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Janita Willumsen
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\documentclass[../main.tex]{subfiles}
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\graphicspath{{\subfix{../images/}}}
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\begin{document}
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\appendix
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\section*{Appendix A}\label{sec:appendix_a}
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Equations given
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\begin{equation}\label{eq:newton_second}
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m \ddot{\mathbf{r}} = \sum_{i} \mathbf{F}_{i}
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\end{equation}
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%
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%
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\begin{equation}\label{eq:e_field_point}
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\mathbf{E} = k_{e} \sum_{j=1}^{n} q_j \frac{\mathbf{r} - \mathbf{r}_{j}}{|\mathbf{r} - \mathbf{r}_{j}|^{3}}
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\end{equation}
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%
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\begin{equation}\label{eq:e_field_potential}
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\mathbf{E} = - \nabla V
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\end{equation}
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\section*{Appendix B}\label{sec:appendix_b}
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Sum of all forces
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\begin{align*}
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sum
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\end{align*}
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We find the physical coordinates from $x(t) = \text{Re} f(t)$ and $y(t) = \text{Im} f(t)$, where $f(t)$ is given by equation \eqref{eq:general_solution}.
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We can rewrite $f(t)$ using the definition as
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\begin{align*}
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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_{+}(\cos{\omega_{+} t + \phi_{+}} - i \sin{\omega_{+} t + \phi_{+}}) \\
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\numberthis \label{eq:general_solution_trig}
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&+ A_{-}(\cos{\omega_{-} t + \phi_{-}} - i \sin{\omega_{-} t + \phi_{-}})
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\end{align*}
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%
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If we rearrange the right side of equation \eqref{eq:general_solution_trig}, we find the physical coordinates
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\begin{align}\label{eq:physical_coord}
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x(t) &= A_{+}(\cos{\omega_{+} t + \phi_{+}}) + A_{-}(\cos{\omega_{-} t + \phi_{-}}) \\
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y(t) &= - A_{+}(i \sin{\omega_{+} t + \phi_{+}}) - A_{-}(i \sin{\omega_{-} t + \phi_{-}})
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\end{align}
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\end{document}
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29
latex/appendix/appendix_a.tex
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29
latex/appendix/appendix_a.tex
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\documentclass[../main.tex]{subfiles}
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\graphicspath{{\subfix{../images/}}}
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\begin{document}
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\section{Equations of electrodynamics and classical mechanics}\label{sec:appendix_a}
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Newton's second law
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\begin{equation}\label{eq:newton_second}
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m \ddot{\mathbf{r}} = \sum_{i} \mathbf{F}_{i}
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\end{equation}%
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%
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Lorentz force
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\begin{equation}\label{eq:lorentz_force}
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\mathbf{F} = q \mathbf{E} + q \mathbf{v} \times \mathbf{B},
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\end{equation}%
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%
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Electric field
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\begin{equation}\label{eq:e_field_potential}
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\mathbf{E} = - \nabla V
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\end{equation} %
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%
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Electric field at a point $\mathbf{r}$
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\begin{equation}\label{eq:e_field_point}
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\mathbf{E} = k_{e} \sum_{j=1}^{n} q_j \frac{\mathbf{r} - \mathbf{r}_{j}}{|\mathbf{r} - \mathbf{r}_{j}|^{3}}
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\end{equation}
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\end{document}
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145
latex/appendix/appendix_b.tex
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latex/appendix/appendix_b.tex
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\documentclass[../main.tex]{subfiles}
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\graphicspath{{\subfix{../images/}}}
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\begin{document}
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\section{Derivation of equations}\label{sec:appendix_b}
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% Problem 1
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\subsection{Equations of motion}\label{sec:eq_motion} %
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First, we need to define the velocity of the particle
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\begin{align*}
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\mathbf{v} \equiv \frac{d \mathbf{r}}{dt} &= \bigg( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \bigg).
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\end{align*} %
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We can rewrite the velocity as $\dot{r} = (\dot{x}, \dot{y}, \dot{z})$, and find the cross product
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\begin{align*}
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q \mathbf{v} \cross \mathbf{B} &=
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q \begin{vmatrix}
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\hat{e}_{x} & \hat{e}_{y} & \hat{e}_{z} \\
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\dot{x} & \dot{y} & \dot{z} \\
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0 & 0 & B_{0}
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\end{vmatrix}
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= q \big( B_{0} \dot{y}, -B_{0} \dot{x}, 0 \big).
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\end{align*} %
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We are considering an ideal Penning traps, where we define the electric potential as
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\begin{align*}
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V(x, y, z) &= \frac{V_{0}}{2 d^{2}}(2z^{2} - x^{2} - y^{2}).
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\end{align*} %
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The relationship between the electric field $\mathbf{E}$ and the electric potential of the field is given by
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\begin{align*}
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\mathbf{E} &= - \nabla V \\
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&= - \bigg( \frac{dV}{dx}, \frac{dV}{dy} \frac{dV}{dz} \bigg) \\
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&= \frac{V_{0}}{d^{2}} \big( x, y, -2z \big).
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\end{align*} %
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We can now express the Lorentz force as
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\begin{align*}
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\mathbf{F} &= q \mathbf{E} + q \mathbf{v} \cross \mathbf{B} \\
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&= \frac{q V_{0}}{d^{2}} \big( x, y, -2z \big) + \big(q B_{0} \dot{y}, -q B_{0} \dot{x}, 0 \big),
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\end{align*} %
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and insert it into Newtons equation \eqref{eq:newton_second}. We get
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\begin{align*}
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\ddot{\mathbf{r}} &= \bigg( \frac{q V_{0}}{m d^{2}} x, \frac{q V_{0}}{m d^{2}} y, -\frac{2 q V_{0}}{m d^{2}} z \bigg) + \bigg(\frac{q B_{0}}{m} \dot{y}, -\frac{q B_{0}}{m} \dot{x}, 0 \bigg),
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\end{align*} %
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which can be written as
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\begin{align*}
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\ddot{x} &= \frac{q V_{0}}{m d^{2}} x + \frac{q B_{0}}{m} \dot{y}, \\
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\ddot{y} &= \frac{q V_{0}}{m d^{2}} y - \frac{q B_{0}}{m} \dot{x}, \\
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\ddot{z} &= -\frac{2 q V_{0}}{m d^{2}} z.
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\end{align*}
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If we define
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\begin{equation*}
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\omega_{0} \equiv \frac{q B_{0}}{m}, \quad \omega_{z}^{2} \equiv \frac{2 q V_{0}}{m d^{2}},
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\end{equation*}
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the equations of motion can be written as
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\begin{align*}
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\ddot{x} &= \frac{1}{2} \omega_{z}^{2} x + \omega_{0} \dot{y}, \\
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\ddot{y} &= \frac{1}{2} \omega_{z}^{2} y - \omega_{0} \dot{x}, \\
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\ddot{z} &= -\omega_{z}^{2} z. \\
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\end{align*}
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\subsection{General solution}\label{sec:eq_general}
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We consider the characteristic equation of a second order differential equation \cite{lindstrom:2016:ch10:5},
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\begin{align*}
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r^{2} + \omega_{z}^{2} &= 0 \\
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r &= \pm \sqrt{- \omega_{z}^{2}}.
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\end{align*} %
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The characteristic equation has two complex roots
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\begin{equation*}
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r_{1} = - i \omega_{z}, \quad r_{2} = i \omega_{z},
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\end{equation*}
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which give us solutions in the general form
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\begin{equation*}
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z = c_{1} e^{i \omega_{z} t} + c_{2} e^{-i \omega_{z} t}.
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\end{equation*}
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In addition, for a complex number $z = a + ib$, we can define $e^{z} \equiv e^{a} (\cos{b} + i \sin{b})$ \cite{lindstrom:2016:ch3}. We can rewrite the general solution as
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\begin{align*}
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c_{1} e^{i \omega_{z} t} + c_{2} e^{-i \omega_{z} t} &= c_{1} (\cos{\omega_{z} t} + i \sin{\omega_{z} t}) \\
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& \quad + c_{2} (\cos{\omega_{z} t} - i \sin{\omega_{z} t} \\
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&= E \cos{\omega_{z} t} + i F \sin{\omega_{z} t}.
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\end{align*} %
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%
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\subsection{Complex function}\label{sec:eq_complex} %
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In sec. \ref{sec:eq_motion} we found the differential equations for $\ddot{x}$ and $\ddot{y}$. To derive a single differential equation, we introduce the complex function $f(t) = x(t) + iy(t)$, which gives us
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\begin{align*}
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0 &= \Big(\ddot{x} - \omega_{0} \dot{y} - \frac{1}{2} \omega_{z}^{2} x \Big) + i \Big(\ddot{y} + \omega_{0} \dot{x} - \frac{1}{2} \omega_{z}^{2} y \Big) \\
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&= \ddot{x} - \omega_{0} \dot{y} - \frac{1}{2} \omega_{z}^{2} x + i\ddot{y} + i\omega_{0} \dot{x} - i \frac{1}{2} \omega_{z}^{2} y \\
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&= \ddot{x} + i\ddot{y} + i\omega_{0} \dot{x} - \omega_{0} \dot{y} - \frac{1}{2} \omega_{z}^{2} x - i \frac{1}{2} \omega_{z}^{2} y. \\
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\end{align*}
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Using the definition $i = \sqrt{-1}$, we can rewrite
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\begin{equation*}
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i \omega_{0} \dot{x} + (-1) \omega_{0} \dot{y} = i \omega_{0} \dot{x} + i^{2} \omega_{0} \dot{y}.
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\end{equation*}
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This gives us a single differential equation
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\begin{align*}
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0 &= \ddot{x} + i\ddot{y} + i\omega_{0} (\dot{x} + i \dot{y}) - \frac{1}{2} \omega_{z}^{2} x - i \frac{1}{2} \omega_{z}^{2} y \\
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&= \ddot{f} + i \omega_{0} \dot{f} - \frac{1}{2} \omega_{z}^{2} f.
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\end{align*}
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%
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\subsection{Physical coordinates}\label{sec:eq_coord}
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We can rewrite eq. \eqref{eq:general_solution}, as
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\begin{align*}
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f(t) &= A_{+}e^{-i(\omega_{+} t + \phi_{+})} + A_{-}e^{-i(\omega_{-} t + \phi_{-})} \\
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&= A_{+}(\cos{(\omega_{+} t + \phi_{+})} - i \sin{(\omega_{+} t + \phi_{+})}) \\
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% \numberthis \label{eq:general_solution_trig}
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& \quad + A_{-}(\cos{(\omega_{-} t + \phi_{-})} - i \sin{(\omega_{-} t + \phi_{-})}).
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\end{align*} %
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%
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\subsection{Upper and lower bounds}\label{sec:upper_lower_bound}
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To obtain the upper and lower bounds of the particle's distance from the origin, we first find an expression for the second norm (?) defined as $|f(t)| = \sqrt{(x(t))^{2} + (y(t))^{2}}$.
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\begin{align*}
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(x(t))^{2} &= \big( A_{+}\cos(\omega_{+} t + \phi_{+}) + A_{-}\cos(\omega_{-} t + \phi_{-}) \big)^{2} \\
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&= A_{+}^{2} \cos^{2}(\omega_{+} t + \phi_{+}) \\
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& \quad + 2 A_{+}A_{-} \cos(\omega_{+} t + \phi_{+})\cos(\omega_{-} t + \phi_{-}) \\
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& \quad + A_{-}^{2}\cos^{2}(\omega_{-} t + \phi_{-}), \\
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\end{align*} %
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\begin{align*}
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(y(t))^{2} &= \big( - A_{+} \sin(\omega_{+} t + \phi_{+}) - A_{-} \sin(\omega_{-} t + \phi_{-}) \big)^{2} \\
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&= A_{+}^{2} \sin^{2}(\omega_{+} t + \phi_{+}) \\
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& \quad + 2 A_{+}A_{-} \sin(\omega_{+} t + \phi_{+})\sin(\omega_{-} t + \phi_{-}) \\
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& \quad + A_{-}^{2} \sin^{2}(\omega_{-} t + \phi_{-}).
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\end{align*} %
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We insert these expressions, and find
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\begin{align*}
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|f(t)| &= \sqrt{(x(t))^{2} + (y(t))^{2}} \\
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&= \sqrt{A_{+}^{2} + 2 A_{+} A_{-} \cos^{2}(\alpha) + A_{-}^{2}},
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\end{align*} %
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where $\alpha = (\omega_{+} - \omega_{-}) t +( \phi_{+} - \phi_{-})$. If we set $\alpha = 0$ we get $\cos(0) = 1$, and obtain the upper bound
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\begin{align*}
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R_{+} &= \sqrt{A_{+}^{2} + 2 A_{+} A_{-} + A_{-}^{2}} \\
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&= \sqrt{(A_{+} + A_{-})^{2}} \\
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&= A_{+} + A_{-}.
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\end{align*}
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If $\alpha = \pi$ we get $\cos(\pi) = -1$, and find the lower bound
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\begin{align*}
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R_{-} &= \sqrt{A_{+}^{2} - 2 A_{+} A_{-} + A_{-}^{2}} \\
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&= \sqrt{(A_{+} - A_{-})^{2}} \\
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&= |A_{+} - A_{-}|.
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\end{align*} %
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%
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\subsection{Bounded solution}\label{sec:bounded_solution}
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\end{document}
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% \begin{align*}
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% \omega_{+} - \omega_{-} &= \frac{\omega_{0} + \sqrt{\omega_{0}^{2} - 2 \omega_{z}^{2}}}{2} - \frac{\omega_{0} - \sqrt{\omega_{0}^{2} - 2 \omega_{z}^{2}}}{2} \\
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% &=
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% \end{align*}
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21
latex/appendix/appendix_c.tex
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latex/appendix/appendix_c.tex
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\documentclass[../main.tex]{subfiles}
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\graphicspath{{\subfix{../images/}}}
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\begin{document}
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\section{Numbers and units}\label{sec:appendix_c}
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\begin{table}[H]
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\centering
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\begin{tabular}[c]{lll}
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Constant & Value & Unit \\
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\hline
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$k_{e}$ (Coulomb) & $1.38935333 \cross 10^{5}$ & $\frac{u (\mu m)^{3}}{(\mu s)^{2} e^{2}}$ \\
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T (Tesla) & $9.64852558 \cross 10^{1}$ & $\frac{u}{(\mu s) e}$ \\
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V (Volt) & $9.64852558 \cross 10^{7}$ & $\frac{u (\mu m)^{2}}{(\mu s)^{2} e}$ \\
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\hline
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\end{tabular}
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\caption{Value of the Coulomb constant ($k_{e}$), and the SI units for magnetic field strength ($T$) and electric potential ($V$). The base units are given by length in micrometre ($\mu m$), time in microseconds ($\mu s$), mass in ($u$), and charge in elementary charge ($e$).}
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\label{tab:constants}
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\end{table}
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\end{document}
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