# Magnetic moment of paramagnetic crystal

Tags:
1. May 18, 2015

### 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. The problem statement, all variables and given/known data

2. Relevant 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

$$F = -Nk_B \ ln \ Z$$

3. The attempt at a solution

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

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

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

Last edited: May 18, 2015
2. May 24, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 25, 2015

### rude man

I would resubmit this to the advanced physics forum.