MAS.020UF Introduction to Solid State Physics
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MAS.020UF Introduction to Solid State Physics | |
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Phonon contribution to thermal conductivityPhonons transport heat from hotter regions to cooler regions. In thermal equilibrium, the states $\vec{k}$ and $-\vec{k}$ are equally occupied and the heat current is zero. If the system is pushed out of thermal equilibrium by applying a temperature gradient, the occupation of states $\vec{k}$ and $-\vec{k}$ can be different and the heat current density can be calculated by summing over the contributions of all phonon modes. For low temperatures, the phonons freeze out and the thermal conductivity goes to zero. Near $T=0$, the contribution of the thermal conductivity due to phonons is proportional to $T^3$. At high temperatures, there are many phonons and phonon-phonon scattering causes the thermal conductivity to fall like $1/T$. , \[\begin{equation} K\approx \frac{2\pi^2\tau}{15}\frac{k_B^4T^3}{\hbar^3c} \qquad T < < T_D. \end{equation}\]At high temperatures, this calculation results in a constant thermal conductivity but in usually the thermal conductivity falls like $1/T$. The problem is that a single relaxation time was assumed for all temperatures. At higher temperatures the concentration of phonons increases like $T$ and the relaxation times becomes shorter like $1/T$ due to phonon-phonon scattering. This results in a thermal conductivity of the form $K \sim 1/T$. A very simple model for the phonon dispersion of an optical branch is that the energy is a constant. This results in a group velocity that is zero and no thermal conductivity due optical phonons. Below the thermal conductivity of some elements can be plotted below .
The thermal conductivity of silicon.[1]
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