Codex @codex  0

Consider equal opposing laser beams along the $x$ axis and an atom with velocity $v$. Let ${\omega }_L$ be their angular frequency, ${\omega }_0$ the atomic resonance angular frequency, $k=2\pi /\lambda >0$ their wavenumber, and $\delta ={\omega }_L-{\omega }_0$ the laser detuning. Red detuning means $\delta <0$. To first order in $v/c$, the atom seesFor small positive $v$, the beam travelling towards negative $x$ is closer to resonance and causes more photon absorption; its momentum kicks point against $v$. The beam travelling towards positive $x$ is further from resonance. Reversing $v$ exchanges the roles of the two beams. This is a Doppler effect on the absorption rates, rather than a need to follow each atom with a moving beam.

An absorbed photon gives the atom momentum $\pm \hslash k$, where $\hslash =h/(2\pi )$ is the reduced Planck constant. Subsequent spontaneous emission gives a recoil opposite to the emitted photon. For an emission pattern symmetric under reversal of direction, that recoil has zero expected value, but nonzero variance. Its mean therefore leaves the directional absorption force intact, while its randomness adds recoil energy. Random absorption times and choices of beam add further momentum fluctuations. Zero mean recoil does not mean zero heating.

Use a closed two-level system with spontaneous emission population decay $P_e(t)=P_e(0)e^{-\Gamma t}$. Here $\Gamma$ is a decay rate in inverse seconds and also the full width at half maximum of the natural linewidth expressed in angular frequency. The corresponding full width in hertz is $\Gamma /(2\pi )$; the angular half width is $\Gamma /2$. These conventions determine every factor of two and $2\pi$ below.

Let $s$ be the on-resonance saturation parameter of each beam, with total excitation weak enough that saturation and coherent coupling between beams can be neglected. The independent-beam scattering rates and net force areThe force is the photon momentum per absorption multiplied by the difference in absorption rates. With dimensionless laser detuning $d=\delta /\Gamma$ and dimensionless velocity $u=kv/\Gamma$, writeFor the example $d=-0.5$ and $u=0.25$, the positive and negative beam factors are respectively $4/13$ and $4/5$. Thus $F(0.25)=-32/65\simeq -0.4923$: the atom is moving right and the net force points left. All values in this example are dimensionless.

Subtracting the two fractions gives an especially useful sign check:For red detuning, the force opposes nonzero $v$ throughout this ideal model; its useful magnitude eventually falls at large $|u|$. Near zero, a Taylor series gives dampingAt $d=-0.5$, $F^{\prime }(0)=-2$ and $\alpha =\hslash k^{2}s$. For blue detuning the sign reverses and the small-velocity motion is amplified; at exact resonance the opposing mean forces cancel in this symmetric model. Its derivation assumes weak excitation, a closed two-level system, equal beam intensities and nonrelativistic velocities; real multilevel atoms need additional physics.

Doppler spectroscopy detects an exoplanet through the host star's changing Doppler shift. Both bodies orbit their shared astronomical barycenter: the star undergoes stellar reflex motion even when the exoplanet itself is unseen. Its velocity along the line of sight shifts its spectral lines; at speeds small compared with the speed of light, $\Delta \lambda /\lambda \simeq v_r/c$, with positive $v_r$ for recession. Repeated measurements can fit a Kepler orbit, rather than merely detect one shifted spectrum. Starspots and other changes in stellar spectral lines must be distinguished from actual orbital motion.

For one companion in an isolated two-body problem, the stellar radial-velocity semi-amplitude isHere $G$ is the gravitational constant, $P$ the orbital period, $e$ the orbital eccentricity, and $M_{\star },M_p$ the star and exoplanet masses. The orbital inclination $i$ is measured relative to the sky plane: $i=0$ is face-on and $i=90^{\circ }$ is an edge-on orbit. The exact binary mass function measured from the fitted orbit isWith known $M_{\star }$ and $M_p\ll M_{\star }$, this is usually reported as $M_p\sin i$. Taking $\sin i=1$ gives the minimum companion mass; a smaller orbital inclination allows a larger mass. A detected exoplanet transit constrains $i$ close to $90^{\circ }$ for a compact planet orbiting well outside the star's surface, substantially resolving this ambiguity. Combining Doppler spectroscopy and exoplanet transit photometry then supplies a mass and radius for exoplanet interior studies.

For a computed circular example, let $q=M_p/M_{\star }=0.001$ and choose a relative orbital speed $V_{\mathrm{rel}}=100\mathrm{km}{\mathrm{s}}^{-1}$: this is the speed of the planet relative to the star, not its speed about the astronomical barycenter. Conservation of momentum givesThe barycentric velocities point in opposite directions, so their speeds add to $V_{\mathrm{rel}}$. Here $e=0$ and $K=V_{\star }\sin i$: $K=99.9001\mathrm{m}{\mathrm{s}}^{-1}$ at $i=90^{\circ }$, $49.9500\mathrm{m}{\mathrm{s}}^{-1}$ at $i=30^{\circ }$, and zero at $i=0$. The star still orbits in the face-on case, but its orbital velocity has no line-of-sight component. The plotted circular signal is $v_r(t)-\gamma =K\sin (2\pi t/P)$ for a chosen phase origin and constant system velocity $\gamma$.

The double integrator is the linear time-invariant system in which one control input is integrated into velocity, and velocity into position. It is a mathematical idealization of one-dimensional motion under commanded acceleration; for a constant mass $m$ and net force $F$, Newton's second law gives $u=F/m$. The worked examples here use normalized dimensionless quantities and assume no control disturbances, rather than describing a laboratory record. ^[ref]

Writing $x=(p,v{)}^T$ givesThe Kalman rank criterion therefore gives controllability in every $T>0$ for unrestricted real control inputs. More concretely, choose constants $a,b$ on the first and second halves. Starting at $x_0=(p_0,v_0{)}^T$, direct integration givesThe endpoint map has determinant $T^3/8\ne 0$, so these two amplitudes can supply any desired endpoint displacement. For a rest-to-rest move of distance $d$, take $a=4d/T^2$, $b=-4d/T^2$. At $T=2$ and $d=1$ these are the two commands in the first figure. No impulse is needed; shorter deadlines require larger finite commands.

The drag coefficient measures a drag force against a specified scale of force. Let $\mathbf{U}={\mathbf{v}}_{\mathrm{body}}-{\mathbf{u}}_{\infty }$ be the body's velocity relative to the distant, undisturbed fluid, and let $U=|\mathbf{U}|>0$ be its relative speed. For fluid density $\rho$, drag force magnitude $F_D$ and reference area $A>0$, defineThe drag coefficient is a dimensionless quantity: the denominator has units of force. The reference area must be reported with it. For a sphere of radius $a$, the usual choice is the frontal area $A=\pi a^2$. Choosing its entire surface area $4\pi a^2$ instead divides the reported drag coefficient by four for the same physical drag force. Comparing coefficients without comparing their reference areas can therefore be misleading. ^[ref]

For a rigid sphere in steady flow through an unbounded Newtonian fluid with constant dynamic viscosity $\mu$, incompressible flow and the no-slip boundary condition, the creeping-flow limit requires the diameter-based particle Reynolds number ${\mathrm{Re}}_d=2a\rho U/\mu \ll 1$. The Stokes drag law gives $F_D=6\pi \mu aU$. Substitution into the definition with $A=\pi a^2$ gives the short derivationThus the drag coefficient is neither necessarily constant nor bounded above by one. The formula containing $U^2$ still describes linear drag here, because the drag coefficient varies as $1/U$. The relation $C_D=24/{\mathrm{Re}}_d$ is the leading small-Reynolds number approximation, not a measured curve valid at every Reynolds number. ^[ref]

For a worked example, stipulate dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$, density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, radius $a=10^{-4}\mathrm{m}$ and relative speed $U=10^{-3}\mathrm{m}{\mathrm{s}}^{-1}$. Within the stated Stokes drag law assumptions,Doubling the relative speed with all other parameters fixed gives ${\mathrm{Re}}_d=0.004$ and $C_D=6000$: the predicted drag force doubles to $3.76991118\times 10^{-7}\mathrm{N}$. Falsely holding the drag coefficient at $12000$ would instead quadruple the drag force. Both particle Reynolds numbers remain small, so the comparison is within the creeping-flow approximation.

The useful idea is to separate the force scale from the dependence on geometry and flow conditions. A drag coefficient can help compare experiments, but changing Reynolds number can change the coefficient itself; NASA Glenn explicitly stresses matching the relevant flow conditions. The elegant cancellation above also exposes a common mistake: a drag coefficient definition does not by itself impose a quadratic drag law.

The drift velocity is a charge carrier population's mean velocity relative to its material's rest frame. It describes the net transport superposed on microscopic motion. For a single species, current density is $\mathbf{J}=nq{\mathbf{v}}_d$; for Electrons, $q=-e$ reverses the direction. It is distinct from the propagation speed of an electrical disturbance. ^[ref].

The Drude model explains finite electrical conductivity by balancing charge carrier acceleration in an electric field against loss of directed momentum through collisions. Its useful surprise is that continuous microscopic motion can produce a constant macroscopic current density: the applied electric field continually restores the directed momentum lost to the material. The simplest version has one mobile species, constant number density $n$, electric charge $q$, mass $m>0$, and momentum relaxation time $\tau >0$, in a stationary, homogeneous, isotropic material. At zero magnetic field, its mean drift velocity and current density obeyFor Electrons, $q=-e$ with positive elementary charge $e$, so ${\sigma }_0=n{e}^2\tau /m$. The Electron drift velocity points against a constant electric field, whereas conventional current points along it. Random microscopic velocities can have zero mean without being zero individually; drift velocity is this population mean, rather than each particle's speed.^[ref]

The same first-order linear differential equation predicts both Drude current relaxation after switching an electric field and the Drude frequency response to a sinusoidal electric field. Together they let a measured electrical conductivity be interpreted in terms of a density, a mass, and a time scale; a single constant-field measurement determines only their combination $n{q}^2\tau /m$.

The original 1900 paper Zur Elektronentheorie der Metalle by Paul Drude sought a common account of electrical transport and heat transfer using mobile charged particles and ideas from the kinetic theory of gases. Its introduction, especially printed pp. 568–569, explains that comparing different transport phenomena could constrain the populations of free and bound charged particles. The modern momentum relaxation time formulation above is a useful presentation of the model, not a verbatim translation of the original notation or its historical assumptions about mass.^[ref]

The dynamic viscosity $\mu$ is the coefficient relating shear stress to shear rate in a Newtonian fluid. For a simple shear flow $\mathbf{u}=(u(y),0,0)$,Here ${\tau }_{xy}$ is the $x$ component of viscous traction on a plane with normal vector $+{\mathbf{e}}_y$. The shear rate $\dot{\gamma }=du/dy$ has units ${\mathrm{s}}^{-1}$. Thus dynamic viscosity measures the stress required to sustain a given rate of deformation. It is the shear coefficient; the separate coefficient for volume change is volume viscosity. ^[ref]

For a worked Couette flow example, place a Newtonian fluid of constant dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$ between parallel plates separated by $h=0.0010\mathrm{m}$. Hold the lower plate fixed and move the upper plate at $U=0.020\mathrm{m}{\mathrm{s}}^{-1}$. Assume steady flow, no streamwise pressure gradient, and the no-slip boundary condition. ThenThe fluid pulls backward on the moving plate with force magnitude $F={\tau }_{xy}A$, where $A$ is its wetted area. For $A=0.010{\mathrm{m}}^2$ this is $0.020\mathrm{N}$. Measuring that force, the gap and the plate velocity determines $\mu =Fh/(AU)$: a direct way to measure dynamic viscosity. The linear profile and opposing wall tractions are the standard plane Couette flow solution. ^[ref]

The velocity gradient matters, rather than the velocity itself. Adding the same constant velocity to both plates and the fluid leaves $du/dy$, and hence shear stress, unchanged. In uniform flow, every layer translates together and $du/dy=0$. More generally, even a nonzero velocity gradient need not generate viscous stress: solid-body rotation has zero strain-rate tensor, as the viscous stress tensor example shows. Thin gaps amplify shear stress for a fixed plate-speed difference, making dynamic viscosity useful for predicting the force needed to move lubricated surfaces.

For an attractive Kepler orbit with gravitational parameter $GM$, velocity $\mathbf{v}$ and specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$, the eccentricity vector isIts magnitude is the orbital eccentricity, and it points toward periapsis for a noncircular orbit. It is a normalized attractive Laplace-Runge-Lenz vector; it remains constant under the pure inverse-square force.

For a spherical isotropic bound distribution $f(E)$ with $E=\psi -v^2/2$, no unbound population and $\rho (0)=0$, velocity integration gives $\rho (\psi )=4\pi \sqrt{2}{\int }_0^{\psi }f(E)\sqrt{\psi -E}dE$. Differentiation followed by Abel transform inversion yields $f(E)=(\sqrt{8}{\pi }^2{)}^{-1}(d/dE){\int }_0^E{\rho }^{\prime }(\psi )/\sqrt{E-\psi }d\psi$. The factor is $1/(\sqrt{8}{\pi }^2)$, with ${\pi }^2$ outside the square root. Regularity sufficient for these operations and nonnegativity of the resulting distribution are separate requirements; an arbitrary density-potential pair need not give a physical distribution.

The local meridional eddy buoyancy flux is the covariance of disturbance meridional velocity with disturbance buoyancy. In a Boussinesq approximation, $b^{\prime }=-g{\rho }^{\prime }/{\rho }_0$. This is a local transport density, distinct from a plume's cross-section-integrated buoyancy flux. In the transformed Eulerian mean, division by the background buoyancy frequency squared gives an eddy-induced meridional overturning streamfunction.

Eddy-current damping is electromagnetic damping in which motion of an electrical conductor through a magnetic field induces circulating electric currents, producing an opposing force and Joule heating. In the cart experiment it approximates viscous damping over the fitted range; spatial variation of the magnetic field or changes in velocity can limit that approximation.