Download Soft Matter Physics, Lecture Notes. Lectures 1 and 2

Soft Matter Physics, Lecture Notes. Lectures 1 and 2
Submitted by: Amir Erez
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 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 :
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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).
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The Radial Distribution Function
Let us consider a solid lattice, where the atoms’ position is rigidly determined.
We can now make a plot of the atom density as a function of x :
In fluids, we have a less ordered structure. In fact, our density function n(x)
now depends on time:
On average, < n(~r) >t = n meaning that the fluid is homogenous.
Instead of using a fixed coordinate distribution function, we can ”sit” on one
molecule and measure the density of the molecules around it, thereby getting a
radial distribution function. We normalize the distribution function by average
density so that it gives 1 when the density is the average density, we can expect
that as r → ∞ we get g(r) → 1.
Let us examine a few examples of radial distribution functions: (we denote σ as
the molecule diameter)
For a real gas, we have to consider short range repulsion and long range Van
Der Waals attraction (more on this below) giving us a different distribution
function:
Note that the repulsion accounts for the climb left of the maximum, whereas
the attraction affects the descent to the right of the maximum.
Fluids have a somewhat different distribution function:
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Figure 1: Ideal gas distribution function
Figure 2: Low density real gas distribution function
The surprising fact is that attractive forces are not required to make the maxima/minima in the plot for liquids. We shall now examine the reason that the
system exhibits a clear structure even in the absense of intermolecular attractive
forces.
To simplify things, we shall examine the case where we have two molecules restricted to moving in one dimention, like two beads on a piece of string of length
L:
We determine the position of A and sum over all the position for B. We can do
the calculation in two ways: either the molecules can penetrate each other, or
they cannot. We begin by assuming they can inter-penetrate.
Therefore A and B are identical, and we can look at either one of them. Following the micro-canonic formalism the probability that A is a certain distance
from the left edge is proportional to the number of microscopic states of the
system. ie. The probability for finding A at position X is the number of possible positions of B, when A is at X.
Figure 3: Fluid distribution function
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When σ < x < 2σ then B cannot penetrate the region between the left side and
A and therefore the number of states of the system:
g(x) ∝ (L − x − 2σ)
Whereas when 2σ < x we can fit B inside the gap therefore:
g(x) ∝ (L − x − 2σ) + (x − 2σ) = L − 4σ
Giving us a plot of the radial distribution function that looks like this:
This demonstrates that we require only a repulsive (hardcore interaction) force
to get a relevant g(r).
Depletion interaction
The depletion interaction force is a force in the sense of statistical physics - that
is, as a derivative of the free energy. Based on the approach outlined above, let
us consider a system composed of a fluid (with polymers) and two plates:
For as long as x > polymer length, then there is no depletion force. However,
when the plates become close x ≈ polymer length, then there are less states to
arrange the system (this is because polymers cannot enter between the plates).
This means the entropy is lower and the free energy is higher. As x → 0 we get
maximal entropy and therefore there is an attractive force that makes the two
plates stick together. This is the depletion interaction.
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Van Der Waals interaction
A quantum mechanical picture of electronic polarizability can be constructed
using the Heitler-London approach. Instead we shall make a simplified quasiclassical derivation for electricly neutral molecules which will give us a result
that is accurate up to a constant prefactor.
The Van Der Waals interaction is the attraction between dipole and induced
dipole. To understand it better we split considerations into three parts:
1) Energy of an induced dipole in an external electric field
2) Interaction energy of dipole-induced dipole pair
3) Two models for polarizability, for non-polar and polar molecules.
1) An induced dipole is subject to an electric force because of the difference in
electric field at its ”poles”:
di = αE = ql
dE
dE
= αE
dr
dr
Integrating this will give the energy of a dipole in an electric field:
Z
1
U = − F dr = − αE 2
2
F = q∆E ≈ ql
2) When the exteral field E is produced by a dipole we can say:
E≈
d
4π0 r3
Thereby giving us:
U ≈−
αd2
(4π0 )2 r6
(1)
3) Electronic polarizability is common to all molecules, both polar and nonpolar. To describe electronic polarizability we consider a model of an atom with
the electron orbiting it in a plane (conservation of angular momentum). The
atom is subject to an external electric field:
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We can find the electron orbit displacement x from force equilibrium:
eE =
e2
e2
x
cos
θ
=
2
2
4π0 R
4π0 R R
By denoting di = ex we get:
di = 4π0 R3 E = αE
α = 4π0 R3
(2)
Note that we can extend this approach to heavy atoms and molecules since the
outer electrons ’see’ a screened nucleus electric charge that is very similar to the
case of hydrogen atom.
Using Eqs. 1 and 2 we can express the Van der Waals energy of two nonpolar molecules as:
−2α2 I
,
(3)
Uvdw (r) =
(4π0 )2 r6
2
where I = 8πe 0 R is the ionization energy of a molecule. Typically, I ≈ 2 ×
10−18 J, so that the VdW energy is of the order of magnitude of thermal energy.
Note that the quantum mechanical calculation will give instead of the ’2’ prefactor, 43 .
We can check whether the result we got for α is reasonable by considering the
electric susceptibility of a medium. The electric susceptibility χ is proportional
to the density of atoms n:
nα
χ=−1=
0
The atom density in liquids or solids is roughly 1/(atom volume) or:
n≈
1
4
3
3 πR
Thereby giving us, for non metals, χ ≈ 3 or ≈ 4. Indeed when the table values
of dielectric constant for nonpolar media is nearly always 2 < < 6.
Mixing energy
When we have two different types of molecules:
Uvdw (r) =
−2α1 α2 I
(4π0 )2 r6
We compare the energy of an unmixed state:
AAABBB
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with that of a mixed state:
AABABB
We can see that the change in energy is:
∆U = UAA + UBB − UAB − UAB = 2Umix
UAA + UBB
− UAB
2
This determines whether mixing is energetically favourable or not. In general
UAA , UBB and UAB stem from Van Der Waals interactions we see that:
Umix =
Umix ∝
2
2
−αA
− αB
1
+ αA αB = − (αA − αB )2 < 0
2
2
Therefore the mixed state energy is higher than the unmixed state. This means
that from energy considerations alone, the molecules will not mix. Entropy considerations dictate maximal mixing so the balance is determined by the temperature of the solution. From this we can conclude that different types of
molecules will not mix in a low enough temperature.
Dipole Polarizability of Polar Molecules
For polar molecules (such as water molecules) there is an additional type of
polarizability: dipole polarizability. Dipole polarizability is typically larger than
electron polarizability and gives media a significantly higher dialectric constant.
For example, for water a dialectric constant is ≈ 80. High polarizability
determines many important properties of water. E.g. salts dissociate easily
in water, since the ion bond energy decreases by a factor of 80 (as compared
to air or vacuum). Therefore the energy price for breaking the ionic bond is
small compared to the entropy gain and this is why at room temperature table
salt (NaCl) will undergo a process of disassociation separating to the two ions.
We now proceed to derive the Van Der Waals interaction energy for polar molecules.
~ without external electric field < d~ >= 0 due to ranFor the molecular dipole d,
domness of orientation. With external electric field we get a preferred direction,
meaning that on average the dipole will be in the direction of the external field:
< d~ >= d~i
~ so using the Bolzman factor we get the
The field-dipole energy is U = −d~E
probability for the dipole to point in the direction θ relative to the direction of
the external electric field:
1 ~~
P (θ) = eβ dE
Z
Z
~~
Z = eβ dE dΩ
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1
d~i =< d~ >=
Z
Z
de cos θeβde ·E·cos θ dΩ
~ τ where β = 1/τ and Taylor expanding:
Taking d~e · E
R
R
2
(de cos θ + βEd2e cos2 Ω)dθ
θdΩ
2 cos
R
di =
= βEde R
(1 + βde E cos θ)dΩ
dΩ
we get the induced dipole:
di =
Or
α=
d2e
E
3kB T
d2e
3kB T
For water, this calculation will give ≈ 105 at room temp. whereas the empiric
value is ≈ 80.
We can immediately see that this is a much higher polarizability than in nonpolar molecules. Plugging this into the general Van Der Waals formula:
Uvdw =
−αd2e
(4π0 )2 r6
We get the interaction enery between two polar molecules:
pol
Uvdw
=
−d4e
(4π0 )2 · 3kB T · r6
Note: For calculating the attractive forces between polar and non-polar molecules, such as oil and water, we assume that the field is created by the polar
molecule (therefore using de for the water) and the non-polar molecule responds
αd2
using α = 4π0 R3 in the formula U = − (4π0 )e2 r6 .
From this, and using mixing energy considerations, we can partially explain the
very good separation between water and oil at room temperature.Later in the
course we shall discuss the behaviour of the hydrophobic tails.
Thermal expansion
It is easily seen that, generally, the potential is not symmetric with respect to
the minimum r0 :
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An increase in temperature results in the system oscillating between two values
of r which are not equidistant from the minimal state r = r0 . On average the
radius r > r0 .
Virial Expansion
One way for accounting interactions in statistical mechanics is the Virial Expansion. We will proceed to expand the free energy (F ) and pressure (P ) to
leading terms in density. Meaning that:
1
1
F = F id + an2 + bn3 + . . .
2
3
1
1
P = nkB T + a1 n2 + b1 n3 + . . .
2
3
(4)
(5)
Where F id and nkB T are the free energy and pressure in ideal gas. The coefficients a, a1 are the coefficients for the n2 terms and are called the Virial
Coefficients.
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