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For surface-fed growth of an accreting sphere with $\mu =\dot{r}>0$, stationary incoming material, and quadratic drag $F_D=C_D{\rho }_a\pi r^2{v}^2/2$, the variable-mass system obeys $\dot{v}+3\mu v/r=g-3C_D{\rho }_a{v}^2/(8\rho r)$. For constant positive parameters, $r\sim \mu t$ and the late balance gives the displayed speed. The omitted acceleration and accretion terms are both $O(t^{-1/2})$, consistent with the balance. Thus this growing sphere has unbounded speed with vanishing acceleration, rather than a finite terminal velocity. This model has a different growth law from the existing accreting raindrop with quadratic drag, whose growth rate is proportional to speed.

A leading fixed-discharge, level-surface approximation to a shallow erodible-bed flow gives perturbation equations $C_x=4U-C$ and $U_t+3\beta U_x=-4U+C$. The displayed Fourier growth rate has negative real part for every nonzero wave number, independently of bedload parameter $\beta$. Bedload changes the propagation speed but not this damping. The zero-wave-number mode is neutral, so absence of growing modes should not be confused with strict asymptotic decay of every spatially uniform perturbation.

For an ideal single transit, let the oscillatory magnetic field occupy a region of length $L$, take the longitudinal speed $v$ constant, and neglect edge fields. Each molecule then experiences one square pulse of duration $\tau =L/v$. This is one contiguous interaction region, distinct from the two separated pulses of Ramsey interferometry. The static magnetic field sets the selected transition gap; the transverse drive sets Rabi angular frequency. A change of RF frequency or static magnetic field changes Rabi detuning.

For an original ideal detector model, suppose every successful transition removes an otherwise accepted molecule and the incident beam occupies the lower state of the selected pair. With accepted no-transition signal $S_0$, constant background $S_{\mathrm{bg}}$, and effective loss contrast $0\le \eta \le 1$, writeFor identical trajectories $\overline{P}=P_{g\to e}(L/v)$, giving an intensity dip where transfer is largest. For real trajectory variation, a normalized joint probability density $w(v,ϵ,{\Omega }^{\prime })$ gives the static-offset modelHere $ϵ$ is the trajectory's transition-frequency offset and ${\Omega }^{\prime }$ its drive strength, each in angular-frequency units. The weighting is for molecules accepted by the apparatus, rather than an assumed unfiltered oven distribution. In particular, designing $\Omega L/v_0=\pi$ does not produce a Rabi pi pulse for every speed: at $0.8v_0$ and $1.2v_0$ the resonant transfer probabilities are respectively ${\sin }^2(5\pi /8)\simeq 0.8536$ and ${\sin }^2(5\pi /12)\simeq 0.9330$.

Spread in transit times, static magnetic fields and drive amplitudes can reduce or smear oscillations without quantum decoherence on any individual trajectory. Even the single-transit square pulse has finite spectral width of order $1/\tau$ in angular frequency. If the magnetic field varies substantially along one path, its Hamiltonian operator changes during the passage and the constant-offset average above must be replaced by a time-ordered exponential. Actual molecules may also have several coupled energy levels; the isolated pair and perfect loss discriminator are explicit idealizations of this model.

Infer a spherically averaged density profile using seismic travel times, normal-mode frequencies, mass, moment of inertia, hydrostatic structure and material constraints. Seismic speeds alone measure elastic-modulus-to-density ratios, so they do not uniquely supply density.

Let a test particle start at rest at radius $r_0$ outside a fixed spherically symmetric source. The spherical shell theorem gives the exterior Newtonian gravitational potential $\Phi (r)=-\mu /r$, where $\mu =GM$ is the source's standard gravitational parameter. Zero initial transverse velocity makes the motion radial. Conservation of mechanical energy per unit mass givesThe minus sign selects inward motion; the positive square root is its speed. For a fall from $r_0=2R$ to the surface $r=R$, that speed is $\sqrt{\mu /R}$. In the limit of release from rest infinitely far away, the familiar free-fall speed is $\sqrt{2\mu /r}$, equal in magnitude to the local escape velocity. A finite starting radius gives a smaller speed.

The arrival time follows from the same first integral. Set $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of nondimensionalization: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu }$ supplies the dimensional time scale.

For asymptotic speed $0\le u<c$, $k=(1-u^2/c^2{)}^{-1/2}$ and the local outward speed obeys $v_{\mathrm{loc}}^2=u^2+(2GM/r)(1-u^2/c^2)$. It tends to $c$ as the horizon is approached from outside, though a static observer cannot actually remain on the horizon.

Ramsey spectroscopy measures a transition frequency by comparing the relative quantum phase acquired between two coherent interactions with the reference oscillator. In a beam, a state selector and apertures admit particular internal quantum states and trajectories; an analyzer and detector measure state-dependent transmission. In an optically prepared atomic clock, optical pumping and fluorescence can perform those functions. The preparation and readout determine which part of the beam contributes; they are not themselves the two coherent interactions.

For an ideal accepted particle, the first resonant half-duration Rabi pi pulse splits the probability amplitude between $|g⟩$ and $|e⟩$. The second combines two indistinguishable alternatives: a transition in the first region followed by no transition in the second, or the reverse. Their relative quantum phase depends on the dark flight and the drive oscillation phases. Adding probability amplitudes gives quantum interference; adding only the two alternatives' probabilities would erase the fringe. “Dark” means that the transverse drive is absent: a static magnetic field and the internal Hamiltonian operator can still act.

The useful separation is between pulse duration and phase-accumulation time. Short regions can share one coherent oscillator while a long drive-free interval resolves small frequency differences. A scan over reference frequency locates a fringe pattern; controlled changes of the second pulse's oscillation phase distinguish that coherent signal from phase-insensitive background. A reference scan must also identify the central fringe before atomic clock frequency lock, because neighboring fringes provide competing lock points.

In an unsoftened plane with number surface density $n_2$ and relative speed $u$, an encounter strip has area $2udtdb$. Thus the velocity diffusion rate is $8G^2{m}^2{n}_2/u\int db/b^2$, dominated by the small-impact cutoff. Setting this cutoff to $Gm/u^2$ and requiring a change of order $u^2$ gives $t_{\mathrm{rel}}\sim u/(Gm{n}_2)$. With $u=\lambda v$, $n_2\sim N/R^2$ and $v^2\sim GNm/R$, this is of order $\lambda R/v$, independent of $N$ at fixed total mass. The point-particle, zero-thickness limit is singular; fixed softening or nonzero thickness changes it. Numerical coefficients are not controlled by extending weak kicks down to order-one deflections.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

For a two-particle collision conserving total momentum, the outgoing speeds in the centre-of-momentum frame are multiplied by the same $q$. The relative kinetic energy is multiplied by $q^2$. For a smooth oblique collision with normal coefficient of restitution $e_n$ and incident relative velocity inclined by $\chi$ to the contact normal, unchanged tangential relative motion gives $q^2=e_n^2{\cos }^2\chi +{\sin }^2\chi$. Thus $q=e_n$ is not a general oblique-impact identity: $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}$. Sticking requires all relative motion to vanish, not merely its normal component.

For a massive free particle with invariant rest mass $m$, speed $v<c$, and Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, its total energy is $E=\gamma m{c}^2$. It includes the rest energy $m{c}^2$ and the relativistic kinetic energy $(\gamma -1)m{c}^2$. Here $c$ is the speed of light. At fixed total energy, a larger speed requires a smaller rest mass.

For a free particle of rest mass $m$, subtract its rest energy $m{c}^2$ from its total energy $\gamma m{c}^2$, where $\gamma$ is the Lorentz factor. The low-speed binomial series gives $K=m{v}^2/2+3m{v}^4/(8c^2)+O(m{v}^6/c^4)$, recovering classical kinetic energy.

Consider an ideal circular vacuum path of laboratory radius $R$, a limit of closely spaced mirrors constraining light around the circle at laboratory speed $c$. The source and detector rotate uniformly together with signed angular velocity $\Omega$, where $|\Omega |R<c$, in flat spacetime. Let $L=2\pi R$ be the laboratory circumference and $A=\pi R^2$ the positive area. The $+$ pulse travels counterclockwise and the $-$ pulse clockwise, independently of the sign of $\Omega$. Both depart from the detector at $t=0$.

Unwrap the circle into a laboratory arc coordinate. The two first-return conditions areConsequently the exact laboratory arrival times and their difference areThe denominators $c\mp \Omega R$ are closing rates in these laboratory equations; they are not local speed of light measurements by a rotating observer. The two arrival spacetime events lie on the same detector world line. Its constant speed is $|\Omega |R$, so proper time along that world line givesThus laboratory coordinate-time and rotating-detector proper time delays agree only to leading order at small $|\Omega |R/c$. The radius and circumference here are laboratory quantities, not lengths obtained by laying comoving rods around the rim. Finite mirror polygons, transparent media, and optical dispersion require their own path and propagation assumptions before using an exact formula.

An exact dimensionless check takes $\Omega R/c=1/5$. Relative to $L/c$, the laboratory arrival times are $5/4$ and $5/6$, giving $\Delta t/(L/c)=5/12$. The detector instead records $\Delta \tau /(L/c)=\sqrt{6}/6\simeq 0.408248$. The leading Sagnac area law would give $\Delta t/(L/c)\simeq 2/5$. These are model values chosen to make the relativistic distinction visible, not attainable parameters of the historical apparatus.

The rotating loop does not constitute a universal measurement of the one-way speed of light. It compares completed circuits at a single readout; assigning a velocity to each remote leg additionally requires a clock synchronization convention. Local Einstein synchronisation around a rotating rim does not close into a single globally synchronized clock network: the accumulated mismatch is another expression of the rotating-path asymmetry. Local comoving vacuum measurements still give $c$ in both directions. Special relativity predicts a nonzero closed-loop delay because the receiver moves during propagation, while no single comoving inertial frame covers the whole circuit.

For positive loop resistance, coupled mechanical and circuit equations relax toward constant sliding speed on sufficiently long rails. The steady current balances the downhill gravitational force, and motional electromotive force balances the resistive voltage. The undamped zero-resistance limit has no relaxation, so its late-time limit is singular.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

A solid cylinder rolling inside a circular loop has centre-path radius $\ell =R-a$, kinetic energy $3m{v}^2/4$, and speed $v^2=4g[h-\ell (1-\cos \theta )]/3$ at angle $\theta$ from the bottom. The normal contact force is smallest at the top. Requiring $v_t^2\ge g\ell$ gives the displayed necessary contact threshold under imposed no-slip rolling. The centre's maximum acceleration on the circle is at the bottom, and at the threshold it is $11g/3$ upward. This is a limiting contact condition, not a guarantee for arbitrary finite static friction or an unspecified approach curvature.

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure $\ell$ from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle $\theta$, the predicted initial speed is $v=\sqrt{g\ell {\sin }^2\theta /\cos \theta }$. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain $r$, or measure the vertical drop $h$ directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time $N$ complete returns to the same marker in the same direction and estimate $P=\Delta t/N$. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute $\sin \theta =r/\ell$ or $\cos \theta =h/\ell$, and check $r^2+h^2\simeq {\ell }^2$ if both were measured. Then compare the measured oscillation period with $P^2=4{\pi }^{2}h/g$. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same $\ell$. Larger $r$ means smaller $h=\sqrt{{\ell }^2-r^2}$, so the ideal oscillation period decreases. A plot of measured $P^2$ against independently measured $h$ should follow a straight line through the origin with slope $4{\pi }^2/g$; conversely, $g=4{\pi }^{2}h/P^2$ estimates gravitational acceleration. Do not infer $h$ from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume $\ell =1.00\mathrm{m}$, $m=0.050\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$, and $\theta =40^{\circ }$. The model gives $r\approx 0.643\mathrm{m}$, $h\approx 0.766\mathrm{m}$, $P\approx 1.756\mathrm{s}$, $v\approx 2.300m/s$, and $T\approx 0.640\mathrm{N}$. Its vertical tension component is $0.4905\mathrm{N}$, equal to the gravitational weight; its inward component is approximately $0.412\mathrm{N}$. These are predictions from chosen parameters, not measured data or measurements of local $g$. The next figure uses the same stipulated $\ell$ and $g$, including illustrative angles $15^{\circ }$, $40^{\circ }$, and $60^{\circ }$.

A planar quadratic saddle $m{\omega }^2(x^2-y^2)/2$ rotated at angular speed $\Omega$ becomes linearly confining for the displayed strict inequality. Its rotating-frame mode frequencies obey ${ϵ}_{\pm }^2={\Omega }^2\pm {\omega }^2$. Equality has a zero-frequency secular mode and is not robust bounded-motion stability. This is a rotating potential model; a rolling ball on an actual surface has additional constraints.

Otto Stern's lecture describes two co-rotating toothed wheels: transit between them must match an advance between openings. He explicitly distinguishes this method from the first historical speed measurement.^[ref]

For an idealized example, let aligned wheels have $N$ equally spaced openings, separation $L$, and rotation rate $f$ in revolutions per second. A particle with constant longitudinal speed $v$ takes $L/v$ between them. The second wheel advances through $fL/v$ revolutions; transmission peaks when that advance is an integer number $k$ of opening pitches:With chosen $N=60$, $f=50\mathrm{Hz}$ and $L=0.080\mathrm{m}$, the first two orders select $240$ and $120\mathrm{m}{\mathrm{s}}^{-1}$. A second transmission maximum therefore need not represent a second source temperature. Finite opening widths admit speed intervals around each order, and an additional measurement is needed to distinguish orders. The ideal expression assumes negligible acceleration and constant rotation rate during a flight, with a fixed relative phase between the wheels.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]