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The order parameter $\mathcal{O}$ in the case of a liquid-gas transition is the density difference $\mathcal{O}=\rho_{liq}-\rho_{gas}$.

But in the case of a liquid-to-solid transition, the order parameter $\mathcal{O}$ is not taken as the density difference $\mathcal{O}=\rho_{sol}-\rho_{liq}$. What is the reason? Is it just because the density difference is too small to measure?

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There is no unique definition of "the" order parameter. However, we would like to use an order parameter that exhibits the full symmetry breaking pattern of the transition. If the transition corresponds to some symmetry $G$ breaking to a smaller symmetry $H$ then we want the order parameter to transform non-trivially under $G$, and exhibit the residual symmetry $H$. In the liquid-gas transition there are no continuous symmetries, and using the density is fine. In liquid-solid we break translational symmetry to a crystallographic symmetry. This requires a more complicated order parameter, like the F-trafo of the density correlator.

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  • $\begingroup$ Dear @Thomas Does the order parameter distinguish two phases between which the transition takes place? $\endgroup$ – SRS Jul 29 '17 at 4:44
  • $\begingroup$ @SRS yes it is designed to distinguish the two phases. Otherwise it would be a poor choice of order parameter. $\endgroup$ – KF Gauss Jul 29 '17 at 4:47
  • $\begingroup$ Yes, it is obviously different. But the point is that we want more, we would like the order parameter to transform non-trivially under broken symmetries, $\endgroup$ – Thomas Jul 29 '17 at 4:48
  • $\begingroup$ How does $\rho_{liq}-\rho_{gas}$ distinguish the gas phase from the liquid phase? It only distinguishes liquid or solid phase from the homogeneous fluid phase that exists above $T_c$. $\endgroup$ – SRS Jul 29 '17 at 4:51
  • $\begingroup$ The order parameter is just the density. Since there are no symmetries, there is nothing special about the order parameter being zero. All we care is that it jumps. In order to construct a Landau functional near the critical endpoint it is useful to subtract the density at the critical point, so that we can expand in powers of the order parameter. $\endgroup$ – Thomas Jul 29 '17 at 5:05
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The reason why the density difference is not used as an order parameter for a liquid to solid transition is that the density does NOT actually describe the difference between a liquid and a solid!

In fact, you could even have a phase transition between a liquid and solid that have exact the same mass density! In some cases, like for water, the solid is less dense than the liquid phase. Density is not the main difference between the two phases.

So what is the actual difference between a solid and liquid?

A solid is characterized by static correlations of atomic positions, while in a liquid the atomic positions are uncorrelated for long-enough time scales. In other words, if you see two atoms in a liquid and wait long enough their positions will change with respect to each other. However, if you do the same in a solid, their positions will stay fixed with respect to each other for basically forever.

Like the other answer states, it is possible to also have additional crystalline order, but that is very rarely the case (normally it is to an amorphous solid). I will discuss the general case below.

The order parameter for the liquid-solid transition would involve something like the atomic position correlation function $G(x,x^{\prime};t-t^{\prime})$, which is basically the probability you will sight an atom in position $x^{\prime}$ at time $t^{\prime}$ if you see an atom at $x$ at time $t$.

For a liquid, the following limit is obeyed for $G(x,x^{\prime};t-t^{\prime})$:

$$\lim_{t\to\infty} G(x,x^{\prime};t-t^{\prime})=0$$

That is to say, there is no correlation between seeing two atoms over long time scales. This makes sense because if you look at two atoms in a liquid and wait long enough, both atoms will have moved to new positions in a random fashion.

However for a solid, the atomic positions are highly correlated. If you see two atoms next to each other today, they will be in the same places tomorrow, the day after, next year, etc. etc. This means that:

$$\lim_{t\to\infty} G(x,x^{\prime};t-t^{\prime}) \neq 0$$

Now that we have a quantity that is zero in one phase, and non-zero in another phase, we have defined the order parameter $\langle \tilde \rho \rangle$:

$$\langle \tilde \rho(x,x^{\prime}) \rangle \equiv \lim_{t\to\infty} G(x,x^{\prime};t-t^{\prime}) \equiv \tilde G(x,x^{\prime})$$

For a crystal, $\langle \tilde \rho(x,x^{\prime}) \rangle$ will have special periodic structure in $x,x^{\prime}$, but in general that isn't the case.

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