3.5.5.4 Energy Flux Equation

The energy flux equation is obtained analogously to the particle current equation starting from equation (3.30) with the weight $\ensuremath{\ensuremath{\mathitbf{X}}}= \ensuremath{{\tau}}\ensuremath{\mathcal{E}}\ensuremath{\ensuremath{\mathitbf{p}}}$

$$\underbrace{\vphantom{\frac{1}{\hbar}}\ensuremath{\ensuremath{\en... ...\langle \ensuremath{\mathcal{E}}\ensuremath{\ensuremath{\mathitbf{p}}} \rangle}$$ (3.66)

The estimation of all tensor valued quantities with their traces as outlined in (3.39) as well as the expansion of the kinetic energy $\ensuremath{\mathcal{E}}$ using the product ansatz (3.40) are applied on equation (3.66). Part (i) is straight-forward

$$\ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\en... ...h{{\tau}}\ensuremath{\mathcal{E}}^2 \ensuremath{\gamma} \rangle \! \rangle} \,.$$ (3.67)

The Poisson bracket in part (ii) has to be expanded using (A.10)

$$\ensuremath{\langle \! \langle \ensuremath{\{\ensuremath{{\tau}}\... ...{\{\ensuremath{\mathcal{E}},\ensuremath{\mathcal{E}}\}} \rangle \! \rangle} \,.$$ (3.68)

The second term vanishes because of (A.8) while the first term is treated according to (3.45) and yields

$$- \ensuremath{\langle \! \langle \ensuremath{\mathcal{E}}^2 \ensu... ...math{\theta_\ensuremath{\ensuremath{\mathitbf{r}}}^\ensuremath{\mathcal{E}}}\,.$$ (3.69)

Application of identity (B.4) to part (iii) and assembly of all three terms results in

$$\ensuremath{\langle \! \langle \ensuremath{\mathcal{E}}\ensuremat... ...emath{{\tau}}\ensuremath{\mathcal{E}}^2 \ensuremath{\gamma} \rangle \! \rangle}$$ (3.70)

$$+ \ensuremath{\langle \! \langle \ensuremath{\mathcal{E}}^2 \ensu... ...uremath{\theta_\ensuremath{\ensuremath{\mathitbf{r}}}^\ensuremath{\mathcal{E}}}$$

$$+ \ensuremath{\mathrm{s}_\nu}\mathrm{q}\ensuremath{\langle \! \la... ...suremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{\tilde{\varphi}} \,.$$

Next, the power-law ansatz is introduced for the relaxation time $\ensuremath{{\tau}}$

$$\ensuremath{\langle \! \langle \ensuremath{\mathcal{E}}\ensuremat... ...al{E}}^{{\ensuremath{r_\nu}}+2} \ensuremath{\gamma} \rangle \! \rangle} \right)$$ (3.71)

$$+ \ensuremath{\tau_0}\left( k_\ensuremath{\mathrm{B}}\ensuremath{... ..._{\mathrm{L}}}} \right)^{{\ensuremath{r_\nu}}} \Bigr\rangle \! \! \Bigr\rangle}$$

$$+ \frac{1}{3} \ensuremath{\tau_0}\left( k_\ensuremath{\mathrm{B}}... ...uremath{\theta_\ensuremath{\ensuremath{\mathitbf{r}}}^\ensuremath{\mathcal{E}}}$$

$$+ \ensuremath{\mathrm{s}_\nu}\mathrm{q}\ensuremath{\tau_0}\left( ... ...suremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{\tilde{\varphi}} \,.$$

For the next steps, parabolic bands and a heated, displaced Maxwellian (3.51) are assumed. The statistical average in part (i) is carried out similarly to (3.53), normalized using (3.54), and reads

$$\ensuremath{\langle \! \langle \ensuremath{\mathcal{E}}^{{\ensure... ...rac{7}{2}}\ensuremath{\Gamma \left({\ensuremath{r_\nu}}+\frac{7}{2}\right)} \,.$$ (3.72)

Thus, part (i) becomes

$$\ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\le... ...m{L}}}\right)^2 \ensuremath{\ensuremath{\mu}_\nu}\frac{m^*}{\mathrm{q}} \right)$$ (3.73)

using the mobility definition (3.55). With the assumption of a $\ensuremath{\ensuremath{\mathitbf{k}}}$ -independent ${\ensuremath{r_\nu}}$ and basic arithmetic operations, the first term of part (ii) is expressed in terms of an effective mass gradient $\ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}m^*$ . The second term is split into a $\ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{T_{\mathrm{L}}}$ and a $\ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}m^*$ term, while the $\ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}m^*$ -term cancels with the third term. Thus, part (ii) finally reads

$$-\frac{3}{2} \frac{m^*}{\mathrm{q}} \ensuremath{\ensuremath{\mu}_... ...{\!r}}}}}\left( k_\ensuremath{\mathrm{B}}\ensuremath{T_{\mathrm{L}}}\right) \,.$$ (3.74)

While the first term of part (iii) is handled straight-forwardly with the statistical average (3.53), the derivative in the second term has to be expressed before processing the statistical average. While the first term contributes with a factor of $\frac{3}{2}$ , the second's contribution is ${\ensuremath{r_\nu}}+ 1$ . Thus, the sum reads

$$\left( {\ensuremath{r_\nu}}+ \frac{5}{2} \right) \ensuremath{\ens... ...nsuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{\tilde{\varphi}}\,.$$ (3.75)

With a definition of the energy flux analogously to (3.58)

$$\ensuremath{\ensuremath{\mathitbf{j}}_\nu^\mathrm{u}}= \ensuremat... ...suremath{\mathcal{E}}\ensuremath{\ensuremath{\mathitbf{p}}} \rangle \! \rangle}$$ (3.76)

the final energy flux for parabolic bands and a heated, displaced Maxwellian reads

$$\ensuremath{\ensuremath{\mathitbf{j}}_\nu^\mathrm{u}}= - \frac{1}... ...{\nu}}\right)^2 \ensuremath{\ensuremath{\mu}_\nu}\frac{m^*}{\mathrm{q}} \right)$$ (3.77)

$$+ \frac{3}{2} \frac{1}{\mathrm{q}} \ensuremath{\ensuremath{\mu}_\... ...bla_{\!r}}}}}\left( k_\ensuremath{\mathrm{B}}\ensuremath{T_{\mathrm{L}}}\right)$$

$$- \ensuremath{\mathrm{s}_\nu}\left( {\ensuremath{r_\nu}}+ \frac{5... ...suremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{\tilde{\varphi}} \,.$$

Introducing the electrochemical potential as defined in Eqs. (3.60) and (3.62) yields

$$\ensuremath{\ensuremath{\mathitbf{j}}_\nu^\mathrm{u}}= - \ensurem... ...nsuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{T_\ensuremath{\nu}}$$ (3.78)

$$\quad - \frac{k_\ensuremath{\mathrm{B}}^2}{\mathrm{q}} \ensuremat... ...h{\nu}}^2 \ensuremath{\ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ln m^*$$

$$\quad - {\ensuremath{r_\nu}}\frac{k_\ensuremath{\mathrm{B}}^2 \en... ...ensuremath{\ensuremath{\mathitbf{\nabla_{\!r}}}}}\ensuremath{T_{\mathrm{L}}}\,.$$

M. Wagner: Simulation of Thermoelectric Devices