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2d
comment A conceptual question related to statistical mechanics
I would also recommend that you have a look at the oldish but beautiful little book "Statistical Mechanics and the Foundations of Thermodynamics" by Anders Martin-Löf (Lecture Notes in Physics 101, 1979). The discussion of the pressure starts on page 23.
Apr
20
comment Why don't we observe spontaneous symmetry restoration in nature?
I agree with the above comments. Nevertheless, one can find (rather restrictive) settings in which such a result actually become a theorem. One is that of ferromagnetic (classical) spin systems, for which, under suitable assumptions, one can prove that, if $T<T'$, then the internal symmetry group at temperature $T$ is necessarily a subgroup of the internal symmetry group at temperature $T'$. See Theorem 4 of Chapter 4 in the book "Group Analysis of Classical Lattice Systems", C. Gruber et al., Lecture Notes in Physics 60, 1977. I am not aware of more recent results, but there should be some.
Mar
22
comment 3-state Potts model - probability of finding a site in state 1
Once you have computed the free energy, just observe that differentiating it with respect to h gives you the density of $1$ in the system. (Just differentiate $N^{-1}\log Z_N$ at finite $N$ to see that.)
Mar
21
comment What does the solid phase in a two-dimensional system with Lennard-Jones potential look like?
@AdiRo : Using precise definitions for the concepts one uses is very important. There are very good reasons for defining them using the thermodynamic limit. Then, of course, one should also say what happens in finite ("real") systems. In particular, these "soft-crystal" phases can and must be studied: they are very interesting and more tractable than the 3d solid phases. However, they should not be called solid.
Mar
21
comment What does the solid phase in a two-dimensional system with Lennard-Jones potential look like?
@AdiRo : He is actually taking a pragmatic position: in finite systems, you see something that looks like a solid phase, so let's call that a solid phase! That might seem reasonable, but is actually conceptually a bad idea: with such a point of view, you can have long-range order in the one-dimensional Ising model, for example! Indeed, for finite systems and sufficiently low temperatures (depending on the system's size), you get a magnetized sample.
Mar
21
comment What does the solid phase in a two-dimensional system with Lennard-Jones potential look like?
@AdiRo: This is even discussed explicitly in the paper you refer to: just read Sections 2 and 3. What the author does is to argue that a new definition of a solid phase is needed. But there is no discussion (it is a mathematical fact) that in two dimensions translation invariance cannot be broken.
Mar
21
comment What does the solid phase in a two-dimensional system with Lennard-Jones potential look like?
@AdiRo : of course not. The proofs in the papers I link to apply just as well to interactions of infinite range. In particular, they apply to Lennard-Jones interactions. In two dimensions, you do not see true solid phases, in the sense that there cannot be positional long-range order. What is possible, however, is to have some kind of "soft crystal" phases, which look like solid locally; this is due to the very slow deformations of the crystal over long distances. All this is completely analogous to the massless (Kosterlitz-Thouless) phase in the 2d XY model.
Mar
16
comment What does the solid phase in a two-dimensional system with Lennard-Jones potential look like?
@Couchyam : Sure, I agree with you: they are the analogues of the massless phase in $O(N)$ models. However they are not considered to be solid phases (the latter are characterized by the breaking of translation invariance), and I doubt that's what the OP had in mind.
Feb
18
comment Why does Triple point exist?
I don't understand why is this downvoted without even leaving a comment? It is clearly not universally true, but it is at least partially correct. However, I would have added a link to the Gibbs phase rule.
Feb
15
comment Difference between theoretical physics and mathematical physics?
@PeterShor : Sure, I agree with you. And let me stress that this was not meant to be derogatory at all. I only wanted to emphasize that many "string theorists" in mathematics departments are generating conjectures (and developing tools to do so) much more often than they prove theorems, and theirs is obviously also a very valuable activity.
Feb
1
comment Relation between the $N$ particle partition function and probability?
I agree with the answer, but disagree with the characterization of the partition function as the most important quantity in statistical physics (even though it is often presented as such in textbooks). This central role should be reserved to the probability measure itself. Indeed, the latter gives you access to much more information (even about macroscopic quantities) than the partition function, which only gives you access to the statistical properties of the macroscopic quantities appearing in the Boltzmann weight: energy, magnetization, etc.).
Jan
22
comment Reference request: 2D conformal field theory and functions on the triangular lattice
You should google "discrete complex analysis". There are many papers introducing these ideas, for example those by Chelkak, Smirnov, etc.
Jan
22
comment Cluster Expansion
Done. I tried to make a readable answer out of the comments, but had only limited time, so there might be typos.
Jan
22
comment Cluster Expansion
Yes you sum over all possible values of $n_\ell$ for each possible values of $\ell$ (so $\{n_\ell\}$ specifies the values of $n_1, n_2, n_3, \ldots$). Then, once these are fixed, you take the product of all the functions $f_\ell(n_\ell)$ (for these specific values of $n_1,n_2,n_3,\ldots$).
Jan
22
comment Cluster Expansion
Sure. Imagine that the only allowed cluster sizes are $\ell=1$ and $\ell=2$ (so that I can write things down explicitely). Let us use the notation $f_\ell(n_\ell) = \frac{1}{n_\ell!}\bigl( \frac{e^{\ell\beta\mu}b_\ell}{\lambda^{3\ell}\ell!} \bigr)^{n_\ell}$. Then your identity becomes $\sum_{\{n_1,n_2\} \in\{0,1,2,\ldots\}^2} \prod_{\ell=1}^2 f_\ell(n_\ell) = \sum_{n_1\geq 0} \sum_{n_2\geq 0} f_1(n_1) f_2(n_2)= \sum_{n_1\geq 0} f_1(n_1) \sum_{n_2\geq 0} f_2(n_2)= \prod_{\ell=1}^2 \sum_{n_\ell\geq 0} f_\ell(n_\ell)$
Jan
21
comment Cluster Expansion
This is just linearity of the sum. Maybe it will become obvious if you look at the following particular case: $\sum_{i,j} a_i b_j = \sum_i \sum_j a_i b_j = \Bigl(\sum_i a_i\Bigr)\Bigl(\sum_j b_j\Bigr)$, which follows by pulling out $a_i$ from the inner sum and then pulling out $\sum_j b_j$ from the outer sum. Now, in your formula $\{n_\ell\}$ denotes the number of clusters of each possible length (so $\sum_{\{n_\ell\}} = \sum_{n_1\geq 0}\sum_{n_2\geq 0}\cdots = \prod_{\ell\geq 1} \sum_{n_\ell\geq 0}$).
Jan
19
comment What are alternative ways to think about transfer matrix as used in Ising model?
@Nathaniel : yes, that's exactly the point. If you denote by $\varphi^{*,1}$ the left Perron-Frobenius eigenvector (associated, of course, to the same eigenvalue $\lambda_1$), then the only thing that changes is the expression for the invariant measure, which becomes $\mu( \sigma) = \varphi^{*,1}_\sigma \varphi^1_\sigma$. (Note also that you could define another Markov chain with transition probabilities $\pi^*(\sigma,\sigma') = \frac{\varphi^{*,1}_{\sigma'}}{\lambda_1\varphi^{*,1}_\sigma} T_{\sigma',\sigma}$; this corresponds to the time-reversed chain (which has the same invariant measure).
Jan
19
comment What are alternative ways to think about transfer matrix as used in Ising model?
@Nathaniel : thanks for your appreciation (and the unexpected bounty). Concerning the nonsymmetric case, I might either add a (short) second part to the answer, in which I point out the relevant changes, or simply point them out in the comments. What would be the best solution?
Dec
29
comment annealed randomness vs quenched randomness
Quenched randomness means that the disorder is sampled first and then equilibrium statistical mechanics is applied to the system's other degrees of freedom with this particular frozen disorder. Annealed means that you average on the disorder at the same time as you take the thermal average over the other degrees of freedom.
Dec
7
comment What is the difference between toy models and normal models?
@Ooker : yes, it is a caricature, keeping enough relevant features to make it useful for the discussion of a particular aspect of the phenomenon, while keeping the model tractable enough to make a detailed analysis possible.