Lagrangian of a free particle in Special Relativity and equivalence between mass and energy

Here is one argument:

  1. OP has already argued that the energy $E$ is of the form $$ E~=~ m_0 \gamma c^2+C, $$ where $C$ is a constant.

  2. In SR, the $4$-momentum $p^{\mu}=(E/c,{\bf p})$ transforms as $4$-vector under Lorentz transformations. In particular, the length-square of the $4$-vector should be an invariant: $$ {\rm const.}~=~\left(\frac{E}{c}\right)^2-{\bf p}^2~=~\left(\frac{m_0 \gamma c^2+C}{c}\right)^2 - (m_0 \gamma{\bf v})^2.$$ It is straightforward to see that this is only possible if the constant $C=0$ is zero.

A way to think about this is the following. Consider you have not one but two particles. For which you can follow the same derivation that was made in order to set the proportionality constant $\alpha$ for each. As we now, it will be related to the mass of each particle (take the case they are different). Now you can see that no matter what constant you add you wont be able to cancel all the constant terms. So the issue remains, there is a piece which when compared to any reference you take, it doesn't go away.

There are other cases where there is more controversy or discussion. If you were to try adding a constant in GR, you will see that the factor $\sqrt{-\det g}$ actually has an impact on the e.o.m.'s.