Relativistic Units

Relativistic Units

From the definition of 4-vectors shown earlier, we know that

[math]\mathbf{R} \equiv \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}= \begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix} \qquad \qquad \mathbf{P} \equiv \begin{bmatrix} p^0 \\ p^1 \\ p^2 \\ p^3 \end{bmatrix}= \begin{bmatrix} \frac{E}{c} \\ p_x \\ p_y \\ p_z \end{bmatrix}[/math]

The 4-vectors and 4-momenta are defined to be in units of distance and momentum and as such must be multiplied or divided respectively by the speed of light to meet this requirement. For simplicity, the units of c can be chosen to be 1. This implies:

[math]c=1=\frac{length}{time}[/math]

[math]\therefore\ length\ units=time\ units[/math]

The relativistic equation for energy

[math]E^2 \equiv m^2c^4+p^2c^2[/math]

[math]\rightarrow E^2 = m^2+p^2[/math]

[math]\therefore\ energy\ units=mass\ units=momentum\ units[/math]

The Planck-Einstein relation and the de Broglie relation can be used to substitute into the relativistic energy equation

[math]E=\hbar \omega \qquad \qquad p=k \hbar[/math]

[math] \hbar^2 \omega^2 = m^2+k^2 \hbar^2[/math]

Since the units of [math]\omega =\frac{1}{time}\ [/math] and the units of [math]k=\frac{1}{distance}[/math]

[math] \mathbf{P} \equiv \begin{bmatrix} p^0 \\ p^1 \\ p^2 \\ p^3 \end{bmatrix}= \begin{bmatrix} \frac{E}{c} \\ p_x \\ p_y \\ p_z \end{bmatrix}= \begin{bmatrix} \frac{\hbar \omega}{c} \\ \hbar k_x \\ \hbar k_y \\ \hbar k_z \end{bmatrix}\rightarrow \mathbf{\frac{P}{\hbar}}\equiv \mathbf{K} \equiv \begin{bmatrix} k^0 \\ k^1 \\ k^2 \\ k^3 \end{bmatrix}= \begin{bmatrix} \frac{\omega}{c} \\ k_x \\ k_y \\ k_z \end{bmatrix}[/math]

[math]\hbar[/math] in SI units is defined as:

[math]\hbar \approx 1.0545718 \times 10^{-34} m \cdot kg \cdot \frac{m}{s}[/math]

Since c is already to be defined as equal to zero, this implies unit of mass must also be equal to one. By convention, the mass of the proton is used

[math]M_{p} \equiv 1.6726219 \times 10^{-27} kg =1[/math]