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In small-amplitude oscillatory shear, expand the measured shear stress asComparison with the definitionof the storage modulus and loss modulus givesand henceThus the measured phase directly gives the loss tangent.

At low angular frequency, a standard viscoelastic fluid has time to relax and responds mainly as a viscous liquid: stress is approximately in phase with strain rate, so $\delta \to \pi /2$ as $\omega \to 0$. At high frequency it cannot relax during one cycle and responds mainly as an elastic solid, so stress is approximately in phase with strain and $\delta \to 0$ as $\omega \to \infty$.

For a material dominated by one viscoelastic relaxation time $\lambda$, the crossover between these regimes occurs when $\omega \lambda$ is of order one. Measure the frequency ${\omega }_c$ at which $G^{\prime }=G^{\prime \prime }$, equivalently $\delta =\pi /4$, and estimateA broad or multi-peaked crossover instead indicates a spectrum of relaxation times.

A resonant interaction requires matching both wavenumber and angular frequency: $k_3=k_1+k_2$ and ${\omega }_3={\omega }_1+{\omega }_2$ for a sum interaction, with corresponding sign changes for differences. A common phase velocity alone does not guarantee that a generated harmonic lies on the dispersion relation.

A quadratic product of plane waves generates sums and differences of their wavenumbers and angular frequencies. A component drives a secular term when its generated pair also satisfies the linear dispersion relation. A solvability condition in the method of multiple scales then determines the coupled amplitude equations on a slow time.

For a point mass moving as $z=L\cos \omega t$ along one fixed axis, the trace-free mass quadrupole moment is $m{L}^2{\cos }^2\omega t\mathrm{diag}(-1/3,-1/3,2/3)$. Its radiating component has twice the source angular frequency. The quadrupole formula gives $⟨P⟩=16G{m}^2{L}^4{\omega }^6/(15c^5)$ for this contribution in the leading slow-motion weak-field approximation. Any physical driver or support has its own quadrupole contribution; this expression is for the prescribed mass alone.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life

For the free-slip plane layer, onset selects the smallest Rayleigh number across all admissible stationary and oscillatory normal modes. The first vertical mode wins because increasing vertical index raises viscous cost and reduces oscillatory admissibility. Oscillatory minimization replaces the right-hand side of the stationary wave-number equation by $P^2\mathrm{Ta}{\pi }^2/(1+P{)}^2$. For the oscillatory minimum also require positive angular frequency squared. Where both selected curves are admissible, at rapid rotation the ratio of the minimized thresholds is $R_{o,c}/R_{s,c}\sim 2P^{4/3}/(1+P{)}^{1/3}$. Equality gives $8P^4-P-1=0$, with positive root about $0.67660$; below it sufficiently rapid rotation can select oscillatory primary onset.