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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567
I was looking at derivations of the kinetic energy and energy/mass equation. Both seem to be derived from the same formula and my question is how is this true? First, the energy/mass derivation:
F = M*A = M*(s/t)/t W = F*s Therefore, E = W = s*M*(s/t)/t E = M*s^2/t^2 Since (s/t) = v we have E = M*v^2 and for v = c we have E=M*C^2
Now, we derive the Kinetic energy equation
a =dv/dt = (dv/ds)*(ds/dt) = v*dv/ds
W = E = Integral(F*ds) = Integral(m*a*ds) E = Integral(m*v*dv/ds)*ds E = Integral(m*v*dv) E = (1/2) * M*v^2
Both methods seem right; yet, they yield different results. There must be some assumption that allows both to be correct - one for v = c and the other for v << c. Anyone want to shed some light on this?
EDIT: I believe I figured it out while I was taking a shower. In the world of kinetic energy, the initial velocity is zero so the total energy involved the average of the initial velocity and final velocity. This results in the 1/2 factor. In the world of light-speed, the velocity is always "c" and there is no averaging. This is the reason for the difference.
Registered Member #32
Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549
In relativity, accelerating something increases its mass. The second derivation doesn't take that into account.
For anything with mass, v never equals c.
To know whether you need to use relativistic theory, or whether simple Newtonian stuff will do, calculate gamma. Y = sqrt(1 / (1 - (v/c) ^ 2)). (That Y is meant to be a greek gamma.) This number is the factor by which relativity does its weirdness. For example, if gamma is 2, time, space, mass, etc will be scaled by a factor of two. If gamma is close to 1, use Newtonian theory.
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Relativity and Kinetic Energy
