The swimming of sperm and other animalcules
B. U. Felderhof,
RWTH Aachen
earlier work (1994) with R. B. Jones, QMC, London
„animalcules“ = an animal so minute in its size, as not to be the
immediate object of our senses
(Encyclopaedia Britannica, first edition, 1771)
=a microscopic animal
(Mrs. Byrne‘s Dictionary of Unusual, Obscure,
and Preposterous Words, 1974)
(Concise Oxford Dictionary, 1974)
discovered by
Antonie van Leeuwenhoek, born Oct. 1632, Delft
died Aug. 1723, Delft
cf.
Johannes Vermeer, born Oct. 1632, Delft
died Dec. 1675, Delft
Antonie van Leeuwenhoek discovered
sperm (about 50 microns long)
protozoa (unicellular)
bacteria (about 1 micron)
reproduce by binary fission
many of these microorganisms move in water by swimming
1-4 van Leeuwenhoek`s sperms
5-8 his dog‘s sperms
In second order perturbation theory one finds steady swimming
velocities
r
U 2 for translation
Ω2
and
rotation.
Simplification: Low Reynolds number hydrodynamics
and point approximation
In low Reynolds number hydrodynamics we can omit the terms
ρ
∂v
∂t
v ⋅∇v
and
and use the Stokes equations
η∇ 2 v − ∇p = − F
∇⋅v = 0
In point approximation
N
F = ∑ K jδ (r − R j )
j =1
We consider polymer structures consisting of N beads centered
at time t at positions
(R1 (t ),..., R N (t ))
{
}
K j (t )
Forces
periodic in time
with period T
Since the equations do not involve a time derivative the forces
determine fluid velocity and pressure instantaneously at any time t.
7
N
Fluid velocity
v (r ) = ∑ T(r − R j ) ⋅ K j
j=1
ˆˆ
1 1 + rr
T(r ) =
8πη r
with Oseen tensor
The instantaneous particle velocities are
N
u j = µ j K j + ∑ T( R j − R k ) ⋅ K k
( j = 1,..., N )
k≠ j
µj =
µj
mobility of bead j
Stokesian dynamics: dR j
1
6πη a j
= u j (R1 (t ),..., R N (t ))
dt
( j = 1,..., N )
The total force is required to vanish
N
∑K
j
=0
j=1
We write the bead positions as a sum of two terms
R j (t ) = S j (t ) + ξ j (t )
where the positions {S j (t )} describe the mean swimming motion
with constant translational velocity U
r
and rotational velocity
Ω
The positions {S j (t )} are solutions of the equations of motion
r
dS j (t )
= U + Ω × (S j (t ) − C(t ))
( j = 1,..., N )
dt
with initial positions
{S
0j
N
= S j (0)} centered at
N
C0 = ∑ a j S 0 j / ∑ a j
j =1
The center of resistance
j =1
C(t )
moves with constant velocity U
C(t ) = C0 + Ut
8
{
}
We require that the displacements ξ j (t )
are periodic in time,
and that their average over a period vanishes
ξj
T
1
= ∫ ξ j (t ) dt = 0
T 0
( j = 1,..., N )
K1 ,..., K N
To first order in the forces
the particle velocities are given by
N
u (t ) = ∑ µ jk ⋅ K k (t )
(1)
j
where
{µ }
jk
k =1
is the mobility matrix for the static structure
(S 01 ,..., S 0 N )
t
By integration over time
ξ j (t ) = ∫ u (1)
j (t ') dt '
Because of the displacements of the beads there is a second order
correction to their velocities given by
N
δ u (t ) = ∑ ξ (1)
j β ( t )Gβαγ (S 0 j − S 0 k ) K kγ ( t )
(2)
jα
( j = 1,..., N )
k≠ j
where G(r ) is the third rank tensor
G (r ) = −
∂
T (r )
∂r
(2)
The corresponding second order flow field δ v (r, t )
can be viewed as being generated by induced forces
that can be calculated from
{δ F
(2)
j
(t )}
N
δ F (t ) = ∑ζ jk ⋅ δ u(2)
k (t )
(2)
j
k =1
Since there is no flow of momentum or angular momentum to infinity,
the polymer must move as a whole such that the actual second order bead
velocities are
r
(2)
( j = 1,..., N )
u(2)
(t ) + ω (2) (t ) × (S 0 j − C0 ) + δ u (2)
j (t ) = u
j (t )
r
with velocities u (2) (t ) and ω (2) (t ) such that the total induced force
and torque vanish.
9
On time average this implies that the swimming velocities
r (2)
r
(2)
(2)
Ω
= ω (2) (t )
U = u (t )
are given by
t
t
U (2) = µ tt ⋅ F St + µ tr ⋅ T St
r (2) t rt St t rr St
Ω = µ ⋅F + µ ⋅T
with Stokes force and torque
N
T = − ∑ ((S0 j − C0 ) × δ Fj(2) )
N
F St = − ∑ δ Fj(2)
St
j =1
j =1
The rate at which energy is dissipated equals
N
D (t ) = ∑ K j (t ) ⋅ u j (t )
j =1
N
D (t ) = ∑ K j (t ) ⋅ u(1)
j (t )
(2)
To second order in the forces
j =1
Average over a period
D (t ) =
(2)
N
∑K
j =1
j
(t ) ⋅ u (1)
j (t )
Define dimensionless efficiencies of swimming as the ratios
ET (ω ) = ηω L2
U (2)
D
(2)
E R (ω ) = ηω L3
r
Ω(2)
D (2)
where L is the size of the polymer.
Then one can compare efficiencies of different strokes.
10
b
Longitudinal mode for structure
Forces
b
1 d
2 d
a
3
K1 (t ), K 2 (t ) e z , K 3 (t ) = −K1 (t ) − K 2 (t )
All motion along z-direction.
z
K1 (t ) = A sin ωt
K 2 (t ) = A sin(ωt + α )
K 3 (t ) = − K1 (t ) − K 2 (t )
Optimal motion in +z-direction for α ≈ −
u1(1)z =
u2(1)z =
u3(1)z
δ u1(2)
z =
δ u2(2)z =
δ u3(2)z =
U
(2)
1
16πη d 2
1
4πη d 2
1
24πη bd
2π
3
[(4d − 3b) K1 + 3bK 2 ]
1
(2d − 3b) K 2
12πη bd
−1
=
[(4d − 3a) K1 + (4d − 6a) K 2 ]
24πη bd
[(ξ1z − ξ3 z ) K1 − (3ξ1z − 4ξ 2 z + ξ3 z ) K 2 ]
[(−ξ1z + 2ξ 2 z − 3ξ3 z ) K1 + (ξ 2 z − ξ3 z ) K 2 ]
1
16πη d 2
[(−ξ1z + ξ3 z ) K1 − 4(ξ2 z − ξ3 z ) K 2 ]
− A2 sin α 4d (56d 2 − 174bd + 135b 2 ) − 3a (88d 2 − 270bd + 189b 2 )
=
192π 2η 2 d 3ω
16bd (2d − 3b) + a(16d 2 − 72bd + 63b 2 )
Ω (2) = 0
D
(2)
A2
=
[ 4(a + b)d − 9ab + b(4d − 3a) cos α ]
24πη abd
α 0 = − arccos
b(4d − 3a)
4(a + b)d − 9ab
11
b
Transverse mode for structure
b
1 d
a
2 d
x
3
z
K1 (t ), K 2 (t ) e x , K 3 (t ) = −K1 (t ) − K 2 (t )
Forces
First order motion along x-direction.
K1 (t ) = A sin ωt
K 2 (t ) = A sin(ωt + α )
K 3 (t ) = − K1 (t ) − K 2 (t )
2π
Optimal motion in +z-direction for α ≈
u1(1)x =
1
3
[ (8d − 3b) K1 + 3bK 2 ]
48πη bd
1
(4d − 3b) K 2
u2(1)x =
24πη bd
−1
u3(1)x =
[(8d − 3a) K1 + (8d − 6a) K 2 ]
48πη ad
δ u1(2)
z =
δ u2(2)z =
1
32πη d 2
1
8πη d 2
δ u3(2)z =
U
(2)
[(−ξ1x + ξ3 x ) K1 + (3ξ1x − 4ξ2 x + ξ3 x ) K 2 ]
[(ξ1x − 2ξ2 x + 3ξ3 x ) K1 − (ξ2 x − ξ3 x ) K 2 ]
1
32πη d 2
[ (ξ1x − ξ3 x ) K1 + 4(ξ 2 x − ξ3 x ) K 2 ]
A2 sin α 2d (224d 2 − 534bd + 297b 2 ) − 3a (164d 2 − 360bd + 189b 2 )
=
768π 2η 2 d 3ω
16bd (2d − 3b) + a (16d 2 − 72bd + 63b 2 )
Ω (2) = 0
D
(2)
A2
=
[8(a + b)d − 9ab + b(8d − 3a) cos α ]
48πη abd
α 0 = arccos
b(8d − 3a )
8(a + b)d − 9ab
12
Similarly for longer linear chains.
Sperm can be modelled as a head of radius a
followed by a tail of beads of radius b.
Instead of specifying N periodic forces one can start by specifying
N-1 first order displacements and calculate the forces and the N-th
displacement from
N
N
K j (t ) = ∑ ζ jk uk(1) (t )
k =1
and
∑K
j =1
j
(t ) = 0
In 3D situations one can specify N-2 displacement vectors
and calculate forces and the last two displacements
from the friction matrix and the condition that total force and torque
vanish.
For a body one can specify shapes S0 and S(t).
Such calculations were performed in
B.U. Felderhof and R. B. Jones, Physica A 202, 94 (1994)
We also considered effect of inertial terms
ρ
∂v
∂t
and
ρ v ⋅∇v
The special case of S0= sphere was studied in
B.U. Felderhof and R. B. Jones, Physica A 202, 119 (1994)
13
Longitudinal
Transverse
MOVIE1