Missing estimate of speed-up and critical temperatures in table.
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Janita Willumsen
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@ -77,13 +77,16 @@
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\clearpage
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\newpage
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% Appendix
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\subfile{sections/appendices}
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\clearpage
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\onecolumngrid
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\bibliographystyle{unsrtnat}
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\bibliography{references}
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% Appendix
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\subfile{sections/appendices}
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\end{document}
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@ -40,12 +40,15 @@ the spin up as visualized in Figure \ref{fig:tikz_neighbor}.
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Using the values estimated for the $2 \times 2$ case, found in \ref{tab:lattice_config},
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we find the partition function
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\begin{align*}
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Z &= 1 \cdot e^{-\beta (-8J)} + 4 \cdot e^{-\beta (0)} + 4 \cdot e^{-\beta (0)} + 2 \cdot e^{-\beta (8J)} \\
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& \quad + 4 \cdot e^{-\beta (0)} 1 \cdot e^{-\beta (-8J)} \\
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Z &= 1 \cdot e^{-\beta (-8J)} + 4 \cdot e^{-\beta (0)} + 4 \cdot e^{-\beta (0)} \\
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& \quad + 2 \cdot e^{-\beta (8J)} + 4 \cdot e^{-\beta (0)} 1 \cdot e^{-\beta (-8J)} \\
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&= 2e^{8 \beta J} + 2e^{-8 \beta J} + 12.
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\end{align*}
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We rewrite the expression using the identity $\cosh(8 \beta J) = 1/2 \big( e^{8 \beta J} + e^{-8 \beta J})$,
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and get
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We rewrite the expression using the identity
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\begin{align*}
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\cosh(8 \beta J) &= 1/2 \big( e^{8 \beta J} + e^{-8 \beta J})
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\end{align*}
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and find
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\begin{align*}
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z &= 4 \cosh (8 \beta J) + 12 \ .
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\end{align*}
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@ -55,12 +58,10 @@ and get
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For a linear function of a stochastic random variable $X$, with a known probability
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distribution, the expected value of $x$ is given by
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\begin{align*}
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\mathbb{E}(aX + b) &= a \cdot \mathbb{E}(X) + b & \text{\cite[p. 131]{springer:2012:modernstat}}
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\langle aX + b \rangle &= a \cdot \langle X \rangle + b & \text{\cite[p. 131]{springer:2012:modernstat}}
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\end{align*}
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In our case the discrete random variable is the spin configuration, and we want
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to find the expected value of the function $E(\mathbf{s})$. Continuing, we will
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use the notation $\langle E \rangle$ for the expectation value of a given variable,
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in this case $E$.
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to find the expected value of the function $E(\mathbf{s})$.
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Both energy per spin and magnetization per spin are functions of $\mathbf{s}$.
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In addition, the number of spins is given as a constant for each lattice. We can
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@ -75,9 +76,10 @@ The same applies to magnetization per spin
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\begin{align*}
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\langle |m| \rangle = \frac{1}{N} \sum_{i=1}^{N} |M(s_{i})| p(s_{i} \ | \ T) \ .
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\end{align*}
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Continuing with the expectation values for a $2 \times 2$ lattice, excluding the terms which give zero we get
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Continuing with the expectation values for a $2 \times 2$ lattice, excluding the terms which result in zero, we get
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\begin{align*}
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\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} \\
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\langle E \rangle &= (-8J) \cdot \frac{1}{Z} e^{8 \beta J} + 2 \cdot (8J) \cdot \frac{1}{Z} e^{-8 \beta J} \\
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& \quad + (-8J) \cdot \frac{1}{Z} e^{8 \beta J} \\
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&= \frac{16J}{Z} \big(e^{-8 \beta J} - e^{8 \beta J}) \\
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&= -\frac{32J \sinh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
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&= -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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@ -90,9 +92,10 @@ and
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&= \frac{4 (2e^{8 \beta J} + 4)}{4(\cosh(8 \beta J) + 3)} \\
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&= \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ .
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\end{align*}
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The squared function
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The squared energy function
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\begin{align*}
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\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} \\
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\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} \\
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& \quad + (-8J)^{2} \cdot \frac{1}{Z} e^{8 \beta J} \\
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&= \frac{128J^{2}}{Z} \big(e^{8 \beta J} + e^{-8 \beta J} \big) \\
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&= \frac{128J^{2} \cosh(8 \beta J)}{4(\cosh(8 \beta J) + 3)} \\
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&= \frac{64J^{2} \cosh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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@ -116,6 +119,7 @@ and
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&= \frac{4(e^{8 \beta J} + 2)^{2}}{(\cosh(8 \beta J) + 3)^{2}} \ .
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\end{align*}
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\section{Heat capacity and magnetic susceptibility}\label{sec:heat_susceptibility}
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To find the heat capacity in Eq. \ref{eq:heat_capacity}, we normalize to heat
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capacity per spin
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@ -149,8 +153,9 @@ and the variance of the total magnetization is given by
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\mathbb{V}(M) &= \mathbb{E}(M^{2}) - [\mathbb{E}(|M|)]^{2} \\
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&= \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}} \\
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&= \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}} \\
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&= \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}} \\
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&= \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}} \\
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&= \frac{4(e^{8 \beta J} + 1) \cdot (e^{8 \beta J} + e^{-8 \beta J})}{(\cosh(8 \beta J) + 3)^{2}} \\
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& \quad + \frac{24(e^{8 \beta J} + 1) - 4(e^{8 \beta J} + 1)^{2}}{(\cosh(8 \beta J) + 3)^{2}}\\
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% &= \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}} \\
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&= \frac{4(3e^{8 \beta J} + e^{-8 \beta J} + 3)}{(\cosh(8 \beta J) + 3)^{2}} \ .
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\end{align*}
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We find the heat capacity
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@ -181,29 +186,30 @@ where the five distinct values are $\Delta E = \{-16J, -8J, 0, 8J, 16J\}$.
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\section{Additional results}\label{sec:additional_results}
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Results of 1 million MC cycles.
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\begin{figure}
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We also did the phase transition experiment using 1 million MC cycles. In Figure \ref{fig:phase_energy_1M}
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we show expected energy per spin, and in Figure \ref{fig:phase_magnetization_1M}
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expected magnetization per spin.
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/energy.pdf}
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\caption{$\langle \epsilon \rangle$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
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\label{fig:phase_energy_1M}
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\end{figure}
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\begin{figure}
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\end{figure} %
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/magnetization.pdf}
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\caption{$\langle |m| \rangle$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
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\label{fig:phase_magnetization_1M}
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\end{figure}
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\begin{figure}
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\end{figure} %
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In Figure \ref{fig:phase_heat_1M} we show heat capacity, and in Figure \ref{fig:phase_susceptibility_1M}
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the magnetic susceptibility.
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/heat_capacity.pdf}
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\caption{$C_{V}$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
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\label{fig:phase_heat_1M}
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\end{figure}
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\begin{figure}
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\end{figure} %
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/10M/susceptibility.pdf}
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\caption{$\chi$ for $T \in [2.1, 2.4]$, $10^{6}$ MC cycles.}
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@ -314,12 +314,13 @@ Monte Carlo cycles when analyzing phase transitions.
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\subsection{Tools}\label{subsec:tools}
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The Ising model and MCMC methods are implemented in C++, and parallelized using
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both \verb|OpenMPI| \cite{gabriel:2004:open_mpi} and \verb|OpenMP| \cite{openmp:2018}. We used the Python library
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\verb|matplotlib| \cite{hunter:2007:matplotlib} to produce all the plots, and
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\verb|seaborn| \cite{waskom:2021:seaborn} to set the theme in the figures. In
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addition, we used \verb|Scalasca| \cite{scalasca} and \verb|Score-P| \cite{scorep}
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to profile and optimize our implementation.
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% Add profiler
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both \verb|OpenMPI| \cite{gabriel:2004:open_mpi} and \verb|OpenMP| \cite{openmp:2018}.
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We used the Python library \verb|matplotlib| \cite{hunter:2007:matplotlib} to produce
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all the plots, \verb|seaborn| \cite{waskom:2021:seaborn} to set the theme in the
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figures. We also used a linear regression method from the Python library \verb|SciPy|.
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To optimize our implementation, we used a profiler tool \verb|Scalasca| \cite{scalasca}
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and \verb|Score-P| \cite{scorep}.
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\end{document}
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@ -3,72 +3,72 @@
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\begin{document}
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\section{Results}\label{sec:results}
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\subsection{Burn-in time}\label{subsec:burnin_time}
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$\boldsymbol{Draft}$
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We decided with a burn-in time parallelization trade-off. That is, we set the
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burn-in time lower in favor of sampling. To take advantage of the parallelization
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and not to waste computational resources. The argument to discard samples generated
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during the burn-in time is ... Increasing number of samples outweigh the ...
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parallelize using MPI. We generated
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samples for the temperature range $T \in [2.1, 2.4]$. Using Fox we generated both
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1 million samples and 10 million samples.
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We started with a lattice size $L = 20$ and considered the temperatures $T_{1} = 1.0 J/k_{B}$,
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and $T_{2} = 2.4 J/k_{B}$, where $T_{2}$ is close to the analytical critical
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temperature given by Equation \eqref{eq:critical_solution}.
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We start with a lattice where $L = 20$, to study the burn-in time, that is the
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number of Monte Carlo cycles necessary for the system to reach an equilibrium.
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We consider two different temperatures $T_{1} = 1.0 J/k_{B}$ and $T_{2} = 2.4 J/k_{B}$,
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where $T_{2}$ is close to the critical temperature. We can use the correlation
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time $\tau \approx L^{d + z}$ to determine time, where $d$ is the dimensionality
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of the system and $z = 2.1665 \pm 0.0012$ \footnote{This value was determined by
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Nightingale and Blöte for the Metropolis algorithm.}
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%
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We show the numerical estimates for temperature $T_{1}$ of $\langle \epsilon \rangle$
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in Figure \ref{fig:burn_in_energy_1_0} and $\langle |m| \rangle$ in Figure
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\ref{fig:burn_in_magnetization_1_0}. For temperature $T_{2}$, the numercal estimate
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of $\langle \epsilon \rangle$ is shown in Figure \ref{fig:burn_in_energy_2_4} and
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$\langle |m| \rangle$ in Figure \ref{fig:burn_in_magnetization_2_4}. The lattice
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is initialized in both an ordered and an unordered state. We observe that for
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$T_{1}$ there is no change in either expectation value with increasing number of
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Monte Carlo cycles, when we start with an ordered state. As for the unordered
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lattice, we observe a change for the first 5000 MC cycles, where it stabilizes.
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The approximated expected energy is $-2$ and expected magnetization is $1.0$,
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which is to be expected for temperature 0f $1.0$. T is below the critical and the
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pdf using $T = 1.0$ result in
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\begin{align*}
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p(s|T=1.0) &= \frac{1}{e^{-\beta \sum E(s)}} e^{-\beta E(s)} \\
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&= \frac{1}{e^{-(1/k_{B}) \sum E(s)}} e^{-(1/k_{B}) E(s)} \ .
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\end{align*}
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To determine the burn-in time, we used the numerical estimates of energy
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per spin for $T_{1}$ in Figure \ref{fig:burn_in_energy_1_0}, and $T_{2}$ in Figure
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\ref{fig:burn_in_energy_2_4}. We also considered the estimates of magnetization
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per spin for $T_{1}$ in Figure \ref{fig:burn_in_magnetization_1_0}, and for $T_{2}$
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in Figure \ref{fig:burn_in_energy_2_4}.
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% Not sure about the relevance of the paragraph below
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We used random numbers to set the initial state and the index of spin to flip, and
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observed different graphs ... However, when we set the seed of the random number
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engine, we could reproduce the results at every run. It is possible to determine
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the burn-in time analytically, using the correlation time given by $\tau \approx L^{d + z}$.
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Where $d$ is the dimensionality of the system and $z = 2.1665 \pm 0.0012$
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\footnote{This value was determined by Nightingale and Blöte for the Metropolis algorithm.}.
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However, when we increased number of Monte Carlo cycles the sampled generated during
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the burn-in time did not affect the mean value (...). % Should add result showing this
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The lattice was initialized in an ordered and an unordered state, for both temperatures. We observed
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no change in expectation value of energy or magnetization for $T_{1}$, when we
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initialized the lattice in an ordered state. As for the unordered initialized
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lattice, we first observed a change in expectation values, and a stabilization around
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$5000$ Monte Carlo cycles. The expected energy per spin is $\langle \epsilon \rangle = -2$
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and the expected magnetization per spin is $\langle |m| \rangle = 1.0$. % add something about what is expected for $T_{1}$ ?
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For $T_{2}$ we observed a change in expectation values for both the ordered and the
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unordered lattice.
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% \begin{align*}
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% p(s|T=1.0) &= \frac{1}{e^{-\beta \sum E(s)}} e^{-\beta E(s)} \\
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% &= \frac{1}{e^{-(1/k_{B}) \sum E(s)}} e^{-(1/k_{B}) E(s)} \ .
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% \end{align*}
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For $T_{2}$ we observe an increase in expected energy per spin $\langle \epsilon \rangle \approx -1.23$,
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and a decrease in expected magnetization per spin $\langle |m| \rangle \approx 0.46$.
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% Burn-in figures
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_1_0.pdf}
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\caption{$\langle |m| \rangle$ as a function of time, for $T = 1.0 J / k_{B}$}
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\caption{Magnetization per spin $\langle |m| \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 1.0 J / k_{B}$}
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\label{fig:burn_in_magnetization_1_0}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_energy_1_0.pdf}
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\caption{$\langle \epsilon \rangle$ as a function of time, for $T = 1.0 J / k_{B}$}
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\caption{Energy per spin $\langle \epsilon \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 1.0 J / k_{B}$}
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\label{fig:burn_in_energy_1_0}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_2_4.pdf}
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\caption{$\langle |m| \rangle$ as a function of time, for $T = 2.4 J / k_{B}$}
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\caption{Magnetization per spin $\langle |m| \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 2.4 J / k_{B}$}
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\label{fig:burn_in_magnetization_2_4}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_energy_2_4.pdf}
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\caption{$\langle \epsilon \rangle$ as a function of time, for $T = 2.4 J / k_{B}$}
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\caption{Energy per spin $\langle \epsilon \rangle$ as a function of time $t$ given by Monte Carlo cycles, for $T = 2.4 J / k_{B}$}
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\label{fig:burn_in_energy_2_4}
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\end{figure}
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\subsection{Probability distribution}\label{subsec:probability_distribution}
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% Histogram figures
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We used the estimated burn-in time as starting time for sampling, and generated
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We used the estimated burn-in time of $5000$ Monte Carlo cycles as starting time, and generated
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samples. To visualize the distribution of energy per spin $\epsilon$, we used histograms
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with a bin size (...). In Figure \ref{fig:histogram_1_0} we show the distribution
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for $T_{1}$, where the majority of the energy per spin is $-2$. The resulting expectation
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with a bin size $0.02$. In Figure \ref{fig:histogram_1_0} we show the distribution
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for $T_{1}$. Where the resulting expectation
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value of energy per spin is $\langle \epsilon \rangle = -1.9969$, with a low variance
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of Var$(\epsilon) = 0.0001$. %
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\begin{figure}[H]
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@ -85,16 +85,15 @@ centered around the expectation value $\langle \epsilon \rangle = -1.2370$. %
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\caption{Histogram $T = 2.4 J / k_{B}$}
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\label{fig:histogram_2_4}
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\end{figure} %
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However, we observe a higher variance of Var$(\epsilon) = 0.0203$. When the temperature
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increase, the system moves from an ordered to an onordered state. The change in
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However, we observed a higher variance of Var$(\epsilon) = 0.0203$. When the temperature
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increased, the system moved from an ordered to an onordered state. The change in
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system state, or phase transition, indicates the temperature is close to a critical
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point.
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\subsection{Phase transition}\label{subsec:phase_transition}
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$\boldsymbol{Draft}$
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% Phase transition figures
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We continue investigating the behavior of the system around the critical temperature.
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We continued investigating the behavior of the system around the critical temperature.
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First, we generated $10$ million samples of spin configurations for lattices of
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size $L \in \{20, 40, 60, 80, 100\}$, and temperatures $T \in [2.1, 2.4]$. We divided the
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temperature range into $40$ steps, with an equal step size of $0.0075$. The samples
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@ -133,7 +132,7 @@ value of heat capacity the lattice size increase.
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\caption{$C_{V}$ for $T \in [2.1, 2.4]$, $10^7$ MC cycles.}
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\label{fig:phase_heat_10M}
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\end{figure} %
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The magnetic susceptibility in Figure \ref{fig:phase_susceptibility_10M}, showed
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The magnetic susceptibility in Figure \ref{fig:phase_susceptibility_10M}, show
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the sharp peak in the same temperature range as that of the heat capacity. Since
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shape of the curve for both heat capacity and the magnetic susceptibility become
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sharper when we increase lattice size, we are moving closer to the critical temperature.
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@ -156,14 +155,14 @@ of lattices of size $L \in \{20, 40, 60, 80, 100\}$ found in Table \ref{tab:crit
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\hline
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$L$ & $T_{c}(L)$ \\
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\hline
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$20$ & $J$ \\
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$40$ & $1$ \\
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$60$ & $J / k_{B}$ \\
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||||
$80$ & $k_{B}$ \\
|
||||
$100$ & $1 / J$ \\
|
||||
$20$ & $ J / k_{B}$ \\
|
||||
$40$ & $ J / k_{B}$ \\
|
||||
$60$ & $ J / k_{B}$ \\
|
||||
$80$ & $ J / k_{B}$ \\
|
||||
$100$ & $ J / k_{B}$ \\
|
||||
\hline
|
||||
\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}
|
||||
\end{table}
|
||||
We used the critical temperatures of finite lattices and the scaling relation in
|
||||
|
||||
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Reference in New Issue
Block a user