Sketch of Einstein's argument for E = mc²

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

\Delta m = \frac{L}{c^2} \quad\Longrightarrow\quad E = mc^2

problems.proof.analysis

Matching coefficients gives $\Delta m = L/c^2$ : the body lost mass $\Delta m$ exactly equal to the emitted light energy $L$ divided by $c^2$ . Since $L$ was an arbitrary amount of emitted energy, the general conclusion is that any energy $E$ corresponds to a mass $m=E/c^2$ , i.e. $E=mc^2$ .

problems.proof.pitfall. This is a simplified sketch of Einstein's original 1905 argument (valid only to leading order in

v/c

), not a fully rigorous derivation at all velocities — modern treatments typically use the four-momentum vector directly for a more general and compact result.