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latex/sections/methods.tex
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\graphicspath{{\subfix{../images/}}}
\begin{document}
\section{Methods}
\section{Theory and methods}
% Problem 1
When we study the Penning traps effect on a particle with a charge $q$, we need to consider the forces acting on the particle. We can use Newton's second law \eqref{eq:newton_second} to determine the position of a particle as a function of time. In addition, we introduce the Lorentz force \eqref{eq:lorentz_force}, which describes the force on the particle. The position can be described by %
%
\begin{equation}\label{eq:newton_lorentz}
m \ddot{\mathbf{r}} = (q \mathbf{E} + q \mathbf{v} \times \mathbf{B}).
\end{equation}
%
Using eq. \eqref{eq:newton_lorentz} we derive the differential equations in sec. \ref{sec:eq_motion}, and can rewrite them as
\begin{align}
\label{eq:motion_x}
\ddot{x} - \omega_{0} \dot{y} - \frac{1}{2} \omega_{z}^{2} x &= 0, \\
\label{eq:motion_y}
\ddot{y} - \omega_{0} \dot{x} - \frac{1}{2} \omega_{z}^{2} y &= 0, \\
\label{eq:motion_z}
\ddot{z} + \omega_{z}^{2} z &= 0,
\end{align} %
%
We find the general solution for eq. \eqref{eq:motion_z}
\begin{equation}\label{eq:eq_general}
z(t) = c_{1} e^{i \omega_{z} t} + c_{2} e^{-i \omega_{z} t},
\end{equation} %
derived in sec. \ref{sec:eq_general}. Continuing, we will use a Calcium ion with a single positive charge. That is, we assume the charge of the particle is $q > 0$.
% Problem 2
Since eq. \eqref{eq:motion_x} and eq. \eqref{eq:motion_y} are coupled, we want to rewrite them as a single differential equation. We derive this in sec. \ref{sec:eq_complex}, and the resulting equation is given by
\begin{equation}\label{eq:single_differential}
\ddot{f} + i \omega_{0} \dot{f} - \frac{1}{2} \omega_{z}^{2} f = 0.
\end{equation}
% Problem 3
Eq. \eqref{eq:single_differential} has a general solution given by
\begin{equation}\label{eq:general_solution}
f(t) = A_{+}e^{-i(\omega_{+} t + \phi_{+})} + A_{-}e^{-i(\omega_{-} t + \phi_{-})}.
\end{equation} %
The amplitude $A_{+}$ and $A_{-}$ are positive, the phases $\phi_{+}$ and $\phi_{-}$ are constant, and the rate is given by
\begin{equation*}
\omega_{\pm} = \frac{\omega_{0} \pm \sqrt{\omega_{0}^{2} - 2 \omega_{z}^{2}}}{2}.
\end{equation*}
We find the physical coordinates at a given time $t$ using
\begin{equation*}
x(t) = \text{Re}f(t), \quad y(t) = \text{Im}f(t),
\end{equation*} %
and eq. \eqref{eq:general_solution}. We can rearrange the right hand side of the derived expression in \ref{sec:eq_coord}, and find the physical coordinates
\begin{align}\label{eq:physical_coord}
x(t) &= A_{+} \cos(\omega_{+} t + \phi_{+}) + A_{-} \cos(\omega_{-} t + \phi_{-}) \\
y(t) &= - A_{+} \sin(\omega_{+} t + \phi_{+}) - A_{-} \sin(\omega_{-} t + \phi_{-})
\end{align} %
% Problem 4
However, to obtain a bound on the particle's movement in the radial direction, we have to ensure that
\begin{align*}
|f(t)| &= \sqrt{(x(t))^{2} + (y(t))^{2}}. \\
\end{align*} %
When $t \rightarrow \infty$, the upper and lower limits are
\begin{align}
R_{+} &= A_{+} + A_{-} \\
R_{-} &= |A_{+} - A_{-}|,
\end{align}
derived in sec. \ref{sec:bounded_solution}.
% Problem 3
To obtain the bounded solution, we need to consider the expression
\begin{equation*}
\alpha = (\omega_{+} - \omega_{-}) t +( \phi_{+} - \phi_{-}).
\end{equation*}
When $t \rightarrow \infty$ the constant phases will not affect the expression, we obtain a bounded solution if we consider the rate such that $|f(t)| < \infty$. That is, we need $\omega_{0}^{2} - 2 \omega_{z}^{2}$ in eq. \eqref{eq:angular_rate} to be real.
\begin{align*}
\omega_{0}^{2} - 2 \omega_{z}^{2} = 0, \quad \omega_{0} > 2 \omega_{z}
\end{align*}
\begin{equation}\label{eq:angular_rate}
\omega_{\pm} = \frac{\omega_{0} \pm \sqrt{\omega_{0}^{2} - 2 \omega_{z}^{2}}}{2}
\end{equation}
We find the solution
Physical properties given by newtons second law \eqref{eq:newton_second}
The particle moves and its position can be determined using newton. where the electric field
\subsection{Units and constants}
Before continuing, we need to define the units we'll be working with.
Since we are working with particles, we need small units to work with so the
numbers we are working with aren't so small that they could potentially lead
to large round-off errors in our simulation. The units that we will use are listed in Table~\ref{tab:units}.
\begin{table}[H]
\begin{center}
\centering
\begin{tabular}[c]{lll}
Dimension & Unit & Symbol \\
Constant & Value & \\
\hline
Length & micrometer & $\mu m$ \\
Time & microseconds & $\mu s$ \\
Mass & atomic mass unit & $u$ \\
Charge & the elementary charge & $e$ \\
$B_{0}$ & $1.00 T$ & $9.65 \cross 10^{1} \frac{u}{(\mu s) e}$ \\
$V_{0}$ & $25.0 mV$ & $2.41 \cross 10^{6} \frac{u (\mu m)^{2}}{(\mu s)^{2} e}$ \\
$d$ & $500 \ \mu m$ & \\
\hline
\end{tabular}
\end{center}
\caption{The set of units we'll be working with.}\label{tab:units}
\caption{Default configuration of the Penning trap, where the value of T and V can be found in table \ref{tab:constants}.}
\label{tab:penning_config}
\end{table}
With these base units, we get
\begin{equation}
k_e = 1.3893533 \cdot 10^5 \frac{u(\mu m)^3}{(\mu s)^2 e},
\end{equation}
and we get that the unit for magnetic field strength (Tesla, $T$) and electric potential (Volt, $V$) are
\begin{equation}
\begin{split}
T &= 9.64852558 \cdot 10^1 \frac{u}{(\mu s) e} \\
V &= 9.64852558 \cdot 10^7 \frac{u (\mu m)^2}{(\mu s)^2 e}. \\
\end{split}
\label{eq:}
\end{equation}
\subsection{Dealing with a multi--particle system}
@@ -230,7 +291,6 @@ and would reduce the required amount of loops down to 4.
\subsection{Relative error and error convergance rate}
\subsection{Tools}
We used matplotlib
The numerical methods implemented in C++, are parallelized using OpenMP \cite{openmp:2018}. We used the Python library matplotlib \cite{hunter:2007:matplotlib} to produce all the plots, and seaborn \cite{waskom:2021:seaborn} to set the theme in the figures.
%\biblio
\end{document}