I read in these papers(1,2) the concept of interface tension. I can't understand its definition. I can hardly imagine there is some tension in a model. Any help will be appreciated.

  • $\begingroup$ The surface tension is defined as the surface-order contribution (per unit area) to the free energy coming from the presence of an interface between two phases. See scholarpedia.org/article/Interface_free_energy for example for a precise definition (in the Ising case, but the Potts case is essentially identical). $\endgroup$ – Yvan Velenik May 7 '13 at 12:53
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    $\begingroup$ I just realized that, in the title of your question, you are asking for the surface tension between the ordered and disordered states. The lack of symmetry between these two phases makes the definition slightly more involved than in the Ising case (or between 2 ordered Potts phases). Nevertheless, the idea remains completely similar, and you should first understand the definition in the link I have given. I can give you a more detailed explanation when symmetry is absent next week (I am at a conference right now...). $\endgroup$ – Yvan Velenik May 9 '13 at 7:30

Symmetric case

I'll first explain the definition of the surface tension between two ordered phases of the Potts model, since the symmetry between the phases simplifies things. I'll work in dimension $3$ to be specific, but everything generalizes in a straightforward way to other dimensions.

So consider the $q$-states Potts model at inverse temperature $\beta$ in the $3$-dimensional cube $\Lambda_n=\{-n,\ldots,n\}^3$. If we consider constant $1$-boundary condition, i.e., fix all spins along the external boundary $\partial\Lambda_n$ to be in state $1$, then the finite-volume free energy of the system can be written as $$ F^1_{\Lambda_n}(\beta) = -\frac1\beta\ln Z^1_{\Lambda_n;\beta} = f_{\rm bulk}(\beta) |\Lambda_n| + f_{\rm wall}(\beta) |\partial\Lambda_n| + o(|\partial\Lambda_n|), $$ where $Z^1_{\Lambda_n;\beta}$ is the partition function, $f_{\rm bulk}(\beta)$ is the bulk free energy per unit volume, $f_{\rm wall}(\beta)$ is the surface-order contribution per unit area resulting from the interaction of the system with the boundary and $o(|\partial\Lambda_n|)$ takes into account higher-order corrections to the finite-volume free energy.

Now, let us consider instead a boundary condition enforcing coexistence of two ordered phases. A natural way of doing that is to use the so-called Dobrushin boundary condition which is constructed as follows: consider a plane $\Pi$ going through $0$ (say, with a normal $\vec m$ making an angle at least $\pi/4$ with the vertical). We then fix each spin along $\partial\Lambda_n$ to state $1$ if it lies above (or on) the plane $\Pi$, and to state $2$ if it lies below. The finite-volume free energy then becomes $$ F^{1,2}_{\Lambda_n}(\beta) = -\frac1\beta\ln Z^{1,2}_{\Lambda_n;\beta} = f_{\rm bulk}(\beta) |\Lambda_n| + f_{\rm wall}(\beta) |\partial\Lambda_n| + \sigma_\beta(\vec m) |\Pi_n| + o(|\partial\Lambda_n|), $$ where $Z^{1,2}_{\Lambda_n;\beta}$ is the new partition function, $|\Pi_n|$ denotes the area of the intersection of $\Pi$ with the cube $[-n,n]^3$. The quantity $\sigma_\beta(\vec m)$ represents the contribution to the finite-volume free energy, per unit area, coming from the presence of the interface between the two phases $1$ and $2$ induced by the boundary condition (each phase occupying, at the macroscopic scale, the corresponding half of the box, with a well-defined planar interface separating them). This is the surface tension between the two ordered phases. It is important to note the following:

  • the bulk free energy $f_{\rm bulk}(\beta)$ is the same for both phases $1$ and $2$;
  • the wall free energy $f_{\rm wall}(\beta)$ is coming from the local interaction between the phase and the wall and is thus also equal to the quantity we introduced in the case of a single phase.

These observations provide a way of extracting the surface tension from the finite-volume free energy: $$ \sigma_\beta(\vec m) \approx \frac1{|\Pi_n|} \bigl( F^{1,2}_{\Lambda_n}(\beta) - F^1_{\Lambda_n}(\beta)\bigr). $$ This leads us to the following definition: $$ \sigma_\beta(\vec m) = -\lim_{n\to\infty} \frac1{\beta|\Pi_n|} \ln\frac{Z^{1,2}_{\Lambda_n;\beta}}{Z^1_{\Lambda_n;\beta}} $$ Remark: The fact that this quantity really plays the role of the thermodynamic surface tension can be seen in many ways. The most compelling might be its appearance when studying equilibrium crystal shapes: consider, to keep things as simple as possible, an Ising model with fixed magnetization $m\in(-m^*(\beta),m^*(\beta))$. It can be shown that, with probability going to $1$ in the thermodynamic limit, the system spontaneously creates a droplet, of deterministic macroscopic shape, of one phase inside the other. The shape can be proved to be the one minimizing the surface tension (defined as above) among all sets of the required volume. More information on that can be found here and there.

Non-symmetric case

Your question is about the surface tension between an ordered phase and the disordered phase of the Potts model at the transition temperature. The situation in that case is made slightly more complicated due to the absence of symmetry between the two phases. It is still true that the bulk free energies of the ordered and disordered phases coincide, but it is no longer the case that the wall free energies coincide, so that we cannot cancel them as easily as we did above. Let me explain the trick used to solve this problem.

Consider a system with two coexisting phases $A$ and $B$, not necessarily related by symmetry. As above, we can write

$$ F^A_{\Lambda_n} = f_{\rm bulk} |\Lambda_n| + f^A_{\rm wall} |\partial\Lambda_n| + o(|\partial\Lambda_n|), $$ $$ F^B_{\Lambda_n} = f_{\rm bulk} |\Lambda_n| + f^B_{\rm wall} |\partial\Lambda_n| + o(|\partial\Lambda_n|), $$ $$ F^{A,B}_{\Lambda_n} = f_{\rm bulk} |\Lambda_n| + f^{A,B}_{\rm wall} |\partial\Lambda_n| + \sigma(\vec m) |\Pi_n| + o(|\partial\Lambda_n|), $$ Now, the thing to observe is that $f^{A,B}_{\rm wall} = \tfrac12(f^A_{\rm wall} + f^B_{\rm wall})$ since each of the two phases is in contact with half of the boundary of the box. Therefore, one way of extracting the surface tension in that case is to use $$ \sigma(\vec m) \approx \frac1{|\Pi_n|} \bigl( F^{A,B}_{\Lambda_n} - \tfrac12 ( F^A_{\Lambda_n} + F^B_{\Lambda_n} )\bigr). $$ This leads to the definition: $$ \sigma(\vec m) = -\lim_{n\to\infty} \frac1{\beta|\Pi_n|} \ln\frac{Z^{A,B}_{\Lambda_n}}{\sqrt{Z^A_{\Lambda_n}Z^B_{\Lambda_n}}}. $$ This applies readily to the ordered/disordered interface in the Potts model. The only thing to explain is what boundary condition to use to induce the disordered phase. It turns out that this role can be played by the free boundary condition, i.e., not putting any spins along the corresponding part of $\partial\Lambda_n$. So, for example, to induce an interface as above, you'd associate $1$-spins to boundary vertices located above the plane $\Pi$ and no spins to vertices located below.

Let me finish by giving you a couple of references about these things in the context you are interested with (but there are many): this one or that one.

  • $\begingroup$ Thank you very much! The link you gave scholarpedia.org/article/Interface_free_energy already explains the Non-symmetric case. $\endgroup$ – hlew May 11 '13 at 12:27
  • $\begingroup$ @hlew: You're right. I should have checked ;) . $\endgroup$ – Yvan Velenik May 11 '13 at 12:30
  • $\begingroup$ I am a master sudent. My supervisor asked me to investigate the three dimensional Potts model using numerical renormalization group method. Do you know some unsolved problems on this model? $\endgroup$ – hlew May 11 '13 at 13:41
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    $\begingroup$ @hlew: from the point of view of a mathematical physicist, there are plenty of open problems concerning the Potts model (including such basic ones as determining the order of the phase transition for the various values of the number of states and of the dimension). About what problem would still be considered open for a theoretical physicist, I unfortunately have no idea. $\endgroup$ – Yvan Velenik May 12 '13 at 9:14
  • $\begingroup$ I know that the phase transion of the ferromanetic $q$-states Potts model is second order for $q=2$ while first order for $q>2$. But I can't find definite result for the antiferromanetic model. Is the order of the phase transition for the antiferromanetic model still unclear? $\endgroup$ – hlew May 12 '13 at 12:35

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