Because when we examine the impulse of particles, m, we have dP = d(mv) = F dt = dm v + m dv. We can make this into an energy relationship by multiplying both sides by c to give us:
F c dt = F ds = dm vc + m c dv = de an increment of energy from the work function F ds = force X distance.
As we are talking about invariant (rest mass) here dm = 0, it is invariant. Which leaves us with mc dv = de, which we integrate from 0 to c the speed of light and from e = 0 to E some total mass energy.
mc INT(dv)|0 to c = INT(de)|0 to E and there you are:
mc [c - 0] = mcc = mc^2 = E = [E - 0] QED
This simply says that the work we can get from mass m is E = mc^2.
Answers & Comments
Verified answer
Because when we examine the impulse of particles, m, we have dP = d(mv) = F dt = dm v + m dv. We can make this into an energy relationship by multiplying both sides by c to give us:
F c dt = F ds = dm vc + m c dv = de an increment of energy from the work function F ds = force X distance.
As we are talking about invariant (rest mass) here dm = 0, it is invariant. Which leaves us with mc dv = de, which we integrate from 0 to c the speed of light and from e = 0 to E some total mass energy.
mc INT(dv)|0 to c = INT(de)|0 to E and there you are:
mc [c - 0] = mcc = mc^2 = E = [E - 0] QED
This simply says that the work we can get from mass m is E = mc^2.
E stands for energy, m for mass, and c for the speed of light,squared is a very large number, the square of the speed of light.
More here http://www.aip.org/history/einstein/emc1_text.htm
This equation shows how mass and energy are very related to each other.
Simple algebraic proof of e=mc^2, on this website.
http://www.adamauton.com/warp/emc2.html