Defining temperature through a derivative of a log of a number of ways-—equations (9-2) an...

Defining temperature through a derivative of a log of a number of ways-—equations (9-2) and (9-3)—may be hard to embrace, but it's built into the Boltzmann distribution and all others. We can at least verify it in the oscillator case. Equation (9-16) comes from equating the average energy Mω0/N of a collection of oscillators to the average predicted by the Boltzmann distribution. According to equation (9-4), the temperature it gives should be. Is it? Reasonable precision demands a fairly large system. Assume the number of particles N is 10,000 and the total energy E is 49,000 ω0. It's easiest to choose energy units in which ω0 = 1, so that, according to equation (9-7), E and M are both 49,000. (a) Calculate the number of particles, Nn, at each level by multiplying N by the exact probabilities of equation (9-9)—using n instead of NF since our interest is the total number at a given level, not a specific particle. Note that with ω0≡1 n is the level's energy, Round each Nn to the nearest integer, then determine the total energy simply by multiplying Nn by n—the number times the energy at each level— and summing over all levels. It might seem that the total should be 49,000. Due to rounding, it won't be, but it should differ by less than 1 %. Entropy is fairly easy to calculate here. The number of ways of rearranging particle labels when N particles are distributed in different boxes/levels is . (See Appendix J for the proof.) Use this to calculate the entropy via equation (9-2). (b) Repeat part (a), changing only M, making it 51,000 instead of 49,000. (c) Calculate the quotient: change in entropy over change in energy. Bearing in mind that we consider a system of N = 10,000 and M ≡ 50,000, how does it compare with equation (9-16)?