Relativistic mechanics

Remarks

In an inertial frame write $\vec v = d\vec x/dt$, $\gamma = 1/\sqrt{1 - \abs{\vec v}^2/c^2}$ and $F = \gamma(\vec f \cdot \vec v / c, \vec f)$ for a three-force $\vec f$. Then the equation of motion becomes $\frac{d}{dt}(\gamma m \vec v) = \vec f$ and $\frac{d}{dt}(\gamma m c^2) = \vec f \cdot \vec v$, because $d\tau = dt/\gamma$. Example: a free particle has $F = 0$, so $p$ is constant and the world line is a straight line; with $\vec v = 0$ it gives $p = (mc, \vec 0)$. Non-example: a four-vector $F$ with $\eta(F, u) \ne 0$ is not an admissible force, because $\frac{d}{d\tau}\eta(p, p) = 2m\,\eta(F, u)$ would change the value $\eta(p, p) = -m^2c^2$, and so the mass. For small speeds $\gamma \approx 1$, and the first equation reduces to Newton's second law $m \, d\vec v/dt = \vec f$.