\section{$\mu \rightarrow eee$ decay}

\subsection{Kinematics}
\label{Kinematics}
The two most prominent charged LFV decays are $\mu \rightarrow e\gamma$ and $\mu \rightarrow eee$.\\
Here the latter is chosen as more diagrams beyond the SM contribute. Namely tree diagrams, Z penguin and box diagrams. This offers the possibility to test more models.\\

Possible ways for the decay $\mu \rightarrow eee$ to occur are shown in Figures \ref{LFV-neutrino_osc}, \ref{LFV-SUSY}, \ref{LFV-tree_lvl}.\\

Still some simplifications are made as it is assumed that only the tree and the photon diagram are relevant. \cite{blondel2013research}\\

This gives us a Lagrangian of:

\begin{equation}
L_{LFV} = \left[\frac{m_\mu}{(\kappa+1)\Lambda^2}\overline{\mu_R}\sigma^{\mu\nu}e_LF_{\mu\nu}\right]_{\gamma-\text{penguin}}+\left[\frac{\kappa}{(\kappa+1)\Lambda^2}(\overline{\mu_L}\gamma^{\mu}e_L)(\overline{e_L}\gamma_\mu e_L)\right]_{\text{tree}}
\end{equation}

If we neglect signal and background we can use momentum conservation as the decay happens rather quickly. As a result the total sum of all particle momenta should be equal to zero:

\begin{equation}
\left\vert \vec{p}_{tot} \right\vert = \left\vert\sum \vec{p}_i \right\vert = 0
\end{equation}

The particles resulting in the decay lie all in a plane. The resulting positrons and electrons are in the energy range of (0-53)MeV.

\subsection{Background events}

\subsubsection{Internal conversions}
The event $\mu \rightarrow eee\nu\nu$ results in the same particles seen by the detector as the event we are searching for\footnote{Neutrinos are invisible to our detector}. As a result it proves to be quite challenging to s$\mu \rightarrow eee\nu\nu$eparate the two.\\
By using momentum conservation, it becomes possible to differentiate the $\mu \rightarrow eee$ and the $\mu \rightarrow eee\nu\nu$ events. In the muon rest frame the total momentum is zero and the energy of the resulting particles is equal to muon rest energy.\\
By reconstructing the energy and momenta of the three $e$ we can check if their momenta add up to zero and their energies equal the muon rest energy. If not we can assume that there are additional neutrinos. This differentiation between the two events is crucial for the experiment as the $\mu \rightarrow eee\nu\nu$ events pose the most serious background for $\mu \rightarrow eee$ decay measurements.\\
As a result, our detector needs a very good energy resolution to consistently make it possible to differentiate between the two events as neutrino energies and momenta are very small.

\subsubsection{Michel decay}

The biggest contributing background however stems from another decay called Michel decay, that is also allowed in the classical SM. As we use a beam of positive muons the corresponding Michel decay looks as follows: $\mu^+ \rightarrow e^+ \nu\bar{\nu}$.\\
Contrary to the events before this one does not produce any em negatively charged particles. This makes these events easily distinguishable from our wanted events. As a result they only enter our data in form of a potential background through wrongly constructed tracks.

\subsubsection{Radiative muon decay}

This is the case where $\mu \rightarrow e^+\gamma\nu\nu$. If the photon produced in this event has high enough energies and creates a matter antimatter pair in the target region ($\gamma \rightarrow e^-e^+$), it can create a similar signature than the searched event. They contribute to the accidental background, as equal to the searched event no neutrinos are produced. To minimize these effects, the material in both the target and detector is minimized and a vertex constraint is applied.

\subsubsection{BhaBha scattering}

Another way how background can get produced is when positrons from muon decays or the beam itself scatter with electrons in the target material. Consequently they share a common vertex and together with an ordinary muon decay it can look similar as our searched $\mu \rightarrow eee$ event. This contributes to the accidental background.

\subsubsection{Pion decays}

Certain pion decays also lead to indistinguishable signature as our searched event, the most prominent being the $\pi \rightarrow eee\nu$ and $\pi \rightarrow \mu\gamma\nu$ decays. The later only produces a similar signature if produced photon converts through pair production to an electron and a positron.\\
However as only a negligible portion will actually contribute to the background, as there is only a small branching fraction and the momenta and energy of the produced particles have to match up with the criteria mentioned in section \ref{Kinematics}.

\subsubsection{Analysis of the background}

The results of simulations indicate that the effect of purely accidental background contributions are small for a high enough energy resolvent detector. \cite{blondel2013research}\\
The most relevant background stems from the $\mu \rightarrow eee\nu\nu$ events. This problem can only be tackled by using a very precise total energy resolution of $\sigma_E = 1MeV$ at the aimed sensitivities.