# Nuclear fission equation

Should not the energy conservation physics equation be

$$E= - mc^2 \text{ instead of } E= m c^2$$

because energy appears simultaneously with mass disappearance in splitting as

$$|\Delta E|= - |\Delta m| c^2?$$

Also what in Einstein's derivation dispenses with the factor $$\frac12$$ from the usual kinetic energy

$$KE= \frac12 m v^2 ?$$

Sorry about the elementary question, it remained with me for longtime.

$$E= mc^2$$ is not an equation that expresses energy conservation. It is an equation that relates the energy content of a physical object to its gravitating/inert mass. So for example, the equation says that, if you have a box filled with say some springs and a cold cup of tea and you open it, compress the springs and heat up the tea, adding some amount of energy $$\Delta E$$ to the system before closing the box again, then the mass of the box will increase by $$\Delta E / c^2$$ .
The classical kinetic energy term arises from the relativistic energy as the leading term in the Taylor approximation that is valid for small momentum/velocity. The factor of 1/2 has its origin in the Taylor expansion of the square root. There are two ways to approach the problem. My preferred view is that the $$m$$ in your equation is the rest mass and the full relativistic energy momentum equation should be:
$$E =\sqrt{m ^2 c^4 + p^2 c^2}$$
where $$p$$ is momentum.
There are also some heretical texts that introduce a so called "relativistic mass" $$m$$ which is $$m = \gamma m_0$$ where $$m_0$$ is the rest mass. In this sacrilegious notation, $$E = m c^2$$ is also valid outside of the rest frame of the object that you are studying and the kinetic energy term arises from a Taylor expansion of the $$\gamma$$ factor.