Kinetic energy (classical vs Einstein)
Contrary to the well known E = m c2, this equation does not represent accurately the mass energy equivalence of Einstein and does not represent the correct expression for the kinetic energy possessed by a point object with mass m. The kinetic energy of a mass m should be:
E = m c2 / √[1 - (v / c)2] - m c2
where c is the speed of light approximately 3.0 x 108 m/s and v is the speed of the object. Note that √[1 - (v / c)2] means the square root of [1 - (v / c)2].
Our question is if Einstein's mass energy equivalence is true, then why is Newton's kinetic energy formula
E = m v2 / 2
proves to be correct in the lab?
To answer this, we start with the Einstein's mass energy equivalence formula. Notice that the expression from the Einstein's formula, i.e. 1 / √[1 - (v / c)2] can be expanded.
Using mathematics, by applying Maclaurin series or using binomial theorem, the series is:
1 / √[1 - (v / c)2] = 1 + (v / c)2 / 2 + (v / c)4 / 4 + (v / c)6 / 8 + ...
For speed v which is small compared to the speed of light such as those which you encounter everyday from the speed of the wind to the fastest F1 car, all the terms in the series with (v / c) raised to a power higher than 2 are small and can be ignored. Hence, our equation for kinetic energy will be reduced to:
E = m c2[1 + (v / c)2 / 2] - m c2
Therefore, E = m v2 / 2
Thus, the kinetic energy equation E = m v2 / 2 only applies for speed v « c.
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