Mass-Energy Equivalence
The assumptions of special relativity lead to the most famous equation in physics: \(E=mc^2\). It derives naturally from the assumption that the speed of light is the same in all frames of reference. Starting from the definition of Kinetic energy:
\[\begin{align*} \frac{dK}{dt} = v\frac{dp}{dt} = vm\frac{d}{dt}(\gamma v) \end{align*}\]Where we have used the relativistic momentum: \(p = \gamma mv\), and \(\gamma = (1-v^2/c^2)^{-1/2}\) is the Lorentz factor:
\[\begin{align*} \frac{d(\gamma v)}{dt} &= \frac{d}{dt} \frac{v}{\sqrt{1-v^2/c^2}} \\ &= \Bigg[\Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-1/2} + \frac{v^2}{c^2} \Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-3/2}\Bigg] \frac{dv}{dt} \\ &= \Bigg[\Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-1/2}\Bigg\{1 + \frac{v^2}{c^2} \Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-1} \Bigg\} \Bigg] \frac{dv}{dt}\\ &= \Bigg[\Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-1/2}\Bigg\{\frac{1 -v^2/c^2}{1 -v^2/c^2} + \frac{v^2/c^2}{1 -v^2/c^2}\Bigg\} \Bigg] \frac{dv}{dt}\\ &= \Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-3/2} \frac{dv}{dt} \end{align*}\]Hence:
\[\begin{align*} \frac{dK}{dt} &= m\Bigg(1 - \frac{v^2}{c^2}\Bigg)^{-3/2} v\frac{dv}{dt} \\ &= \frac{d}{dt} \frac{m c^2}{\sqrt{1-v^2/c^2}} = \frac{d}{dt} \Big(\gamma m c^2\Big) \end{align*}\]Where we have used the fact that:
\[\begin{align*} \frac{d}{dt} \frac{c^2}{\sqrt{1-v^2/c^2}} = \Big(1 - \frac{v^2}{c^2}\Big)^{-3/2} v\frac{dv}{dt} \end{align*}\]Integrating both sides, knowing that the kinetic energy is zero at rest (\(\gamma=1\) when \(v=0\)), we get:
\[\begin{align*} K &= \gamma m c^2 + C \\ &= \gamma m c^2 - mc^2 \end{align*}\]We can recover the classic Newtonian kinetic energy \(K = \frac{1}{2} m v^2\) by taking \(v<<c\):
\[\begin{align*} K &= \gamma m c^2 - mc^2 \\ &= mc^2\Bigg(\frac{1}{\sqrt{1-v^2/c^2}} - 1\Bigg) \\ &= mc^2\Bigg(1 + \frac{1}{2} \frac{v^2}{c^2} + ... - 1\Bigg) \\ &\approx \frac{1}{2} m v^2 \end{align*}\]We can interpret \(K= \gamma m c^2 - mc^2\) as \(K = T - E\) where \(T\) is the total energy and \(E\) is the rest energy. Hence, we have:
\[E = m c^2\]