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Missing estimate of speed-up and critical temperatures in table.

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Janita Willumsen 2023-12-04 10:08:43 +01:00
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7
latex/ising_model.tex
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@ -77,13 +77,16 @@
\clearpage \clearpage
\newpage \newpage
% Appendix
\subfile{sections/appendices}
\clearpage
\onecolumngrid \onecolumngrid
\bibliographystyle{unsrtnat} \bibliographystyle{unsrtnat}
\bibliography{references} \bibliography{references}
% Appendix
\subfile{sections/appendices}
\end{document} \end{document}

56
latex/sections/appendices.tex
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@ -40,12 +40,15 @@ the spin up as visualized in Figure \ref{fig:tikz_neighbor}.
Using the values estimated for the $2 \times 2$ case, found in \ref{tab:lattice_config}, Using the values estimated for the $2 \times 2$ case, found in \ref{tab:lattice_config},
we find the partition function we find the partition function
\begin{align*} \begin{align*}
Z &= 1 \cdot e^{-\beta (-8J)} + 4 \cdot e^{-\beta (0)} + 4 \cdot e^{-\beta (0)} + 2 \cdot e^{-\beta (8J)} \\ Z &= 1 \cdot e^{-\beta (-8J)} + 4 \cdot e^{-\beta (0)} + 4 \cdot e^{-\beta (0)} \\
& \quad + 4 \cdot e^{-\beta (0)} 1 \cdot e^{-\beta (-8J)} \\ & \quad + 2 \cdot e^{-\beta (8J)} + 4 \cdot e^{-\beta (0)} 1 \cdot e^{-\beta (-8J)} \\
&= 2e^{8 \beta J} + 2e^{-8 \beta J} + 12. &= 2e^{8 \beta J} + 2e^{-8 \beta J} + 12.
\end{align*} \end{align*}
We rewrite the expression using the identity $\cosh(8 \beta J) = 1/2 \big( e^{8 \beta J} + e^{-8 \beta J})$, We rewrite the expression using the identity
and get \begin{align*}
\cosh(8 \beta J) &= 1/2 \big( e^{8 \beta J} + e^{-8 \beta J})
\end{align*}
and find
\begin{align*} \begin{align*}
z &= 4 \cosh (8 \beta J) + 12 \ . z &= 4 \cosh (8 \beta J) + 12 \ .
\end{align*} \end{align*}
@ -55,12 +58,10 @@ and get
For a linear function of a stochastic random variable $X$, with a known probability For a linear function of a stochastic random variable $X$, with a known probability
distribution, the expected value of $x$ is given by distribution, the expected value of $x$ is given by
\begin{align*} \begin{align*}
\mathbb{E}(aX + b) &= a \cdot \mathbb{E}(X) + b & \text{\cite[p. 131]{springer:2012:modernstat}} \langle aX + b \rangle &= a \cdot \langle X \rangle + b & \text{\cite[p. 131]{springer:2012:modernstat}}
\end{align*} \end{align*}
In our case the discrete random variable is the spin configuration, and we want In our case the discrete random variable is the spin configuration, and we want
to find the expected value of the function $E(\mathbf{s})$. Continuing, we will to find the expected value of the function $E(\mathbf{s})$.
use the notation $\langle E \rangle$ for the expectation value of a given variable,
in this case $E$.
Both energy per spin and magnetization per spin are functions of $\mathbf{s}$. Both energy per spin and magnetization per spin are functions of $\mathbf{s}$.
In addition, the number of spins is given as a constant for each lattice. We can In addition, the number of spins is given as a constant for each lattice. We can
@ -75,9 +76,10 @@ The same applies to magnetization per spin
\begin{align*} \begin{align*}
\langle |m| \rangle = \frac{1}{N} \sum_{i=1}^{N} |M(s_{i})| p(s_{i} \ | \ T) \ . \langle |m| \rangle = \frac{1}{N} \sum_{i=1}^{N} |M(s_{i})| p(s_{i} \ | \ T) \ .
\end{align*} \end{align*}
Continuing with the expectation values for a $2 \times 2$ lattice, excluding the terms which give zero we get Continuing with the expectation values for a $2 \times 2$ lattice, excluding the terms which result in zero, we get
\begin{align*} \begin{align*}
\langle E \rangle &= (-8J) \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J) \cdot \frac{1}{Z} e^{-8 \beta J} + (-8J) \cdot \frac{1}{Z} e^{8 \beta J} \\ \langle E \rangle &= (-8J) \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J) \cdot \frac{1}{Z} e^{-8 \beta J} \\
& \quad + (-8J) \cdot \frac{1}{Z} e^{8 \beta J} \\
&= \frac{16J}{Z} \big(e^{-8 \beta J} - e^{8 \beta J}) \\ &= \frac{16J}{Z} \big(e^{-8 \beta J} - e^{8 \beta J}) \\
&= -\frac{32J \sinh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\ &= -\frac{32J \sinh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
&= -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ , &= -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
@ -90,9 +92,10 @@ and
&= \frac{4 (2e^{8 \beta J} + 4)}{4(\cosh(8 \beta J) + 3)} \\ &= \frac{4 (2e^{8 \beta J} + 4)}{4(\cosh(8 \beta J) + 3)} \\
&= \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ . &= \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ .
\end{align*} \end{align*}
The squared function The squared energy function
\begin{align*} \begin{align*}
\langle E^{2} \rangle &= (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J)^{2} \cdot \frac{1}{Z} e^{-8 \beta J} + (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} \\ \langle E^{2} \rangle &= (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J)^{2} \cdot \frac{1}{Z} e^{-8 \beta J} \\
& \quad + (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} \\
&= \frac{128J^{2}}{Z} \big(e^{8 \beta J} + e^{-8 \beta J} \big) \\ &= \frac{128J^{2}}{Z} \big(e^{8 \beta J} + e^{-8 \beta J} \big) \\
&= \frac{128J^{2} \cosh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\ &= \frac{128J^{2} \cosh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
&= \frac{64J^{2} \cosh(8 \beta J)}{\cosh(8 \beta J) + 3} \ , &= \frac{64J^{2} \cosh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
@ -116,6 +119,7 @@ and
&= \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \ . &= \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \ .
\end{align*} \end{align*}
\section{Heat capacity and magnetic susceptibility}\label{sec:heat_susceptibility} \section{Heat capacity and magnetic susceptibility}\label{sec:heat_susceptibility}
To find the heat capacity in Eq. \ref{eq:heat_capacity}, we normalize to heat To find the heat capacity in Eq. \ref{eq:heat_capacity}, we normalize to heat
capacity per spin capacity per spin
@ -149,8 +153,9 @@ and the variance of the total magnetization is given by
\mathbb{V}(M) &= \mathbb{E}(M^{2}) - [\mathbb{E}(|M|)]^{2} \\ \mathbb{V}(M) &= \mathbb{E}(M^{2}) - [\mathbb{E}(|M|)]^{2} \\
&= \frac{8e^{8 \beta J} + 8}{\cosh(8 \beta J) + 3} - \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\ &= \frac{8e^{8 \beta J} + 8}{\cosh(8 \beta J) + 3} - \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\
&= \frac{(8(e^{8 \beta J} + 1)) \cdot (\cosh(8 \beta J) + 3) - 4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\ &= \frac{(8(e^{8 \beta J} + 1)) \cdot (\cosh(8 \beta J) + 3) - 4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\
&= \frac{4(e^{8 \beta J} + 1) \cdot (e^{8 \beta J} + e^{-8 \beta J}) + 24(e^{8 \beta J} + 1) - 4(e^{8 \beta J} + 1)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \\ &= \frac{4(e^{8 \beta J} + 1) \cdot (e^{8 \beta J} + e^{-8 \beta J})}{(\cosh(8 \beta J) + 3)^{2}} \\
&= \frac{4e^{2(8 \beta J)} + 4e^{8 \beta J} 4e^{0} + 4e^{-8 \beta J} 24e^{8 \beta J} + 24 - 4e^{2(8 \beta J)} - 16e^{8 \beta J} - 16}{(\cosh(8 \beta J) + 3)^{2}} \\ & \quad + \frac{24(e^{8 \beta J} + 1) - 4(e^{8 \beta J} + 1)^{2}}{(\cosh(8 \beta J) + 3)^{2}}\\
% &= \frac{4e^{2(8 \beta J)} + 4e^{8 \beta J} 4e^{0} + 4e^{-8 \beta J} 24e^{8 \beta J} + 24 - 4e^{2(8 \beta J)} - 16e^{8 \beta J} - 16}{(\cosh(8 \beta J) + 3)^{2}} \\
&= \frac{4(3e^{8 \beta J} + e^{-8 \beta J} + 3)}{(\cosh(8 \beta J) + 3)^{2}} \ . &= \frac{4(3e^{8 \beta J} + e^{-8 \beta J} + 3)}{(\cosh(8 \beta J) + 3)^{2}} \ .
\end{align*} \end{align*}
We find the heat capacity We find the heat capacity
@ -181,29 +186,30 @@ where the five distinct values are $\Delta E = \{-16J, -8J, 0, 8J, 16J\}$.
\section{Additional results}\label{sec:additional_results} \section{Additional results}\label{sec:additional_results}
Results of 1 million MC cycles. We also did the phase transition experiment using 1 million MC cycles. In Figure \ref{fig:phase_energy_1M}
\begin{figure} we show expected energy per spin, and in Figure \ref{fig:phase_magnetization_1M}
expected magnetization per spin.
\begin{figure}[H]
\centering \centering
\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/energy.pdf} \includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/energy.pdf}
\caption{$\langle \epsilon \rangle$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.} \caption{$\langle \epsilon \rangle$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
\label{fig:phase_energy_1M} \label{fig:phase_energy_1M}
\end{figure} \end{figure} %
\begin{figure}[H]
\begin{figure}
\centering \centering
\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/magnetization.pdf} \includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/magnetization.pdf}
\caption{$\langle |m| \rangle$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.} \caption{$\langle |m| \rangle$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
\label{fig:phase_magnetization_1M} \label{fig:phase_magnetization_1M}
\end{figure} \end{figure} %
In Figure \ref{fig:phase_heat_1M} we show heat capacity, and in Figure \ref{fig:phase_susceptibility_1M}
\begin{figure} the magnetic susceptibility.
\begin{figure}[H]
\centering \centering
\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/heat_capacity.pdf} \includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/heat_capacity.pdf}
\caption{$C_{V}$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.} \caption{$C_{V}$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
\label{fig:phase_heat_1M} \label{fig:phase_heat_1M}
\end{figure} \end{figure} %
\begin{figure}[H]
\begin{figure}
\centering \centering
\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/susceptibility.pdf} \includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/susceptibility.pdf}
\caption{$\chi$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.} \caption{$\chi$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}

13
latex/sections/methods.tex
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@ -314,12 +314,13 @@ Monte Carlo cycles when analyzing phase transitions.
\subsection{Tools}\label{subsec:tools} \subsection{Tools}\label{subsec:tools}
The Ising model and MCMC methods are implemented in C++, and parallelized using The Ising model and MCMC methods are implemented in C++, and parallelized using
both \verb|OpenMPI| \cite{gabriel:2004:open_mpi} and \verb|OpenMP| \cite{openmp:2018}. We used the Python library both \verb|OpenMPI| \cite{gabriel:2004:open_mpi} and \verb|OpenMP| \cite{openmp:2018}.
\verb|matplotlib| \cite{hunter:2007:matplotlib} to produce all the plots, and We used the Python library \verb|matplotlib| \cite{hunter:2007:matplotlib} to produce
\verb|seaborn| \cite{waskom:2021:seaborn} to set the theme in the figures. In all the plots, \verb|seaborn| \cite{waskom:2021:seaborn} to set the theme in the
addition, we used \verb|Scalasca| \cite{scalasca} and \verb|Score-P| \cite{scorep} figures. We also used a linear regression method from the Python library \verb|SciPy|.
to profile and optimize our implementation. To optimize our implementation, we used a profiler tool \verb|Scalasca| \cite{scalasca}
% Add profiler and \verb|Score-P| \cite{scorep}.
\end{document} \end{document}

99
latex/sections/results.tex
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@ -3,72 +3,72 @@
\begin{document} \begin{document}
\section{Results}\label{sec:results} \section{Results}\label{sec:results}
\subsection{Burn-in time}\label{subsec:burnin_time} \subsection{Burn-in time}\label{subsec:burnin_time}
$\boldsymbol{Draft}$ We started with a lattice size $L = 20$ and considered the temperatures $T_{1} = 1.0 J/k_{B}$,
We decided with a burn-in time parallelization trade-off. That is, we set the and $T_{2} = 2.4 J/k_{B}$, where $T_{2}$ is close to the analytical critical
burn-in time lower in favor of sampling. To take advantage of the parallelization temperature given by Equation \eqref{eq:critical_solution}.
and not to waste computational resources. The argument to discard samples generated
during the burn-in time is ... Increasing number of samples outweigh the ...
parallelize using MPI. We generated
samples for the temperature range $T \in [2.1, 2.4]$. Using Fox we generated both
1 million samples and 10 million samples.
We start with a lattice where $L = 20$, to study the burn-in time, that is the To determine the burn-in time, we used the numerical estimates of energy
number of Monte Carlo cycles necessary for the system to reach an equilibrium. per spin for $T_{1}$ in Figure \ref{fig:burn_in_energy_1_0}, and $T_{2}$ in Figure
We consider two different temperatures $T_{1} = 1.0 J/k_{B}$ and $T_{2} = 2.4 J/k_{B}$, \ref{fig:burn_in_energy_2_4}. We also considered the estimates of magnetization
where $T_{2}$ is close to the critical temperature. We can use the correlation per spin for $T_{1}$ in Figure \ref{fig:burn_in_magnetization_1_0}, and for $T_{2}$
time $\tau \approx L^{d + z}$ to determine time, where $d$ is the dimensionality in Figure \ref{fig:burn_in_energy_2_4}.
of the system and $z = 2.1665 \pm 0.0012$ \footnote{This value was determined by % Not sure about the relevance of the paragraph below
Nightingale and Blöte for the Metropolis algorithm.} We used random numbers to set the initial state and the index of spin to flip, and
% observed different graphs ... However, when we set the seed of the random number
We show the numerical estimates for temperature $T_{1}$ of $\langle \epsilon \rangle$ engine, we could reproduce the results at every run. It is possible to determine
in Figure \ref{fig:burn_in_energy_1_0} and $\langle |m| \rangle$ in Figure the burn-in time analytically, using the correlation time given by $\tau \approx L^{d + z}$.
\ref{fig:burn_in_magnetization_1_0}. For temperature $T_{2}$, the numercal estimate Where $d$ is the dimensionality of the system and $z = 2.1665 \pm 0.0012$
of $\langle \epsilon \rangle$ is shown in Figure \ref{fig:burn_in_energy_2_4} and \footnote{This value was determined by Nightingale and Blöte for the Metropolis algorithm.}.
$\langle |m| \rangle$ in Figure \ref{fig:burn_in_magnetization_2_4}. The lattice However, when we increased number of Monte Carlo cycles the sampled generated during
is initialized in both an ordered and an unordered state. We observe that for the burn-in time did not affect the mean value (...). % Should add result showing this
$T_{1}$ there is no change in either expectation value with increasing number of
Monte Carlo cycles, when we start with an ordered state. As for the unordered The lattice was initialized in an ordered and an unordered state, for both temperatures. We observed
lattice, we observe a change for the first 5000 MC cycles, where it stabilizes. no change in expectation value of energy or magnetization for $T_{1}$, when we
The approximated expected energy is $-2$ and expected magnetization is $1.0$, initialized the lattice in an ordered state. As for the unordered initialized
which is to be expected for temperature 0f $1.0$. T is below the critical and the lattice, we first observed a change in expectation values, and a stabilization around
pdf using $T = 1.0$ result in $5000$ Monte Carlo cycles. The expected energy per spin is $\langle \epsilon \rangle = -2$
\begin{align*} and the expected magnetization per spin is $\langle |m| \rangle = 1.0$. % add something about what is expected for $T_{1}$ ?
p(s|T=1.0) &= \frac{1}{e^{-\beta \sum E(s)}} e^{-\beta E(s)} \\ For $T_{2}$ we observed a change in expectation values for both the ordered and the
&= \frac{1}{e^{-(1/k_{B}) \sum E(s)}} e^{-(1/k_{B}) E(s)} \ . unordered lattice.
\end{align*} % \begin{align*}
% p(s|T=1.0) &= \frac{1}{e^{-\beta \sum E(s)}} e^{-\beta E(s)} \\
% &= \frac{1}{e^{-(1/k_{B}) \sum E(s)}} e^{-(1/k_{B}) E(s)} \ .
% \end{align*}
For $T_{2}$ we observe an increase in expected energy per spin $\langle \epsilon \rangle \approx -1.23$,
and a decrease in expected magnetization per spin $\langle |m| \rangle \approx 0.46$.
% Burn-in figures % Burn-in figures
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_1_0.pdf} \includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_1_0.pdf}
\caption{$\langle |m| \rangle$ as a function of time, for $T = 1.0 J / k_{B}$} \caption{Magnetization per spin $\langle |m| \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 1.0 J / k_{B}$}
\label{fig:burn_in_magnetization_1_0} \label{fig:burn_in_magnetization_1_0}
\end{figure} \end{figure}
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\includegraphics[width=\linewidth]{../images/burn_in_time_energy_1_0.pdf} \includegraphics[width=\linewidth]{../images/burn_in_time_energy_1_0.pdf}
\caption{$\langle \epsilon \rangle$ as a function of time, for $T = 1.0 J / k_{B}$} \caption{Energy per spin $\langle \epsilon \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 1.0 J / k_{B}$}
\label{fig:burn_in_energy_1_0} \label{fig:burn_in_energy_1_0}
\end{figure} \end{figure}
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_2_4.pdf} \includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_2_4.pdf}
\caption{$\langle |m| \rangle$ as a function of time, for $T = 2.4 J / k_{B}$} \caption{Magnetization per spin $\langle |m| \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 2.4 J / k_{B}$}
\label{fig:burn_in_magnetization_2_4} \label{fig:burn_in_magnetization_2_4}
\end{figure} \end{figure}
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\includegraphics[width=\linewidth]{../images/burn_in_time_energy_2_4.pdf} \includegraphics[width=\linewidth]{../images/burn_in_time_energy_2_4.pdf}
\caption{$\langle \epsilon \rangle$ as a function of time, for $T = 2.4 J / k_{B}$} \caption{Energy per spin $\langle \epsilon \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 2.4 J / k_{B}$}
\label{fig:burn_in_energy_2_4} \label{fig:burn_in_energy_2_4}
\end{figure} \end{figure}
\subsection{Probability distribution}\label{subsec:probability_distribution} \subsection{Probability distribution}\label{subsec:probability_distribution}
% Histogram figures % Histogram figures
We used the estimated burn-in time as starting time for sampling, and generated We used the estimated burn-in time of $5000$ Monte Carlo cycles as starting time, and generated
samples. To visualize the distribution of energy per spin $\epsilon$, we used histograms samples. To visualize the distribution of energy per spin $\epsilon$, we used histograms
with a bin size (...). In Figure \ref{fig:histogram_1_0} we show the distribution with a bin size $0.02$. In Figure \ref{fig:histogram_1_0} we show the distribution
for $T_{1}$, where the majority of the energy per spin is $-2$. The resulting expectation for $T_{1}$. Where the resulting expectation
value of energy per spin is $\langle \epsilon \rangle = -1.9969$, with a low variance value of energy per spin is $\langle \epsilon \rangle = -1.9969$, with a low variance
of Var$(\epsilon) = 0.0001$. % of Var$(\epsilon) = 0.0001$. %
\begin{figure}[H] \begin{figure}[H]
@ -85,16 +85,15 @@ centered around the expectation value $\langle \epsilon \rangle = -1.2370$. %
\caption{Histogram $T = 2.4 J / k_{B}$} \caption{Histogram $T = 2.4 J / k_{B}$}
\label{fig:histogram_2_4} \label{fig:histogram_2_4}
\end{figure} % \end{figure} %
However, we observe a higher variance of Var$(\epsilon) = 0.0203$. When the temperature However, we observed a higher variance of Var$(\epsilon) = 0.0203$. When the temperature
increase, the system moves from an ordered to an onordered state. The change in increased, the system moved from an ordered to an onordered state. The change in
system state, or phase transition, indicates the temperature is close to a critical system state, or phase transition, indicates the temperature is close to a critical
point. point.
\subsection{Phase transition}\label{subsec:phase_transition} \subsection{Phase transition}\label{subsec:phase_transition}
$\boldsymbol{Draft}$
% Phase transition figures % Phase transition figures
We continue investigating the behavior of the system around the critical temperature. We continued investigating the behavior of the system around the critical temperature.
First, we generated $10$ million samples of spin configurations for lattices of First, we generated $10$ million samples of spin configurations for lattices of
size $L \in \{20, 40, 60, 80, 100\}$, and temperatures $T \in [2.1, 2.4]$. We divided the size $L \in \{20, 40, 60, 80, 100\}$, and temperatures $T \in [2.1, 2.4]$. We divided the
temperature range into $40$ steps, with an equal step size of $0.0075$. The samples temperature range into $40$ steps, with an equal step size of $0.0075$. The samples
@ -133,7 +132,7 @@ value of heat capacity the lattice size increase.
\caption{$C_{V}$ for $T \in [2.1, 2.4]$, $10^7$ MC cycles.} \caption{$C_{V}$ for $T \in [2.1, 2.4]$, $10^7$ MC cycles.}
\label{fig:phase_heat_10M} \label{fig:phase_heat_10M}
\end{figure} % \end{figure} %
The magnetic susceptibility in Figure \ref{fig:phase_susceptibility_10M}, showed The magnetic susceptibility in Figure \ref{fig:phase_susceptibility_10M}, show
the sharp peak in the same temperature range as that of the heat capacity. Since the sharp peak in the same temperature range as that of the heat capacity. Since
shape of the curve for both heat capacity and the magnetic susceptibility become shape of the curve for both heat capacity and the magnetic susceptibility become
sharper when we increase lattice size, we are moving closer to the critical temperature. sharper when we increase lattice size, we are moving closer to the critical temperature.
@ -156,14 +155,14 @@ of lattices of size $L \in \{20, 40, 60, 80, 100\}$ found in Table \ref{tab:crit
\hline \hline
$L$ & $T_{c}(L)$ \\ $L$ & $T_{c}(L)$ \\
\hline \hline
$20$ & $J$ \\ $20$ & $ J / k_{B}$ \\
$40$ & $1$ \\ $40$ & $ J / k_{B}$ \\
$60$ & $J / k_{B}$ \\ $60$ & $ J / k_{B}$ \\
$80$ & $k_{B}$ \\ $80$ & $ J / k_{B}$ \\
$100$ & $1 / J$ \\ $100$ & $ J / k_{B}$ \\
\hline \hline
\end{tabular} \end{tabular}
\caption{Estimated critical temperatures for lattices $L \times L$.} \caption{Estimated critical temperatures for lattices $L \times L$, where $L$ denote the lattice size.}
\label{tab:critical_temperatures} \label{tab:critical_temperatures}
\end{table} \end{table}
We used the critical temperatures of finite lattices and the scaling relation in We used the critical temperatures of finite lattices and the scaling relation in