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Comment by eesmith | Hacker News Reader

Again, Einstein's full equation is E² = (p c)² + (mc²)². You left out the momentum term, which means your equation assumes v=0, in which case K.E. = 0 is trivially valid.

The momentum term is also important as massless photons still have kinetic energy, which cannot be derived from classical Newtonian mechanics.

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I don't need Einstein's equation. Momentum is already assumed in kinetic energy and kinetic energy assumes massive bodies. This is the reason for using mass in the formula. I have no idea why you want the kinetic energy to be 0. The equation assumes a moving body. Thus v cannot be equal to zero.

You need Einstein's equation to have a "striking analogy".

You used Einstein's equation for v=0; the rest-mass/energy equivalence.

If you use Einstein's full equation, there is no "striking analogy."

Your definition of kinetic energy excludes photons, which have kinetic energy and momentum but no mass.

In my opinion, it is not Einstein's equation. It's Newton's equation applied to the speed of light. It's not v=0 but v=c. It just so happens that mc² is also the potential energy of a body at rest.

Yes, a photon's energy is all kinetic but so what? Newton's equation specified massive bodies.

Again, the 1/2 is BS but the winner of this debate is still to be determined.

A photon has no mass.

Newton's equation specified massive bodies.

Ergo, Newton's equation does not apply to photons.

Einstein's equation does apply to photons.

Ergo, Einstein's equation is not Newton's equation.

For velocities much less than the speed of light, Einstein's equation is well approximated by Newton's equation.

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