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A line defect in a crystal structure. Its Burgers vector measures the closure failure of a circuit around the line. Motion can produce permanent plasticity without simultaneously breaking every bond across a whole plane.

For line tangent $\mathbf{t}$, a screw dislocation has Burgers vector parallel to $\mathbf{t}$, an edge dislocation has it perpendicular, and a mixed dislocation has both components.

Motion of a screw segment from one allowed glide plane to another containing the same Burgers vector.

Conservative motion within a plane containing both the dislocation line and its Burgers vector. For a nonscrew line their cross product determines the plane normal; a screw line can belong to more than one such plane.

The energy criterion associated with Frederick Charles Frank states that, for dislocations of comparable length with approximately common line-energy factors, a reaction conserving Burgers vector is energetically favourable when the displayed inequality holds. The rule is an energy estimate, not a proof that the geometry or kinetic barriers permit the reaction. Edge/screw character, elastic anisotropy and changed lengths refine it.

A dislocation whose Burgers vector is not a full translation of the perfect crystal, so its glide can leave a planar stacking fault. Vector conservation and the actual slip plane constrain reactions and dissociation.

A Shockley partial dislocation in an FCC slip plane has Burgers vector of the family $a⟨112⟩/6$. Its vector must lie in the specified plane, and the two vectors must sum to the perfect dislocation vector.

For the first plane normal ${\mathbf{n}}_1=[\bar{1}11]$, one allowed dissociation isThe vectors add to $a[303]/6$, and ${\mathbf{n}}_1\cdot [211]=-2+1+1=0$, ${\mathbf{n}}_1\cdot [1\bar{1}2]=-1-1+2=0$.

For ${\mathbf{n}}_2=[111]$,Their sum is $a[\bar{3}30]/6$, and both have zero dot product with $[111]$. Interchanging leading and trailing labels does not change these vector identities.

Each perfect vector has squared magnitude $a^2/2$, whereas the sum of the two partial squared magnitudes is $a^2/3$. The line-energy estimate proportional to $G{b}^2$ therefore favors separation, opposed by the stacking-fault energy of the ribbon between the partials.

Use the two partials on each plane as obtained above. The four pairwise sums, in units of $a/6$, areThe shortest resultant gives the largest line-energy reduction in the $G{b}^2$ approximation:The initial summed squared magnitude is $a^2/3$ and the final value is $a^2/18$, a sixfold reduction in that part of the energy estimate.

The line of intersection of the two original planes is parallel toIts dot product with the resultant Burgers vector is zero, so the junction is edge in character. A plane containing both line and Burgers vector has normal $[100]$, rather than an FCC close-packed $\{111\}$ glide normal. The resulting stair-rod dislocation is therefore sessile for ordinary glide and can form a Lomer-Cottrell lock. It obstructs further dislocation motion and promotes work hardening.

This energetic comparison assumes comparable line lengths and elastic prefactors. Geometry, line character and stacking-fault geometry affect a complete junction-energy calculation, but the small resultant identifies the favorable candidate requested by the vector model.

The nickel-rich $\gamma$ matrix is a disordered substitutional solid solution on a face-centered cubic lattice. The ordered ${\gamma }^{\prime }$ phase has the L12 structure, approximately $\mathrm{N}{\mathrm{i}}_3\mathrm{Al}$: aluminium occupies cube corners and nickel occupies face centres. Its occupied-site geometry is often called ordered FCC, but its chemically labelled Bravais lattice is primitive cubic, with a four-atom basis. A face-centre translation exchanges chemical sublattices and is not a symmetry of the ordered crystal.

Each Al atom has twelve nearest-neighbour Ni atoms at Euclidean distance $a/\sqrt{2}$. Each Ni atom has twelve nearest neighbours too, but these are four Al and eight Ni. The chemical distinction matters for slip. In disordered $\gamma$, the shortest perfect Burgers vector isFor the ordered primitive cubic translation lattice, the shortest order-preserving perfect vector isThis does not imply that $⟨100⟩$ slip must dominate. A vector must lie in a favourable glide plane and have an accessible core structure and activation stress. Matrix dislocations commonly glide on $\{111\}⟨110⟩$ systems. A single $a⟨110⟩/2$ passage through ${\gamma }^{\prime }$ leaves an antiphase boundary; a second matching passage restores order, giving total vector $a⟨110⟩$. Such a superdislocation has a longer total vector than $a⟨100⟩$, yet its constituent dislocations can move on the favourable close-packed planes and interact with the matrix. Paired motion balances line and antiphase-boundary energies, so the shortest abstract lattice translation alone does not select the active slip system.^[ref]

The exponential in the Peierls stress makes a large core-width-to-Burgers vector ratio $w/b$ favourable for easy dislocation glide. Densely packed slip planes often have wider cores because the shear distortion can spread over more atomic rows, reducing the lattice-periodic barrier. Close-packed directions provide short lattice translations and hence small allowed Burgers vectors; their line energies, approximately proportional to $G{b}^2$, are also relatively low. These effects favour dense planes and directions, but the relation between packing and core width is a physical argument rather than a consequence of substituting arbitrary independent $w,b$ into the formula. Body-centred cubic metals have densely packed directions without a truly close-packed plane, and their screw-core behaviour can require additional qualifications.

Burgers vector conservation gives two candidate reactions, depending on the relative signs of the approaching segments:Initially $b_1^2+b_2^2=3a^2/2$. The sum product has $b^2=a^2$, while the difference product has $b^2=2a^2$. Thus the $a[001]$ product is favoured by Frank's rule; the $a[1\bar{1}0]$ product is disfavoured. The result is not annihilation, because the two supplied vectors are not opposite. The same ordering survives a simple character correction: with $\nu =0.30$, line energy proportional to $G{b}^2[{\cos }^2\chi +{\sin }^2\chi /(1-\nu )]$ gives the two parent segments a combined coefficient $1.786a^2$, whereas any character of the sum product has coefficient between $a^2$ and $1.429a^2$ and any character of the difference product at least $2a^2$, comparing equal lengths and common logarithmic factors. A favourable dislocation junction can obstruct further glide and harden the crystal. Its actual mobility requires the junction line and allowed glide plane: the second screw can occupy several $\{110\}$ planes containing $[\bar{1}11]$, so geometry is not unique. For instance, its $(101)$ plane intersects $(10\bar{1})$ along $[010]$; an $a[001]$ junction on that line would glide geometrically on $(100)$, not either original plane. Dislocation cross-slip, junction unzipping or bypass can compete under load. An energy inequality alone is not a universal proof of a permanently sessile product.

The cosine of the angle between Burgers vector and line direction is $1/3$, so the angle is $70.53^{\circ }$: the dislocation is mixed. Both vectors must lie in its dislocation glide plane. Their cross product is proportional toso the plane is $(10\bar{1})$, a member of $\{110\}$. The screw component magnitude is $a/(2\sqrt{3})$ and the edge component magnitude is $a\sqrt{2/3}$.

Irradiation softening occurs when irradiation removes pre-existing strengthening obstacles faster than it creates new ones. It is particularly plausible in an initially hardened alloy, rather than an initially soft annealed reference.

In a material strengthened by cold working, the starting dislocation density is high. Irradiation-generated point defects can assist dislocation climb, rearrangement and annihilation, reducing the contribution of work hardening. A schematic estimate iswhere $\mu$ is the shear modulus, $b$ the magnitude of the Burgers vector, ${\rho }_d$ the dislocation density, and the last two terms represent precipitate and irradiation-obstacle strengthening. It is an illustrative decomposition, not an exact additive law for every alloy. A sufficiently large decrease in the pre-existing contributions can exceed the increase in $\Delta {\sigma }_{\mathrm{irr}}$.

In a material strengthened by precipitation hardening, ballistic mixing can dissolve or disorder the original obstacles. Radiation-enhanced diffusion can change their composition or accelerate coarsening; increasing obstacle spacing can reduce strength. In an ordered alloy, irradiation-induced loss of order can also remove a strengthening contribution. None of these outcomes is inevitable: irradiation can instead create new strengthening precipitates.

The required conditions therefore include an appropriate initial microstructure, enough displacement dose to modify it, and a temperature and displacement dose rate at which obstacle removal or recovery can compete with new defect accumulation. Compare with an unirradiated specimen held for the same time at the same temperature to separate ordinary thermal aging from an irradiation effect.

A drop in yield strength or hardness need not restore ductility: cavities, helium embrittlement or localization of dislocation glide can coexist with softening. Conversely, an initially annealed steel may show strong irradiation hardening under the same exposure conditions.

An energetic neutron can collide with a lattice atom and create a primary knock-on atom. That recoil transfers energy to other atoms in a displacement cascade. Some atoms leave their normal lattice sites, creating Frenkel pairs: a vacancy and an extra atom at an interstitial position.

Many nearby defects recombine during the cascade and its thermal relaxation. The surviving point defects are therefore fewer than a simple count of all displacement events. Some defects are already clustered within the cascade; others migrate and aggregate later.

A collection of vacancies can collapse into a platelet bounded by a vacancy-type dislocation loop. An aggregation of self-interstitial atoms can create an extra atomic platelet bounded by an interstitial-type dislocation loop. The boundary is a line defect because a circuit around it has a nonzero Burgers vector. The allowed Burgers vectors and habit planes depend on the material's crystal structure and fault energies.

Loops can grow by absorbing mobile point defects, shrink by emitting or absorbing the opposite defect type, and interact with existing dislocations. Some remain faulted, while suitable loops can unfault or become part of a larger dislocation network. Repeated cascades and subsequent defect transport therefore create a population of extended defects even in material with a low initial dislocation density.

This describes the main displacement-damage mechanism. Ionizing energy deposited in a material does not automatically create the same defect population as a neutron recoil cascade; recoil spectrum, displacement thresholds and material response matter.

A lattice-friction stress for dislocation glide, associated with Rudolf Peierls and Frank Nabarro. A simple model gives ${\tau }_P\sim 3G\exp (-2\pi w/b)$, where $w$ is core width and $b$ the Burgers vector magnitude. Spreading slip over many atomic rows reduces the periodic energy barrier; narrower cores are harder to move in this model.

An FCC partial dislocation with Burgers vector $a⟨112⟩/6$ lying in a close-packed $\{111\}$ plane. Two suitable Shockley vectors can sum to an FCC perfect vector $a⟨110⟩/2$.

A sessile partial junction associated with intersecting stacking faults. A typical FCC resultant has vector $a⟨110⟩/6$. Its Burgers vector and junction line can define a plane outside the ordinary $\{111\}$ glide family.