 Solution

A three-phonon normal process conserves crystal wavevector without a reciprocal-lattice shift, ${\mathbf{k}}_1+{\mathbf{k}}_2={\mathbf{k}}_3$. An Umklapp process instead hasThe crystal can absorb the reciprocal-lattice momentum; after reduction to the first Brillouin zone the phonon drift transporting heat is relaxed. A normal process redistributes phonons while preserving their total crystal momentum, so it is not by itself the same resistive mechanism. energy must also be conserved in either process.

At low temperature, large-wavevector phonons are rare, suppressing Umklapp scattering. Boundary and defect scattering then limit the mean free path; with a nearly constant boundary-limited path, ${\kappa }_{\mathrm{ph}}\simeq (C_{\mathrm{ph}}/V)v\ell /3$ grows as $T^3$. As temperature increases, heat capacity rises but Umklapp scattering shortens $\ell$, often producing a electrical conductivity maximum. At sufficiently high temperature a simple anharmonic model has $C_{\mathrm{ph}}$ nearly constant and ${\tau }_U^{-1}\propto T$, giving ${\kappa }_{\mathrm{ph}}\propto T^{-1}$. These are regime estimates, not a universal law for every material or size.