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Entropy Change at a Demagnetization Broadened First Order Transition

  • Syed Q. A. Shah,
  • Balamurugan Balasubramanian,
  • Christian Binek

摘要

The literature on the thermodynamics of magnetism tends to obscure the nature of the magnetic field used in thermodynamic potentials and relations. It is often implied that H is a homogeneous internal magnetic field of a homogeneously magnetized body. In this framework, isothermal first order phase transitions are associated with a magnetization discontinuity, ΔM, at a critical field Hc with Hc = 0 for a ferromagnet and Hc \(\ne \) 0, e.g., for antiferromagnets with metamagnetic and spin flop transitions. The idealized description in terms of H is in sharp contrast to experiments performed on real samples where magnetic stray-fields are present. Here, the applied magnetic field, Ha, is experimentally controlled and the first order phase transition is continuous spreading out over a regime of coexisting phases. This work shows, with and without reference to a thermodynamic potential that, for the special case \(H={H}_{a}-{D}_{eff}M\) , the Maxwell relation \({\mu }_{0} {\left(\partial M/\partial T\right)}_{H}= {\left(\partial s/\partial H\right)}_{T}\) implies validity of \({\mu }_{0} {\left(\partial M/\partial T\right)}_{{H}_{a}}={\left(\partial s/\partial {H}_{a}\right)}_{T}\) . We show for magnetometry data of a Gd single crystal that a fictitious internal field \(H={H}_{a}-{D}_{eff}(T) M\) transforms M versu Ha isotherms to M vs H with a discontinuity at H = 0. This formal transformation comes at the price that Deff is temperatureTemperature dependent, rejecting the notion of a purely geometry dependent demagnetizing factor. The unjustified assumption of a homogeneous internal field in the mixed phase gives rise to the incorrect result \({\left(\partial s/\partial {H}_{a}\right)}_{T}=0\) . Magnetization isotherms in Gd are virtually hysteresis free and domain states of the mixed phase are equilibrium states with well-defined entropy quantifiable via \({\mu }_{0} {\left(\partial M/\partial T\right)}_{{H}_{a}}={\left(\partial s/\partial {H}_{a}\right)}_{T}\) .