Download Lecture 12. Life in the low Reynolds

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 )
Gravity:
w g V
z
mg
Buoyant force:  w g V
Net force: m−w V  g ≡m net g
Profile of particle density
mg
c  z =c 0 e
m net g z
−
kBT r
[sedimentation ( 沉降 ) equilibrium in gravity]
0
Suspension of globular protein
In a 4cm test tube
Myoglobin ( 肌红蛋白 ): mnet≈4 kg/mol
z ∗≡k B T r / mnet g≈60 m
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 )
Centripetal ( 向心 ) acceleration:
ω
2
 r
Equivalence principle
m
Centrifugal ( 离心 ) force
f =m  r
Centrifugal potential
r
Boltzmann law==>
1
2 2
U r =−∫ f dr =− m r
2
c r =c 0 e
2
2 2
m  r /2 k B T r
(sedimentation equilibrium in centrifuge)
Problem: derive the above equation from Nernst-Planck formula
Sedimentation time scale( 沉降系数 )
●
Sedimentation velocity
m net g
v drift =

●
Depends on g
NOT intrinsic property of particles
Sedimentation time scale
s=v drift / g =m net /
unit: Svedbergs (10-13s)
Problem: Find the relation between s and the mass of an ideal polymer
s=m net /
=6   w R g
mnet =m−w V
2Rg
〈
1
R =
N
2
g
V ∝ N : number of monomers
N
〉 〈 ∑
1

R
−
R

=
∑ n c
N
n=1
2
N
N
n=1
1
Rn −
N
2
1
=
R
−
R
n
m
2∑ 
2 N n ,m
2
⇒ Rg ∝ N
1/ 2
∑ Rm
m=1
〈  R − R  〉=∣n−m∣b
n
〉
2
s ∝m0.44
〉
2
Gaussian chain:
2
〈
N
⇒ m net ∝m
m
2
s ∝m
0.5
⇒ R g = Nb /6 for large N
1/ 2
∝m
⇒ s ∝m
0.5
[Data: Meyerhoff & Schultz(1952)]
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  D=k B T
Stokes relation
∝ corn
k BT
D∝
 corn
  ∞ ⇒ D 0
Thus 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
y
y+dy
x
v  y
z
Uniform flow along x
dv
 f =−
dy
y
v dv
τfdxdz
v
Force applied on y+dy layer by y layer
(Newtonian viscous formula)
Coefficient of viscosity
Discussion: guess the above formula from simple analysis.
Viscous critical force
●
Dimensional analysis for Newtonian fluid
-1
[η]: Pa s=kg m s
-1
[ρ]: kg m-3
We cannot construct a dimensionless quantity from η and ρ.
Hence no intrinsic distinction between “thick” and “thin” fluids.
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
[ ]
−1 −1 2
2
kg m s 

-2
:
=
kg
m
s
−3

kg m
2
f crit = /
(dimension of force)
If an external force f < fcrit, the fluid can be
called “thick”. Friction will quickly damp out
inertial effects. Flow is dominated by friction.
≃1 nN
Typical scale of forces inside cells is in pN range.
Friction rules the world of the inner cell!
Reynolds number
●
R
v
A ball moves in water
●
Dimensional analysis
[η]: Pa s=kg m-1 s-1
[ρ]: kg m-3
[R]: m
[v]: m s-1
[ ]
 Rv
: Dimensionless number

Reynolds number
 Rv
ℜ≡

[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
R
v
2
1 2
4R v
at =2 R⇒ a= 2 =
2
R
t
inertial term of water= mass X acceleration ~ R
friction between the ball and water:
f
frict
3
2
2
v / R= R v
2
= v=6   R v~ R v
inertial term  Rv
=
=ℜ

friction
The inertial term can be safely omitted if ℜ≪1
●
Physical meaning II
Kinetic energy of water
3
E K ~ R v
R
v
2
Friction between the ball and water
f
frict
= v=6   R v~ R v
Dissipative work done by the friction in displacement R:
W frict = f
2
frict
R~ R v
EK
 Rv
=
=ℜ

W frict
●
Physical meaning III
f
frict
~ R v
2
f crit = /
ℜ≪1 ⇒ f
frict
= f ext
f frict  Rv
~
=ℜ

f crit
(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:
3
(1)
 Rv 10 ×30×10
8
ℜ=
=
=3×10
≫1
−3

10
(2)
 Rv 10 ×10 ×30×10 
−5
ℜ=
=
=3×10
≪1
−3

10
3
−6
−6
Bacteria live in the world of low Reynolds number
where the viscous friction rules their motions!
Time reversal properties
●
Time reversal operation (t→-t)
Dynamical equation
t→-t
Time reversal
dynamical equation
Time reversal Invariance
solution
so
lu
ti o
n
solution
Equation of motion
t→-t
Time reversal
Equation of motion
●
Example I: falling ball in gravity
z
Dynamical equation
solution
2
d z
=−g
2
dt
v0
0
t→-t
mg
Equation of motion
so
l
Time reversal
dynamical equation
2
d z
=−g
2
dt
Time reversal Invariance!
ut
io
n
solution
1 2
z t =v 0 t− g t
2
t→-t
Time reversal
equation of motion
z r t=z −t 
1 2
=−v0 t− g t
2
●
Example II: diffusion equation
2
∂c
∂ c
=D 2
∂t
∂x
t→-t
1
−x /4 Dt
c  x , t=
e
 4  Dt
2
solution
NO
T
a
so
l
ut
t→-t
io
n
2
∂c
∂ c
− =D 2
∂t
∂x
solution
NOT time-reversal invariant
c r  x ,t =c  x ,−t
1
x / 4 Dt
=
e
i  4  Dt
2
●
Viscous friction rule is not time-reversal invariant
Dynamical
equation
R
f
A ball in highly viscous fluid
dx
 =f
dt
NOT time-reversal invariant
dx
Time reversal
dynamical equation − dt = f
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
Simple elastic solid
du
=−G
dy
(Hooke relation)
u
y
Time-reversal invariant!
(displacement)
x
Newtonian fluid
 
dv
d du
 f =− =−
dy
dy dt
NOT time-reversal invariant!
v
y
x
§12.3 Biological applications
Swimming of microorganisms
●
Strictly reciprocating ( 往复 ) motion
Time interval t
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.
Time interval t'
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.
Friction force of body from water:
f b=−0 u
Speed of paddles relative to water:
u p =−vu
Friction force of paddles from water:
f p =−1 u p=−1  u−v
Force balance:
f b f p=0 ⇒ u=1 v /0 1 
Displacement of body after time t:
 x=u t=1 vt / 01 
Friction force of body from water:
'
b
f =−0 −u '=0 u '
Speed of paddles relative to water:
'
p
u =v ' −u ' =v ' −u '
Friction force of paddles from water:
'
p
'
p
f =−1 u =−1  v '−u ' 
Force balance:
'
b
'
p
f  f =0 ⇒ u ' =1 v ' /0 1 
Displacement of body after time t':
 x ' =−u ' t '=−1 v ' t ' /01 
Constraint of relative speed in a period t+t':
the paddles should return to their original positions on the body
vt−v ' t ' =0
Net displacement of the body in a period t+t':
 x x ' =1 vt / 01 −1 v ' t ' / 01
=1 vt−v ' t ' / 01 =0
Scallop theorem: Strictly reciprocating motion won’t work for
swimming in the low-Reynolds world [Purcell (1977) Am. J. Phys.]
Question: What other options does a microorganism have?
The required motion must be periodic, so that it can be repeated.
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.
有效行程
复位行程
Typical motion of cilia.
Periodic but NOT reciprocal!
Understand intuitively the trick to generate the net motion
and then
Quickly retract
paddles
v'
Quickly extend
paddles
time t
time t'
u'
u
v
Key: viscous friction coefficient of paddles
are proportional to their lengths
'
1
'
1
net displacement=[1 /0 1 − /0 ]vt
'
(Homework?)
'
=0 vt 1−1/ 01 01 0
●
Bacterial flagella
钩
55nm
推进器
衬套
16nm
Peptidoglycan( 肽聚糖 )
螺栓
41nm
Mechanism of flagellar propulsion ( 鞭毛推进机理 )
df
df
v
v
Key: although v is perpendicular to z, df can have z component. Why?
f
f┴=-ζ┴v┴
f║ =-ζ║v║
only if  ⊥ =∥⇒ f = f ⊥ f ∥ antiparallel v
Generally , ⊥ ∥
⇒ f = f ⊥  f ∥ not antiparallel v
v║
v┴
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
torsion
rotation is very large ⇒ bacteria =
0
rotation
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
Dimer of 2 pairs of paddles
Strictly reciprocal
Single pair: strictly reciprocal
=>No net motion
Dimer: nonreciprocal
=> Net motion is possible
No many-scallop theorem [Lauga and Bartolo (2008) PRE]
Blood flow in capillary ( 毛细血管 )
●
Model
3
≃10 kg/ m
R≃10  m
3
−3
v≃10 cm /s
≃10 Pa s
3
ℜ=
blood vessel
−5
 Rv 10 ×10 ×0.1
=
=1
−3

10
Reynolds's study suggests
laminar pipe flow for ℜ10
Profile of v(r)
Constraints:
v  R=0 non−slip 
v 0∞
3
●
Force balance equation
Pressure contributes a force:
df p = 2 r dr  p
Viscous force from inner layer fluid drags forward the fluid shell:
dv r 
df in =−  [2  r L]
dr
note : dv / dr0
L---length
Viscous force from outer layer fluid pulls backward the fluid shell:
∣
dv r ' 
df out = [2  rdr L]
dr ' r ' = rdr
[
2
dv  r  d v  r 
= [2 r dr  L]

dr
2
dr
dr
]
2
rp dv
d v
df p df in df out =0 ⇒
 r 2 =0
 L dr
dr
●
Flow profile and flow rate
Problem: prove that the general solution is
2
v  r= AB ln r −r p /4  L
Problem: prove that
2
B=0 and A= pR /4  L
Flow profile
2
2
p  R −r 
v  r=
4 L
Flow rate (Hagen–Poiseuille relation)
4
R p
4
Q=∫ v dA=2  ∫ v r rdr=
∝R
8 L
Blood vessels can efficiently regulate flow with only slight dilations or contractions!
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?
前导链
后随链
2R
ω
Y-shaped junction
●
Estimate the effect of friction
Constraints:
v  R= R and v ∞=0
v(r)
r+dr
Torsion from the fluid inner r that drags
the fluid shell rotating counter-clockwise:
r
R
dv r 
2 dv r
T in =−
2  rL  r=−2  Lr
dr
dr
ω
Note:
dv
0
dr
Torsion from the fluid inner r+dr that
drags the fluid shell rotating clockwise:
∣
dv r ' 
T out =2 L  rdr 
dr ' r ' = rdr
2
 
2 dv
T in T out =0 ⇒ d r
=0
dr
⇒ v  r=
R
r
2
and
dv
R
=− 2
dr
r
2
Friction torsion between the rod and fluid:

dv  r
T  R=−2   L r
dr
2
∣
2
=2   LR 
r =R
Book: 4π, which is correct?
The rod rotates 2π for opening each helical turn. The work of friction
2
2
W frict =T R×2 =4  LR 
DNA polymerase synthesizes new DNA in E. coli at a rate ~1000bp/s
=
2
1000 bp/ s
×
≈600 rad /s
turn 10.5 bp/ turn
10-3 Pa s
2
<~10μm
2
Problem: please estimate R.
1nm
−22
⇒ W frict =4  LR ≈2.4×10
J=0.06 k B T r
●
DNA helicase ( 解旋酶 )
Helicase walks along the DNA in front of the polymerase,
unzipping the double helix as it goes along.
Energy source: ATP etc.
1 ATP ≡ 20 kBTr
(in physiological condition)
W frict =0.06 k B T r ≪20 k B T r
The energy lost of viscous friction can be negligible!
Cytoplasmic streaming ( 胞浆流动 )
chloroplasts
leaf
indifferent zone
node
close-up of
indifferent zone
internodal
cell
chloroplasts
indifferent zone
endoplasm
vacuole
Chara corallina
珊瑚轮藻
an internodal cell
close-up of indifferent
( 中性 ) zone (3D view)
[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
2
2
L
L
2
t
=
≃
〈 x 〉=2 D t ⇒ Time for molecules diffusing L: D 2 D D
L
t
=
Time for molecules reach L due to flow:
f
U
t D UL
Peclet number: Pe = =
tf
D
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 =
UL
=30≫1 ⇒ flow dominates the transport.
D
●
Physical description
Flow of streaming:
2
 ∇ u=∇ p , ∇⋅u=0
(stokes equation)
Fluid velocity
The change of concentration of materials:
2
c t u⋅∇ c=D ∇ c (advection-diffusion equation)
Note: if you stand a frame S' with the velocity of u, you will observe
2
c t ' =D ∇ ' c
For large Peclet number, diffusion can be neglected, thus
c t u⋅∇ c=0 (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 ,  H J  r , 
Parallel to sprial
In radial direction
3
pitch/R
Flow stems from surface, then passes the inner (vacuole),
Max at pitch/R=3
and reaches the other points in surface again. Helpful to This ratio corresponds to
material transport across the vacuole.
max nutrient ( 营养 ) uptake
rate from environment
§ Summary & further reading
Summary
●
●
●
Suspension
–
“Suspension never settles out ( 沉降 )”
–
Sedimentation time scale: s=m net /
Viscous liquid
–
Hard to mix a viscous liquid
–
Friction rules the world of the cell!
–
Critical viscous force: f crit = / 
2
Reynolds number: ℜ= Rv/
–
inertial term / friction
–
kinetic energy/dissipative work of friction
–
external force/critical viscous force
●
●
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
 R4 p
Q=
8 L
–
Flow rate:
–
Blood vessels can efficiently regulate flow with only
slight dilations or contractions
DNA replication fork
2
2
–
Dissipative work of friction: W frict =4   LR ≈0.06 k B T r
–
Energy lost of viscous friction can be negligible:
W frict ≪ E ATP ≈20 k B T r
Cytoplasmic streaming
UL
number: Pe = D
–
Peclet
–
Physical description:
2
 ∇ u=∇ p , ∇⋅u=0
2
c t u⋅∇ c=D ∇ 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)