# Confusion regarding the Boltzman distribution as it applies to paramagnetism

1. Oct 31, 2011

### FluxPin

I'm currently in an introductory thermal physics course, and today we
learned about the Boltzmann factor and how to use it to calculate
various quantities with the corresponding probability distribution. All was going well until we got to the subject of paramagnetism.

We supposed a system of independent magnetic spins, pointing either
parallel or anti-parallel to the Magnetic field given by $\vec{B}$. In other
words, there are N spins each with magnetic moment $\vec{μ}$ in contact with a
thermal reservoir of temperature T.

I understand that each spin has a potential energy given by
E$_{n}$=$\pm$$\vec{μ}$$\vec{B}$ (A basic E&M result). However, I DON'T really understand how we can relate this
quantity to the Boltzmann distribution: P$_{n}$= e^($\frac{-E_{n}}{kT}$)/Z where Z is the partition function which normalizes the distribution.

How can the thermal energy of the reservoir (kT) can be related to the
purely magnetic state of each magnetic spin? I'm clearly missing
something since this makes no sense to me. Is energy also going into
the thermal energy of these spins and we are just not considering it?
I'm quite confused by this concept and any clarification would be most
appreciated.

Last edited: Oct 31, 2011
2. Oct 31, 2011

### marcusl

Think about it in the other order: the spins don't determine the reservoir, but rather the reservoir sets the temperature of the magnetic material, allowing you to calculate the magnetization. Electrons can be in two states (aligned and anti-aligned) having energies +/-E_n; the Boltzmann distribution tells you the fractions of total electrons, P_1=N_1/N and P_2=N_2/N, that are in each state. The net magnetization is then
$$M=N(P_1-P_2)\mu=(N_1-N_2)\mu.$$

3. Oct 31, 2011

### FluxPin

I understand that the magnetic moments do not determine the reservoir. What I'm having a conceptual issue with is why the temperature of the magnetic material (which I understand is completely dependent on the reservoir) can be related to its magnetization (i.e. the probability of a moment aligning itself along the field) with the Boltzmann distribution.
How does the temperature of the material determine its magnetization?

Last edited: Oct 31, 2011
4. Oct 31, 2011

### Bill_K

The temperature is not causing the magnetization, it's partially preventing it. If the temperature was absolute zero, all the dipoles would sit happily in the lower energy state and the system would be fully magnetized. Thermal agitation kicks a fraction of them to the higher energy state, reducing the magnetization.

5. Oct 31, 2011

### FluxPin

Okay, all the dipoles being in their lowest energy level at absolute zero makes. The thing I still don't get is how the thermal energy of the reservoir puts the individual dipoles into their higher energy magnetic state. That is, he mechanism by which they are "kicked" into the higher energy state.

Last edited: Oct 31, 2011