The fluorite structure has Ce on the corner and face-centre sites of a face-centered cubic lattice. Oxygen occupies all eight positions $(x,y,z)$ with each coordinate independently $1/4$ or $3/4$. In projection, there are four oxygen positions $(1/4,1/4),(1/4,3/4),(3/4,1/4),(3/4,3/4)$, each representing two heights. Accounting for shared boundary sites gives four Ce and eight O per conventional unit cell.

The Weiss zone law gives $u+w=0$ and $u-v=0$, so a common direction is $[11\bar{1}]$. Its physical vector is $(a,a,-c)$, while $[100]$ is parallel to $(a,0,0)$. Their dot product therefore givesTaking a Euclidean angle directly between the index triples would incorrectly treat the tetragonal axes as equal in length. Reversing the chosen common direction gives the supplementary oriented angle.

In the tetragonal crystal system, physical squared distances use $a^2(\Delta x^2+\Delta y^2)+c^2\Delta z^2$. Two Ti–O distances from the corner Ti have length $\sqrt{2}ua$; the other four have length $\sqrt{2a^2(1/2-u{)}^2+c^2/4}$. Equating both to $r$ givesThus, within the equal-bond model,The rutile structure remains tetragonal and its octahedral angles remain distorted; these numbers are not a claim about measured ambient-pressure lattice parameters.

Each Ti has coordination number six, with oxygen neighbours forming a distorted octahedron. Each oxygen has coordination number three, with titanium neighbours forming a planar triangle. For oxygen at $(u,u,0)$, the nearest Ti positions can be taken as $(0,0,0)$ and $(1/2,1/2,\pm 1/2)$; all three lie in the plane $x=y$. The triangle is not generally equilateral. Equal Ti–O bond lengths in the following calculation do not make every angle ideal.

Reduce the supplied fractional coordinates modulo one. The rutile structure has Ti at $(0,0)$ with height $z=0$ and at $(1/2,1/2)$ with height $z=1/2$. The oxygen projections are $(0.30,0.30),(0.70,0.70)$ at $z=0$, and $(0.80,0.20),(0.20,0.80)$ at $z=1/2$. Include equivalent corner sites to show the boundary of a complete unit cell; the cell edges are $a$ along $[100]$ and $[010]$.

TODO — user-deferred biology. This biological question is intentionally left unsolved under the current scope.

TODO — user-deferred biology. This biological question is intentionally left unsolved under the current scope.

TODO — user-deferred biology. This biological question is intentionally left unsolved under the current scope.

TODO — user-deferred biology. This biological question is intentionally left unsolved under the current scope.

TODO — user-deferred biology. This biological question is intentionally left unsolved under the current scope.

Irradiation both creates dislocation loops and supplies mobile point defects that allow existing dislocations to climb, rearrange and annihilate. At low initial dislocation density, net production can dominate; at high initial density, sink absorption and recovery can dominate. A simple illustrative rate equation isThus both initial states approach ${\rho }_{\mathrm{ss}}=A/B$ in this fixed-condition model. Here $D$ is irradiation dose, not a diffusion coefficient. The asymptote reflects a dynamic balance, not the cessation of damage. It need not be the same for different alloys, temperatures, dose rates or applied stresses.

The annealed sample initially has relatively few obstacles to dislocation glide. After its elastic response it yields, undergoes appreciable plastic deformation and develops strain hardening. Irradiation produces dislocation loops and defect clusters that obstruct gliding dislocations, usually raising the yield strength. In a simple dispersed-obstacle model the strengthening scales as $\Delta \sigma \sim M\alpha Gb\sqrt{Nd}$, with obstacle density $N$ and diameter $d$.

Irradiated material often has less uniform plastic elongation and weaker subsequent hardening because slip localizes into channels that have cleared irradiation obstacles. Earlier necking or fracture can then reduce ductility. The elastic slope is much less affected than the yield stress in this schematic comparison; the exact strength and elongation depend on dose and temperature.

Energetic neutron collisions create displacement cascades and Frenkel pairs: displaced atoms become self-interstitials, leaving vacancies. During migration, some opposite defects recombine, while others cluster or reach sinks. Interstitials can condense into extra atomic planes bounded by dislocation loops; vacancy clustering can instead give missing-plane loops. Their subsequent growth is driven by defect absorption and dislocation climb.

Randomly oriented loops do not select a macroscopic shear direction in an initially isotropic polycrystal. That does not prove that every external dimension remains unchanged: interstitial and vacancy loops have opposite relaxation-volume contributions, and an imbalance can change the volume. Macroscopic void swelling requires the additional vacancy-cavity mechanism; it must not be identified automatically with the mere presence of loops. Texture, stress-biased absorption and unequal loop populations can also give anisotropic dimensional change.

The number and spacing of primary slip planes remain fixed, so the length projected on their normal is invariant: $l_0\cos {\varphi }_0=l_1\cos {\varphi }_1$. The initial and boundary loading directions are $[514]$ and $[211]$, respectively. With normal $[1\bar{1}1]$,ThereforeThis is engineering plastic strain in the ideal geometric model; its corresponding logarithmic strain is $\ln (4/\sqrt{7})$. Elastic extension and changes in slip resistance are neglected, as required by this construction.