Natural units

In the realm of high-energy particle physics, standard SI units are cumbersome. Instead, physicists utilize Natural Units, a system based on setting fundamental constants of nature to unity. This simplifies complex equations (like the Dirac equation) and allows all physical quantities to be expressed in powers of energy (GeV).

[Image of natural units conversion table particle physics]

Derivation of Physical Quantities in Natural Units

The foundation of this system rests on two definitions:

1. Speed of light: \(c = 1\)
2. Reduced Planck's constant: \(\hbar = 1\)

Consequently, dimensions of Length \([L]\) and Time \([T]\) become identical, and both are inversely proportional to Energy \([E]\).

1. Mass (\(M\))

We start with Einstein's mass-energy equivalence:

\[ E = mc^2 \]

Setting \(c = 1\):

\[ E = m \]

This implies mass has the same dimension as energy.

\[ [M] = \text{GeV} \]

2. Time (\(T\)) and Length (\(L\))

From the quantum mechanical relation between energy and frequency (inverse time):

\[ E = \hbar \omega \approx \frac{\hbar}{t} \]

Setting \(\hbar = 1\):

\[ E \sim \frac{1}{t} \implies [t] = [E]^{-1} \]

For length, we use \(x = ct\). Since \(c=1\), length and time share dimensions:

\[ [T] = \text{GeV}^{-1} \] \[ [L] = \text{GeV}^{-1} \]

Conversion Note: \(1 \text{ GeV}^{-1} \approx 0.197 \text{ fm}\).

3. Area (\(A\))

Area is physically defined as Length squared (\(L^2\)). This unit is crucial for scattering cross-sections (\(\sigma\)).

\[ [A] = [L]^2 = (\text{GeV}^{-1})^2 \]

\[ [A] = \text{GeV}^{-2} \]

4. Force (\(F\))

Force can be defined as the change in energy over distance (Work relation \(W = F \cdot d\)):

\[ F = \frac{dE}{dx} \]

Dimensionally:

\[ [F] = \frac{[E]}{[L]} = \frac{\text{GeV}}{\text{GeV}^{-1}} = \text{GeV}^{1 - (-1)} \]

\[ [F] = \text{GeV}^{2} \]

5. Electric (\(\mathcal{E}\)) and Magnetic (\(\mathcal{B}\)) Fields

In particle physics (Heaviside-Lorentz units), the energy density \(u\) of the electromagnetic field is:

\[ u = \frac{1}{2}(\mathcal{E}^2 + \mathcal{B}^2) \]

First, we find the dimension of energy density (Energy per unit Volume):

\[ [u] = \frac{[E]}{[L]^3} = \frac{\text{GeV}}{(\text{GeV}^{-1})^3} = \frac{\text{GeV}}{\text{GeV}^{-3}} = \text{GeV}^4 \]

Since the field squared must equal the energy density:

\[ [\mathcal{E}]^2 = [u] = \text{GeV}^4 \] \[ [\mathcal{E}] = \sqrt{\text{GeV}^4} \]

Since \(c=1\), Electric and Magnetic fields share the same dimension:

\[ [\mathcal{E}] = \text{GeV}^{2} \] \[ [\mathcal{B}] = \text{GeV}^{2} \]

Summary Table

Quantity	Natural Unit Dimension
Mass	\(\text{GeV}\)
Length / Time	\(\text{GeV}^{-1}\)
Area (Cross Section)	\(\text{GeV}^{-2}\)
Force / Fields	\(\text{GeV}^{2}\)