PH3-SM (PHY3032)
Soft Matter Physics
Lecture 3
Potential Energy in Condensed Matter and its
Response to Mechanical Stress
18 October, 2011
See Jones’ Soft Condensed Matter, Chapt. 2; Ashcroft and Mermin, Ch. 20
In the previous lecture:
Interaction Potentials: w = -Cr -n
• If n <3, molecules interact with all others in the system of size, L.
If n >3, molecules interact only with the “nearer” neighbours.
• Gravity: negligible at the molecular level. W(r) = -Cr -1
• Coulombic: relevant for salts, ionic liquids and charged molecules.
W(r) = -Cr -1
• van der Waals’ Interactions: three types; usually quite weak;
cause attraction between ANY two molecules. W(r) = -Cr -6
• Covalent bonds: usually the strongest type of bond; directional
forces - not described by a simple potential.
• Hydrogen bonding: stronger than van der Waals bonds; charge
attraction resulting from unshielded proton of H.
Last Lecture:
• Discussed polar molecules and dipole moments (Debye
units) and described charge-dipole and dipole-dipole
interactions.
+Q
-
u
-
+
+
u
u
-
+
• Discussed polarisability of molecules (electronic and
orientational) and described charge-nonpolar, polarnonpolar, and dispersive (London) interactions.
+Q
a
a
-
u
+
• Summarised ways to measure polarisability.
a
a
Summary
Type of Interaction
Charge-charge
In vacuum: =1
Interaction Energy, w(r)
Q 1Q 2
Coulombic
4  o r
_ Qu cos 
Dipole-charge
4  o r
2
_Q u
2
2
6 ( 4  o ) kTr
2
4
_ u 1 u 2 f ( 1 ,  2 ,  )
2
2
4  o r
Dipole-dipole
2
_ u1 u 2
2
3( 4  o ) kTr
2
Charge-nonpolar
Keesom
2
2 ( 4  o ) r
2
4
_ u a (1 + 3 cos  )
2
2 ( 4  o ) r
2
_u a
2
( 4  o ) r
2
Nonpolar-nonpolar
6
_Q a
2
Dipole-nonpolar
3
Dispersive
6
6
Debye
w (r ) = _
3
a o 2h
4 ( 4  o ) 2 r 6
Cohesive Energy
• Def’n.: Energy, E, required to separate all molecules in
the condensed phase or energy holding molecules in the
condensed phase.
• We found previously (Lecture 2, slide 17) for a single
molecule, and with n>3:
4 C r
E 
( n  3 )s
n3
We can write r = number of molecules per unit volume 
s -3, where s is the molecular diameter. So, for van der
Waals’ interactions with n = 6:
E 
4 C r
3s
3

4 C
3s
6
1/2 to avoid double
counting!
• For one mole, Esubstance = (1/2)NAE
• Esubstance = sum of heats of melting + vaporisation.
• Predictions agree well with experiment!
Boiling Point
• At the boiling point, TB, for a liquid, the thermal energy of a
monoatomic molecule (3 degrees of freedom), 3/2 kTB, will
exactly equal the energy of attraction between molecules.
• Of course, the strongest attraction will be between the
“nearest neighbours”, rather than pairs of molecules that are
farther away.
• The interaction energy for van der Waals’ interactions is of the
form, w(r) = -Cr -6. If molecules have a diameter of s, then the
shortest centre-to-centre distance will likewise be s.
• Thus the boiling point is approximately:
TB 
w (s )
3 k
2

2C
3ks
6
Comparison of Theory and Experiment
London equation
Non-polar
Note that ao and C increase with s.
(P  a
V
2
)(V  b )  RT
(Per mole, n = 1)
E mole ~
N A 4 C
2 3s
6
TB =
w (r )
3 k
2
Evaluated at close
contact where r = s.
Additivity of Interactions
Molecule
Mol. Wt.
H
H
C-C
H
H
Ethane: C2H6
u (D)
H
TB(°C)
Dispersive only
H
30
0
-89
H
C=O
H
Formaldehyde: CH2O
Keesom + dispersive
30
2.3
-21
H
H
C-O-H
H
Methanol: CH3OH
H-bonding + Keesom + dispersive
32
1.7
64
Lennard-Jones Potential
• To describe the total interaction energy (and hence the
force) between two molecules at a distance r, a pair
potential is used.
• The pair potential for isolated molecules that are affected
only by van der Waals’ interactions can be described by a
Lennard-Jones potential:
w(r) = +B/r12 - C/r6
• The -ve r -6 term is the attractive v.d.W. contribution
• The +ve r -12 term describes the hard-core repulsion
stemming from the Pauli-exclusion. 12 is a
mathematically-convenient exponent with no physical
significance!
• The two terms are additive.
L-J Potential for Ar
London Constant for Ar is calculated to be C = 4.5 x 10-78 Jm6
We can guess that B = 10-134 Jm12
Actual s  0.3 nm
(Our guess for B is too large!)
wmin  -5 x 10-22 J
Compare to:
(3/2)kTB= 2 x 10-21 J
(m)
Ar boiling
point:
TB = 87 K
(x 10-9 )
Intermolecular Force for Ar Pair
Attractive force
Very short-range force!
Equ’m
F = dw/dr
s
Repulsive force
(m)
r
Comparison of Force and Potential Energy for Ar Pair
F= dw/dr
(m)
(m)
Weak Nano-scale Forces Can be Measured with an
Atomic Force Microscope
The AFM probe is
exceedingly sharp so that
– in principle - only a few
atoms are at its tip!
Sensitive to forces on the order of nano-Newtons.
Tips for Scanning Probe Microscopy
F
The tip is on a cantilever, which typically has a
spring constant on the order of k = 10 N/m.
Radius of curvature ~ 10 nm
Ideally, one of the atoms at the tip
is slightly above the others.
Modelled as a simple spring:
F = kz
where z is the deflection in the vertical
direction.
AFM tips from NT-MDT. See www.ntmdt.ru
Tip/Sample Interactions: Function of Distance
h
Physical contact
between tip and
surface
Measuring Attractive Forces at the Nano-Scale
Tip deflection 
B = “jump” to
contact
C
0
A = approach
Force
C
A
B
C = contact
E
D
Vertical position
D = adhesion
E = pull-off
Measuring Force of Attraction to a
Polymer Surface
200
Pushing on AFM
probe tip
Force /nN
100
trace
retrace
0
-100
-200
Pulling on the AFM
probe tip
-300
-400
0
2
4
Distance/ m
6
8
Imaging with the AFM Tip
The AFM tip is held at a constant distance from the surface - or a constant
force is applied - as it scanned back and forth.
Surface Force Apparatus
Mica has an
atomistically
smooth surface.
A piezoelectric moves the arm up by a known amount. Force on
the mica is determined by measuring the distance between the
mica and knowing the arm’s stiffness.
www.fisica.unam.mx/liquids/tutorials/surface.html
Distance between mica
sheets is measured with
interferometry.
L-J Potential in Molecular Crystals
Noble gases, such as Ar and Xe, form crystalline solids (called
molecular crystals) that are held together solely by the dispersive
energy. We will analyse them as a very simple form of “soft matter.”
In molecular crystals, the pair potential for neighbouring atoms (or
molecules) is written as
 s 
w  4  
 r 
12
s  
  
r  
6
The molecular diameter in the gas state is s. Note that when r =
s, then w = 0.
 is a bond energy (related to the London constant), such that
w(r) = -  when r is at the equilibrium spacing of r = ro.
Lennard-Jones Potential for
Molecular Pairs in a Crystal
+
w(r)
-
-
s
ro
r
L-J Potential in Molecular Crystals
The minimum of the potential is found from the first derivative of
the potential. Also corresponds to the point where F = 0.
 12 s
 F  0  4  
13
dr
r

dw
12
6s 
 7 
r 
6
We can solve this expression for r to find the pair’s
equilibrium spacing, ro:
1
ro = 2
6s
= 1 . 12 s
To find the minimum energy in the potential, we can evaluate
it when r = ro:
 s
1/ 6
W ( 2 s )  4    1
  2 6 s




12
 s
  1
 2 6s




6

1 1 

 
  4
 4 2 

Variety of Atomic Spacings in Cubic Crystals
The particular crystal structure (FCC, BCC, etc.) defines the distances
between nearest neighbours, 2nd neighbours, 3rd neighbours, etc., and it
defines the number of neighbours at each distance.
6 nearest
neighbours; 12
second nearest
8 nearest
neighbours; 6
2nd nearest; 12
3rd nearest
12 nearest neighbours; 6
second nearest; 24 3rd
nearest
Image from: http://www.uccs.edu/~tchriste/courses/PHYS549/549lectures/figures/cubes.gif
Potential Energy of an Atom in a Molecular Crystal
• For each atom/molecule in a molecular crystal, we need to sum
up the interaction energies between all pairs (assuming
additivity of the potential energies).

• The total cohesive energy per atom is Wtot = 1/2 S w(r) since
r
each atom in a pair “owns” only 1/2 of the interaction energy.
• As shown already, the particular crystal structure (FCC, BCC,
etc.) defines the distances of neighbours, 2nd neighbours, 3rd
neighbours, etc., and it defines the number of neighbours at
each distance.
• This geometric information that is determined by the crystal
structure can be described by constants, known as the lattice
sums: A12 and A6 (where the 12 and 6 represent the two terms
of the L-J potential.)
• For FCC crystals, A12 = 12.13 and A6= 14.45. There are different
values for BCC, SC, etc.
Cohesive Energy of Atoms in a Molecular Crystal
So, for a pair we write the interaction potential as:
 s 
w  4  
 r 
12
s  
6
  
r  
For each atom in a molecular crystal, however, we write that
the cohesive energy is:
W tot
12
6

s
s
 
  
 2   A12    A6   
 r 
 r  

From the first derivative, we can find the equilibrium spacing
1
for an FCC crystal:
6
 2 A12 
ro  
 s  1 . 09 s
 A6 
We notice that the molecules are slightly closer together in a
crystal compared to when they are in an isolated pair (ro=1.12 s).
Cohesive Energy of Atoms in a Molecular Crystal
We can evaluate Wtot(r) when r = ro to find for an FCC crystal:
W TOT
FCC
-
ε A6
2
 - 8 .6 ε
2 A12
This expression represents the energy holding an atom/molecule
within the molecular crystal. Its value is only 8.6 times the
interaction energy for an isolated pair.
This result demonstrates that the dispersive energy is operative
over fairly short distances, so that most of the interaction energy is
contributed by the nearest neighbours. (In an FCC crystal, each
atom has 12 nearest neighbours, but the equivalent of 8.6 pairs of
atoms (17.2) contribute to the energy!)
Connection between the macroscopic and the atomic?
st
A
F
L
The Young’s modulus, Y, relates tensile stress and strain:
s
Y = t
t
How does the
interatomic force,
F, relate to the
macroscopic Y?
Y
t
Model for Elastic Modulus of Molecular Crystals
We can model the interatomic force using a spring with a spring constant, k.
The force, F, to separate two atoms in the crystal is:
F = k(r - ro). At equilibrium, r = ro and F = 0.
ro
ao
ao
F
ro
The tensile stress st is defined as a force acting per unit area, so that:
st 
F

A
k ( r - ro )
ro
2
The tensile strain t is given as the change in length as a result of the
r  ro
r
stress:
 

t
ro
ro
Y can thus be expressed in terms of atomic interactions:
k ( r - ro )
Y 
st
t

ro
( r - ro )
ro
2

k
ro
What is k?
Elastic Modulus of Molecular Crystals
+
Wtot
The Young’s modulus is
sometimes known simply
as the elastic modulus.
F 
dW tot
s
ro
-8.6
r
+
F = 0 when r = ro
dr
F
k 
dF
dr
r  rO
-
ro
r
Linear region around ro is
approximated as: F = k(r - ro).
Elastic Modulus of Molecular Crystals
Force to separate atoms is the derivative of the potential:
 - 12 A12 s 12 6 A6s 6 
 F  2 


13
7
dr
r
r


dW
So, taking the derivative again to find k:
dF
dr

 2 

13( 12 ) A12 s
But we already know that:
Re-arranging, we see that:
r
7 ( 6 ) A 6s
12
14
 2 A12
ro  
 A6

r




 A6
s  ro 
 2 A 12
8
6



1
6
s




1
6
We will therefore make a substitution for s when finding k.
Elastic Modulus of Molecular Crystals
 13( 12 ) A12 s 12 7 ( 6 ) A6s 6 
dF
 2 


14
8
dr
r
r


with
 A6
s  ro 
 2 A 12




1
6
To find k, we now need to evaluate dF/dr when r = ro.
Combining the constants to create new constants, C1 and C2, and
setting r = ro, we can write:

k  2 

C 1 ro
ro
14
12
6
C 2 ro 
C1 - C 2

2

(
)

8
2
ro 
ro
Finally, we find the Young’s modulus to be:
Y 
k
ro

2 (C1 - C 2 )
ro
3
As ro3 can be considered an atomic volume, we see that the
modulus can be considered an energy density, directly related to
the pair interaction energy.
Response of Condensed Matter to Shear Stress
A
F
y
A
ss =
F
A
How does soft matter respond to shear stress?
When exposed to a shear stress, the response of condensed
matter can fall between two extremes: Hookean (solid-like)
or Newtonian (liquid-like)
Elastic Response of Hookean Solids
The shear strain gs is given by the angle  (in units of radians).
ss =
A
x
y
A
F

F
A
gs = ~
x
y
The shear strain gs is linearly related to the shear stress by the shear
ss
modulus, G:
gs =
G
No time-dependence in the response to stress. Strain is
instantaneous and constant over time.
Viscous Response of Newtonian Liquids
The top plane moves at a constant velocity, v, in response to a shear
stress:
F
ss =
A
x
A
F
v
y
v =
A
x
t
There is a velocity gradient (v/y) normal to the area. The viscosity h relates the
shear stress, ss, to the velocity gradient.
s s= h
v
y
=h
x
t y
h has S.I. units of Pa s.
The shear strain increases by a constant amount over a time interval, allowing
g
us to define a strain rate:
-1
g 
t
Units of s
The viscosity can thus be seen to relate the shear stress to the shear rate:
x
x
g
ss h
h
h
 h g
t y
y t
t
Hookean Solids vs. Newtonian Liquids
Hookean Solids:
Newtonian Liquids:
s s  Gg
ss h
dg
 h g
dt
Many substances, i.e. “structured liquids”, display both types of
behaviour, depending on the time scale. At short time scales, the
response is solid-like. At longer time scales, the response is liquid
like.
This type of response is called “viscoelastic”.
Examples of viscoelastic systems include colloidal dispersions
and melted polymers.
The simple Maxwell model assumes that the elastic and viscous
responses act in series and are additive.
Example of Viscoelasticity
Video: Viscoelastic gel
Maxwell Model of Viscoelasticity
Spring:
Elastic
element
Dashpot:
ss
Viscous
element
The model says that the elastic and viscous
elements act in series, such that the shear
stress, ss, is the same for both of them.
The elastic shear strain, ge, and the viscous
shear strain, gv, are additive:
γ  γe  γv
As ss is the same for both elements, we have:
dg v
s s  GM g e  h
dt
For a constant applied shear stress, the total strain can be
ss
ss
written as:
g (t ) 

t
GM
h
We define a Maxwell relaxation time: t = h/GM
Stress Relaxation after a Step Strain
g
Constant shear strain applied
time
ss
Hookean
viscoelastic solid
s s  g G eq
viscoelastic liquid
Newtonian
t
time
If viscoelastic, then the stress relaxes over time as molecules re-arrange.
Stress Relaxation after a Step Strain
For a viscoelastic liquid, the shear stress relaxation in the
Maxwell model is described as:
s s (t )  G M ge
-t
t
t is the relaxation time
For typical solids, t is exceedingly large: t  1012 s, such
that there is no observed relaxation – just a Hookean
response. For melted polymers, however, t  1 s.
From here, we can define a time-dependent stress relaxation
t
modulus:
s s (t )
t
G (t ) 
g
 GM e
For a viscoelastic solid, the equilibrium shear modulus
(non-zero) is defined as:
G eq  lim( t   ) G ( t )
Response of Soft Matter to a Constant Shear Stress
Apply a constant shear stress, ss, and observe the time
dependence of the shear strain, g(t)
In this experiment, we find the creep compliance:
J(t) = 1/G(t)= g(t)/ss.
s
s
Recalling that: g ( t )  s  s t
GM
h
1
t
We see in the Maxwell model that:
J (t ) 

GM
η
Response of Soft Matter to a Constant Shear Stress
J (t ) 
g s (t )
ss
(Pa-1)
Slope:
d
Jeq
(
g
dt s s
)
g
ss

1
h
Viscous response
(strain increases
linearly over time)
Elastic response
(provides initial and recoverable strain)
t
Steady-state compliance = Jeq= 1/Geq
In the Maxwell model, Geq = h/t  so, t = hJeq and Jeq = t/h
Viscoelastic Response after Stress is Removed
J (t ) 
g s (t )
ss
Jeq
(Pa-1)
Elastic strain is
recovered
Jeq
Constant stress,
ss, is applied
Stress removed
Viscous flow is
permanent
t
Viscosity of Soft Matter Often Depends
on the Shear Rate
h
ss
Newtonian:
h
(simple liquids
like water)
g s
h
ss
Shear thinning
or thickening:
g s
h
g s
g s
An Example of Shear Thickening
Future lectures will
explain how polymers
and colloids respond
to shear stress.
Video: shear thickening
Problem Set 1
1. Noble gases (e.g. Ar and Xe) condense to form crystals at very low temperatures. As the atoms do not undergo any
chemical bonding, the crystals are held together by the London dispersion energy only. All noble gases form crystals having
the face-centred cubic (FCC) structure and not the body-centred cubic (BCC) or simple cubic (SC) structures. Explain why
the FCC structure is the most favourable in terms of energy, realising that the internal energy will be a minimum at the
equilibrium spacing for a particular structure. Assume that the pairs have an interaction energy, u(r), described as

s 
u ( r )  2   A12  
r 

12
s  
 A6    ,
r  
6
where r is the centre-to-centre spacing between atoms. The so-called "lattice sums", An, are given below for each of the three
cubic lattices.
SC
BCC
FCC
A6
8.40
12.25
14.45
A12
6.20
9.11
12.13
Then derive an expression for the maximum force required to move a pair of Ar atoms from their point of contact to an infinite
separation.
2. (i) Starting with an expression for the Coulomb energy, derive an expression for the interaction energy between an ion of
charge ze and a polar molecule at a distance of r from the ion. The dipole moment is tilted by an angle  with relation to r, as
shown below.
ze
r

(ii) Evaluate your expression for a Mg2+ ion (radius of 0.065 nm) dissolved in water (radius of 0.14 nm) when the water
dipole is oriented normal to the ion and when the water and ion are at the point of contact. Express your answer in units of
kT. Is it a significant value? (The dipole moment of water is 1.85 Debye.)
3. Show that 1 kJ mole-1 = 0.4 kT per molecule at 300 K.
Problem Set 2
1. Calculate the energy required to separate two atoms from their equilibrium spacing ro to a very large
distance apart. Then calculate the maximum force required to separate the atoms. The pair potential is
given as w(r) = - A/r6 + B/r12, where A = 10-77 Jm6 and B = 10-134 Jm12. Finally, provide a rough estimate of
the modulus of a solid composed of these atoms.
2. The latent heat of vaporisation of water is given as 40.7 kJ mole-1. The temperature dependence of the
viscosity of water h is given in the table below.
(i) Does the viscosity follow the expectations of an Arrhenius relationship with a reasonable activation
energy?
(ii) The shear modulus G of ice at 0 C is 2.5 x 109 Pa. Assume that this modulus is comparable to the
instantaneous shear modulus of water Go and estimate the characteristic frequency of vibration for water, .
Temp (C)
h (10-4 Pa s)
0
17.93
10
13.07
20
10.02
30
7.98
40
6.53
Temp (C)
h (10-4 Pa s)
60
4.67
70
4.04
80
3.54
90
3.15
100
2.82
50
5.47
3. In poly(styrene) the relaxation time for configurational rearrangements t follows a Vogel-Fulcher law
given as
t = to exp(B/T-To),
where B = 710 C and To = 50 C. In an experiment with an effective timescale of texp = 1000 s, the glass
transition temperature Tg of poly(styrene) is found to be 101.4 C. If you carry out a second experiment
with texp = 105 s, what value of Tg would be obtained?
Typical Relaxation Times
For solids, t is exceedingly large: t 1012 s
For simple liquids, t is very small: t 10 -12 s
For soft matter,t takes intermediate values. For instance, for
melted polymers, t 1 s.
Slime movie
Relaxation Time in a Creep Experiment
•In a “creep” experiment, a constant stress is applied for a fixed period of
time and then released.
•The strain is recorded while the stress is applied and after it is released.
•The instantaneous (elastic) response is modeled as a spring giving an elastic
modulus, Y.
•The time-dependent (viscous flow) response is modeled as a “dashpot”
giving a viscosity, h.
Stress,
Strain,
s
Y
h

0
Time
0
Time
Response of Soft Matter to a Constant Shear
Stress: Viscoelasticity
1
G (t )
=
g (t )
ss
Slope:
1
d
Go
(
gs
dt s s
)=
g s
ss
t is the “relaxation time”
We see that 1/Go  (1/h)t
t
Hence, viscosity can be approximated as h  Got
t
=
1
h
Modelling of Creep and Relaxation
Maxwell model: Spring and dashpot in series
s
Instantaneous elastic
deformation is followed by
flow. Elastic deformation is
recovered but not the flow.
 
s
Y

s
h
t
Voigt-Kelvin model: Spring and dashpot in parallel
s
Anelastic deformation is
followed by recovery with an
exponential dependence on
t = h/Y, which is called the
relaxation time.
Deformation:
 
s 
  Yt   s 
  t 
 1  exp 
    1  exp 

Y 
h
Y
t





Total Creep and Recovery
Constant stress applied
Displacement
Anelastic
Elastic
Flow
Spring and dashpot
acting both in series
Time
and in parallel.
Strain is described by the sum of the
Maxwell and Voigt-Kelvin models.
Creation of a New Surface Leads to
a “Thermodynamic” Adhesion Force
F
Surface area
increases when
tip is removed.
R
G
is the surface tension (energy) of the tip and the surface assumed here to be equal.
Work of adhesion:
W =
G dA
∫
W 
L
GLSA
L
 G dA
W = GLVA + GSVA - GLSA
GLVA
Work per unit area, W:
S
W  GLV+ GSV - GSL
GSVA
GLV
S

When liquid (L) and solid (S) are
separated, two new interfaces with the
vapour (V) are created.
Work per unit area, W:
W  GLV (1  cos  )
Young-Dupré Equation
GSL
GSV  GSL  GLV cos 
GSV  GSL  GLV cos 
W  GLV  GLV cos 
GSV
The Capillary Force
Pressure is required to bend a
surface with a surface tension, G
 P = G[
F
1
1
r1
r2
]≈
G
r1
F = PA ≈
G
r1
Max. Capillary Force:
2  Rd
F  4RG cos
With G = 0.072 N/m for water and R = 10
nm, F is on the order of 10-8 -10-9 N!
Force for Polymer Deformation
20
Indent
Pull
10
Force/ nN
0
-10
-20
-30
-40
-50
0
1
2
3
4
5
6
Distance/ m
7
8
9
Hookean Solids vs. Newtonian Liquids
Hookean Solids:
Newtonian Liquids:
σ  Gγ
σ η
dγ
 η γ
dt
Many substances, i.e. “structured liquids”, display both types of
behaviour, depending on the time scale: solid-like on short timescales and liquid-like on longer time-scales.
Examples include colloidal dispersions and melted polymers.
This type of response is called “viscoelastic”. The simplest
model of viscoelasticity is the Maxwell model, which assumes
that the viscous and elastic elements act in a series. Under a
constant shear stress, the shear strains are additive.