Microbial locomotion 18.S995 - L24-26 dunkel@mit.edu Why microbial - PowerPoint PPT Presentation

Microbial locomotion 18.S995 - L24-26 dunkel@mit.edu

Why microbial 5 ㎛ 10 ㎛ hydrodynamics ? • micro-machines • hydrodynamic propulsion • > 50% global biomass • gut flora, biofilms, ... • global food web • > 50% global carbon fixation 30 ㎛ 100 ㎛ � Whitman et al (1998) PNAS Guasto et al (2012) Annu Rev Fluid Mech

Reynolds numbers Re = ρ UL = UL µ ν dunkel@math.mit.edu

Turbulence

Swimming at low Reynolds number R � UL ρ / η ⇥ 1 Geoffrey Ingram Taylor James Lighthill 0 = µ ⌥ 2 u � ⌥ p + f , 0 = ⌥ · u . + time-dependent BCs Edward Purcell Shapere & Wilczek (1987) PRL

Zero-Re flow

E.coli (non-tumbling HCB 437) Drescher, Dunkel, Ganguly, Cisneros, Goldstein (2011) PNAS dunkel@math.mit.edu

Bacterial motors movie: V. Kantsler ~20 parts 20 nm source: wiki Berg (1999) Physics Today Chen et al (2011) EMBO Journal dunkel@math.mit.edu

Chlamydomonas alga 10 ㎛ 10 ㎛ ~ 50 beats / sec speed ~100 μ m/s Goldstein et al (2011) PRL dunkel@math.mit.edu

Volvox carteri 200 ㎛ 10 ㎛ Chlamydomonas reinhardtii dunkel@math.mit.edu

Stroke Sareh et al (2013) J Roy Soc Interface dunkel@math.mit.edu

Volvox carteri beating frequency 25Hz Sareh et al (2013) J Roy Soc Interface dunkel@math.mit.edu

Meta-chronal waves Brumley et al (2012) PRL dunkel@math.mit.edu

Dogic lab (Brandeis) dunkel@math.mit.edu

Volvox carteri somatic cell cilia 200 ㎛ daughter colony Drescher et al (2010) PRL dunkel@math.mit.edu

• How can Volvox perform phototaxis? (discussed later) dunkel@math.mit.edu

Swimming at low Reynolds number R � UL ρ / η ⇥ 1 Geoffrey Ingram Taylor James Lighthill 0 = µ ⌥ 2 u � ⌥ p + f , 0 = ⌥ · u . + time-dependent BCs Edward Purcell Shapere & Wilczek (1987) PRL

Superposition of singularities 2x stokeslet = stokeslet symmetric dipole rotlet -F F F p ( r ) = ˆ r · F 4 π r 2 + p 0 v i ( r ) = (8 π µ ) − 1 [ δ ij + ˆ r i ˆ r j ] F j r r − 1 r − 2 r − 2 flow ~ ‘pusher’

Volvox Chlamy swimming speed ~ 100 ㎛ /sec swimming speed ~ 50 ㎛ /sec PIV 100 ㎛ ⌃ � � �� v ∼ 1 /r 2 ⇧ ⌅⇧� ⌃ ⇤ ⌥ ⌫ � ⌦ � / ⌃ � � / ⇤ / ⇤ ⌥ ⇥ / ⌃�⌥ 3D : ⌃ � ⇤ � � � ⇣⌃ �⌥� ⌥ ⇥ ⇥ / � � ⌥�⌃ ⇤ � � ⌥⇤ ⌅ v ∼ 1 /r � 2D : ⇣ � / ⇤ / ⇥ / ⇤ ⌃ � ⇥ � ⇤ ... no dipoles ! Drescher et al (2010) PRL Guasto et al (2010) PRL

E.coli (non-tumbling HCB 437) eed V 0 = 22 ± 5 µ m/s. A A = � F r = r h i r . ˆ d ) 2 − 1 | r | , th � = 1 . 9 µ m u ( r ) = 3( ˆ r , ˆ 8 πη , ˆ regions, we obtai | r | 2 rce F = 0 . 42 pN. weak ‘pusher’ dipole Drescher, Dunkel, Ganguly, Cisneros, Goldstein (2011) PNAS

Twitching motility Type-IV Pili

Twitching motility Pseudomonas

Amoeboid locomotion

5.1 Navier-Stokes equations Consider a fluid of conserved mass density % ( t, x ), governed by continuity equation @ t % + r · ( % u ) = 0 , (5.1) where u ( t, x ) is local flow velocity. According to standard hydrodynamic theory, the

5.1 Navier-Stokes equations Consider a fluid of conserved mass density % ( t, x ), governed by continuity equation @ t % + r · ( % u ) = 0 , (5.1) where u ( t, x ) is local flow velocity. According to standard hydrodynamic theory, the dynamics of u is described by the Navier-Stokes equations (NSEs) % [ @ t u + ( u · r ) u ] = f � r p + r · ˆ (5.2) T, where p ( t, x ) the pressure in the fluid, ˆ T ( t, x ) the deviatoric 2 ) stress-energy tensor of the fluid, and f ( t, x ) an external force-density field. A typical example of an external force f ,

5.1 Navier-Stokes equations Consider a fluid of conserved mass density % ( t, x ), governed by continuity equation @ t % + r · ( % u ) = 0 , (5.1) where u ( t, x ) is local flow velocity. According to standard hydrodynamic theory, the dynamics of u is described by the Navier-Stokes equations (NSEs) % [ @ t u + ( u · r ) u ] = f � r p + r · ˆ (5.2) T, where p ( t, x ) the pressure in the fluid, ˆ T ( t, x ) the deviatoric 2 ) stress-energy tensor of the fluid, and f ( t, x ) an external force-density field. A typical example of an external force f , that is also relevant in the biological context, is the gravitational force f = % g , (5.3) where g ( t, x ) is the gravitational acceleration field.

Considering a Cartesian coordinate frame, Eqs. (5.1) and (5.2) can also be rewritten in the component form @ t % + r i ( % u i ) = 0 , (5.4a) F i � @ i p + @ j ˆ % ( @ t u i + u j @ j u i ) = (5.4b) T ji . To close the system of equations (5.4), one still needs to (i) fix the equation of state p = p [ % , . . . ] , (ii) choose an ansatz the symmetric stress-energy tensor T = ( ˆ ˆ T ij [ % , u , . . . ]) , (iii) specify an appropriate set of initial and boundary conditions. 2 ‘deviatoric’:= without hydrostatic stress (pressure); a ‘full’ stress-energy tensor ˆ σ may be defined by σ ij := � p δ ij + ˆ ˆ T ij .

In the case of a homogeneous fluid with 3 Simplifications @ t % = 0 and r % = 0 , (5.5) the associated flow is incompressible (isochoric) r · u = 0 . (5.6)

In the case of a homogeneous fluid with 3 Simplifications @ t % = 0 and r % = 0 , (5.5) the associated flow is incompressible (isochoric) r · u = 0 . (5.6) A Newtonian fluid is a fluid that can, by definition, be described by ˆ T ij := � ( r · u ) � ij + µ ( @ i u j + @ j u i ) . (5.7) where � the first coe ffi cient of viscosity (related to bulk viscosity), and µ is the second coe ffi cient of viscosity (shear viscosity). Thus, for an incompressible Newtonian fluid, the Navier-Stokes system (5.4) simplifies to 0 = r · u , (5.8a) �r p + µ r 2 u + f . % [ @ t u + ( u · r ) u ] = (5.8b)

The SI physical unit of dynamic viscosity µ is the Pascal ⇥ second Dynamic viscosity [ µ ] = 1 Pa · s = 1 kg / (m · s) (5.9) If a fluid with a viscosity µ = 1 Pa · s is placed between two plates, and one plate is pushed sideways with a shear stress of one pascal, it moves a distance equal to the thickness of the layer between the plates in one second. The dynamic viscosity of water ( T = 20 � C) is µ = 1 . 0020 ⇥ 10 � 3 Pa · s.

The SI physical unit of dynamic viscosity µ is the Pascal ⇥ second Dynamic viscosity [ µ ] = 1 Pa · s = 1 kg / (m · s) (5.9) If a fluid with a viscosity µ = 1 Pa · s is placed between two plates, and one plate is pushed sideways with a shear stress of one pascal, it moves a distance equal to the thickness of the layer between the plates in one second. The dynamic viscosity of water ( T = 20 � C) is µ = 1 . 0020 ⇥ 10 � 3 Pa · s. Below we will be interested in comparing viscous and inertial Kinematic viscosity forces. Their ratio is characterized by the kinematic viscosity ⌫ , defined as ⌫ = µ [ ⌫ ] = m 2 / s % , (5.10) The kinematic viscosity of water with mass density % = 1 g / cm 3 is ⌫ = 10 � 6 m 2 / s = 1 mm 2 /s = 1 cSt.

5.2 Stokes equations 5.2.1 Motivation Consider an object of characteristic length L , moving at absolute velocity U = | U | through (relative to) an incompressible, homogeneous Newtonian fluid of constant viscosity µ and constant density % . The object can be imagined as a moving boundary (condition), which induces a flow field u ( t, x ) in the fluid. The ratio of the inertial (dynamic) pressure % U 2 and viscous shearing stress µU/L can be characterized by the Reynolds number 4 ' % U 2 /L R = | % ( @ t u + ( u · r ) u ) | µU/L 2 = UL % = UL ⌫ . (5.11) | µ r 2 u | µ For example, when considering swimming in water ( ⌫ = 10 − 6 m 2 / s), one finds for fish or humans: R ' 10 6 , L ' 1 m , U ' 1 m / s ) whereas for bacteria: R ' 10 − 5 . L ' 1 µ m , U ' 10 µ m / s )

5.2 Stokes equations 5.2.1 Motivation Consider an object of characteristic length L , moving at absolute velocity U = | U | through (relative to) an incompressible, homogeneous Newtonian fluid of constant viscosity µ and constant density % . The object can be imagined as a moving boundary (condition), which induces a flow field u ( t, x ) in the fluid. The ratio of the inertial (dynamic) pressure % U 2 and viscous shearing stress µU/L can be characterized by the Reynolds number 4 ' % U 2 /L R = | % ( @ t u + ( u · r ) u ) | µU/L 2 = UL % = UL ⌫ . (5.11) | µ r 2 u | µ For example, when considering swimming in water ( ⌫ = 10 − 6 m 2 / s), one finds for fish or humans: R ' 10 6 , L ' 1 m , U ' 1 m / s ) whereas for bacteria: R ' 10 − 5 . L ' 1 µ m , U ' 10 µ m / s ) If the Reynolds number is very small, R ⌧ 1, the nonlinear NSEs (5.8) can be approx- imated by the linear Stokes equations 5 0 = r · u , (5.12a) µ r 2 u � r p + f . 0 = (5.12b) initial and boundary conditions, such as for example ( u ( t, x ) = 0 , as | x | ! 1 . (5.13) p ( t, x ) = p ∞ ,