ph3-sm (phy3032) soft matter physics lecture 3 potential energy in condensed matter and its response...
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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
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.
In the previous lecture:
Last Lecture:• Discussed polar molecules and dipole moments (Debye
units) and described charge-dipole and dipole-dipole interactions.
• Discussed polarisability of molecules (electronic and orientational) and described charge-nonpolar, polar-nonpolar, and dispersive (London) interactions.
• Summarised ways to measure polarisability.
+Q +- +
+
-
-
+Q +-
u u u
u
SummaryType of Interaction Interaction Energy, w(r)
Charge-charge rQQ
o421 Coulombic
Nonpolar-nonpolar 62
2
443
r
hrw
o
o
)(_=)(
Dispersive
Charge-nonpolar 42
2
42 rQ
o )(_
Dipole-charge24 r
Qu
ocos_
42
22
46 kTruQ
o )(_
Dipole-dipole
62
22
21
43 kTruu
o )(_
Keesom
321
22
21
4 rfuu
o ),,(_
Dipole-nonpolar
62
2
4 ru
o )(_
Debye
62
22
4231
ru
o )()cos+(_
In vacuum: =1
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:
We can write = number of molecules per unit volume -3, where is the molecular diameter. So, for van der Waals’ interactions with n = 6:
• For one mole, Esubstance = (1/2)NAE• Esubstance = sum of heats of melting + vaporisation.• Predictions agree well with experiment!
3)3(
4
nn
CE
63 3
4
3
4
CC
E 1/2 to avoid double counting!
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 , then the shortest centre-to-centre distance will likewise be .
• Thus the boiling point is approximately:
63
2
23
)(
k
C
k
wTB
Comparison of Theory and Experiment
63
42
CNE A
mole ~
Evaluated at close contact where r = .
k
rwTB
23
)(=Note that o and C increase with .
Non-polar
London equation
RTbVV
aP ))(( 2
(Per mole, n = 1)
Additivity of InteractionsMolecule Mol. Wt. u (D) TB(°C)
Ethane: C2H6 30 0 -89
Formaldehyde: CH2O 30 2.3 -21
Methanol: CH3OH 32 1.7 64
C=OH
H
C-O-HH
HH
C-CH
HH
H
HH Dispersive only
Keesom + dispersive
H-bonding + Keesom + dispersive
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 ArLondon Constant for Ar is calculated to be C = 4.5 x 10-78 Jm6
We can guess that B = 10-134 Jm12
wmin -5 x 10-22 J
Compare to: (3/2)kTB= 2 x 10-21 J
Actual 0.3 nm
(Our guess for B is too large!)
(m)
Ar boiling point: TB = 87 K
(x 10-9 )
Intermolecular Force for Ar Pair
F = dw/dr
(m)
Very short-range force!
Repulsive force
Equ’m
Attractive force
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.
AFM tips from NT-MDT. See www.ntmdt.ru
Tips for Scanning Probe Microscopy
Radius of curvature ~ 10 nm
Ideally, one of the atoms at the tip is slightly above the others.
The tip is on a cantilever, which typically has a spring constant on the order of k = 10 N/m.
Modelled as a simple spring:
F = kz
where z is the deflection in the vertical direction.
F
Tip/Sample Interactions: Function of Distance
Physical contact between tip and surface
h
Measuring Attractive Forces at the Nano-Scale
A = approach
B = “jump” to contact
C = contact
D = adhesion
E = pull-off
Tip deflection Force
Vertical position
AB
C
D
EC0
Measuring Force of Attraction to a Polymer Surface
Pulling on the AFM probe tip
Pushing on AFM probe tip
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.
www.fisica.unam.mx/liquids/tutorials/surface.html
Distance between mica sheets is measured with
interferometry.
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.
Surface Force ApparatusMica has an
atomistically smooth surface.
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
The molecular diameter in the gas state is . Note that when r = , 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.
612
4rr
w
Lennard-Jones Potential for Molecular Pairs in a Crystal
rw(r)
+
-
-
ro
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.
We can solve this expression for r to find the pair’s equilibrium spacing, ro:
1212 61
.==or
To find the minimum energy in the potential, we can evaluate it when r = ro:
7
6
13
12 61240
rrF
drdw
21
41
422
4)2(6
61
12
61
6/1W
Variety of Atomic Spacings in Cubic Crystals
Image from: http://www.uccs.edu/~tchriste/courses/PHYS549/549lectures/figures/cubes.gif
8 nearest neighbours; 6
2nd nearest; 12 3rd nearest
12 nearest neighbours; 6 second nearest; 24 3rd
nearest
6 nearest neighbours; 12 second nearest
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.
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 r
w(r) since 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
6
6
12
122r
Ar
AWtot
So, for a pair we write the interaction potential as:
We notice that the molecules are slightly closer together in a crystal compared to when they are in an isolated pair (ro=1.12 ).
From the first derivative, we can find the equilibrium spacing for an FCC crystal:
09.12 6
1
6
12
AA
ro
For each atom in a molecular crystal, however, we write that the cohesive energy is:
612
4rr
w
Cohesive Energy of Atoms in a Molecular CrystalWe can evaluate Wtot(r) when r = ro to find for an FCC crystal:
εA
AεW FCC
TOT 6.8 -2
-12
26
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!)
FA
L
The Young’s modulus, Y, relates tensile stress and strain:
ttY
=
Connection between the macroscopic and the atomic?
Y
t
t
How does the interatomic force, F, relate to the macroscopic Y?
Model for Elastic Modulus of Molecular CrystalsWe 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.
The tensile stress t is defined as a force acting per unit area, so that:
2
)-(
A
F
o
ot
r
rrk
Fro
The tensile strain t is given as the change in length as a result of the stress:
o
o
ot r
rr
r
r
ooro
o
o
o
o
o
t
t
r
k
rrr
rrrk
Y )-(
)-(2
What is k?
Y can thus be expressed in terms of atomic interactions:
rWtot
+
-
-8.6ro
Elastic Modulus of Molecular Crystals
Orrdr
dFk
F = 0 when r = ro
rF
+
-
ro
dr
dWF tot
The Young’s modulus is sometimes known simply as the elastic modulus.
Linear region around ro is
approximated as: F = k(r - ro).
Elastic Modulus of Molecular CrystalsForce to separate atoms is the derivative of the potential:
7
66
13
1212 612-
2r
A
r
AF
dr
dW
So, taking the derivative again to find k:
8
66
14
1212 )6(7)1213(
2r
A
r
A
dr
dF
But we already know that: 6
1
6
122
A
Aro
Re-arranging, we see that: 61
12
6
2
A
Aro
We will therefore make a substitution for when finding k.
Elastic Modulus of Molecular Crystals
To find k, we now need to evaluate dF/dr when r = ro.
8
66
14
1212 )6(7)1213(
2r
A
r
A
dr
dF
Combining the constants to create new constants, C1 and C2, and setting r = ro, we can write:
)-
(2-2 221
8
62
14
121
oo
o
o
o
r
CC
r
rC
r
rCk
Finally, we find the Young’s modulus to be:
321 )-(2
oo r
CC
r
kY
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.
with6
1
12
6
2
A
Aro
Response of Condensed Matter 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)
How does soft matter respond to shear stress?
A
A
y
F
AF
s =
Elastic Response of Hookean Solids
No time-dependence in the response to stress. Strain is instantaneous and constant over time.
The shear strain s is given by the angle (in units of radians).
The shear strain s is linearly related to the shear stress by the shear modulus, G:
Gs
s
=
A
A
y
FAF
s =
yx
s
~=
x
Viscous Response of Newtonian Liquids
AF
s =
A
A
y
Fx
tx
v
=
There is a velocity gradient (v/y) normal to the area. The viscosity relates the shear stress, s, to the velocity gradient.
ytx
yv
s
==
The viscosity can thus be seen to relate the shear stress to the shear rate:
ttyx
ytx
s
The top plane moves at a constant velocity, v, in response to a shear stress:
v
has S.I. units of Pa s.
The shear strain increases by a constant amount over a time interval, allowing us to define a strain rate:
t Units of s-1
Hookean Solids vs. Newtonian Liquids
Hookean Solids: Gs
Newtonian Liquids: dt
ds
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.
Examples of viscoelastic systems include colloidal dispersions and melted polymers.
This type of response is called “viscoelastic”.
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 ViscoelasticitySpring: Elastic element
Dashpot: Viscous element
The model says that the elastic and viscous elements act in series, such that the shear stress, s, is the same for both of them.
The elastic shear strain, e, and the viscous shear strain, v, are additive:
ve γγγ
We define a Maxwell relaxation time: = /GM
As s is the same for both elements, we have:
dt
dG v
eMs
For a constant applied shear stress, the total strain can be written as: t
Gt s
M
s
)(
s
Stress Relaxation after a Step Strain
time
Constant shear strain applied
If viscoelastic, then the stress relaxes over time as molecules re-arrange.
eqs G viscoelastic solid
viscoelastic liquid
s
time
Hookean
Newtonian
t
Ms eGt-
)(
For a viscoelastic liquid, the shear stress relaxation in the Maxwell model is described as:
is the relaxation time
Stress Relaxation after a Step Strain
For a viscoelastic solid, the equilibrium shear modulus (non-zero) is defined as:
)()lim( tGtGeq
From here, we can define a time-dependent stress relaxation modulus:
t
Ms eG
ttG
)()(
For typical solids, is exceedingly large: 1012 s, such that there is no observed relaxation – just a Hookean response. For melted polymers, however, 1 s.
Response of Soft Matter to a Constant Shear StressApply a constant shear stress, s, and observe the time dependence of the shear strain, (t)
In this experiment, we find the creep compliance: J(t) = 1/G(t)= (t)/sRecalling that: t
Gt s
M
s
)(
We see in the Maxwell model that:η
t
GtJ
M
1
)(
Response of Soft Matter to a Constant Shear Stress
t
s
s ttJ
)(
)(
Elastic response
Viscous response
(provides initial and recoverable strain)
(strain increases linearly over time)
Slope:
1
)( ssdt
d
Jeq
Steady-state compliance = Jeq= 1/Geq
In the Maxwell model, Geq = / so, = Jeq and Jeq = /
(Pa-1)
Viscoelastic Response after Stress is Removed
t
s
s ttJ
)(
)(
Jeq
(Pa-1)
Constant stress, s, is applied
Stress removed
Jeq
Elastic strain is recovered
Viscous flow is permanent
Viscosity of Soft Matter Often Depends on the Shear Rate
Newtonian:
(simple liquids like water)
s
Shear thinning or thickening:
s
s s
s s
An Example of Shear Thickening
Future lectures will explain how polymers and colloids respond to shear stress.
Video: shear thickening
Problem Set 11. 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
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 14.45A12 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.
(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.
,2)(6
6
12
12
rA
rAru
r
ze
Problem Set 21. 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 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) 0 10 20 30 40 50(10-4 Pa s) 17.93 13.07 10.02 7.98 6.53 5.47
Temp (C) 60 70 80 90 100(10-4 Pa s) 4.67 4.04 3.54 3.15 2.82
3. In poly(styrene) the relaxation time for configurational rearrangements follows a Vogel-Fulcher law given as
= o exp(B/T-To),
where B = 710 C and To = 50 C. In an experiment with an effective timescale of exp = 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 exp = 105 s, what value of Tg would be obtained?
Typical Relaxation TimesFor solids, is exceedingly large: 1012 s
For simple liquids, is very small: 10 -12 s
For soft matter, takes intermediate values. For instance, for melted polymers, 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, .
Stress,
Time
Strain,
Time
Y
0 0
Response of Soft Matter to a Constant Shear Stress: Viscoelasticity
t
s
ttG
)(=
)(1
oG1
Slope:
1
==)(s
s
s
s
dtd
We see that 1/Go (1/)
is the “relaxation time”
Hence, viscosity can be approximated as Go
Modelling of Creep and Relaxation
Maxwell model: Spring and dashpot in seriesInstantaneous elastic deformation is followed by flow. Elastic deformation is recovered but not the flow.
Voigt-Kelvin model: Spring and dashpot in parallel
Anelastic deformation is followed by recovery with an exponential dependence on = /Y, which is called the relaxation time.
tY
t
YYt
Yexp1exp1Deformation:
Total Creep and Recovery
Time
Displacement
Strain is described by the sum of the Maxwell and Voigt-Kelvin models.
Elastic
Anelastic
Flow
Constant stress applied
Spring and dashpot acting both in series
and in parallel.
R
Creation of a New Surface Leads to a “Thermodynamic” Adhesion Force
F
is the surface tension (energy) of the tip and the surface - assumed here to be equal.
Work of adhesion:
Surface area increases when tip is removed.
∫= dAW
S
L
LSA
L
SSVA
LVA
dAW
When liquid (L) and solid (S) are separated, two new interfaces with the vapour (V) are created.
W = LVA + SVA - LSA
Work per unit area, W:
WLV+ SV - SL
SV
LV
SL
cosLVSLSV cosLVSLSV
Work per unit area, W:
)cos1( LVWYoung-Dupré Equation cosLVLVW
121≈]
11[=
P
RdPAF
2≈=1
Pressure is required to bend a surface with a surface tension,
F
Max. Capillary Force: F 4R cos
With = 0.072 N/m for water and R = 10 nm, F is on the order of 10-8 -10-9 N!
The Capillary Force
Force for Polymer Deformation
Hookean Solids vs. Newtonian Liquids
Hookean Solids: γGσ
Newtonian Liquids: γηdt
γdησ
Many substances, i.e. “structured liquids”, display both types of behaviour, depending on the time scale: solid-like on short time-scales 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.