1. Introduction
Intrinsically rotating structures are recurring features in driven systems such as molecular motors (Sumino et al. Reference Sumino, Nagai, Shitaka, Tanaka, Yoshikawa, Chaté and Oiwa2012), cilia in living organisms (Cartwright, Piro & Tuval Reference Cartwright, Piro and Tuval2004), sperm cell colonies (Riedel, Kruse & Howard Reference Riedel, Kruse and Howard2005) and artificial robotic particles (Scholz et al. Reference Scholz, Ldov, Pöschel, Engel and Löwen2021) – all of which display fluid-like behaviour over mesoscopic to macroscopic scales. The effective hydrodynamics of these so-called chiral active fluids is governed not only by the transport of mass and linear momentum, but also by the dynamics of spin angular momentum. The presence of this additional degree of freedom fundamentally modifies the constitutive relations between stresses and strain rates, allowing for antisymmetric components in the viscosity tensor, commonly referred to as odd viscosity (Avron Reference Avron1998; Fruchart, Scheibner & Vitelli Reference Fruchart, Scheibner and Vitelli2023). Such terms are permitted by the Onsager–Casimir reciprocity relations (Onsager Reference Onsager1931a , Reference Onsagerb ; Casimir Reference Casimir1945) when the orientation of the intrinsic spin angular momentum is included as an additional degree of freedom. Odd viscosity can lead to unconventional flow patterns (Khain et al. Reference Khain, Scheibner, Fruchart and Vitelli2022; Lier Reference Lier2024), affect turbulence (Chen et al. Reference Chen, de Wit, Fruchart, Toschi and Vitelli2024; de Wit et al. Reference de Wit, Fruchart, Khain, Toschi and Vitelli2024), and, in certain cases, endow the fluid with topological properties (Souslov et al. Reference Souslov, Dasbiswas, Fruchart, Vaikuntanathan and Vitelli2019; Lou et al. Reference Lou, Yang, Ding, Liu, Chen, Zhou, Ye, Podgornik and Yang2022). Although odd viscosity has been recognised for decades – albeit known under different names, such as transverse viscosity (Beenakker, Coope & Snider Reference Beenakker, Coope and Snider1971) and gyroviscosity (Chang & Callen Reference Chang and Callen1992) – recent discoveries have sparked a renewed interest in the study of odd fluids. Notable examples of such studies are the experimental realisation of odd viscosity in graphene (Berdyugin et al. Reference Berdyugin2019) and in a colloidal fluid of magnetic cubes (Soni et al. Reference Soni, Bililign, Magkiriadou, Sacanna, Bartolo, Shelley and Irvine2019).
Biology and soft matter provide various possibilities for chiral active fluids. Due to the length scales involved in such systems, the creeping-flow (low Reynolds number) regime is of special interest, and has recently been the subject of intense study. In particular, (quasi-)two-dimensional odd systems have been investigated for their flow properties and resulting drag forces (Ganeshan & Abanov Reference Ganeshan and Abanov2017; Lier et al. Reference Lier, Duclut, Bo, Armas, Jülicher and Surówka2023; Daddi-Moussa-Ider, Vilfan & Hosaka Reference Daddi-Moussa-Ider, Vilfan and Hosaka2025). In three-dimensional odd fluids, work has focused on e.g. a full classification of the types of odd viscosities (Khain et al. Reference Khain, Scheibner, Fruchart and Vitelli2022), odd Stokesian dynamics (Yuan & Olvera de la Cruz Reference Yuan and Olvera de la Cruz2023), microswimmers suspended in odd fluids (Hosaka et al. Reference Hosaka, Chatzittofi, Golestanian and Vilfan2024), the Lorentz reciprocal theorem (Hosaka, Golestanian & Vilfan Reference Hosaka, Golestanian and Vilfan2023), and the motion of suspended microparticles of various shapes (Khain et al. Reference Khain, Fruchart, Scheibner, Witten and Vitelli2024). Furthermore, the full Green’s function of an unbounded odd fluid has been computed analytically, including the full grand mobility matrix of a sphere suspended in it (Everts & Cichocki Reference Everts and Cichocki2024a , Reference Everts and Cichockib ).
Despite significant recent progress, our understanding of microhydrodynamics in three-dimensional chiral active fluids remains limited compared with the extensive results available for ordinary Stokesian fluids (Happel & Brenner Reference Happel and Brenner1983; Kim & Karrila Reference Kim and Karrila2013). Here, we address this gap by deriving exact results for steady incompressible three-dimensional chiral active flows at low Reynolds number. First, we establish a theorem for viscous dissipation in odd fluids by generalising the Helmholtz theorem (Helmholtz Reference Helmholtz1868), which expands upon our previous analysis of the viscous dissipation due to a translating and rotating passive sphere in a fluid with odd viscosity (Everts & Cichocki Reference Everts and Cichocki2024a , Reference Everts and Cichockib ). Second, we compute the stress response to a localised point-force density, a quantity frequently required in computational techniques such as the boundary element method (Pozrikidis Reference Pozrikidis1992). Third, we provide an intuitive derivation for the singularity representation of the flow and pressure fields around suspended (solid) particles, complementing more formal and rigorous analyses (Brenner Reference Everts and Cichocki1964a , Reference Brennerb , Reference Brenner1966). Finally, we derive exact closed-form expressions for the flow and pressure around a translating and rotating sphere in a chiral active fluid using the singularity representations. For the translating sphere, we also perform an in-depth analysis of the flow streamlines for arbitrary directions of the fluid’s spin momentum. Here, we go beyond the fluid flow results from Everts & Cichocki (Reference Everts and Cichocki2024a , Reference Everts and Cichockib ) where closed-form expressions were only presented for spheres translating parallel to the fluid’s spin-angular momentum. Finally, we show that our results are fully consistent with our generalised Helmholtz theorem.
2. Viscous dissipation in systems with odd viscosity
For steady Newtonian fluids in the creeping-flow regime, two fundamental results emerge from analysing the viscous dissipation rate. The first result concerns the uniqueness of the solution to the incompressible Stokes equations. The second states that among all incompressible (i.e. solenoidal) flows under specified boundary conditions, the flow satisfying the linear momentum balance has the minimal dissipation rate (Kim & Karrila Reference Kim and Karrila2013). The latter result is known as the Helmholtz minimum dissipation theorem. We aim to generalise these key results to arbitrary stress–strain constitutive relations, including those for chiral active fluids, to assess the role of odd viscosity in viscous dissipation.
Consider an incompressible chiral active fluid with a spatially constant spin angular momentum density
$\boldsymbol{\ell }$
, fluid velocity
$\boldsymbol{v}(\boldsymbol{r})$
, and pressure
$p(\boldsymbol{r})$
. The corresponding stress tensor is
$\boldsymbol{\sigma }(\boldsymbol{r})=-p(\boldsymbol{r})\, \unicode{x1D644}+\boldsymbol{\sigma }^{{V}}(\boldsymbol{r})$
. In the framework of linear irreversible thermodynamics, the viscous part of the stress tensor satisfies the constitutive relation
$\boldsymbol{\sigma }^{{V}}(\boldsymbol{r})=\boldsymbol{\eta }:\boldsymbol{\nabla }\boldsymbol{v}(\boldsymbol{r})$
, with
$\boldsymbol{\eta }$
being the viscosity tensor. Here, we define the double contraction as
$[ \unicode{x1D63C}: \unicode{x1D63D}]_{\alpha \ldots \beta } = A_{\alpha \ldots \lambda \nu }B_{\nu \lambda \ldots \beta }$
, with summation implied over repeated indices. Greek indices range over all Cartesian coordinates, and for the moment, we do not put any restriction on the spatial dimension. In this work, we also make use of the single contraction defined as
$[ \unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D63D}]_{\alpha \ldots \beta } = A_{\alpha \ldots \lambda }B_{\lambda \ldots \beta }$
, and we use carets to denote unit vectors; e.g. for an arbitrary vector
$\boldsymbol{w}$
, we write
$\hat {\boldsymbol{w}}=\boldsymbol{w}/|\boldsymbol{w}|$
. Microscopic time reversibility constrains the form of
$\boldsymbol{\eta }$
via the Onsager–Casimir reciprocal relations
$\eta _{\alpha \beta \lambda \nu }({\boldsymbol{\ell }})=\eta _{\lambda \nu \alpha \beta }(-{\boldsymbol{\ell }})$
(Casimir Reference Casimir1945; de Groot & Mazur Reference de Groot and Mazur1954).
The microscopic origin for a non-zero
$\boldsymbol{\ell }$
in chiral active fluids requires the presence of active sources of torque in the fluid. However, Markovich & Lubensky (Reference Markovich and Lubensky2021) have shown that when
$\boldsymbol{\sigma }^{{V}}$
is interpreted as the viscous stress tensor for the total linear momentum, it necessarily has to be symmetric, i.e.
$\sigma _{\alpha \beta }^{{V}}(\boldsymbol{r})=\sigma _{\beta \alpha }^{{V}}(\boldsymbol{r})$
. An alternative justification is provided by Banerjee et al. (Reference Banerjee, Souslov, Abanov and Vitelli2017), who interpret
$\boldsymbol{\sigma }^{{V}}$
as the stress tensor for the centre-of-mass momentum. They show that antisymmetric stresses can be neglected when
$\boldsymbol{\ell }$
acquires a spatially constant value in a non-equilibrium steady state. Therefore, we assume that there are effectively no intrinsic sources of torque in the fluid. It follows that
$\boldsymbol{\sigma }^{{V}}(\boldsymbol{r})=\boldsymbol{\eta }(\boldsymbol{\ell }): \unicode{x1D65A}(\boldsymbol{r})$
with rate of strain tensor
$e_{\alpha \beta }(\boldsymbol{r})=[\partial _\alpha v_\beta (\boldsymbol{r})+\partial _\beta v_\alpha (\boldsymbol{r})]/2$
. Furthermore, from the incompressibility condition, it follows that
$\eta _{\lambda \lambda \alpha \beta }(\boldsymbol{\ell })=0$
. Combining this with the Onsager–Casimir relations, we obtain
$\eta _{\alpha \beta {\lambda }\lambda }(\boldsymbol{\ell })=0$
.
Let
$\mathcal{V}$
be a fluid domain with boundary
$\mathcal{S}$
, with stick-boundary condition on all surfaces. Taking the above symmetry considerations into account, the equations governing steady fluid flow at low Reynolds number are given by the balance of linear momentum and the incompressibility condition
\begin{equation} \boldsymbol{\nabla }\boldsymbol{\cdot }[\boldsymbol{\eta }(\boldsymbol{\ell }): \unicode{x1D65A}(\boldsymbol{r})]-\boldsymbol{\nabla }p(\boldsymbol{r})=-\boldsymbol{f}(\boldsymbol{r}), \quad \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{v}(\boldsymbol{r})=0, \quad \boldsymbol{r}\in \mathcal{V}, \end{equation}
respectively, with
$\boldsymbol{v}(\boldsymbol{r})$
given for
$\boldsymbol{r}\in \mathcal{S}$
, and
$\boldsymbol{f}(\boldsymbol{r})$
a body-force density. We decompose the viscosity tensor for a general odd fluid as
$\boldsymbol{\eta }=\boldsymbol{\eta }^{{S}}+\boldsymbol{\eta }^{{A}}$
, where
$\eta _{\alpha \beta \lambda \nu }^{{S}}(\boldsymbol{\ell })=\eta _{\lambda \nu \alpha \beta }^{{S}}(\boldsymbol{\ell })$
and
$\eta _{\alpha \beta \lambda }\nu ^{{A}}(\boldsymbol{\ell })=-\eta _{\lambda \nu \alpha \beta }^{{A}}(\boldsymbol{\ell })$
. The tensor
$\boldsymbol{\eta }^{{A}}$
contains all the odd-viscosity coefficients. The total dissipated power due to viscous effects is
\begin{equation} \dot {E}=\int _{\mathcal{V}}\textrm{d}V\, \boldsymbol{\sigma }^{{V}}(\boldsymbol{r}):\boldsymbol{\nabla }\boldsymbol{v}(\boldsymbol{r})=\int _{\mathcal{V}}\textrm{d}V\, \unicode{x1D65A}(\boldsymbol{r}):\boldsymbol{\eta }^{{S}}: \unicode{x1D65A}(\boldsymbol{r}). \end{equation}
Note that for systems in local thermodynamic equilibrium (implied in this construction),
$ \unicode{x1D65A}(\boldsymbol{r}):\boldsymbol{\eta }^{{S}}: \unicode{x1D65A}(\boldsymbol{r})\geqslant 0$
holds for any
$ \unicode{x1D65A}(\boldsymbol{r})$
, which follows from the second law of thermodynamics (de Groot & Mazur Reference de Groot and Mazur1962).
2.1. Uniqueness of solutions
Let
$(\boldsymbol{v},p)$
and
$(\boldsymbol{v}^{\prime},p^{\prime})$
be two solutions of (2.1) with the same boundary condition,
$\boldsymbol{v}^{\prime}(\boldsymbol{r})=\boldsymbol{v}(\boldsymbol{r})$
for
$\boldsymbol{r}\in \mathcal{S}$
. Furthermore, denote their corresponding stress tensors and strain rate tensors as
$(\boldsymbol{\sigma }, \unicode{x1D65A})$
and
$(\boldsymbol{\sigma }^{\prime}, \unicode{x1D65A}^{\prime})$
, respectively. Consider the following steps for the viscous dissipation of their difference fields:
\begin{align} \Delta \dot {E}&:= \int _{\mathcal{V}}\textrm{d}V\,\big[e_{\alpha \beta }^{\prime}(\boldsymbol{r})-e_{\alpha \beta }(\boldsymbol{r})\big]\eta _{\alpha \beta \lambda \nu }^{{S}} \big[e_{\lambda \nu }^{\prime}(\boldsymbol{r})-e_{\lambda \nu }(\boldsymbol{r})\big] \nonumber \\ &= \int _{\mathcal{V}}\textrm{d}V\,\big[\partial _\alpha v^{\prime}_{\beta }(\boldsymbol{r})-\partial _\alpha v_{\beta }(\boldsymbol{r})\big]\underbrace {\big\{\eta _{\alpha \beta \lambda \nu } \big[e_{\lambda \nu }^{\prime}(\boldsymbol{r})-e_{\lambda \nu }(\boldsymbol{r})\big]-\big[p^{\prime}(\boldsymbol{r})-p(\boldsymbol{r})\big]\delta _{\alpha \beta}\big\}}_{=\sigma ^{\prime}_{\alpha \beta }(\boldsymbol{r})-\sigma _{\alpha \beta }(\boldsymbol{r})}\nonumber \\ &=\int _{\mathcal{S}}\textrm{d}S\, \hat {n}_\alpha \big[ v^{\prime}_{\beta }(\boldsymbol{r})- v_{\beta }(\boldsymbol{r})\big]\big[\sigma _{\alpha \beta }^{\prime}(\boldsymbol{r})-\sigma _{\alpha \beta }(\boldsymbol{r})\big]\nonumber \\ &\quad {}-\int _{\mathcal{V}}\textrm{d}V\,\big[v^{\prime}_{\beta }(\boldsymbol{r})- v_{\beta }(\boldsymbol{r})\big]\big[\partial _\alpha \sigma _{\alpha \beta }^{\prime}(\boldsymbol{r})-\partial _\alpha \sigma _{\alpha \beta }(\boldsymbol{r})\big]=0, \end{align}
with
$\hat{\boldsymbol{n}}$
a unit normal pointing out of the fluid. The second equality follows from the incompressibility and the symmetric stress conditions, and
$( \unicode{x1D65A}^{\prime}- \unicode{x1D65A}):\boldsymbol{\eta }^{{A}}:( \unicode{x1D65A}^{\prime}- \unicode{x1D65A})=0$
. The third equality follows from a partial integration, and the last equality follows from the equality of
$\boldsymbol{v}$
and
$\boldsymbol{v}^{\prime}$
on
$\mathcal{S}$
and using (2.1). We conclude that
$ \unicode{x1D65A}^{\prime}= \unicode{x1D65A}$
, and from the boundary condition it follows that
$\boldsymbol{v}^{\prime}=\boldsymbol{v}$
throughout
$\mathcal{V}$
. Therefore, we have shown that (2.1) admits a unique solution for
$\boldsymbol{v}$
. Our uniqueness theorem generalises the result of Lapa & Hughes (Reference Lapa and Hughes2014), which was restricted to two-dimensional flows with a particular form of
$\boldsymbol{\eta }$
.
2.2. Helmholtz minimum dissipation theorem for anisotropic odd fluids
From (2.2), it is often mistakenly concluded that odd viscosity does not contribute to viscous dissipation. This is, however, incorrect, because generally speaking,
$ \unicode{x1D65A}(\boldsymbol{r})$
can depend on the components of
$\boldsymbol{\eta }^{{A}}$
. In fact, it was shown for a passive translating sphere in an odd fluid that odd viscosity leads to higher viscous dissipation than in ordinary Stokes flow (Everts & Cichocki Reference Everts and Cichocki2024a
). Here, we will make statements for the general case.
The standard Helmholtz minimum dissipation theorem compares the viscous dissipation rate of an incompressible flow that satisfies the linear momentum balance with that of an arbitrary divergence-free flow field under the same boundary conditions. An analogous approach can be used to compare incompressible flows with and without odd viscosity, provided that their even viscosity components are identical. Consider a solution
$(\boldsymbol{v},p)$
to (2.1), and let
$(\boldsymbol{v}^{(0)},p^{(0)})$
be a reference system without odd viscosity which solves (2.1) with
$\boldsymbol{\eta }^{{A}}=0$
and corresponding viscous dissipation rate
$\dot {E}^{(0)}$
. Furthermore,
$\boldsymbol{v}^{(0)}=\boldsymbol{v}$
on
$\mathcal{S}$
. Denote the strain rate tensor and stress tensor by
$(\boldsymbol{\sigma }, \unicode{x1D65A})$
and
$(\boldsymbol{\sigma }^{(0)}, \unicode{x1D65A}^{(0)})$
, respectively. Following steps similar to those leading to (2.3), we find
\begin{align} \int _{\mathcal{V}}\textrm{d}V \left [e_{\alpha \beta }(\boldsymbol{r})-e_{\alpha \beta }^{(0)}(\boldsymbol{r})\right ]\eta _{\alpha \beta \lambda \nu }^{{S}}\,e_{\lambda \nu }^{(0)}(\boldsymbol{r})=0. \end{align}
Next, we compute the viscous energy dissipation:
Equality is achieved only when
$ \unicode{x1D65A}= \unicode{x1D65A}^{(0)}$
. We conclude that adding odd viscous effects to a system with given boundary conditions will always increase viscous dissipation, unless the velocity fields are unaffected by odd viscosity. However, in the latter case, pressure fields can differ; see the example discussed in § 6.
3. Description of a model fluid with odd viscosity and stress response
The simplest
$\boldsymbol{\eta }$
with non-trivial
$\boldsymbol{\ell }$
that satisfies the Onsager–Casimir symmetry and produces a symmetric stress tensor is
\begin{eqnarray} \eta _{\alpha \beta \lambda \nu }(\hat{\boldsymbol{\ell }})&=&\eta _{\textit{s}}\left (\delta _{\alpha \lambda }\delta _{\nu \beta }+\delta _{\alpha \nu }\delta _{\lambda \beta }-\dfrac {2}{3}\delta _{\alpha \beta }\delta _{\lambda \nu }\right )\nonumber \\ &&-\eta _{\textit{o}}\hat {\ell }_\kappa \big(\epsilon _{\kappa \alpha \lambda }\delta _{\nu \beta }+\epsilon _{\kappa \alpha \nu }\delta _{\lambda \beta }+\epsilon _{\kappa \beta \lambda }\delta _{\nu \alpha }+\epsilon _{\kappa \beta \nu }\delta _{\lambda \alpha}\big). \end{eqnarray}
Here,
$\eta _{\textit{s}}$
is the dynamic shear viscosity, and
$\eta _{\textit{o}}$
is the odd-viscosity coefficient that quantifies the magnitude of
$\boldsymbol{{\ell}}$
through the relation
$\boldsymbol{\ell } = 4\eta _{\textit{o}}\boldsymbol{\hat \ell }$
, following the convention of Markovich & Lubensky (Reference Markovich and Lubensky2021). The tensors
$\delta _{\alpha \beta },\epsilon _{\alpha \beta \nu }$
represent the Kronecker delta and the Levi-Civita symbol, respectively. The corresponding stress tensor is
\begin{equation} \sigma _{\alpha \beta }(\boldsymbol{r})=-p(\boldsymbol{r})\,\delta _{\alpha \beta }+2\eta _{\textit{s}}\,e_{\alpha \beta }(\boldsymbol{r})-2\eta _{\textit{o}}\hat {\ell }_\kappa [\epsilon _{\kappa \alpha \lambda }\,e_{\lambda \beta }(\boldsymbol{r})+\epsilon _{\kappa \beta \lambda }\,e_{\lambda \alpha }(\boldsymbol{r})], \end{equation}
which depends on the fluid velocity solely through
$ \unicode{x1D65A}(\boldsymbol{r})$
. Inserting (3.1) in (2.1), we find
\begin{equation} \eta _{\textit{s}}{\nabla} ^2\boldsymbol{ v}(\boldsymbol{r})-\boldsymbol{\nabla }\tilde p(\boldsymbol{r})+\eta _{\textit{o}}\big(\hat{\boldsymbol{\ell }}\boldsymbol{\cdot }\boldsymbol{\nabla }\big)[\boldsymbol{\nabla }\times \boldsymbol{ v}(\boldsymbol{r})]=-\boldsymbol{ f}(\boldsymbol{r}), \quad \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{ v}(\boldsymbol{r})=0. \end{equation}
Here, we introduced the effective pressure
$\tilde {p}(\boldsymbol{r})=p(\boldsymbol{r})+2\eta _{\textit{o}}\hat{\boldsymbol{\ell }}\boldsymbol{\cdot }[\boldsymbol{\nabla }\times \boldsymbol{ v}(\boldsymbol{r})]$
. In two spatial dimensions, the third term in (3.3) can be absorbed into the pressure, so
$\eta _{\textit{o}}$
does not contribute to
$\boldsymbol{v}(\boldsymbol{r})$
. It then follows from (2.5) and the ordinary Stokes form for
$\boldsymbol{\eta }^{{S}}$
(first line in (3.1)) that
$\eta _{\textit{o}}$
does not contribute to
$\dot {E}$
. This is a well-known result for this type of constitutive relation (Banerjee et al. Reference Banerjee, Souslov, Abanov and Vitelli2017). In contrast, in three (or higher) spatial dimensions, odd viscosity can contribute to the viscous dissipation. Here, we focus on the three-dimensional case.
In our analysis, the fundamental solution to (3.3) is essential, which is defined as the response of
$\boldsymbol{v}(\boldsymbol{r})$
and
$\tilde {p}(\boldsymbol{r})$
to a point-force density
$\boldsymbol{f}(\boldsymbol{r})=\boldsymbol{F}_0\,\delta (\boldsymbol{r})$
. This defines the Green’s tensor
$ \unicode{x1D642}(\boldsymbol{r})$
and pressure vector
$\boldsymbol{Q}(\boldsymbol{r})$
via
$\boldsymbol{v}(\boldsymbol{r})= \unicode{x1D642}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{F}_0$
and
$\tilde {p}(\boldsymbol{r})=\boldsymbol{Q}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{F}_0$
. The Green’s functions can be explicitly computed (Everts & Cichocki Reference Everts and Cichocki2024a
). We define an orthonormal triad of spherical basis vectors
$\{\hat{\boldsymbol{r}},\hat{\boldsymbol{\theta }},\hat{\boldsymbol{\phi }}\}$
, with
$\hat{\boldsymbol{\phi }}=(\hat{\boldsymbol{\ell }}\times \hat{\boldsymbol{r}})/|\hat{\boldsymbol{\ell }}\times \hat{\boldsymbol{r}}|$
and
$\hat{\boldsymbol{\theta }}=\hat{\boldsymbol{\phi }}\times \hat{\boldsymbol{r}}$
, and the auxiliary variable
$s=\gamma\, |\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|$
, where
$\gamma =\eta _{\textit{o}}/\eta _{\textit{s}}$
. With this notation, we find
\begin{align} \unicode{x1D642}(\boldsymbol{r}) &=\frac {\varLambda (s)}{4\pi \eta _{\textit{s}}r[1+\varLambda (s)]}\left [ \unicode{x1D644}+\varLambda (s)\,\hat {\boldsymbol{r}}\hat {\boldsymbol{r}}+s \varLambda (s)\,\big(\hat{\boldsymbol{r}}\hat{\boldsymbol{\phi }}-\hat{\boldsymbol{\phi }}\hat{\boldsymbol{r}}\big) -\left [1-\varLambda (s)\right ]\hat{\boldsymbol{\phi }}\hat{\boldsymbol{\phi }}\right ],\end{align}
\begin{align} \boldsymbol{Q}(\boldsymbol{r}) &=\frac {1}{4\pi r^2}\hat{\boldsymbol{r}},\end{align}
with
$\varLambda (s)=(1+s^2)^{-1/2}$
. Observe that (3.5) is of the Stokes form, and that
$\hat{\boldsymbol{r}}\hat{\boldsymbol{\phi }}-\hat{\boldsymbol{\phi }}\hat{\boldsymbol{r}}$
is the rotation operator around the
$\hat{\boldsymbol{\theta }}$
axis. For the transformations
$(\boldsymbol{r},\hat{\boldsymbol{\ell }})\rightarrow (\boldsymbol{r},-\hat{\boldsymbol{\ell }})$
and
$(\boldsymbol{r},\hat{\boldsymbol{\ell }})\rightarrow (-\boldsymbol{r},\hat{\boldsymbol{\ell }})$
, we have that
$\hat{\boldsymbol{\phi }}\rightarrow -\hat{\boldsymbol{\phi }}$
. We have the following symmetry properties:
\begin{align} G_{\alpha \beta }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big) &=G_{\beta \alpha }\big(\boldsymbol{r};-\hat{\boldsymbol{\ell }}\big), \end{align}
\begin{align} G_{\alpha \beta }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)=G_{\alpha \beta }\big(-\boldsymbol{r};\hat{\boldsymbol{\ell }}\big), & \quad Q_\alpha \big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)=-Q_\alpha \big(-\boldsymbol{r};\hat{\boldsymbol{\ell }}\big). \end{align}
Equation (3.6) follows from the Onsager–Casimir relations, while (3.7) follows from the defining differential equation for
$ \unicode{x1D642}(\boldsymbol{r})$
and
$\boldsymbol{Q}(\boldsymbol{r})$
. These relations can also be deduced from the explicit expressions (3.4) and (3.5). We define the corresponding rate of strain tensor
$\boldsymbol{\varTheta }(\boldsymbol{r})$
and stress tensor
$\boldsymbol{\varSigma }(\boldsymbol{r})$
through the relations
$ \unicode{x1D65A}(\boldsymbol{r})=\boldsymbol{\varTheta }(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{F}_0$
and
$\boldsymbol{\sigma }(\boldsymbol{r})=\boldsymbol{\varSigma }(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{F}_0$
. We find
\begin{equation} \varSigma _{\alpha \beta \nu }(\boldsymbol{r})=-\delta _{\alpha \beta } P_{\nu }(\boldsymbol{r})+2\eta _{\textit{s}} \varTheta _{\alpha \beta \nu }(\boldsymbol{r})+2\eta _{\textit{o}}\hat {\ell }_{\kappa }[\epsilon _{\kappa \alpha \lambda }\,\varTheta _{\lambda \beta \nu }(\boldsymbol{r})+\epsilon _{\kappa \beta \lambda }\,\varTheta _{\lambda \alpha \nu }(\boldsymbol{r})], \end{equation}
where
\begin{equation} \boldsymbol{P}(\boldsymbol{r})=\boldsymbol{Q}(\boldsymbol{r})-2\eta _{\textit{o}}\hat{\boldsymbol{\ell }}\boldsymbol{\cdot }[\boldsymbol{\nabla }\times \unicode{x1D642}(\boldsymbol{r})], \end{equation}
with
$[\boldsymbol{\nabla }\times \unicode{x1D642}(\boldsymbol{r})]_{\alpha \beta }=\epsilon _{\alpha \nu \lambda }\,\partial _{\nu } G_{\lambda \beta }(\boldsymbol{r})$
. We have the symmetry relations
\begin{equation} \varSigma _{\alpha \beta \nu }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)=\varSigma _{\beta \alpha \nu }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big), \quad \varSigma _{\alpha \beta \nu }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)=-\varSigma _{\alpha \beta \nu }\big(-\boldsymbol{r};\hat{\boldsymbol{\ell }}\big). \end{equation}
Expressions for the elements of
$\boldsymbol{\varSigma }(\boldsymbol{r})$
are listed in Appendix A and are a crucial novelty of this work.
4. Friction problem for a single sphere and its singularity representation
Consider a sphere of radius
$a$
in an unbounded fluid described by (3.3) for
$r\gt a$
and
$\boldsymbol{f}(\boldsymbol{r})=\boldsymbol{0}$
. Now (3.3) is supplemented by the boundary conditions
\begin{equation} \boldsymbol{v}(a\hat {\boldsymbol{r}})=\boldsymbol{U}+\boldsymbol{\varOmega }\times (a\hat {\boldsymbol{r}}), \quad \boldsymbol{v}(\boldsymbol{r})\rightarrow \boldsymbol{v}_\infty (\boldsymbol{r}), \ \tilde {p}(\boldsymbol{r})=\tilde {p}_\infty (\boldsymbol{r}) \quad (r\rightarrow \infty ). \end{equation}
Here, the solid-body motion of the sphere is characterised by translational velocity
$\boldsymbol{U}$
and rotational velocity
$\boldsymbol{\varOmega }$
, and we included ambient flow and pressure fields
$\boldsymbol{v}_\infty (\boldsymbol{r})$
and
$\tilde {p}_\infty (\boldsymbol{r})$
, respectively. Our goal is to find the singularity representation for
$\boldsymbol{v}(\boldsymbol{r})$
and
$\tilde {p}(\boldsymbol{r})$
, with corresponding stress tensor
$\boldsymbol{\sigma }(\boldsymbol{r})$
. To find this expression, we define an auxiliary flow field
$\boldsymbol{v}_0(\boldsymbol{r})$
with stress tensor
$\boldsymbol{\sigma }_0(\boldsymbol{r})$
. Two such flows defined in a volume
$\mathcal{V}$
bounded by a closed surface
$\mathcal{S}$
are related by the Lorentz reciprocal theorem for odd fluids (Hosaka et al. Reference Hosaka, Golestanian and Vilfan2023):
\begin{align} &\int _{\mathcal{S}}\textrm{d}S\, \boldsymbol{v}\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\big[\boldsymbol{\sigma }_0\big(\boldsymbol{r};-\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\hat{\boldsymbol{n}}\big]-\int _{\mathcal{V}}\textrm{d}V\, \boldsymbol{v}_0\big(\boldsymbol{r};-\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\big[\boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{\sigma }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)\big] \nonumber \\ &=\int _{\mathcal{S}}\textrm{d}S\, \boldsymbol{v}_0\big(\boldsymbol{r};-\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\big[\boldsymbol{\sigma }\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\hat{\boldsymbol{n}}\big]-\int _{\mathcal{V}}\textrm{d}V\, \boldsymbol{v}\big(\boldsymbol{r};\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\big[\boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{\sigma }_0\big(\boldsymbol{r};-\hat{\boldsymbol{\ell }}\big)\big]. \end{align}
The form of this theorem is the same as in ordinary Stokes flow (Lorentz Reference Lorentz1896; Masoud & Stone Reference Masoud and Stone2019), with the important difference that
$\boldsymbol{v}_0$
satisfies (3.3) with
$\hat{\boldsymbol{\ell }}$
replaced by
$-\hat{\boldsymbol{\ell }}$
(as indicated by the second argument). For the auxiliary flow field, we replace
$\boldsymbol{v}_0(\boldsymbol{r};-\hat{\boldsymbol{\ell }})$
by
$ \unicode{x1D642}(\boldsymbol{r}^{\prime}-\boldsymbol{r};-\hat{\boldsymbol{\ell }})$
. Since we impose stick-boundary conditions, the double-layer contribution can be eliminated using the same steps as for ordinary Stokes flow; see § 2.4.2 in Kim & Karrila (Reference Kim and Karrila2013). We find that the induced forces picture for a rigid body with surface
$\mathcal{S}_{\textit{p}}$
in a chiral active fluid is identical to the one for ordinary Stokes flow (Mazur & Bedeaux Reference Mazur and Bedeaux1974):
\begin{equation} \boldsymbol{u}(\boldsymbol{r})-\boldsymbol{v}^\infty (\boldsymbol{r})=\int _{{\mathcal{S}_{\textit{p}}}} \textrm{d}S_{\boldsymbol{r}^{\prime}}\, \unicode{x1D642}\big(\boldsymbol{r}-\boldsymbol{r}^{\prime};\hat{\boldsymbol{\ell }}\big)\boldsymbol{\cdot }\boldsymbol{f}_{{ind}}(\boldsymbol{r}^{\prime}), \end{equation}
where
$\boldsymbol{f}_{{ind}}=\boldsymbol{\sigma }(\boldsymbol{r})\boldsymbol{\cdot }\hat{\boldsymbol{n}}|_{\boldsymbol{r}\in \mathcal{S}_{\textit{p}}}$
, with
$\hat{\boldsymbol{n}}$
pointing towards the particle. Note that in Kim & Karrila (Reference Kim and Karrila2013),
$ \unicode{x1D642}$
is located on the right-hand side of
$\boldsymbol{f}_{{ind}}$
. For the odd case, commuting it to the left-hand side changes
$-\hat{\boldsymbol{\ell }}$
to
$+\hat{\boldsymbol{\ell }}$
, using (3.7). Equation (4.3) is also valid inside the particle, where for a sphere,
$\boldsymbol{u}(\boldsymbol{r})=\boldsymbol{U}+\boldsymbol{\varOmega }\times \boldsymbol{r}$
for
$r\lt a$
, and
$\boldsymbol{u}(\boldsymbol{r})=\boldsymbol{v}(\boldsymbol{r})$
for
$r\gt a$
. Furthermore,
$\boldsymbol{f}_{{ind}}(\boldsymbol{r})$
has the first few moments given by
with
$\boldsymbol{F}$
the force,
$\boldsymbol{T}$
the torque, and
$ \unicode{x1D64E}$
the stresslet of the particle acting on the fluid. The overbracket defines the symmetric traceless part of a tensor. For later use, we also define the grand friction tensor components as
\begin{equation} \begin{pmatrix} \boldsymbol{F}\\ \boldsymbol{T}\\ \unicode{x1D64E} \end{pmatrix} = \begin{pmatrix} \boldsymbol{\zeta }^{{tt}} & \boldsymbol{\zeta }^{{tr}} & \boldsymbol{\zeta }^{{td}}\\ \boldsymbol{\zeta }^{{rt}} & \boldsymbol{\zeta }^{{rr}} & \boldsymbol{\zeta }^{{rd}}\\ \boldsymbol{\zeta }^{{dt}} & \boldsymbol{\zeta }^{{dr}} & \boldsymbol{\zeta }^{{dd}} \end{pmatrix} \begin{pmatrix} \boldsymbol{U}-\boldsymbol{U}^\infty \\ \boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty \\ - \unicode{x1D640}^\infty , \end{pmatrix}, \end{equation}
where the first moments of
$\boldsymbol{v}^\infty (\boldsymbol{r})$
give a contribution from constant ambient flow
$\boldsymbol{U}^\infty$
, a rotating ambient velocity
$\boldsymbol{\varOmega }^\infty$
, and a linear straining field
$ \unicode{x1D640}^\infty$
.
For a sphere, (4.3) simplifies to
\begin{equation} \boldsymbol{u}(\boldsymbol{r})-\boldsymbol{v}^\infty (\boldsymbol{r})=a^2\int _{S^2} \textrm{d}^2\hat {\boldsymbol{r}}^{\prime} \, \unicode{x1D642}\big(\boldsymbol{r}-a\hat {\boldsymbol{r}}^{\prime}\big)\boldsymbol{\cdot }\boldsymbol{{f}}_{{ind}}\big(\hat {\boldsymbol{r}}^{\prime}\big), \end{equation}
with
$S^2$
the two-dimensional unit sphere. For the case
$r\geqslant a$
, we take the sources in the centre, and perform a multipole expansion
\begin{equation} \unicode{x1D642}\big(\boldsymbol{r}-a\hat {\boldsymbol{r}}^{\prime}\big)=\left [1-a\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\frac {\partial }{\partial \boldsymbol{r}}+\frac {a^2}{2}\left (\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\frac {\partial }{\partial \boldsymbol{r}}\right )^2+\cdots \right ] \unicode{x1D642}(\boldsymbol{r})=:{\rm e}^{-a\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\boldsymbol{\nabla }}\, \unicode{x1D642}(\boldsymbol{r}), \end{equation}
which results in
\begin{align} \boldsymbol{v}(\boldsymbol{r})-\boldsymbol{v}^\infty (\boldsymbol{r}) &=a^2\int _{S^2} \textrm{d}^2\hat {\boldsymbol{r}}^{\prime}\, {\rm e}^{-a\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\boldsymbol{\nabla }}\, \unicode{x1D642}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{{f}}_{{ind}}(\hat {\boldsymbol{r}}^{\prime}), \end{align}
\begin{align} \tilde {p}(\boldsymbol{r})-\tilde {p}^\infty (\boldsymbol{r}) &=a^2\int _{S^2} \textrm{d}^2\hat {\boldsymbol{r}}^{\prime}\, {\rm e}^{-a\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\boldsymbol{\nabla }}\,\boldsymbol{Q}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{{f}}_{{ind}}(\hat {\boldsymbol{r}}^{\prime}). \end{align}
To find the singularity representation of
$\tilde {p}(\boldsymbol{r})$
, we used that
$\boldsymbol{Q}(\boldsymbol{r})$
is of the same form as for a Newtonian fluid. We note that for more complicated boundary conditions (e.g. droplets immersed in an odd fluid), (4.8) contains not only the single-layer contributions, but also terms from the hydrodynamic double layer. Such contributions can be computed using the results in Appendix A.
5. Translating sphere in constant ambient flow
5.1. Singularity representation for a translating sphere
We first consider (4.8) for the case with
$\boldsymbol{\varOmega }=0$
and a constant ambient flow field
$\boldsymbol{v}^\infty (\boldsymbol{r})=\boldsymbol{U}^\infty$
. As an ansatz, we assume that this case is described by a constant
$\boldsymbol{{f}}_{{ind}}(\hat{\boldsymbol{r}})$
. Using (4.4) we find then that
$\boldsymbol{F}=4\pi a^2\boldsymbol{{f}}_{{ind}}$
and (4.8a
) reduces to
\begin{equation} \boldsymbol{v}(\boldsymbol{r})-\boldsymbol{U}^\infty =\frac {1}{4\pi }\int _{S^2} \textrm{d}^2\hat {\boldsymbol{r}}^{\prime}\, {\rm e}^{-a\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\boldsymbol{\nabla }}\, \unicode{x1D642}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{F}=:\mathcal{L}_0 \unicode{x1D642}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{F}. \end{equation}
Note that within this ansatz, it follows from (4.4) that
$\boldsymbol{T}=0$
and
$ \unicode{x1D64E}=0$
, therefore there is no translational–rotational and translational–dipolar coupling (i.e.
$\boldsymbol{\zeta }^{{tr}}=\boldsymbol{\zeta }^{{rt}}=0$
and
$\boldsymbol{\zeta }^{{td}}=\boldsymbol{\zeta }^{{dt}}=0$
). Thus we can write
\begin{equation} \boldsymbol{v}(\boldsymbol{r})-\boldsymbol{U}^\infty =\mathcal{L}_0 \unicode{x1D642}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{\zeta }^{{tt}}\boldsymbol{\cdot }(\boldsymbol{U}-\boldsymbol{U}^\infty ), \end{equation}
where
$\boldsymbol{\zeta }^{{tt}}$
is the translational–translational friction tensor. The
$\mathcal{L}_0$
operator can be explicitly found by the angular integration in (5.1):
\begin{equation} \mathcal{L}_0=\frac {1}{4\pi }\int _{S^2} \textrm{d}^2\hat {\boldsymbol{r}}^{\prime}\, {\rm e}^{-a\hat {\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\boldsymbol{\nabla }}=\sum _{n=0}^\infty \frac {a^{2n}}{(2n+1)!}\big({\nabla} ^2\big)^n=:j_0(\rm i\mathcal{D}), \end{equation}
with
$\mathcal{D}^2=a^2{\nabla} ^2$
, and
$j_n$
the
$n$
th-order spherical Bessel function of the first kind.
The operator series acting on
$ \unicode{x1D642}(\boldsymbol{r})$
can be evaluated using Fourier methods (Everts & Cichocki Reference Everts and Cichocki2024a
):
\begin{equation} \mathcal{L}_0 \unicode{x1D642}(\boldsymbol{r})=\frac {1}{\eta _{\textit{s}}}\int \frac {\textrm{d}^3\boldsymbol{k}}{(2\pi )^3}\, j_0(ka)\,{\rm e}^{{\rm i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r}}\,\frac { \unicode{x1D644}-\hat {\boldsymbol{k}}\hat {\boldsymbol{k}}+\gamma \big(\hat {\boldsymbol{k}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat {\boldsymbol{k}}\big)}{k^2\big[1+\gamma ^2\big(\hat {\boldsymbol{k}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)^2\big]}. \end{equation}
To show that the initial ansatz of a constant
$\boldsymbol{{f}}_{{ind}}$
is correct, we observe that (5.1) satisfies (3.3). We thus need to check only whether the boundary conditions are satisfied. Indeed, in Everts & Cichocki (Reference Everts and Cichocki2024a
), it was shown that
$\mathcal{L}_0 \unicode{x1D642}(a\hat {\boldsymbol{r}})$
is independent of
$\hat{\boldsymbol{r}}$
and, therefore, the boundary condition
$\boldsymbol{v}(a\hat {\boldsymbol{r}})=\boldsymbol{U}$
can be satisfied when we identify
$\boldsymbol{\zeta }^{{tt}}=[\mathcal{L}_0 \unicode{x1D642}(a\hat {\boldsymbol{r}})]^{-1}$
. We conclude that (5.2) solves the boundary-value problem and that
$\boldsymbol{{f}}_{{ind}}$
is indeed constant. Explicit evaluation of
$\boldsymbol{\zeta }^{{tt}}$
gives
\begin{equation} \boldsymbol{\zeta }^{{tt}}=24\pi \eta _{\textit{s}}a\Bigg [\frac {R(\gamma )\big( \unicode{x1D644}-{\boldsymbol{\hat {{\ell }}}\boldsymbol{\hat{\ell }}}\big)+S(\gamma )\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)}{R(\gamma )^2+S(\gamma )^2}+\frac {{\boldsymbol{\hat{\ell }}\boldsymbol{\hat{\ell }}}}{T(\gamma )}\Bigg ]. \end{equation}
Here, we defined the following functions of the parameter
$\gamma =\eta _{\textit{o}}/\eta _{\textit{s}}$
:
\begin{align} R(\gamma )=\gamma ^2[f(\gamma )-g(\gamma )]+4, \quad S(\gamma )=2\gamma f(\gamma ),\quad T(\gamma )=2\gamma ^2[f(\gamma )+g(\gamma )]+4, \end{align}
with
\begin{equation} f(\gamma )=\frac {3}{\gamma ^2}\left (\frac {\arctan {\gamma }}{\gamma }-1\right )=-1+O\big(\gamma ^2\big),\quad {g(\gamma )=\frac {1+f(\gamma )}{\gamma ^2}=\frac {3}{5}+O\big(\gamma ^2\big)}. \end{equation}
Equations (5.2) and (5.5) form the singularity solution for a translating sphere. The first-order approximate result for
$\gamma \ll 1$
found by expanding (5.5) is consistent with results found in Khain et al. (Reference Khain, Scheibner, Fruchart and Vitelli2022) and Yuan & Olvera de la Cruz (Reference Yuan and Olvera de la Cruz2023). Furthermore, since
$({\nabla} ^2)^n\boldsymbol{Q}(\boldsymbol{r})=0$
for
$n\geqslant 2$
, we find for the effective pressure using (4.8b
)
\begin{equation} \tilde {p}(\boldsymbol{r})-\tilde {p}^\infty =\boldsymbol{Q}(\boldsymbol{r})\boldsymbol{\cdot }\boldsymbol{\zeta }^{{tt}}\boldsymbol{\cdot }(\boldsymbol{U}-\boldsymbol{U}^\infty ), \end{equation}
with
$\tilde {p}^\infty$
a constant. Note that for
$\gamma =0$
, we find the Oseen tensor
$ \unicode{x1D642}(\boldsymbol{r})|_{\gamma =0}=(8\pi \eta _{\textit{s}}r)^{-1}( \unicode{x1D644}+\hat {\boldsymbol{r}}\hat {\boldsymbol{r}})$
, and since
$({\nabla} ^2)^n \unicode{x1D642}(\boldsymbol{r})|_{\gamma =0}=0$
for
$n\geqslant 2$
, we find the well-known singularity representations for ordinary Stokes flow:
\begin{align} & \boldsymbol{v}(\boldsymbol{r})|_{\gamma =0}-\boldsymbol{U}^\infty =6\pi \eta _{\textit{s}}a\left (1+\frac {a^2}{6}{\nabla} ^2\right ) \unicode{x1D642}(\boldsymbol{r})|_{\gamma =0}\boldsymbol{\cdot }(\boldsymbol{U}-\boldsymbol{U}^\infty ), \end{align}
\begin{align} & p(\boldsymbol{r})|_{\gamma =0}-p^\infty =\frac {3a}{2\eta _{\textit{s}} r^2}\hat {\boldsymbol{r}}\boldsymbol{\cdot }(\boldsymbol{ U}-\boldsymbol{U}^\infty ). \end{align}
5.2. Explicit evaluation of the fluid velocity field around a translating sphere
Now we consider
$\boldsymbol{v}(\boldsymbol{r})$
for
$\gamma \neq 0$
. Since the singularity representation (5.1) solves the boundary-value problem of a translating sphere, we can obtain an explicit form for
$\boldsymbol{v}(\boldsymbol{r})$
by evaluating (5.4) for all
$r\gt a$
. The details are presented in Appendix B. The final result is
\begin{align} \boldsymbol{v}(\boldsymbol{r})={}&\frac {6}{\gamma ^2\,T(\gamma )}\left \{\left [\frac {1+\gamma ^2}{\gamma }\mathcal{M}(\boldsymbol{r};\gamma )-\frac {a}{r}\right ]\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\ell }}+\mathcal{N}(\boldsymbol{r};\gamma )\,\big(\gamma \hat{\boldsymbol{\phi }}\hat{\boldsymbol{\ell }}-\hat{\boldsymbol{\rho }}\hat{\boldsymbol{\ell }}\big)\right \}\boldsymbol{\cdot }(\boldsymbol{U}-\boldsymbol{U}^\infty ) \nonumber \\ &{}+\frac {6}{\gamma ^2\big[R(\gamma )^2+S(\gamma )^2\big]}\Bigg (\mathcal{O}(\boldsymbol{r};\gamma )\left [R(\gamma )\,\hat{\boldsymbol{\rho }}\hat{\boldsymbol{\rho }}+S(\gamma )\,\hat{\boldsymbol{\rho }}\hat{\boldsymbol{\phi }}\right ]\nonumber \\ &{}+\mathcal{N}(\boldsymbol{r};\gamma )\left \{[\gamma S(\gamma )-R(\gamma )]\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\rho }}-[\gamma R(\gamma )+S(\gamma )]\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\phi }}\right \} \\ &{}-\left [\frac {1}{2}\mathcal{O}(\boldsymbol{r};\gamma )+\frac {1-\gamma ^2}{2\gamma }\mathcal{M}(\boldsymbol{r};\gamma )-\frac {a}{2 r}\right ]\left [R(\gamma )\,\big(\unicode{x1D644}-{\boldsymbol{\hat{\ell }}\boldsymbol{\hat{\ell }}}\big)+S(\gamma )\,\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\right ]\nonumber \\ &{}-\left [\mathcal{M}(\boldsymbol{r};\gamma )-\frac {a\gamma }{r}\right ]\left [R(\gamma )\,\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)-S(\gamma )\,\big(\unicode{x1D644}-{\boldsymbol{\hat{\ell} }\boldsymbol{\hat{\ell }}}\big)\right ] \Bigg )\boldsymbol{\cdot }(\boldsymbol{U}-\boldsymbol{U}^\infty ), \nonumber \end{align}
with the dimensionless functions
\begin{align} &\mathcal{M}(\boldsymbol{r};\gamma )=\textrm{arcsin}\left [\frac {1}{\mathcal{R}_+(\boldsymbol{r};\gamma )}\right ],\nonumber \\ &\mathcal{N}(\boldsymbol{r};\gamma )=\frac {a}{r\,|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|}\left [\textrm{sgn}\big(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\sqrt {1-\mathcal{R}_-(\boldsymbol{r};\gamma )^2}-\big(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\right ], \\ &\mathcal{O}(\boldsymbol{r};\gamma )=\frac {a^2}{r^2\,|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|^2}\left \{\gamma \sqrt {\mathcal{R}_+(\boldsymbol{r};\gamma )^2-1}\left [1-\sqrt {1-\mathcal{R}_-(\boldsymbol{r};\gamma )^2}\right ]^2-\frac {r}{ a}\big(1-|\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}|\big)^2\right \} \nonumber \end{align}
defined in terms of
\begin{align} \mathcal{R}_\pm (\boldsymbol{r};\gamma )=\frac {r}{2a}\frac {\mathcal{A}_+\pm \mathcal{A}_-}{\sin \psi }, \quad \mathcal{A}_\pm (\boldsymbol{r};\gamma )=\sqrt {\big(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)^2\cos ^2\psi +\left (|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|\pm \frac {a}{r}\sin \psi \right )^2}, \end{align}
and
$\psi =\textrm{arctan}(\gamma )$
. Moreover, we have adopted a cylindrical basis
$\{\hat{\boldsymbol{\rho }},\hat{\boldsymbol{\phi }},\hat{\boldsymbol{\ell }}\}$
with
$\hat{\boldsymbol{\rho }}=\hat{\boldsymbol{\phi }}\times \hat{\boldsymbol{\ell }}$
. Note that
$\hat{\boldsymbol{\rho }}\rightarrow \hat{\boldsymbol{\rho }}$
under the transformation
$\hat{\boldsymbol{\ell }}\rightarrow -\hat{\boldsymbol{\ell }}$
. The first line in (5.10) corresponds to the flow field for
$\boldsymbol{U}\parallel \boldsymbol{\hat \ell }$
, which was already found in previous work (Everts & Cichocki Reference Everts and Cichocki2024a
). However, in this work, the closed-form expressions for the case
$\boldsymbol{U}\perp \boldsymbol{\hat \ell }$
were not presented, although the solution was sketched in the Supplemental Material. Here, we have simplified this calculation by rewriting certain integrals using identities of Bessel functions, and explicitly carried out this computation until the end. Our new result (5.10) is valid for arbitrary directions of
$\boldsymbol{U}$
and
$\boldsymbol{\hat \ell }$
, and it is worth noting that it is an analytical, exact closed-form expression for a fully three-dimensional flow field around a sphere in a chiral active fluid. Furthermore, (5.10) can be expanded for small
$\gamma$
and
$\boldsymbol{U}^\infty=\boldsymbol{0}$
as
\begin{align} \boldsymbol{v}(\boldsymbol{r})={}&\left [\frac {3a}{4r}\left ( \unicode{x1D644}+\hat {\boldsymbol{r}}\hat {\boldsymbol{r}}\right )+\frac {a^3}{4r^3}\left ( \unicode{x1D644}-3\hat {\boldsymbol{r}}\hat {\boldsymbol{r}}\right )\right ]\boldsymbol{\cdot }\boldsymbol{U}\nonumber \\&{}-\frac {3a}{8r}\left (1-\frac {a^2}{r^2}\right )\gamma \left [\boldsymbol{\hat \ell }\times \boldsymbol{U}-2(\hat {\boldsymbol{r}}\times \boldsymbol{U})\big(\hat {\boldsymbol{r}}\boldsymbol{\cdot }\boldsymbol{\hat \ell }\big)-\hat {\boldsymbol{r}}\boldsymbol{\cdot }\big(\boldsymbol{\hat \ell }\times \boldsymbol{U}\big)\hat {\boldsymbol{r}}\right ]+{O}(\gamma ^2). \end{align}
The first term corresponds to the ordinary Stokes solution of a Newtonian fluid (Kim & Karrila Reference Kim and Karrila2013), and the second term corresponds to the solution first calculated by Khain et al. (Reference Khain, Scheibner, Fruchart and Vitelli2022).
Representative streamlines of the fluid velocity field
$\boldsymbol{v}(\boldsymbol{r})$
around a spherical particle translating with velocity
$\boldsymbol{U}$
in the absence of ambient flow scaled to
$U=|\boldsymbol{U}|$
. All plots are generated using the exact analytical solution (5.10). (a) Stokes flow without odd viscosity (
$\gamma =\eta _{\textit{o}}/\eta _{\textit{s}}=0$
). (b–d) Odd viscous flow for
$\gamma =3$
at different relative orientations of
$\boldsymbol{U}$
and
$\boldsymbol{\hat \ell }$
: (b)
$\boldsymbol{U}\parallel \boldsymbol{\hat \ell }$
, (c)
$\sphericalangle (\boldsymbol{U},\boldsymbol{\hat \ell })=45^\circ$
and (d)
$\boldsymbol{U}\perp \boldsymbol{\hat \ell }.$
Note that the flows in (a) and (b) are cylindrically symmetric around the
$\boldsymbol{U}$
axis. All streamlines are directed from bottom to top.
Figure 1. Long description
The image shows four panels labeled (a) through (d), each depicting streamlines of fluid velocity around a spherical particle. The streamlines are color-coded to represent the magnitude of the velocity scaled to a specific value. Panel (a) illustrates Stokes flow without odd viscosity, showing cylindrically symmetric streamlines around the axis, directed from bottom to top. Panels (b), (c), and (d) depict odd viscous flow for different relative orientations of the particle’s velocity and spin angular momentum. Panel (b) shows the streamlines for a specific orientation, while panels (c) and (d) show different orientations, each with unique flow patterns. The streamlines in all panels are directed from bottom to top, highlighting the impact of odd viscosity on fluid dynamics.
Projections of
$\boldsymbol{v}(\boldsymbol{r})$
around a spherical particle translating with velocity
$\boldsymbol{U}$
in an odd viscous fluid. The anisotropy axis
$\boldsymbol{\hat \ell }$
is fixed in the
$+z$
direction, with the orientation of
$\boldsymbol{U}$
being varied with respect to
$\boldsymbol{\hat \ell }$
. (a,b) Projections of the streamlines of
$\boldsymbol{v}(\boldsymbol{r})$
onto the
$xz$
-plane. The colours are a measure for the out-of-plane component of
$\boldsymbol{v}(\boldsymbol{r})$
in the
$y$
direction scaled to
$U=|\boldsymbol{U}|$
. (c,d) Projections of the streamlines of
$\boldsymbol{v}(\boldsymbol{r})$
onto the
$xy$
-plane, with the colours being a measure for
$|\boldsymbol{v}(\boldsymbol{r})|$
. Here, (a) and (c) are for
$\gamma =1$
, whereas (b) and (d) are for
$\gamma =3$
.
Figure 2. Long description
A heat map displays streamlines around a spherical particle translating with velocity in an odd viscous fluid. The anisotropy axis is fixed in the z-direction, with the orientation of the translational velocity varying with respect to this axis. The projections of the streamlines of velocity onto the x-z plane are shown, with colors indicating the out-of-plane component of velocity in the z direction scaled to U. Another set of projections shows streamlines of velocity onto the x-y plane, with colors representing the magnitude of velocity divided by the magnitude of the translational velocity. Panels (a) and (c) are for one orientation, while panels (b) and (d) are for another orientation.
Evaluating (5.8), we find for the effective pressure
\begin{equation} \tilde {p}(\boldsymbol{r})-\tilde {p}^\infty =\frac {6\eta _{\textit{s}}a}{r^2}\left \{\frac {|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|\left [R(\gamma )\,\hat{\boldsymbol{\rho }}+S(\gamma )\,\hat{\boldsymbol{\phi }}\right ] }{R(\gamma )^2+S(\gamma )^2}+\frac {\big(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\hat{\boldsymbol{\ell }}}{T(\gamma )}\right \}\boldsymbol{\cdot }(\boldsymbol{U}-\boldsymbol{U}^\infty ). \end{equation}
Since the velocity field is affected by
$\eta _{\textit{o}}$
, it follows from (2.5) that a translating sphere dissipates more energy in the odd model fluid than in a Newtonian fluid. The same result was obtained by Everts & Cichocki (Reference Everts and Cichocki2024a
) by explicitly computing
$\dot {E}=\boldsymbol{\zeta }^{{tt}}: (\boldsymbol{U}-\boldsymbol{U}^\infty )(\boldsymbol{U}-\boldsymbol{U}^\infty )$
, which is larger than the Stokesian dissipation
$6\pi \eta _{\textit{s}}a\,|\boldsymbol{U}-\boldsymbol{U}^\infty |^2$
for all
$\gamma \gt 0$
and directions of
$\boldsymbol{U}-\boldsymbol{U}^\infty$
with respect to
$\hat{\boldsymbol{\ell }}$
.
Using (5.10), we visualise the streamlines of
$\boldsymbol{v}(\boldsymbol{r})$
for
$\gamma =3$
in figures 1(b–d), and compare them with ordinary Stokes flow
$(\gamma =0)$
; see figure 1(a). The details of the flow and the effect of changing
$\gamma$
can be seen in projections onto two orthogonal planes in figures 2(a–d). A distinct feature of odd viscosity is the emergence of azimuthal flows, which are most pronounced for
$\boldsymbol{U}\parallel \hat{\boldsymbol{\ell }}$
(figure 1
b) where the flow is cylindrically symmetric. These azimuthal patterns can still be seen when
$\boldsymbol{U}$
is not parallel to
$\boldsymbol{\hat \ell }$
, but with a tilted axis, as shown in figures 1(c,d) and in the projections figures 2(a,b). Such patterns were first seen for a point-force response parallel to
$\hat{\boldsymbol{\ell }}$
; see Khain et al. (Reference Khain, Scheibner, Fruchart and Vitelli2022). The flows for a general direction of the point force follow from (3.4).
6. Rotating sphere
6.1. Solution for a rotating sphere without using the singularity representation
Consider now a rotating sphere without translation (
$\boldsymbol{U}=\boldsymbol{0}$
) in an ambient flow field
$\boldsymbol{v}^\infty (\boldsymbol{r})=\boldsymbol{\varOmega }^\infty \times \boldsymbol{r}$
. Before evaluating the singularity representation for this case, we note that there is an alternative solution method (Hosaka et al. Reference Hosaka, Chatzittofi, Golestanian and Vilfan2024) by transforming (3.3) to
\begin{equation} \eta _{\textit{s}}{\nabla} ^2\boldsymbol{ v}(\boldsymbol{r})-\boldsymbol{\nabla }\hat p(\boldsymbol{r})-\eta _{\textit{o}}{\nabla} ^2\big[\boldsymbol{ v}(\boldsymbol{r})\times \hat{\boldsymbol{\ell }}\big]=\boldsymbol{0}, \quad \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{ v}(\boldsymbol{r})=0, \end{equation}
with modified effective pressure (Yuan & Olvera de la Cruz Reference Yuan and Olvera de la Cruz2023)
\begin{equation} \hat {p}(\boldsymbol{r})=\tilde {p}(\boldsymbol{r})-\eta _{\textit{o}}\hat{\boldsymbol{\ell }}\boldsymbol{\cdot }[\boldsymbol{\nabla }\times \boldsymbol{v}(\boldsymbol{r})]. \end{equation}
We make the following important observation. The general solution for ordinary Stokes flow (
$\eta _{\textit{o}}=0$
) is given by the Lamb solution (Lamb Reference Lamb1924), which can be classified according to the irreducible representations of the rotational group by the multipoles
$\boldsymbol{v}_{lm\sigma }^-(\boldsymbol{r})$
(singular at the origin) and
$\boldsymbol{v}_{lm\sigma }^+(\boldsymbol{r})$
(singular at infinity), with
$l=1,2,\ldots$
,
$m=-l,\ldots ,l$
and
$\sigma =0,1,2$
. See Cichocki, Felderhof & Schmitz (Reference Cichocki, Felderhof and Schmitz1988) for explicit expressions of
$\boldsymbol{v}_{lm\sigma }^\pm (\boldsymbol{r})$
. We find that
\begin{align} {\nabla} ^2 \boldsymbol{v}_{lm\sigma }^-(\boldsymbol{r})=\boldsymbol{0}, \quad \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{v}_{lm\sigma }^-(\boldsymbol{r})=0, \quad \textrm{for} \quad \sigma =1,2, \end{align}
\begin{align} {\nabla} ^2 \boldsymbol{v}_{lm\sigma }^+(\boldsymbol{r})=\boldsymbol{0}, \quad \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{v}_{lm\sigma }^+(\boldsymbol{r})=0, \quad \textrm{for} \quad \sigma =0,1, \end{align}
i.e. they are the general solutions to the ordinary Stokes equations with constant pressure. Clearly, these multipoles are also solutions to (6.1) with
$\hat {p}(\boldsymbol{r})$
constant. Therefore, (6.3) constitutes all solutions from the ordinary Stokes equations that are also solutions to the odd Stokes equations (3.3) with
$\boldsymbol{f=0}$
. If supported by the appropriate boundary conditions, it follows from our generalised Helmholtz theorem in § 2.2 that such flows have no contribution from odd viscosity to the viscous dissipation. The solution by Hosaka et al. (Reference Hosaka, Chatzittofi, Golestanian and Vilfan2024) is an explicit example of this. Namely, the
$\boldsymbol{v}(\boldsymbol{r})$
that satisfies the boundary condition of a rotating sphere can be constructed from
$\boldsymbol{v}_{1m1}^-(\boldsymbol{r})$
, to give the rotlet solution
$\boldsymbol{v}(\boldsymbol{r})={-a^3(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )\times \boldsymbol{\nabla }(1/r )}$
– the same as in an ordinary Stokes flow (Kim & Karrila Reference Kim and Karrila2013). Since
$ \unicode{x1D640}^\infty =0$
, we have
$\boldsymbol{T}=\boldsymbol{\zeta }^{{rr}}\boldsymbol{\cdot }(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )$
and
$ \unicode{x1D64E}=\boldsymbol{\zeta }^{{dr}}\boldsymbol{\cdot }(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )$
. It follows from the corresponding stress tensor that
\begin{equation} \boldsymbol{\zeta }^{{rr}}={8\pi \eta _{\textit{s}}a^3}\left [ \unicode{x1D644}-\frac {\gamma }{2}\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big) \right ], \quad \boldsymbol{\zeta }^{{dr}}=4\pi \eta _{\textit{s}}a^3\gamma \unicode{x1D668}, \end{equation}
where
$s_{\alpha \beta \nu }=(\hat {\ell }_{\alpha }\delta _{\beta \nu }+\hat {\ell }_{\beta }\delta _{\alpha \nu })/2-(1/3)\delta _{\alpha \beta }\hat {\ell }_\nu$
. This is different from Newtonian fluids, where rotation of a suspended solid body does not produce a stresslet. Furthermore, from symmetry we also find
$\boldsymbol{\zeta }^{{rd}}$
, since
${\zeta }_{\alpha \beta \lambda }^{{rd}}(\hat{\boldsymbol{\ell }})={\zeta }_{\lambda \alpha \beta }^{{dr}}(-\hat{\boldsymbol{\ell }})$
. Taking a reference fluid with
$\eta _{\textit{o}}=0$
, we indeed conclude from (2.5) that the viscous dissipation is the same as in ordinary Stokes flow. The same conclusion is obtained from
$\dot {E}=\boldsymbol{\zeta }^{{rr}}:(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )$
with the
$\boldsymbol{\zeta }^{{rr}}$
from (6.4).
Although this solution strategy is elegant, we cannot find
$\boldsymbol{\zeta }^{{dd}}$
nor can we describe
$\boldsymbol{v}(\boldsymbol{r})$
in the presence of a linear ambient shear flow using this method. Furthermore, if multiple odd viscosities (Khain et al. Reference Khain, Scheibner, Fruchart and Vitelli2022) are present, then this method breaks down as well. Therefore, it is important to find the general singularity representation for this type of boundary-value problem.
6.2. Solution from singularity representation and rotational dissipation
To construct the singularity representation for the rotating sphere, we assume an induced force density that is linear in
$\hat {\boldsymbol{r}}$
, i.e.
$\boldsymbol{{f}}_{{ind}}(\hat {\boldsymbol{r}})= \unicode{x1D646}\boldsymbol{\cdot }\hat {\boldsymbol{r}}$
, with a constant tensor
$ \unicode{x1D646}$
. Generally,
$ \unicode{x1D646}$
can be decomposed as
\begin{equation} K_{\alpha \beta }=K^{(1)}\delta _{\alpha \beta }+K^{(2)}_\lambda \epsilon _{\lambda \alpha \beta }+K_{\alpha \beta }^{(3)}, \end{equation}
with a constant scalar
$K^{(1)}$
, vector
$\boldsymbol{K}^{(2)}$
, and symmetric traceless tensor
$ \unicode{x1D646}^{(3)}$
. These quantities are determined below. Without loss of generality, we can set
$K^{(1)}=0$
due to the incompressibility condition. Using (4.4), we find
$\boldsymbol{F}=\boldsymbol{0}$
,
$\boldsymbol{T}=-(8/3)\pi a ^3\boldsymbol{K}^{(2)}$
and
$ \unicode{x1D64E}=(4/3)\pi a^3 \unicode{x1D646}^{(3)}$
. Then (4.8) reduces to
\begin{equation} [\boldsymbol{v}(\boldsymbol{r})-\boldsymbol{\varOmega }^\infty \times \boldsymbol{r}]_\alpha =-[\mathcal{L}_1\partial _\nu G_{\alpha \beta }](\boldsymbol{r})\left (\tfrac {1}{2}\epsilon _{\nu \beta \lambda }T_\lambda +S_{\nu \beta }\right ), \end{equation}
with the formal operator expression
\begin{equation} \mathcal{L}_1\boldsymbol{\nabla }=-\frac {3}{4\pi a}\int _{S^2}\textrm{d}^2\hat{\boldsymbol{r}}^{\prime} \hat{\boldsymbol{r}}^{\prime}\, {\rm e}^{-a\hat{\boldsymbol{r}}^{\prime}\boldsymbol{\cdot }\boldsymbol{\nabla }}. \end{equation}
The operator
$\mathcal{L}_1$
is found by performing the angular integration:
\begin{equation} \mathcal{L}_1=\sum _{n=0}^\infty \frac {6(n+1)}{(2n+3)!}a^{2n}\big({\nabla} ^2\big)^n=:\frac {3}{\rm i\mathcal{D}}\,j_1(\rm i\mathcal{D}). \end{equation}
Since there is no linear ambient shear flow, we find
\begin{equation} [\boldsymbol{v}(\boldsymbol{r})-\boldsymbol{\varOmega }^\infty \times \boldsymbol{r}]_\alpha =-[\mathcal{L}_1\partial _\nu G_{\alpha \beta }](\boldsymbol{r})\left (\frac {1}{2}\epsilon _{\nu \beta \lambda }\zeta ^{{rr}}_{\lambda \kappa }+\zeta ^{{dr}}_{\nu \beta \kappa }\right )[\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty ]_\kappa . \end{equation}
Everts & Cichocki (Reference Everts and Cichocki2024a
,
Reference Everts and Cichockib
) showed that
$[\mathcal{L}_1\boldsymbol{\nabla }\unicode{x1D642}](a\hat {\boldsymbol{r}})\sim \hat {\boldsymbol{r}}$
, therefore all linear-type boundary conditions can be satisfied. In particular, (6.4) can be constructed, although the procedure is more complicated than the one in § 6.1. However, with this procedure, we can obtain results for a stationary sphere in general linear ambient flow problems, such as linear shear flow.
For the effective pressure, we find using (4.8b ) that
\begin{equation} \tilde {p}(\boldsymbol{r})-\tilde {p}^\infty =-\big(\boldsymbol{\nabla }\boldsymbol{Q}:\boldsymbol{\zeta }^{{dr}}\big)\boldsymbol{\cdot }(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty ), \end{equation}
where we have used that
$\boldsymbol{\nabla }\times \boldsymbol{Q}(\boldsymbol{r})=\boldsymbol{0}$
. Furthermore,
$\tilde {p}^\infty$
(without argument) is a constant. It is instructive to consider the
$\gamma =0$
case. We then find the well-known results (Kim & Karrila Reference Kim and Karrila2013)
\begin{equation} \boldsymbol{v}(\boldsymbol{r})-\boldsymbol{\varOmega }^\infty \times \boldsymbol{r}=-4\pi \eta _{\textit{s}}a^3\left [\boldsymbol{\nabla }\times \unicode{x1D642}(\boldsymbol{r})\right ]|_{\gamma =0}\boldsymbol{\cdot }(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty ), \quad p(\boldsymbol{r})|_{\gamma =0}=\tilde {p}^\infty , \end{equation}
where we used that the Oseen tensor is symmetric. Computing the curl of the Oseen tensor, we find the rotlet velocity field
$-a^3(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )\times \boldsymbol{\nabla }(1/r )$
.
To show that the same solution holds for
$\gamma \neq 0$
using (6.9), we use
\begin{equation} [\mathcal{L}_1\boldsymbol{\nabla }\unicode{x1D642}]({\boldsymbol{r}})=\frac {3}{a}\int \frac {\textrm{d}^3\boldsymbol{ k}}{(2\pi )^3}\, {\rm e}^{{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r}}\,j_1(ka)\,\textrm{i}\hat {\boldsymbol{k}}\,\frac { \unicode{x1D644}-\hat {\boldsymbol{k}}\hat {\boldsymbol{k}}+\gamma \big(\hat {\boldsymbol{k}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat {\boldsymbol{k}}\big)}{\eta _{\textit{s}}k^2\big[1+\gamma ^2\big(\hat {\boldsymbol{k}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)^2\big]}. \end{equation}
In contrast with the translating sphere, we do not need an explicit expression for
$[\mathcal{L}_1\boldsymbol{\nabla }\unicode{x1D642}]({\boldsymbol{r}})$
. Insertion of (6.12) into (6.9) with friction tensors (6.4) results in
\begin{align} \boldsymbol{v}(\boldsymbol{r})-\boldsymbol{\varOmega }^\infty \times \boldsymbol{r}&=-\dfrac {3a^3}{2\pi ^2}(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )\times \boldsymbol{\nabla }\underbrace {\int \textrm{d}^3\boldsymbol{k}\, \frac {j_1(ka)}{ka}\frac {{\rm e}^{{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r}}}{k^2}}_{=2\pi ^2/(3r)}\nonumber \\ &=-a^3(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty )\times \boldsymbol{\nabla }\left (1/r\right ), \end{align}
where we used that
$r\gt a$
for computing the integral. Equation (6.13) coincides with the result from § 6.1. Using (6.10), we find for the effective pressure that
\begin{equation} \tilde {p}(\boldsymbol{r})-\tilde {p}^\infty =-\frac {\eta _{\textit{o}}a^3}{r^3}\big[\hat{\boldsymbol{\ell }}-3\big(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\hat{\boldsymbol{r}}\big]\boldsymbol{\cdot }(\boldsymbol{\varOmega }-\boldsymbol{\varOmega }^\infty ), \end{equation}
which corresponds to a constant
$\hat p$
. The simplification that occurs for a rotating sphere does not apply to the case of a stationary sphere in an ambient linear shear flow. In this case,
$[\mathcal{L}_1\boldsymbol{\nabla }\unicode{x1D642}]({\boldsymbol{r}})$
needs to be computed explicitly in the fluid region, even though the results for
$\boldsymbol{\zeta }^{{dr}}$
and
$\boldsymbol{\zeta }^{{dd}}$
are known, as was (indirectly) presented by Everts & Cichocki (Reference Everts and Cichocki2024a
,
Reference Everts and Cichockib
). We will present the explicit expressions for the flow field for this case in our future work.
6.3. Generalised odd-viscosity model
The lack of an odd-viscosity contribution to the viscous dissipated power is not a generic feature of a rotating sphere. A fluid with a different type of odd viscosity than in (3.1), or a fluid with more than just one odd-viscosity coefficient, generally has a
$\boldsymbol{v}(\boldsymbol{r})$
that is different from the Stokes flow of a Newtonian fluid. The reason is that a general
$\boldsymbol{\eta }^{{A}}$
cannot be absorbed into a suitable redefinition of the pressure. Therefore, rotating (passive) spheres in general chiral active fluids dissipate more energy than in a fluid with
$\boldsymbol{\eta }^{{A}}=0$
. The only exceptions are fluids described by the
$\boldsymbol{\eta }$
given in (3.1).
To illustrate this, we consider a minimally extended model of the viscosity tensor with two odd-viscosity coefficients
\begin{equation} \eta _{\alpha \beta \lambda \nu }(\boldsymbol{\hat \ell }) =\eta _{\textit{s}}\big (\delta _{\alpha \lambda }\delta _{\nu \beta }+\delta _{\alpha \nu }\delta _{\lambda \beta }-\tfrac {2}{3}\delta _{\alpha \beta }\delta _{\lambda \nu }\big ) + \eta _{\textit{o}}^{(1)}\, c^{(1)}_{\alpha \beta \lambda \nu }(\boldsymbol{\hat \ell }) + \eta _{\textit{o}}^{(2)}\, c^{(2)}_{\alpha \beta \lambda \nu }(\boldsymbol{\hat \ell }), \end{equation}
where
\begin{align} c^{(1)}_{\alpha \beta \lambda \nu } (\boldsymbol{\hat \ell }) &=\hat {\ell }_\kappa \big(\epsilon _{\kappa \alpha \lambda }\hat \ell _\beta \hat \ell _\nu +\epsilon _{\kappa \alpha \nu }\hat \ell _\beta \hat \ell _\lambda +\epsilon _{\kappa \beta \lambda }\hat \ell _\alpha \hat \ell _\nu +\epsilon _{\kappa \beta \nu }\hat \ell _\alpha \hat \ell _\lambda \big), \end{align}
\begin{align} c^{(2)}_{\alpha \beta \lambda \nu }(\boldsymbol{\hat \ell }) &=\hat {\ell }_\kappa \Big [\epsilon _{\kappa \alpha \lambda }\big(\hat \ell _\beta \hat \ell _\nu -\delta _{\beta \nu }\big)+\epsilon _{\kappa \alpha \nu }\big(\hat \ell _\beta \hat \ell _\lambda -\delta _{\beta \lambda }\big) +\epsilon _{\kappa \beta \lambda }\big(\hat \ell _\alpha \hat \ell _\nu -\delta _{\alpha \nu }\big) \nonumber\\ &\quad +\epsilon _{\kappa \beta \nu }\big(\hat \ell _\alpha \hat \ell _\lambda -\delta _{\alpha \lambda }\big)\Big ]. \end{align}
Furthermore,
$\eta _{\textit{o}}^{(1)}$
and
$\eta _{\textit{o}}^{(2)}$
are odd-viscosity coefficients. Inserting (6.15) into the linear momentum balance (2.1) for
$\boldsymbol{f=0}$
gives
\begin{align} \eta _s{\nabla} ^2\boldsymbol{v}(\boldsymbol{r}) & -\boldsymbol{\nabla }\hat p(\boldsymbol{r}) -\left (\eta _{\textit{o}}^{(1)}+2\eta _{\textit{o}}^{(2)}\right ){\nabla} ^2\big[\boldsymbol{v}(\boldsymbol{r})\times \boldsymbol{\hat \ell }\big] \nonumber\\ & +2\left (\eta _{\textit{o}}^{(1)}+\eta _{\textit{o}}^{(2)}\right )\big(\boldsymbol{\hat \ell }\boldsymbol{\cdot }\boldsymbol{\nabla }\big)^2\big[\boldsymbol{v}(\boldsymbol{r})\times \boldsymbol{\hat \ell }\big] = \boldsymbol0, \end{align}
where the effective pressure takes the form
$\hat p(\boldsymbol{r})=p(\boldsymbol{r})-\eta _{\textit{o}}^{(1)}\boldsymbol{\hat \ell }\boldsymbol{\cdot }[\boldsymbol{\nabla }\times \boldsymbol{v}(\boldsymbol{r})]$
. This extended model reduces to the one odd-viscosity model from (3.1), when
$\eta _{\textit{o}}^{(2)}=-\eta _{\textit{o}}^{(1)}=\eta _{\textit{o}}$
. The additional term
$(\boldsymbol{\hat \ell }\boldsymbol{\cdot }\boldsymbol{\nabla })^2[\boldsymbol{v}(\boldsymbol{r})\times \boldsymbol{\hat \ell }]$
can manifestly not be absorbed into a modified pressure. We conclude that for
$\eta _{\textit{o}}^{(2)}\neq -\eta _{\textit{o}}^{(1)}$
, the solutions to the ordinary Stokes equations listed in (6.3) are not solutions to (6.17). In particular, this observation implies that the rotlet is not a solution. It follows from our generalised Helmholtz theorem that generally there are contributions of
$\eta _{\textit{o}}^{(1)}$
and
$\eta _{\textit{o}}^{(2)}$
to the viscous dissipation rate of a rotating sphere, and the value of the dissipated power is higher than for a Newtonian fluid.
7. Conclusions
In summary, we have shown that the creeping flow equations for a general viscosity tensor admit a unique solution for the fluid flow. Furthermore, we have proven a general Helmholtz theorem showing that odd fluids generally dissipate more energy than equivalent fluids without odd viscosity, unless both systems share the same fluid velocity field. An example of higher viscous dissipation due to odd effects is a translating sphere in an odd fluid with viscosity tensor (3.1), whereas an example of equal dissipation is the rotating sphere. For both systems, we have derived exact singularity representations of the velocity and pressure fields, for which explicit closed-form expressions were obtained. As in previous works (Khain et al. Reference Khain, Scheibner, Fruchart and Vitelli2022; Everts & Cichocki Reference Everts and Cichocki2024a
), we retrieve the axial flow fields for a translating sphere (not just restricted to small
$\eta _{\textit{o}}$
), and have discussed how the flow field is altered when the direction of spin angular momentum is not aligned with the translation direction of the particle. Compared to a Newtonian fluid, the rotating sphere in an odd fluid has only a modified pressure, but the same type of rotlet flow field. Finally, we derived an exact expression for the stress tensor of the fundamental solution, which can be used to numerically compute mobility (or equivalently, friction) tensors of arbitrarily shaped particles subjected to general boundary conditions. Our results are important for the development of odd microhydrodynamics, microswimmers suspended in odd fluids, and odd Brownian motion. In future work, we will focus on the flow fields produced by a particle in an ambient linear straining flow, for which the dipolar–dipolar sector takes an important role.
Acknowledgements
We thank A. Vilfan and S. Čopar for insightful discussions.
Funding
We acknowledge funding from the National Science Centre, Poland, within the OPUS LAP grant no. 2024/55/I/ST3/00998.
Declaration of interests
The authors report no conflict of interest.
Appendix A
In this appendix, we list the tensor components for
$\boldsymbol{\varSigma }(\boldsymbol{r})$
defined in (3.8), which can be determined from (3.4). The strain rate tensor is (using the short-hand notation
$\varLambda =\varLambda (s)$
)
with
$\boldsymbol{b}=(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }})/{|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|}\,\hat{\boldsymbol{\theta }}$
. Furthermore, we have
\begin{align} 4\pi \eta _{\textit{s}}[\boldsymbol{\nabla }\times \unicode{x1D642}(\boldsymbol{r})]={}&\frac {\varLambda }{r^2}\Bigg (\big(\hat{\boldsymbol{\theta }}\hat{\boldsymbol{\phi }}-\hat{\boldsymbol{\phi }}\hat{\boldsymbol{\theta }}\big)-\frac {\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}}{|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|}s^2\varLambda ^2\big(\hat{\boldsymbol{r}}\hat{\boldsymbol{\phi }}-\hat{\boldsymbol{\phi }}\hat{\boldsymbol{r}}\big)\nonumber \\ &{}-\frac {s\varLambda }{1+\varLambda }\left \{\hat{\boldsymbol{\theta }}\hat{\boldsymbol{r}}+\hat{\boldsymbol{r}}\hat{\boldsymbol{\theta }}+\frac {\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}}{|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|}\left [\hat{\boldsymbol{\theta }}\hat{\boldsymbol{\theta }}-\hat{\boldsymbol{\phi }}\hat{\boldsymbol{\phi }}+\varLambda (\varLambda +1)\big(\unicode{x1D644}-\hat{\boldsymbol{\theta }}\hat{\boldsymbol{\theta }}\big)\right ]\right \}\Bigg ), \end{align}
which leads to the explicit form
\begin{equation} \boldsymbol{P}(\boldsymbol{r})=\frac {1}{4\pi r^2}\left [2\varLambda ^3\big(1+\gamma ^2\big)\big(\hat{\boldsymbol{r}}+s\hat{\boldsymbol{\phi }}\big)-\hat{\boldsymbol{r}}\right ]. \end{equation}
Appendix B
In this appendix, we give details on evaluating (5.4). We write
\begin{equation} \mathcal{L}_0 \unicode{x1D642}(\boldsymbol{r})=\textrm{Tr}[ \unicode{x1D63D}(\boldsymbol{r})]\, \unicode{x1D644}- \unicode{x1D63D}(\boldsymbol{r})+\gamma \boldsymbol{\epsilon }\boldsymbol{\cdot }\unicode{x1D63D}(\boldsymbol{r})\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}, \end{equation}
where the tensor
$ \unicode{x1D63D}(\boldsymbol{r})$
is given by
\begin{equation} \unicode{x1D63D}(\boldsymbol{r})=\frac {1}{\eta _{\textit{s}}}\int \frac {\textrm{d}^3\boldsymbol{k}}{(2\pi )^3}\, {\rm e}^{{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r}}\,j_0(ka)\,\frac {\hat {\boldsymbol{k}}\hat {\boldsymbol{k}}}{k^2\big[1+\gamma ^2\big(\hat {\boldsymbol{k}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)^2\big]}. \end{equation}
We define a cylindrical coordinate system by
$k_x=k_\perp \cos k_\phi$
,
$k_y=k_\perp \sin k_\phi $
,
$k_z=\boldsymbol{k}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}$
and
$x=\rho \cos \phi$
,
$y=\rho \sin \phi$
,
$z=z$
. Using these coordinates, (B2) reduces to
The integrals over
$k_z$
and
$k_\phi$
can be evaluated explicitly by contour integration and using results from Gradshteyn & Ryzhik (Reference Gradshteyn and Ryzhik2014), respectively. See Everts & Cichocki (Reference Everts and Cichocki2024a
) for further details. The result is
\begin{align} \unicode{x1D63D}(\boldsymbol{r}) &=\int _0^\infty \frac {\textrm{d}k_\perp }{{4\pi \eta _{\textit{s}}\gamma ^2}}\Bigg \{L_{0}(k_\perp ,z)\,\frac {J_1(k_\perp \rho )}{k_\perp \rho }\hat{\boldsymbol{\phi }}\hat{\boldsymbol{\phi }}+L_0(k_\perp ,z)\left [J_0(k_\perp \rho )-\frac {J_1(k_\perp \rho )}{k_\perp \rho }\right ]\hat{\boldsymbol{\rho }}\hat{\boldsymbol{\rho }}\nonumber \\&\quad {}-L_1(k_\perp ,z)\,J_1(k_\perp \rho )\,\big(\hat{\boldsymbol{\rho }}\hat{\boldsymbol{\ell }}+\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\rho }}\big) -L_2(k_\perp ,z)\,J_0(k_\perp \rho )\,\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\ell }}\Bigg \}, \end{align}
with
\begin{align} L_{m}(k_\perp ,z)=[\textrm{sgn}(z)]^m\left [{\rm e}^{-k_\perp \cos \psi\, |z|}\,j_0(k_\perp a\sin \psi )(\cos \psi )^{m-1}- {\rm e}^{-k_\perp |z|}\right ], \!\quad\! m=0,1,2.\\[-12pt]\nonumber \end{align}
Each term in (B4) can be explicitly computed using the following integrals (Gradshteyn & Ryzhik Reference Gradshteyn and Ryzhik2014):
\begin{align} & \int _0^\infty \textrm{d}k_\perp \, {\rm e}^{-k_\perp |z|}\,\frac {J_1(k_\perp \rho )}{ k_\perp \rho }=\dfrac {\sqrt {\rho ^2+z^2}-|z|}{\rho ^2},\end{align}
\begin{align} & \int _0^\infty \textrm{d}k_\perp \, {\rm e}^{-k_\perp |z|}\,J_m(k_\perp \rho )=\frac {\left (\sqrt {\rho ^2+z^2}-|z|\right )^m}{\rho ^m\sqrt {\rho ^2+z^2}}, \quad m=0,1,\ldots ,\end{align}
\begin{align} & \int _0^\infty \textrm{d}k_\perp \, {\rm e}^{-k_\perp |z| \cos \psi }\,J_0(k_\perp \rho )\,j_0(k_\perp a\sin \psi )=\frac {1}{a\sin \psi }\textrm{arcsin}\left [\frac {1}{\mathcal{R}_+(\boldsymbol{r};\gamma )}\right ],\end{align}
\begin{align} & \int _0^\infty \textrm{d}k_\perp \, {\rm e}^{-k_\perp |z|\cos \psi }\,J_1(k_\perp \rho )\,j_0(k_\perp a\sin \psi )= \frac {1}{\rho }\left [1-\sqrt {1-\mathcal{R}_-(\boldsymbol{r};\gamma )^2}\right ],\end{align}
\begin{align} & \int _0^\infty \textrm{d}k_\perp \, {\rm e}^{-k_\perp |z|\cos \psi }\,\frac {J_1(k_\perp \rho )}{ k_\perp \rho }\,j_0(k_\perp a\sin \psi )={}\frac {1}{2a\sin \psi }\textrm{arcsin}\left [\frac {1}{\mathcal{R}_+(\boldsymbol{r};\gamma )}\right ] \nonumber\\ &\qquad +\, \frac {a\sin \psi }{2\rho ^2}\sqrt {\mathcal{R}_+(\boldsymbol{r};\gamma )^2-1}\left [1-\sqrt {1-\mathcal{R}_-(\boldsymbol{r};\gamma )^2}\right ]^2, \end{align}
where the quantities
$\mathcal{R}_\pm$
are defined in (5.12) with the identification
$\rho =r|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}|$
and
$z=r(\hat{\boldsymbol{r}}\boldsymbol{\cdot }\hat{\boldsymbol{\ell }})$
. Furthermore, we used that
$\rho \gt 0$
and
$\cos [\psi (\gamma )]\gt 0$
. Insertion of (B6)–(B10) into (B4) and (B5) gives
$ \unicode{x1D63D}(\boldsymbol{r})$
. From
$ \unicode{x1D63D}(\boldsymbol{r})$
and (B1), it follows that
with
$\mathcal{M}(\boldsymbol{r};\gamma )$
,
$\mathcal{N}(\boldsymbol{r};\gamma )$
and
$\mathcal{O}(\boldsymbol{r};\gamma )$
defined in (5.11).
It is instructive to evaluate (B11) on the surface of a sphere with radius
$a$
. It is straightforward to check that
\begin{equation} \mathcal{R}_+(a\hat{\boldsymbol{r}};\gamma )=\frac {1}{\sin \psi },\quad \mathcal{R}_-(a\hat{\boldsymbol{r}};\gamma )=\big|\hat{\boldsymbol{r}}\times \hat{\boldsymbol{\ell }}\big|, \end{equation}
and therefore
$\mathcal{M}(a\hat{\boldsymbol{r}};\gamma )=\textrm{arctan}(\gamma )$
and
$\mathcal{N}(a\hat{\boldsymbol{r}};\gamma )=\mathcal{O}(a\hat{\boldsymbol{r}};\gamma )=0$
, where we used that
$\gamma \gt 0$
. Direct substitution of these results in (B11) gives
\begin{equation} \mathcal{L}_0 \unicode{x1D642}(a\boldsymbol{\hat{\boldsymbol{r}}})=\frac {1}{24\pi \eta _{\textit{s}}a}\left [R(\gamma )\,\big(\unicode{x1D644}-\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\ell }}\big)+T(\gamma )\,\hat{\boldsymbol{\ell }}\hat{\boldsymbol{\ell }}-S(\gamma )\,\big(\boldsymbol{\epsilon }\boldsymbol{\cdot }\hat{\boldsymbol{\ell }}\big)\right ], \end{equation}
which equals the translational–translational mobility tensor
$\boldsymbol{\mu }^{{tt}}=[\boldsymbol{\zeta }^{{tt}}]^{-1}$
, as it should be.
v(r)
U
U=|U|
γ=ηo/ηs=0
γ=3
U
ℓ^
U∥ℓ^
∢(U,ℓ^)=45∘
U⊥ℓ^.
U
v(r)
U
ℓ^
+z
U
ℓ^
v(r)
xz
v(r)
y
U=|U|
v(r)
xy
|v(r)|
γ=1
γ=3
Open access
