What makes me chuckle is Smith's
    (E)^2 = (pc)^2 + (mc^2)^2
that he calls "correct".

If we take  E= mc^2 as a given and equal to the rest mass,
(let me know if you want a proof of this) and give the mass
a push so that it has some kinetic energy 1/2 mv^2,
Then this simulates a particle in an accelerator, and the total is
    E = mc^2 + 1/2 mv^2 

	[
	  Note to reader:
	     Relativistic energy is defined as
		E = mc^2/sqrt(1-v^2/c^2)
	]

Squaring,
    E^2 = (mc^2 + 1/2 mv^2)^2
        = (mc^2 + 1/2 mv^2)(mc^2 + 1/2 mv^2)
        = m^2c^4 + m^2c^2 v^2 + 1/4 m^2v^4
Defining  p = mv ,

	[
	  Note to reader:
	     Relativistic momentum is defined as
		p = mv/sqrt(1-v^2/c^2)
	]

    E^2 = (mc^2)^2  + (pc)^2  + (1/2mv^2)^2

Smith conveniently drops off the 1/4 m^2v^4,
which, if v almost equals c, would be quite a lot of energy.
The reason for doing this is to include gamma
so that they can make for the missing energy in any real
experiment involving a particle.
And of the course the actual speed, v, cannot be directly
measured and is calculated from the equations, thus
"proving" SR.

	[
	  Note to reader:
	     With the given definitions it can easily be shown that
		(E)^2 = (pc)^2 + (mc^2)^2
	]