Blackholes Forum Message Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem
 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora By The Book! Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Jim Bergquist on January 9, 2001 10:00:39 UTC

From "On the Electrodynamics of Moving Bodies" by Albert Einstein in The Principle of Relativity by Einstein et al., p. 63

"We will now determine the kinetic energy of the electron. If an electron moves from rest at the origin of co-ordinates of the system K along the axis of X under the action of an electrostatic force X, it is clear that the energy withdrawn from the electrostatic field has the value SeXdx [S for the integral sign]. As the electron is to be slowly accelerated, and consequently may not give off any energy in the form of radiation, the energy withdrawn from the electrostatic field must be put down as equal to the energy of motion W of the electron. Bearing in mind that during thr whole process of motion which we are considering, the first of the equations (A) applies, we therefore obtain

W = SeXdx = m S(0->v)B^2vdv [S(0->v) for "integral from 0 to v", B for beta]

= m c^2 {1/(1-v^2/c^2)^(1/2) - 1}.

Thus, when v = c, W becomes infinite. Velocities greater than that of light have--as in our previous results--no possibility of existence."

In modern notation with rest mass m0, relativistic mass m, and kinetic energy T this translates to:

T = m c^2 - m0 c^2

or E = m c^2 = m0 c^2 + T

So m c^2 contains both the rest mass and kinetic energy terms.