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An equilibrium has asymptotic stability when it is Lyapunov stable and every trajectory starting sufficiently nearby converges to it. Stability controls how far a small perturbation can wander; attraction supplies the eventual return. The definition does not require an exponential decay rate.

For $\dot{x}=-x^3$, the solution from $x_0$ is $x(t)=x_0/\sqrt{1+2x_0^{2}t}$. Its magnitude never increases and tends to zero, so the origin is asymptotically stable. The decay is algebraic rather than exponential, showing why exponential stability is a stronger condition. A center equilibrium provides a different comparison: neighboring trajectories stay close but circulate indefinitely instead of returning to the origin. These distinctions help interpret a measured response whose amplitude changes slowly or whose observation interval is too short to establish its eventual limit.

Stereoscopy supplies depth because different cameras see different projections of the same track. A functional reconstruction first matches the same interaction vertex and track segments across views. A calibrated lens and window model maps each measured image coordinate back to a ray inside the liquid. Refraction at the windows must be included. Intersecting the rays, or finding their closest agreement when measurements have errors, estimates three-dimensional coordinates. Several points then determine a trajectory, rather than only a projected arc.

An original ideal-camera model makes the geometry explicit. Take two parallel pinhole cameras with centers $(0,0,0)$ and $(b,0,0)$, focal length $f$, and optical axes along $z$. Use virtual image planes in front of the pinholes so that image coordinates have the same orientation as the scene. A point $(X,Y,Z)$ then givesSubtracting the horizontal image coordinates gives disparity $d=u_L-u_R=fb/Z$, henceFor stipulated $b=0.20\mathrm{m}$, $f=0.050\mathrm{m}$, $u_L=0.010\mathrm{m}$, $u_R=0.005\mathrm{m}$ and $v_L=0.002\mathrm{m}$, the reconstructed point is $(0.40,0.08,2.0)\mathrm{m}$. These are ideal optical inputs, not dimensions of Alvarez's apparatus. This model omits refraction and lens distortion; actual calibration replaces its simple ray geometry. Since $|dZ/dd|=fb/d^2$, the same disparity error gives worse depth precision for distant points.

The Lorentz force then supplies a momentum scale. With magnetic field $B$ uniform and negligible stopping power over a fitted segment, bubble-chamber track readout gives $p_{\perp }=|q|Br$. The local three-dimensional helix tangent supplies the missing direction: if $\alpha$ is its angle to the field, $p=p_{\perp }/\sin \alpha$. In practical units,Here $e$ is the elementary charge and $c$ the speed of light. A short, almost straight arc poorly constrains its radius of curvature; a visually sharp track can therefore have poor momentum precision.

A clearance bound between trajectory samples converts finitely many separation checks into a guarantee over a continuous trajectory, using a justified bound on how quickly clearance can change. Clear endpoints alone are insufficient: a foot or another robot part can approach an obstacle and move away again between them. The primary planning formulation requires constraints along the entire path.^[ref]

Let nominal clearance $c(s)$ have Lipschitz constant $L$ on a checked interval $[s_i,s_{i+1}]$ of length $h$. If both endpoint clearances are at least $c_{\min }$, then every intermediate point is within $h/2$ of an endpoint. ConsequentlyThis follows directly from $c(s)\ge c(s_i)-L|s-s_i|$ using the nearer endpoint. A uniform additional worst-case geometry or execution allowance $e$ gives actual clearance at least $c_{\min }-Lh/2-e$.

A tighter interval certificate retains the two different endpoint values, $a=c(s_i)$ and $b=c(s_{i+1})$. Writing $u=s-s_i$ gives two lower cones,If $L>0$ and $|a-b|\le Lh$, the cones meet inside the interval. Their common height yieldsIf the endpoint difference exceeds $Lh$, the proposed Lipschitz bound and exact endpoint values are inconsistent; simply applying the formula does not repair that inconsistency. With $L=0$, clearance is constant.

For an original example, take $h=0.10\mathrm{s}$, $L=0.20\mathrm{m}{\mathrm{s}}^{-1}$, $a=18\mathrm{mm}$ and $b=26\mathrm{mm}$. The simpler endpoint-minimum bound gives $8\mathrm{mm}$; the two-cone bound gives $12\mathrm{mm}$. Subtracting a $4\mathrm{mm}$ execution allowance leaves a certified $8\mathrm{mm}$ actual gap, sufficient for a requested $5\mathrm{mm}$ margin. These are lower bounds, not assertions that the trajectory reaches their values.

A counterexample explains why the allowance is necessary. On $[0,1]$, $c(s)=1-4s(1-s)$ has endpoint values one but touches zero at the midpoint. Checking only those endpoints misses the contact. Its derivative is bounded in magnitude by four, so a sufficiently fine sample grid with that valid bound can expose the missing guarantee.

The constant must match the geometry and parameter units. For a point moving relative to a fixed closed obstacle set, the distance to that set changes by at most the point’s displacement: the triangle inequality gives $|d_O(x)-d_O(y)|\le \Vert x-y\Vert$. A speed bound therefore supplies a time-based clearance bound. A rotating body requires a bound on every occupied point, such as the angular clearance bound for a rotating body; terrain-height clearance across a discontinuous edge cannot use a finite slope bound across that jump.

Adaptive clearance sampling subdivides intervals that fail the chosen certificate, concentrating computation where the margin or speed bound is least favorable. Failure of a conservative certificate means the interval needs better bounds or further checking; it does not itself prove a collision. For a proposed robot-foot test, validate the motion and geometry error bounds in one coordinate frame and compare the certified interval minima with dense independent measurements. Use the robust swing-foot clearance margin allowances without counting the same uncertainty twice.

For a map $T:X\to X$ and a given dynamical observable $f$, the additive cohomological equation asks for a function $u$ satisfyingWhen such $u$ exists, $f$ is a difference accumulated between consecutive states. Summing along a trajectory makes the intermediate terms cancel:Thus bounded $u$ forces these sums to remain bounded, and any periodic orbit of period $n$ must have sum zero. These are necessary conditions, not a universal existence theorem.

For the circle rotation $T(x)=x+\alpha (\mathrm{mod}1)$, try $f(x)=e^{2\pi ikx}$ and $u(x)=c{e}^{2\pi ikx}$. Then $c=(e^{2\pi ik\alpha }-1{)}^{-1}$ when the denominator is nonzero. If it vanishes, this mode is an obstruction; if it is very small, the coefficient is large. This explains how a simple equation leads to delicate regularity questions in dynamical systems.

In 2014, Gabriele Rosi and colleagues used atom interferometry: two rubidium-87 clouds prepared by laser cooling in vacuum, $328\mathrm{mm}$ apart, received shared three-pulse Raman transitions spaced by $160\mathrm{ms}$. Fluorescence readout and paired-fringe ellipse fitting yielded differential wave phase. Twenty-four tungsten alloy cylinders, about $516\mathrm{kg}$ total, moved between C and F. Cloud subtraction rejected shared vibration; configuration subtraction isolated the source change.^[ref]

Calibration used timed flight and light-sheet crossings; mirror tilt compensated Earth rotation. Wave vector reversal rejected unwanted shifts. Modeling included source Newtonian gravitational potential, atom trajectories, cloud distributions and moving supports. Source positioning, mass density variation and cloud geometry limited the reported combined relative standard uncertainty to $150$ parts per million.^[ref]

To understand the subtraction without reproducing that full analysis, consider an ideal atom interferometer in uniform vertical acceleration during each pulse sequence. Let $k_{\mathrm{eff}}$ be its signed effective vertical wave vector component and $T$ the pulse spacing. Its acceleration-dependent wave phase is $\varphi \simeq k_{\mathrm{eff}}g{T}^2$; labels U and L denote upper and lower clouds. The double difference is thenA background field identical in C and F cancels even if it differs between the two clouds. In a fixed-geometry, linear-response model, write $S=G\mathcal{K}$, where $\mathcal{K}$ is calculated from the source mass density, calibrated pulse sequence and atom trajectories. Thus $G_{\mathrm{inferred}}=S/\mathcal{K}$. Measuring wave phase accurately is one part of the task; calculating $\mathcal{K}$ accurately is another. The real nonuniform field and finite clouds require the trajectory-dependent analysis described above rather than this uniform-field approximation.

Combined pivot and axis uncertainty in rotational clearance bounds the occupied-point error of a rotating robot when its modeled pivot and axis direction are both uncertain. Adding separate bounds is valid only after specifying the same physical points, coordinate frame and endpoint quantity. A rigid rotation is represented by a rotation matrix, obtainable from a unit axis and angle through Rodrigues rotation formula.^[ref]

Let the nominal initial point and pivot be $\hat{p},\hat{c}$, with $r=\hat{p}-\hat{c}$ and $\Vert r\Vert \le \hat{R}$. Let actual values be $p=\hat{p}+\delta p$ and $c=\hat{c}+e$, where $\Vert \delta p\Vert \le {ϵ}_p$ and $\Vert e\Vert \le {ϵ}_c$. The actual and nominal rotations $Q,\hat{Q}$ have the same known principal angle $\theta \in [0,\pi ]$, while their unit axis directions differ by at most ${\alpha }_{\max }$. Expanding the actual endpoint and subtracting the nominal one gives the exact identityThe three terms isolate initial-point error, pivot error and direction-induced rotation error without assigning one error twice.

An orthogonal matrix preserves length, and $\Vert I-Q{\Vert }_2=2\sin (\theta /2)$. The axis-direction result in rotation-axis direction uncertainty in clearance bounds givesThereforeThe radius in this expression is the nominal $\hat{R}$, because the initial-point perturbation has already been accounted for by $Q\delta p$. Replacing it by an inflated radius is safe but may count some uncertainty conservatively twice. The pivot term uses the actual $Q$, whose rotation angle is shared and known in this model.

For an original tool-motion example, take $\hat{R}=0.50\mathrm{m}$, $\theta =90^{\circ }$, ${\alpha }_{\max }=1^{\circ }$, ${ϵ}_c=1\mathrm{mm}$ and ${ϵ}_p=0.2\mathrm{mm}$. The respective allowances are approximately $0.200$, $1.414$ and $12.341\mathrm{mm}$, totaling $E\approx 13.955\mathrm{mm}$. If every nominal occupied point is at least $20\mathrm{mm}$ from a fixed, exactly known obstacle, the actual clearance is at least $20-E\approx 6.045\mathrm{mm}$, provided point correspondence and the enclosure hold for the entire body at that orientation.

That last inference follows from the triangle inequality for set distance. An additional obstacle-position allowance must be subtracted separately when the obstacle is uncertain. A final-orientation calculation alone does not establish a bound along the intervening trajectory; robot motion planning constrains the whole continuous path.^[ref]

Shared-frame cancellation in rotational clearance uncertainty explains why a joint model can be tighter: translating the robot, pivot and obstacle coordinate descriptions together changes none of their relative clearances. A physical pivot displacement with a fixed obstacle is different. A proposed rotary-stage experiment fits pivot and direction in one frame, withholds fiducial observations at intermediate angles, and checks measured point errors against the bound. Uncertain rotation angle or changing structural deformation requires a further model; neither is included silently in ${\alpha }_{\max }$.

Use local east, north, and up unit vectors $({\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U)$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. In a spherical-Earth approximation at geographical latitude $\phi$, the angular velocity has components $\mathbf{\Omega }=(0,\Omega \cos \phi ,\Omega \sin \phi )$. Expanding the cross product for relative velocity $(v_E,v_N,v_U)$ givesThese are Coriolis acceleration components, not the total acceleration. For horizontal motion, the horizontal part is $(f{v}_N,-f{v}_E)$, where the Coriolis parameter is $f=2\Omega \sin \phi$. Thus northward motion in the northern hemisphere has an eastward Coriolis acceleration, to the right of its velocity; in the southern hemisphere the horizontal turning is to the left. At the equator $f=0$, but the full Coriolis acceleration need not vanish: vertical or eastward motion still couples through $\Omega \cos \phi$.

For a numerical example, take the Earth rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, geographical latitude $\phi =45^{\circ }$, and an initially northward relative velocity ${\mathbf{v}}_{\mathrm{rot}}=20{\mathbf{e}}_N\mathrm{m}{\mathrm{s}}^{-1}$. Then

For a local, short-time estimate with no other horizontal force and zero initial eastward velocity, hold the initial northward velocity in the leading term: $\Delta x_E\simeq \frac{1}{2}fu{t}^2$. At $t=10\mathrm{s}$ this gives about $0.1031\mathrm{m}$ eastward. It assumes nearly fixed local axes and geographical latitude, negligible changes of height, and $|ft|\ll 1$. It is not a complete wind trajectory: pressure gradients, friction, and evolving velocity must enter an atmospheric equation of motion. The effective gravity in a rotating frame already combines ordinary gravity with the steady terrestrial centrifugal acceleration.

The Coriolis force is the apparent force proportional to a particle's velocity relative to a rotating reference frame. For constant angular velocity $\mathbf{\Omega }$ and particle mass $m$,Here $\mathbf{v}$ in the title formula means ${\mathbf{v}}_{\mathrm{rot}}$, not the velocity measured in an inertial frame. The Coriolis force belongs to the chosen reference frame: a particle stationary in that rotating reference frame has no Coriolis force, even though it can have a nonzero velocity in the laboratory. Other reference frames assign different angular velocities and relative velocities, and hence different Coriolis forces to the same motion.

The cross product makes the Coriolis force perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}_{\mathrm{rot}}$, with magnitude $2m\Omega |{\mathbf{v}}_{\mathrm{rot}}|\sin \alpha$, where $\alpha$ is the angle between them. Apply the right-hand rule to $\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}$ and then reverse its direction. Motion parallel to the rotation axis gives zero Coriolis force; reversing the relative velocity reverses the Coriolis force.

The appeal of this description is practical: axes attached to the Earth or a turning apparatus can be kept fixed on the page, provided Newton's second law includes the extra terms. A free particle in rotating coordinates shows exactly how a straight laboratory trajectory can become curved, while Coriolis acceleration in local Earth coordinates gives a useful terrestrial scale. The rotating-table demonstration of Coriolis deflection makes the two viewpoints visible in one experiment.

For a trajectory with incoming and outgoing straight asymptotes, the deflection angle is the angle between its limiting incoming and outgoing velocity directions. For an attractive hyperbolic orbit of eccentricity $e$ and impact parameter $d$, $\sin (\delta /2)=1/e$ and $\tan (\delta /2)=GM/(d{v}^2)$, where $v$ is the incoming speed at infinity. The incoming velocity points opposite to the incoming radial direction; ignoring that reversal gives the wrong angle.

In the same uniform stationary sound model, let $\tau$ be emission time, $t$ reception time, and ${\mathbf{x}}_s(\tau )$ and ${\mathbf{x}}_o(t)$ source and receiver positions. The direct-path propagation condition and its direction areThe unit vector $\mathbf{n}$ points from the source's old emission position to the receiver's reception position. With velocities ${\mathbf{v}}_s=d{\mathbf{x}}_s/d\tau$ and ${\mathbf{v}}_o=d{\mathbf{x}}_o/dt$, taking a derivative of the propagation condition givesHere the dot products project onto the actual propagation direction. The last equality counts source wave phase cycles at reception for constant emitted frequency $f_s$. It assumes both speeds are below $c$, a unique direct arrival and smooth motion; interpreting a measured short-window frequency spectrum additionally requires little frequency change within that window. The source velocity must be evaluated at emission, not at reception. This delayed geometry matters when the direction changes.

For an original pass-by model, a source follows ${\mathbf{x}}_s(\tau )=(v\tau ,0)$ and a stationary receiver is at $(0,b)$ with $b>0$. The source passes its closest position at emission time $\tau =0$. ThenThese parametric equations give received frequency against reception time. For $\tau <0$ the shift is positive, for $\tau >0$ negative, and the unshifted emission at $\tau =0$ reaches the receiver at $t=b/c$. Thus zero shift occurs after the source's geometric closest passage in the material's reference frame. Increasing $b$ spreads the transition over a longer interval even though $v$ is unchanged. Far before and after passage, the frequency approaches $f_{s}c/(c-v)$ and $f_{s}c/(c+v)$ respectively.

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

A dynamical observable is a function $\phi :X\to \mathbb{R}$ or $\mathbb{C}$ that assigns a measurable quantity to each state of a dynamical system. If the dynamics is $T:X\to X$, the observed sequence from state $x$ is $\phi (x),\phi (Tx),\phi (T^{2}x),\dots$. The state space may be high dimensional even when this record is scalar.

For a harmonic oscillator with state $(q,p)$, position $\phi (q,p)=q$ is one dynamical observable; energy is another. Position generally oscillates, while energy is constant along an ideal conservative trajectory. Composition $\phi \mapsto \phi \circ T$ describes evolution of the measured quantity instead of evolution of states. Its time average, $n^{-1}{\sum }_{j=0}^{n-1}\phi (T^{j}x)$, is the object compared with a space integral in ergodic theory.

A dynamical system combines a state space with a rule for evolution in time. A discrete rule sends a complete state $x_n$ to $x_{n+1}=F(x_n)$, while a continuous model can specify $\dot{x}=f(x)$. The state must include enough information for the chosen rule: a harmonic oscillator needs both position and velocity, because equal positions with opposite velocities have different immediate futures.

The same model can be studied through several questions. An equilibrium point stays fixed, a periodic orbit repeats, and a basin of attraction describes which initial states approach a selected long-term behavior. Changing a parameter can cause a bifurcation. These qualitative properties can be useful even when no explicit formula for every trajectory is available. Deterministic evolution also requires well-defined forward solutions; merely writing a differential equation does not establish existence or uniqueness of its initial value problem.

Testing requires independently constrained force laws, mass assignments, initial conditions and a common family of inertial frames. Defining the net force separately for every observation by $\mathbf{F}=m\mathbf{a}$ cannot test Newton's second law. With interaction parameters fixed, the same equation predicts new trajectories. Likewise, zero net force implies uniform motion within the second-law framework, but the existence and empirical identification of its inertial frames are extra content. Pairwise Newton's third law adds a constraint not supplied by the individual second-law equations.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

On the punctured plane, the vector field $\dot{x}=(1/r-1)x-y$, $\dot{y}=(1/r-1)y+x$ has the displayed polar equations. Its exact trajectories satisfy $r(t)=1+(r(0)-1)e^{-t}$ and $\theta (t)=\theta (0)+t$. Every forward trajectory approaches the unit circle, rotating counterclockwise, from inside or outside according to its initial radius. The circle is an attracting limit cycle with least period $2\pi$. The origin is not an equilibrium: the original vector field is undefined there. The singularity prevents an application of a whole-plane Bendixson-Dulac criterion to this periodic orbit.

Consider a ball moving horizontally with no net horizontal force in the laboratory. Treat its horizontal motion as that of a free particle: ${\mathbf{r}}_{\mathrm{lab}}=(ut,0)$ with constant $u>0$. Axes rotating counterclockwise at constant angular velocity $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ about the launch point assign it the positionThis is an exact rotation matrix transformation, with the axes coinciding at $t=0$. Taking $u=0.25\mathrm{m}{\mathrm{s}}^{-1}$, $\Omega =0.8\mathrm{rad}{\mathrm{s}}^{-1}$, and $0\le t\le 3\mathrm{s}$ gives the two views below. At $t=3\mathrm{s}$, the laboratory position is $(0.75,0)\mathrm{m}$, but the rotating position is approximately $(-0.5530,-0.5066)\mathrm{m}$. Both assign the same distance $0.75\mathrm{m}$ from the origin.

The entire rotating acceleration must include the centrifugal acceleration as well as the Coriolis acceleration. To see the balance explicitly, write $c=\cos \Omega t$ and $s=\sin \Omega t$. Direct differentiation givesThe two apparent contributions areAt launch $\mathbf{r}=0$, so only the Coriolis acceleration contributes to the initial bending. Later the centrifugal acceleration also contributes; the full curved trajectory cannot be explained by keeping only the Coriolis force. In particular, $|{\mathbf{v}}_{\mathrm{rot}}{|}^2=u^2(1+{\Omega }^2{t}^2)$ grows even though the Coriolis force does no work: the rotating-frame kinetic energy increase is supplied by work from the centrifugal force. The laboratory kinetic energy remains $m{u}^2/2$.

An orbit around an isolated source constrains its standard gravitational parameter $\mu =GM$ rather than independently measuring the source's mass and the gravitational constant. In the test particle approximation, a circular orbit of radius $a$ and orbital period $T$ satisfiesThe left equation equates the required centripetal acceleration to the gravitational acceleration. More generally, a Kepler orbit depends on the same $\mu$. In the exact relative two-body problem, replace $GM$ by $G(M+m)$.

For any positive dimensionless number $q$, replacing $G$ by $qG$ and $M$ by $M/q$ leaves $GM$ and every predicted test particle trajectory unchanged. This is the gravitational mass degeneracy: more observations of the same orbit can refine $GM$ without separating its factors. A laboratory source whose mass is calibrated independently supplies the missing information.

Consequently an astronomical mass expressed in kilograms as $M=\mu /G$ inherits measurement uncertainty from $G$, even if $\mu$ is known much more precisely. To first order,When the relative standard uncertainty of $\mu$ is negligible, that of $M$ is approximately $u(G)/G$. Conversely, the ratio of two source masses inferred using the same $G$ is $M_1/M_2={\mu }_1/{\mu }_2$, so this common factor cancels. That distinction explains how excellent orbit determination can coexist with less precise absolute astronomical masses.

A version of the GRW model whose material events are collapse centres in spacetime. A macroscopic object is represented by a pattern of such events, not by particle trajectories between flashes. Exact flash locations coexist with the absence of a position assigned at every instant.

Heisenberg's microscope is a thought experiment, not a report of a performed measurement. In Werner Heisenberg's Nobel lecture, The development of quantum mechanics, delivered on 11 December 1933, the imagined electron is illuminated with very short-wavelength light and observed through a high-resolution microscope. A scattered photon entering the observer's eye would, through Compton scattering, eject the electron from its original path. The passage challenges a classical trajectory picture; its microscope is hypothetical. The lecture date is 1933 even though the award is the Nobel Prize in Physics in 1932.^[ref]

The following idealized geometry makes the tradeoff visible. Let the collected light have wavelength $\lambda$ and let the microscope objective accept outgoing directions within a half-angle $\alpha$ about the vertical axis. Take $x$ transverse to that axis and hold the incoming photon momentum fixed. A diffraction-limited estimate of transverse resolution has scale $\delta x\sim \lambda /\sin \alpha$. The unobserved outgoing direction admits transverse photon momentum components between approximately $-(h/\lambda )\sin \alpha$ and $+(h/\lambda )\sin \alpha$, neglecting the angular variation of the scattered wavelength. By conservation of momentum, uncertainty in that component leaves uncertainty in the transverse recoil of the electron:Here $h=2\pi \hslash$ is the Planck constant; the two $\delta$ symbols denote heuristic resolution and recoil scales, not the standard deviations in the Heisenberg uncertainty relation. Increasing the acceptance angle improves the estimated resolution while broadening the allowed recoil range. Numerical factors depend on the resolution criterion and angular distribution; this scaling argument cannot supply the sharp constant $\hslash /2$.

The thought experiment makes the interaction needed to obtain an image memorable. The preparation uncertainty theorem has broader use: it applies to the prepared quantum state even when the two observables are measured on separate copies. No claim that Werner Heisenberg built this gamma ray apparatus is needed to understand either argument.