both sides. The quantity(2m)dm/dt can be recognized as the time derivative of m2, and(2mv). d(mv)/dt is the time derivative of (mD)2. So, Eq.( 15 15)is the same as If the derivatives of two quantities are equal, the quantities themselves differ at most by a constant, say C. This permits us to write (15.17) 'e need to define the constant C more explicitly. Since Eq. (15. 17)must be true for all velocities, we can choose a special case where v =0, and say that in this case the mass is mo. Substituting these values into Eq.(15. 17) gives We can now use this value of C in Eq(15.17), which becomes Dividing by c and rearranging terms gives m2(1-2/c2)=m from which we get This is the formula(15. 1), and is exactly what is necessary for the agr remen tween mass and energy in Eq(15. 12) Ordinarily these energy changes represent extremely slight changes in mass, because most of the time we cannot generate much energy from a given amount of material; but in an atomic bomb of explosive energy equivalent to 20 kilotons of TNT, for example, it can be shown that the dirt after the explosion is lighter by I gram than the initial mass of the reacting material, because of the energy that was eleased, i.e., the released energy had a mass of l gram, according to the relationship AE= A(mc). This theory of equivalence of mass and energy has been beautifully verified by experiments in which matter is annihilated-converted totally to energy An electron and a positron come together at rest, each with a rest mass mo. When they come together they disintegrate and two gamma rays emerge, each with the measured energy of moc2. This experiment furnishes a direct determination of the energy associated with the existence of the rest mass of a particl