Page 397 - Electrical Properties of Materials
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Thermodynamical treatment                         379

               In view of eqn (14.16) the Gibbs free energy of the two phases must be
            equal at the critical magnetic field, H c , that is,

                                    G s (H c )= G n (H c ).          (14.21)

            Substituting eqns (14.19) and (14.20) into eqn (14.21) we get
                                                  2
                                             1
                                  G n = G s (0) + μ 0 H V.           (14.22)
                                             2    c
               Comparison of eqns (14.19) and (14.22) clearly shows that at a given
            temperature (below the critical one) the conditions are more favourable for the
            superconducting phase than for the normal phase, provided that the magnetic
            field is below the critical field. There are three cases:

            (i)         If H < H c  then G n > G s (H).              (14.23)
            (ii)        If H > H c  then G n < G s (H).              (14.24)
            (iii)        If H = H c  then G n = G s (H c ).          (14.25)

               Now our substance will prefer the phase for which the Gibbs free energy is
            smaller; that is in case (i) it will be in the superconducting phase, in case (ii) in
            the normal phase, and in case (iii) just in the process of transition.
               If the transition takes place at temperature, T +dT, and magnetic field, H c +
            d H c , then it must still be valid that
                                   G s +dG s = G n +dG n ,           (14.26)

            whence

                                       dG s =dG n .                  (14.27)

            This, using eqn (14.15), leads to

                         –S s dT – μ 0 VM s d H c =–S n dT – μ 0 VM n d H c .  (14.28)

            But, as suggested before,
                                           and  M n = 0,             (14.29)
                                 M s =–H c
            and this reduces eqn (14.28) to

                                                 d H c
                                  S n – S s =–μ 0 VH c  .            (14.30)
                                                  dT
               The latent heat of transition may be written in the form,

                                      L = T(S n – S s ),             (14.31)

            and with the aid of eqn (14.30) this may be expressed as

                                               d H c
                                    L =–μ 0 TVH c  .                 (14.32)
                                                dT
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