Energy-Momentum Relation

Energy-Momentum Relation

Energy and momentum unite into a 4-vector in special relativity. The energy-momentum relation is the relativistic counterpart of $E = p^2/2m$:

$E^2 = (pc)^2 + (mc^2)^2$

This relation is Lorentz-invariant — every observer agrees on it, regardless of their relative motion. Unlike $E = \gamma mc^2$ or $p = \gamma mv$, it does not reference the particle's velocity at all.

Special Cases

• At rest ($p = 0$): recovers $E = mc^2$
• Photon ($m = 0$): gives $E = pc$, so light carries momentum proportional to its energy

Your Task

Implement:

• energy_from_momentum(m, p) — computes $E = \sqrt{(pc)^2 + (mc^2)^2}$
• momentum_from_energy(m, E) — inverts to get $p = \sqrt{E^2 - (mc^2)^2} / c$

Use $c = 299792458.0$ m/s defined inside each function.