Paramagnetism — Question 5

Question 5

Magnetic free energy yields the mean moment and the weak-field half factor

N independent two-state moments at fixed temperature have magnetic Helmholtz free energy $F(B)=-Nk_BT\ln[2\cosh(\mu B/(k_BT))]$. Ignore field-source energy; F refers only to the moments.

1. Differentiate F to recover the total mean moment.

2. Expand F(B)-F(0) through order B squared and express it in terms of the linear-response total moment.

Question 5 - Solution

Magnetic free energy yields the mean moment and the weak-field half factor

(a) Method and calculation

$$m_{\mathrm{tot}}=-\left(\frac{\partial F}{\partial B}\right)_T=Nk_BT\tanh x\frac{\mu}{k_BT}.$$$$\boxed{m_{\mathrm{tot}}=N\mu\tanh x}.$$

(b) Interpretation and further calculation

$$\ln\cosh x=\tfrac12x^2+O(x^4)\Rightarrow\boxed{\Delta F=-\frac{N\mu^2B^2}{2k_BT}+O(B^4)}.$$$$m_{\mathrm{lin}}=\frac{N\mu^2B}{k_BT}\Rightarrow\boxed{\Delta F\simeq-\tfrac12m_{\mathrm{lin}}B}.$$

Physical check

The half factor reflects the field-dependent thermal mean moment; using a fixed-dipole energy for that mean would double the quadratic free-energy change.