# Lec31 - The second term is the Newtonian kinetic energy The...

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The second term is the Newtonian kinetic energy. The first term has dimensions of energy, but it is not zero even when the speed goes down to zero. We therefore interpret mc 2 as an energy associated with an object at rest. We therefore define E := γ mc 2 to be the total relativistic energy , so that E 2 = p 2 c 2 + m 2 c 4 . We justify this definition below. In the rest frame of the object, and only in that rest frame, p = 0, so that E RF = mc 2 . Because E 2 p 2 c 2 = m 2 c 4 is an equation describing an invariant, i.e., E 2 p 2 c 2 = E 2 p 2 c 2 , di ff erent observers will disagree on measurements of E, p , but will always agree on this particular combination thereof, in complete analogy with the spacetime interval, and hence all observers agree on the mass m of a particle. Relativistic dynamics: kinetic energy To show that our definition of E makes sense and is not arbitrary, we next show that E is the sum of the rest energy, and what we naturally call the relativistic kinetic energy. We define the force to be the rate of change of momentum, f i := dp i / dt , where the time t is in some reference frame. Then the kinetic energy of an object accelerated by force f is K = u u =0 f dx = u u =0 dp dt dx = u u =0 dp dt dx dt dt = u u =0 u dp Next, calculate dp : dp = d ( γ mu ) = d mu 1 u 2 /c 2 = m 1 u 2 c 2 3 2 du Therefore, K = u u =0 mu 1 u 2 c 2 3 2 du = mc 2 1 1 u 2 c 2 1 = mc 2 ( γ 1) E mc 2 as indeed one would expect, or E = mc 2 + K . The law of conservation of energy applies to the total energy of a closed system, so that E is the conserved quantity. Therefore, part of the total energy can be transferred from K to the rest energy and vice versa. 106

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Example : Inelastic collision of two balls. Let m 1 , m 2 be two balls with initial relativistic kinetic energies K 1 , K 2 , respectively. Then, before the collision, E 1 = m 1 c 2 + K 1 and E 2 = m 2 c 2 + K 2 . The total energy of the system before the collision is therefore
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