Model statement & free energy Laplace’s method Phase transitions The tight-knit community model 1 Model statement & free energy 2 Laplace’s method 3 Free energy, phase transitions 1 / 24 Model statement & free energy Laplace’s method Phase transitions Who talks to whom in Gridhamshire? Neighbours talk about politics: won’t vote independently? Could look at all pairs (i, j) of (nearest) neighbours σi σj = +1 means they vote the same way σi σj = −1 means they vote opposite But maybe everyone talks to everyone else (facebook, . . . )? To measure how likely voters are to “align” opinions, consider E(σ) = 1 N X σi σj all pairs i,j Larger E: more tendency for voters to align Assume we know both hM (σ)i and hE(σ)i Maximum entropy distribution P (σ) of voting patterns? 2 / 24 Model statement & free energy Laplace’s method Phase transitions Calculation of free energy First rewrite E(σ) = 1 2N P i6=j σi σj Can be expressed in terms of M : E(σ) = M 2 (σ)/(2N ) − 1/2 Maximum entropy distribution: P (σ) = (1/Z)eλ1 M (σ)+λ2 E(σ) Abbreviate λ1 ≡ h, λ2 ≡ J So need to calculate partition function X 2 Z= ehM (σ)+(J/2N )M (σ)−J/2 σ and then free energy F = − ln Z 3 / 24 Model statement & free energy Laplace’s method Phase transitions Counting states Exponent doesn’t factorize But get same contribution from every σ with same M (σ) How many states for given M ? (N + M )/2 voters vote +1, (N − M )/2 voters −1 So number of states = binomial coefficient: X 2 N Z= ehM +(J/2N )M −J/2 (N + M )/2 M where M = −N , −N + 2, . . . , N − 2, N Better than before: a sum with N + 1 rather than 2N terms Can we simplify further for large N ? 4 / 24 Model statement & free energy Laplace’s method Phase transitions Converting to an integral for large N Can write sum for Z in terms of normalized poll lead m = M/N : X 2 N Z= eN [hm+(J/2)m ]−J/2 N (1 + m)/2 m where m = −1, −1 + 2/N , . . . , 1 − 2/N , 1 Think of numbers being summed as heights of rectangles P Rectangle width 2/N , so (2/N ) × m . . . = Riemann sum For N → ∞, rectangles become narrow, approach integral: Z N 1 2 N Z≈ eN [hm+(J/2)m ]−J/2 dm 2 −1 N (1 + m)/2 Subtlety: integrand also depends on N ; replacement of sum by integral for large N can be shown to work nonetheless 5 / 24 Model statement & free energy Laplace’s method Phase transitions Simplifying the integrand for large N Want to simplify factorials N and N (1 ± m)/2 are all large for large N (except at endpoints m = ±1, but these turn out not to matter) So can use Stirling n! ≈ Z≈ N 2 Z 1 −1 Here s(m) = − √ 2πn(n/e)n to get 2 eN s(m) 2 p eN [hm+(J/2)m ]−J/2 dm 2πN (1 − m2 ) 1+m ln 2 1+m 2 − 1−m ln 2 1−m 2 6 / 24 Model statement & free energy Laplace’s method Phase transitions Interpretation of s(m) We defined s(m) = − 1+m ln 2 1+m 2 − 1−m ln 2 1−m 2 This is the entropy of a voter making decisions with probabilities P (σi = ±1) = (1 ± m)/2 So as before, entropy can be seen as counting number of states: to exponential accuracy N = eN s(m) N (1 + m)/2 Empirical probabilities are [N (1 ± m)/2]/N = (1 ± m)/2 for σ = ±1 as found above by inspection 7 / 24 Model statement & free energy Laplace’s method Phase transitions The tight-knit community model 1 Model statement & free energy 2 Laplace’s method 3 Free energy, phase transitions 8 / 24 Model statement & free energy Laplace’s method Phase transitions Laplace’s method: Motivation Summarizing so far, have for large N r Z N 1 ˜ Z ≈ e−J/2 (1 − m2 )−1/2 e−N f (m) dm 2π −1 with f˜(m) = −hm − (J/2)m2 − s(m) How do we do the integral, again for large N ? ˜ Exponential e−N f (m) is largest at the minimum of f˜(m) Factor N ⇒ drops very quickly away from the minimum So integrand is sharply peaked around m = m∗ where m∗ = argminm f˜(m) is position of minimum Intuition: large is easier; large N ⇒ small fluctuations in m 9 / 24 Model statement & free energy Laplace’s method Phase transitions Laplace’s method Method for doing integrals of form IN = Rb a dx w(x)e−N g(x) Idea: for large N integral is dominated by values of x near (global) minimum of g(x) Call position of minimum x∗ = argminx g(x) p ∗ Result: IN = 2π/[N g 00 (x∗ )]w(x∗ )e−N g(x ) Leading term: −(1/N ) ln IN → g(x∗ ) = minx g(x) for N →∞ Informally: IN = e−N minx g(x) to exponential accuracy We’ve assumed that a < x∗ < b: if minimum is at end points, exponential term in IN is the same but prefactors are different 10 / 24 Model statement & free energy Laplace’s method Phase transitions Application: Stirling approximation Claim: for large n, n! ≈ √ 2πn(n/e)n (lhs/rhs → 1 for n → ∞) For normalized log factorials, need only leading term: 1 ln n! ≈ ln(n) − 1 n Formally, lhs − rhs → 0 for n → ∞ Derivation by Laplace’s method: 11 / 24 Model statement & free energy Laplace’s method Phase transitions Back to our partition function We had Z ≈ e−J/2 [N/(2π)]1/2 Laplace’s method gives us R1 −1 (1 ˜ − m2 )−1/2 e−N f (m) dm ˜ Z ≈ e−J/2 [N/(2π)]1/2 {2π/[N f˜00 (m∗ )]}1/2 [1−(m∗ )2 ]−1/2 e−N f (m ∗) So free energy per voter f = F/N = −(1/N ) ln Z is f≈ 1 [J + ln(f˜00 (m∗ )) + ln(1 − (m∗ )2 )] + f˜(m∗ ) 2N All terms except last go to zero for N → ∞ So for large N everything is reduced to a minimization problem: f = f˜(m∗ ) = min f˜(m) m 12 / 24 Model statement & free energy Laplace’s method Phase transitions The tight-knit community model 1 Model statement & free energy 2 Laplace’s method 3 Free energy, phase transitions 13 / 24 Model statement & free energy Laplace’s method Phase transitions Final expression for free energy Tight-knit community model with many voters (large N ) Free energy per voter is f (h, J) = minm f˜(m, h, J) with J f˜(m, h, J) = −hm − m2 2 1+m 1+m 1−m 1−m ln + ln + 2 2 2 2 As we vary h and J, can get phase transitions E.g. one minimum may split in two Or one minimum becomes lower than other 2 − s(m) is consistent Form of f˜(m, h, J) = −hm − (J/2)m P with general relation F/N = − C λ c=1 c hfc (x)i/N − S[P ]/N 14 / 24 Model statement & free energy Laplace’s method Phase transitions Meaning of f˜(m) Look at the distribution of M By counting states with same M as before, 1 2 N P (M ) = ehM +(J/2N )M −J/2 (N + M )/2 Z Convert to probability density p(m) of m = M/N : P (M ) = (2/N )p(m) for large N To exponential accuracy, ˜ ˜ p(m) = eN [f (h,J)−f (m,h,J)] ∝ e−N f (m,h,J) So f˜(m, h, J) determines directly the distribution of m 15 / 24 Model statement & free energy Laplace’s method Phase transitions Mean and variance of m ˜ e−N f (m,h,J) To p(m) ∝ Laplace method can apply same argument as in Narrow peak around minimum m∗ of f˜(m, h, J) Expanding around m∗ shows p(m) ∝ e−N (∂ 2 f˜/∂m2 )(m−m∗ )2 /2 where ∂ 2 f˜/∂m2 is evaluated at m∗ (and the given h, J) This is a Gaussian with mean and variance hmi = m∗ , h(∆m)2 i = 1/[N (∂ 2 f˜/∂m2 )] Variance ∝ 1/N as for independent voters: large is easier Low curvature of f˜ at minimum gives larger variance 16 / 24 Model statement & free energy Laplace’s method Phase transitions Plots of f˜(m, h, J) and p(m) For h = 0 7 ~ f p(m) J = 0.4 J = 0.6 J = 0.8 J=1 J = 1.2 J = 1.4 -0.3 -0.4 -0.5 Gridhamshire, N = 400 6 5 4 3 -0.6 2 -0.7 1 -0.8 -1 0 -0.5 0 -1 1 0.5 m 0 -0.5 1 0.5 m Continuous transition to collective voting at J = 1 17 / 24 Model statement & free energy Laplace’s method Phase transitions Effect of small changes in party preference h = -0.03 h=0 h = 0.03 -0.5 -0.5 -0.5 -0.6 -0.6 -0.6 -0.7 -0.7 -0.7 -1 -0.5 -0.5 0 0.5 1 -1 -0.5 -0.5 0 0.5 1 -1 -0.5 -0.6 -0.6 -0.6 -0.7 -0.7 -0.7 -1 -0.5 -0.5 0 0.5 1 -1 -0.5 -0.5 0 0.5 1 -1 -0.5 -0.6 -0.6 -0.6 -0.7 -0.7 -0.7 -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 -1 J=0.8 -0.5 0 0.5 1 J=1 -0.5 0 0.5 1 J=1.2 -0.5 0 0.5 1 Overall party preference encoded by h Even small changes in h can have large effects on m 18 / 24 Model statement & free energy Laplace’s method Phase transitions Mean normalized poll lead So far: mean poll lead given by position m∗ of peak in p(m) Equivalently, position of minimum in f˜ Condition ∂ f˜/∂m = 0 can be written as m∗ = tanh(Jm∗ + h) Self-consistency equation Non-interacting case J = 0: m∗ = tanh(h) as before For J > 1 can have up to 3 different solutions; need to choose the one with lowest f˜ 19 / 24 Model statement & free energy Laplace’s method Phase transitions Dependence on constituency size hmi vs h for J = 1.1, N = 23 , . . . , 210 0.5 -0.2 -0.1 0.1 0.2 -0.5 Position of lowest minimum in f˜ jumps at h = 0 if J > 1: abrupt transition Transition only emerges formally for N → ∞; otherwise hmi is smooth function of h 20 / 24 Model statement & free energy Laplace’s method Phase transitions Subtleties at h = 0 When no overall bias for either party (h = 0) and strong voter alignment (J > 1), p(m) has two peaks of equal height Peak positions are ±m∗ if m∗ is positive solution of m∗ = tanh(Jm∗ ) But hmi = 0 by symmetry (also true for any finite N at h = 0) Only case where we need to distinguish m∗ and hmi Interpretation: half of all constituencies have m ≈ m∗ , the other half m ≈ −m∗ 21 / 24 Model statement & free energy Laplace’s method Phase transitions Finding hmi from free energy From p(m) we found hmi = m∗ (except at h = 0, J > 1) Should get same from hM i = −∂F/∂h, i.e. hmi = −∂f /∂h This is true: if minimum of f˜ is at m∗ (h, J), then using f (h, J) = f˜(m∗ (h, J), h, J) find hmi = −∂f /∂h = m∗ (h, J) 22 / 24 Model statement & free energy Laplace’s method Phase transitions Variance h(∆m)2 i from free energy We know h(∆M )2 i ≡ hM 2 i − hM i2 = −∂ 2 F/∂h2 = ∂hM i/∂h So h(∆m)2 i = (1/N )∂hmi/∂h = (1/N )∂m∗ (h, J)/∂h Differentiating condition ∂ f˜/∂m = 0 for m∗ (h, J) w.r.t. h gives 1 h(∆m)2 i = N ∂ 2 f˜/∂m2 Again this agrees with result obtained from p(m) 23 / 24 Model statement & free energy Laplace’s method Phase transitions Phase diagram Summarize behaviour of tight-knit community as f’n of h and J 24 / 24

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