**Índice**

Introduction <br
Muon telescope
Rate of cosmic muons
Measuring the cosmic muon flux

**Abstract:**

Our cosmos is full of radiation.
Its composition is made of massless particles (photons) and different mass particles such as protons, electrons, positrons
and heavier nuclei.
Their observation is a consequence of being accelerated with striking energies on the cosmos accelerator, at singular supernova regions.

Muons are smoking guns of these primary cosmic particles.
They are relatively short lived particles that are generated on interactions of primaries on the top of the atmosphere, ten kilometers above earth surface.
Its detection upon earth is due to its high energy that allows its survival for kilometers as was explained by the Einstein relativity.
On this hands-on project we are going to use a bi-scintilator telescope to detect muons and to measure its rate (number of muons per second) at different
angular inclinations.
From those measurements we will be able to answer the following qustions:

- is the flux isotropic? if not which angular law can be derived?
- can we estimate the vertical flux?
- what flux is expected at very large angles ($\theta \rightarrow 90^{\circ}$)?

Muons are unstable particles, with a lifetime of the order of $2 \, \mu$s, and a mass 200 times the mass of the electron. They are created in the interactions primary of cosmic rays, which the majority are protons, with the atmosphere atoms. Given their high energy (GeV) and the consequent time dilation in the Earth framework, muons have a large probability of reaching the ground before decaying.

At the ground, muons can be detected using solid scintillation detectors (plastics). Whenever a muon crosses the scintillator, it interacts with the medium, loosing a small amount of energy that is converted into light. This light can afterwards be detected by photomultipliers (PMT).

In our case, we have a system of two scintillator detectors coupled to photon detectors - PMTs. The PMT window is directly in contact with the scintillator to collect all the produced photons in scintillator due to the muon passage.

The dimensions of the scintillators are the following: $2.5 \times 2.5 \times 1 \, cm^3$.

To build a telescope, able to observe muons arriving from a given direction, we can align the axis of the two scintillators and separate them by some distance.

The detection of a muon occurs when there is a luminous signal in both PMTs within a time window (time coincidence). To measure the rate of cosmic muons that cross both detectors one has only to count the number of times that we get a binary signal in the detectors coincident in time.

\begin{equation} Rate \equiv \frac{dN}{dt} = \frac{\Delta N}{\Delta t} \end{equation}

The luminous signal emanating from the scintillation is collected by the PMTs where it is converted into an electric signal. These two electrical signals coming from both detectors are introduced in an electronics board for data acquisition and a time coincidence of the two signal is applied.

The electrical signals are in reality the voltage signals. The photomultiplier is a source of electric current (electric charge moving due to an electric field inside the photomultiplier), and the cable connecting the photomultiplier to the oscilloscope has an impedance of 50 Ohms. Therefore, in this system one obtains a voltage of the order of milliVolts, where the precise value depends on the light signal generated in the scintillator.

The muon flux measures the number of muons inciding at a given direction, per unit of solid angle, per unit of time (per second, for instance), per unit of surface (perpendicular surface) and per unit of energy. That's what we can call a differential flux.

\begin{equation} Flux = \frac{dN}{dA_{\perp} \, d\Omega \, dE \, dt} \propto f(E, \cos\theta) \end{equation}

We are detecting muons at earth without measuring their energy. Why? Because our telescope is very simple does not allow any energy measurement. Is there any idea how could we do it?

So, it means we are integrating our muon flux over all energies. This way, the energy dependency cand be dropped on the differential flux.

Therefore we need:

- a rate measurement, which means, counting a number of events in certain time window
- an estimation of the detector area and solid angle, $G$

this is called the*geometrical detector acceptance*

**geometrical detector acceptance**

The concept of geometrical acceptance, corresponds to the detector area times the solid angle available for muon detection.
\begin{equation}
G = \int d\Omega \, \vec{n} \, \cdot \, d\vec{A}
\end{equation}
where,

$\vec{n}$, observation direction

$d\vec{A}$, detector area element

But, geometrical acceptance intrinsically depends on angular flux. Why?

Look above from your detector and any direction you look has a differential solid angle associated ($d\Omega$).
A very inclined particle will see a very small area of detector...

**acceptance calculation**

Try to compute the geometrical acceptance of one of the scintillators.

Estimate the acceptance geometrical reduction due to the fact we require two signals.

**flux**

$$ \Phi = \frac{Rate}{Acceptance} $$

- Particle Data Group: cosmic rays review
- Muon Flux Measurements at the Davis Campus of the Sanford Underground Research Facility with the Majorana Demonstrator Veto System (2016)
- International Cosmic Day (DESY)

Fernando Barão, P. Assis, R. Conceição (Feb, 2019)