Accelerator Physics/Units

Units

Most of the time, accelerator physics uses SI units for its properties. The most well known exceptions are:

1. Particle energy, momentum and mass
2. Magnetic flux density

Particle energy

For convenience, accelerator physicists would like to know the particle's kinetic energy by measuring the potential difference between the distance that the particle traverses. An electron (or a particle with the same charge) gains 1 eV in kinetic energy after traversing a potential difference of +1 Volt. Therefore, $1\mathrm{e}\mathrm{V}\approx 1.6021\times 10^{-19}\mathrm{j}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}$

Modern acceleration devices can achieve very large potential difference between a gap. Therefore, the units are scaled accordingly so that

$1\mathrm{T}\mathrm{e}\mathrm{V}=10^3\mathrm{G}\mathrm{e}\mathrm{V}=10^6\mathrm{M}\mathrm{e}\mathrm{V}=10^9\mathrm{k}\mathrm{e}\mathrm{V}=10^{12}\mathrm{e}\mathrm{V}$

For instance, the Center of Mass energy of the ${\mu }^{+}$and ${\mu }^{-}$beams in a muon collider design can reach 6 TeV. The fixed target neutrino experiment LBNF uses 120 GeV Proton on Target (POT).

Particle momentum and mass

The total energy of a particle is the addition of its rest energy and kinetic energy: $E_{tot}=E_0+E_{kin}$. The rest energy $E_0$ is defined as $E_0=m{c}^2$ where $m$ is the mass of the particle and $c$ is the speed of light. Thereafter, particle mass, instead of using the standard unit kg, often uses the unit $\mathrm{e}\mathrm{V}/{\mathrm{c}}^{\mathrm{2}}$: $1\mathrm{e}\mathrm{V}/{\mathrm{c}}^{\mathrm{2}}\approx 1.783\times 10^{-36}kg$.

For instance, the mass of an electron is $9.11\times 10^{-31}$kg, or $0.511\mathrm{M}\mathrm{e}\mathrm{V}/{\mathrm{c}}^{\mathrm{2}}=5.11\times 10^5\mathrm{e}\mathrm{V}/{\mathrm{c}}^{\mathrm{2}}$. In order to acquire a kinetic energy of the rest energy of an electron, one has to accelerate it within a potential gap of 0.511 MV!

Regarding the particle momentum, physicists commonly use $\mathrm{e}\mathrm{V}/\mathrm{c}$ as the unit, because of Einstein's equation $E_{tot}^2=m^2{c}^4+c^2{p}^2$. Therefore, one can calculate the momentum of the above electron as follows:

$p=\sqrt{(2\times 0.511{)}^2-0.511^2\mathrm{M}\mathrm{e}{\mathrm{V}}^{\mathrm{2}}/{\mathrm{c}}^{\mathrm{2}}}=0.885\mathrm{M}\mathrm{e}\mathrm{V}/\mathrm{c}$

Magnetic flux density

The SI unit for magnetic flux density, or $B$, is $\mathrm{T}$, or Tesla. In many cases, scientists do also use Gauss, where $1\mathrm{k}\mathrm{G}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{s}=10^3\mathrm{G}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{s}=10^{-1}\mathrm{T}.$