# The mass-energy equivalence for rest mass

It is clear that the kinetic energy can be derived as $(m-m_0)c^2$. However, why do we say that $m_0c^2$ is the rest mass energy? It seems that this mass-energy equivalence for rest mass is just a easy convention but cannot be proven, i.e. we can allocate any value to the rest mass energy. Is that true?

• Word of warning: never, never, never use the "relativistic mass" as a single symbol. Rest mass is $m$, "relativistic mass" is $\gamma m$. Hiding the $\gamma$ factor in with the $m$ is something physicists did about a century ago, before they knew any better, and it only leads to confusion. – user10851 Jan 15 '15 at 11:05
• I'd go further and suggest that you purge any mention of "relativistic mass" from your brain. The concept is unnecessary and leads to confusion. It was abandoned many decades ago, but appears to persist. I'm not sure why. People must be reading old books. And perhaps writing new books based on old outdated books? – garyp Jan 15 '15 at 15:06

The mass energy equivalence is proved every day in nuclear reactors around the world. When a $^{235}$U nucleus fissions the amount of energy given off is exactly equivalent to the mass change multiplied by $c^2$.
The most precise test I know of was done in 2005 by a combined team from MIT, NIST and ILL in 2005. They found that $E$ differs from $mc^2$ by at most 0.0000004, or four-tenths of 1 part in 1 million.
$m_{0}c^{2}$ is the energy of the particle in the frame where the particle is at rest, that is the frame where its momentum is zero. Thats why its called rest mass energy. If you look for the expression for the kinetic energy of the photon you will find that there is no such analog "rest mass energy" term in this case because you cant find any inertial frame where the photon is at rest because you will need to find a frame that moves at speed of light and that frame does not exist
• The question wasn't about the rest mass, but about why it stores an energy of $m_0 c^2$. – Sofia Jan 15 '15 at 10:36