The photon, for example ,has energy $E=h\nu$, but $m_0=0$ When it is one particle under observation it is the mass that you will find in the table of masses for particles, and can be considered the building block of matter as we know it except for particles at the limit of $m_0=0$. $m_0^2=E^2-p^2$ where $E$ is the total energy of a system of particles and $p$ the total momentum. Where $m_0$ is called the rest mass of a particle or a system and is an invariant quantity under four dimensional transformations, characterizing the Energy_momentum four vector of the system. $m_r=E/c^2$ or equivalently $m_r=\gamma\times m_0$ Secondly the mass itself in your formula is the relativistic massĪnd can be defined for all particles and particle systems whether moving or at rest: So as others have observed, this formula is a mathematical convenience, it follows the conservation laws that are imposed by the differential equations that describe all systems in the microcosm, and should absolutely hold. So it is an equation that gives an equivalence of one variable to another. In the formula that puzzles you, $E = mc^2$, the mass is a variable too. To start with when we use relativistic equations we are in the realm of particle/nuclear physics, one way or another. I will have a go at this since it came up once more.
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