Lecture 12. Life in the low
Reynolds-number world
Zhanchun Tu (涂展春 )
Department of Physics, BNU
Email: [email protected]
Homepage: www.tuzc.org
Main contents
● Friction in fluids
● Low Reynolds-number world
● Biological applications
§12.1 Friction in fluids
Suspension ( 悬浮液 )
Suspension of globular protein
0
z
mg
w g V
Gravity: mg Buoyant force: w g V
Net force: m−w V g≡mnet g
Profile of particle density
c z =c0 e−
mnet g z
k B T r
[sedimentation ( 沉降 ) equilibrium in gravity]
Myoglobin (肌红蛋白 ): mnet
≈4 kg/mol
z∗≡k B T r /mnet g≈60 m
In a 4cm test tube c z /c0=e−0.04 /60=0.999
Thus the suspension never settles out (沉降 ). Called colloid
Homework: Please estimate the thickness of PM2.5 in our atmosphere
Centrifuge (离心机 )
r
ω
m
Centripetal (向心 ) acceleration: 2 r
Equivalence principle
Centrifugal (离心 ) force f =m2 r
Centrifugal potential
U r =−∫ f dr=−12
m2 r2
Boltzmann law==> c r =c0em
2 r2/2 k BT r
(sedimentation equilibrium in centrifuge)
Problem: derive the above equation from Nernst-Planck formula
Sedimentation time scale(沉降系数 )● Sedimentation velocity
v drift=mnet g
Depends on g
NOT intrinsic property of particles
● Sedimentation time scale
s=vdrift / g=mnet /
unit: Svedbergs (10-13s)
Problem: Find the relation between s and the mass of an ideal polymer
mnet=m−w V
2Rg
s=mnet /
V ∝N : number of monomers⇒mnet∝m
=6w Rg
Rg2=⟨ 1
N ∑n=1
N
Rn−Rc2⟩=⟨ 1
N ∑n=1
N
Rn−1N ∑
m=1
N
Rm2
⟩=
1
2 N 2∑n ,m
N
⟨Rn−Rm2⟩
Gaussian chain: ⟨Rn−Rm2⟩=∣n−m∣b2
⇒ Rg2=Nb2
/6 for large N
⇒ Rg∝N 1 /2∝m1 /2
⇒ s∝m0.5[Data: Meyerhoff & Schultz(1952)]
s∝m0.5
s∝m0.44
Hard to mix a viscous liquid● Experiment: ink in corn syrup (玉米糊 )
slowly
● Question:
2nd law==>mix. Is 2nd law violated in the above experiment?
Einstein relation
Stokes relation
D=k B T
∝corn
D∝k BT
corn
∞⇒ D0Thus the blob of ink cannot change much in a short time due to diffusion, and no evident mixing phenomenon
Slowly Stirring (搅拌 ) causes an organized motion, in which successive layers of fluid simply slide over each other. Such a fluid motion is called laminar flow (层流 ).
§12.2 Low Reynolds-number world
Newtonian fluid
Uniform flow along x
x
y
zv y y
y+dy
τfdxdz v
vdv
Force applied on y+dy layer by y layer
f =−dvdy
(Newtonian viscous formula)
Coefficient of viscosity
Discussion: guess the above formula from simple analysis.
Viscous critical force● Dimensional analysis for Newtonian fluid
[η]: Pa s=kg m-1 s-1 [ρ]: kg m-3
Hence no intrinsic distinction between “thick” and “thin” fluids.
We cannot construct a dimensionless quantity from η and ρ.
We cannot construct a dimension of length from η and ρ.
Newtonian fluid “has no intrinsic length scale”, or is “scale invariant”.
Macroscopic experiments can tell us something relevant to a microscopic organism.
● Viscous critical force
[ 2
]: kg m−1 s−12
kg m−3 (dimension of force)
f crit=2/
If an external force f < fcrit
, the fluid can be
called “thick”. Friction will quickly damp out inertial effects. Flow is dominated by friction.
Typical scale of forces inside cells is in pN range.
Friction rules the world of the inner cell!
≃1 nN
= kg m s-2
Reynolds number
R
v
A ball moves in water
● Dimensional analysis
[η]: Pa s=kg m-1 s-1 [ρ]: kg m-3
[v]: m s-1[R]: m
ℜ≡ R v
[ R v ]: Dimensionless number
● Reynolds number
[Reynolds (1880s)]
● Physical meaning I
Time for a ball moving 2R: t=2 R /v
The same volume of water moves 2R in time t,the mean acceleration is estimated as
12
at2=2 R⇒ a=
4 R
t 2 =v2
R
~ R3v2
/R =R2 v2
R
v
inertial term of water= mass X acceleration
friction between the ball and water: f frict= v=6 R v~ R v
inertial termfriction
=Rv
=ℜ
The inertial term can be safely omitted if ℜ≪1
● Physical meaning II
R
v Friction between the ball and water
f frict= v=6 R v~ R v
Kinetic energy of water
E K~R3 v2
Dissipative work done by the friction in displacement R:
W frict= f frict R~ R2 v
EK
W frict
= Rv
=ℜ
● Physical meaning III
f frict~ R v
f crit=2/
f frict
f crit
~ Rv
=ℜ
ℜ≪1⇒ f frict= f ext (omit the inertial term)
Force applied on water by the ball,also=the external pulling force on the ball
Thus f ext≪ f crit for ℜ≪1
● Problem: Calculate the Reynolds numbers – (1) A 30 m whale, swimming in water at 10m/s
– (2) A 1 μm bacterium, swimming at 30 μm/s
Solutions:
(1) ℜ=Rv
=103
×30×10
10−3=3×108
≫1
(2) ℜ=Rv
=103
×10−6×30×10−6
10−3=3×10−5
≪1
Bacteria live in the world of low Reynolds number where the viscous friction rules their motions!
Time reversal properties
● Time reversal operation (t→-t)
solution Equation of motion
t→-t t→-t
Time reversaldynamical equation
Time reversalEquation of motion
solution
Dynamical equation
Time reversal Invariance
solution
● Example I: falling ball in gravity
mg
v0
z
0
Dynamical equation
d 2 z
dt 2=−g
Equation of motion
z t =v0t−12
g t2solution
Time reversaldynamical equation
d 2 zdt 2
=−g
Time reversalequation of motion
zrt =z −t
=−v0 t−12
g t 2
t→-tt→-t
solution
Time reversal Invariance!
solution
● Example II: diffusion equation
∂ c∂t
=D∂
2 c
∂ x2solution
t→-t
−∂c∂ t
=D∂
2 c
∂ x2
c x , t =1
4 Dte−x2
/4 Dt
solutioncrx ,t =c x ,−t
=1
i4 Dtex2 /4 Dt
t→-t
NOT time-reversal invariant
NOT a solution
● Viscous friction rule is not time-reversal invariant
R
A ball in highly viscous fluid
f
Time reversaldynamical equation −
dxdt= f
Dynamicalequation
dxdt
= f
NOT time-reversal invariant
Something irreversible in the motion ruled by friction!
Discussion: Does this conclusion contradict the experiment of ink drop in corn syrup?
● Newtonian fluid & simple elastic solid
f =−dvdy
=−ddy
dudt
Newtonian fluid
NOT time-reversal invariant!
Simple elastic solid
=−Gdudy
u
x
y
v
x
y
Time-reversal invariant!
(displacement)
(Hooke relation)
§12.3 Biological applications
Swimming of microorganisms● Strictly reciprocating (往复 ) motion
a. the paddles (桨 ) move backward in speed v relative to the body==>the forward motion of the body in speed u relative to water.
b. the paddles move forward in speed v' relative to the body==>==> the backward motion of the body in speed u' relative to water.
c. repeating a.
Time interval t Time interval t'
Friction force of body from water:
f b=−0u
Speed of paddles relative to water:
u p=−vu
Friction force of paddles from water:
f p=−1 u p=−1u−v
Force balance:
f b f p=0⇒u=1 v /01
Displacement of body after time t: x=u t=1 vt /01
Friction force of body from water:
f b'=−0−u ' =0u '
Speed of paddles relative to water:
u p'=v '−u ' =v '−u '
Friction force of paddles from water:
f p'=−1 u p
'=−1v '−u '
Force balance:f b
' f p
'=0⇒u '=1v ' /01
Displacement of body after time t':
x '=−u ' t '=−1 v ' t ' /01
Net displacement of the body in a period t+t':
Constraint of relative speed in a period t+t':
the paddles should return to their original positions on the body
vt−v ' t '=0
x x '=1 vt /01−1 v ' t ' / 01
=1vt−v ' t ' /01=0Scallop theorem: Strictly reciprocating motion won’t work for swimming in the low-Reynolds world [Purcell (1977) Am. J. Phys.]
The required motion must be periodic, so that it can be repeated.
Question: What other options does a microorganism have?
It can’t be of the reciprocal type described above.
● Ciliary propulsion (纤毛推进 )
Many cells use cilia to generate net thrust (推力 ). Each cilium contains internal filaments and motors which create an overall bend in the cilium.
Periodic but NOT reciprocal!
有效行程 复位行程
Typical motion of cilia.
Understand intuitively the trick to generate the net motion
v
u
v'
u'time t
Quickly retractpaddles
and then
time t'Quickly extendpaddles
Key: viscous friction coefficient of paddles are proportional to their lengths
net displacement=[1/01−1'/01
']vt
=0 vt 1−1'/0101
'0
(Homework?)
● Bacterial flagella
衬套
推进器
钩
螺栓
55nm
16nm
41nm
Peptidoglycan(肽聚糖 )
Mechanism of flagellar propulsion (鞭毛推进机理 )
v v
dfdf
Key: although v is perpendicular to z, df can have z component. Why?
v
v┴
v║
f┴=-ζ┴v┴
f║ =-ζ║v║
f
Generally ,⊥∥
only if ⊥=∥⇒ f = f ⊥ f ∥ antiparallel v
⇒ f = f ⊥ f ∥ not antiparallel v
Discussion: prove that for a long flagellum
∫ d f ∥z
Discussion: There exists torsion around z although the total force along z. How can bacteria resist rotating themselves around z axis?
Option 1: the bacteria are rough enough (i.e., large surface area), such that
rotation is verylarge⇒bacteria=torsionrotation
0
Option 2: the bacteria have even number of flagella. A pair of flagella have the opposite rotation. The torsion is canceled out while force
along z axis is double.
Option 3: ...... (maybe)
● Dimer (二聚体 ) of two reciprocating motion
Single pair of paddles
Strictly reciprocal
=>No net motion
Dimer of 2 pairs of paddles
Single pair: strictly reciprocal
Dimer: nonreciprocal
=> Net motion is possible
No many-scallop theorem [Lauga and Bartolo (2008) PRE]
Blood flow in capillary (毛细血管 )● Model
blood vessel
ℜ=Rv
=103
×10−5×0.1
10−3=1
≃103 kg/m3 R≃10 m
v≃10 cm /s ≃10−3 Pa s
Reynolds's study suggests
laminar pipe flow for ℜ103
Profile of v(r) Constraints:
v R=0 non−slip
v 0∞
● Force balance equation
Pressure contributes a force: df p=2 r dr p
Viscous force from inner layer fluid drags forward the fluid shell:
df out=[2rdr L]dv r '
dr ' ∣r '= rdr
df in=−[2 r L]dv r
dr
Viscous force from outer layer fluid pulls backward the fluid shell:
= [2rdr L][ dv r dr
d 2v r
dr2dr]
df pdf indf out=0⇒rp L
dvdr
rd 2 v
dr2=0
note: dv /dr0L---length
● Flow profile and flow rate
Problem: prove that the general solution is
v r=AB ln r−r2 p /4 L
Problem: prove that
B=0 and A=pR2/4 L
Flow profile
v r=p R2
−r2
4 L
Flow rate (Hagen–Poiseuille relation)
Q=∫ v dA=2∫ v r rdr= R4 p8 L
∝R4
Blood vessels can efficiently regulate flow with only slight dilations or contractions!
前导链后随链
Viscous force at DNA replication fork
● Question on DNA replication
Since the two single strands cannot
pass through each other, the original
must continually rotate (arrow).
Would frictional force resisting this
rotation be enormous?
Y-shaped junctionω
2R
● Estimate the effect of friction
R
r+dr
r
v(r)
ω
Constraints:
v R=R and v ∞=0
Torsion from the fluid inner r that drags the fluid shell rotating counter-clockwise:
T in=−dv r
dr2 rL r=−2 Lr2 dv r
dr
Note: dvdr
0
Torsion from the fluid inner r+dr that drags the fluid shell rotating clockwise:
T out=2 L rdr 2 dv r '
dr ' ∣r '= rdr
T inT out=0⇒dr2 dvdr =0 ⇒ v r=
R2
rand
dvdr
=−R2
r2
T R=−2 L r2 dv rdr ∣r=R
=2 LR2
Friction torsion between the rod and fluid:
Book: 4π, which is correct?
The rod rotates 2π for opening each helical turn. The work of friction
W frict=T R×2=42 LR2
DNA polymerase synthesizes new DNA in E. coli at a rate ~1000bp/s
=2turn
×1000 bp/s
10.5 bp/ turn≈600 rad /s
⇒W frict=42 LR2
≈2.4×10−22 J=0.06 k B T r
10-3 Pa s <~10μm 1nm
Problem: please estimate R.
● DNA helicase (解旋酶 )
Helicase walks along the DNA in front of the polymerase,
unzipping the double helix as it goes along.
1 ATP ≡ 20 kBT
r
Energy source: ATP etc.
W frict=0.06 k B T r≪20 k B T r
The energy lost of viscous friction can be negligible!
(in physiological condition)
Cytoplasmic streaming (胞浆流动 )
Chara corallina珊瑚轮藻
leaf
node
internodalcell
an internodal cell
close-up of indifferent zone
chloroplasts
chloroplasts
close-up of indifferent (中性 ) zone (3D view)
endoplasm
indifferent zone
indifferent zone
vacuole
[PNAS 105(2008) 3663]
● Intracellular (细胞内 ) material transport: Flow dominate or diffusion dominate?
Characteristic length: L Characteristic velocity of flow: U
Diffusion constant: D
⟨ x2⟩=2 D t⇒Time for molecules diffusing L: tD=
L2
2 D≃
L2
D
Time for molecules reach L due to flow: t f=LU
Peclet number: Pe=t D
t f
=ULD
If Pe >>1, molecules reach L more quickly due to flow than diffusion.
Thus, flow dominates the transport.
If Pe <<1, then diffusion dominates the transport.
In an internodal cell: L=1mm, U=30 μm/s, D=10-5 cm2/s.
Pe=ULD
=30≫1⇒ flow dominates the transport.
● Physical description
Flow of streaming: ∇2u=∇ p , ∇⋅u=0
Fluid velocity
The change of concentration of materials:
c tu⋅∇ c=D∇2 c (advection-diffusion equation)
Note: if you stand a frame S' with the velocity of u, you will observe
c t '=D ∇ ' 2c
For large Peclet number, diffusion can be neglected, thus
c tu⋅∇ c=0
(stokes equation)
(advection equation)
If only we have solved the stokes equation and obtain u(x,y,z,t),
then we can obtain c(x,y,z,t).
● Spiral flow
u=v r ,HJ r ,
Parallel to sprial
In radial direction
Flow stems from surface, then passes the inner (vacuole), and reaches the other points in surface again. Helpful to material transport across the vacuole.
pitch/R3
This ratio corresponds to max nutrient (营养 ) uptake rate from environment
Max at pitch/R=3
§ Summary & further reading
Summary● Suspension
– “Suspension never settles out (沉降 )”
– Sedimentation time scale: s=mnet /
● Viscous liquid– Hard to mix a viscous liquid
– Friction rules the world of the cell!
– Critical viscous force:
● Reynolds number:– inertial term / friction
– kinetic energy/dissipative work of friction
– external force/critical viscous force
ℜ= Rv/
f crit=2/
● Time reversal properties– 2nd Newtonian law: time-reversal invariant
– Viscous friction rule: NOT time-reversal invariant
– Simple elastic solid: time-reversal invariant
– Viscous fluid: NOT time-reversal invariant
● Swimming of microorganisms: – Strictly reciprocating motion won’t work
– Periodic but NOT reciprocal motion can work: Ciliary propulsion & flagellar propulsion
● Blood flow in capillary– Flow rate:
– Blood vessels can efficiently regulate flow with only slight dilations or contractions
Q= R4 p8 L
● DNA replication fork– Dissipative work of friction:
– Energy lost of viscous friction can be negligible:
W frict=42 LR2
≈0.06 k BT r
W frict≪EATP≈20 k B T r
● Cytoplasmic streaming– Peclet number:
– Physical description:
Pe=ULD
∇2u=∇ p , ∇⋅u=0
c tu⋅∇ c=D∇2 c
Further reading
● E. M. Purcell, Life at low Reynolds number, Am. J. Phys. 45, 3 (1977)
● H. C. Berg, Motile Behavior of Bacteria, Physics Today, January (2000)
● R. E. Goldstein, I. Tuval, and J. van de Meent, Microfluidics of cytoplasmic streaming and its implications for intracellular transport, PNAS 105, 3663 (2008)