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 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Thanks Bruce... Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Jim Bergquist on February 17, 2001 10:37:49 UTC

Thanks Bruce. I realized that what I had said was not quite correct after I had logged off, but I was trying to keep things simple. It was also rather late.

Mass can be defined as a resistance of a body to its change in motion as given in the definition of the inertial force f = m a = m dv/dt. This means that force that is needed to produce a given acceleration or equivalently, rate of change in velocity, is proportional to the mass of the body. The change in kinetic energy (dT) is equal to the work (dW) done on the body:

dT = dW = m a·dx = m dv/dt·dx = m dx/dt·dv = m v·dv

which can be integrated to give

T = ½ m v² [the superscript is a 2]

This is the classical relation connecting mass, energy and velocity. Here the m is the rest mass of relativity. The relativistic equivalent to this expression is

T = m c² - m0 c² [= E(v) - E(v=0)]

where m is now the relativistic mass and m0 is the rest mass and where

m = m0/sqrt(1-v²/c²)

The quantity that m0 is divided by comes from the Lorentz transformation which is used to convert quantities from a stationary frame of reference to a moving one. The relativistic mass is converted in the same way that time is. Note that since only the square of v is involved, the result does not depend on the direction of travel. We get the same increase in mass whether the object is traveling towards us or away from us. This differs from the way that light is "Doppler" shifted by a moving source.

For small v, m is approximated by

m = m0 (1 + ½ v²/c² + ...)

if this is substituted into T above we get:

T = m0 c² (1 + ½ v²/c² + ...) - m0 c² = ½ m0 v² + ...

So for small v we get the classical results for the kinetic energy.