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I propose this : As long as the detailed balance is respected, the system can be studied with equilibrium thermodynamic. Therefore we can write the free energy of the solution, for exemple $F=A \log(A) + B \log(B) + S \log(S) + \chi_{A S} A S + \chi_{B S} B S + \chi_{A B} A B$ ($S$ the solvent, $\chi$ the interaction parameters), plug the equilibrium ...

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In thermodynamic equilibrium, the solidification process can be tracked using the phase diagram of water and salt. One example (from wikipedia) is: It is fairly straightforward as a binary phase diagram. Above 0C, adding NaCl to water results in complete dissolution until somewhere above 26.3 wt.%. At that point, trying to stir more in will result in ...

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To grasp the relevant physics at a sloppy level, perhaps you simply need a few examples. You know a concept is commonly constructed by the manner you refer to it together with other concepts. Symmetry breaking usually results in ground state degeneracy and long range order. Order parameter field aids you in identifying degenerate sectors with the symmetries ...

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To answer your second question, yes, condensation typically requires nucleation. Supersaturated vapours like the one you describe are the basis of the cloud chambers that used to be used as particle detectors. The energetic particles passing through the vapor would ionize molecules, and these ions would act as nucleation sites.

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For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)).

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Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ...

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I guess you can use the Van der Waals equation and some estimates of the molecule volume and intermolecular attractive forces. The parameters of the critical point depend on these characteristics, so you need to assess them.

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An order parameter distinguishes two different phases (or orders). In one phase the order parameter is zero and in another phase it is non-zero. It does not have to be macroscopic. For example, in the BCS theory of superconductivity the order parameter is called the gap $\Delta$. It can be interpreted as the binding energy of a Cooper pair, namely two ...

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It's because normal ice, ice Ih, is less dense than liquid water. Ice Ih forms hexagonal crystals. The bonds in that crystalline structure make the water molecules slightly further apart than they are in the liquid form at the same pressure. That water expands on freezing makes water resist freezing as pressure increases. This in turn makes the fusion point ...

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My previous comments are almost ok, your actual problem seems to be that you do not average over the magnetization. You are measuring <cos(theta)>, which is the average of m_x. So just change m = magnetization_cossin(); to magnetization_cossin(&mx, &my); where you define void magnetization_cossin(double* mx, double* my) { int x, y, z; ...

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I think that you are really interested in the $q$-state clock model, which is similar to the Potts model, and is defined as follows. Fix an integer $q\geq2$. For each $i\in\mathbb{Z}^d$, let $$\theta_i \in \bigl\{\frac{2\pi}{q} k\,:\, k\in\{0,1,\ldots,q-1\}\bigr\},$$ and define the spin at site $i$ by $$\mathbf{S}_i = (\cos\theta_i,\sin\theta_i) .$$ The ...

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