Magnetic moment of paramagnetic crystal

[Je m'appelle]

Hello, I've been having some trouble with a paramagnetism problem from my Statistical Mechanics class textbook (F. Mandl, Statistical Physics, 2nd edition, p. 25). The problem is as follows

1. Homework Statement

2. Homework Equations

1. The temperature parameter

\displaystyle{ \beta = \frac{1}{k_B T} }
where k_B is Boltzmann's constant

2. The partition function

Z = \displaystyle{ \sum_{r} e^{-\beta E_{r}} }
for the energy of eigenstate r

3. The Boltzmann distribution

\displaystyle{ p_r = \frac{1}{Z} e^{-\beta E_r} }
for the energy of eigenstate r

4. The net magnetic moment

\displaystyle{ M = \frac{N}{\beta} \left( \frac{\partial ln Z}{\partial B} \right)_{\beta} }5. The energy

E = -M B

6. The entropy

S = k_{B} \ ln \ \Omega

7. The Helmholtz free energy

F = E - TS

or

F = -Nk_B \ ln \ Z

The Attempt at a Solution

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My difficulty arises when trying to deduce M from "a minimum Helmholtz free energy", what exactly does this imply?

EDIT: Alright, can anyone verify this? I decided to take the helmholtz free energy, F as a function of the field B and then derive it with respect to it, and since it's a minimum the derivative must equal zero, also keeping in mind that since the system is in a heat bath the temperature is constant, so

\frac{dF}{dB} = \frac{dE}{dB} - T \frac{dS}{dB} - S \frac{dT}{dB} = 0

\frac{dE}{dB} = \frac{d}{dB}(-MB) = T\frac{dS}{dB}

By setting the statistical weight of the microstate, \Omega, of the crystal with N ions that can be oriented in two ways (parallel and antiparallel) to the applied magnetic field, one finds that the entropy can be written as

S = k_B N (2 cosh(x) - xtanh(x))

where x \equiv \left( \frac{\mu B}{k_B T} \right), and the former equation becomes

-M = T\frac{d}{dB} \left( k_B N (2 cosh(x) - xtanh(x)) \right)

M = - N \mu \left(2 tanh(x) + xsech^2(x) \right)

 

[rude man]

I would resubmit this to the advanced physics forum.