Problem — Proof of the energy-momentum invariant

Problem

Starting from the definitions $E = \gamma mc^2$ and $p = \gamma m v$, prove the invariant relation $E^2 - (pc)^2 = (mc^2)^2$. Show explicitly that for a photon ($m=0$) it implies $E = pc$.

Solution

Starting point.

$E = \gamma m c^2, \qquad p = \gamma m v, \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

Computing $E^2 - (pc)^2$.

$E^2 - (pc)^2 = (\gamma m c^2)^2 - (\gamma m v\, c)^2 = \gamma^2 m^2 c^4 - \gamma^2 m^2 v^2 c^2$

Factor out $\gamma^2 m^2 c^4$:

$E^2 - (pc)^2 = \gamma^2 m^2 c^4\left(1 - \frac{v^2}{c^2}\right)$

Using the definition of $\gamma$. Since $\gamma^2 = \dfrac{1}{1 - v^2/c^2}$, the factor $\left(1 - v^2/c^2\right)$ exactly cancels $\gamma^2$:

$E^2 - (pc)^2 = \gamma^2 m^2 c^4 \cdot \frac{1}{\gamma^2} = m^2 c^4$

$\ev{E^2 - (pc)^2 = (mc^2)^2}$

The right-hand side depends only on the mass $m$ (and on $c$), not on the velocity: it is therefore the same in every inertial frame, i.e. an invariant.

Photon case ($m = 0$). Setting $m = 0$:

$E^2 - (pc)^2 = 0 \quad\Rightarrow\quad E^2 = (pc)^2 \quad\Rightarrow\quad \ev{E = pc}$

The photon, having no mass, has energy and momentum related simply by $E = pc$ and travels at speed $c$.

Links

Topics: Special relativity Concepts: Spacetime invariant · Mass-energy equivalence Skills: Symbolic setup