2.3.2.3 Isotropic Distribution Function

Eqns. (2.76) to (2.78) all contain a statistical average of a symmetric tensor (see Appendix A) of the form $\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{p}}} \ensuremath{\otimes}\ensuremath{\boldsymbol{\mathrm{p}}} \rangle}$ which will be evaluated in this section. The distribution function can be decomposed into a symmetric and an anti-symmetric part

$$f(\ensuremath{\boldsymbol{\mathrm{k}}}) = f_\mathrm{S}(\ensuremat... ...ldsymbol{\mathrm{k}}}) + f_\mathrm{A}(\ensuremath{\boldsymbol{\mathrm{k}}}) \ .$$ (2.79)

In this section we assume that the symmetric part is isotropic

$$f(\ensuremath{\boldsymbol{\mathrm{k}}}) = f_\mathrm{S}(\vert\ensu... ...bol{\mathrm{k}}}\vert) + f_\mathrm{A}(\ensuremath{\boldsymbol{\mathrm{k}}}) \ .$$ (2.80)

This is a special case of the diffusion approximation [21, p.49] which will be explained in more detail for the MAXWELL distribution in Section 2.3.3.1 on page . By using this assumption the statistical average of the tensor can be written as

$$\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{p}}} \ensurem... ...\ensuremath{\boldsymbol{\mathrm{k}}}\vert) \,\, \ensuremath{\mathrm{d}}^3 k}\ .$$ (2.81)

For symmetry reasons all elements outside the trace vanish. For instance, the element

$$\ensuremath{\langle k_x \, k_y \rangle}= \ensuremath{\iiint\limit... ...ath{\mathrm{d}}k_x \, \ensuremath{\mathrm{d}}k_y \, \ensuremath{\mathrm{d}}k_z}$$ (2.82)

evaluates to zero because of the integral

$$\int_{-\infty}^{\infty} k_x \, f_\mathrm{S} \left( \sqrt{k_x^2 + k_y^2 + k_z^2} \right) \,\, \mathrm{d} k_x = 0 \ .$$ (2.83)

Since the distribution function is assumed to be isotropic, the integrals determining the elements of the trace all evaluate to a common value $J$

$$\ensuremath{\langle k_l \, k_l \rangle}= \ensuremath{\iiint\limit... ...ensuremath{\mathrm{d}}k_z}= J \ , \textcolor{lightgrey}{.......}l = x, y, z \ .$$ (2.84)

The value of $J$ can be evaluated by the simple transformation

$$\ensuremath{\langle k_x \, k_x \rangle}= \ensuremath{\langle k_y \, k_y \rangle}= \ensuremath{\langle k_z \, k_z \rangle} = J \ ,$$ (2.85)

$$\ensuremath{\langle k_x^2 \rangle}+ \ensuremath{\langle k_y^2 \ra... ...math{\langle k_z^2 \rangle}= \ensuremath{\langle k_x^2 + k_y^2 + k_z^2 \rangle} = 3 \, J \ ,$$ (2.86)

$$\frac{\ensuremath{\langle k^2 \rangle}}{3} = J \ .$$ (2.87)

Therefore, the statistical averages of the tensors are diagonal with all diagonal elements being equal:

$$\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{p}}} \ensuremath{\otimes}\ensuremath{\boldsymbol{\mathrm{p}}} \rangle} = \frac{{\ensuremath{\langle p^2 \rangle}}}{3} \, \ensuremath{\widetilde{\delta}}\ ,$$ (2.88)

$$\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{p}}} \ensuremath{\otimes}\ensuremath{\boldsymbol{\mathrm{p}}} \, \mathcal{E} \rangle} = \frac{{\ensuremath{\langle \mathcal{E}\, p^2 \rangle}}}{3} \, \ensuremath{\widetilde{\delta}}\ .$$ (2.89)

By inserting eqns. (2.88) and (2.89) into eqns. (2.76) to (2.78) one gets

$$\ensuremath{\boldsymbol{\mathrm{\phi}}}_1: \textcolor{lightgrey}{.......} \frac{2}{3} \, \ensuremath{\ensuremath{\ensuremath{\boldsymbol{\mathrm{\nabla}}}}\, \ensuremath{\langle \mathcal{E} \rangle}} + q \, \ensuremath{\boldsymbol{\mathrm{E}}}\, \ensuremath{\langle 1 \rangle} = - m \, \frac{\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{v}}} \rangle}}{\tau_m} \ ,$$ (2.90)

$$\ensuremath{\boldsymbol{\mathrm{\phi}}}_3: \textcolor{lightgrey}{.......} \frac{1}{3} \, \ensuremath{\ensuremath{\ensuremath{\boldsymbol{\mathrm{\nabla}}}}\, \ensuremath{\langle v^2 \, \mathcal{E} \rangle}} + \frac{5}{3} \, \frac{\mathrm{q}}{m} \, \ensuremath{\boldsymbol{\mathrm{E}}}\, \ensuremath{\langle \mathcal{E} \rangle} = - \frac{\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{v}}} \, \mathcal{E} \rangle}}{\tau_S} \ ,$$ (2.91)

$$\ensuremath{\boldsymbol{\mathrm{\phi}}}_5: \textcolor{lightgrey}{.......} \frac{1}{3} \, \ensuremath{\ensuremath{\ensuremath{\boldsymbol{\mathrm{\nabla}}}}\, \ensuremath{\langle v^2 \, v^2 \, \mathcal{E} \rangle}} + \frac{7}{3} \, \frac{\mathrm{q}}{m} \, \ensuremath{\boldsymbol{\mathrm{E}}}\, \ensuremath{\langle v^2 \, \mathcal{E} \rangle} = - \frac{\ensuremath{\langle \ensuremath{\boldsymbol{\mathrm{v}}} \, v^2 \, \mathcal{E} \rangle}}{\tau_K} \ .$$ (2.92)

Note that the divergences of the tensors simplify to gradients of scalars.

M. Gritsch: Numerical Modeling of Silicon-on-Insulator MOSFETs PDF