Mass-Energy Equivalency

Question

We call $E=mc^2$ the Mass-Energy Equivalency because it equates mass and energy together. But, by that same logic, shouldn't we call $E=\frac{1}{2}(mv^2)$, the equation of kinetic energy in Newtonian mechanics as the Mass-Energy Equivalence ?

Answer 1

The equation $E=mc^2$ equates rest energy to mass. There is a third symbol in this equation that represents the speed of light, but this is a universal constant. One can always select physical units such that this constant attains value unity. Regardless the system of units selected, up to a numerical proportionality constant, the equation $E=mc^2$ identifies the mass of a system as the energy observed from a center-of-mass frame. Hence the term mass-energy equivalence.

The equation $E=\frac{1}{2}mv^2$ is entirely different in character. It contains three symbols that represent physical quantities, and relates kinetic energy to the product of mass and velocity squared. If you want you can refer to this equation as squaredvelocitymass-kineticenergy equivalence, but that is a bit of a mouthful, and it is only valid in the low-velocity (Newtonian) approximation.

Answer 2

The formula $E=mc^2$ is ambiguous - you can take it as either $$ E_0 = mc^2 $$ relating rest energy and invariant mass, or as $$ E = m_rc^2 $$ relating total energy and relativistic mass $m_r=\gamma m$.

Now, as $$ \gamma = 1 + \frac 12 \left(\frac vc\right)^2 + \mathcal O\left(\left(\frac vc\right)^4\right) $$ we have $$ E = mc^2 + \frac 12 mv^2 + \mathcal O\left(\left(\frac vc\right)^2\right) $$

This means the Newtonian kinetic energy relation is inconsistent with either interpretation of $E=mc^2$: It does not contribute to rest mass and is only a small correction to the total energy for non-relativistic velocities (and incorrect for relativistic ones).

As a side note, we also have $$ \gamma = \sqrt{1+\gamma^2\left(\frac vc\right)^2} $$ and thus $$ E = \sqrt{(mc^2)^2 + (\gamma mvc)^2} = \sqrt{(mc^2)^2 + (pc)^2} $$ which can we re-written to $$ \left(\frac Ec\right)^2 - p^2 = (mc)^2 $$

This is the invariant 'length' of the 4-vector $p^\mu = (E/c,\vec p)$.

The concept of relativistic mass has fallen out of favour among physicists: It's just another name for the total energy (up to constant factors) and thus the time component of a 4-vector, whereas the invariant mass (aka rest mass or just mass) is an invariant scalar.