Microbial
locomotion
18.S995 - L13
dunkel@mit.edu
Why microbial hydrodynamics ?
5㎛
10 ㎛
• micro-machines
• hydrodynamic propulsion • > 50% global biomass • gut flora, biofilms, ...
• global food web
• > 50% global carbon
fixation
30 ㎛
Whitman et al (1998) PNAS
100 ㎛
!
Guasto et al (2012) Annu Rev Fluid Mech
lds Numbers in Biology
Reynolds numbers
number is dimensionless group that characterizes the ratio o
fined as
⇥U L
UL
Re =
=
µ
density of the medium the organism is moving through; µ is t
; is the kinematic viscosity; U is a characteristic velocity of
stic length scale. When we discuss swimming biological organ
eatures that are moving through water (or through a fluid with
hose of water). This means that the material properties µ and
ber is roughly determined by the size of the organism.
e characteristic size of the organism and the characteristic sw
rule-of-thumb, the characteristic locomotion velocity, U , in bi
y U L/second e.g. for people L 1 m and we move
at U 1
dunkel@math.mit.edu
Turbulence
ertial (dynamic) pressure ⇤U 2 and viscous shearing str
µU/L can be characterized by the Reynolds number4
Swimming at low Reynolds number
R ⌅ U L⇤/µ = U L/⇥.
Example:
Swimming in water with ⇥ = 10
6
m2 /s
• fish/human: L ⌅ 1 m, U ⌅ 1 m/s ⇧ R ⌅ 106 .
R
• bacteria: L ⌅ 1 µm, U ⌅ 10 µm/s ⇧ R ⌅ 10
U L⇥/ ⇥ 1
Geoffrey Ingram Taylor
5
James Lighthill
If the Reynolds number is very small, R ⇥ 1, t
NSE (8) can be approximated by the Stokes equation
0 = µ ⌥2 u
0 = ⌥ · u.
⌥p + f ,
(10
(10
These equations+must
still be endowed
time-dependent
BCs with appropri
initial and boundary conditions, such as ,e.g.,6
Edward Purcell
u(t, x) = 0,
as
Shapere
& Wilczek
PRL
p(t, x)
= p⇥(1987)
,
|x| ⇤⌃ .
(1
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
Berg (1999) Physics Today
source: wiki
Chen et al (2011) EMBO Journal
dunkel@math.mit.edu
Chlamydomonas alga
10 ㎛
~ 50 beats / sec
Goldstein et al (2011) PRL
10 ㎛
speed ~100 μm/s
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?
Lecture 17 or 18
dunkel@math.mit.edu
ertial (dynamic) pressure ⇤U 2 and viscous shearing str
µU/L can be characterized by the Reynolds number4
Swimming at low Reynolds number
R ⌅ U L⇤/µ = U L/⇥.
Example:
Swimming in water with ⇥ = 10
6
m2 /s
• fish/human: L ⌅ 1 m, U ⌅ 1 m/s ⇧ R ⌅ 106 .
R
• bacteria: L ⌅ 1 µm, U ⌅ 10 µm/s ⇧ R ⌅ 10
U L⇥/ ⇥ 1
Geoffrey Ingram Taylor
5
James Lighthill
If the Reynolds number is very small, R ⇥ 1, t
NSE (8) can be approximated by the Stokes equation
0 = µ ⌥2 u
0 = ⌥ · u.
⌥p + f ,
(10
(10
These equations+must
still be endowed
time-dependent
BCs with appropri
initial and boundary conditions, such as ,e.g.,6
Edward Purcell
u(t, x) = 0,
as
Shapere
& Wilczek
PRL
p(t, x)
= p⇥(1987)
,
|x| ⇤⌃ .
(1
Superposition of singularities
2x stokeslet =
symmetric dipole
stokeslet
rotlet
-F
F
r̂ · F
p(r) =
+ p0
2
4⇥r
(8⇥µ) 1
vi (r) =
[ ij + r̂i r̂j ]Fj
r
flow ~
r
1
F
r
2
‘pusher’
r
2
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swimming speed ~ 100 ㎛/sec
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<
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⇧
64
✓
74
<
8⌦
Volvox Chlamy
swimming
PHYSICAL
PRL 105,
168101 speed
(2010) ~ 50 ㎛/sec
PIV
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al (2010)
PRL
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2
FIG. 4 (color
online).
Timeand
azimuthally-averaged
3D : v 1/r
from velocity vectors (blue [dark gray]). The spiraling nea
velocity field.
A color
scheme
indicates
flow speed magnit
2D
:
v
1/r
model: flagellar thrust is distributed among two Stokes
arrows), whose sum balances drag on the cell body (cen
separate colors in the inset, compared to results from the
... no dipoles !
flow may be important [30]. We are currently investi
Guasto
et similar
al (2010)
PRL
whether
conclusions
hold for the flow field a
Results
w
of itswe
“puller”
image.
d. field
Atwalls,
distances
rfocused
<6µ
mon
theadipole
model
overestimates
the
bacterial
flow
field.
(E)
Experimentally
measured flow to
plane 50 µm from the top and bottom
tancesurfaces
2 µm parallel
to sample
the wall. chamber,
(F) Best fitand
force-dipole
model,
(G) residual
field.
Notewhere
the
Bacterial
cell
body,
of the
recorded
∼ 2 and
terabytes
of flow
4
flowmovie
fieldResults
ofdata.
an E. In
coli
“pusher”
decays
much faster,
when
swims close
thecule
surface,
fortothe
length
of t
theflow
mea
this
data
we
identified
∼
10
events when
(non-tumbling HCB 437)rarea bacterium
achieved
by
fittin
theByTo
measured
andminisbest-fit
force
cellsBacterial
swam in the
for > surfaces.
1.5 s.
tracking
decays
labeled,
n
flowfocal
fieldplane
far from
resolve the
the
at
variable
locatio
fluidcule
tracers
in
each
of
the
rare
events,
relating
their
position
of
the
c
decays
of
the
flow
speed
u
with
flow
field
created
by
individual bacteria, we tracked gfp- sion of
flu
m surfaces.
To
resolve
the
minisfield (r >field
8 µm).
and labeled,
velocity to
the
position
and
orientation
of
the
bacterium,
dis
non-tumbling
E.
coli
as
they
swam
through
a
suspenof
the
cell
body
(Fig.
1D)
illustr
walls,
we
dividual
bacteria,
tracked
gfp- overforce
the measured
andaverage
best-fit
dipole we
field
The
the 1C).
specific
fitting
and performing
anwe
ensemble
all tracers,
re- (Fig.
Howeve
sionswam
ofdecays
fluorescent
tracer
particles.
For
measurements
farcharacteristic
from
field
displays
the
1
dipole
length
ℓ =o
of
the
flow
speed
u
with
distance
r
from
thesurfaces
center
i
as
they
through
a
suspensolved
the
time-averaged
flow
field
in
the
E.
coli
swimming
he minismeasure
walls,
we
focused
on
a
plane
50
µm
from
the
top
and
bottom
value
of
F
is cons
movie
dat
However,
the
force
dipole
flow
sig
oftothe
cell
(Fig.
1D) illustrate
that
the
flow
down
0.1%
of body
the mean
swimming
speed V0 =
22 ±
5 measured
ticles.
For
measurements
far
from
ckedplane
gfpcell force
bod
surfaces
of
the
sample
chamber,
and
recorded
∼
2
terabytes
of
2
and
resistive
µm/s.
As E.
colidisplays
rotate
about
their
swimming
direction,
their
cells
field
the
characteristic
1/r4 decay
oftoa the
forceside
dipole.
measured
flow
ofswam
thel
a 50
suspenµmmovie
from
the
top
and
bottom
for
the
data.
In field
this in
data
we dimensions
identified ∼is10cylindrically
rare events when
note that
in
the b
time-averaged
flow
three
fluid
trace
However,
the
force
dipole
flow
significantly
overestimates
the
cell1.5body,
where
the
flow
magnit
far from
achieved
ber,
and
recorded
∼
2
terabytes
of
cells
swam
in
the
focal
plane
for
>
s.
By
tracking
the
µm behind the ce
symmetric.measured
Our
measurements
capture
all
components
of
this
flow to when
the side of
the
cell
body,ofand
behind
the
4
and
veloci
didentified
bottom
for
the
length
the
flagellar
bun
at
varia
fluid
drag
on
the
∼
10
rare
events
fluid
tracers
in
each
of
the
rare
events,
relating
their
position
cylindrically symmetric flow, except the azimuthal flow due to
cell
body,
where
the
flow
magnitude
u(r)
is
nearly
constant
and
perfo
abytes
of
field
(r f
achieved
by
fitting
two
opposite
and
velocity
to
the
position
and
orientation
of
the
bacterium,
the rotation
of the
cell
about its the
body axis. The topology of
ne for
> 1.5fors.
By
tracking
the
length
of the
flagellar
bundle.
The
force
dipole
fit
was
solved
the
nts
when
the
spec
the
measured
flow
field
(Fig.
1A)
is
the
same
as
that
of
a
and
performing
an
ensemble
average
over
all
tracers,
we
reat
variable
locations
along
the
sw
are events, relating
their
position
Bacterial
flowdow
fiel
achieved
by1B),
fitting
two
opposite
force
monopoles
(Stokeslets)
dipole
le
plane
kingforce
the
dipole
flowtime-averaged
(Fig.
defined
by
solved
the
flow
field
in
the
E.
coli
swimming
field
(r
>
8
µm).
From
the
best
and
orientation
of
the
bacterium,
dipole
flow
descri
at
variable
locations
along
the
swimming
direction
to
the
far
value
of
position
µm/s.
As
plane down to 0.1% of the mean swimming
speed
V
=
22
±
5
0
the
specific
fitting
routines
fi
with good and
accura
h
i
e
average
over
all
tracers,
we
refield
(r
>
8
µm).
From
the
best
fit,
which
is
insensitive
to
and resi
A E. coli
r direction, their time-avera
ℓF swimming
acterium,
µm/s.
As
2rotate about their
ˆ
thisµm
approximatio
u(r)in= the
3(r̂.coli
d) −
1 r̂, routines
A=
, r̂fitting
= length
, regions,
[ℓ
1 ]= we
dipole
1.9
and
dip
ow
field
E.
swimming
2
the
specific
fitting
and
obtain
the
note
tha2
|r|
8πηdimensions
|r| is cylindrically
s, we retime-averaged
flow field in three
symmetric
a
wall.
Focusing
value
F is Fconsistent
with
opt
dipole
length
1.9±µm
and
dipole
force
= of
0.42
pN.
This
µm
beh
mean
swimming
speed
V0 ℓ==22
5 capture
symmetric.
Our
measurements
allof
components
this
wimming
and applying
the
cylindrica
and
resistive
force
theory
calculat
fluid
dra
of
Fforce,
is consistent
with
optical
trap
measurements
[45]
where
F isvalue
the dipole
ℓ the
distance
separating
the force
ut
direction,
their
symmetric
flow,
except
the
azimuthal
flow due
to the
= their
22
±cylindrically
5swimming
resulted
in
a
sligh
rotati
note
thatThe
in
the
best
fit,
the
cell
and
resistive
force
theory
calculations
[46].
It is
interesting
to
η the
viscosity
the
fluid,
dˆ the
orientation
vector
theof
flow
field
struc
the
rotation
ofisof
the
cell
about
itsunit
body
axis.
topology
on, pair,
their
three
dimensions
cylindrically
the measu
(swimming
direction)
the best
bacterium,
and
rbehind
the
distance
surfaces,
the
note
thatflow
inofthe
fit, 1A)
theµm
cell
drag
Stokeslet
isfrom
0.1
the measured
field
(Fig.
is the
same
as that
oflocated
aof
the
center
the
cell
ndrically
nts
capture
all
components
of
this
force
dipo
Bacteria
vector
relative
to
the
center
of
the
dipole.
Yet
there
are
some
ity
of
a
no-slip
sur
µm
behind
the
center
of
the
cell
body,
possibly
reflecting
the
force
dipole
flow
(Fig.
1B),
defined
by
fluid
drag
on
the
flagellar
bundle
tsexcept
of
this
flow
due
Dunkel,
Ganguly,
Cisneros,
Goldstein
(2011) to
PNAS
Fig. differences
1.Drescher,
Averagethe
flow
fieldazimuthal
created
by a single
freely-swimming
bacterium.
(A)
Experimentally measured flow field far from a surface. Stream lines indicate local
direction offl
dipole
close
to
the
cell
body
as
shown
by
the
residual
of
outward
streamlin
fluid
drag
on
the
flagellar
bundle.
flow. (B) Best fit force-dipole model, and (C) residual flow field, obtained by subtracting the best-fit dipole from the experimentally measured field. The presence of the flagella
E.coli
w due to
weak ‘pusher’ dipole
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
dynamics of u is described by the Navier-Stokes equations (NSEs)
% [@t u + (u · r)u] = f
rp + r · T̂ ,
(5.2)
where p(t, x) the pressure in the fluid, T̂ (t, x) the deviatoric2 ) 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 % + ri (%ui ) = 0,
% (@t ui + uj @j ui ) = Fi
@i p + @j T̂ji .
To close the system of equations (5.4), one still needs to
(5.4a)
(5.4b)
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 % + ri (%ui ) = 0,
% (@t ui + uj @j ui ) = Fi
@i p + @j T̂ji .
(5.4a)
(5.4b)
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
T̂ = (T̂ij [%, u, . . .]),
(iii) specify an appropriate set of initial and boundary conditions.
2
‘deviatoric’:= without hydrostatic stress (pressure); a ‘full’ stress-energy tensor ˆ may be defined by
ˆij :=
p
ij
+ T̂ij .
Simplifications In the case of a homogeneous fluid with
@t % = 0
and
3
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
T̂ij :=
(r · u)
ij
+ µ (@i uj + @j ui ).
(5.7)
where the first coefficient of viscosity (related to bulk viscosity), and µ is the second
coefficient of viscosity (shear viscosity). Thus, for an incompressible Newtonian fluid, the
Navier-Stokes system (5.4) simplifies to
0 = r · u,
% [@t u + (u · r)u] =
rp + µr2 u + f .
(5.8a)
(5.8b)
Dynamic viscosity The SI physical unit of dynamic viscosity µ is the Pascal⇥second
[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
where the first coefficient of viscosity (related to bulk viscosity), and µ is the second
coefficient of viscosity (shear viscosity). Thus, for an incompressible Newtonian fluid, the
Navier-Stokes system (5.4) simplifies to
0 = r · u,
% [@t u + (u · r)u] =
rp + µr2 u + f .
(5.8a)
(5.8b)
Dynamic viscosity The SI physical unit of dynamic viscosity µ is the Pascal⇥second
[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.
Kinematic viscosity Below we will be interested in comparing viscous and inertial
forces. Their ratio is characterized by the kinematic viscosity ⌫, defined as
µ
⌫= ,
%
[⌫] = m2 /s
The kinematic viscosity of water with mass density % = 1 g/cm3 is ⌫ = 10
1 mm2 /s = 1 cSt.
3
(5.10)
6
m2 /s =
By virtue of the conservation law (5.1), a homogeneous material is always incompressible, but in
general the converse is not true.
5.2
5.2.1
Stokes equations
5.2
Stokes equations
Motivation
Motivation
Consider5.2.1
an object
of characteristic length L, moving at absolute velocity U = |U | through
(relativeConsider
to) an incompressible,
homogeneous
Newtonian
of constant
viscosity
and
an object of characteristic
length
L, movingfluid
at absolute
velocity
U = |U |µthrough
constant(relative
density to)
%. The
object can be homogeneous
imagined as Newtonian
a moving boundary
(condition),
which
an incompressible,
fluid of constant
viscosity
µ and
2
density
object
canThe
be imagined
as a inertial
moving boundary
which
induces aconstant
flow field
u(t, %.
x) The
in the
fluid.
ratio of the
(dynamic)(condition),
pressure %U
induces
a flowstress
field u(t,
x) can
in the
The ratioby
of the inertial
(dynamic)
and viscous
shearing
µU/L
befluid.
characterized
Reynolds
number4pressure %U 2
4
and viscous shearing stress µU/L can be characterized
by
the
Reynolds
number
2
R=
|%(@t u + (u · r)u)|
%U /L
U L%
UL
2
'
=
=
|%(@2t u + (u · r)u)|
%U
/L
U L% .U L
2
|µr
u|
µU/L
µ
⌫=
R=
'
=
.
|µr2 u|
µU/L2
µ
⌫
(5.11)
(5.11)
For example, when considering swimming in water (⌫ = 10 6 m2 /s),
one finds for fish or
For example, when considering swimming in water (⌫ = 10 6 m2 /s), one finds for fish or
humans:
humans:
L ' 1 m, U ' 1 m/s
)
R ' 106 , 6
whereas whereas
for bacteria:
for bacteria:
L ' 1 m, U ' 1 m/s
L ' 1 µm,
' 10U µm/s
L 'U
1 µm,
' 10 µm/s )
)
R ' 10 ,
10'510
. 5.
) R 'R
If the Reynolds
numbernumber
is veryissmall,
R ⌧ 1,
NSEsNSEs
(5.8)(5.8)
can can
be approxIf the Reynolds
very small,
R the
⌧ 1,nonlinear
the nonlinear
be approx5
imated by
the linear
equations
imated
by theStokes
linear Stokes
equations 5
0 = 0r =
· u,r · u,
2
=2 uµ rrp
u +rp
0 = 0µ r
f. + f.
(5.12a)
(5.12a)
(5.12b)
(5.12b)
four equations
determine
the unknown
four unknown
functions
(u, p).
However,
The fourThe
equations
(5.12) (5.12)
determine
the four
functions
(u, p).
However,
to to
such solutions,
these equations
stillendowed
be endowed
appropriate
uniquelyuniquely
identify identify
such solutions,
these equations
must must
still be
withwith
appropriate
initial
and boundary
conditions,
for example
initial and
boundary
conditions,
such assuch
for as
example
(
(
u(t, x)u(t,
= 0,x) = 0,
as
! 1.
(5.13)
|x| !|x|1.
(5.13)
p(t, x) = p1 , as
p(t, x) = p1 ,
Note that, by neglecting the explicit time-dependent inertial terms in NSEs, the time-
Note that,
by neglecting the explicit time-dependent inertial terms in NSEs, the timedependence of the flow is determined exclusively and instanteneously by the motion of the
5.2.2 Special solutions
5.2.2 Special solutions
Oseen solution Consider the Stokes equations (5.12) for a point-force
Oseen solution Consider the Stokes equations (5.12) for a point-force
f (x) = F (x).
f (x) = F (x).
In this case, the solution with standard boundary conditions (5.13) reads66
In this case, the solution with standard boundary conditions (5.13) reads
F j xj
F
ui (x) = Gij (x) Fj ,
p(x) = j xj 3 + p1 ,
ui (x) = Gij (x) Fj ,
p(x) = 4⇡|x|3 + p1 ,
4⇡|x|
where
tensor
wherethe
theGreens
Greensfunction
function G
Gijij isis given
given by
by the
the Oseen
Oseen tensor
✓
◆
✓
◆
11
x
x
xiixjj ,
GGij (x)
=
+
ij
,
ij (x) = 8⇡µ |x|
ij + |x|22
8⇡µ |x|
|x|
which
whichhas
hasthe
theinverse
inverse
GGjk11(x)
= 8⇡µ|x|
jk (x) = 8⇡µ|x|
asascan
canbe
beseen
seenfrom
from
GGijijGGjkjk11 =
=
=
=
=
=
=
=
✓
✓
✓
✓
jk
jk
◆
◆
x
xkk
xjj x
,
2 ,
2
2|x|
2|x|
◆✓
xiixj
+
ij
ij +
jk
|x|2
xxiixkk
xi xk
+
ik
ik
2|x|
|x|2
2|x|22
xxiixkk
xi xk
+
ik
ik
2|x|22 2|x|2
2|x|
ik..
ik
(5.14)
(5.14)
(5.15a)
(5.15a)
(5.15b)
(5.15b)
(5.16)
(5.16)
◆
◆
xjj xkk
2|x|22
2|x|
xiixjj xxjjxxkk
2|x|22
|x|22 2|x|
(5.17)
(5.17)
Stokessolution
solution(1851)
(1851) Consider
Consider aa sphere
sphere of radius a, which
Stokes
which at
at time
time tt isislocated
locatedatatthe
the
origin,X(t)
X(t)==0,0,and
and moves
moves at
at velocity
velocity U
U (t). The corresponding
origin,
corresponding solution
solution ofof the
theStokes
Stokes
77
equationwith
withstandard
standardboundary
boundary conditions
conditions (5.13)
(5.13) reads
equation
reads
2|x|2
=
2|x|2
ik .
(5.17)
Stokes solution (1851) Consider a sphere of radius a, which at time t is located at the
origin, X(t) = 0, and moves at velocity U (t). The corresponding solution of the Stokes
equation with standard boundary conditions (5.13) reads7

✓
◆
✓
◆
3 a
xj xi
1 a3
xj xi
ui (t, x) = Uj
+
3 2 ,
(5.18a)
ji +
ji
4 |x|
|x|2
4 |x|3
|x|
3 Uj x j
p(t, x) =
µa
+ p1 .
(5.18b)
2 |x|3
If the particle is located at X(t), one has to replace xi by xi
Eqs. (5.18). Parameterizing the surface of the sphere by
Xi (t) on the rhs. of
a = a sin ✓ cos ex + a sin ✓ sin ey + a cos ✓ ez = ai ei
6
7
Proof by insertion.
Proof by insertion.
76
https://www.boundless.com/physics/
2|x|2
=
2|x|2
ik .
(5.17)
Stokes solution (1851) Consider a sphere of radius a, which at time t is located at the
origin, X(t) = 0, and moves at velocity U (t). The corresponding solution of the Stokes
equation with standard boundary conditions (5.13) reads7

✓
◆
✓
◆
3 a
xj xi
1 a3
xj xi
ui (t, x) = Uj
+
3 2 ,
(5.18a)
ji +
ji
4 |x|
|x|2
4 |x|3
|x|
3 Uj x j
p(t, x) =
µa
+ p1 .
(5.18b)
2 |x|3
If the particle is located at X(t), one has to replace xi by xi
Eqs. (5.18). Parameterizing the surface of the sphere by
Xi (t) on the rhs. of
a = a sin ✓ cos ex + a sin ✓ sin ey + a cos ✓ ez = ai ei
6
7
Proof by insertion.
Proof by insertion.
where ✓ 2 [0, ⇡],
76
2 [0, 2⇡), one finds that on this boundary
u(t, a(✓, )) = U ,
3µ
p(t, a(✓, )) =
Uj aj (✓, ) + p1 ,
2 a2
(5.19a)
(5.19b)
corresponding to a no-slip boundary condition on the sphere’s surface. The O(a/|x|)contribution in (5.18a) coincides with the Oseen result (5.15), if we identify
F = 6⇡ µa U .
(5.20)
The prefactor = 6⇡ µa is the well-known Stokes drag coefficient for a sphere.
The O[(a/|x|)3 ]-part in (5.18a) corresponds to the finite-size correction, and defining
the Stokes tensor by
✓
◆
2
1 a
xj xi
Sij = Gij +
3
,
(5.21)
ji
24⇡µ |x|3
|x|2
2|x|2
=
2|x|2
ik .
(5.17)
Stokes solution (1851) Consider a sphere of radius a, which at time t is located at the
origin, X(t) = 0, and moves at velocity U (t). The corresponding solution of the Stokes
equation with standard boundary conditions (5.13) reads7

✓
◆
✓
◆
3 a
xj xi
1 a3
xj xi
ui (t, x) = Uj
+
3 2 ,
(5.18a)
ji +
ji
2
|x|this
4 |x|3
|x|
where ✓ 2 [0, ⇡], 2 [0, 2⇡), one4 |x|
finds that on
boundary
3 Uj x j
p(t, x) = u(t,
µa 3 ))
+ p=
(5.18b)
1. U ,
(5.19a)
2 a(✓,
|x|
3µ
p(t, a(✓, )) =
Uj aj (✓, ) + p1 ,
(5.19b)
If the particle is located at X(t), one has2to
a2 replace xi by xi Xi (t) on the rhs. of
Eqs. (5.18). Parameterizing the surface of the sphere by
corresponding to a no-slip boundary condition on the sphere’s surface. The O(a/|x|)contribution in (5.18a)
the
Oseen
identify
a = acoincides
sin ✓ cos with
ex +
a sin
✓ sinresult
ey +(5.15),
a cos ✓ifewe
z = ai e i
6
7
Proof by insertion.
Proof by insertion.
F = 6⇡ µa U .
(5.20)
The prefactor = 6⇡ µa is the well-known Stokes drag coefficient for a sphere.
The O[(a/|x|)3 ]-part in (5.18a) corresponds
76 to the finite-size correction, and defining
the Stokes tensor by
✓
◆
1 a2
xj xi
Sij = Gij +
3 2 ,
(5.21)
ji
24⇡µ |x|3
|x|
we may rewrite (5.18a) as8
ui (t, x) = Sij Fj .
5.3
(5.22)
Golestanian’s swimmer model
This part is copied (with very minor adaptations) from the article of Golestanian and
Ajdari [GA07], for their excellent discussion is difficult, if not impossible, to improve.
5.3.1
Three-sphere swimmer: simplified analysis
ui (t, x) = Sij Fj .
5.3
(5.22)
Golestanian’s swimmer model
This part is copied (with very minor adaptations) from the article of Golestanian and
Ajdari [GA07], for their excellent discussion is difficult, if not impossible, to improve.
5.3.1
Three-sphere swimmer: simplified analysis
As a minimal model of a low Reynolds number swimmer, consider three spheres of radii
ai (i = 1, 2, 3) that are separated by two arms of lengths L1 and L2 . Each sphere exerts
a force Fi on, and experiences a force Fi from, the fluid that we assume to be along the
swimmer axis. In the limit ai /Lj ⌧ 1, we can use the Oseen tensor (5.15) to relate the
forces and the velocities as
(5.23a)
v2
(5.23b)
v3
8
F1
F2
F3
+
+
,
6⇡µa1 4⇡µL1 4⇡µ(L1 + L2 )
F1
F2
F3
=
+
+
,
4⇡µL1 6⇡µa2 4⇡µL2
F1
F2
F3
=
+
+
.
4⇡µ(L1 + L2 ) 4⇡µL2 6⇡µa3
v1 =
For arbitrary sphere positions X(t), replace x ! x
(5.23c)
X(t).
77
other
‘minimal’ swimmer
ui (t, x) = Sij Fj .
5.3
(5.22)
Golestanian’s swimmer model
This part is copied (with very minor adaptations) from the article of Golestanian and
Ajdari [GA07], for their excellent discussion is difficult, if not impossible, to improve.
5.3.1
Three-sphere swimmer: simplified analysis
As a minimal model of a low Reynolds number swimmer, consider three spheres of radii
ai (i = 1, 2, 3) that are separated by two arms of lengths L1 and L2 . Each sphere exerts
a force Fi on, and experiences a force Fi from, the fluid that we assume to be along the
swimmer axis. In the limit ai /Lj ⌧ 1, we can use the Oseen tensor (5.15) to relate the
forces and the velocities as
(5.23a)
v2
(5.23b)
v3
8
F1
F2
F3
+
+
,
6⇡µa1 4⇡µL1 4⇡µ(L1 + L2 )
F1
F2
F3
=
+
+
,
4⇡µL1 6⇡µa2 4⇡µL2
F1
F2
F3
=
+
+
.
4⇡µ(L1 + L2 ) 4⇡µL2 6⇡µa3
v1 =
For arbitrary sphere positions X(t), replace x ! x
(5.23c)
X(t).
Note that in this simple one dimensional case, the tensorial structure of the hydrodynamic
Green’s function (Oseen tensor) does not enter the calculations as all the forces and velocities are
parallel to each other and to the position vectors. The swimming velocity of the
Swim-speed
77 namely
whole object is the mean translational velocity,
1
V0 = (v1 + v2 + v3 ).
3
(5.24)
We are seeking to study autonomous net swimming, which requires the whole system to
be force-free
(i.e. constraint
there are no external forces acting on the spheres). This means that the
Force-free
above equations are subject to the constraint
F1 + F2 + F3 = 0.
(5.25)
Eliminating F2 using Eq. (5.25), we can calculate the swimming velocity from Eqs. (5.23a),
(5.23b), (5.23c), and (5.24) as
V = (v1 + v2 + v3 ).
3
(5.24)
We are seeking to study autonomous net swimming, which requires the whole system to
be force-free (i.e. there are no external forces acting on the spheres). This means that the
above equations are subject to the constraint
F1 + F2 + F3 = 0.
(5.25)
Eliminating F2 using Eq. (5.25), we can calculate the swimming velocity from Eqs. (5.23a),
(5.23b), (5.23c), and (5.24) as
✓
◆
✓
◆ ✓
◆
1
1
1
3
1
1
F1
V0 =
+
+
3
a
a2
2 L1 + L2 L 2
6⇡µ
✓ 1
◆
✓
◆ ✓
◆
1
1
1
3
1
1
F3
+
,
(5.26)
3
a3 a2
2 L1 + L2 L 1
6⇡µ
where the subscript 0 denotes the force-free condition. To close the system of equations,
we should either prescribe the forces (stresses) acting across each linker, or alternatively
the opening and closing motion of each arm as a function of time. We choose to prescribe
the motion of the arms connecting the three spheres, and assume that the velocities
L̇1 = v2
L̇2 = v3
v1 ,
v2 ,
(5.27a)
(5.27b)
are known functions. We then use Eqs. (5.23a), (5.23b), (5.23c), and (5.25) to solve for F1
and F3 as a function of L̇1 and L̇2 . Putting the resulting expressions for F1 and F3 back
in Eq. (5.26), and keeping only terms in the leading order in ai /Lj consistent with our
original scheme, we find the average swimming velocity to the leading order.
5.3.2
Swimming velocity
The above calculations yield a lengthy expression summarized in Eq. (B.1) of the Appendix.
This result (B.1) is suitable for numerical studies of swimming cycles with arbitrarily large
and F3 as a function of L̇1 and L̇2 . Putting the resulting expressions for F1 and F3 back
in Eq. (5.26), and keeping only terms in the leading order in ai /Lj consistent with our
original scheme, we find the average swimming velocity to the leading order.
5.3.2
Swimming velocity
The above calculations yield a lengthy expression summarized in Eq. (B.1) of the Appendix.
This result (B.1) is suitable for numerical studies of swimming cycles with arbitrarily large
deformations. For the simple case where all the spheres have the same radii, namely
a = a1 = a2 = a3 , Eq. (5.26) simplifies to
"
!
!#
a
L̇2 L̇1
L̇1 L̇2
+2
,
(5.28)
V0 =
6
L1 + L2
L2 L1
plus terms that average to zero over a full swimming
cycle. Equation (5.28) is also valid
78
plus terms that average to zero over a full swimming cycle. Equation (5.28) is also valid
for arbitrarily large deformations.
for arbitrarily large deformations.
We can also consider relatively small deformations and perform an expansion of the
We can also consider relatively small deformations and perform an expansion of the
swimming velocity to the leading order. Using
swimming velocity to the leading order. Using
L1
L1
L2
L2
=
=
=
=
`1 + u1 (t),
`1 + u1 (t),
`2 + u2 (t),
`2 + u2 (t),
(5.29)
(5.29)
(5.30)
(5.30)
in Eq. (B.1), and expanding to the leading order in ui /`j , we find the average swimming
in Eq. (B.1), and expanding to the leading order in ui /`j , we find the average swimming
velocity
velocityasas
K
K (u1 u̇2 u̇1 u2 ),
VV0 =
(5.31)
u̇1 u2 ),
(5.31)
0 = 2 (u1 u̇2
2
where

where
33aa1 aa2 aa3
11
11
11
1
2
3
K
(5.32)
K == (a1 + a2 + a3 )22 `22 +
+ `22 (`1 + `2 )22 ..
(5.32)
(a1 + a2 + a3 ) `11 `22 (`1 + `2 )
InInthe
full cycle.
cycle. Note
Note
theabove
aboveresult,
result,the
theaveraging
averaging isis performed
performed by
by time
time integration
integration over
over aa full
that
because they
they are
are full
full
thatterms
termsproportional
proportionalto
touu11u̇u̇11,,uu22u̇u̇22,, and
and uu11u̇u̇22+
+u̇u̇11uu22 are
are eliminated
eliminated because
time
shows that
that the
the
timederivatives
derivativesand
andthey
they average
average out
out to
to zero
zero in
in aa cycle.
cycle. Equation
Equation (5.31)
(5.31) shows
swimmer can achieve higher velocities if it can maximize this area by introducing sufficient
phase di↵erence between the two deformation cycles (see below).
5.3.3
Harmonic deformations
As a simple explicit example, consider harmonic deformations of the two arms, with identical frequencies ! and a mismatch in phases,
u1 (t) = d1 cos(!t + '1 ),
u2 (t) = d2 cos(!t + '2 ).
(5.33)
(5.34)
The average swimming velocity from Eq. (5.31) reads
V0 =
K
d1 d2 ! sin('1
2
'2 ).
(5.35)
This result shows that the maximum velocity is obtained when the phase di↵erence is
⇡/2, which supports the picture of maximizing the area covered by the trajectory of the
swimming cycle in the parameter space of the deformations. A phase di↵erence of 0 or ⇡,
for example, will create closed trajectories with zero area, or just lines.
79
5.3.4
Force-velocity relation and stall force
The
e↵ect of
an external force or
load on the
efficiency
the swimmer can be easily studied
5.3.4
Force-velocity
relation
and
stall offorce
within the linear theory of Stokes hydrodynamics. When the swimmer is under the e↵ect
of external
an external
force
load
on the
efficiency
of the swimmer
can be easily studied
ofThe
an e↵ect
applied
force
F ,orEq.
(5.25)
should
be changed
as
within the linear theory of Stokes hydrodynamics. When the swimmer is under the e↵ect
+ F3 =
(5.36)
1 + F2should
of an applied external force F , Eq. F(5.25)
beF.changed as
Following through the calculations ofF1Sec.
+ F25.3.1
+ F3above,
= F. we find that the following changes
(5.36)
take place in Eqs. (5.23a), (5.23b), (5.23c), and (5.24):
Following through the calculations of Sec. 5.3.1 above, we find that the following changes
F
take place in Eqs. (5.23a),
(5.23c),
v1 7! (5.23b),
v1
, and (5.24):
(5.37)
4⇡µL1
F
F ,
v1 7! v1
(5.37)
v2 7! v2
(5.38)
4⇡µL1,
6⇡µa2
F
F ,
v2 7! v2
(5.38)
v3 7! v3
(5.39)
6⇡µa2 ,
4⇡µL2
✓F
◆
v3 7! v3 1
(5.39)
1,
1
1
4⇡µL
V 7! V
+
+
(5.40)
2
◆ F.
31 ✓6⇡µa
4⇡µL
4⇡µL
2
1
2
1
1
1
V
!
7
V
+
+
F.
(5.40)
These lead to the changes
3 6⇡µa2 4⇡µL1 4⇡µL2
✓
◆
1
1
These lead to the changes
L̇1 7! L̇1 ✓
(5.41)
◆ F,
6⇡µa
4⇡µL
1 2
1 1◆
✓
L̇1 7! L̇1
(5.41)
1 2 4⇡µL
1 1 F,
6⇡µa
L̇2 7! L̇2 ✓
(5.42)
◆ F,
4⇡µL
6⇡µa
1 2
1 2
L̇2 7! L̇2
F,
(5.42)
4⇡µL
6⇡µa
in Eq. (B.1), which together with correction
coming
from2 Eq. (5.40) leads to the average
2
swimming
velocity
in Eq. (B.1),
which together with correction coming from Eq. (5.40) leads to the average
F
swimming velocity
V (F ) = V0 +
,
(5.43)
18⇡µa
F R
V (F ) = V0 +
,
(5.43)
to the leading order, where aR is an e↵ective (renormalized)
hydrodynamic radius for the
18⇡µaR
1
three-sphere
swimmer.
To
the
zeroth
order,
we
have
a
=
the for
general
R
1 + a2 + a3 ) for
to the leading order, where aR is an e↵ective (renormalized)3 (a
hydrodynamic
radius
the
1
case
and thereswimmer.
are a large
number
of order,
correction
terms
at higher
wegeneral
should
three-sphere
To the
zeroth
we have
aR =
(a + a2orders
+ a3 ) that
for the
3 1
keep
order
to be
in our perturbation
reporting
caseinand
there
areconsistent
a large number
of correction theory.
terms atInstead
higheroforders
that the
we lengthy
should
L̇2 7! L̇2
(5.40)
F, + 4⇡µL u+2 (t)
(5.42) + '
F, = dF.
(5.41)
cos(!t
).
2
2
6⇡µa
4⇡µL
22
11
2
6⇡µa
4⇡µL
2
2 ✓
◆
1
1
These lead to the changes
L̇
!
7
L̇
F,
(5.42)
2 Eq.✓(5.40)
2 (5.43)
◆
h together
with correction
coming
from
leads
to
the
average
,
The average
swimming
velocity
from
Eq.
(5.31)
reads
4⇡µL21 6⇡µa21
µaR
L̇1 7! L̇1
F,
(5.41)
6⇡µa2 from
4⇡µL
in Eq. (B.1), which together with correction coming
Eq.
(5.40)
leads
to
the
average
1
✓
◆ K
F
malized) swimming
hydrodynamic
radius
for
the
velocity
1
1
V (F ) = V0 +
,
(5.43)
V
=
d
d
'2 ).
1
L̇
!
7
L̇
F,
0
1
2 ! sin('1 (5.42)
2
2
F
e aR = 3 (a1 + a2 + a3 ) for the
general
18⇡µa
R
2
2
V (F ) = V0 +4⇡µL2 6⇡µa
,
(5.43)
18⇡µa
R
erms
at higher
that(renormalized)
we should
r, where
aR in
is Eq.
anorders
e↵ective
hydrodynamic
radius
theto the average
(B.1),
which
together
with correction
coming from Eq.
(5.40)for
leads
to the leading This
order, where
aR shows
is an e↵ective
(renormalized)
hydrodynamic
radius
for
the
result
that
the
maximum
velocity
is
obtained
when the
1
swimming
velocity
heory.
Instead
of
reporting
the
lengthy
1
mer. To the
zeroth
order,
we
have
a
=
(a
+
a
+
a
)
for
the
general
R order,
2 a
3 (a1 + a2 + a3 ) for the general
three-sphere swimmer. To the zeroth
3 1we have
R =
F
3of maximizing the area covered by t
⇡/2,
which
supports
the
picture
V
(F
)
=
V
+
,
(5.43)
pression
for
a
=
a
=
a
=
a,
which
0
a large number
of2 correction
terms ofatcorrection
higher terms
orders
should
1 there
case and
are a3large number
higher we
orders
that we should
18⇡µaRat that
swimming
cycle
inInstead
the parameter
space
of
the deformations.
A phase
keep
order
to
be consistenttheory.
in our perturbation
theory.
Instead
of reporting
the lengthy
consistent
intoinour
perturbation
of
reporting
the
lengthy
the leading order, where aR is an e↵ective (renormalized) hydrodynamic radius for the
expression
for for
the general
case, will
we provide
theclosed
expression
for1 a1 = a2 =
a3 =zero
a, which
example,
create
trajectories
with
area, or just line
swimmer.
To the zeroth
order,
we
have
a
=
(a
+
a
+
a
)
for
the general
R
1
2
3
general ◆
case,
we
provide
the
expression
for
a
=
a
=
a
=
a,
which
2 three-sphere
33
1
2
1
1 readscase
a and there
1 are a large number of correction terms at higher orders that we should
.consistent
(5.44)
✓
◆2
2
keep
in
order
to
be
in
our
perturbation
theory.
the lengthy
a 1
1 Instead
a of reporting
1
L1 L 2
2 (L11+ =
L2 )1 + 1 ✓+ 1 + 1◆
.a = a,
(5.44)
2
expression
for
the
general
case,
we
provide
the
expression
for
a
=
a
=
which
2
1
2
3
aR 1
a L
2 (L1 + L2 )
1
1
1
a1 1L2 L11 + L2 2a L1 1L2
reads
which
have
been expected based
+ could
+ The
+
.have been
(5.44)
2
force-velocity
relation
given
in
Eq.
(5.43),
which
could
expected based
a L 1 L 2 L1 + L2 2 L1 L 2
2 (L✓1 + L2 )◆2
79
1
1
1 yields
1 a stall1 force a 1
1
a
1
on linearity of hydrodynamics,
=
+
+
+
.
(5.44)
2
a
a
L
L
L
+
L
2
L
L
2
(L
+
L
)
ty relation given in Eq.
(5.43), which
have
been
expected
R
1
2 could
1
2
1
2
1based
2
F
=
18⇡µa
V
.
(5.45)
s
R
0
V0 .
(5.45)
The force-velocity
relation given in Eq. (5.43), which could have been expected based
odynamics, yields
a stall force
Using
the
zerothof order
expression yields
for thea stall
hydrodynamic
radius, one can see that this is
on
linearity
hydrodynamics,
force
namic radius,
can
see
that
this
is
equal toone
the
Stokes
force
exerted
on
spheres moving with a velocity
V0 .
Fs = 18⇡µaR V0 . theFthree
(5.45)
18⇡µaR V0 .
(5.45)
s =
es moving with a velocity V .
VL̇17!7!V L̇1
4⇡µL
6⇡µa 3
0
80 hydrodynamic
rder expression
hydrodynamic
radius,
one can see
that
is that
Usingfor
the the
zeroth
order expression for
the
radius,
onethis
can see
equal
the three
Stokes force
exerted
on the three
movingVwith
a velocity V0 .
force exerted
ontothe
spheres
moving
withspheres
a velocity
0.
80
80
this is
5.3.5
Power consumption and efficiency
Because we know the instantaneous values for the velocities and the forces, we can easily
calculate Power
the powerconsumption
consumption in and
the motion
of the spheres through the viscous fluid.
5.3.5
efficiency
The rate of power consumption at any time is given as
Because we know the instantaneous values for the velocities and the forces, we can easily
P = F1 v1 +inF2the
v2 +motion
F3 v3 =ofF1the
( L̇spheres
(L̇2 ),
(5.46)
calculate the power consumption
the viscous fluid.
1 ) + F3 through
The rate of power consumption at any time is given as
where the second expression is the result of enforcing the force-free constrain of Eq. (5.25).
= F1 vand
+ aF3function
v3 = F1 ( ofL̇L̇
F3 (L̇
L̇22,),one finds for a1 (5.46)
Using the expressionsPfor
and
= a2 =
1 + FF
23v2as
1 )1 +
a3 = a
where the second expression is theresult of enforcing the force-free constrain of Eq. (5.25).
1 a of L̇1a and L̇22, one finds for a1 = a2 =
Using the expressions for F1 and F3 as aa function
P = 4⇡µa 1 +
+
L̇1 +
a3 = a
L 1 2 L2 L1 + L2


1 a
a
a
2
1+ a + a
4⇡µa 1 a
L̇
22 +
P = 4⇡µa 1 + 2 L1 L2+ L1 + L2 L̇1 +
 L 1 2 L2 L1 + L2

1 a
1 a
5
a
1
a
a
a
4⇡µa 1
+
(5.47)
2 L̇1 L̇2 .
4⇡µa 1
+
+
L̇
+
2
L
2
L
2
L
+
L
2
2
2 L1 1 L2 2 L1 + L12

1 a
1 a
5
a as
We can now define a Lighthill
4⇡µa 1hydrodynamic efficiency
+
L̇1 L̇2 .
(5.47)
2 L 1 2 L2 2 L1 + L 2
2
18⇡µaR V0
µL ⌘
,
(5.48)
We can now define a Lighthill hydrodynamic
efficiency
as
P
2
for which we find to the leading order 18⇡µaR V0
µL ⌘
,
P
2
2
9 aR
K (u1 u̇2 u̇1 u2 )
for which we find to the leading
µL =order
,
8 a C1 u̇21 + C2 u̇22 + C3 u̇1 u̇2
where
µL =
9 aR
K 2 (u1 u̇2
2
u̇1 u2 )
2
(5.48)
(5.49a)
2
,
(5.49a)
Using the expressions for F1 and F3 as a function of L1 and L2 , one finds for a1 = a2 =
the second expression is the result of enforcing the force-free constrain of Eq. (5.25).
a3 = where
a
Using the expressions for F1 and F3 as a function of L̇1 and L̇2 , one finds for a1 = a2 =

a3 = a
1 a
a
a
P = 4⇡µa 1 +
+
L̇21 +
La1 12 L
L1a + L2 2
a2

P = 4⇡µa 1 +
+
L̇ +
1L1a 2 La2 L1 + aL2 1 2
4⇡µa 1
+a + a
L̇2 +
1
a
4⇡µa 1 2 L1 + L2 + L1 + L2L̇22 +
12 La1 L12 a L1 +
5 L2 a

4⇡µa 1
L̇1 L̇2 .
(5.47)
1 a
1 a +5
a
2 L2+ 2 L1 + L2L̇1 L̇2 .
4⇡µa 1 2 L1
(5.47)
2 L1
2 L2
2 L1 + L 2
We can
now define a Lighthill hydrodynamic efficiency as
We can now define a Lighthill hydrodynamic efficiency as
22
18⇡µa
18⇡µaRRVV0 0
µµLL ⌘
⌘
,,
PP
(5.48)(5.48)
for which
we find
leadingorder
order
for which
we find
to to
thethe
leading
2
2
9 aR
K 2 (u1 u̇2 u̇1 u2 ) 2
K (u1 u̇2 u̇1 u2 ) ,
µ L = 9 aR
µL = 8 a C1 u̇21 + C2 u̇22 + C3 u̇1 u̇2 ,
2
2
8 a C1 u̇1 + C2 u̇2 + C3 u̇1 u̇2
(5.49a)
(5.49a)
where
where
a
`a1
C1 = 1 + 1 a
C2 = 1 ` 1
12 `a1
C2 = 1 1 a
C3 = 1 2 ` 1
12 `a1
C1 = 1 +
C3 = 1
1a
a
+
,
1
a
a
2 `2 `1 + `2
+ a
,
a
+2 `2+ `1 + `, 2
`2a `1 + a`2
+1 a+ 5 a ,
`2 + `1 + `2 .
21`2a 2 5`1 + `a2
+
(5.49b)
(5.49b)
(5.49c)
(5.49c)
(5.49d)
.
(5.49d)
It is interesting to note that for harmonic
a single
frequency, Eq. (5.49a)
2 `deformations
2 `2 2with
`1 +
`2
1
is independent of the frequency and scales like a2 d2 /`4 , which reflects the generally low
It is interesting
to efficiency
note thatoffor
withIna single
frequency,
Eq. to
(5.49a)
hydrodynamic
lowharmonic
Reynolds deformations
number swimmers.
this case,
it is possible
5.4
5.4 Dimensionality
Dimensionality
5.4 Dimensionality
We
We saw
saw above
above that,
that, in
in 3D,
3D, the
the fundamental
fundamental solution
solution to the Stokes equations for a point
We
above
that,
in
3D,by
the
fundamental
solution to the Stokes equations for a point
force
at
isis given
the
Oseen
forcesaw
at the
the origin
origin
given
by
the
Oseen solution
solution
force at the origin is given by the Oseen solution
Fjjxjj
uuii(x)
p(x)
(x) =
=G
Gijij(x)
(x) FFjj ,,
p(x) = F x 33 + p1 ,
(5.50a)
j j
4⇡|x|
ui (x) = Gij (x) Fj ,
p(x) =
+ p1 ,
(5.50a)
4⇡|x|3
where
where
✓
✓
◆
where
xiixjj ◆
11 ✓
Gijij(x)
(x) =
=
(5.50b)
G
,
ij + x x22
ij
1 |x|
i j
8⇡µ
|x|
8⇡µ
|x|
Gij (x) =
,
(5.50b)
ij +
8⇡µ |x|
|x|2
interesting to
to compare
compare this
this result
result with
with corresponding
corresponding 2D solution
ItIt isis interesting
It is interesting to compare this result with corresponding 2D solution
Fjxxj
F
(x) =
= JJijij(x)F
(x)Fjj ,,
= Fj xj 22 + @1
(5.51a)
uuii(x)
pp =
x = (x, y)
1,
j j
2⇡|x|
2⇡|x|
ui (x) = Jij (x)Fj ,
p=
+ @1 ,
x = (x, y)
(5.51a)
2⇡|x|2
where
where
✓ ◆
where

✓
1
|x|◆ xii xj
1 
|x|
✓
J
(x)
=
ln
+ x x22
(5.51b)
Jijij(x) = 1
ij ln |x|
ij
i
j
4⇡µ
a
|x|
a +
Jij (x) = 4⇡µ
(5.51b)
ij ln
4⇡µ
a
|x|2
with aa being
being an
an arbitrary
arbitrary constant
constant fixed
fixed by
by some
some intermediate flow normalization condiwith
with
being
an(5.51)
arbitrary
constant
fixed
by
some
intermediate
flow normalization
condition. a
Note
that
(5.51)
decays
much more
more slowly
slowly
than
(5.50), implying
that hydrodynamic
tion.
Note
that
decays
much
than
tion.
Note that
(5.51)
decays much
slowlystronger
than (5.50),
that
hydrodynamic
interactions
in 2D
2D
freestanding
filmsmore
are much
much
stronger
than implying
in 3D bulk
solutions.
interactions
in
freestanding
films
are
interactions
2D(5.51)
freestanding
films
are much
than in
3D bulk we
solutions.
To verify
verifyin
that
(5.51)
indeed
solution
of stronger
the 2D Stokes
equations,
first note that
To
that
isis indeed
aa solution
of
the
To verify that (5.51) is indeed a solution of the 2D Stokes equations, we first note that
generally
generally
generally
xjj
1/2
1/2
1/2
1/2
@
|x|
=
@
(x
x
)
=
x
(x
x
)
=
(5.52a)
j
j
i
i
j
i
i
@j |x| = @j (xi xi ) = xj (xi xi )
xj
|x|
1/2
1/2
@j |x| = @j (xi xi ) = xj (xi xi )
=
(5.52a)
x
|x|
j
n
n/2
(n+2)/2
j
|x| n =
= @@jj(x
(xiixxii)) n/2 =
= nx
nxjj(x
(xiixxii)) (n+2)/2 =
= n
n xn+2
(5.52b)
@@jj|x|
..
(5.52b)
n+2
j
|x|
n
n/2
(n+2)/2
|x|
@j |x|
= @j (xi xi )
= nxj (xi xi )
= n n+2 .
(5.52b)
|x|
From this,
this, we
we find
find
From
From this, we find
✓
◆
✓
◆
F
F
x
x
F
x
x
Fjjxjjxii = Fjj ✓
xjjxii ◆
ii
= F
2
2
(5.53)
ij
@@iipp =
2
=
2
(5.53)
F 2
F x x4
F 2 ij
x x2
@j |x|
n
= @j (xi xi )
n/2
=
nxj (xi xi )
(n+2)/2
=
n
xj xi
|x|2
◆
xj
.
|x|n+2
(5.52b)
From this, we find
@i p =
Fi
2⇡|x|2
2
F j xj xi
Fj
=
2⇡|x|4
2⇡|x|2
✓
ij
2
(5.53)
and
@k Jij
1
=
@k
4⇡µ

1
=
4⇡µ

1
=
4⇡µ

✓
◆
|x|
xi xj
+
ij ln
a
|x|2
✓
◆
1
xi xj
@k |x| + @k
ij
|x|
|x|2
✓
xk
xj
xi
+
+
ij
ik
jk
|x|2
|x|2
|x|2
xi xj xk
2
|x|4
◆
.
(5.54)
82
To check the incompressibility condition, note that

✓
1
xi
xj
@i Jij =
+
+
ij
ii
4⇡µ
|x|2
|x|2
✓
1
xj
xj
xj
=
+
2
+
4⇡µ
|x|2
|x|2 |x|2
= 0,
xi
xi xj xi
ji
|x|2
2|x|4
◆
xj
2 2
|x|
◆
(5.55)
which confirms that the solution (5.51) satisfies the incompressibility condition r · u = 0.
Moreover, we find for the Laplacian

@k
xk
xj
xi
xi xj xk
@k @k Jij =
+
+
2
ij
ik
jk
4⇡µ
|x|2
|x|2
|x|2
|x|4

✓
◆
✓
◆
✓
◆
✓
◆
1
xk
xj
xi
xi xj xk
1
xj
xj
xj
xj
(5.55)
= 0,
+
2
+
2
2
2
2
2
4⇡µ
|x|
|x|
|x|
|x|
= solution
0,
which confirms that the
(5.51) satisfies the incompressibility condition r · u(5.55)
= 0.
Moreover, we find for the Laplacian
which confirms that the solution (5.51) satisfies the incompressibility condition r · u = 0.
@k for thexLaplacian
xj
xi
xi xj xk
Moreover, we find
k
@k @k Jij =
+
+
2
ij
ik
jk
4⇡µ 
|x|2
|x|2
|x|2
|x|4
◆ xj
✓ xi ◆ xi xj xk✓
◆
✓
◆
@k 
xk✓
1
x
x
x
x
x
x
@k @k Jij =
+ k ik 2 + jk j2 2
i
i j k
ij
2
4
= 4⇡µ
@
+
@
+
@
2@
|x|
|x|
|x|
|x|
ij k
ik k
jk k
k
4⇡µ 
✓ |x|2 ◆
✓ |x|2 ◆
✓ |x|2 ◆
✓ |x|4 ◆
✓
◆
✓
1 
xk
xj✓
x◆i
xi xj xk ◆
1
xi xk
=
+ jk @x
2@k ik
k xkik @k
jk
ij @k kk 2 x+
k j xk 2
2
4
2 4|x| + jk
= 4⇡µ
|x| 2 4 + |x|
|x|2 4
ij
ik
4⇡µ 
|x| ◆
|x| ◆
|x| ◆
✓ |x|2
✓ |x|2
✓ |x|2
✓
◆
1
xk xk
xj xk
xi xk
kk
jk
ik
x
x
x
x
x
x
x
x
x
x
2
+
2
+
2
=
i jk k
i j k k jk
ij ik j 2 k
ik i j kk2
4
4
4⇡µ 2
|x|4 + |x|4 4 +
|x|
|x|
|x|2
|x|4
4
6
|x|
|x|
|x|
✓ |x|
◆

✓ik xj xk xi jk◆xk ✓ xi xj kk
◆
✓
◆
xi xj xk xk
1
2 + 1
xi xj
2
+ ij 4 xj4xi
ij
4
4
6
=
2
+
2
+
2
|x|
|x|
|x|
ij |x| 2
4⇡µ 
|x|2 ◆ ✓ |x|2
|x|4 ◆ ✓ |x|2
|x|4 ◆
✓ |x|
✓ 2
◆j xi
1
1
x
xi xj
ij
ij
x
x
x
x
x
x
x
x
=
2
+
2
+
2
j
i
i
j
i
j
i
j
ij
2+
+2 2 4 |x|42 4 |x|4
4⇡µ 2
|x|
|x|
|x|2
|x|4
4
4
|x|
|x|
|x| ◆
✓ |x|
✓
xj xi xi◆
x
xx
xx
1
x+j xi j + 2 i j 4 i j
2 ij
4
=
(5.56)
|x|42 |x|
|x|4
|x|4
2
4
2⇡µ ✓ |x|
|x| ◆
1
xj xi
ij
=
2 4
(5.56)
2
Hence, by comparing
with
(5.53),
2⇡µ |x|
|x| we see that indeed
Hence, by comparing with
indeed
@i p (5.53),
+ µ@k @we
= that
@i p +
µ@k @k Jij Fj = 0.
k ui see
(5.57)
µ@k @2D
= @i p + µ@k @khas
Jij Fjbeen
= 0. confirmed experimentally
(5.57)
The di↵erence between @3D
i p +and
k ui hydrodynamics
for Chlamydomonas algae [GJG10, DGM+ 10].
The di↵erence between 3D and 2D hydrodynamics has been confirmed experimentally
for Chlamydomonas algae [GJG10, DGM+ 10].
5.5
Force dipole and dimensionality
In the absence
must satisfy the force-free constraint.
5.5
Forceof external
dipoleforces,
andmicroswimmers
dimensionality
5.5
Force dipole and dimensionality
for Chlamydomonas algae [GJG10, DGM+ 10].
In the absence of external forces, microswimmers must satisfy the force-free constraint.
This
realization
is aand
force-dipole
flow, which provides a very good approximation
5.5simplest
Force
dipole
dimensionality
for the mean flow field generated by an individual bacterium [DDC+ 11] but not so much
absence
of +external
forInanthealga
[DGM
10]. forces, microswimmers must satisfy the force-free constraint.
This simplest realization is a force-dipole flow, which provides a very good approximation
To construct a force dipole, consider two opposite point-forces
F + = F = F ex
+
for the mean flow field +generated by an individual bacterium [DDC 11] but not so much
located
x = ±`ex . Due to linearity of the Stokes equations the total flow at
for an at
algapositions
[DGM+ 10].
To construct a force dipole, consider two opposite point-forces F + = F = F ex
located at positions x+ = ±`ex . Due to linearity of the Stokes equations the total flow at
83
some point x is given by
83
ui (x) =
⇥
=
= [
ij (x
ij (x
ij (x
x+ ) Fj+ +
x+ )
`ex )
ij (x
x ) Fj
⇤ +
x ) Fj
ij (x
+
ij (x + `ex )] Fj
where ij = Jij in 2D and ij = Gij in 3D. If |x|
`, we can Taylor expand
` = 0, and find to leading order
⇥ +
⇤
ui (x) ' [ ij (x)
xk @k ij (x) xk @k ij (x) Fj+
ij (x)]
+
=
2x+
k [@k ij (x)] Fj
(5.58)
ij
near
(5.59)
2D case Using our above result for @k Jij , and writing x+ = `n and F + = F n with
|n| = 1, we find in 2D

✓
◆
x+
x
x
x
x
x
x
k
j
i
i j k
+
k
ui (x) =
+
+
2
F
ij
ik
jk
j
2⇡µ
|x|2
|x|2
|x|2
|x|4
✓
◆
F`
xk nk
xj nj
xi
nk xi xj xk nj
=
ni
+
n
+
n
n
2
i
k k
2⇡µ
|x|2
|x|2
|x|2
|x|4
and, hence,
+
algae [GJG10,
DGM
10].
In for
theChlamydomonas
absence of external
forces,
microswimmers
must satisfy the force-free constraint.
This simplest realization is a force-dipole flow, which provides a very good approximation
for5.5
the mean
flow dipole
field generated
by an individual bacterium [DDC+ 11] but not so much
Force
and
dimensionality
for an alga [DGM+ 10].
+
InTo
theconstruct
absence ofaexternal
forces, microswimmers
satisfypoint-forces
the force-free F
constraint.
force dipole,
consider twomust
opposite
= F = F ex
This simplest
realization
force-dipole
flow, which provides a very good
approximation
located
at positions
x+ is=a±`e
equations
the total flow at
x . Due to linearity of the Stokes
+
for the mean flow field generated by an individual bacterium [DDC 11] but not so much
some
point
x is
is given
given
by
+
some
point
for an
alga x[DGM
10]. by
+
To construct a force dipole, consider two opposite
= F = F ex
+ point-forces F
+
83
+
+
u
(x)
=
(x
x
)
F
+
(x
x
)
F
ij
(xlinearity
x )F
+ Stokes
x ⇤) Fj jthe total flow at
located at positions x+u=ii(x)
±`ex=
. Due⇥ijij
to
ofj jthe
ij (x equations
⇥ (x x++ )
=
=
x )
ijij(x
= [[ ijij(x
(x 83`e
`exx))
=
⇤) F++
x
x ) Fj j
++
(x
+
`e
)]
F
ij
x
ij (x + `ex )] F j
(x
ijij(x
j
(5.58)
(5.58)
where ijij =
= JJijij in
in 2D
2D and
and ijij =
=G
Gijij in
in 3D.
3D. IfIf |x|
|x| `,`, we
wecan
canTaylor
Taylorexpand
expand ijijnear
near
where
= 0,
0, and
and find
find to
to leading
leading order
order
`` =
⇥⇥ ++
⇤ ⇤ ++
(x) '
' [[ ijij(x)
(x)
(x)] xxk@@kk ijij(x)
(x) xxk@@k k ijij(x)
(x) FFj j
uuii(x)
ijij(x)]
k
k
+
++
= 2x
2x+
[@
(x)]
F
(5.59)
k
ij
=
[@
(x)]
F
(5.59)
k
k
ij
jj
k
2D case Using
Using our
our above
above result
result for
for @@kkJJijij,, and
and writing
writing xx++ ==`n
`nand
andFF++==FFnnwith
with
|n| = 1, we find
find in
in 2D
2D
✓✓
◆◆
+ 
+
x
x
x
x
x
x
x
xkk
xkk
xj j
xi i
xi xi j xj k k
++
(x) =
=
+
+
2
F
uii(x)
+
+
2
F
ij
ik
jk
ij
ik
jk
j
j
2⇡µ
|x|22
|x|22
|x|2 2
|x|4 4
2⇡µ
|x|
|x|
|x|
|x|
✓
✓
◆◆
FF``
xxkknnkk
xxjjnnj j
xxi i
nnk x
kx
ix
jx
kn
ix
jx
kn
jj
=
=
nnii 22 ++nni i 22 ++nnkknnkk 2 2 22
2⇡µ
|x|
|x|
|x|
|x|4 4
2⇡µ
|x|
|x|
|x|
|x|
and, hence,
where x̂ = x/|x|.
x/|x|.
⇤⇤
FF`` ⇥⇥
22
u(x)
2(n
u(x) =
=
2(n· ·x̂)
x) 11 x̂x̂
2⇡µ|x|
2⇡µ|x|
(5.60)
(5.60)
where x̂ = x/|x|.
3D case To compute the dipole flow field in 3D, we need to compute the partial derivatives of the Oseen tensor
Gij (x) =
1
(1 + x̂i x̂j ) ,
8⇡µ|x|
x̂k =
xk
.
|x|
(5.61)
Defining the orthogonal projector (⇧ik ) for x̂k by
⇧ik :=
x̂i x̂k ,
ik
(5.62)
we have
xk
= x̂k ,
|x|
xk xi
⇧ik
ik
=
=
,
3
|x|
|x|
|x|
1
=
(x̂i ⇧nk + x̂k ⇧ni ) ,
|x|
@k |x| =
@k x̂i
@n ⇧ik
and from this we find
(5.63a)
(5.63b)
(5.63c)
84
x̂k

Gij +
(⇧ik x̂j + ⇧jk x̂i )
|x|
|x|2

=
x̂k ij + x̂j ik + x̂i jk 3x̂k x̂i x̂j .
|x|2
@k Gij =
(5.64)
Inserting this expression into (5.59), we obtain the far-field dipole flow in 3D
F` ⇥
2
u(x) =
3(n
·
x̂)
4⇡µ|x|2
⇤
1 x̂.
(5.65)
As shown in Ref. [DDC+ 11], Eq. (5.65) agrees well with the mean flow-field of a bacterium.
Upon comparing Eqs. (5.60) and (5.65), it becomes evident that hydrodynamic inter-
Results
w
of itswe
“puller”
image.
d. field
Atwalls,
distances
rfocused
<6µ
mon
theadipole
model
overestimates
the
bacterial
flow
field.
(E)
Experimentally
measured flow to
plane 50 µm from the top and bottom
tancesurfaces
2 µm parallel
to sample
the wall. chamber,
(F) Best fitand
force-dipole
model,
(G) residual
field.
Notewhere
the
Bacterial
cell
body,
of the
recorded
∼ 2 and
terabytes
of flow
4
flowmovie
fieldResults
ofdata.
an E. In
coli
“pusher”
decays
much faster,
when
swims close
thecule
surface,
fortothe
length
of t
theflow
mea
this
data
we
identified
∼
10
events when
(non-tumbling HCB 437)rarea bacterium
achieved
by
fittin
theByTo
measured
andminisbest-fit
force
cellsBacterial
swam in the
for > surfaces.
1.5 s.
tracking
decays
labeled,
n
flowfocal
fieldplane
far from
resolve the
the
at
variable
locatio
fluidcule
tracers
in
each
of
the
rare
events,
relating
their
position
of
the
c
decays
of
the
flow
speed
u
with
flow
field
created
by
individual bacteria, we tracked gfp- sion of
flu
m surfaces.
To
resolve
the
minisfield (r >field
8 µm).
and labeled,
velocity to
the
position
and
orientation
of
the
bacterium,
dis
non-tumbling
E.
coli
as
they
swam
through
a
suspenof
the
cell
body
(Fig.
1D)
illustr
walls,
we
dividual
bacteria,
tracked
gfp- overforce
the measured
andaverage
best-fit
dipole we
field
The
the 1C).
specific
fitting
and performing
anwe
ensemble
all tracers,
re- (Fig.
Howeve
sionswam
ofdecays
fluorescent
tracer
particles.
For
measurements
farcharacteristic
from
field
displays
the
1
dipole
length
ℓ =o
of
the
flow
speed
u
with
distance
r
from
thesurfaces
center
i
as
they
through
a
suspensolved
the
time-averaged
flow
field
in
the
E.
coli
swimming
he minismeasure
walls,
we
focused
on
a
plane
50
µm
from
the
top
and
bottom
value
of
F
is cons
movie
dat
However,
the
force
dipole
flow
sig
oftothe
cell
(Fig.
1D) illustrate
that
the
flow
down
0.1%
of body
the mean
swimming
speed V0 =
22 ±
5 measured
ticles.
For
measurements
far
from
ckedplane
gfpcell force
bod
surfaces
of
the
sample
chamber,
and
recorded
∼
2
terabytes
of
2
and
resistive
µm/s.
As E.
colidisplays
rotate
about
their
swimming
direction,
their
cells
field
the
characteristic
1/r4 decay
oftoa the
forceside
dipole.
measured
flow
ofswam
thel
a 50
suspenµmmovie
from
the
top
and
bottom
for
the
data.
In field
this in
data
we dimensions
identified ∼is10cylindrically
rare events when
note that
in
the b
time-averaged
flow
three
fluid
trace
However,
the
force
dipole
flow
significantly
overestimates
the
cell1.5body,
where
the
flow
magnit
far from
achieved
ber,
and
recorded
∼
2
terabytes
of
cells
swam
in
the
focal
plane
for
>
s.
By
tracking
the
µm behind the ce
symmetric.measured
Our
measurements
capture
all
components
of
this
flow to when
the side of
the
cell
body,ofand
behind
the
4
and
veloci
didentified
bottom
for
the
length
the
flagellar
bun
at
varia
fluid
drag
on
the
∼
10
rare
events
fluid
tracers
in
each
of
the
rare
events,
relating
their
position
cylindrically symmetric flow, except the azimuthal flow due to
cell
body,
where
the
flow
magnitude
u(r)
is
nearly
constant
and
perfo
abytes
of
field
(r f
achieved
by
fitting
two
opposite
and
velocity
to
the
position
and
orientation
of
the
bacterium,
the rotation
of the
cell
about its the
body axis. The topology of
ne for
> 1.5fors.
By
tracking
the
length
of the
flagellar
bundle.
The
force
dipole
fit
was
solved
the
nts
when
the
spec
the
measured
flow
field
(Fig.
1A)
is
the
same
as
that
of
a
and
performing
an
ensemble
average
over
all
tracers,
we
reat
variable
locations
along
the
sw
are events, relating
their
position
Bacterial
flowdow
fiel
achieved
by1B),
fitting
two
opposite
force
monopoles
(Stokeslets)
dipole
le
plane
kingforce
the
dipole
flowtime-averaged
(Fig.
defined
by
solved
the
flow
field
in
the
E.
coli
swimming
field
(r
>
8
µm).
From
the
best
and
orientation
of
the
bacterium,
dipole
flow
descri
at
variable
locations
along
the
swimming
direction
to
the
far
value
of
position
µm/s.
As
plane down to 0.1% of the mean swimming
speed
V
=
22
±
5
0
the
specific
fitting
routines
fi
with good and
accura
h
i
e
average
over
all
tracers,
we
refield
(r
>
8
µm).
From
the
best
fit,
which
is
insensitive
to
and resi
A E. coli
r direction, their time-avera
ℓF swimming
acterium,
µm/s.
As
2rotate about their
ˆ
thisµm
approximatio
u(r)in= the
3(r̂.coli
d) −
1 r̂, routines
A=
, r̂fitting
= length
, regions,
[ℓ
1 ]= we
dipole
1.9
and
dip
ow
field
E.
swimming
2
the
specific
fitting
and
obtain
the
note
tha2
|r|
8πηdimensions
|r| is cylindrically
s, we retime-averaged
flow field in three
symmetric
a
wall.
Focusing
value
F is Fconsistent
with
opt
dipole
length
1.9±µm
and
dipole
force
= of
0.42
pN.
This
µm
beh
mean
swimming
speed
V0 ℓ==22
5 capture
symmetric.
Our
measurements
allof
components
this
wimming
and applying
the
cylindrica
and
resistive
force
theory
calculat
fluid
dra
of
Fforce,
is consistent
with
optical
trap
measurements
[45]
where
F isvalue
the dipole
ℓ the
distance
separating
the force
ut
direction,
their
symmetric
flow,
except
the
azimuthal
flow due
to the
= their
22
±cylindrically
5swimming
resulted
in
a
sligh
rotati
note
thatThe
in
the
best
fit,
the
cell
and
resistive
force
theory
calculations
[46].
It is
interesting
to
η the
viscosity
the
fluid,
dˆ the
orientation
vector
theof
flow
field
struc
the
rotation
ofisof
the
cell
about
itsunit
body
axis.
topology
on, pair,
their
three
dimensions
cylindrically
the measu
(swimming
direction)
the best
bacterium,
and
rbehind
the
distance
surfaces,
the
note
thatflow
inofthe
fit, 1A)
theµm
cell
drag
Stokeslet
isfrom
0.1
the measured
field
(Fig.
is the
same
as that
oflocated
aof
the
center
the
cell
ndrically
nts
capture
all
components
of
this
force
dipo
Bacteria
vector
relative
to
the
center
of
the
dipole.
Yet
there
are
some
ity
of
a
no-slip
sur
µm
behind
the
center
of
the
cell
body,
possibly
reflecting
the
force
dipole
flow
(Fig.
1B),
defined
by
fluid
drag
on
the
flagellar
bundle
tsexcept
of
this
flow
due
Dunkel,
Ganguly,
Cisneros,
Goldstein
(2011) to
PNAS
Fig. differences
1.Drescher,
Averagethe
flow
fieldazimuthal
created
by a single
freely-swimming
bacterium.
(A)
Experimentally measured flow field far from a surface. Stream lines indicate local
direction offl
dipole
close
to
the
cell
body
as
shown
by
the
residual
of
outward
streamlin
fluid
drag
on
the
flagellar
bundle.
flow. (B) Best fit force-dipole model, and (C) residual flow field, obtained by subtracting the best-fit dipole from the experimentally measured field. The presence of the flagella
E.coli
w due to
weak ‘pusher’ dipole