Merge branch 'coryab/latex' into develop
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latex/main.pdf
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latex/main.pdf
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@ -74,4 +74,14 @@
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publisher = {Farleia Forlag},
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publisher = {Farleia Forlag},
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year = {2018},
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year = {2018},
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pages = {162--163}
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pages = {162--163}
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}
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}
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@article{rk4_method,
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author = "Morten Hjorth-Jensen",
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title = "Computational Physics, Lecture Notes Fall 2015",
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journal = "Department of Physics, University of Oslo",
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year = "2015",
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chapter = "8.4",
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pages = "250--252",
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}
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@ -49,11 +49,6 @@ Since \eqref{eq:motion_x} and \eqref{eq:motion_y} are coupled, we want to rewrit
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\end{align*}
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\end{align*}
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Physical properties given by newtons second law \eqref{eq:newton_second}
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Physical properties given by newtons second law \eqref{eq:newton_second}
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\begin{equation}\label{eq:general_solution}
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\begin{equation}\label{eq:general_solution}
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f(t) = A_{+}e^{-i(\omega_{+} t + \phi_{+})} + A_{-}e^{-i(\omega_{-} t + \phi_{-})}
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f(t) = A_{+}e^{-i(\omega_{+} t + \phi_{+})} + A_{-}e^{-i(\omega_{-} t + \phi_{-})}
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@ -66,4 +61,230 @@ The particle moves and its position can be determined using newton. where the el
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\subsection*{Tools}
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\subsection*{Tools}
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We used matplotlib
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We used matplotlib
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\end{document}
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\subsection{Units and constants}
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Before continuing, we need to define the units we'll be working with.
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Since we are working with particles, we need small units to work with so the
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numbers we are working with aren't so small that they could potentially lead
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to large round-off errors in our simulation. The units that we will use are listed in Table~\ref{tab:units}.
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\begin{table}[H]
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\begin{center}
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\begin{tabular}[c]{lll}
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Dimension & Unit & Symbol \\
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\hline
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Length & micrometer & $\mu m$ \\
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Time & microseconds & $\mu s$ \\
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Mass & atomic mass unit & $u$ \\
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Charge & the elementary charge & $e$ \\
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\hline
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\end{tabular}
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\end{center}
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\caption{The set of units we'll be working with.}\label{tab:units}
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\end{table}
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With these base units, we get
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\begin{equation}
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k_e = 1.3893533 \cdot 10^5 \frac{u(\mu m)^3}{(\mu s)^2 e},
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\end{equation}
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and we get that the unit for magnetic field strength (Tesla, $T$) and electric potential (Volt, $V$) are
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\begin{equation}
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\begin{split}
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T &= 9.64852558 \cdot 10^1 \frac{u}{(\mu s) e} \\
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V &= 9.64852558 \cdot 10^7 \frac{u (\mu m)^2}{(\mu s)^2 e}. \\
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\end{split}
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\label{eq:}
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\end{equation}
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\subsection{Dealing with a multi--particle system}
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For a multi-particles system, we need to modify $\vb{F}$ to account for the
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force of other particles in the system acting upon each other. To do that, we
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add another term to $\vb{F}$
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\begin{equation}
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\vb{F}_i(t, \vb{v}_i, \vb{r}_i) = q_i \vb{E}(t, \vb{r}_i) + q_i \vb{v}_i \cross \vb{B} - \vb{E}_p(t, \vb{r}_i),
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\end{equation}
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where $i$ and $j$ are particle indices and
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\begin{equation}
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\vb{E}_p(t, \vb{r}_i) = q_i k_e \sum_{j \neq i}
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q_j \frac{\vb{r_i} - \vb{r_j}}{\left| \vb{r_i} - \vb{r_j} \right|^3}.
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\label{eq:}
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\end{equation}
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Newton's second law for a particle $i$ is then
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\begin{equation}
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\frac{d^2\vb{r}_i}{dt^2} = \frac{\vb{F}_i\left(t, \frac{d\vb{r}_i}{dt}, \vb{r_i}\right)}{m_i},
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\label{eq:newtonlaw2}
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\end{equation}
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We can then rewrite the second order ODE from equation~\ref{eq:newtonlaw2}
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into a set of coupled first order ODEs.
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We now rewrite Newton's second law of motion as
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\begin{equation}
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\begin{split}
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\frac{d\vb{r}_i}{dt} &= \vb{v}_i \\
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\frac{d\vb{v}_i}{dt} &= \frac{\vb{F}_i(t, \vb{v}_i, \vb{r}_i)}{m_i}.
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\end{split}
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\label{eq:coupled}
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\end{equation}
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\subsection{Forward Euler}
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For a particle $i$, the forward Euler method for a coupled system is
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expressed as
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\begin{equation}
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\begin{split}
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\vb{r}_{i,j+1} &= \vb{r}_{i,j} + h \cdot \frac{d\vb{r}_{i,j}}{dt} = \vb{r}_{i,j} + h \cdot \vb{v}_{i,j} \\
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\vb{v}_{i,j+1} &= \vb{v}_{i,j} + h \cdot \frac{\vb{v}_{i,j}}{dt} = \vb{v}_{i,j} + h \cdot \frac{\vb{F}\left( t_{j}, \vb{v}_{i,j}, \vb{r}_{i,j} \right)}{m},
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\end{split}
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\end{equation}
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for particle $i$ where $j$ is the current time step of the particle,
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$m$ is the mass of the particle, and $h$ is the step length.
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When dealing with a multi-particle system, we need to ensure that we do not
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update the position of any particles until every particle has calculated their
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next step. An easy way of doing this is to create a copy of all the particles,
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then update the copy, and when all the particles have calculated their next
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step, simply replace the particles with the copies.
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Algorithm~\ref{algo:forwardeuler} provides an overview on how that can be achieved. % Make this better
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\begin{figure}[H]
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\begin{algorithm}[H]
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\caption{Forward Euler method}
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\label{algo:forwardeuler}
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\begin{algorithmic}
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\Procedure{Evolve forward Euler}{$particles, dt$}
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\State $N \leftarrow \text{Number of particles in } particles$
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\State $a \leftarrow \text{Calculate } \frac{\vb{F_i}}{m_i} \text{ for each particle in } particles$
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\For{ $i = 1, 2, \ldots , N$ }
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\State $particles_i.\vb{r} \leftarrow particles_i.\vb{r} + dt \cdot particles_i.\vb{v}$
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\State $particles_i.\vb{v} \leftarrow particles_i.\vb{v} + dt \cdot a_i$
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\EndFor
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\end{figure}
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\subsection{4th order Runge-Kutta}
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For a particle $i$, we can express the 4th order Runge-Kutta (RK4) method as
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\begin{equation}
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\begin{split}
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\vb{v}_{i,j+1} &= \vb{v}_{i,j} + \frac{h}{6} \left( \vb{k}_{\vb{v},1,i}
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+ 2\vb{k}_{\vb{v},2,i} + 2\vb{k}_{\vb{v},3,i} + \vb{k}_{\vb{v},4,i}
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\right) \\
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\vb{r}_{i,j+1} &= \vb{r}_{i,j} + \frac{h}{6} \left( \vb{k}_{\vb{r},1,i}
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+ 2\vb{k}_{\vb{r},2,i} + 2\vb{k}_{\vb{r},3,i} + \vb{k}_{\vb{r},4,i}
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\right),
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\end{split}
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\end{equation}
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where
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\begin{equation}
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\begin{split}
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\vb{k}_{\vb{v},1,i} &= \frac{\vb{F}_i(t_j, \vb{v}_{i,j},
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\vb{r}_{i,j})}{m} \\
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\vb{k}_{\vb{r},1,i} &= \vb{v}_{i,j} \\
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\vb{k}_{\vb{v},2,i} &= \frac{\vb{F}_i(t_j+\frac{h}{2}, \vb{v}_{i,j}
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+ h \cdot \frac{\vb{k}_{\vb{v},1,i}}{2}, \vb{r}_{i,j}
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+ h \cdot \frac{\vb{k}_{\vb{r},1,i}}{2})}{m} \\
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\vb{k}_{\vb{r},2,i} &= \vb{v}_{i,j}
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+ h \cdot \frac{\vb{k}_{\vb{v},1,i}}{2} \\
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\vb{k}_{\vb{v},3,i} &= \frac{\vb{F}_i(t_j+\frac{h}{2}, \vb{v}_{i,j}
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+ h \cdot \frac{\vb{k}_{\vb{v},2,i}}{2}, \vb{r}_{i,j}
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+ h \frac{\cdot \vb{k}_{\vb{r},2,i}}{2})}{m} \\
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\vb{k}_{\vb{r},3,i} &= \vb{v}_{i,j}
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+ h \cdot \frac{\vb{k}_{\vb{v},2,i}}{2} \\
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\vb{k}_{\vb{v},4,i} &= \frac{\vb{F}_i(t_j+h, \vb{v}_{i,j}
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+ h \cdot \vb{k}_{\vb{v},3,i}, \vb{r}_{i,j}
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+ h \cdot \vb{k}_{\vb{r},3,i})}{m} \\
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\vb{k}_{\vb{r},4,i} &= \vb{v}_{i,j}
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+ h \cdot \frac{\vb{k}_{\vb{v},1,i}}{2}.
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\end{split}
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\end{equation}
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In order to find each $\vb{k}_{\vb{r},i}$ and $\vb{k}_{\vb{v},i}$,
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we need to first compute all $\vb{k}_{\vb{r},i}$ and $\vb{k}_{\vb{v},i}$
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for all particles, then we can update the particles in order to compute
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$\vb{k}_{\vb{r},i+1}$ and $\vb{k}_{\vb{v},i+1}$. In order for the algorithm
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to work, we need to save a copy of each particle before starting so that we
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can update the particles correctly for each step.
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This approach would require 8 loops to be able to complete the calculation since
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we cannot update the particles until after all $\vb{k}$ values have been
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computed, however if we create a temporary array that holds particles, we can
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put the updated particles in there, and then use that array in the next loop,
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and would reduce the required amount of loops down to 4.
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\begin{figure}
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\begin{algorithm}[H]
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\caption{RK4 method}
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\label{algo:rk4}
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\begin{algorithmic}
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\Procedure{Evolve RK4}{$particles, dt$}
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\State $N \leftarrow \text{Number of particles inside the Penning trap}$
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\State $orig\_p \leftarrow \text{Copy of particles}$
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\State $tmp\_p \leftarrow \text{Array of particles of size }N$
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\State $\vb{k}_{\vb{r}} \leftarrow \text{2D array of vectors of size } 4 \cross N$
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\State $\vb{k}_{\vb{v}} \leftarrow \text{2D array of vectors of size } 4 \cross N$
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\For{ $i = 1, 2, \ldots, N$ }
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\State $\vb{k}_{\vb{r},1,i} \leftarrow particles_i.\vb{v}$
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\State $\vb{k}_{\vb{v},1,i} \leftarrow \frac{\vb{F}_i}{m_i}$
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\State $tmp\_p_i.\vb{r} \leftarrow orig\_p_i.\vb{r}
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+ \frac{dt}{2} \cdot \vb{k}_{\vb{r},1,i}$
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\State $tmp\_p_i.\vb{v} \leftarrow orig\_p_i.\vb{v}
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+ \frac{dt}{2} \cdot \vb{k}_{\vb{v},1,i}$
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\EndFor
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\State $particles \leftarrow tmp\_p$ \Comment{Update particles}
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\For{ $i = 1, 2, \ldots, N$ }
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\State $\vb{k}_{\vb{r},2,i} \leftarrow particles_i.\vb{v}$
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\State $\vb{k}_{\vb{v},2,i} \leftarrow \frac{\vb{F}_i}{m_i}$
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\State $tmp\_p_i.\vb{r} \leftarrow orig\_p_i.\vb{r}
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+ \frac{dt}{2} \cdot \vb{k}_{\vb{r},2,i}$
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\State $tmp\_p_i.\vb{v} \leftarrow orig\_p_i.\vb{v}
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+ \frac{dt}{2} \cdot \vb{k}_{\vb{v},2,i}$
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\EndFor
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\State $particles \leftarrow tmp\_p$ \Comment{Update particles}
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\For{ $i = 1, 2, \ldots, N$ }
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\State $\vb{k}_{\vb{r},3,i} \leftarrow particles_i.\vb{v}$
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\State $\vb{k}_{\vb{v},3,i} \leftarrow \frac{\vb{F}_i}{m}$
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\State $tmp\_p_i.\vb{r} \leftarrow orig\_p_i.\vb{r} + dt \cdot \vb{k}_{\vb{r},3,i}$
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\State $tmp\_p_i.\vb{v} \leftarrow orig\_p_i.\vb{v} + dt \cdot \vb{k}_{\vb{v},3,i}$
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\EndFor
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\State $particles \leftarrow tmp\_p$ \Comment{Update particles}
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\For{ $i = 1, 2, \ldots, N$ }
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\State $\vb{k}_{\vb{r},4,i} \leftarrow particles_i.\vb{v}$
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\State $\vb{k}_{\vb{v},4,i} \leftarrow \frac{\vb{F}}{m}$
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\State $tmp\_p_i.\vb{r} \leftarrow orig\_p_i.\vb{r} + \frac{dt}{6}
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\cdot \left( \vb{k}_{\vb{r},1,i} + \vb{k}_{\vb{r},2,i}
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+ \vb{k}_{\vb{r},3,i} + \vb{k}_{\vb{r},4,i} \right)$
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\State $tmp\_p_i.\vb{v} \leftarrow orig\_p_i.\vb{v} + \frac{dt}{6}
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\cdot \left( \vb{k}_{\vb{v},1,i} + \vb{k}_{\vb{v},2,i}
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+ \vb{k}_{\vb{v},3,i} + \vb{k}_{\vb{v},4,i} \right)$
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\EndFor
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\State $particles \leftarrow tmp\_p$ \Comment{Final update}
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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$\vb{F}$ in the algorithm does not take any arguments as it uses the
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velocities and positions of the particles inside the array $particles$ to
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calculate the total force acting on particle $i$.
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\end{figure}
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\subsection{Testing the simulation}
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\subsection{Relative error and error convergance rate}
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%\biblio
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\end{document}
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