Simple Nature - Light and Matter

talking about relativity, so an object at rest has E = mc 2 , not E = 0as we’d assume in classical physics.Suppose we start accelerating the object with a constant force.A constant force means a constant rate of transfer of momentum,but p = mγv approaches infinity as v approaches c, so the objectwill only get closer and closer to the speed of light, but never reachit. Now what about the work being done by the force? The forcekeeps doing work and doing work, which means that we keep onusing up energy. Mass-energy is conserved, so the energy beingexpended must equal the increase in the object’s mass-energy. Wecan continue this process for as long as we like, and the amount ofmass-energy will increase without limit. We therefore conclude thatan object’s mass-energy approaches infinity as its speed approachesthe speed of light,E → ∞ when v → c .Now that we have some idea what to expect, what is the actualequation for the mass-energy? As proved in section 7.3.4, it isE = mγc 2 .self-check DVerify that this equation has the two properties we wanted.Answer, p. 924KE compared to mc 2 at low speeds example 18⊲ An object is moving at ordinary nonrelativistic speeds. Compareits kinetic energy to the energy mc 2 it has purely because of itsmass.⊲ The speed of light is a very big number, so mc 2 is a huge numberof joules. The object has a gigantic amount of energy becauseof its mass, and only a relatively small amount of additionalkinetic energy because of its motion.Another way of seeing this is that at low speeds, γ is only a tinybit greater than 1, so E is only a tiny bit greater than mc 2 .The correspondence principle for mass-energy example 19⊲ Show that the equation E = mγc 2 obeys the correspondenceprinciple.⊲ As we accelerate an object from rest, its mass-energy becomesgreater than its resting value. Classically, we interpret this excessmass-energy as the object’s kinetic energy,K E = E(v) − E(v = 0)= mγc 2 − mc 2= m(γ − 1)c 2 .⊲Section 7.3 Dynamics 417