Microswimmer

A microswimmer is a microscopic body that propels itself through a fluid. At small Reynolds number its motion is governed by Stokes flow, and self-propulsion must respect the force-free and torque-free conditions.

Two coaxial opposite-handed helices linked by an internal motor share a translation speed but counterrotate. In an additive local-drag model, a length ratio $n$ multiplies the second helix's diagonal resistances by $n$ and reverses its coupling. With ${\Omega }_1-{\Omega }_2=\omega$, force and torque balance give $U=-2nBD\omega /[AD(1+n{)}^2-B^2(1-n{)}^2]$.

As the length ratio of an opposite-handed reaction helix tends to infinity, its translation and rotation tend to zero as $1/n$, while the finite first helix rotates at almost the full motor rate. Small motions of the long rotor balance finite force and torque. The common propulsion vanishes in this limit of the additive model.

If the second rotor has zero drag, it cannot supply reaction torque. The single remaining helix must be both force-free and torque-free; a positive-definite resistance matrix forces its translation and rotation to vanish. The relative motor motion is absorbed by the zero-resistance rotor.

Equal local resistance magnitudes and opposite coupling signs give ${\Omega }_1=\omega /2$, ${\Omega }_2=-\omega /2$ and $U=-B\omega /(2A)$. Translation-induced torques cancel, while propulsive forces add because handedness and rotation both reverse.

For a flagellum with axial resistance matrix of a slender helix and a head with resistances $A_0=6\pi \mu a$, $D_0=8\pi \mu a^3$, neglect interactions between the parts. A motor with relative angular velocity $\omega$ givesThe head rotates at $\Omega -\omega$. A large head creates excessive translational resistance; a small head supplies too little rotational resistance, allowing the head to counterrotate while the flagellum barely moves relative to the fluid.

Suppose a small spherical head co-translates with the large-scale flagellar flow and rotates at $\Omega -\omega$, while the local ambient rotation is approximated by $\Omega$. Its force is neglected and its torque on the fluid is $-8\pi \mu a^3\omega$. Balance this against the axial resistance matrix of a slender helix, with local coefficient $C=4\pi \mu /|\log ϵ|$, to obtain the displayed swimming speed and motor power $8\pi \mu a^3{\omega }^2$. This is an entrainment approximation, distinct from treating the head as an isolated sphere in stationary fluid.

When head translation is weakly resisted compared with the flagellum but head rotation is strongly resisted, $A_0\ll A$ and $D_0\gg D$. Then $U\simeq -B\omega /A=\omega b\sin \varphi \cos \varphi /(1+{\cos }^2\varphi )$. Maximizing this expression gives the displayed pitch and speed for the chosen handedness and positive $\omega$.

A squirmer is an idealized spherical microswimmer propelled by a prescribed tangential surface slip velocity. The Lorentz reciprocal theorem for Stokes flow determines its rigid translation and rotation without solving its complete exterior flow.

For a sphere of radius $a$ with prescribed surface slip $u^{\prime }$ and zero external torque, the Lorentz reciprocal theorem for Stokes flow giveswhere $n$ points from the sphere into the fluid. Use an auxiliary rotating sphere in Stokes flow with arbitrary angular velocity $\hat{\Omega }$ and traction $-3\mu \hat{\Omega }\times n$. The actual torque-free cross-work vanishes. Substituting the boundary velocity $\Omega \times an+u^{\prime }$ into the other cross-work leaves $-8\pi \mu a^3\hat{\Omega }\cdot \Omega -3\mu \hat{\Omega }\cdot \int n\times u^{\prime }=0$. A possible translation contributes zero because the auxiliary net force is zero.

A spherical squirmer can rotate without disturbing the surrounding fluid if its prescribed surface slip exactly cancels the velocity of rigid rotation: $u^{\prime }=-\Omega \times an$. The laboratory-frame boundary velocity is then zero. Uniqueness of Stokes flow with decay at infinity gives identically zero exterior velocity. In particular $u^{\prime }=b_0\sin \theta e_{\varphi }$ gives $\Omega =-(b_0/a)e_z$. The cancellation concerns the total fluid boundary velocity, not the solid material's angular velocity.

A spherical squirmer with tangential surface slip velocity $(I-nn)(A+Bn)$, where $B$ is a symmetric second-rank tensor and a traceless second-rank tensor, translates with $U=-2A/3$ and does not rotate in an unbounded quiescent fluid. With $S=x\cdot Bx$ its exterior Stokes flow isThis follows from the Unscaled Papkovich–Neuber representation using harmonic functions as potentials $\Phi =-a^{2}Bx/(2r^3)$ and $\chi =a^{3}A\cdot x/(3r^3)+a^{4}S/(2r^5)$. Matching the surface coefficients proves the formula. The two modes generate a potential dipole and a stresslet, respectively.

The surface slip velocity is the fluid velocity relative to the rigid motion of an active particle at its boundary. For a spherical force-free, torque-free swimmer of radius $a$ with slip ${\mathbf{u}}_s$,

The Taylor swimming sheet is an infinite two-dimensional microswimmer whose prescribed traveling deformation drives Stokes flow. A transverse wave $y_s=ϵ\sin (x-t)$ of small amplitude swims opposite to its direction of propagation, with speed $U={ϵ}^2/2+O({ϵ}^4)$ when lengths are scaled by inverse wavenumber and velocities by wave speed. Its biharmonic stream function for planar Stokes flow has first-order part ${\psi }_1=(1+y)e^{-y}\sin (x-t)$.

A small transverse Taylor swimming sheet in a Brinkman medium obeys $({\nabla }^2-A^2){\nabla }^2\psi =0$ with $A=\alpha /k$. Its no-slip material velocity is imposed on the moving sheet. The matrix's preferred frame and the mean force-free condition must be included when finding the second-order swimming speed.

For $s=\sqrt{1+(\alpha /k{)}^2}$, the leading work per projected area on one side of a transverse Brinkman swimming sheet is $\mu b^{2}k{\omega }^{2}s(s+1)/2$. Two fluid sides double it. Work includes viscous dissipation and drag against the matrix, giving a greater fixed-stroke cost than in the unscreened fluid.

The leading fixed-stroke speed of a transverse Brinkman swimming sheet is enhanced over the pure-fluid value by $\sqrt{1+(\alpha /k{)}^2}$. It follows from the mean displaced-boundary velocity and zero mean traction. The expansion is at fixed screening parameter and does not assert an enhancement at fixed available motor power.

With $s=\sqrt{1+A^2}$, a dimensionless decaying first-order transverse sheet has $f=(s{e}^{-y}-e^{-sy})/(s-1)$, $f(0)=1$, $f^{\prime }(0)=0$, and pressure $-s(s+1)e^{-y}\cos (x-t)$. The coincident-root limit is $(1+y)e^{-y}$, the ordinary transverse mode of a Taylor swimming sheet.

A transversely waving Taylor swimming sheet can have a Navier slip boundary condition relating surface tangential velocity to shear and a slip length $\gamma$. The small-amplitude parameters are $ϵ=k{y}_0$ and $\delta =k\gamma$. The first-order velocity is unchanged by slip, but evaluation of shear at the displaced surface changes the second-order mean velocity.

At second order, the displaced-boundary Navier slip boundary condition gives $({\psi }_2{)}_Y-\delta [({\psi }_2{)}_{YY}-({\psi }_2{)}_{XX}]=(1+2\delta ){\sin }^2(X-\tau )$ at $Y=0$. Its mean shear is zero in a bounded half-space flow, so the Navier-slip Taylor swimming sheet dimensionless speed is $U^{\ast }={ϵ}^2(1+2\delta )/2+O({ϵ}^4)$. The sheet travels opposite to the wave. This fixed-slip asymptotic expansion predicts a speed enhancement factor $1+2\delta$.

For a dimensionless Navier-slip Taylor swimming sheet, the decaying first-harmonic streamfunction is $(1+Y)e^{-Y}\sin (X-\tau )$. Its tangential surface velocity and surface shear both vanish at $Y=0$. Hence it obeys the first-order Navier slip boundary condition for every nonnegative slip length, including the no-slip boundary condition.

For tangential displacement $ϵa\sin (k(x-t))$, the first-order decaying streamfunction is $-kay{e}^{-ky}\cos (k(x-t))$. It obeys tangential material velocity $-ka\cos (k(x-t))$ and zero normal velocity on the reference boundary. Its second-order mean swimming coefficient is $-k^2{a}^2/2$.

For normal displacement $ϵb\sin (k(x-t))$, the first-order decaying streamfunction is $b(1+ky)e^{-ky}\sin (k(x-t))$. It obeys zero tangential material velocity and normal velocity $-kb\cos (k(x-t))$ on the reference boundary. Its second-order mean swimming coefficient in the convention of far-field velocity $U{\mathbf{e}}_x$ is $k^2{b}^2/2$.

A Taylor expansion of the no-slip boundary condition about the flat sheet makes the second-order mean tangential velocity $⟨u_2(0)⟩=-⟨y_1{\partial }_y{u}_1(0)⟩$. The mean mode of Stokes flow is linear in height when no pressure gradient is imposed. In an unbounded fluid, bounded velocity excludes mean shear; in a confined fluid, the force-free condition excludes it. The remaining mean velocity is uniform and equals the swimming speed in the sheet frame. Thus the leading speed can be found from the first-order field without solving the oscillatory second-order field.

The second-order mean no-slip boundary condition correction contains products of first-order deformation modes and velocity gradients. Distinct integer wavenumbers have zero averaged mixed products. Their separate mean longitudinal and transverse contributions therefore add, even though mixed products create nonzero oscillatory flow modes.

When longitudinal and transverse sheet waves share unit phase speed but have different wavenumbers $k_{\parallel },k_{\perp }$, their second-order swimming coefficient is $(k_{\perp }^2{b}^2-k_{\parallel }^2{a}^2)/2$. Thus their leading propulsion cancels at $b/a=k_{\parallel }/k_{\perp }$. This statement concerns the small-amplitude leading order.

For a transverse Taylor swimming sheet below a flat no-slip boundary condition at height $d$, the first-order amplitude $f(y)$ satisfies $({\partial }_y^2-1{)}^{2}f=0$, $f(0)=1$, $f^{\prime }(0)=f(d)=f^{\prime }(d)=0$. Writing $\Delta ={\sinh }^{2}d-d^2$ givesThe mean boundary velocity determines Taylor-sheet swimming speed, yieldingThis is greater than the unbounded value $1/2$ for every $d>0$. In a narrow gap it scales as $3/d^2$, with the small-amplitude calculation requiring $ϵ\ll d$ as well as $ϵ\ll 1$. Taylor's swimming sheet near a soft boundary recovers this rigid-wall limit while studying how compliance changes propulsion.

The leading far field of many force-free microswimmers is an axisymmetric force dipole, or stresslet,

Two equal parallel axial point stresslets with positive strength $S$ and separation $\ell$ have relative velocity $S/(2\pi \mu {\ell }^2)$. Identical self-propulsion cancels in their relative motion. The solution is the displayed cubic separation law while the far-field approximation is valid.

The angular factor of an axial stresslet is linear in the symmetric traceless rank-two tensor $\mathbf{ee}-I/3$. Hence averaging a collection of directors amounts to averaging this tensor. For uniformly distributed in-plane directions, $⟨\mathbf{ee}⟩=(I-\mathbf{nn})/2$, so $⟨S(\mathbf{ee}-I/3)⟩=-(S/2)(\mathbf{nn}-I/3)$. The averaged director is normal to the plane and its signed strength is reversed and halved.

A stresslet whose centre follows a circle and whose director stays tangent has a leading far-field mean stresslet centred at the circle centre. Orientation averaging of an axisymmetric stresslet gives normal-axis strength $-S/2$. Thus a pusher microswimmer becomes a leading mean puller microswimmer, and conversely. The radius and traversal frequency enter higher-order spatial corrections or averaging time, but not the leading dipole strength.

Opposite points on the orbit have opposite displacement and the same stresslet director up to sign. Their paired fields are $[G(\mathbf{r}-R\mathbf{p},\mathbf{e})+G(\mathbf{r}+R\mathbf{p},\mathbf{e})]/2$, so odd displacement powers cancel. Since a stresslet decays as $r^{-2}$, the first correction to its far-field orbit average of a tangent stresslet is $O(S{R}^2/(\mu r^4))$. The finite-radius mean need not be an exact point stresslet.

Opposite forces $\pm F\mathbf{e}$ applied to the fluid at $\pm (a/2)\mathbf{e}$ have no net force or torque. Their leading Stokeslet difference is $-Fa(\mathbf{e}\cdot \nabla )[J(\mathbf{r})\mathbf{e}]$, equal to the axial stresslet with strength $S=Fa$. Positive $S$ drives outward flow on the axis and inward flow in the equatorial plane, corresponding to a pusher microswimmer; negative $S$ gives a puller microswimmer.

A pusher microswimmer generates an extensile force-dipole flow, expelling fluid along its swimming axis and drawing fluid inward from its sides. In the displayed convention it has $\mathcal{P}>0$.

A puller microswimmer generates a contractile force-dipole flow, drawing fluid inward along its swimming axis and expelling it sideways. In the displayed convention it has $\mathcal{P}<0$.

Reflecting a force dipole's position and normal orientation component across a flat clean free surface produces an image whose superposition enforces no penetration and zero tangential traction. Its flow at the swimmer causes both normal drift and reorientation.

For equal stresslets with height $h$ and axial separation $\ell \gg h$, the neighbour image doubles the leading axial repulsion. Its normal contribution is smaller than own-image attraction by order $(h/\ell {)}^3$, so leading height dynamics are unchanged. The approximation also neglects weak neighbour-induced orientation corrections.

A positive-strength parallel point stresslet below a flat shear-free impermeable plane is attracted by its identical reflected image. Its height satisfies $\dot{h}=-S/(32\pi \mu h^2)$ and formally reaches zero in finite time. Negative strength reverses the drift. Finite-body near-contact physics lies outside this singularity prediction.