Add to method and conclusion.
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# Ising model related literature
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# Ising model related literature
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@misc{hj:2015:comp_phys,
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author = {Morten Hjorth-Jensen},
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howpublished = {\url{https://raw.githubusercontent.com/CompPhysics/ComputationalPhysics/master/doc/Lectures/lectures2015.pdf}},
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title = {Computational Physics, Lecture Notes Fall 2015},
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year = {2015}
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}
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@misc{britannica:2023:ferromagnetism,
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@misc{britannica:2023:ferromagnetism,
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author = {Britannica, The Editors of Encyclopaedia},
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author = {Britannica, The Editors of Encyclopaedia},
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title = {ferromagnetism},
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title = {ferromagnetism},
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@ -2,5 +2,18 @@
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\begin{document}
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\begin{document}
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\section{Conclusion}\label{sec:conclusion}
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\section{Conclusion}\label{sec:conclusion}
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% Draft based on abstract
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We have used the Ising model to study the behavior in ferromagnets, when undergoing
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a phase transition near a critical temperature. We generated samples using the
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Markov chain Monte Carlo method, while utilizing methods of parallelization.
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Finding the burn-in time to be approx. 3000 Monte Carlo cycles. For temperature
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$T = 1.0 J / k_{B}$ we found a propability distrobution with an expected mean
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value of $\mu \approx -1.9969$ and variation $\sigma^{2} = 0.0001$. Whereas the
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pdf close to the critical temperature is $\mu \approx -1.2370$ and variation
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$\sigma^{2} = 0.0203$. We estimated the expected energy and magnetization per spin,
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in addition to the heat capacity and susceptibility. Using the values from
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finite sized lattices, we approximated the critical temperature of an infinite
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sized lattice. Using linear regression, we numerically estimated $T_{c}$ $T_{C}(L = \infty) \approx 2.2695$
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which is close to the analytical solution $T_{C}(L = \infty) \approx 2.269 J/k_{B}$
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found by Lars Onsager.
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\end{document}
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\end{document}
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@ -80,7 +80,8 @@ have two spins oriented up the total energy have two possible values, as shown i
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conditions.}
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conditions.}
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\label{tab:lattice_config}
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\label{tab:lattice_config}
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\end{table}
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\end{table}
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We use the analytical values, found in Table for both for lattices where $L = 2$ and $L > 2$.
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We use the analytical values, found in Table for both for lattices where $L = 2$
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and $L > 2$.
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However, to compare the quantities for lattices where $L > 2$, we find energy
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However, to compare the quantities for lattices where $L > 2$, we find energy
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per spin given by
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per spin given by
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@ -94,6 +95,7 @@ and magnetization per spin given by
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\label{eq:magnetization_spin}
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\label{eq:magnetization_spin}
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\end{equation}
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\end{equation}
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\subsection{Statistical mechanics}\label{subsec:statistical_mechanics}
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\subsection{Statistical mechanics}\label{subsec:statistical_mechanics}
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When we study ferromagnetism, we have to consider the probability for a microstate
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When we study ferromagnetism, we have to consider the probability for a microstate
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$\mathbf{s}$ at a fixed temperature $T$. The probability distribution function
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$\mathbf{s}$ at a fixed temperature $T$. The probability distribution function
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@ -103,13 +105,19 @@ $\mathbf{s}$ at a fixed temperature $T$. The probability distribution function
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\label{eq:boltzmann_distribution}
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\label{eq:boltzmann_distribution}
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\end{equation}
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\end{equation}
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known as the Boltzmann distribution. This is an exponential distribution, where
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known as the Boltzmann distribution. This is an exponential distribution, where
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$\beta$ and $Z$ are given by
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$\beta$ is given by
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\begin{align*}
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\begin{equation}
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\beta =& \frac{1}{k_{B}} \ , &
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\beta = \frac{1}{k_{B}} \ ,
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Z &= \sum_{\text{all possible } \mathbf{s}} e^{-\beta E(\mathbf{s})} \ , \\
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\label{eq:beta}
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\end{align*}
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\end{equation}
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and $k_{B}$ is the Boltzmann constant. $Z$ is a normalizing factor of the pdf,
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where and $k_{B}$ is the Boltzmann constant. $Z$ is a normalizing factor of the
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known as the partition function, which we derive in Appendix \ref{sec:partition_function}
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pdf, given by
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\begin{equation}
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Z = \sum_{\text{all possible } \mathbf{s}} e^{-\beta E(\mathbf{s})} \ ,
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\label{eq:partition}
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\end{equation}
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and is known as the partition function. We derive $Z$ in Appendix \ref{sec:partition_function},
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which gives us
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\begin{equation*}
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\begin{equation*}
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Z = 4 \cosh (8 \beta J) + 12 \ .
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Z = 4 \cosh (8 \beta J) + 12 \ .
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\end{equation*}
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\end{equation*}
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@ -117,27 +125,28 @@ Using the partition function and Eq. \eqref{eq:boltzmann_distribution}, the pdf
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of a microstate at a fixed temperature is given by
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of a microstate at a fixed temperature is given by
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\begin{equation}
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\begin{equation}
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p(\mathbf{s} \ | \ T) = \frac{1}{4 \cosh (8 \beta J) + 12} e^{-\beta E(\mathbf{s})} \ .
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p(\mathbf{s} \ | \ T) = \frac{1}{4 \cosh (8 \beta J) + 12} e^{-\beta E(\mathbf{s})} \ .
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\label{eq:pdf}
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\end{equation}
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\end{equation}
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% Add something about why we use the expectation values?
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% Add something about why we use the expectation values?
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We derive the analytical expressions for expectation values in Appendix.
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We derive the analytical expressions for expectation values in Appendix.
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\ref{sec:expectation_values}. We find the expected total energy
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\ref{sec:expectation_values}. We find the expected total energy
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\begin{equation*} %\label{eq:energy_total_first}
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\begin{equation}\label{eq:energy_total_result}
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\langle E \rangle = -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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\langle E \rangle = -\frac{8J \sinh(8 \beta J)}{\cosh(8 \beta J) + 3} \ ,
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\end{equation*}
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\end{equation}
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and the expected energy per spin
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and the expected energy per spin
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\begin{equation*} %\label{eq:energy_spin_first}
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\begin{equation}\label{eq:energy_spin_result}
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\langle \epsilon \rangle = \frac{-2J \sinh(8 \beta J)}{ \cosh(8 \beta J) + 3} \ .
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\langle \epsilon \rangle = \frac{-2J \sinh(8 \beta J)}{ \cosh(8 \beta J) + 3} \ .
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\end{equation*}
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\end{equation}
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We find the expected absolute total magnetization
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We find the expected absolute total magnetization
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\begin{equation*} %\label{eq:magnetization_total_first}
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\begin{equation}\label{eq:magnetization_total_result}
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\langle |M| \rangle = \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ ,
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\langle |M| \rangle = \frac{2(e^{8 \beta J} + 2)}{\cosh(8 \beta J) + 3} \ ,
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\end{equation*}
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\end{equation}
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and the expected magnetization per spin
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and the expected magnetization per spin
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\begin{equation*} %\label{eq:magnetization_spin_first}
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\begin{equation}\label{eq:magnetization_spin_result}
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\langle |m| \rangle = \frac{e^{8 \beta J} + 1}{2( \cosh(8 \beta J) + 3)} \ .
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\langle |m| \rangle = \frac{e^{8 \beta J} + 1}{2( \cosh(8 \beta J) + 3)} \ .
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\end{equation*}
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\end{equation}
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We will also determine the heat capacity
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We also need to determine the heat capacity
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\begin{equation}
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\begin{equation}
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C_{V} = \frac{1}{k_{B} T^{2}} (\mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2}) \ ,
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C_{V} = \frac{1}{k_{B} T^{2}} (\mathbb{E}(E^{2}) - [\mathbb{E}(E)]^{2}) \ ,
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\label{eq:heat_capacity}
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\label{eq:heat_capacity}
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@ -180,6 +189,12 @@ Boltzmann constant we derive the remaining units, which can be found in Table
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\subsection{Phase transition and critical temperature}\label{subsec:phase_critical}
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\subsection{Phase transition and critical temperature}\label{subsec:phase_critical}
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% P9 critical temperature
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% P9 critical temperature
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When a ferromagnetic material is heated, it will change at a macroscopic level.
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Based on a $2 \times 2$ lattice, we can show that the total energy is equal to the
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energy where all spins have the orientation up \cite[p. 426]{hj:2015:comp_phys}.
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Increasing the temperature of the external field, the Ising model move from an
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ordered to an unordered phase. At the critical temperature the heat capacity $C_{V}$,
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and the magnetic susceptibility $\chi$ diverge \cite[p. 431]{hj:2015:comp_phys}.
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\subsection{The Markov chain Monte Carlo method}\label{subsec:mcmc_method}
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\subsection{The Markov chain Monte Carlo method}\label{subsec:mcmc_method}
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Markov chains consist of a sequence of samples, where the probability of the next
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Markov chains consist of a sequence of samples, where the probability of the next
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