The Phases of Hard Sphere Systems Tutorial for Spring 2015 ICERM - PowerPoint PPT Presentation

The Phases of Hard Sphere Systems Tutorial for Spring 2015 ICERM workshop “Crystals, Quasicrystals and Random Networks” Veit Elser Cornell Department of Physics

outline: • order by disorder • constant temperature & pressure ensemble • phases and coexistence • ordered-phase nucleation • infinite pressure limit • what is known in low dimensions • the wilds of higher dimensions

eliminating time: ergodic hypothesis sphere centers: x 1 , x 2 , . . . , x N arbitrary property: θ ( x 1 , x 2 , . . . , x N )

eliminating time: ergodic hypothesis sphere centers: x 1 , x 2 , . . . , x N arbitrary property: θ ( x 1 , x 2 , . . . , x N ) Z T θ = 1 ¯ time average: dt θ ( x 1 ( t ) , . . . , x N ( t )) T 0

eliminating time: ergodic hypothesis sphere centers: x 1 , x 2 , . . . , x N arbitrary property: θ ( x 1 , x 2 , . . . , x N ) Z T θ = 1 ¯ time average: dt θ ( x 1 ( t ) , . . . , x N ( t )) T 0 Z h θ i = dµ θ ( x 1 , . . . , x N ) configuration average:

eliminating time: ergodic hypothesis sphere centers: x 1 , x 2 , . . . , x N arbitrary property: θ ( x 1 , x 2 , . . . , x N ) Z T θ = 1 ¯ time average: dt θ ( x 1 ( t ) , . . . , x N ( t )) T 0 Z h θ i = dµ θ ( x 1 , . . . , x N ) configuration average: ¯ θ = h θ i ergodic hypothesis:

dµ = d D x 1 · · · d D x N e − β H 0 ( x 1 ,...,x N ) Gibbs measure: ( x 1 , . . . , x N ) ∈ A N A ⊂ R D “box” domain:

dµ = d D x 1 · · · d D x N e − β H 0 ( x 1 ,...,x N ) Gibbs measure: ( x 1 , . . . , x N ) ∈ A N A ⊂ R D “box” domain: absolute temperature: 1 / β = k B T sphere diameter: a

dµ = d D x 1 · · · d D x N e − β H 0 ( x 1 ,...,x N ) Gibbs measure: ( x 1 , . . . , x N ) ∈ A N A ⊂ R D “box” domain: absolute temperature: 1 / β = k B T sphere diameter: a Hamiltonian: ⇢ 0 if 8 i 6 = j k x i � x j k > a H 0 ( x 1 , . . . , x N ) = 1 otherwise

constant temperature & pressure ensemble environment pV = work performed on constant pressure environment V in expanding box to volume V T , p H ( x 1 , . . . , x N ; V ) = H 0 ( x 1 , . . . , x N ) + pV

dimensionless variables & parameters x i = a ˜ x i V = a D ˜ V

dimensionless variables & parameters x i = a ˜ x i V = a D ˜ V p ˜ β p V = ˜ V p = ( k B T/a D ) ˜ p

rescaled constant T & p ensemble measure: dµ ( p ) = d D x 1 · · · d D x N dV e − H 0 − pV

rescaled constant T & p ensemble measure: dµ ( p ) = d D x 1 · · · d D x N dV e − H 0 − pV configurations: ( x 1 , . . . , x N ) ∈ A ( V ) N A ( V ) = R D / ( V 1 /D Z ) D (periodic box) 8 i 6 = j k x i � x j k > 1

rescaled constant T & p ensemble measure: dµ ( p ) = d D x 1 · · · d D x N dV e − H 0 − pV configurations: ( x 1 , . . . , x N ) ∈ A ( V ) N A ( V ) = R D / ( V 1 /D Z ) D (periodic box) 8 i 6 = j k x i � x j k > 1 dimensionless pressure (parameter): p specific volume (property): v = h V i /N

p = 2 . 00 v = 1 . 73

p = 2 . 00 v = 1 . 73

p = 20 . 00 v = 1 . 02

p = 20 . 00 v = 1 . 02

108 spheres ( D = 3) √ v = 1 / 2

φ = | B D (1 / 2) | packing fraction: v

quiz: hard sphere phase diagram crystal crystal p p p gas crystal gas gas T T T

quiz: hard sphere phase diagram crystal crystal p p p gas crystal gas gas T T T p ( k B /a 3 ) T a ≈ 4˚ (krypton) p = ˜ A p ≈ 10 ˜ p ( k B /a 3 ) ≈ 20 atm / K ˜

thermal equilibrium Here we are, trapped in the amber of the moment. — Kurt Vonnegut • configuration sampling via Markov chains, small transitions • Markov sampling is not unlike time evolution • diffusion process: slow for large systems

approaching equilibrium 108 spheres p = 10 − 5 ˙ 10 − 6 10 − 7 10 − 8 ∆ p ˙ ∆ t = ˙ p > 0 φ > 0

approaching equilibrium 108 spheres p = 10 − 5 ˙ 10 − 6 10 − 7 10 − 8 p ∗ φ 1 φ 2 ∆ p ˙ ∆ t = ˙ p > 0 φ > 0

quiz: between phases Suppose we sample hard spheres (3D) at fixed volume , corresponding to an intermediate packing fraction (0.52), what should we expect to see in a typical configuration? A. gas phase B. crystal phase C. quantum superposition of gas and crystal D. none of the above

quiz: between phases Suppose we sample hard spheres (3D) at fixed volume , corresponding to an intermediate packing fraction (0.52), what should we expect to see in a typical configuration? A. gas phase B. crystal phase C. quantum superposition of gas and crystal D. none of the above

phase coexistence At the coexistence pressure, gas and crystal subsystems are in equilibrium with each other. We can accommodate a range of volumes by changing the relative fractions of the two phases. φ 1 = 0 . 494 gas { p ∗ = 11 . 564 φ 1 < φ < φ 2 gas crystal φ 2 = 0 . 545 crystal

thermodynamics Z Z dµ (0) e − pV number of microstates: Z ( p ) = dµ ( p ) =

thermodynamics Z Z dµ (0) e − pV number of microstates: Z ( p ) = dµ ( p ) = s ( p ) = k B entropy per sphere: N log Z ( p )

thermodynamics Z Z dµ (0) e − pV number of microstates: Z ( p ) = dµ ( p ) = s ( p ) = k B entropy per sphere: N log Z ( p ) free energy per sphere: F ( p ) = − Ts ( p )

thermodynamics Z Z dµ (0) e − pV number of microstates: Z ( p ) = dµ ( p ) = s ( p ) = k B entropy per sphere: N log Z ( p ) free energy per sphere: F ( p ) = − Ts ( p ) identity: v = 1 N h V i = � 1 dp log Z ( p ) = � 1 d ds N k B dp

s gas ( p ∗ ) = s cryst ( p ∗ ) s gas s ( p ) s cryst p ∗ v gas ds v = − k − 1 ∆ v B dp v cryst p ∗

phase equilibrium: the ice-9 problem … suppose, young man, that one Marine had with him a tiny capsule containing a seed of ice-nine, a new way for the atoms of water to stack and lock, to freeze . If that Marine threw that seed into the nearest puddle … ? — Kurt Vonnegut (Cat’s Cradle) • Phase coexistence implies there is an entropic penalty, a negative contribution from the interface, proportional to its area. • The interfacial penalty severely inhibits the spontaneous formation of the ordered phase, especially near the coexistence pressure. “critical nucleus”

size of critical nucleus Assume interface-surface entropy is isotropic, so the critical nucleus is spherical:

size of critical nucleus Assume interface-surface entropy is isotropic, so the critical nucleus is spherical: s nuc = ( s cryst − s gas ) V nuc − σ A nuc v

size of critical nucleus Assume interface-surface entropy is isotropic, so the critical nucleus is spherical: s nuc = ( s cryst − s gas ) V nuc − σ A nuc v s cryst ( p ∗ + ∆ p ) − s gas ( p ∗ + ∆ p ) = k B ( − v cryst + v gas ) ∆ p = k B ( − ∆ v ) ∆ p > 0

size of critical nucleus Assume interface-surface entropy is isotropic, so the critical nucleus is spherical: s nuc = ( s cryst − s gas ) V nuc − σ A nuc v s cryst ( p ∗ + ∆ p ) − s gas ( p ∗ + ∆ p ) = k B ( − v cryst + v gas ) ∆ p = k B ( − ∆ v ) ∆ p > 0 V nuc = (4 π / 3) R 3 A nuc = 4 π R 2

size of critical nucleus Assume interface-surface entropy is isotropic, so the critical nucleus is spherical: s nuc = ( s cryst − s gas ) V nuc − σ A nuc v s cryst ( p ∗ + ∆ p ) − s gas ( p ∗ + ∆ p ) = k B ( − v cryst + v gas ) ∆ p = k B ( − ∆ v ) ∆ p > 0 V nuc = (4 π / 3) R 3 A nuc = 4 π R 2 s nuc R c

R c = 3( σ /k B )( v/ ∆ v ) 1 s nuc ( R c ) = 0 ⇒ ∆ p e − V nuc /v 0 exponentially small nucleation rate:

R c = 3( σ /k B )( v/ ∆ v ) 1 s nuc ( R c ) = 0 ⇒ ∆ p e − V nuc /v 0 exponentially small nucleation rate: The critical nucleus would in any case have to be at least as large to contain most of a kissing sphere configuration, thus making the spontaneous nucleation rate decay exponentially with dimension.

infinite pressure limit ✓ ◆ ds − k − 1 = v min lim B dp p →∞

infinite pressure limit ✓ ◆ ds − k − 1 = v min lim B dp p →∞ s ( p ) /k B ∼ − v min p + C

infinite pressure limit ✓ ◆ ds − k − 1 = v min lim B dp p →∞ s ( p ) /k B ∼ − v min p + C Z ( p ) ∼ e − N ( v min p + C )

infinite pressure limit ✓ ◆ ds − k − 1 = v min lim B dp p →∞ s ( p ) /k B ∼ − v min p + C Z ( p ) ∼ e − N ( v min p + C ) In three dimensions we know there are many sphere packings — stacking sequences of hexagonal layers — with exactly the same 𝑤 min . The constant C depends on the stacking sequence.

free volume: disks

free volume: disks Allowed positions of central disk when surrounding disks are held fixed.

free volume: spheres φ = 0 . 45 hexagonal hexagonal cubic AB … AC …

hexagonal stacking B C A “hexagonal close packing” (hcp) : … ABAB … “face-centered cubic” (fcc) : … ABCABC …

densest & most probable sphere packing p →∞ ∆ s ( p ) /k B = C fcc − C hcp ≈ 0 . 001164 lim prob(fcc) N = 1000 : prob(hcp) ≈ 3 . 2 fcc and hcp stackings are the extreme cases; other stackings have intermediate probabilities.