In high energy physics and astro-particle physics it is very common to use the natural units where \(\hbar = c = 1\). The correspondence between natural units and physical units can be established by use of
In these units there is then only one fundamental dimension:
Energy and momentum, usually expressed in GeV
Time and space are GeV\(^{-1}\)
Tutorial I: Working with Units
Along these notes you are going to find different tutorials in python dealing with will help to develop your programing skills in and solve some numerical problems. In this case we are going to see an easy way to work with different units which is uing the module units that exist for example in astropy:
import astropy.units as ufrom astropy import constants as constfrom IPython.display import display, MarkdownM_Earth =5.97E24* u.kgM_Sun =1.99E30* u.kgM_MW =1E12* M_Sun#By adding the quantity u.kg you can print directly the mass in Kgdisplay(Markdown (f"Mass Earth is: {M_Earth}"%M_Earth))
Mass Earth is: 5.97e+24 kg
Note that the variables defined above already have their units attached to them, so when you make a print or similar statement it will provide as well the units as a string.
We the radius we can calculate the mean density of Earth and Sun. We will show that units are preserved along calculations:
import numpy as npvol_sphere =lambda r: 4*np.pi/3*r**3rho_Sun = M_Sun / vol_sphere(R_Sun)rho_Earth = M_Earth / vol_sphere(R_Earth)#A unit can be changed calling the .to(u.unit) methoddisplay(Markdown (rf"$\rho_{{\oplus}}$ = {rho_Earth.to(u.g/u.cm**3):.2f}"))display(Markdown (rf"$\rho_{{\odot}}$ = {rho_Sun:.2f}"))
\(\rho_{\oplus}\) = 5.51 g / cm3
\(\rho_{\odot}\) = 1412.12 kg / m3
We can use this module to make different transformations of uings, for example from light years to meters:
ly =1* u.lyrdisplay(Markdown (f"Number of seconds for light to travel from Sun to Earth: {1./const.c.to(u.AU/u.s):.2e}"))display(Markdown (f"Meters in a light year: {ly.to(u.m):.2e}"))
Number of seconds for light to travel from Sun to Earth: 4.99e+02 s / AU
Meters in a light year: 9.46e+15 m
Assuming that the Galaxy is roughly a disk 50 kpc in diameter and 500 pc thick we can now calculate its density:
V_Gal = np.pi * (25000*u.pc)**2*500*u.pcdisplay(Markdown (f"Volume of the Milky Way is approximately: {V_Gal.to(u.m**3):.2e}"))M_Gal =1E12* M_Sunrho_Gal = M_Gal / V_Galdisplay(Markdown (f"Average density of Milky Way is {rho_Gal.to(u.g/u.cm**3):.2e}"))
Volume of the Milky Way is approximately: 2.88e+61 m3
Average density of Milky Way is 6.90e-23 g / cm3
Larmor Radius and Rigidity
Larmor radius, or gyroradius, \(r_L\), is the radius of the orbit of a charged particle moving in a uniform, perpendicular magnetic field, 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
In cosmic ray physics, one often sees references in the literature to the rigidity of a particle, defined as:
\[
R \equiv r_L B c = \frac{pc}{Ze}.
\]
:rightfinger: Note that the rigidity, \(R\) has units of Volts.
Superposition Model
Another concept that we will use frecuently in cosmic-ray physics is the superpositoin model. In the superposition model, a nucleus with mass \(A\) and energy \(E(A)\) is considered as \(A\) independent nucleons with energy \(E_0\). In a spallation process the energy per nucleon is approximately cosnserved therefore:
Cross sections, number density, lifetime, and interaction lengths
The cross-section of a reaction is a very important parameter. It can be considered as the effective area for a collision between a target and a projectile. The cross-section of an interaction depends on interaction force, the energy of the particle, etc…
Cross-section is typically measured in surface, \(\mathrm{cm}^2\) or “barns”:
\[ 1 \mathrm{ barn} = 10^{-24} \mathrm{cm}^2\]
The unit barn comes from the expression “big as a barn” as in the past physisits saw with surprise that interactions were more frequent than expected, and they thought the nucleus was in fact bigger than they thought… big as a barn.
If a flux of projectile particles are crossing a volume of target particles with cross section \(\sigma_N\) then the disapperance of flux will be proportional to the initial number, the length travelled and number of target particles:
\[ \mathrm{d} I = -I n \sigma_N \mathrm{d} x ,\]
where \(n\) is the number density, ie, the number of particles per volume unit:
\[ n = \frac{N}{V} \]
note that the number density is related with the mass density as:
\[n = \frac{N_A}{M} \rho_m = \frac{\rho_m}{m_N}\]
where \(N_A\) is the avogadro number, \(M\) is the total mass of a mol and \(m_N\) is the mass of is the mass of a single particles N making up the volume. Solving the equation above we have:
\[I = I_0 e^{-\frac{x}{n\sigma_N}}\]
where we can define:
\[\lambda = \frac{1}{n\sigma_N},\]
as the interaction length. Likewise if projectile particles are travelling at speed \(v\), the length travelled can be expressed as \(\mathrm{d}x = v \mathrm{d} t\) giving a similar expression with a time constant:
\[\tau = \frac{1}{n v\sigma_N}\]
Known as the lifetime. If several processes are taking place, we need to replace \(n\sigma_N\) as \(\sum n_i\sigma_i\), which gives: