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In a fixed square coil beside a long wire, neglect self-inductance compared with resistance. A changing wire current induces coil current proportional to $dI/dt$. Forces on the two perpendicular sides cancel, and the unequal parallel-side fields give magnitude ${\mu }_{0}Na|Ii|[1/s-1/(s+a)]/(2\pi )$. When $I=I_0\sin \omega t$, the signed force alternates at twice the drive frequency and has zero cycle mean, while Joule dissipation is positive with mean ${\mathcal{M}}^2{\omega }^2{I}_0^2/(2R)$. A fixed coil has no mechanical work despite its time-dependent force.

Let a conducting loop of area A rotate about z with normal $(\cos \omega t,\sin \omega t,0)$ in a fixed field $B\hat{\mathbf{x}}$. Neglect inductance so $I=(BA\omega /R)\sin \omega t$. Put $m_0=B{A}^2\omega /R$. Its vector magnetic moment isThe static component does not radiate. The remaining component rotates at twice the mechanical frequency and has $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$. Thus $\overline{P}=2{\mu }_0{B}^2{A}^4{\omega }^6/(3\pi c^3{R}^2)$. Treating the same sinusoidal current as a fixed-axis moment underestimates this rotating geometry by a factor of eight. The energy source is mechanical rotation, with additional ohmic dissipation and magnetic braking.

Substitution into the wave equation gives ${\partial }_x^2\psi +{\partial }_y^2\psi =-(k_x^2+k_y^2)\psi$ and ${\partial }_t^2\psi =-{\omega }^2\psi$. Therefore ${\omega }^2=c^2(k_x^2+k_y^2)$. The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges requireNeither integer may vanish, because that would make the entire displacement zero.

The element's mass is $\rho dxdy$, and its speed is $-\omega A\sin (k_{x}x)\sin (k_{y}y)\sin (\omega t)$. Squaring and multiplying by half its mass gives the requested energy density. The two sine-squared spatial integrals are $a/2$ and $b/2$, soThere is no dissipation. The total mechanical energy equals the maximum kinetic energy, givingEquivalently, integrate the small-slope tension energy $\rho c^2|\nabla \psi {|}^2/2$ and use the dispersion relation.

For $b=4a/5$, the two lowest modes are $(p,q)=(1,1)$ and $(2,1)$:The fundamental has one antinode at $(a/2,b/2)$ and no interior nodal line. The next mode has a node along $x=a/2$ and opposite-sign antinodes at $(a/4,b/2)$ and $(3a/4,b/2)$. Every boundary is a node.

The square fundamental has $f=c\sqrt{2}/(2\beta a)$. Equating it to $f_{11}$ givesAt the same peak displacement amplitude and frequency, time-averaged kinetic energy is proportional to membrane area. ThereforeThe reciprocal ratio is $41/40$ if the rectangle is placed first.

The Stokes–Einstein relation connects thermal diffusion with dissipative mobility. For a dilute suspension of isolated rigid spheres of hydrodynamic radius $a$ in an unbounded Newtonian fluid of viscosity $\eta$,where $\zeta$ is translational drag coefficient, $D$ diffusion coefficient and $T$ absolute temperature. Stokes drag supplies $\zeta =6\pi \eta a$ under no-slip, continuum, low-Reynolds-number conditions. The Einstein diffusion relation $D=k_{B}T/\zeta$ is more general than that particular spherical drag law.

A short equilibrium argument explains the connection. Let a weak external potential $U$ exert force $\mathbf{F}=-\nabla U$, so drift velocity is $\mathbf{F}/\zeta$. The number flux is $\mathbf{j}=n\mathbf{F}/\zeta -D\nabla n$. A Boltzmann distribution $n\propto e^{-U/(k_{B}T)}$ has $\nabla n=n\mathbf{F}/(k_{B}T)$; requiring zero equilibrium flux fixes $D=k_{B}T/\zeta$. thermal fluctuations and viscous dissipation are therefore linked, rather than independent adjustable effects. The original theory also relates diffusion to measured random displacements.^[ref]

On times long compared with momentum relaxation, an isotropic diffusing sphere hassingle-particle tracking can measure the slope. Alternatively, dilute single-scattering dynamic light scattering measures a field correlation $g_1(q,t)=e^{-D{q}^{2}t}$; the corresponding intensity-correlation excess is proportional to $e^{-2D{q}^{2}t}$. The scattering wavenumber and the distinction between field and intensity decay must be retained.

A Stokes-Einstein experimental consistency test independently measures temperature and solvent viscosity, uses a narrow distribution of inert spheres, checks decay rate proportional to $q^2$ or displacement variance proportional to time, and extrapolates toward infinite dilution. Subtract drift, account for wall-enhanced drag, and verify that aggregation is absent. Tracking additionally needs spatial calibration and correction for finite exposure and localization precision. At finite concentration, collective diffusion is not automatically the single-particle self-diffusion in the equation. A hydrodynamic radius is not generally identical to a polymer radius of gyration; nonspherical particles, slip or viscoelastic solvents require a different mobility model.

A viscoelastic material stores part of the work of deformation and dissipates part. In the linear regime its stress depends on the strain history, not just the instantaneous strain or shear strain rate. For simple shear, a causal relaxation modulus $G(t)$ gives $\sigma (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. A sudden strain then probes stress relaxation, while a constant imposed stress probes material creep; these are different experiments.

For steady sinusoidal strain use the convention $\gamma (t)={\gamma }_0\cos \omega t$ and $e^{i\omega t}$ complex amplitudes. The complex viscoelastic modulus isso $\sigma (t)={\gamma }_0[G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t]$. The storage modulus $G^{\prime }$ is the in-phase elastic component. The loss modulus $G^{\prime \prime }$ multiplies the component in phase with shear strain rate; stress leads strain by $\delta =atan2(G^{\prime \prime },G^{\prime })$. The loss tangent is $\tan \delta =G^{\prime \prime }/G^{\prime }$ when $G^{\prime }>0$.

Multiplying stress by shear strain rate and averaging gives $⟨P⟩/V=\frac{1}{2}\omega G^{\prime \prime }{\gamma }_0^2$. Hence the dissipated energy per oscillatory shear cycle isThe elastic contribution returns its work over the cycle. A passive material has nonnegative dissipation, which fixes the sign of the loss term in the chosen Fourier convention. This complex viscoelastic modulus is a response function, not the mathematical absolute value of a complex number.

A Maxwell model places a spring of shear modulus $G$ in series with a dashpot of viscosity $\eta$, so the strains add at common stress:The Maxwell fluid dynamic modulus isAt $\omega \tau \ll 1$ the response is predominantly viscous, $G^{\prime \prime }\simeq \eta \omega$ and $G^{\prime }\simeq G(\omega \tau {)}^2$. At $\omega \tau \gg 1$ it approaches the elastic plateau $G^{\prime }=G$. The two moduli cross, and $G^{\prime \prime }$ reaches its maximum, at $\omega \tau =1$. A Kelvin-Voigt model instead has the spring and dashpot in parallel, giving $G^{\ast }=G+i\eta \omega$ and no fluid-like zero-frequency terminal flow. Real materials often need a spectrum of relaxation times rather than one Maxwell time.

Oscillatory rheometry measures stress amplitude and phase while imposing a small strain. An oscillatory-shear amplitude sweep first identifies a linear regime; an oscillatory-shear frequency sweep then measures $G^{\prime }$ and $G^{\prime \prime }$. Repeat at controlled temperature, suppress evaporation, check wall slip and sample aging, and correct instrument inertia at high frequency. Frequency-dependent structure and a nonlinear response cannot be inferred reliably by fitting one spring and dashpot to one point.

A physical sheet under uniform isotropic tension per unit length T and with mass per unit area $\rho$ has small transverse displacement satisfying ${\partial }_t^2\psi =c^2{\nabla }^2\psi$, where $c^2=T/\rho$. Fixed edges impose zero displacement. This mechanical membrane model neglects bending stiffness, large slopes and dissipation.