diff --git a/latex/main.pdf b/latex/main.pdf
index c6f27db..1ce8a25 100644
Binary files a/latex/main.pdf and b/latex/main.pdf differ
diff --git a/latex/main.tex b/latex/main.tex
index 4bd63f2..f9c63f7 100644
--- a/latex/main.tex
+++ b/latex/main.tex
@@ -63,6 +63,9 @@
 % Introduction
 \subfile{sections/introduction}
 
+% Theory
+\subfile{sections/theory}
+
 % Methods
 \subfile{sections/methods}
 
diff --git a/latex/sections/methods.tex b/latex/sections/methods.tex
index 42348f6..125bd37 100644
--- a/latex/sections/methods.tex
+++ b/latex/sections/methods.tex
@@ -3,63 +3,6 @@
 
 \begin{document}
 \section{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. The sum of all forces acting on the particle, is given by the Lorentz force \eqref{eq:lorentz_force}.
-\begin{equation}\label{eq:lorentz_force}
-    \mathbf{F} = q \mathbf{E} + q \mathbf{v} \times \mathbf{B},
-\end{equation}
-We can use Newton's second law \eqref{eq:newton_second} to determine this sum by 
-\begin{align*}
-    \ddot{\mathbf{r}} &= \frac{1}{m} \sum_{i} \mathbf{F}_{i} \\
-    &= \frac{1}{m} (q \mathbf{E} + q \mathbf{v} \times \mathbf{B}) \\
-    &= \frac{q}{m} \big(\frac{V_{0}}{d^{2}} (x, y, -2z) + (B_{0}\dot{y}, -B_{0}\dot{x}, 0) \big),
-\end{align*}
-where the complete derivation can be found in \ref{sec:appendix_b}. We can now write the particle's position 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}
-% define w_0 og w_z i appendix?
-In addition, we can find the general solution for eq. \eqref{eq:motion_z}, when we consider the characteristic equation of a second order differential equation \cite{lindstrom:2016:ch10:5}. 
-\begin{align*}
-    r^{2} + \omega_{z}^{2} &= 0 \\
-    r &= \pm \sqrt{- \omega_{z}^{2}} = \mp i \omega_{z},
-\end{align*}
-two complex roots which gives us solutions in the form of
-\begin{equation}\label{eq:all_diff_sol}
-    z = c_{1} e^{i \omega_{z} t} + c_{2} e^{-i \omega_{z} t},
-\end{equation}
-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 \eqref{eq:all_diff_sol} as
-\begin{align*}
-    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}) + c_{2} (\cos{\omega_{z} t} - i \sin{\omega_{z} t} \\
-    &= E \cos{\omega_{z} t} + i F \sin{\omega_{z} t}
-\end{align*}
-%
-Since \eqref{eq:motion_x} and \eqref{eq:motion_y} are coupled, we want to rewrite it as a single differential equation. We can obtain this by introducing $f(t) = x(t) + iy(t)$,
-\begin{align*}
-    (\ddot{x} - \omega_{0} \dot{y} - \frac{1}{2} \omega_{z}^{2} x) + i (\ddot{y} + \omega_{0} \dot{x} - \frac{1}{2} \omega_{z}^{2} y) &= 0 \\
-    \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 &= 0 \\
-    \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 &= 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 &= 0 \text{where $i \omega_{0} \dot{x} + (-1) \omega_{0} \dot{y} = i \omega_{0} \dot{x} + i^{2} \omega_{0} \dot{y}$} \\
-    \ddot{f} + i \omega_{0} \dot{f} - \frac{1}{2} \omega_{z}^{2} f &= 0
-\end{align*}
-
- 
-Physical properties given by newtons second law \eqref{eq:newton_second}
-\begin{equation}\label{eq:general_solution}
-    f(t) = A_{+}e^{-i(\omega_{+} t + \phi_{+})} + A_{-}e^{-i(\omega_{-} t + \phi_{-})}
-\end{equation}
-The particle moves and its position can be determined using newton. where the electric field
-
-\subsection*{Algorithm}
-
-
-\subsection*{Tools}
-We used matplotlib
 
 \subsection{Units and constants}
 
@@ -286,5 +229,8 @@ and would reduce the required amount of loops down to 4.
 
 \subsection{Relative error and error convergance rate}
 
+\subsection{Tools}
+We used matplotlib
+
 %\biblio
 \end{document}
diff --git a/latex/sections/theory.tex b/latex/sections/theory.tex
new file mode 100644
index 0000000..e8d43c8
--- /dev/null
+++ b/latex/sections/theory.tex
@@ -0,0 +1,61 @@
+\documentclass[../main.tex]{subfiles}
+\graphicspath{{\subfix{../images/}}}
+
+\begin{document}
+\section{Theory}
+% 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. The sum of all forces acting on the particle, is given by the Lorentz force \eqref{eq:lorentz_force}.
+\begin{equation}\label{eq:lorentz_force}
+    \mathbf{F} = q \mathbf{E} + q \mathbf{v} \times \mathbf{B},
+\end{equation}
+We can use Newton's second law \eqref{eq:newton_second} to determine this sum by 
+\begin{align*}
+    \ddot{\mathbf{r}} &= \frac{1}{m} \sum_{i} \mathbf{F}_{i} \\
+    &= \frac{1}{m} (q \mathbf{E} + q \mathbf{v} \times \mathbf{B}) \\
+    &= \frac{q}{m} \big(\frac{V_{0}}{d^{2}} (x, y, -2z) + (B_{0}\dot{y}, -B_{0}\dot{x}, 0) \big),
+\end{align*}
+where the complete derivation can be found in \ref{sec:appendix_b}. We can now write the particle's position 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}
+% define w_0 og w_z i appendix?
+In addition, we can find the general solution for eq. \eqref{eq:motion_z}, when we consider the characteristic equation of a second order differential equation \cite{lindstrom:2016:ch10:5}. 
+\begin{align*}
+    r^{2} + \omega_{z}^{2} &= 0 \\
+    r &= \pm \sqrt{- \omega_{z}^{2}} = \mp i \omega_{z},
+\end{align*}
+two complex roots which gives us solutions in the form of
+\begin{equation}\label{eq:all_diff_sol}
+    z = c_{1} e^{i \omega_{z} t} + c_{2} e^{-i \omega_{z} t},
+\end{equation}
+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 \eqref{eq:all_diff_sol} as
+\begin{align*}
+    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}) + c_{2} (\cos{\omega_{z} t} - i \sin{\omega_{z} t} \\
+    &= E \cos{\omega_{z} t} + i F \sin{\omega_{z} t}
+\end{align*}
+%
+Since \eqref{eq:motion_x} and \eqref{eq:motion_y} are coupled, we want to rewrite it as a single differential equation. We can obtain this by introducing $f(t) = x(t) + iy(t)$,
+\begin{align*}
+    (\ddot{x} - \omega_{0} \dot{y} - \frac{1}{2} \omega_{z}^{2} x) + i (\ddot{y} + \omega_{0} \dot{x} - \frac{1}{2} \omega_{z}^{2} y) &= 0 \\
+    \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 &= 0 \\
+    \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 &= 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 &= 0 \text{where $i \omega_{0} \dot{x} + (-1) \omega_{0} \dot{y} = i \omega_{0} \dot{x} + i^{2} \omega_{0} \dot{y}$} \\
+    \ddot{f} + i \omega_{0} \dot{f} - \frac{1}{2} \omega_{z}^{2} f &= 0
+\end{align*}
+
+ 
+Physical properties given by newtons second law \eqref{eq:newton_second}
+\begin{equation}\label{eq:general_solution}
+    f(t) = A_{+}e^{-i(\omega_{+} t + \phi_{+})} + A_{-}e^{-i(\omega_{-} t + \phi_{-})}
+\end{equation}
+The particle moves and its position can be determined using newton. where the electric field
+
+
+
+%\biblio
+\end{document}