How does rest energy $E=mc^2$ follow just from special relativity?

Question

I'm reading Friedman and Susskind's Special Relativity and Classical Field Theory.

They define the Lagrangian of a free particle

$$\mathcal L = -mc^2\sqrt{1-{v^2\over c^2}}$$

and then derive the corresponding Hamiltonian to be

$$H = \frac{mc^2}{\sqrt{1-{v^2\over c^2}}} .$$

Then they note that in the non-relativistic limit

$$v \ll c \\H \to mc^2+{1\over 2}mv^2.$$

Also, for $v=0$, $H = mc^2$. They then identify this $mc^2$ as "energy of assembly" of the particle.

Now this SE post's answers suggest that the process of conversion of mass into energy follows from particle physics. But particle physics can't be just explained by the two postulates of special relativity! How can then this "rest energy" be derived from just postulating the free particle Lagrangian? What is so non-trivial about it?

Answer 1

You are fundamentally correct, but are mixing up what's necessary for each. Special relativity is both necessary and sufficient to construct the idea of rest energy, but it is merely necessary for constructing a model of conversion of rest energy to other forms of energy. In order to do that, you need particle physics, and the only way we know how to construct a good model of particle physics is quantum field theory.

To elaborate a little: in special relativity, you can derive the rest energy of a particle as you have shown above! Special relativity is the source of the idea of rest energy. However, in special relativity, there is no concept of particle creation and destruction; there is no concept of converting this energy into anything else. You need quantum field theory, the marriage of special relativity and quantum mechanics, to do that.

Check out this question for more information.

Answer 2

Written as $ΔE = (Δm)c^2$ the $ΔE$ is the increase in the kinetic energy of a mass being accelerated, and the $Δm$ is the corresponding increase in the inertial mass (as predicted by special relativity).

Answer 3

One of the first steps in developing the kinematics and dynamics of special relativity is to extend three-dimensional momentum to a four dimensional four-vector. This is done by multiplying the four-velocity $(\gamma,\gamma{\bf v})$ by a constant, $m$, with the dimensions of energy (in natural units with c equal to 1). The invariant length squared of the momentum four-vector is $E^2-{\bf p}^2=m^2$. When this equation is reduced in the limit $p<<E$, all the non-relativistic formulas up show. The relation $E^2-{\bf p}^2=m^2$ shows that $m$ is the invariant length of the four-vector, and I hope most people agree should be called the 'invariant mass'. It also shows that $m$ and $E$ are two quite different things. $E$ is the fourth component of the four-vector and $m$ is its invariant length. $m$=$E$ in the rest system, but that doesn't make them the same thing.