Place a normal Stokeslet below a plane and an equal, oppositely directed image at the reflected point above it. The normal velocity of their sum is odd across the plane and the tangential velocity is even. The plane therefore has zero normal velocity and zero normal derivative of tangential velocity, giving zero tangential stress. For a sphere of radius $a$ translating towards that plane at speed $U$ from distance $d\gg a$, the leading radial surface flow is $3aUrd/[2(r^2+d^2{)}^{3/2}]$, directed outward. The image's incident flow at the sphere changes its required force by relative order $a/d$, so the first correction to surface velocity at fixed $r/d$ is $O(U(a/d{)}^2)$; the isolated sphere's potential-dipole correction occurs only at third order.

A normal point force $F\mathbf{n}$ at height $d$ above a plane with a no-slip boundary condition is balanced by an image Stokeslet $-F\mathbf{n}$, a stresslet of tensor strength $S$, and a source dipole of moment $\mathbf{M}$ at the reflected point. Here the stresslet convention is $u_i=-S_{jk}{\partial }_k{G}_{ij}$, with $G$ the free-space Stokes tensor. Substitution on the plane cancels both normal and tangential velocity. The original and image forces cancel their $r^{-1}$ terms, while the image stresslet cancels the remaining $r^{-2}$ force-dipole term; the generic fixed-angle far field is $O(r^{-3})$.

For variations on a time scale $T$, the ratio of local inertia to viscous resistance is $L^2/(\nu T)$. Steady Stokes flow at each instant requires both $UL/\nu \ll 1$ and $L^2/(\nu T)\ll 1$. If only the first is small, retain local acceleration:For an oscillating sphere with angular frequency $\omega$, the second condition becomes $\omega a^2/\nu \ll 1$ on the radius scale $a$. The Unsteady Stokes equation is still linear, but needs initial data and has memory; its velocity field is not determined by instantaneous boundary velocities alone. See E. J. Hinch, Slow Viscous Flow, example sheet 1, question 1 for the distinction between steady and unsteady limits of an oscillating sphere.

Two plane internal gravity waves can each have zero Stokes drift but give a nonzero drift through interference. The displacement of one wave need not be perpendicular to the other wave's wave vector. For upward waves with wave vectors $(\kappa ,-\kappa )$ and $(-\kappa ,-\kappa )$ and vertical amplitudes $W$, the cross-wave drift is $(0,-2\kappa W^2\cos (2\kappa x)/\omega )$ with $\omega =N/\sqrt{2}$. It has zero spatially averaged vertical flux. It cannot on its own satisfy an impermeable oscillating boundary: the second-order Eulerian mean velocity supplies the necessary normal correction there.

For queue-class pairs $a$, external arrival rates ${\nu }_a$, and routing probabilities $p_{ab}$, the effective throughputs satisfy ${\lambda }_a={\nu }_a+{\sum }_b{\lambda }_b{p}_{ba}$. In an open network with routing spectral radius less than one, $\lambda =(I-P^T{)}^{-1}\nu$ is the unique nonnegative solution. Summing gives ${\sum }_a{\lambda }_a{p}_{a0}={\sum }_a{\nu }_a$, balancing total entrances and exits.

Connect multiclass single-server queues by Markovian routing. Each node must either have common exponential service parameters across its classes or use a symmetric service discipline. If the traffic equations for a multiclass queueing network give loads ${\rho }_j={\sum }_r{\lambda }_{jr}/{\mu }_{jr}<1$, its equilibrium word law is the product of the individual queue laws. Queue populations are independent at a fixed equilibrium time. A proof reverses routing with $p_{ba}^{\ast }={\lambda }_a{p}_{ab}/{\lambda }_b$, external arrivals with ${\nu }_a^{\ast }={\lambda }_a{p}_{a0}$, and exits with $p_{a0}^{\ast }={\nu }_a/{\lambda }_a$, and interchanges the insertion and service position rules. Weight ratios verify each reversed transition, and the traffic equations match total rates. This is a quasireversibility construction. A standard ordered-queue formulation is described in Walton's paper on multiclass queueing networks, Section 2.

A closed migration process is a continuous-time Markov chain of populations moving between colonies with a conserved total number of individuals. A colony containing $k$ individuals has departure rate ${\varphi }_j(k)$, with ${\varphi }_j(0)=0$, and each departure chooses its next colony using a fixed routing matrix. Its colony populations can represent stages of a service system, including an idle stage.

// Target: probability-and-statistics.bigb

The useful infinite-horizon path space for a stable queue isFor $\delta ,\sigma >0$, the workload functional $R(f)={\sup }_{t\ge 0}(\sigma f(t)-\delta t)$ is finite and continuous. Fix $f$, and choose $T\ge 1$ so that $|\sigma f(t)|\le \delta t/4$ for $t\ge T$. If $\Vert g-f{\Vert }_{\mathrm{sl}}<\delta /(4\sigma )$, then $\sigma g(t)-\delta t<0$ for $t\ge T$. Both suprema can thus be taken over $[0,T]$, whereUniform convergence on bounded time intervals alone does not give this continuity: a pulse escaping to later times can produce a large workload while tending locally to zero.

Local time records how much a stochastic process visits a level. Its normalization depends on the process: when the state space is a discrete space, it is occupation time at a state, while local time of a semimartingale is an occupation density weighted by quadratic variation. Brownian local time is a special case.

For an Itô process $d{X}_t=b_{t}dt+{\sigma }_{t}d{B}_t$ and a twice differentiable function $f$,The second-order term reflects the nonzero quadratic variation of Brownian motion.

A Prony brake measures shaft power using a resisting band: its force difference gives torque $\tau =r(F_1-F_2)$, and rotation gives $P=\tau \omega$, where $\omega$ is angular velocity. The Stirling engine experiment above uses this independent load measurement to distinguish shaft output from gas-cycle work.

For a spatially homogeneous minimally coupled scalar field with zero scalar potential, the stress-energy tensor has $\rho =P={\dot{\varphi }}^2/2$. Its field equation is $\ddot{\varphi }+3H\dot{\varphi }=0$, so $a^3\dot{\varphi }$ is constant and its energy density decreases as $a^{-6}$. Thus its homogeneous stress is precisely that of a stiff fluid. Spatial gradients would introduce directional stresses, so this equivalence requires homogeneity.

Thermosteric sea level change is the thermal expansion part of steric sea level change. For a uniform layer of depth $H$, conservation of mass gives $\rho H=(\rho +\Delta \rho )(H+\Delta \eta )$, hence $\Delta \eta \simeq -H\Delta \rho /\rho =H\alpha \Delta T$.

The optical fiber acceptance angle is the largest launch half-angle admitted by the specified guide model. For a flat entrance face perpendicular to the optical axis, external refractive index $n_0$, and a straight step-index optical fiber, Snell's law at the entrance givesAt the core–cladding interface, the critical angle for total internal reflection $i_c$ satisfiesFor a meridional ray, strict total internal reflection requires $i>i_c$. At the limiting geometry $i=i_c$, the refracted light ray is tangent to the interface, soThis identifies the numerical aperture with an entrance cone under these assumptions. With $n_0=1$ for air, the conventional inclusive boundary condition isfor a nonnegative launch angle $\theta$; a practical launch stays inside the boundary. If $\mathrm{NA}>n_0$, this ideal model imposes no further restriction within the forward hemisphere, and $\arcsin (\mathrm{NA}/n_0)$ must not be used. The light ray must also intersect the optical fiber core: a suitable angle alone does not correct a misplaced beam.

For an original example, take $n_1=1.50$, $n_2=1.48$ and air at the entrance. ThenThe internal limiting angle is only ${\varphi }_{\max }\simeq 9.37^{\circ }$, while the critical angle for total internal reflection is $i_c\simeq 80.63^{\circ }$. A $10^{\circ }$ launch has $\varphi \simeq 6.65^{\circ }$ and $i\simeq 83.35^{\circ }$, so it satisfies total internal reflection. A $20^{\circ }$ launch has $\varphi \simeq 13.18^{\circ }$ and $i\simeq 76.82^{\circ }$, so it fails this test.

This cone is a geometrical optics result for meridional rays. A single-mode optical fiber requires the spatial overlap of the incident electromagnetic wave with its guided normal mode; its numerical aperture does not impose an exact angular cutoff on arbitrary beams. Diffraction, finite cladding, and bends require a physical optics description.^[ref]

Using an axisymmetric hydrodynamic mass flux function, set $s=s(\psi )$, $\ell =R{u}_{\varphi }=\ell (\psi )$ and $\epsilon =\epsilon (\psi )$. The azimuthal vorticity is ${\omega }_{\varphi }=-R\nabla \cdot (\nabla \psi /(\rho R^2))$. The meridional projection of the Crocco form of the unsteady Euler equation gives the displayed equation. The mass density also satisfies the Bernoulli relation $\epsilon =|\nabla \psi {|}^2/(2{\rho }^2{R}^2)+{\ell }^2/(2R^2)+w+\Phi$, so an equation of state is part of closing the system.

A magnetized stellar wind extracts angular momentum through matter transport and Maxwell stress tensor. The Alfvén radius supplies an effective lever arm larger than the stellar radius, with geometry factors absorbed into an averaged lever arm. Dynamo action couples the torque to rotation by supplying the magnetic field. For a simplified radial wind of fixed speed, $R_A^2$ scales as $B_{\ast }^2{R}_{\ast }^4/({\mu }_0{\dot{M}}_w{v}_w)$, so unsaturated $B_{\ast }\propto \Omega$ gives a torque proportional to $-{\Omega }^3$. Magnetic topology and wind thermodynamics can alter this scaling.

An episode of particularly strong mass loss near the end of Asymptotic giant branch evolution, capable of removing the remaining envelope rapidly compared with its prior evolution. Radiation acting on dust can transfer momentum to the gas; pulsations help lift material into cooler dust-forming layers. Envelope removal terminates sustained hydrogen-burning shell and helium-burning shell activity.

In spherical stellar structure, a shell of radius $r$, thickness $dr$ and mass density $\rho$ has mass $dm=4\pi r^2\rho dr$. Integrating outward gives the enclosed mass, $m(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$.

The regular $n=5$ Lane-Emden equation solution has no finite first zero. Its Lane-Emden mass formula gives $m(\xi )=4\pi {\rho }_c{\alpha }^3{\xi }^3/[3(1+{\xi }^2/3{)}^{3/2}]$, so its infinite-radius total mass is $4\pi \sqrt{3}{\rho }_c{\alpha }^3$. Its enclosed mean mass density is ${\rho }_c(1+{\xi }^2/3{)}^{-3/2}$ and tends to zero. A finite externally confined core instead requires a specified truncation pressure.