Simple Nature - Light and Matter

will the the one that maximizes the entropy. Notice that the dependenceof the entropy on the masses m 1 and m 2 occurs in termsthat are entirely separate from the energy terms. If we want tomaximize S with respect to E 1 (with E 2 = E − E 1 by conservationof energy), then we differentiate S with respect to E 1 and set itequal to zero. The terms that contain the masses don’t have anydependence on E 1 , so their derivatives are zero, and we find thatthe molecular masses can have no effect on the energy sharing.Setting the derivative equal to zero, we have0 = ∂ (n 1 k ln V + n 2 k ln V + 3 ∂E 1 2 n 1k ln m 1 + 3 2 n 2k ln m 2+ 3 2 n 1k ln E 1 + 3 )2 n 2k ln(E − E 1 ) + . . .= 3 (2 k n1− n )2E 1 E − E 10 = n 1− n 2E 1 E − E 1n 1= n 2.E 1 E 2In other words, each gas gets a share of the energy in proportionto the number of its atoms, and therefore every atom gets, onaverage, the same amount of energy, regardless of its mass. Theresult for the average energy per atom is exactly the same as foran unmixed gas, ¯K = (3/2)kT .EquipartitionExample 20 is a special case of a more general statement calledthe equipartition theorem. Suppose we have only one argon atom,named Alice, and one helium atom, named Harry. Their total kineticenergy is E = p 2 x/2m + p 2 y/2m + p 2 z/2m + p ′ 2x /2m ′ + p ′ 2y /2m ′ +p ′ 2z /2m ′ , where the primes indicate Harry. We have six terms thatall look alike. The only difference among them is that the constantfactors attached to the squares of the momenta have different values,but we’ve just proved that those differences don’t matter. Inother words, if we have any system at all whose energy is of theform E = (. . .)p 2 1 + (. . .)p2 2 + . . ., with any number of terms, theneach term holds, on average, the same amount of energy, 1 2kT . Wesay that the system consisting of Alice and Harry has six degrees offreedom. It doesn’t even matter whether the things being squaredare momenta: if you look back over the logical steps that went intothe argument, you’ll see that none of them depended on that. In asolid, for example, the atoms aren’t free to wander around, but theycan vibrate from side to side. If an atom moves away from its equilibriumposition at x = 0 to some other value of x, then its electricalenergy is (1/2)κx 2 , where κ is the spring constant (written as theGreek letter kappa to distinguish it from the Boltzmann constant k).We can conclude that each atom in the solid, on average, has 1 2kT of322 Chapter 5 Thermodynamics