The total number of microstates is given by N N N So we can calculate the

# The total number of microstates is given by n n n so

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The total number of microstates is given by N ! N ! N ! So we can calculate the entropy via : S k ln S k ln N ! ln N ! ln N ! Now we use Stirling’s approximation (Guenault Appendix 2) which says ln x ! x ln x x ; this is important you should remember this. PH261 – BPC/JS 1997 Page 2.7 Hence S k N ln N N ln N N ln N where we have used . Into this we substitute for and since we need S in terms of E for differentiation to get temperature. N N N N N S k N ln N 1 2 N E / ln 1 2 N E / 1 2 N E / ln 1 2 N E / Now we use the definition of statistical temperature: to obtain the temperature: 1 / T S / E N V 1 T k 2 ln N E / N E / (You can check this differentiation with Maple or Mathematica if you wish!) Recalling the expressions for and , the temperature expression can be written: N N 1 T k 2 ln N N which can be inverted to give the ratio of the populations as N N exp 2 / kT Or, since we find N N N N N exp / kT exp / kT exp / kT N N exp / kT exp / kT exp / kT 16 14 12 10 8 6 4 2 0 0.0 0.2 0.4 0.6 0.8 1.0 N / N N / N kT / Temperature variation of up and sown populations This is our final result. It tells us the fraction of particles in each of the two energy states as a function of temperature. This is the distribution function. On inspection you see that it can be written rather concisely as n N exp / kT z PH261 – BPC/JS 1997 Page 2.8 where the quantity z z exp / kT exp / kT is called the (single particle) partition function . It is the sum over the possible states of the factor . This distribution among energy levels is an example of the Boltzmann distribution. exp / kT This distribution applies generally(to distinguishable particles) where there is a whole set of possible energies available and not merely two as we have considered thus far. This problem we treat in the next section. In general z states exp / kT where the sum is taken over all the possible energies of a single particle. In the present example there are only two energy levels. In the general case is the average number of particles in the state of energy (In the present special case of only two levels and are uniquely determined). To remind you, an example of counting the microstates and evaluating the average distribution for a system of a few particles with a large number of available energies is given in Guenault chapter 1. The fluctuations in about this average value are small for a sufficiently large number of particles. (The 1/ factor) n n 1 n 2 n N 2.4.2 Magnetisation of magnet Curie’s law S 1 / 2 We can now obtain an expression for the magnetisation of the spin 1/2 paramagnet in terms of temperature and applied magnetic field. The magnetisation (total magnetic moment) is given by: M N N We have expressions for and , giving the expression for M as N N M N exp / kT exp / kT exp / kT exp / kT N tanh / kT or, since , the magnetisation in terms of the magnetic field is B M N tanh B kT 3.0 2.0 1.0 0.0 0.0 0.2 0.4 0.6 0.8 1.0 B / kT M / N saturation M N 2 B kT linear region magnetisation of paramagnet PH261 – BPC/JS 1997 Page 2.9 The general behaviour of magnetisation on magnetic field is nonlinear. At low fields the magnetisation starts linearly but at higher fields it saturates as all the moments become aligned.  #### You've reached the end of your free preview.

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