs. (B6) and (B7), we obtain




1
1
K1 (λr)1
1
K1 (κr)B1 (r) =K0 (κr) +
+
,
−
−
+
2π ηs
κr
(κr)2
2π (ηs + ηd )
λr
(λr)2




1
2
2
1
2K1 (κr)
2K1 (λr)−K0 (κr) −+.
B2 (r) =+K0 (λr) +
−
2π ηs
κr
(κr)2
2π (ηs + ηd )
λr
(λr)2

(B7)

(B8)

(B9)

APPENDIX C: GENERALIZED LORENTZ RECIPROCAL THEOREM

The Lorentz reciprocal theorem gives a relation regarding the resistance of finite-size bodies moving in a passive fluid [31,32].
Here we generalize this theorem for a 2D compressible fluid with finite odd viscosity. Let unprimed and primed symbols
represent the variables for any two arbitrary types of flows satisfying the following equations:∂ j σi j + bi = 0,

∂ j σij + bi = 0,

(C1)

where b and b are the arbitrary body force densities and the associated velocity fields are given by v and v , respectively. In
the above, σi j is the stress tensor in Eq. (3) that also includes the 2D isotropic pressure term, − δi j (similarly for σij ). The
divergence of σij v j − σi j v j becomes [40]





∂i [(σS,ij + σA,i j )v j ] − ∂i [(σS,i j − σA,i j )v j ] = v j b j − v j b j −∂jvj +∂ j v j + 2v j ∂i σA,i j ,

(C2)

where
1
ηo (∂ j vi∗ + ∂i v ∗j + ∂ ∗j vi + ∂i∗ v j ),2with σi j = σS,i j + σA,i j . Integrating the above equation over the fluid area A, we obtain the integral identity as
ds ni (σS,i j + σA,i j )v j − ds ni (σS,i j − σA,i j )v j
C
C









= − dA v j b j + dA v j b j + dA ∂ j v j − dA ∂ j v j − 2 dA v j ∂i σA,i j ,σS,i j = − δi j + (ηd − ηs )δi j ∂k vk + ηs (∂ j vi + ∂i v j ),AA

σA,i j =

A

A

(C3)

(C4)

A

where C denotes the curve bounding the area A and the unit vector n is directed into that area. We note that Eq. (C4) is notinvariant under the exchange of the unprimed and primed variables when ηo = 0. When ηo = 0, however, Eq. (C4) reduces tothe Lorentz reciprocal theorem for a 2D compressible fluid [40].
APPENDIX D: DERIVATION OF EQ. (24)Here we derive the boundary integral equation in Eq. (24). When
unprimed and primed flows:∂ j σi j = 0,== 0, we consider the following two types of∂ j σij + Fi δ(R − R ) = 0.

(D1)

Suppose that a circular rigid disk is moving in the fluid area A, Eq. (C4) with the use of Eq. (D1) becomes [30,31]







ds(R) ni (σS,i j − σA,i j )v j =ds(R) ni (σS,i j + σA,i j )v j − Fj dA(R) v j δ(R − R ),
Cd

Cd

A

042610-9(D2)

HOSAKA, KOMURA, AND ANDELMANPHYSICAL REVIEW E 103, 042610 (2021)

FIG. 7. Sketch of the circular curve, Cd , and the unspecified curve, Cu , with the accompanying unknown force distribution, f, while A isthe fluid area bounded by both Cd and Cu (white area). The curves, Cd and Cu , are parameterized by the vectors R and R , respectively, and thetwo arrows represent the direction of the line integral. In the sketch, we have |R | > |R| only for presentation purposes. In actual calculations|R| is used, the two curves overlap with each other.
where the condition |R |where Cd is the circular curve bounding the moving disk, as schematically depicted in Fig. 7. The notations, ds(R) and dA(R),indicate that R is the integration variable. Assuming a nonslip boundary condition at the disk perimeter and using the form of
the point-force solution, vi (R) = Gi j (R − R )Fj , we obtain [31]dA(R) v (R)δ(R − R ) = −ds(R) f j (R)G j (R − R ),
(D3)
A

Cu

where Cu denotes the domain of the unknown force distribution, f, and we have assumed that the force distribution can be definedas f j = ni (σS,i j − σA,i j ). Interchanging R and R , we finally obtain Eq. (24).
APPENDIX E: DERIVATION OF EQS. (26) and (27)

We derive the translational friction tensor  and the rotational friction coefficient in Eqs. (26) and (27), respectively. If weassume |R ||R| [22], then the right-hand side of Eq. (24) can be expanded up to first order in R as∂G ji (R)Ui + i jk j Rk ≈ −.
(E1)ds(R ) f j (R ) G ji (R) − Rk∂RkCuIntegrating Eq. (E1) over the circular disk perimeter, Cd , parametrized by R, we obtain the relation between the velocity and theforce as1Ui = −
ds(R ) f j (R )ds(R) G ji (R) = −[(C1 + C2 /2)δi j − C3 i j ]Fjd ,(E2)
2π R Cu
Cd
where Fid = Cu ds(R ) fi (R ) is the force acting on the disk. In the above, the integrals of the odd terms in R vanish because of
the symmetry of the disk. Hence, we obtain Eq. (26).
Next, we multiply both sides of Eq. (E1) by R and integrate over Cd∂G ji (R)
π R3 i j j =ds(R ) Rk f j (R )ds(R) R
.
(E3)
∂Rk
Cu
Cd
We further multiply both sides of the above equation by in and obtain
∂C1 C2
1∂C3−
nk j −
nki i j
ds(R ) Rk f j (R ).
n =
2R
∂R
R
∂R
Cu

(E4)

On the right-hand side, the term with C3 represents the radial pressure acting on the disk [1]. As this term does not contribute to
the torque, it vanishes and the relation between the angular velocity and the torque becomes
1 ∂C1 C2
−
Td ,
=
(E5)
2R ∂R
R

where Ti d = i jk Cu ds(R ) Rj fk (R ) is the torque on the disk. Hence, we obtain Eq. (27).

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