Derivation of Energy-Mass Equivalence: Total energy = kinetic energy $+ mc^2$

Question

How do you derive the energy mass equivalence just from special relativity? To be exact, in this video, https://www.youtube.com/watch?v=KZ8G4VKoSpQ, at around 23 minutes in, he claims that the total energy is equal to the Kinetic energy $+ mc^2$, which I don't understand how.

Answer 1

he claims that the total energy is equal to the Kinetic energy + mc^2, which I don't understand how.

The total energy of a free particle (no potential energy) is defined as the kinetic energy plus the rest energy $mc^2$, where $m$ is the rest mass.

The total energy, in terms of the rest mass $m$ is: $$ E = \gamma m c^2\;, $$ where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ and $v$ is the velocity.

We then define the kinetic energy as: $$ KE = E - mc^2 = (\gamma - 1)mc^2\;, $$

This definition make sense because in the non-relativistic limit ($v<<c$), we have: $$ \gamma \approx 1 + \frac{v^2}{2c^2}\;. $$

So that, in the non-relativistic limit, we also have: $$ KE \approx (1 + \frac{v^2}{2c^2} - 1)mc^2 = \frac{1}{2}mv^2\;, $$ as expected.

---

If you are interested in where $$ E = \gamma m c^2 $$ comes from to begin with, you can consider the possible classical actions for a free relativistic particle.

There is only one relativistic invariant that could make up the action. We must have $$ S = \alpha \int ds\;, $$ where $ds$ is the world line and $\alpha$ is a constant (to be determined below). So we have $$ S = \alpha \int dt \sqrt{\eta_{\mu\nu}\dot{x}^\mu\dot{x}^\nu}\;. $$

In other words the Lagrangian is: $$ L = \alpha \sqrt{c^2 - v^2}\;, $$ which must reduce to $\frac{1}{2}mv^2$ when $v<<c$.

This means that: $$ \alpha = -mc $$

This means that: $$ L = -mc^2\sqrt{1 - \frac{v^2}{c^2}} = -mc^2/\gamma\;. $$

Now we can turn the usual crank to get the Hamiltonian (another name for the energy).

First, we see that (by definition): $$ p \equiv \frac{\partial L}{\partial v} = mv\gamma $$

So we have (by the usual relation between the Lagrangian and the Hamiltonian): $$ E = pv - L = \gamma m (v^2 + \frac{c^2}{\gamma^2}) = \gamma m c^2\;. $$

And, again, to be clear, I am using the symbol $m$ to mean the rest mass.

Answer 2

If you take the total energy, like it is usually done as $E=mc^2$ the kinetic energy ist $KE=mc^2-m_0c^2$ where m is the relativistic mass $m_0$ the mass of a particle without kinetic energy. The m in the video ist what is usually called $m_0$