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Using Metropolis algorithm, Curie temperature was calculated for square lattices with different sizes 4x4, 8x8, 16x16 and 32x32. Here Curie temperature was estimated as temperature of maximum of heat capacity.It appeared Curie temperature is decreasing with increasing a lattices size. But why? I think this is due to neglect of long-range fluctuations. But I don't quite understand this. Heat capacity for 2D Ising model with square lattices

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    $\begingroup$ Because exact answer for $T_c$ is valid only in $N\rightarrow\infty$ limit. I assume that you have fixed spin-spin interaction by unity ($J=1$), so the exact answer $T\approx 2.269$ and you can achieve this result with large size of lattices. You observe finite size effects $\endgroup$ Commented Dec 1, 2023 at 11:28
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    $\begingroup$ Thanks for comment. Yes, I actually have fixed spin-spin interaction by unity. But what the physical reasoning or intuition behind getting incorrect T_c is for finite size lattices? $\endgroup$
    – John
    Commented Dec 1, 2023 at 11:44
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    $\begingroup$ @ArtemAlexandrov consider to spell this out as an answer. $\endgroup$ Commented Dec 1, 2023 at 12:06
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    $\begingroup$ Phase-transitions are are technically only sharply defined in the so-called Thermodynamic limit, as Artem also points out. Outside that limit, there are no sharp phase-transitions and the transition temperature is not as sharply defined. The keyword to read up on is Thermodynamic limit, you can start here: en.wikipedia.org/wiki/Thermodynamic_limit I hope somebody will write out a more detailed answer. $\endgroup$
    – Heidar
    Commented Dec 1, 2023 at 12:47
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    $\begingroup$ One should also mention that the shift in the critical temperature for finite systems can be analyzed precisely (finite size scaling). See, for instance, this answer. $\endgroup$ Commented Dec 2, 2023 at 9:29

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Why numerical simulation gives seemingly wrong result?

The exact expression for $T_c$ is given by $$T_c=\frac{2J}{\ln\left(\sqrt{2}+1\right)}\approx 2.269\,J,$$ where we assume that $k_B=1$. This expression can be obtained by Krammers-Wannier duality and to find $T_c$ we do not imply thermodynamic limit, $N\rightarrow\infty$.

However, when we compute observables (like magnetization, heat capacity and so on) directly from the partition function, we imply $N\rightarrow\infty$ limit. It is encoded in our choice of boundary conditions (typically, for 1D and 2D Ising models periodic boundary conditions are used): only in $N\rightarrow\infty$ different boundary conditions will give the same answers for observables and critical exponents.

When we simulate 2D Ising model with finite number of spins, the observables do not diverge and behave like $$C\sim L^{\alpha/\nu},\quad m\sim L^{-\beta/\nu},$$ where we consider $L\times L$ square lattice. It is caused from the fact that for finie $N$ the correlation length is not divergent and here $\nu$ is the critical exponent related to correlation length.

One way to capture and analyze such effects is so-called Binder cumulants method (original paper). I am not a specialist in lattice-related computations, so I cannot give more.

Physical reasoning

All phase transitions exist only in thermodynamic limit, i.e. $N\rightarrow\infty$. The partition function $\mathcal{Z}$ is the sum over all possible microstates of system and each term of this sum is analytic function, so for finite number of microstates the partition is analytic function. It implies that there is no any singular behavior. But infinite sum of analytic functions can becomes non-analytic and here singularities appear, i.e. phase transitions take place. It is so-called Lee-Yang theorem. The detailed discussion can be found in book "Scale Invariance: From Phase Transitions to Turbulence" by A. Lesne and M. Lagues.

Simplest possible example

The simplest possible example to see it is molecular zipper model. Consider the model of polymer, where we have $N$ bonds and each bond can be closed or open and has energy $\epsilon$ if open. If bond is open, it can be in one of $g$ states. The partition function is $$ \mathcal{Z}(\beta)=\sum_{n=0}^{N}g^ne^{-\beta\epsilon n}=\frac{1-\left(ge^{-\beta\epsilon}\right)^{N+1}}{1-ge^{-\beta\epsilon}}.$$ The average number of open bonds is $$ \langle n \rangle = \frac{1}{\mathcal{Z}}\sum_{n=0}^{N}n e^{-\beta\epsilon n}=-\frac{1}{\epsilon\mathcal{Z}}\frac{\partial\mathcal{Z}}{\partial\beta}=-\frac{1}{\epsilon }\frac{\partial\ln\mathcal{Z}}{\partial\beta}=-\frac{1}{\epsilon}\frac{\partial F}{\partial\beta}=\\=-\frac{1}{\epsilon}\frac{\partial}{\partial\beta}\left\{\ln\left[1-\left(g^{-\beta\epsilon}\right)^{N+1}\right]-\ln\left[1-ge^{-\beta\epsilon}\right]\right\}=\\=-\frac{ge^{-\beta\epsilon}}{ge^{-\beta\epsilon}-1}+\frac{(N+1)\left(ge^{-\beta\epsilon}\right)^{N+1}}{\left(ge^{-\beta\epsilon}\right)^{N+1}-1}.$$ Let $\kappa=ge^{-\beta\epsilon}$, so $$ \langle n\rangle = \frac{\kappa}{1-\kappa}-\frac{(N+1)\kappa^{N+1}}{1-\kappa^{N+1}}.$$ Next, we can measure the fraction of opened bonds, $\langle n\rangle/N$. And it is straightforward to see that $$ \lim\limits_{N\rightarrow\infty}\frac{\langle n\rangle}{N}=\frac{\kappa}{1-\kappa},$$ so the fraction is divergent at $\kappa_c=1$ but only in $N\rightarrow\infty$ limit. It can be also seen from plots.

enter image description here

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  • $\begingroup$ Artem Alexandrov, the second plot is obviously the derivative $\langle n\rangle/N$ respect to k - $\frac{\partial(\langle n\rangle/N)}{\partial k}$ ? You cited a model system as an example, but the question was, what are the physical reasons for temperature decreasing with increasing a lattices size? $\endgroup$ Commented Dec 7, 2023 at 12:49
  • $\begingroup$ @AlekseyDruggist , 1) phase transitions exist in thermodynamic limit, there are no discontinuities in observables, how we can speak about $T_c$, if everything is continuous? 2) physical reason: observables have different behavior for finite $N$, 3) technical solution: use Binder cumulants approach or see Yvan Velenik answer (in comments). Plots for molecular zipper are presented to emphasize that phase transition occurs only in $N\rightarrow\infty$ limit. What should I add? $\endgroup$ Commented Dec 7, 2023 at 14:32
  • $\begingroup$ @ Artem Alexandrov: "...so the fraction is divergent at κc=1 but only in N→∞ limit ..." - this is an incorrect statement, I would recommend making an edit $\endgroup$ Commented Dec 9, 2023 at 0:02

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