Paramagnetism — Question 4

Section 32-9 · Physics pages 5338–5339

Question 4 Ideal adiabatic demagnetization preserves the spin entropy parameter For N independent two-state moments, the spin entropy is S = NkB[ln(2 cosh x) −x tanh x], where x = µB/(kBT) > 0. The sample is thermally isolated and the field is lowered reversibly. Neglect all non-spin heat capacity and interactions. B T reversible cooling Schematic; labels, not drawing lengths, specify the data. (a) Show that constant entropy requires constant x, hence a relation between B and T. (b) If initially B=2.0 T and T=1.5 K, find T at B=0.20 T. Explain why the ideal formula alone cannot establish an attainable zero-temperature state.
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Question 4 - Solution Ideal adiabatic demagnetization preserves the spin entropy parameter B T reversible cooling (a) Method and calculation dS dx = −NkBx sech2 x < 0 (x > 0). dS = 0 ⇒dx = 0 ⇒B/T = constant . (b) Interpretation and further calculation Tf = TiBf/Bi = (1.5)(0.20/2.0) = 0.150 K . B →0 ⇒T →0 only within the ideal spin model. Physical check Residual interactions and non-spin heat capacities invalidate the ideal extrapolation at sufficiently small fields.

Question 4

Ideal adiabatic demagnetization preserves the spin entropy parameter

For N independent two-state moments, the spin entropy is S=NkB[ln⁡(2cosh⁡x)−xtanh⁡x]S=Nk_B[\ln(2\cosh x)-x\tanh x], where x=μB/(kBT)>0x=\mu B/(k_BT)>0. The sample is thermally isolated and the field is lowered reversibly. Neglect all non-spin heat capacity and interactions.

Diagram shown in the original SVG worksheet.

  1. Show that constant entropy requires constant x, hence a relation between B and T.

  2. If initially B=2.0 T and T=1.5 K, find T at B=0.20 T. Explain why the ideal formula alone cannot establish an attainable zero-temperature state.

Question 4 - Solution

Ideal adiabatic demagnetization preserves the spin entropy parameter

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dSdx=−NkBxsech⁡2x<0(x>0).\frac{dS}{dx}=-Nk_Bx\operatorname{sech}^2x<0\quad(x>0).dS=0⇒dx=0⇒B/T=constant.dS=0\Rightarrow dx=0\Rightarrow\boxed{B/T=\text{constant}}.

(b) Interpretation and further calculation

Tf=TiBf/Bi=(1.5)(0.20/2.0)=0.150K.\boxed{T_f=T_iB_f/B_i=(1.5)(0.20/2.0)=0.150\,\mathrm K}.B→0⇒T→0only within the ideal spin model.B\to0\Rightarrow T\to0\ \text{only within the ideal spin model}.

Physical check

Residual interactions and non-spin heat capacities invalidate the ideal extrapolation at sufficiently small fields.