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 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. Discussed polarisability of molecules (electronic and orientational) and described charge-nonpolar, polar-nonpolar, and dispersive (London) interactions. Summarised ways to measure polarisability. +Q + - u u u a a +Q - + a a u

Type of Interaction Interaction Energy, w(r) Summary Type of Interaction Interaction Energy, w(r) In vacuum: e=1 Charge-charge Coulombic Dipole-charge Dipole-dipole Keesom Charge-nonpolar Dipole-nonpolar Debye Nonpolar-nonpolar Dispersive

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 r = number of molecules per unit volume  s -3, where s 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! 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 s, then the shortest centre-to-centre distance will likewise be s. • Thus the boiling point is approximately:

Comparison of Theory and Experiment London equation Evaluated at close contact where r = s. Non-polar (Per mole, n = 1) Note that ao and C increase with s.

Additivity of Interactions Molecule Mol. Wt. u (D) TB(°C) C-C H Dispersive only Ethane: C2H6 30 0 -89 C=O H Keesom + dispersive Formaldehyde: CH2O 30 2.3 -21 C-O-H H H-bonding + Keesom + dispersive Methanol: CH3OH 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 Actual s  0.3 nm Ar boiling point: TB = 87 K London 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 s  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 Attractive force Very short-range force! Equ’m F = dw/dr s Repulsive force r (m)

Comparison of Force and Potential Energy for Ar Pair F= dw/dr

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 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. Radius of curvature ~ 10 nm Ideally, one of the atoms at the tip is slightly above the others. 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 A = approach B = “jump” to contact C = contact D = adhesion E = pull-off Tip deflection  Force C C A B E D Vertical position

Measuring Force of Attraction to a Polymer Surface Pushing on AFM probe tip Pulling on the 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.

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. Distance between mica sheets is measured with interferometry. www.fisica.unam.mx/liquids/tutorials/surface.html

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 s. Note that when r = s, then w = 0. e is a bond energy (related to the London constant), such that w(r) = - e when r is at the equilibrium spacing of r = ro.

Lennard-Jones Potential for Molecular Pairs in a Crystal + w(r) -e ro r s -

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: To find the minimum energy in the potential, we can evaluate it when r = ro:

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. 8 nearest neighbours; 6 2nd nearest; 12 3rd nearest 12 nearest neighbours; 6 second nearest; 24 3rd nearest 6 nearest neighbours; 12 second 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 Sr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 So, for a pair we write the interaction potential as: For each atom in a molecular crystal, however, we write that the cohesive energy is: From the first derivative, we can find the equilibrium spacing for an FCC crystal: 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: 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!)

F Y Connection between the macroscopic and the atomic? st A et L The Young’s modulus, Y, relates tensile stress and strain: How does the interatomic force, F, relate to the macroscopic Y?

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. F ro ao ro ao The tensile stress st is defined as a force acting per unit area, so that: The tensile strain et is given as the change in length as a result of the stress: Y can thus be expressed in terms of atomic interactions: What is k?

Elastic Modulus of Molecular Crystals Wtot + - -8.6e ro s The Young’s modulus is sometimes known simply as the elastic modulus. F = 0 when r = ro r F + - ro 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: So, taking the derivative again to find k: But we already know that: Re-arranging, we see that: We will therefore make a substitution for s when finding k.

Elastic Modulus of Molecular Crystals with 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: Finally, we find the Young’s modulus to be: 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 y 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 q (in units of radians). A F Dx y q A The shear strain gs is linearly related to the shear stress by the shear modulus, 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: v A y F Dx There is a velocity gradient (v/y) normal to the area. The viscosity h relates the shear stress, ss, to the velocity gradient. h 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: Units of s-1 The viscosity can thus be seen to relate the shear stress to the shear rate:

Hookean Solids vs. Newtonian Liquids 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 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: Dashpot: Viscous element ss As ss is the same for both elements, we have: For a constant applied shear stress, the total strain can be written as: We define a Maxwell relaxation time: t = h/GM

Stress Relaxation after a Step Strain g Constant shear strain applied time viscoelastic solid viscoelastic liquid t ss time Hookean Newtonian 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: 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 modulus: For a viscoelastic solid, the equilibrium shear modulus (non-zero) is defined as:

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. Recalling that: We see in the Maxwell model that:

Response of Soft Matter to a Constant Shear Stress Viscous response Slope: (Pa-1) (strain increases linearly over time) Jeq Steady-state compliance = Jeq= 1/Geq Elastic response (provides initial and recoverable strain) t In the Maxwell model, Geq = h/t  so, t = hJeq and Jeq = t/h

Viscoelastic Response after Stress is Removed Elastic strain is recovered Jeq (Pa-1) Jeq Constant stress, ss, is applied Viscous flow is permanent t Stress removed

Viscosity of Soft Matter Often Depends on the Shear Rate ss Newtonian: (simple liquids like water) h ss h or thickening: Shear thinning h

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 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 q 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. r q ze

Problem Set 2 t = to exp(B/T-To), 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, n. Temp (C) 0 10 20 30 40 50 h (10-4 Pa s) 17.93 13.07 10.02 7.98 6.53 5.47 Temp (C) 60 70 80 90 100 h (10-4 Pa s) 4.67 4.04 3.54 3.15 2.82 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, s Time h Y Strain, e Time

Response of Soft Matter to a Constant Shear Stress: Viscoelasticity Slope: t t is the “relaxation time” t We see that 1/Go  (1/h)t Hence, viscosity can be approximated as h  Got

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. 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:

Total Creep and Recovery Constant stress applied Displacement Anelastic Elastic Flow Spring and dashpot acting both in series and in parallel. Time Strain is described by the sum of the Maxwell and Voigt-Kelvin models.

Creation of a New Surface Leads to a “Thermodynamic” Adhesion Force Surface area increases when tip is removed. G is the surface tension (energy) of the tip and the surface - assumed here to be equal. Work of adhesion:

L W = GLVA + GSVA - GLSA L S S GLVA GLSA GSVA GSV q GLV GSL Work per unit area, W: W = GLV+ GSV - GSL L GLVA GLSA S GSVA S GSV q GLV GSL When liquid (L) and solid (S) are separated, two new interfaces with the vapour (V) are created. Work per unit area, W: Young-Dupré Equation

F The Capillary Force F  4pRG cosq Pressure is required to bend a surface with a surface tension, G F F  4pRG cosq Max. Capillary Force: 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

Hookean Solids vs. Newtonian Liquids 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.