Now let us look at some further uences of relativistic change of mass Consider the motion of the molecules in a small tank of gas. When the gas is m0(1-u2/c2)-12=m(1+b2/c2+器/c4+…) see clearly from the formula that the series converges rapidly when v is small d the terms after the first two or three are negligible So we can write m (15.11) in which the second term on the right expresses the increase of mass due to mo- lecular velocity. When the temperature increases the u increases proportionately, so we can say that the increase in mass is proportional to the increase in tempera- ture. But since fmo2is the kinetic energy in the old-fashioned Newtonian sense increase in kinetic energy divided by e?, or Am a(KEy/ci as is equal to the can also say that the in 15-9 Equivalence of mass and energy The above observation led Einstein to the suggestion that the mass of a body can be expressed more simply than by the formula(15. 1), if we say that the mass is equal to the total energy content divided by c2. If Eq.( 15. 11)is multiplied by c the result is Here, the term on the left expresses the total energy of a body, and we recognize the last term as the ordinary kinetic energy. Einstein int ed the large constant term, moc, to be part of the total energy of the body, an intrinsic energy known Let us follow out the consequences of assuming, with Einstein, that the energy of a body always equals me. As an interesting result, we shall find the formula(15. 1) for the variation of mass with speed, which we have merely assumed up to now. We start with the body at rest, when its energy is moc 2. Then we apply a force to the body, which starts It moving and gives it kinetic energy; therefore, since the energy has increased the mass has increased-this is implicit in the original assumption. So long as the force continues, the energy and the mass both continue to increase. We have already seen( Chapter 13)that the rate of change of energy with time equals the force times the velocity, or (1513) We also have(Chapter 9, Eq 91)that F= d(mu)/dt. When these relations are put together with the definition of E, Eq(15 13)becomes (15.14) We wish to solve this equation for m. to do this we first use the mathematical trick of multiplying both sides by 2m, which changes the equation to c2(2m) (1515) We need to get rid of the derivatives, which can be accomplished by integrating