By Marcus Müller, Friederike Schmid (auth.), Christian Holm, Kurt Kremer (eds.)
This sequence offers severe stories of the current and destiny developments in polymer and biopolymer technology together with chemistry, actual chemistry, physics and fabrics technology. it truly is addressed to all scientists at universities and in who desire to maintain abreast of advances within the issues lined.
Impact issue score: continuously no 1 in Polymer technology. additional info in addition to the digital model of the complete content material on hand at: www.springerlink.com
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Extra info for Advanced Computer Simulation Approaches for Soft Matter Sciences II
Müller · F. Schmid ﬁelds WA can be derived: ⎛ φA∗ (r) = φ¯ A ⎝1 – ⎞ gA (q)WAq eiqr ⎠ q=0 ⎛ 1 = φ¯ A ⎝1 – V (88) ⎞ gA (q)WA (r ) eiq(r–r ) ⎠ dd r q=0 A similar expression is valid for φB∗ (r). Here g(q) = (2/x2 )(exp[– x] – 1 + x) with x ≡ R2e q2 /6 denotes the Debye function. Hence, the Fourier transform of KAA can be identiﬁed with the single chain structure factor: KAA (r, r ) =– 1 δφA∗ (r) = φ¯ A δWA (r ) V KAA (q) = φ¯ A gA (q) , gA (q) eiq(r–r ) (89) q=0 ˜ Eq. 84, one oband a similar expression holds KBB .
Our simulation data are 46 M. Müller · F. Schmid Fig. 9 Anisotropy parameters F¯2 (solid) and F¯4 (dashed) vs. χN for different homopolymer volume fractions ΦH at the order/disorder transition. Left: From Monte Carlo Simulations (EP theory) Right: From Complex Langevin simulations. After  consistent with the assumption that the transition is continuous or weakly ﬁrst order. In particular, we have established that the miscibility gap between the ordered and the disordered phase is very small (data not shown).
Relatively few Fourier components are often sufﬁcient to characterize the structure of a phase satisfactorily. In Fourier space, Eq. 65 turns into a matrix equation ∂q(k, t) =– ∂t Ak,k (t)q(k , t) (66) k,k with R2e 2 k δk,k + W(k, k ) (67) 6 and the initial condition q(k, 0) = δ(k, 0). The formal solution of this equation is Ak,k = [exp(– At)]k,k q(k , 0) . q(k, t) = (68) k It can be evaluated by diagonalizing the matrix A. , one has no errors due to a time discretization. Unfortunately, the efﬁciency drops rapidly if the number of Fourier components and hence the dimension of the matrix A is very large.