\documentstyle[12pt,epsfig]{article}
\title{\begin{flushright}
               {\small V Physics Summer Program, Mainz-Valencia\\
                               $\,$ \\
                               $\,$ \\}
        \end{flushright}
	\sc{In the Search of the Decay}
	$K^0_S\rightarrow\mu^+\mu^-\gamma$}
\author{P\'erez D\'iaz, Pablo Jos\'e}
\date{September, 2002}

\begin{document}
\maketitle

  \begin{abstract}
    Our object is to find the decay $K^0_S\rightarrow\mu^+\mu^-\gamma$, study the cuts we can make to remove background, and if possible
    calculate the branching fraction of this particular channel.
  \end{abstract}

%-------------------------------------------------------------------------------------------------------------------------------
\section{\sc{Neutral Kaons}}
Hadrons are built by the associaton of quarks, and when two quarks associate we say that it is
a Meson. Kaons are one kind of mesons with some particular quarks, and there are several types of them. One of the first divisions we
can make is between the ones
that have charge ($K^\pm$) and the ones that are neutral ($K^0$), which we are interested in. $K^0$ is composed by a
$\bar{s}$ quark and $d$ quark, while its antiparticle $\bar{K}^0$ is made by an $s$ quark and $\bar{d}$ quark.

These two particles are only eigenstates of the strong interaction, but not of the combined operation CP\footnote{
$CP|K^0\rangle=|\bar{K}^0\rangle$}. However, a mixture of them can be a \emph{CP} eigenstate, thus we get the states $K_1$ and $K_2$:
$$K_1=\frac{1}{\sqrt{2}}\Big(|K^0\rangle+|\bar{K}^0\rangle\Big) (CP=+1)$$
$$K_2=\frac{1}{\sqrt{2}}\Big(|K^0\rangle-|\bar{K}^0\rangle\Big) (CP=-1)$$

But we cannot observe directly in nature $K_1$ and $K_2$, instead we observe two kinds of neutral kaons, ones that live very long,
and others that have much less mean lifetime, we
call these physical states $K^0_L$ and $K^0_S$.

To have $CP$ conserved, as the main decay channels of this particles are the pions,
we have got that $K_1\rightarrow\pi\pi$ and $K_2\rightarrow\pi\pi\pi$. This second decay is much less probable than the first
decay, because it leaves much less free energy, so it was thought that $K_1$ was $K^0_S$ and $K_2$ was $K^0_L$. However, it was found in
1964\cite{CP} that $K^0_L$ could also decay into two pions. This is mainly $CP$ indirect violation by mixing
procedures.

So we can try to build a combination of $K_1$ and $K_2$ to get the physical states of the kaons\footnote{That is, $K^0_L$ and $K^0_S$}
(\emph{mass eigenstates} are called) and it can be found that:
$$K^0_S=\frac{1}{\sqrt{1+|\epsilon|^2}}\Big(|K_1\rangle+\epsilon|K_2\rangle\Big)$$
$$K^0_L=\frac{1}{\sqrt{1+|\epsilon|^2}}\Big(\epsilon|K_1\rangle+|K_2\rangle\Big)$$

The difference between $K^0_L$ and $K^0_S$ are their lifetimes and the decay channels they have (see tables \ref{lifetimes}
and \ref{channels}).

\begin{table}[!h]
\caption{\sc{$K_{L,S}$ Lifetime Comparison}}
\label{lifetimes}
\begin{center}
\begin{tabular}{|c|c|c|} \hline
& $K_S$ & $K_L$ \\ \hline
$\tau$ & $0.8935\times10^{-10} s$ & $5.17\times10^{-8} s$\\ \hline
$c\tau$ & 2.6786 cm & 15.51 m\\ \hline
\end{tabular}
\end{center}
\end{table}

%-------------------------------------------------------------------------------------------------------------------------------
\section{\sc{$K^0_{L,S}$ Decays}}
$K_L$ and $K_S$\footnote{We will suppose that we talk about neutral kaons and sometimes omit the $^0$} mostly decay
into the channels shown in table \ref{channels}. My study in particular will be about $K^0_S\rightarrow\mu^+\mu^-\gamma$,
 which is one of the possibilities of the more general and
unlikely\footnote{That is why most of them do not appear in the table} decay $K_{S,L}\rightarrow\gamma\gamma$, where each one of these
gammas can be real or virtual. Processes leading to the production of real or virtual photons can be seen in the figures \ref{feynman1} and
\ref{feynman2}, and all the possible combinations of this decay are shown in table \ref{twophoton}.

\begin{table}[!h]
\caption{\sc{$K^0_{L,S}$ Main Decay Channels}}
\label{channels}
\begin{center}
\begin{tabular}{|c|c||c|c|}
\hline
$K^0_S\rightarrow$ & \% & $K^0_L\rightarrow$ & \%\\ \hline\hline
$\pi^+\pi^-$ & 68.61 & $3\pi^0$ & 21.13\\ \hline
$\pi^0\pi^0$ & 31.39 & $\pi^+\pi^-\pi^0$ & 12.55\\ \hline
$\pi^+\pi^-\gamma$ & $1.78\times10^{-3}$ & $\pi^\pm\mu^\mp\nu_\mu (K^0_{\mu3})$ & 27.18\\ \hline
&& $\pi^\pm e^\mp\nu_e (K^0_{e3})$ & 38.78\\ \hline
\end{tabular}
\end{center}
\end{table}

\begin{table}[!h]
\caption{\sc{Two Photon Decay}}
\label{twophoton}
\begin{center}
\begin{tabular}{|c|c|}
\hline
\multicolumn{2}{|c|}{$K_{S,L}\rightarrow$}\\ \hline
$\gamma\gamma$ & both real photons\\ \hline
$e^+e^-\gamma$ & $1\times\gamma^{(*)}\rightarrow e^+e^-$\\ \hline
$\mu^+\mu^-\gamma$ & $1\times\gamma^{(*)}\rightarrow\mu^+\mu^-$\\ \hline
$e^+e^-e^+e^-$ & $2\times\gamma^{(*)}\rightarrow e^+e^-$\\ \hline
$\mu^+\mu^-\mu^+\mu^-$ & $2\times\gamma^{(*)}\rightarrow\mu^+\mu^-$\\ \hline
$\mu^+\mu^-e^+e^-$ & $1\times\gamma^{(*)}\rightarrow\mu^+\mu^-+1\times\gamma^{(*)}\rightarrow e^+e^-$\\ \hline
\end{tabular}
\end{center}
\end{table}

\begin{figure}[!h]
\caption{\sc{$K\rightarrow\gamma\gamma$ Feynman Diagram}}
\label{feynman1}
\begin{center}
\includegraphics[width=12cm]{p-p.eps}
\end{center}
\end{figure}

\begin{figure}[!h]
\caption{\sc{$K\rightarrow\gamma\gamma^*$ Feynman Diagram}}
\label{feynman2}
\begin{center}
\includegraphics[width=12cm]{v-v.eps}
\end{center}
\end{figure}

%-------------------------------------------------------------------------------------------------------------------------------
\section{\sc{The NA48 Experiment}}

The NA48 Experiment is located at the CERN facilities in Geneva. It uses the Super Proton Synchrotron (SPS) to accelerate protons up to
400 GeV and make the colide to a 2 mm diameter, 400 mm long rod of beryllium. The intensity we get now is $5\times10^{10}$ protons per
burst\footnote{Aprox. 5 seconds}.

Downstream the target, the beam enters the field of a strong,
vertical sweeping magnet, the gap of which is filled with tungsten-alloy inserts containing a passage for the neutral beam. This passage is
shaped to absorb the remaining primary protons at a point where they are separated form the neutral beam, and to intercept the curved
trajectories of all charged secondary particles from the target.

The magnet is followed by a steel collimator, which is fitted with further precisely-bored inserts, ranging from a beam defining aperture
of 3.6 mm at 4.8 m from the target to a final diameter of 6.0 mm a the exit, 6.0 m after the target, where the $K_S$ beam enters the decay
volume. The overall length of the $K_S$ collimator is close to 1 $c\tau_{K_S}$\cite{datos}.

The main components of the detector consist of a high resolution liquid krypton calorimeter (LKr) and a magnetic spectrometer. To pre-trigger
on charged decays, a plastic scintillator charged hodoscope (CHOD) is used. The hadron calorimeter is used to keep the pre-trigger rate low
enough to strobe the Level II charged trigger and to also reconstruct hadronic energy.

A set of seven anti-counter rings (AKL) surrounding the vacuum decay tank is used to veto charged and neutral particles escaping from
the fiducial region. A muon filter at the end of the hall provides a two-muon trigger.

Summarizing, with our experimental setup we can measure the momentum and energy of a track, the energy of neutral particles (that do not
leave track) and we can also say if a track is from a muon or not. A picture of the detector can be seen in figure \ref{detector}.

\begin{figure}[!h]
\caption{\sc{NA48 Experiment}}
\label{detector}
\begin{center}
\includegraphics[width=12cm]{detector.epsi}
\end{center}
\end{figure}

%-------------------------------------------------------------------------------------------------------------------------------
\section{\sc{Background Sources}}
We have to deal with the background in our analysis. When the beam impacts the target several particles appear, and they can decay into
many different channels. The most dangerous background is the most similar to our decay, that is a positive muon, a negative one and a
photon. All the decays that can fit into this profile have to be removed.

\begin{table}[!h]
\caption{\sc{$K^0_{L,S}$ Important Background Sources}}
\begin{center}
\begin{tabular}{|c|c|}
\hline
$K_S\rightarrow\pi^+\pi^-$ & $K_S\rightarrow\pi^+\pi^-\gamma$\\ \hline
$K_L\rightarrow\pi^+\pi^-\pi^0$ & $K_S\rightarrow\pi^\pm e^\mp\nu_e$\\ \hline
$K^0_{\mu3}$ & $K^0_{e3}$\\ \hline
$\Lambda\rightarrow p\pi^-$ & $\Lambda\rightarrow p\pi^-\gamma$\\ \hline
\end{tabular}
\end{center}
\end{table}

\begin{itemize}
\item $K_S\rightarrow\pi^+\pi^-$ and $K_S\rightarrow\pi^+\pi^-\gamma$ decays. This two decays are really one\footnote{It is usual to
consider there is a photon only if it has a certain quantity of energy, because $K_S\rightarrow\pi^+\pi^-$ also emits low energy
bremsstralung photons}. The problem is that pions can decay to muon plus neutrino
($\pi\rightarrow\mu\nu$) and then this channel becomes very similar to a $\mu\mu\gamma$ decay. Also, the muon detector can misidentify one
or two of the pions, and this effect added to the pion decay makes this background dangerous. We also have the photons, that in this case
will come from bremsstrahlung.
\item $K_L\rightarrow\pi^+\pi^-\pi^0$ The $\pi^0$ of this channel almost always decay inmediatly to $\gamma\gamma$, so we have the photon and
the two tracks, but we lose the other photon, so when we build the $\mu\mu\gamma$ invariant mass it gives us a peak at 0.35 GeV. The pions
have also to decay to muons or be misidentified.
\item $K_{\mu3}$ Here we already have a muon, and only one pion has to decay or be misidentified. However, there is no photon here, but
it can be an accidental one (from other decay, for example).
\item $\Lambda$ \emph{decays}. Due to the lifetime of $\Lambda$ particle, which is very similar\footnote{$2.632\times10^{-10}$ s} to
the $K_S$, we also have to deal with this background, but it is by far not the most important.
\end{itemize}

%-------------------------------------------------------------------------------------------------------------------------------
\section{\sc{Cuts Applied}}

\subsection{Muon detector}
In the final stage of our experimental setup we have a muon filter, which identifies whether if the track is a muon or not.
As the two tracks that we have are muons, we will require in our analysis that the detector found a muon in the both tracks
(of course, the detector is not perfect and we misidentify some).

\subsection{Energy over Momentum}
The spectrum of the ratio energy/momentum (see figure \ref{eop}) in a track depends of the particle that leaves the track. In principle
it should be around 1 for high energy particles\footnote{$E=\sqrt{p^2+m^2}\approx p$ if $p\gg m$}
, but of course we do not detect all the energy of the track in our calorimeter. For example, muons interact very little
with matter, and they go through the calorimeter without depositing any energy or very few, so we have very low E over P (Eop) ratio.

However, pions are so massive that they produce such a big shower of particles when they enter the calorimeter that part of the shower can
remain undetected because it is outside the calorimeter. Pions will give us a very wide peak ranging from low values to 1.

We have also electrons, which produce small showers in our calorimeter, so we can get all the energy that it had; it will give us a peak
around 1.

In our analysis we apply a cut that requires an $Eop<0.1$ for an event to be counted. This is a very efective cut that removes all the
electrons, and lots of pions also (but we have many, many pions).

\begin{figure}[!hp]
\caption{\sc{E over P Spectrum}}
\label{eop}
\begin{center}
\includegraphics[width=12cm]{eop.eps}
\end{center}
\end{figure}

\subsection{Momenta of the Tracks}
We require that the ratio of the momentum of the positive and the negative track must be less than 3. This cuts very little of our events,
and removes most of the $\Lambda\rightarrow p\pi^-$ and $\Lambda\rightarrow p\pi^-\gamma$, because in this kind of events the momentum is
taken mostly by the photon (see figure \ref{ratiop}).

We also cut on the minimum (for high efficiency of the detector reasons) and maximum (to get rid of $\Lambda$ decays) values of the momenta.

\begin{figure}[!hp]
\caption{\sc{Ratio of Momenta of Positive and Negative Tracks}}
\label{ratiop}
\begin{center}
\includegraphics[width=12cm]{ratiop.eps}

\small{The upper histogram is real data, the middle one is data from $K_S\rightarrow\mu^+\mu^-\gamma$ Montecarlo, and the last is
$\Lambda\rightarrow p\pi^-$ Montecarlo}
\end{center}
\end{figure}

\subsection{Invariant Masses Cuts}
We can select two tracks and a photon and build invariant masses for the biggest sources of background: $K\rightarrow\pi^+\pi^-$ and
$K\rightarrow\pi^+\pi^-\gamma$. Then we will obtain a peak plus a flat distribution, and we delete the peak, which are the events
corresponding to $K\rightarrow\pi^+\pi^-$ and $K\rightarrow\pi^+\pi^-\gamma$. This cuts lots of these events (see figure \ref{cortepi}).

\begin{figure}[!hp]
\caption{\sc{Effect of $K\rightarrow\pi^+\pi^-\gamma$ Invariant Mass Cut}}
\label{cortepi}
\begin{center}
\includegraphics[width=12cm]{cortepi.eps}

\small{The upper histogram is without cut, and the lower one making a cut in 0.01 GeV around 0.5 GeV $\pi^+\pi^-\gamma$ Invariant Mass}
\end{center}
\end{figure}

\subsection{Existence of a Photon}
In our decay we have a photon, and as it is a neutral particle, it leaves no track in the drift chambers: we only have the information
of the clusters seen in the LKr for that event. What we do is extrapolate the position of the track in the LKr, and then look for
clusters (showers) which are at more than 10 cm away from both tracks extrapolated positions.

Then we make a cut in the time of hit to eliminate the accidental photons from other decays, and we also cut on the energy of the photon
found, because we have bremsstrahlung photons and they do not belong to our analysis. In figure \ref{eg} can be seen how effective this
cut is. For example good cuts can be made at 10 GeV, but it is advisable to make them at 15 GeV, because our acceptance do not falls so
much (table \ref{acceptances}) and we delete a lot of background.

\begin{figure}[!hp]
\caption{\sc{Energy of the Photon Found}}
\label{eg}
\begin{center}
\includegraphics[width=12cm]{eg.eps}

\small{The upper histogram is real data, the middle one is data from $K_S\rightarrow\mu^+\mu^-\gamma$ Montecarlo, and the last is
$K_L\rightarrow\mu^+\mu^-\gamma$ Montecarlo}
\end{center}
\end{figure}

\subsection{Transversal Momentum}
This is a cut we use to identify missing products of decay. As we know the momenta of the two tracks and of the photon, we can build a total
momentum of the sum of the decay products, and compare it with the direction of the kaon if it did not decay. If we had identified all
the products of decay, in principle these two vectors should have the same direction, but if that decay had more products, the directions would
not be the same.

The cut just consists in calculating the total momentum vector of the decay products and projecting it into the kaon direction, if it is too
big we say that we missed some decay product, and that cannot be our reaction.

This cut is particularly good to surpress background coming from decaying pions, such as from
$K_S\rightarrow\pi^+\pi^-\gamma\rightarrow\mu^+\nu\mu^-\bar{\nu}\gamma$. That is because the neutrinos remain undetected, while they take
part of the momentum of the decaying pion, changing then the direction the muon detected.

\subsection{Ratio of $\mu^+\mu^-\gamma$ and $\mu^+\mu^-$ Transversal Momentum}
Our reaction is a kaon decaying into two photons, one of them virtual, that goes into a pair of muons. If we build the transversal
momentum of the three decay products the expected value should be small. However, if we build it only for the two muons pair, we are
doing it for the virtual photon, which of course has a lot of transversal momentum.

So, the ratio $\frac{P_T(\mu\mu\gamma)}{P_T(\mu\mu)}$ should be very small, as we see in the figure \ref{ratio} this cut is very
effective. It removes mainly decays which do not come from the type $K\rightarrow\gamma\gamma^*$, that is, decays whose photon come from
bremmstralung (and it is usually emited in the same direction af the decay product) such as $K\rightarrow\pi^+\pi^-\gamma$.

\begin{figure}[!hp]
\caption{\sc{Ratio} $\frac{P_T(\mu\mu\gamma)}{P_T(\mu\mu)}$}
\label{ratio}
\begin{center}
\includegraphics[width=12cm]{ratio.eps}

\small{Upper histogram: real data (after cut in 10 GeV gamma energy). Lower histogram: $K_S\rightarrow\mu^+\mu^-\gamma$ Montecarlo }
\end{center}
\end{figure}

\subsection{$\mu^+\mu^-\gamma$ Invariant Mass Cut}
As we see in $K_S\rightarrow\mu^+\mu^-\gamma$ Montecarlo (figure \ref{prueba}) after all the cuts we should be able to build an invariant
mass for our decay using the two tracks and the photon found. MC gives us a peak around 0.5 GeV, as we expected. That should be our final
cut: build invariant masses for all the events that passed the filters and cut around the 0.5 GeV peak.

\begin{figure}[!hp]
\caption{$K_S\rightarrow\mu^+\mu^-\gamma$ \sc{Invariant Mass}}
\label{prueba}
\begin{center}
\includegraphics[width=12cm]{prueba.eps}
\end{center}
\end{figure}

\subsection{Other Cuts}
We require that the Closest Distance of Aproach (CDA) of the reconstructed vertex must be less than 3 cm. We also require a number of
tracks equal to two.

%-------------------------------------------------------------------------------------------------------------------------------
\section{\sc{$K_S\rightarrow\mu^+\mu^-\gamma$ and $K_L\rightarrow\mu^+\mu^-\gamma$} Comparison}
The decay $K_L\rightarrow\mu^+\mu^-\gamma$ has already been studied by NA48\cite{jorg} and Fermilab\cite{fermilab} and satisfactory results
were obtained. But our interest is in $K_S\rightarrow\mu^+\mu^-\gamma$, which in principle should have only the difference of
the lifetime and the new decay channels.

However, during our analysis we found more differences derived from these two points. In previous studies of $K_L\rightarrow\mu^+\mu^-\gamma$
by NA48 another target placed much far away from the detector was used, thus eliminating all the $K_S$. In our analysis we use
another target, much more nearer, and we have both $K_S$ and $K_L$, with their particular background too.

One of the first effects we may note is that the decaying vertex position is strongly related with the energy of the decaying particle:

$$Z_{vertex}=\gamma\beta c\tau_S\approx E/m\times c\tau_S$$

Thus, the higher the energy of the particle, the greater the distance of decay. We already said that there was a 1 $\tau_S$ lenght collimator
after the target, and all the particles that decay inside it have very few probabilities of being detected, because they hit the collimator
walls from \emph{inside}. That is not such a big problem for $K_L$, which has such a long lifetime compared with the collimator lenght
, but it is important for $K_S$, because only the most energetic kaons will decay after the
collimator, and then be seen. In table \ref{zvertex} can be seen this effect. In principle we should have a nice exponential, but we see no
events before one lifetime, then it increases (these are the decays inside the collimator \footnote{Of which only some decay in the collimator
direction and then are seen}) and after the collimator we have the typical exponential of decay.

\begin{figure}[!ht]
\caption{\sc{Position of Decay Vertex in Units of $\tau_S$}}
\label{zvertex}
\begin{center}
\includegraphics[width=12cm]{zvertex.eps}
\end{center}
\end{figure}

This is the first difference in this analysis: \emph{the greater energy of the decaying particle}. Every energy related measure
will increase proporcionally with it: the transversal momentum, the momenta of the tracks, the energy of the photons (see figure
\ref{eg})\ldots

Also, to reconstruct the direction of the kaon we need the vertex of decay and the target position, but as they are very near
to each other, that direction cannot be well stablished. That \emph{resolution effect}, combined with the greater energy
will lead to less efficiency of transversal
momentum and ratio\footnote{That is $\frac{P_T(\mu\mu\gamma)}{P_T(\mu\mu)}$ Ratio if anything else said} cuts.

In figure \ref{trans} we can see the much greater dispersion of these two variables for the $K_S$ than for the $K_L$. That can also be
seen in figures \ref{comppt} and \ref{compratio}.

\begin{figure}[!ht]
\caption{\sc{Transversal Momentum Vs. Ratio Dispersion Diagram}}
\label{trans}
\begin{center}
\includegraphics[width=12cm]{trans.eps}

\small{Upper diagram: Montecarlo for $K_L$. Lower diagram: $K_S$ Montecarlo }
\end{center}
\end{figure}

\begin{figure}[!ht]
\caption{\sc{Comparison of Transversal Momentum}}
\label{comppt}
\begin{center}
\includegraphics[width=12cm]{comppt.eps}

\small{Upper histogram: Montecarlo for $K_L$. Lower histogram: $K_S$ Montecarlo }
\end{center}
\end{figure}

\begin{figure}[!ht]
\caption{\sc{Comparison of Ratio}}
\label{compratio}
\begin{center}
\includegraphics[width=12cm]{compratio.eps}

\small{Upper histogram: Montecarlo for $K_L$. Lower histogram: $K_S$ Montecarlo }
\end{center}
\end{figure}

In table \ref{acceptances} we can observe how our acceptance falls because of these effects. It can be seen to that $K_L$ is a little
more sensible to increase in gamma energy cuts that $K_S$ (as we expected).

\begin{table}[!ht]
\caption{\sc{Acceptances and Cuts Made}}
\label{acceptances}
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$P_T<$ & $(Ratio)^2<$ & \emph{Gamma Energy} $<$ & $K_S$ MC & $K_L$ MC\\ \hline
$\sqrt{0.002}$ GeV & 0.03 & 8 GeV \tiny{Cuts made in \cite{jorg}}& 5.52\% & 12.10\%\\ \hline
0.2 GeV & 0.1 & 10 GeV & 16.37\% & 20.99\%\\ \hline
0.3 GeV & 0.2 & 10 GeV & 23.11\% & 22.76\%\\ \hline
0.2 GeV & 0.1 & 15 GeV & 15.05\% & 17.58\%\\ \hline
0.3 GeV & 0.2 & 15 GeV & 21.54\% & 19.22\%\\ \hline
\end{tabular}
\end{center}
\end{table}

%-------------------------------------------------------------------------------------------------------------------------------
\section{Conclusion}
Although we have not been able to see a peak in the invariant mass diagram (see figure \ref{example1} as an example of histogram) several
conclusions can be made:

\begin{itemize}
\item There is still no implemented Muon-Vecto for Montecarlo analysis, so we cannot observe in a controllated way how our Muon-Veto
system works.
\item The analysis is much more difficult than for $K_L\rightarrow\mu^+\mu^-\gamma$ because of the combined background and the resolution
and energy effects explained before.
\item In principle a peak at the invariant mass should be seen, but it is not. Especulations about why can be made, but none of them has
yet proven to be satisfactory.
\end{itemize}

\begin{figure}[!h]
\caption{\sc{Example of Invariant Mass Diagram}}
\label{example1}
\begin{center}
\includegraphics[width=12cm]{example1.eps}

\small{The peak around 0.35 GeV belongs to $K_L\rightarrow\pi^+\pi^-\pi^0$ events. We would expect a peak at 0.5 GeV but nothing is seen}
\end{center}
\end{figure}

%-------------------------------------------------------------------------------------------------------------------------------
\begin{thebibliography}{99}
\bibitem{CP} Cristenson \emph{et al}, Phys. Rev. Lett. 13, 138, 1964
\bibitem{datos} R. Batley \emph{et al.}, \emph{A high sensitivity investigation of $K_S$ and neutral hyperon decays using a modified $K_S$
beam (Addendum 2 to P253)}, CERN/SPSC 2002-002, 1999
\bibitem{eegamma} V. Fanti \emph{et al.}, \emph{Measurement of the decay rate and form factor parameter alpha(K*) in the decay $K_L
\rightarrow e^+e^-\gamma$}, Phys. Lett. B, 458:553-563, 1999
\bibitem{jorg} J. Scheidt, \emph{Messung des Zerfalls $K_L\rightarrow\mu^+\mu^-\gamma$}, Johannes Gutenberg Universit\"at, Mainz, 1996
\bibitem{fermilab} G. Breese, \emph{A measurement of the branching ratio and form factor of $K_L\rightarrow\mu^+\mu^-\gamma$}, University
of Chicago, 2000
\end{thebibliography}

\end{document}




