Why E = mc2
Here we have a tube with two projectiles and a chemical charge between them.
When we burn the chemical charge the projectiles fly apart with equal and opposite momenta mv and -vm, each with KE = ½mv2. We deduce the potential energy stored in the powder charge to be PE = mv2. Momentum is conserved, -vm + mv = 0.
By changing the symbol 'v' to the symbol 'c' to represent the speed of the projectiles, we have potential energy PE = mc2 in the chemical charge.
Thus Einstein's E = mc2 is actually derived from Newtonian Mechanics.
Total energy = mc2 + ½mv2
Total energy squared:
E2 = (mc2 + ½mv2)2
= (mc2 + ½mv2)(mc2 + ½mv2)
= m2c4 + m2c2v2 [ + ¼m2v4]
Since p2= m2v2 by definition of p = mv,
E2 = m2c4 + p2c2 [ +¼p2v2]
The lower corner is missing so that relativists can divide by (gamma = 1/ ) and always get the answer they desire to fit the data. It goes hand in glove with the relativist's lookup table:
No matter what the energy is, there is always an unmeasured v of 0.99xxxxx that matches the data within the limits of experimental error, and why you can't disprove the theory from anything you do with particles. The relativists manufacture the velocity to get the result they seek.