Kinematic effects

We introduce proper time \(\tau\) as \[\begin{equation} \require{physics} \dd{\tau} = \frac{\dd{t}}{\gamma(\vb{v})} % \thinspace , \end{equation}\] in which \(\gamma(\vb{v})\) is called the Lorentz factor: \[\begin{equation} \gamma(\vb{v})^2 = \qty(1 - \frac{||\vb{v}||^2}{c^2} )^{-1/2} % \thinspace . \end{equation}\] We then find that the derivative of the space-time distance with respect to the proper time is equal to the light speed: \[\begin{equation} \dd{s} = c \dd{\tau} % \thinspace . \end{equation}\]

The \(4\)-velocity \(u^\mu\) is defined as: \[\begin{equation} u^\mu % = \dv{x^\mu}{\tau} % = \gamma \begin{pmatrix} c \\ \vb{v} \end{pmatrix} \end{equation}\] and the \(4\)-momentum as: \[\begin{equation} p^\mu % = m u^\mu % = \gamma \begin{pmatrix} mc \\ m \vb{v} \end{pmatrix} = \begin{pmatrix} E/c \\ \vb{p} \end{pmatrix} \thinspace , \end{equation}\] in which \(\vb{p}\) is the relativistic 3-momentum: \[\begin{equation} \vb{p} = \gamma m \vb{v} % \thinspace , \end{equation}\] and \(E\) is the relativistic energy: \[\begin{equation} E = \gamma m c^2 % \thinspace . \end{equation}\]

We can then immediately derive the relativistic energy-momentum relation: \[\begin{equation} \label{eq:squared_relativistic_energy} E^2 = c^2 \norm{\vb{p}}^2 + m^2 c^4 % \thinspace , \end{equation}\] which predicts massive (i.e. \(m>0\)) particles with positive and negative energies.

We should furthermore note that for the \(4\)-position, we have: \[\begin{equation} u_{\mu} u^{\mu} = c^2 % \thinspace . \end{equation}\]