Simple Nature - Light and Matter

Expressing γ as ( 1 − v 2 /c 2) −1/2and making use of the approximation(1 + ɛ) p ≈ 1 + pɛ for small ɛ, we have γ ≈ 1 + v 2 /2c 2 ,soK E ≈ m(1 + v 2= 1 2 mv 2 ,2c 2 − 1)c2which is the classical expression. As demanded by the correspondenceprinciple, relativity agrees with classical physics atspeeds that are small compared to the speed of light.7.3.3 ⋆ The energy-momentum four-vectorStarting from E = mγ and p = mγv, a little algebra allows oneto prove the identitym 2 = E 2 − p 2 .We can define an energy-momentum four-vector,p = (E, p x , p y , p z ) ,and the relation m 2 = E 2 − p 2 then arises from the inner productp·p. Since E and p are separately conserved, the energy-momentumfour-vector is also conserved.Energy and momentum of light example 20Light has m = 0 and γ = ∞, so if we try to apply E = mγ andp = mγv to light, or to any massless particle, we get the indeterminateform 0·∞, which can’t be evaluated without a delicate andlaborious evaluation of limits as in problem 11 on p. 439.Applying m 2 = E 2 − p 2 yields the same result, E = p, much moreeasily. This example demonstrates that although we encounteredthe relations E = mγ and p = mγv first, the identity m 2 = E 2 − p 2is actually more fundamental.Mass-energy, not energy, goes in the energy-momentum fourvectorexample 21When we say that something is a four-vector, we mean that itbehaves properly under a Lorentz transformation: we can drawsuch a four-vector on graph paper, and then when we changeframes of reference, we should be able to measure the vector inthe new frame of reference by using the new version of the graphpapergrid derived from the old one by a Lorentz transformation.If we had used the energy E rather than the mass-energy E toconstruct the energy-momentum four-vector, we wouldn’t havegotten a valid four-vector. An easy way to see this is to considerthe case where a noninteracting object is at rest in some frameof reference. Its momentum and kinetic energy are both zero. If418 Chapter 7 Relativity