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Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.

For a free surface at $x_3=0$ with solid occupying $x_3>0$, let $(q_0,a_0)$ be a unit-amplitude incident upgoing mode. The reflected field consists of the three downgoing propagating or depth-decaying modes:Frequency and horizontal slowness are common to all modes, by translation invariance of the surface. Zero surface traction imposes the three equationsHence the free-surface anisotropic reflection amplitudes arewhenever $B$ is invertible. This specifies mode conversion as well as reflection. If $B$ is singular, a nonzero outgoing homogeneous surface mode exists; these are surface-wave poles, including Rayleigh waves in the isotropic case. At such parameters one must use the appropriate limiting radiation solution and check solvability instead of assuming a unique finite inverse.

Put $\mu ={\mu }^0$, $\lambda ={\kappa }^0-2\mu /3$, and $L=\lambda +2\mu ={\kappa }^0+4\mu /3$. Assume positive bulk modulus and shear modulus. For a unit vector $\xi$, direct contraction of the isotropic elastic stiffness tensor gives the acoustic tensor and its inverse:These are its transverse and longitudinal orthogonal projections. Symmetrizing the definition of the directional elastic strain Green operator consequently givesSubstitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as distributions. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ givesThe first identity holds as a locally integrable distribution. Since the second derivative of $|s|$ is $2\delta (s)$, taking two derivatives in the first identity and four in the second givesThus an explicit complete isotropic elastic strain Green kernel isAll derivatives here are distributional derivatives, which retain the contribution at the origin.

For the requested contraction, ${\tilde{\Gamma }}_{ijkk}={\xi }_i{\xi }_j/L$, or equivalently trace the complete kernel and use $\Delta r=2/r$. ThereforeMore explicitly, its distribution form iswhere the Cauchy principal value uses spherical exclusions about the origin. The coefficient of the Dirac delta can be checked by integrating by parts over a punctured ball: ${\partial }_i{\partial }_j(r^{-1})$ has contact term $-4\pi {\delta }_{ij}\delta /3$. Taking the trace, the principal-value part has zero trace and $\Delta (r^{-1})=-4\pi \delta$, soKeeping only the ordinary kernel away from the origin would incorrectly give zero for this double trace.

Now extend the elastic polarization by zero outside $V$. Since the shear modulus is unchanged, the stiffness contrast acts only on the spherical strain:Hence $\tau =(\kappa -{\kappa }^0)\mathrm{tr}\epsilon$ is its scalar coefficient. Relative to the same external loading, the elastic strain Green operator gives $\epsilon ={\epsilon }^0-\Gamma \ast (\tau I)$. Taking the trace and using the contact identity reduces this nonlocal equation toSolving the scalar equation gives the constant-shear bulk-modulus inclusion formulaUsing the single contraction instead of the double trace givesFinally eliminate the actual trace using the scalar formula:The reference trace ${\epsilon }_{kk}^0$ in this last formula is essential and is what the original PDF prints; the converted TeX loses its superscript. The second derivatives are interpreted distributionally, or with their principal-value and contact terms explicitly included. With ordinary regularity they give the corresponding pointwise formula at regular interior points.

Write $B$ for the filament bending modulus and $C$ for the filament torsional stiffness, so that the energy per unit length is $(B{\kappa }^2+C{\Omega }^2)/2$. For one circular filament,where $I_1$ is the second moment of area, $J_1$ the polar moment, $G_s$ the shear modulus, and $\nu$ Poisson's ratio. Numerical twisting constants cannot be fixed by Young's modulus alone without specifying $\nu$.

For a filled sliding bundle, $N\sim (R/r{)}^2$. Filaments share the centerline curvature but slide instead of accumulating the additional axial strains of a bonded cross-section. Their bending energies add: $B\sim N{B}_1\sim E{r}^2{R}^2$. If end constraints impose the same material twist on every filament, their torsional energies also add: $C\sim G_s{r}^2{R}^2$.

For a bonded, one-filament-thick circular tube, the wall thickness $h$ is of order $r$. Its moments are $I\sim \pi R^{3}h$ and $J_p\sim 2\pi R^{3}h$, giving $B\sim Er{R}^3$ and $C\sim G_{s}r{R}^3$. For a bonded filled rod, $I\sim \pi R^4/4$ and $J_p\sim \pi R^4/2$, giving $B\sim E{R}^4$ and $C\sim G_s{R}^4$. Packing fractions and the convention for tube radius affect constants, not these powers.

There is an important qualification to bending and torsion of filament bundles: free axial sliding does not specify whether each filament can also spin freely. If both spin and slip are unconstrained, the common-material-twist calculation is not forced by the geometry. A helix of small twist rate has centerline curvature of order $\rho {\Omega }^2$, so its bending energy starts at ${\Omega }^4$, not ${\Omega }^2$. A nonzero linear collective torsional stiffness then requires additional end, contact, or spin constraints.

With vertical shear modulus ${\rho }^{\prime }{\beta }_z^{\prime 2}$ and horizontal shear modulus ${\rho }^{\prime }{\beta }_x^{\prime 2}$, force balance therefore gives the anisotropic shear-horizontal wave equationFor $u_y=Y(z)e^{i(kx-\omega t)}$, the layer vertical wavenumber and substrate decay exponent areThe free top still gives $Y=A\cos (qz)$; the substrate still has $Y=B{e}^{-\kappa (z-h)}$. Welded displacement continuity and the now anisotropic traction continuity giveHence the anisotropic Love-wave dispersion relation isEquivalently, writing $c=\omega /k$,This replaces the isotropic speed in the layer equation by separate horizontal and vertical speeds and replaces the layer traction modulus by ${\rho }^{\prime }{\beta }_z^{\prime 2}$. No additional SH/P-SV coupling arises for the aligned symmetry axes and constitutive law here.

A rigid base still imposes $q_n={\chi }_n/h$, but nowThusThe vertical speed sets the resonance frequencies, while the horizontal speed sets the high-frequency limits and $c_n{U}_n={\beta }_x^{\prime 2}$.

For the elastic substrate, put ${ϵ}_x={\beta }_x^{\prime }/\beta$, $\zeta ={\rho }^{\prime }{\beta }_x^{\prime }{\beta }_z^{\prime }/(\rho {\beta }^2)$, $W=\omega h/{\beta }_z^{\prime }$ and $K=kh{\beta }_x^{\prime }/{\beta }_z^{\prime }$. The exact parameterization isIt reproduces the isotropic analysis with ${\beta }^{\prime }$ replaced by ${\beta }_z^{\prime }$ in the frequency scale, by ${\beta }_x^{\prime }$ in the velocities, and by the two effective contrast parameters above. In particular the true cutoffs are ${\omega }_{c,n}=n\pi {\beta }_z^{\prime }/[h\sqrt{1-{ϵ}_x^2}]$, the group minima occur near ${\omega }_{n,z}$, and a sufficient rigid-limit validity range isFor comparable horizontal and vertical layer speeds, a bounded density ratio and fixed mode index, the minimum estimate becomes ${\delta }_{\min }\sim ({\zeta }^2{\chi }_n/2{)}^{1/3}$ and $U_{\min }/{\beta }_x^{\prime }\sim 3(\zeta /{\chi }_n{)}^{1/3}/2^{2/3}$. If their ratio is extreme, that more specific minimum-depth estimate additionally needs $\zeta /{\delta }_{\min }\gg {ϵ}_x$; the exact parameterization and load criterion remain valid without it. As in the isotropic case, neither a slow horizontal nor a slow vertical speed alone justifies the rigid boundary at cutoff or within the strong-dispersion transition.

Choose depth $z$ positive downwards and harmonic time dependence $e^{-i\omega t}$. At normal incidence, P waves and the two polarizations of S waves are independent scalar problems: a P wave uses the longitudinal modulus $\rho {\alpha }^2$, and an S wave uses the shear modulus $\rho {\beta }^2$. In a uniform layer with wave speed $c$ and seismic impedance $q=\rho c$, write velocity and compressive traction asAt a perfectly bonded interface the velocity and traction are continuous. For incident velocity wave amplitude one from medium 1 into medium 2, these two conditions are $1+r_v=t_v$ and $q_1(1-r_v)=q_2{t}_v$. ThusReflection is determined by seismic impedance, not by speed alone. A step to larger seismic impedance reverses the reflected velocity/displacement field waveform; the pressure reflection coefficient has the opposite sign. Its transmitted pressure coefficient is $t_p=2q_2/(q_1+q_2)$. Equal seismic impedance eliminates the reflection even when speed and wavelength change. At a free surface traction vanishes and velocity reflects with coefficient $+1$; at a rigid surface velocity reflects with coefficient $-1$.

The mean elastic-wave energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=q(|A{|}^2-|B{|}^2)/2$ for positive real $q$. Therefore the reflected and transmitted energy fractions areA displacement field transmission coefficient greater than one is compatible with conservation of energy; its flux must be weighted by the receiving seismic impedance.

For many layers, use the seismic layer transfer matrix. A layer of thickness $h_j$ has wave phase thickness ${\delta }_j=\omega h_j/c_j$ andThis follows either by eliminating $A,B$ between the two faces or by integrating $V^{\prime }=i\omega P/(\rho c^2)$ and $P^{\prime }=i\omega \rho V$. For the complete stack put $M=M_N\cdots M_1=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. With upper and lower half-space impedances $q_a,q_b$, an outgoing lower wave satisfies $P_b=q_b{V}_b$. Solving this condition gives the input seismic impedanceHere $T$ is lower-face velocity divided by upper-face incident velocity, including the propagation wave phase. For lossless layers and real frequency, $|R{|}^2+(q_b/q_a)|T{|}^2=1$. This supplies a sensitive check on any spreadsheet or numerical implementation; it is not correct simply to multiply the individual interface transmissions and discard the elastic waves.

The physical reason is repeated wave interference. In one layer between media 0 and 2, let $r_{ij},t_{ij}$ denote the velocity coefficients above and $\delta =\omega h/c_1$. The first transmitted wave has wave amplitude $t_{01}{t}_{12}{e}^{i\delta }$. Every extra round trip supplies $r_{10}{r}_{12}{e}^{2i\delta }$. Summing the geometric series givesIn the time domain the successive arrivals are delayed by $2h/c_1$, producing reverberation. In the frequency domain the same delayed terms produce resonant maxima and minima. For a layer between identical half-spaces of impedance $q_0$,At $\delta =n\pi$ the layer transmits all the incident energy: the internally elastic waves cancel the direct reflection. Halfway between these resonances, a large impedance contrast gives strong reflection. For unequal exterior impedances, a quarter-wave layer with $q_1^2=q_a{q}_b$ transforms its lower load to $Z_{\mathrm{in}}=q_1^2/q_b=q_a$, again giving zero reflection at its design frequency. This matching is narrow-band.

Periodic alternation produces a periodic seismic multilayer stop band. For a cell of two layers with total thickness $d$, the cell matrix has reciprocal eigenvalues $e^{\pm iKd}$, soWhen the absolute value of this expression exceeds one, the infinite periodic medium has no propagating cell mode; a long finite stack has exponentially small transmission. The energy is reflected, not absorbed. Pass bands instead contain oscillatory transmission peaks, becoming more numerous as the stack length increases. Disorder removes the exact cell periodicity but not multiple scattering or frequency-selective interference.

For a concrete reproducible investigation, take $q_H/q_L=4$, equal one-way travel times $\tau$ in both layer types, exterior impedance $q_L$, and frequency coordinate $\nu =2\omega \tau /\pi$. At $\nu =1$ each layer is a quarter wavelength; one cell has diagonal matrix entries $-4,-1/4$. For $N$ cells,It is about $0.22145$ for one cell, $6.10\times 10^{-5}$ for four cells and $9.31\times 10^{-10}$ for eight cells. The infinite-cell stop band is $2\arcsin (0.8)/\pi <\nu <2-2\arcsin (0.8)/\pi$. The accompanying calculation also shows the first transmitted arrivals for a single high-impedance slab: their wave amplitudes are $0.64(0.36{)}^n$ at times $(2n+1)\tau$. Their sum is one, consistent with unit zero-frequency transmission. These calculations illustrate both the spectrum and the pulse train without assuming access to the course's spreadsheet.

Finally, as $\omega \to 0$ for a fixed finite stack, every $M_j\to I$, so the response approaches that of a direct interface between the exterior half-spaces. Equal exterior impedances give $R\to 0,T\to 1$. For wavelengths much longer than a repeating cell, expansion of $M_j$ to first order instead gives an effective medium with averaged mass density ${\rho }_{\mathrm{eff}}=\sum h_j{\rho }_j/d$ and harmonic-mean modulus $M_{\mathrm{eff}}^{-1}=\sum h_j/M_j^{\mathrm{mod}}/d$, where $M_j^{\mathrm{mod}}={\rho }_j{c}_j^2$. Its speed is $\sqrt{M_{\mathrm{eff}}/{\rho }_{\mathrm{eff}}}$, not generally the arithmetic mean of the layer speeds. Attenuation makes the energy balance an inequality, while for a fixed oblique horizontal slowness the scalar SH calculation uses $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$. Oblique P-SV waves require coupled displacement field and traction matrices because of mode conversion at a planar elastic interface. Thick evanescent layers require scattering or impedance recursion rather than exponentially ill-conditioned transfer-matrix products.

Use the quadratic spectral flux $S_z=-\frac{1}{2}\mathrm{Re}(V^{\ast }T)$. For positive real material mass density and shear modulus, differentiation of the velocity–traction system givesBecause $k$ is real, $\mu p_{\beta }^2=\rho -\mu k^2/{\omega }^2$. If $v=\mathrm{Im}\omega >0$, then $\mathrm{Re}(i{\omega }^{\ast })=v$ and $\mathrm{Re}(-i\omega \mu p_{\beta }^2)=v(\rho +\mu k^2/|\omega {|}^2)$. This proves complex-frequency SH flux monotonicity:Strictness holds for a nontrivial wavefield: if $V,T$ vanished together at one point, uniqueness of the first-order system would make the entire solution zero. The zero solution is the necessary exception to the literal strict inequality. Across a bonded material boundary $V,T$, and hence $S_z$, are continuous; integrating the negative derivative gives strict decrease across any positive-depth interval as well.

Where the SH input impedance is finite, $S_z=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. If $\mathrm{Re}Z(z_1)>0$, then $S_z(z_1)>0$ and all shallower fluxes are larger and positive. They cannot have $V=0$, since that would make $S_z=0$. Thus $\mathrm{Re}Z(z)>0$ for every $z<z_1$. Likewise, negative real SH input impedance at $z_0$ gives negative flux throughout $z>z_0$, excludes velocity zeros there and implies $\mathrm{Re}Z(z)<0$ for every $z>z_0$.

In a uniform region let $q=\mu p_{\beta }$ on the causal vertical wavenumber branch. The constant Riccati solutions are $Z=\pm q$, and integrating the one-way equation givesSince $\mathrm{Re}q>0$, the plus solution carries positive downward flux and the minus solution negative upward flux. Writing $k_{\beta }=a+ib$ with $b>0$, their moduli are proportional to $e^{-bz}$ and $e^{bz}$ respectively: each decays in its direction of energy flow.

For complex frequency the displayed quadratic form is the spectral flux diagnostic specified in the question; a genuinely periodic time average is recovered on the real-frequency axis. The decrease here comes from the upper-half-plane convergence parameter, not from material dissipation. At real frequency in a lossless layer the corresponding flux is constant.

In infinitesimal elasticity and plasticity in continuum mechanics, decompose the symmetric strain as $e=e^e+e^p$. For incompressible isotropic elasticity, ${\sigma }^{\prime }=2G{e}^e$, with $G$ the shear modulus and both strains traceless. The associated flow rule for the Von Mises yield criterion has ${\dot{e}}^p=\Lambda {\sigma }^{\prime }$ after absorbing a positive factor into $\Lambda \ge 0$. Plane strain has $e_{33}=0$, soIf ${\sigma }_{33}^{\prime }=0$ initially, it remains zero by this homogeneous scalar differential equation. Then $e_{33}^e=0$ and ${\dot{e}}_{33}^p=0$, so $e_{33}^p=0$ is consistent with zero initial plastic component. This proves consistency, not a claim that arbitrary prestressed plane-strain histories must have zero separate components. The out-of-plane shear components can consistently be zero by the same argument.

Write $p=-\mathrm{tr}\sigma /3$, so ${\sigma }_{33}=-p$. The proposed in-plane parameterization hasIts squared contraction is $2k^2({\sin }^{2}2\varphi +{\cos }^{2}2\varphi )=2k^2$, proving the yield condition identically. Its principal deviatoric stresses are $k,-k,0$. Define the two slip line families by tangent unit vectorsThus $\alpha$ lines have tangent angle $\varphi$ and $\beta$ lines have tangent angle $\varphi +\pi /2$. Their normal extensions from the associated plastic strain rate vanish, while their shear stresses have maximum magnitude $k$; they are not the principal stress directions.

In the absence of body force, differentiate the stress parameterization in the two equilibrium equations. One obtainsThe symmetric matrix multiplying $2k\nabla \varphi$ has eigenvectors $a,b$ with eigenvalues $+1,-1$. Taking its contraction with these vectors provesHence the Hencky stress relations areConstants can differ between lines, and these signs depend on the specified negative shear-stress convention.

Place the notch tip at the origin, with the opening towards negative $x_1$ and tensile loading in the transverse direction. For a usual reentrant notch, $0\le \gamma \le \pi /2$. The upper face has tangent angle ${\vartheta }_f=\pi -\gamma$. Choose its tangent $t=(\cos {\vartheta }_f,\sin {\vartheta }_f)$ directed away from the tip and outward material normal $n=(-\sin {\vartheta }_f,\cos {\vartheta }_f)$ pointing into the notch. Direct rotation of the stress gives the normal and tangential traction componentsTake the notch faces traction-free, as intended by the notch problem. Then the cosine vanishes. Select the tensile branch $\sin (2\varphi -2{\vartheta }_f)=1$, for which the face-tangential normal stress is $2k$. Thus the upper face has $\boxed{p=-k,{\varphi }_f=\pi /4-\gamma (\mathrm{mod}\pi )}$. Choosing the opposite branch would describe compression rather than the tensile field.

Construct the centred slip-line fan at a traction-free V-notch to connect this surface state to the forward symmetry axis. Near the upper face there is a constant-stress region. It joins a fan bounded by raysInside the fan choose $\beta$ lines radial and $\alpha$ lines circular about the tip, so their angle parameter is $\varphi =\vartheta -\pi /2$. Along a circular $\alpha$ line the first invariant givesThe second invariant is automatically constant along each radial $\beta$ line because $p,\varphi$ are independent of radius. Thus this is an actual equilibrium fan, not just an assumed transport of a surface value. At ${\vartheta }_2$ it matches the free-face constant state. At ${\vartheta }_1$ it matches a forward constant-stress region withReflect the upper field across the symmetry axis to obtain the lower field. The forward region has zero shear, respects this symmetry, and contains the positive $x_1$ axis. There $\sin 2\varphi =-1$, and henceThis value holds in the fully plastic notch-tip region described by the field. It is not a uniform stress assertion for the entire axis of an arbitrary finite specimen, including any elastic or far-field region.

For $\omega \tau \ll 1$,andThus the material flows at sufficiently low frequency: its static shear modulus vanishes and its terminal response is viscous. The dashpot in the Maxwell branch allows its spring stress to relax, so there is no permanent elastic path through the parallel network.

The fact that both $G^{\prime }$ and $G^{\prime \prime }$ tend to zero is not evidence of a force-free material: stress at fixed shear rate tends to ${\eta }_0\dot{\gamma }$. At higher frequency the spring stores energy, but a finite-frequency elastic response does not make the long-time material a solid.

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.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.