Proton Speed at the LHC

Reletivistic Kinematics:

\begin{equation} \gamma = \frac{E}{m} = \frac{1}{\sqrt{1-\beta^2}} \end{equation}

An nice aproximation valid for \(\beta \sim 1\):

\begin{equation} \left(\frac{1}{\gamma}\right)^2 = 1-\beta^2 = (1-\beta)(1+\beta) \approx 2(1-\beta) \end{equation}

or

\begin{equation} (1-\beta) = \frac{m^2}{2E^2} \end{equation}

For a proton with \(m = 1\,\text{GeV}\) at the LHC with \(E = 13\,\text{TeV}\) protons:

\begin{equation} \frac{E}{m} = \frac{13\,\text{TeV}}{1\,\text{GeV}} = 13000 \end{equation} \begin{equation} (1-\beta) = \frac{1}{2 \times 13000^2} \approx 2\times10^{-9} \end{equation}

So

\begin{equation} \beta = 0.999999997 \end{equation}