Page 241 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers
P. 241

Interacting Subsystems
                                                                 y
                             origin (a scattering centre) on the sphere, say  , makes at x
                                                                              with respect
                                                                            0
                             to the negative x-axis, then that phonon has a velocity component along
                             the x-axis of  vcos θ  . Like the temperature, the internal energy is also
                             assumed to vary only along the x-axis.  This means that the phonons
                             arrive at  x   with an internal energy density of ux –(  lcos θ)  . The aver-
                                     0                                0
                             age energy current in the x-direction is now the solid angle integral of the
                                                                         (
                             the product of velocity with energy density  v cos θ ux –  lcos θ)  . To
                                                                   p       0
                                                                     θ
                             obtain the differential solid angle as a function of   we take the ratio of
                             the differential surface area to the total spherical area, i.e.,
                                            ( 2πr) rsin θ)dθ  1
                                                 (
                                            ---------------------------------------- =  ---sin θdθ  (7.2)
                                                    2
                                                 4πr         2
                             Thus we have to evaluate
                                              π                  1
                                                       (
                                        q =    v cos θ ux –  lcos θ)---sin θdθ
                                         x    0 ∫  p    0        2
                                                                                   (7.3)
                                                 v p 1
                                                 2 ∫
                                                               d
                                               =  -----  – 1  µux –(  0  lµ) µ
                             Note the change of variables  µ =  cos θ   and the new integration limits.
                             We next expand the energy density   about the position x  . We drop the
                                                         u
                                                                           0
                                                                       ⁄
                             quadratic and higher terms to obtain  u =  u  +  ∂u ∂x  µl  , and insert
                                                                 x 0      x 0
                             this into (7.3) and evaluate
                                         v 1       ∂u    2         v l ∂u
                                                                      p
                                         ---
                                    q =  2 – ∫  1   u  x 0 µ +  ------  µ l d µ =  0 +  -------------  (7.4)
                                                           
                                     x
                                                    ∂x
                                                                     3 ∂x
                                                       x 0                x 0
                                                             (
                                                                            ⁄
                                                         ⁄
                                                                 ⁄
                             Remembering that  ∂u ∂x⁄  =  ( ∂u ∂T) ∂T ∂x) =  c ∂T ∂x  , we can
                                                                         v
                             relate the derived to the phenomenological equation
                                                ∂T    v l ∂T
                                                       p
                                          q =  – κ------- =  -------c -------   and hence  (7.5a)
                                                         v
                                                ∂x    3   ∂x
                                                  1       1 2
                                              κ =  ---v lc =  ---v τ c            (7.5b)
                                                             p p v
                                                       v
                                                    p
                                                  3       3
                238          Semiconductors for Micro and Nanosystem Technology
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