Download Introduction NOTES AND PROBLEM SET 1

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Oleg Krichevsky
Soft Matter Physics
I.
NOTES AND PROBLEM SET 1
due to June 7, 2009
Introduction
This course deals with Complex Fluids. Complex Fluids are typically solutions of some large
molecules (e.g. polymers) or supramolecular structures (e.g. micelles or bilayers, see below)
in ordinary liquids.
Thus ordinary liquids are the basis of Complex Fluids, and we will start by describing
the structure and dynamics of ordinary liquids.
This already is complicated enough,
since we can construct a good theory only when we have a small parameter and there is
no such parameter for description of the structure of liquids. In solids e.g., free energy
is dominated by the potential energy of interactions between molecules which favors
symmetric (crystal) structures, while entropy contribution causes only minor distortions
to the perfect symmetry (atom vibrations or phonons, vacancies etc). The behavior of
gases, vice versa is dominated by the kinetic energy and the entropy contributions with
interaction energy being a small parameter. However, in liquids none of the terms in the
free energy is negligible in comparison to other and thus there is no free parameter and
therefore no exact theory of liquids. There are ingenious guesses and we will get acquainted
with some of them and learn the basics of liquid theory.
Further on in the course we will populate the ordinary liquids with polymers and
other supramolecular structures and thus study the statistics and dynamics of Complex
Fluids.
Soap Molecules
Soap molecules can be described by a hydrophobic (oily) tail ( such as CH2 −CH2 −CH2 − )
and a hydrophilic head. The head can be hyrophilic because of dissociation of two ions - the
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Soft Matter Physics
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ionic bond electrostatic force is screened because of the water’s high dialectric constant and
this makes entropy gain the significant contribution to the free energy. (More on this below).
Therefore when we put soap molecules in water they will tend to fill the surface
with tails outside and heads in the water. This is called a monolayer :
Adding more soap after the air-water interface is full will give self assembled structures,
called spherical micelles where all the tails are inside (not touching the water) and the
heads our outside (in contact with the water):
However the setup above is not always possible. When the ’tail’ takes too much volume,
such as when there is more than one ’tail’, then the closed structures are not possible and
instead we get bilayers:
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These bilayers can be closed to form vesicles. Note that bilayers are self assembled, meaning
that they are equilibrium structures. Vesicles are not. However, once formed, vesicles are
very stable (the energy barriers to ”open” the vesicles into bilayers are large).
A.
StatMech basics
We consider classical systems only: those of N molecules in volume V with states defined
by {~pi , ~ri }N couples and the Hamiltonian
N
X
p2i
+ U (~r1 , ~r2 , ..., ~rN )
H({~pi , ~ri }N ) =
2m
i=1
(1.1)
We further assume pairwise, spherically symmetric interactions between molecules
u(~r1 , ~r2 ) = u(|~r2 − ~r1 |) = u(r12 ):
U (~r1 , ~r2 , ..., ~rN ) =
N X
N
X
i=1 j=i+1
N
u(rij ) =
1X
u(rij )
2 i,j=1
(1.2)
i6=j
One of the major goals of theory of liquids is to describe the structure and properties of
liquids from the knowledge of the average molecular density n = N/V and of the pair
interaction energy u(r).
Averages in StatMech. The ensemble average of any measurable quantity A is defined as:
Z 0
Z
1
hAi =
A(Λ)P (Λ)dΛ =
A({~pi , ~ri }N )P ({~pi , ~ri }N )d~p1 d~r1 ...d~pN d~rN , (1.3)
N !(2π~)3N
where P (Λ) ≡ P ({~pi , ~ri }N ) is the probability density to find the system in a given state, the
tag in the first integral reminds that only unique states in phase space Λ should be counted.
Changing to integration over d~p1 d~r1 ...d~pN d~rN in the second integral leads to counting each
Oleg Krichevsky
Soft Matter Physics
unique state N ! number of times, since molecules are indistinguishable. The factor
4
1
N!
takes
care of this multiple counting. The factor (2π~)3N ensures smooth transition to quantum
statistics.
Normalization of probability density:
Z
Z 0
1
P (Λ)dΛ =
P ({~pi , ~ri }N )d~p1 d~r1 ...d~pN d~rN = 1,
N !(2π~)3N
Entropy. The general definition for entropy S in any ensemble is:
Z 0
S = −hln P i = −
P (Λ) ln P (Λ)dΛ =
Z
1
=−
P ({~pi , ~ri }N ) ln P ({~pi , ~ri }N )d~p1 d~r1 ...d~pN d~rN ,
N !(2π~)3N
(1.4)
(1.5)
Energy E of the system is a shortcut for an average energy:
Z 0
Z
1
H(Λ)P (Λ)dΛ =
E = hHi =
H({~pi , ~ri }N )P ({~pi , ~ri }N )d~p1 d~r1 ...d~pN d~rN ,
N !(2π~)3N
(1.6)
Microcanonical ensemble is defined by E = Const, V = Const, N = Const. Main quantity
to calculate is the number of available states N :
N (E, V, N ) =
X
1
(1.7)
H({~
pi ,~
ri }N )=E
Probabilities of all available states are postulated to be equivalent. Then
S(E, V, N ) = lnN (E, V, N )
dS =
dU
τ
+ τp dV − µτ dN,
(1.8)
(1.9)
∂S
)V,N is the definition of thermodynamic temperature τ = 1/β = kB T .
where 1/τ = ( ∂E
The chemical potential µ is defined in the context of Grand Canonical ensemble and the
corresponding differential appears in Eq. 1.9 for completeness. All main thermodynamic
functions (including µ) can be found from the partial derivatives of S (Eq. 1.9).
Second Law of Thermodynamics. Entropy can only increase in a closed system and is at
maximum in thermodynamic equilibrium.
Canonical ensemble is defined by τ = Const, V = Const, N = Const. Main quantity to
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Soft Matter Physics
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calculate is the partition function ZN :
ZN (τ, V, N ) =
X
{~
pi ,~
ri }N
Z
1
=
e−βH({~pi ,~ri }N ) d~p1 d~r1 ...d~pN d~rN =
3N
N !(2π~)
Z
n3N
n3N
q
q
−βU (~
r1 ,...~
rN )
QN ,
(1.10)
e
d~r1 ...d~rN =
=
N!
N!
|
{z
}
−βH({~
pi ,~
ri }N )
e
QN
mτ 1/2
where Eq. 1.1 was substituted, integration over {d~pi } was performed, nq = ( 2π~
is
2)
quantum density and QN is configurational integral. Probability density of a microscopic
state {~pi , ~ri }N of the system is given by Boltzmann distribution:
P ({~pi , ~ri }N ) =
e−βH({~pi ,~ri }N )
ZN
(1.11)
Energy and free energy F = E − Sτ are determined from ZN through:
∂
E = − ∂β
ln ZN
(1.12)
F = −τ ln ZN
(1.13)
dF = −Sdτ − pdV + µdN
(1.14)
The rest of the thermodynamic functions can be found then from the partial derivatives of
F (Eq. 1.14). Free energy is at minimum in the system of constant volume and number of
molecules in thermodynamic equilibrium with heat reservoir.
Grand Canonical ensemble is defined by τ = Const, V = Const, µ = Const. Main quantity
to calculate is the grand partition function Ξ:
Ξ(τ, V, µ) =
X
e
β(µN −H({~
pi ,~
ri }N ))
=
∞
X
eβµN ZN
(1.15)
N =0
N,{~
pi ,~
ri }N
Probability density of a microscopic state (N, {~pi , ~ri }N ) of the system is given by Gibbs
distribution:
P (N, {~pi , ~ri }N ) =
eβ(µN −H({~pi ,~ri }N ))
Ξ
(1.16)
The average number of molecules hN i in the system and the grand thermodynamic potential
Ω = F − µN = −pV are determined from Ξ through:
∂
hN i = τ ∂µ
ln Ξ
(1.17)
Ω = −τ ln Ξ
(1.18)
dΩ = −Sdτ − pdV − N dµ
(1.19)
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Soft Matter Physics
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The rest of the thermodynamic functions can be found then from the partial derivatives of
Ω (Eq. 1.19). Grand thermodynamic potential is at minimum in the system of constant
volume in the thermodynamic equilibrium with a reservoir of heat and molecules.
B.
Molecular interactions
Short range repulsion at the distances ∼ molecular diameter σ due to the overlap of electron
orbits of the two molecules. The repulsion can be approximated by the hard core potential:

 0, r > σ
uhc (r) =
(1.20)
 ∞, r < σ
Long range attraction decaying roughly as ∝ r−6 stems from the dipole-induced dipole
interactions, called Van der Waals (VdW) interactions:
uV dW (r) = −
αd2
,
(4π0 )2 r6
(1.21)
where d~ is an instantaneous molecular dipole moment and α is molecular polarizability
~ a molecule assumes an induced dipole moment d~i =
(defined such that in external field E
~
αE).
We distinguish two types of molecular polarizability:
• The polarizability of non-polar molecules is determined by the distortion of electron
orbits in the external field (electronic polarizability). Within quasi-classical approach
the instantaneous dipole moment d~ can be associated with the instantaneous inhomogeneity of electron density: d ≈ eR, where R = σ/2 is molecular radius and e is
electron charge. Then the electronic polarizability is estimated as:
αel ∼ 4π0 R3
(1.22)
• Polar molecules (such as molecules of water, ethanol etc) have an additional type of
polarizability due to the preferential orientation of their native dipole moment d~ along
~ = hd~Ei/|
~ E|
~ (dipole polarizability):
the external field, so that d~i = hdi
αd =
d2
3τ
(1.23)
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Soft Matter Physics
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In general, total interaction potential between two molecules is the sum of short-range
repulsion and long-range attraction. Since hard core potential is not smooth, for modeling
it is often replaced by some other steep repulsive potential, e.g. Lennard-Jones interaction
(repulsion + VdW attraction) is:
σ 12
u(r) = 4
r
−
σ 6 r
,
(1.24)
where is the binding energy of two molecules in potential minimum.
Virial expansion is the low density (nσ 3 1) expansion of the free energy(or of many other
thermodynamic functions, e.g. pressure) of gas of interacting molecules:


1
1


F = τ V n(ln(n/nq ) − 1) + an2 + bn3 + ...
|
{z
} 2
3
(1.25)
F of ideal gas
, where a, b, ... are second, third,... virial coefficients respectively.
The second virial coefficient is calculated from the intermolecular interaction potential u(r)
as follows:
Z
a=
1−e
−βu(r)
Z
d~r = 4π
∞
1 − e−βu(r) r2 dr
(1.26)
0
For the gas of hard spheres a =
C.
4π 3
σ .
3
Distribution functions
Radial distribution function g(r) is defined as a ratio of an average density of molecules at
a distance r from a certain (can be any) molecule in the system to the average density of
molecules n = N/V . In general, g(r → ∞) = 1, g(r < σ) ≈ 0. In liquids, g(r) oscillates up
to the distances of several σ.
(l)
l-particle distribution function gN (~r1 , ~r2 , ..., ~rl ) is proportional to the probability density to
have any l molecules (in the system of N molecules) in the defined positions ~r1 , ~r2 , ..., ~rl . In
the canonical ensemble:
(l)
nl gN (~r1 , ~r2 , ..., ~rl )
N!
=
(N − l)!QN
Z
d~rl+1 d~rl+2 , ..., d~rN e−βU (~r1 ,~r2 ,...,~rN ) ,
(l)
(1.27)
where the normalization factor nl is added to have gN dimensionless and to remove trivial
dependence on density.
Oleg Krichevsky
Soft Matter Physics
The simplest distribution function is pair distribution function:
Z
N (N − 1)
2 (2)
d~r3 d~r4 , ..., d~rN e−βU (~r1 ,~r2 ,...,~rN )
n gN (~r1 , ~r2 ) =
QN
8
(1.28)
In uniform isotropic fluid the pair distribution function is equivalent to the radial distribution
function:
(2)
(2)
gN (~r1 , ~r2 ) = gN (|~r1 − ~r2 |) = g(r12 )
(1.29)
(l)
Yvon-Born-Green (YBG) hierarchy is a set of exact equations relating gN (~r1 , ~r2 , ..., ~rl ) dis(l+1)
tribution to gN
(~r1 , ~r2 , ..., ~rl+1 ), molecular interaction potential and density. The first equa-
tion in the hierarchy relates the pair distribution function to the triplet distribution function:
Z
∂
∂
∂ (2)
(2)
(3)
gN (~r1 , ~r2 ) = gN (~r1 , ~r2 )
u(~r1 , ~r2 ) + n gN (~r1 , ~r2 , ~r3 )
u(~r1 , ~r3 )d~r3
(1.30)
−τ
∂~r1
∂~r1
∂~r1
In order to solve the YBG hierarchy approximations (closures) are required. Kirkwood’s
superposition approximation is:
(3)
(2)
(2)
(2)
gN (~r1 , ~r2 , ~r3 ) = gN (~r1 , ~r2 )gN (~r1 , ~r3 )gN (~r2 , ~r3 )
(1.31)
This together with Eq. 1.30 leads to a closed equation for g(r).
D.
Fluid Dynamics basics
f
-
-
v
-
FIG. 1: Simple shear
Force acting on a surface of unit area in a viscous fluid is decomposed into components
normal to the surface, i.e. pressure, and tangential to the surface, called shear stress σ. In
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Soft Matter Physics
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general, the mechanical stress is a tensor, whose diagonal components characterize pressure
and non-diagonal components characterize shear stress. For common fluids, the shear stress
is proportional to the gradient of velocity field, with viscosity defined as a coefficient of
proportionality, such as in Fig. 1:
f
dvx
v
= σxy = η
=η
S
dy
h
(1.32)
Viscosity causes energy losses, which calculated per unit time and per unit volume are
proportional to viscosity and to the square of velocity gradient. E.g. in Fig. 1:
2
v 2
∆E
fv
dv
=−
= −η
= −η
∆t∆V
Sh
h
dh
(1.33)
Navier-Stokes equation is the main equation of fluid dynamics. It is essentially just an
application of the second law of Newton to the unit volume portion of fluid:
v
ρ d~
= −∇p + η∇2~v + (sometimes volume forces, e. g. ρ~g )
dt
∂~v
∂t
+ (~v ∇)~v = − ∇p
+ ν∇2~v ,
ρ
(1.34)
where ρ is fluid density, ν = η/ρ is kinematic viscosity and p is the pressure.
Reynolds number is an estimate of the relative importance of the nonlinear term in the
Navier-Stokes equation (1.34):
Re =
v∇v
v 2 /l
vl
vlρ
≈
=
=
,
2
2
ν∇ v
νv/l
ν
η
(1.35)
where v and l are respectively the characteristic velocities and characteristic sizes of the
objects.
In colloidal physics, the objects are so small (typically < 1µm), that Re 1 and nonlinear
term in Eq. 1.34 is neglected in most of the cases.
Navier-Stokes equation has to be complemented by the continuity equation, which for incompressible fluid is just
div ~v = 0
(1.36)
and by the boundary conditions, which typically are the non-slip conditions, i.e. the velocity
of the liquid layer adjacent to the solid surface is equal to the velocity of that surface.
For qualitative understanding of fluid dynamics problems one has to realize that:
• The motion of an object in fluid leads to gradients in velocity, pressure and shear
stress at the distances of about the object size.
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Soft Matter Physics
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• Pressure field propagates with the speed of sound: infinitely fast for colloidal-type
problems
• Shear stress propagates in a diffusion-like manner with a ”diffusion” coefficient of ν
(see Eq. 1.34), so that within time t the shear stress propagates to the distances of
∼ (νt)1/2 .
E.g. in the case of a flat solid slab performing periodic motion along the surface of
liquid, similar to Fig. 1, the mechanical stress is pure shear which propagates to the
depth of ∼ (νT )1/2 ∼ (ν/ω)1/2 , where T and ω are respectively the period and the
frequency of the motion.
• Viscosity is a momentum transport phenomenon: the momentum flux (per unit area
and time) is proportional to the velocity gradient. The consequence of this is that the
hydrodynamic interactions are extremely long ranged, decaying as ∝ 1/r
Drag force on a body moving through the fluid:
FD ∼ σL2 ∼ η Lv L2 ∼ ηLv
F~D = −γ~v = − µ~v ,
(1.37)
where L is the characteristic size of the body, γ is friction coefficient (dependent on the
shape and dimensions of the body and fluid viscosity) and µ = γ −1 is mobility.
In particular, for a sphere: γ = 6πηR.
E.
Surface waves
Gravitational waves: one has to consider gravitational potential energy, kinetic energy of
the fluid and viscous losses.
• Deep water waves (wavelength λ H depth): motion is confined to the surface layer
of lambda in depth; all of the viscous losses happen there; the characteristic velocity
of the fluid is ∼ the surface elevation/depreciation speed.
• Shallow water waves (λ H): motion occurs throughout the depth of the liquid; the characteristic velocity is much larger (by ∼ λ/H factor) than surface elevation/depreciation velocity due to continuity; the largest velocity gradient and hence
the main losses occur in a thin (∼ (ν/ω)1/2 ) layer near the bottom.
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Soft Matter Physics
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Capillary waves: the potential energy comes from surface tension. For weak perturbations
of the surface h(x, y) = h(~r⊥ ) the change in surface energy is:
" 2 #
2
Z
Z
2
α
∂h
α
∂h
∂h
=
∆Fs ≈
dxdy
+
d~r⊥
2
∂x
∂y
2
∂~r⊥
(1.38)
Otherwise the treatment is similar to gravitational waves. In general, both capillary and
gravitational effects have to be taken into account.
Surfactant molecules change the conditions on the surface to (almost) sticky boundary
conditions. Hence, the viscous losses occur in a thin layer near the surface (∼ (ν/ω)1/2 ).
Problem set 1 (7 obligatory problems and 1 optional)
1. Two hard spheres (with diameters σ) are fixed at the distance L. Other 3 similar
spheres are free to move along the line connecting the first two spheres. Determine the
dependence of average density of the spheres on the distance from the leftmost sphere
(similar system with 4 spheres in total was considered in the class).
2. A water molecule, guess what, consists of two O-H bonds at ∼ 105o angle and ∼ 0.095
nm distance between O and H. Water density is ≈ 1g/cm3 and molar mass is 18 g/mol.
• Estimate the electric susceptibility of water. Electric susceptibility is related to
the molecular polarizability of the material by: χ = nα/0 . Compare the result
to the table value. Would you consider the agreement to be good? What are the
possible reasons for the discrepancy?
• Estimate the parameters of Lennard-Jones potential for water molecules.
• As most of other substances water expands with increasing temperature (for water
this is true in fact only above 4o C). Estimate the volume expansion coefficient of
water defined as γ =
∆V
.
V ∆T
3. Consider a real gas with the following interaction potential between the molecules:

 −ψ(r), r > σ
uhc (r) =
(1.39)
 ∞,
r<σ
Assume ψ(r) positive and much smaller than τ and decaying to 0 at ∞. Obtain the
expression for the second virial coefficient for this case. Show that it crosses 0 at a
certain temperature.
Oleg Krichevsky
Soft Matter Physics
12
4. Use the formal definitions of ensemble averaging (Eq. 1.3) and of g(r) (Eq. 1.28 and
Eq. 1.29) to show that in a homogeneous system with pair-wise interactions Eq. 1.2:
the average total energy of interactions per molecule is equal to:
Z
Z
n
hU i
=
u(r)g(r)d~r = 2πn u(r)g(r)r2 dr
N
2
(1.40)
Explain the result.
5. Use of Virial Theorem in Statistical Mechanics: Virial of a system of N bodies (particles) is defined in the Classical Mechanics as:
V=
N
X
~ri · F~i ,
(1.41)
i=1
where ~ri and F~i are respectively the position and the total force acting on the ith body
(particle). The virial theorem of Classical Mechanics states that (under very general
assumptions - consult any class. mech. textbook) the time averaged values of the
virial and of the kinetic energy K of the system are related by:
V̄ = −2K̄
(1.42)
The proof of virial theorem is rather straightforward and goes roughly like that:
Z
Z
N
N
1 t 0X
1 t 0X
0
0
~
dt
~ri (t ) · Fi (t ) = lim
dt
~ri (t0 )mi~r¨i (t0 ) =
V̄ = lim
t→∞ t 0
t→∞ t 0
i=1
i=1
1
= [integrate by parts] = − lim
t→∞ t
Z
t
0
dt
0
N
X
mi |~r˙i (t0 )|2 = −2K̄
i=1
• Show that in the context of Statistical Mechanics, for a system with volume V ,
pressure p and temperature T the virial theorem can be restated as:
* N
+
X d
pV
1
=1−
~ri U (~r1 , ~r2 , ...~rN ) ,
N kB T
3N kB T i=1 d~ri
(1.43)
where as before U (~r1 , ~r2 , ...~rN ) is the internal potential energy of interactions
between particles.
Hint: Split forces on particles into external and internal to the system; assume
that the external forces enter through the outside pressure on the system surface
only (i.e. that there are no bulk external forces, such as gravity); use equipartition
theorem and common sense.
Oleg Krichevsky
Soft Matter Physics
• Show that in a homogeneous system with pair-wise interactions:
Z ∞
p
2πn
du 3
=1−
r dr
g(r)
nkB T
3kB T 0
dr
13
(1.44)
The notation is the same as in the Problem 4.
• Another derivation of second virial coefficient: Use the previous equation and the
definition of g(r) to show that low density expansion of pressure in any system
is:
p
1
= 1 + an,
nkB T
2
(1.45)
where a is already defined second virial coefficient Eq. 1.26. Why are second
virial coefficients of free energy expansion and pressure expansion are related at
all? (recall Thermo II)
• Use virial theorem to show that for any density, the pressure of a system of
particles interacting though hard core repulsion only can be expressed through:
2π 3
p
=1+
nσ g(σ) = 1 + 4φ g(σ),
nkB T
3
(1.46)
where σ is particle core diameter and g(σ) is the value of radial distribution
function at the separation equal to σ. φ = (1/6)πnσ 3 is the fraction of total
volume occupied by particles (called volume fraction). Show that for low densities
this and the previous expressions give equivalent results up to a second virial
coefficients (included).
Hint: You can overcome some mathematical difficulties of differentiating sharp
functions by introducing a new function y(r) = g(r) exp(βu(r)). This function is
expected to be ”smoother” than g(r). Why?
6. This problem is not obligatory but you may want to give it a try to see whether you
understood well the derivation of second virial coefficient.
• Show that the third virial coefficient in Eq. 1.25 is related to molecular interaction
potential u(r) through
Z
1
b=
d~r1 d~r2 1 − e−βu(r1 ) 1 − e−βu(r2 ) 1 − e−βu(|~r1 −~r2 |)
2
(1.47)
Oleg Krichevsky
Soft Matter Physics
14
or in a more symmetric form:
Z
1
b=
d~r1 d~r2 d~r3 1 − e−βu(|~r1 −~r2 |) 1 − e−βu(|~r1 −~r3 |) 1 − e−βu(|~r2 −~r3 |)
2V
(1.48)
• Estimate or, if you are courageous, calculate the value of b for the gas of hard
spheres of diameter σ
7. Consider thermal fluctuations of the surface of water.
• First, assume the fluctuations are limited by surface energy only. Use Eq. 1.38
to show that mean-square fluctuation of the surface diverges as a log-function of
the system size:
kB T L
ln ,
2πα a
where L is the size of the system and a is molecular size. Ln is a very slowly
hh2 i =
diverging function. Use characteristic values for water and calculate the meanp
square fluctuation hh2 i for the systems of 10 nm, 1 cm, 1 km, the World Ocean
in size.
Hint: In the expression for free energy Eq. 1.38 the surface elevations at different
positions are coupled through spatial derivatives
∂h
.
∂~
r⊥
This makes a direct use
of Stat. Mech. apparatus impossible. We had a similar problem in Thermo I
(where?). Like in Thermo I, we change from coupled variables h(~r⊥ ) to normal
modes by expressing the energy through harmonic waves - Fourier transforms
h(~q⊥ ) exp(i~q⊥ ~r). Prove that the energy of surface corrugations ∆Fs can be expressed as a sum of independent terms from different harmonic waves. You will
see that these terms are quadratic in the amplitudes of the waves h(~q⊥ ). The set
of these amplitudes is the set of independent variables and you can use equipartition theorem to calculate their ensemble-averaged values. This approach works
when the energy is quadratic in some variable or/and in its spatial derivatives.
• Show that the gravitational field restricts further the fluctuations. The energy of
the fluctuating interface in the gravitational field is:
"
#
2
Z
α
∂h
ρg 2
∆F =
d~r⊥
+ h (~r⊥ )
2
∂~r⊥
α
Explain the second term. Derive and estimate hh2 i for this case.
Oleg Krichevsky
Soft Matter Physics
=
P
+
q~
P
squeezing
15
q~
bending
FIG. 2: Decomposition of the fluctuations of soap film into the normal modes
8. Soap films consist of two layers of surfactant (soap) molecules separated by a slab of
water(see Fig.2). The fluctuations of soap films can be decomposed into two types
of modes: bending and squeezing. In the bending modes the two surfaces of the
soap film move in phase (so that the thickness of the film does not change, but the
central plane fluctuates), while in the squeezing modes they move out of phase (so
that the thickness fluctuates but not the central plane), like pictured in Fig. 2. For
the wave lengths much larger than the average film thickness λ = 2π/q h and for
small fluctuations δh h , the temporal behavior of bending and squeezing modes is
characterized by the following dispersion relations:
• Bending modes: oscillate with frequency
s
r
2α q 2 ν 2
2α
ω≈q
−
≈q
ρh
4
ρh
(1.49)
and relax exponentially with time with the characteristic rate Γ = q 2 η/(2ρ) =
q 2 ν/2 ,
• Squeezing modes are overdamped, i.e. they relax faster than oscillate. Their
relaxation rate is
h3 q 2
Γ=
24η
d2 U (h)
2
αq + 2
,
dh2
(1.50)
where U (h) is the interaction potential between the soap layers normalized per
unit film area. There are several sources of interaction between the soap layers,
Oleg Krichevsky
Soft Matter Physics
16
e.g. the ionization of soap molecules and the resulting electrostatic interaction.
Another reason is Van der Waals interaction. Any interaction between the soap
layers will of course depend on the distance between the layers, i.e. film thickness
h.
Derive the above dispersion relations for both squeezing and bending modes (up to
a numeric coefficient). Assume that the soap molecules are immobile in the plane of
the film and move only perpendicular to the film surface. Neglect the size of soap
molecules in comparison to film thickness. If you have difficulties taking into account
the interactions between film surfaces, start by assuming no interactions, obtain the
formula and then try to figure out how to take interactions into account.