\section{Evaluation}

\subsection{Time Calibration}
All our measurements have been done through a time to amplitude converter, which through a digital converter, produces a histogram of all measured times. This needs to be calibrated to be anyhow useful. To calibrate, two set of measurements have been done with a clock. 


\begin{figure}[H]
\begin{center}
\includegraphics[width=0.75\textwidth]{picture/TimeCal_Clock.png}
\caption{Calibration with counter}
\label{clock}
\end{center}
\end{figure}

The first measurement (figure \ref{clock}) was done with a counter and gave us $10ns$ intervals but the measured peaks are not very stable.


\begin{figure}[H]
\begin{center}
\includegraphics[width=0.75\textwidth]{picture/TimeCal_Cable.png}
\caption{Calibration with cable}
\label{cable-clock}
\end{center}
\end{figure}

The second measurement was done with cables as a delay (figure \ref{cable-clock}). For that we used $8ns$ cables. This time the peaks appear to be much more sharp and we decided to use this to calibrate the bins of the histogram.\\
One interval is bigger due to the use of a $10ns$ cable by accident.\\
Using a quick fit of 7 gaussians, we could determine precisely the position of the peak and we put it against the real time delay.


\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/Calibration_Cable.png}
\caption{Slope of calibration}
\label{fig.calib}
\end{center}
\end{figure}


A linear fit (figure \ref{fig.calib}) gave us the slope of the line and at the same moment this will be the factor to transform the bin number in relative time, since the $0s$ is not defined.

\begin{center}
$t_{rel} = n^{\circ}_{bin} \cdot a_{slope} + b$
\end{center}

We then get the following value:
\begin{center}

$a_{slope} = 0.02443 ns$
\end{center}

We applied this factor to all of our future calculations. At this point we assumed that we do not need to consider errors.
\newpage

\subsection{Positron lifetime in Aluminium}
\subsubsection{Simple fit}

A set of measurements have been done with the aluminium frame in order to find the starting point of the process. For that we used a gaussian to fit the measured lifetime distribution with the least square method. 

\begin{equation}
f_{gauss}(t) = A \cdot \exp^{-\frac{(t-\mu)^2}{2\sigma^2}} + const
\end{equation}

This fit (figure \ref{fig.gau_alu}) gives us the following results:

\begin{center}
$A = 1990(10)$\\
$\mu = 17.122(1) ns$ \\
$\sigma = 0.301(1) ns$
\end{center}

We then quickly see that the function does not quite fit the data. We see a clear structure in the residual (figure \ref{fig.res_alu}) and a little tail on the right side of the gaussian. The last is an evidence for a small life time of the positron diffusing through aluminium frame.\\
We note that all the errors shown here have been given by the least square method we used on python.

\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/FittedAlu.png}
\caption{Fitted gaussian}
\label{fig.gau_alu}
\end{center}
\end{figure}

\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/ResidualsAlu.png}
\caption{Residual}
\label{fig.res_alu}
\end{center}
\end{figure}

\subsubsection{Advanced fit}
Since a simple gaussian was not enough to fit the the data correctly, we decided to use a combination of distributions (see formula \ref{eq.spec_distrib_A}). \\

\begin{equation}
F(t) = g_{\text{gauss}}(t) + a_{\text{convol}}(t) + \text{const.}
\label{eq.spec_distrib_A}
\end{equation}

where $g_{\text{gauss}}(t)$ is a simple gaussian distribution (formula \ref{eq.gauss})

\begin{equation}
g_{\text{gauss}}(t) = G e^{-\frac{(t-\mu_{g})^2}{2\sigma_{g}^2}}
\label{eq.gauss}
\end{equation}

and the distribution $a_{\text{convul}}(t)$ is a convulsion between a gaussian and exponential distribution (formula \ref{eq.convul}). We used this to equate for the fact that the positron diffuses through the metal before it annihilates.

\begin{equation}
a_{convol}(t) = A e^{-\frac{1}{\tau}\left((t-\mu) - \frac{\sigma^2}{2\tau}\right)} \cdot \left(1 + erf\left(\frac{(t-\mu)-{\sigma^2}{\frac{1}{\tau}}}{\sqrt{2}\sigma}\right)\right)
\label{eq.convul}
\end{equation}

This distribution gave us a much cleaner fit on the data (figure \ref{fig.gau_alu_s}).

\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/FittedAlu_s.png}
\caption{Fitted distribution}
\label{fig.gau_alu_s}
\end{center}
\end{figure}

\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/ResidualsAlu_s.png}
\caption{Residual}
\label{fig.res_alu_s}
\end{center}
\end{figure}
\newpage
The residuals looks almost perfect. We get diverse values from this fit.
\begin{center}
\begin{tabular}{r l}
&Convolution 1 \\
$A = $&$4501(283)$\\
$\tau = $&$0.134(10) ns $\\
$\mu = $&$16.994(7) ns $\\
$\sigma = $&$0.228(3) ns$\\
\\
&Gaussian \\
$A = $&$267(29)$\\
$\mu = $&$17.098(81) ns$\\
$\sigma = $&$0.692(62) ns$
\end{tabular}
\end{center}

Also here, the errors have been calculated by the least square method on python. A good indication if our fit now works, is to take a look at the $\chi^2$-value. In our case its $\chi^2 = 1.081$, which is very good. This confirmed our fit.\\
Even though no positronium forms in aluminium, we see a small lifetime, due to the positron diffusing through the aluminium before it annihilates with an electron of the electron gas.

\subsection{Lifetime in POM}
For the lifetime distribution in a polymer basically reused the same function as developed above (formula \ref{eq.spec_distrib_POM}) and just added another convoluted gaussian (2 particles decaying: para- and ortho-positronium) and fit the data.

\begin{equation}
F(t) = g_{\text{gauss}}(t) + a_{\text{convol}}(t) + b_{\text{convol}}(t) + \text{const.}
\label{eq.spec_distrib_POM}
\end{equation}

\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/FittedPOM.png}
\caption{Fitted distribution}
\label{fig.gau_pom_s}
\end{center}
\end{figure}

Also here we can describe precisely our data with this distribution. From the fit we then get the following result:

\begin{center}
\begin{tabular}{r l r l}
&Convolution 1 & &Convolution 2\\
$A = $ & $12141(201)$ & $B = $ & $973(35)$ \\
$\tau = $&$0.246(5) ns $&$\tau = $&$1.800(34) ns $\\
$\mu = $&$16.998(2) ns $&$\mu =  $&$16.597(9) ns$\\
$\sigma = $&$0.208(2) ns$&$\sigma = $&$0.180(7) ns$\\
\\
&Gaussian & & \\
$G =$ & $608(32)$\\
$\mu = $&$17.279(23) ns$ & &\\
$\sigma = $&$0.809(19) ns$ & &
\end{tabular}
\end{center}

\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\textwidth]{picture/ResidualsPOM.png}
\caption{Residual}
\label{fig.res_pom_s}
\end{center}
\end{figure}

This is again a very clean result. The value $\chi^2 = 1.081$ also confirmed a good fit.\\
This time we see clearly a difference in lifetime. So we can recognize the para-positronium with a lifetime $\tau=0.246(5)ns$ and the ortho-positronium with $\tau = 1.800(34)ns$. This values are coherent with what we expected.\\
\\
At this point we want to mention that the goal of the measurement with the aluminium frame was to define the starting point of the process. The position should then be used as a constant parameter in the fit. In the end we decided that this wasn't necessary since the position of the gaussians in all three fits (figures \ref{fig.gau_alu}, \ref{fig.gau_alu_s} and \ref{fig.res_pom_s}) agrees with very little variation. Nevertheless we tried some fits while holding the parameters $\mu$ and $\sigma$ fixed. The result of these fits agreed on previous results, so we do not show the result here.\\
Again, we want to remember that all the errors on the values are given by the least square method.
