# Basic Concepts and Notations

## Natural Units

In particle and astroparticle physics it is very common to use the **natural or Planck units,** where $$\hbar = c = k\_B =  1$$. The idea of the natural units system is to reduce the number of fundamental constants to the necessary minimum and simplify calculations. This implies getting rid of the constants serving for unit conversions and the capability to reduce the dimension of differing quantities with to one dimensional quantity. In the natural system we have that:

$$
\[length] = \[time] = \[energy]^{-1} = \[mass]^{-1}.
$$

The Planck constant serves as conversion between time and energy:

$$
\hbar = \frac{ 6.63 \times 10^{-34}}{2\pi} \sim 1.055 \cdot 10^{-34} {\rm J;s} = 1,
$$

which means that $$1 \rm ;s^{-1} = 1 \times 10^{-34} \rm J = 6.6 \times 10^{-16} \rm eV$$ and therefore  $$1;\rm{GeV} = 1.52 \times 10^{15};{\rm{ns}^{-1}}$$

The speed of light serves as conversion between time and length:

$$
c = 3 \times 10^{10} \rm cm/s = 1,
$$

which means that $$1 \rm ;s = 3 \times 10^{10} \rm cm$$ and combining with the expression above we can relate length with energy as:

$$
1 \rm cm^{-1} = 3 \times 10^{10} \times 6.6 \times 10^{-16} \rm eV \sim 2 \times 10^{-5} \rm eV
$$

 Finally the Boltzmann constant serves for conversion between energy and temperature:

$$
k\_b =8 \times 10^{-5} {\rm eV; ^\circ K^{-1}} = 1,
$$

 and therefore  $$1 ^\circ {\rm K} = 8.6 \times 10^{-5} {\rm eV}$$.

The permeability of vacuum satisfies the relation $$4\pi \epsilon\_0 = 1$$ so that the Coulomb force is simplified to $$F = \frac{q\_1 q\_2}{r^2}$$. For the fine structure constant, given by $$\alpha = \frac{e^2}{4\pi \epsilon\_0 \hbar c} = \frac{1}{137} \Rightarrow e = \sqrt{\frac{1}{137}} = 0.085$$. 

Another example concerns the magnetic field. Consider circular motion of a particle of charge $$q = e$$ in a plane perpendicular to the direction of the B-field: $$q vB = m \frac{v^2}{R} \Rightarrow B = \frac{mv}{eR}= \frac{Energy}{eRv}.$$  Hence, $$1 , \rm Tesla = 10^4 , \rm Gauss =  \frac{ 1 eV}{e}\frac{s}{m^2} = \frac{1 eV \times 3 \times 10^{10} \rm cm \times 2\times 10^{-5 } eV}{0.0085 \times 10^4\rm} = 7.05 \times 10^2 \rm eV^2$$, where we used what awritten above $$1 \rm ;cm^{-1} = 2 \times 10^{-5} \rm eV$$ and  $$1 \rm ;s = 3 \times 10^{10} \rm cm$$.

So, as you can see, in these units there is only one fundamental dimension, energy, and energy and momentum are expressed in $$\rm{GeV}$$ while time and space are expressed in $$\rm GeV^{-1}$$. More generally, we can express :

* mass as $$\[M] = \frac{\[E]}{\[c^2]}$$
* length as $$\[L] = \frac{\[\hbar]}{\[E]}$$ (e.g. from E = hc/$$\lambda$$)
* time as $$\[T] = \frac{\[\hbar]}{\[E]}$$ (eg. from the Heisenberg indetermination principle $$\Delta E \Delta t  \ge \hbar$$)

So that in general the dimension of a quantity $$\alpha$$ are:

$$
\[\alpha]= \[M]^p \[L]^q \[T]^r = \[E]^{p-q-r},
$$

  and

| Symbol    | Value in NU | Value in MKS                            | V\[LT$$^{-1}$$]       |
| --------- | ----------- | --------------------------------------- | --------------------- |
| $$\hbar$$ | 1           | $$1.054 \times 10^{-34} {\rm ; Js}$$    | \[ML$$^2$$T$$^{-1}$$] |
| E         | 1           | $$1 {\rm ;eV}$$                         | \[ML$$^2$$T$$^{-1}$$] |
| c         | 1           | $$2.998 \times 10^8 {\rm ; m; s^{-1}}$$ | \[LT$$^{-1}$$]        |

{% hint style="info" %}
**Example 1**: Mass $$\[M]  =  \frac{\[MLT]}{\[LT]^2}$$=$$\frac{\[E]}{\[c^2]} = \frac{10^9 \times 1.602 \times 10^{-19}}{(2.998 \times 10^8)^2}$$$$\Rightarrow 1.782 \times 10^{-27}$$ kg = 1 GeV

**Example 2**: Length  $$\[L]  =  \frac{\[ML^2T^{-1}]\[LT^{-1}]}{\[ML^2T^{-1}]}$$=$$\frac{\hbar c}{E} =\frac{\hbar c}{Mc^2}  =\frac{1.055 \times 10^{-34} \times 2.998 \times 10^{8}}{1.602 \times 10^{-10}}= 0.197 \rm; fm$$$$\Rightarrow 0.197 \rm ; fm$$ = 1 GeV$$^{-1}$$

**Example 3:** Time  $$\[T]  =  \frac{\[ML^2T^{-1}]}{\[ML^2T^{-2}]} = \frac{\hbar}{E} = \frac{\hbar}{Mc^2} =$$$$\frac{1.055 \times 10^{-34} }{1.602 \times 10^{-10}}= 6.586\times 10^{-25} s =$$ 1 GeV$$^{-1}$$
{% endhint %}

## Vector Notation

Along this book we will use the traditional notation of using Greek indices running over 0, 1, 2, 3 and roman indices ($$i, j, k..$$) to indicate only the spatial three-dimensional components. Three-vectors will be indicated in **bold**, while four-vectors will be represented in capital letters as:

$$
X^{\mu} = (x^0, \bold{x})
$$

 

## Larmor Radius and Rigidity

One important definition specially when discussing about cosmic ray propagation is the **Larmor radius**, or **gyroradius.** The gyroradius, usually denoted as $$r\_L$$, is the radius of the orbit of a charged particle moving in a uniform, perpendicular magnetic field. This radius is obtained by simply equating the Lorentz force with the centripetal force. It is defined as:

$$
q v B = \frac{mv^2}{r\_L} \rightarrow r\_L = \frac{p}{ZeB},
$$

where $$p$$ has replaced $$mv$$ in the classical limit. However, this also holds for the relativistic generalization by considering $$p$$ to be the relativistic 3-momentum. There are several adaptations of this formula, tuned to units natural to various scenarios. One such is:

$$
r\_L \simeq 1 ;{\rm kpc} \left(\frac{p}{10^{18};\rm{eV}\cdot{c}}\right)\left(\frac{1}{Z}\right)\left(\frac{\mu\rm{ G}}{B}\right).
$$

We will see later that in cosmic-ray physics one often sees references in the literature to the **rigidity** of a particle, defined as:

$$
R \equiv r\_L B c = \frac{pc}{Ze}
$$

there is a good reason for this, as we will see in next chapters. 

{% hint style="warning" %}
:point\_right: Note that rigidity, $$R$$, has units of **Volts!**
{% endhint %}

## Superposition model

Another concept that we will use frequently in cosmic-ray physics is the *superposition model*. The superposition model tells us that a nucleus with mass $$A$$ and energy$$E(A) = A E\_0$$ is considered as $$A$$ independent nucleons with energies $$E\_0$$. Why is this useful? Well, as we will see, in a spallation process, i.e. when a nucleus gets smashed and fragmented in two nucleus like this:

$$
A + p \rightarrow A\_1 + A\_2,
$$

 since the energy per nucleon is more or less conserved we can write things like: 

$$
\begin{alignedat}{2} \&E(A) &= \&A E\_0,\ \&E(A\_1) &= \&A\_1 E\_0,\ \&E(A\_2) &= \&A\_2 E\_0 \end{alignedat}
$$

This will help us to simplify the models of cosmic rays air-showers. 

##