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Local Equilibrium Description
                             6.2.1 Irreversible Fluxes and Thermodynamic Forces
                             The distribution function that locally deviates from its equilibrium value
                                                                   ,,
                                                                 (
                                                        ,,
                             is given by  f kx t,,(  ) =  f ( kx t) +  δf kx t)  . Once we have
                                                     0
                                 ,,
                             δf kx t)   it is possible to calculate all the current densities introduced
                               (
                             in 6.1.4 as moments of δf kx t,,(  )  . We insert the expansion of the distri-
                             bution function into the BTE (6.2) with the scattering term given by a
                             relaxation time approximation (6.23) and obtain
                               ∂  (  ,,   F                 (  ,,      δf kx t)
                                                                           ,,
                                                                         (
                                          ----∇ f kx t,,(
                                                                           τ
                                t ∂  f kx t) +  —  k  ) +  v∇ f kx t) =  – -------------------------  . (6.25)
                                                          x
                             The deviation is assumed to be small and thus to a first approximation its
                             contribution to the streaming motion term is neglected. We obtain
                                                                          (
                              ∂  f (  ,,  ----∇ f (  ,,  v∇ f (  ,,     δf kx t)
                                                                            ,,
                                          F
                                                                            τ
                               t ∂  0  kx t) +  —  k 0  kx t) +  x 0  kx t) =  – -------------------------  (6.26)
                             Inserting (6.24) in (6.26) we can immediately solve for  δf kx t,,(  )  . To
                             this end we calculate the effect of the streaming motion operator on the
                             Fermi distribution as given by (6.24), where the gradient in k-space gives
                                                        (
                                                     ∂ f E k() x t)
                                                            ,,
                                                       0
                                             ,,
                                       ∇ f ( kx t) =  -------------------------------------∇ E k()  .  (6.27)
                                        k 0
                                                                   k
                                                          ∂
                                                           E
                             The gradient in real space is
                                             ∂ f E k() x t,,(  )  E –  µ T∂  1 ∂ µ
                                                0
                               ∇ f ( kx t) =  -------------------------------------k T –  ------------------- –  ---------------  (6.28)
                                      ,,
                                 x 0
                                                                k T ∂
                                                           B
                                                  ∂
                                                                         k T x∂
                                                   E
                                                                   2
                                                                     x
                                                                          B
                                                                 B
                             The partial derivative with respect to time gives
                                     (
                                  ∂ f E k() x t)  ∂ f E k() x t,,(  )  E –  µ T  ∂ µ
                                          ,,
                                                                     ∂
                                    0
                                                    0
                                  ------------------------------------- =  ------------------------------------- –  ------------------- –  ------  (6.29)
                                        t ∂            ∂ E        T    t ∂  t ∂
                             Solving (6.26) for δf   we obtain
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