= 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 has
$$
\boxed{\mathbf k_1+\mathbf k_2=\mathbf k_3+\mathbf G,\qquad \mathbf G\ne0}.
$$
The 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, $\kappa_{\rm ph}\simeq(C_{\rm ph}/V)v\ell/3$ 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 $C_{\rm ph}$ nearly constant and $\tau_U^{-1}\propto T$, giving $\kappa_{\rm ph}\propto T^{-1}$. These are regime estimates, not a universal law for every material or size.
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