Energy–momentum relation

Remarks

The four-momentum is $p = (E/c, \vec p)$, and its Minkowski square is $\eta(p, p) = -m^2 c^2$, where $\eta = \operatorname{diag}(-1, 1, 1, 1)$ is the metric of Minkowski space. Writing out $-E^2/c^2 + \abs{\vec p}^2 = -m^2c^2$ gives the relation. For $m > 0$ one can check it directly: $E = \gamma m c^2$ and $\vec p = \gamma m \vec v$ give $E^2 - \abs{\vec p}^2c^2 = \gamma^2 m^2 c^4 (1 - \abs{\vec v}^2/c^2) = m^2 c^4$. For a photon, $m = 0$ and the relation reduces to $E = \abs{\vec p} c$. For a particle at rest it gives $E = mc^2$. For small momentum, $E = mc^2 \sqrt{1 + \abs{\vec p}^2/(m^2c^2)} \approx mc^2 + \abs{\vec p}^2/(2m)$, the rest energy plus the Newtonian kinetic energy. The same relation holds for a system, with $m$ replaced by its invariant mass and $E$, $\vec p$ by its total energy and momentum.

Going deeper

For a student

The relation is the statement that the four-momentum lies on the mass shell, the hyperboloid $\eta(p, p) = -m^2c^2$ with $p^0 > 0$. It is used constantly in collision kinematics, for example to find the energy threshold of a reaction from the invariant mass of the initial state. In quantum mechanics it leads, with $E \to i\hbar \partial_t$ and $\vec p \to -i\hbar\nabla$, to the Klein–Gordon equation.