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A sphere of radius $a$ touching both faces of a symmetric groove of opening angle $\varphi$ rotates instantaneously about the line joining its contacts. The centre is a distance $h=a\sin (\varphi /2)$ from this line, so rolling without slipping means $v=h\omega$. For a homogeneous sphere, $K=m{v}^2/2+I{\omega }^2/2=m{v}^2[1+2/(5{\sin }^2(\varphi /2))]/2$. Differentiating this kinetic energy with respect to distance down an incline gives the displayed acceleration. Both contact forces do no work for fixed faces and no slip.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

Let the atomic beam travel along $x$, with velocity $v>0$, and let $z$ measure displacement along the dominant magnetic field gradient. For a fixed effective magnetic moment component ${\mu }_z$, the magnetic dipole potential energy is $U\simeq -{\mu }_z{B}_z$. Taking its negative partial derivative givesA spatially uniform magnetic field can produce torque and Larmor precession, but cannot produce this translational separation. In a quantum state following a local internal energy eigenvalue $E_n(B)$, the more general expression is $F_z=-{\partial }_z{E}_n$, with effective magnetic moment $-\partial E_n/\partial B$. Treating that moment as constant is an approximation with a physical regime, not an extra measured fact. Assume the local magnetic field direction stays nearly fixed across the sampled beam, and that internal states follow it without transitions, as in the adiabatic theorem. The scalar gradient model describes the beam's neighborhood; it omits the full magnetic field's transverse components and fringe regions.

Assume a constant $G$ over magnet length $\ell$, followed by a field-free drift $D$. Neglect transverse motion at entry and changes in longitudinal velocity. Newton's second law gives transverse acceleration $a_z={\mu }_{z}G/M$ for whole-atom mass $M$. The magnet transit time is $t_m=\ell /v$, and the drift time is $t_d=D/v$. The magnet supplies both a displacement and an exit transverse velocity:The $\ell D$ contribution is motion with the acquired transverse velocity, not continued acceleration outside the magnet. For the plate-at-exit geometry, set $D=0$. For a nonzero entrance transverse velocity $w$, add $w(\ell +D)/v$. These terms show why initial collimation and flight geometry matter when inferring a magnetic moment from a deposit.

Choose hypothetical parameters $M=1.79\times 10^{-25}\mathrm{kg}$, $G=300\mathrm{T}{\mathrm{m}}^{-1}$, $\ell =0.035\mathrm{m}$, $D=0.100\mathrm{m}$, and $v=500\mathrm{m}{\mathrm{s}}^{-1}$. Approximate the two effective magnetic moments by ${\mu }_z=\pm {\mu }_B$, where the Bohr magneton is ${\mu }_B\simeq 9.274\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$. Then $|a_z|\simeq 1.55\times 10^4\mathrm{m}{\mathrm{s}}^{-2}$, $t_m=70\mu \mathrm{s}$ and $t_d=200\mu \mathrm{s}$. The positive displacement is about $0.0381\mathrm{mm}$ inside the magnet plus $0.2176\mathrm{mm}$ during drift: $d=0.2557\mathrm{mm}$, with full separation $2d=0.5114\mathrm{mm}$. These chosen values illustrate the model and are not reconstructed 1922 measurements.

The exit transverse velocity is only about $1.09\mathrm{m}{\mathrm{s}}^{-1}$, small compared with the chosen forward velocity; this checks the small-deflection assumption. Halving $G$ halves $d$, doubling $v$ divides $d$ by four, and setting $D=0$ removes most of this example's separation. A real pole has fringe magnetic fields and a position-dependent gradient; these require a more complete trajectory calculation.

Take downward as positive and let gravity have magnitude $g$. For a rigid sphere of density ${\rho }_p$ in otherwise stationary fluid of density ${\rho }_f$, its weight minus buoyancy is $\frac{4}{3}\pi a^3({\rho }_p-{\rho }_f)g$. At terminal velocity, this balances the Stokes drag law:This signed downward settling velocity is negative for a buoyantly rising sphere and zero for equal densities. For fixed density contrast and dynamic viscosity, the submerged weight grows as $a^3$ while linear drag at fixed speed grows as $a$; their balance gives the settling-velocity radius scaling $U_s\propto a^2$. Meanwhile,Thus doubling the radius multiplies the predicted Stokes settling speed by four and the particle Reynolds number by eight. This makes the formula useful for separating particles by size, while also providing its own check against extrapolation to large particles.

For a computed water example, take ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and $a=10.0\mu \mathrm{m}$. These are specified rounded inputs rather than a claim of exact water properties at every temperature. They giveThe diameter is $20.0\mu \mathrm{m}$, and both the net downward weight and the upward drag force are $6.16\times 10^{-11}\mathrm{N}$. At this terminal velocity, falling $1.00\mathrm{mm}$ takes $3.06\mathrm{s}$. At $a=20.0\mu \mathrm{m}$ the prediction is $1.308\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.05232$; at $a=50.0\mu \mathrm{m}$ it is $8.175\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.8175$, which is no longer a convincing small-Reynolds number case. The figure labels these as predictions of the Stokes settling velocity formula and shows where its inertia check fails.

The calculation describes the eventual terminal velocity, not the release transient governed by particle acceleration and possibly the Unsteady Stokes equation. Walls, hydrodynamic interactions, departures from a rigid sphere, or a different surface condition require a different resistance. For sufficiently small particles, Brownian motion can obscure individual settling trajectories; the formula still specifies the gravitational drift within its assumptions. The balancing argument already appears in article 43 of ^[ref].

For a twice continuously differentiable regular curve describing motion, let $v=\Vert r^{\prime }(t)\Vert >0$, $T=r^{\prime }/v$, and let $N$ be the principal normal vector where the curvature $\kappa$ is positive. Differentiating the velocity vector $r^{\prime }=vT$ givesThe tangential component changes the speed; the normal component changes the direction of the velocity vector. A particle can therefore have nonzero acceleration at constant speed, as in circular motion with constant speed. The formula follows from $dT/dt=(dT/ds)(ds/dt)=\kappa vN$ and the product rule. When $\kappa =0$, the normal contribution vanishes even though the principal normal vector is undefined there. ^[ref]

In dimensionless coordinates, take $r(t)=(t,t^2)$. At $t=1$,The resulting vector components areBoth increasing speed and turning contribute to the same acceleration.

The tangential component is the orthogonal projection of a vector onto the tangent space. For a plane curve, it is $(w\cdot T)T$; for a hypersurface, it is $w-(w\cdot n)n$. The scalar tangential component of acceleration is $\dot{v}$, which can be positive, zero or negative as the speed increases, remains constant or decreases.

Terminal velocity is a constant limiting velocity approached when a velocity-dependent resistance balances the other applied forces. A falling body may keep moving while its net force and acceleration tend to zero. For downward motion in a stationary medium, its limiting speed is set by balancing gravitational weight against drag force, including buoyancy if appreciable. The balance, rather than a universal numerical value, is the defining idea. ^[ref]

The terminal velocity under linear drag gives a simple example with an explicit approach to the limit. Real falling objects need an appropriate resistance model: linear drag is useful for slow motion of a small sphere in Stokes flow, whereas many faster falling objects require quadratic drag. Changing mass also changes the question: the existing accreting raindrop with quadratic drag has a speed-radius attractor of an accreting raindrop, rather than the fixed-mass balance used below.

Consider a point mass of fixed mass $M>0$ released from rest in a stationary medium. Choose downward as positive, assume constant gravitational acceleration $g>0$, neglect buoyancy, and impose linear drag $F_d=-bv$ with constant linear friction coefficient $b>0$. Here $v$ is the signed downward velocity component; it equals speed during this descent. Newton's second law becomesThe terminal velocity balances the two forces, and the time constant $\tau$ controls how quickly that balance is approached:For the illustrative parameters $M=1\mathrm{kg}$, $b=2kg/s$ and $g=9.8m/{\mathrm{s}}^2$, the time constant is $0.5\mathrm{s}$ and the terminal velocity is $4.9m/s$ downward. At $t=1\mathrm{s}=2\tau$, the downward velocity is $4.9(1-e^{-2})m/s\simeq 4.23686m/s$ and the downward acceleration is $9.8e^{-2}m/{\mathrm{s}}^2\simeq 1.32629m/{\mathrm{s}}^2$. This is a toy linear drag model, not a claim that every $1\mathrm{kg}$ falling object has this resistance law.

For a proof outline, subtract the equilibrium value: $u=v-v_T$ obeys $du/dt=-u/\tau$. The exponential decay solution and the initial condition give $u=-v_T{e}^{-t/\tau }$; one derivative gives the acceleration. Every time constant multiplies the remaining gap $v_T-v$ by $e^{-1}$. The exact limit is reached only as $t\to \infty$. This makes the model useful for estimating a settling duration without mistaking vanishing acceleration for vanishing velocity.

Newton already analyzed vertical motion under uniform gravity and resistance proportional to velocity in Book II, Section I, Proposition III of the Principia. His geometric construction includes a greatest downward velocity, illustrating that a limiting rate can emerge from balancing a driving force and resistance. ^[ref]

A third-law interaction pair consists of the two forces belonging to the same mutual interaction, with their agent and recipient exchanged. The names action-reaction pair do not imply a delay: in the instantaneous model, the two forces occur together. Test a proposed pair by naming both bodies and the interaction, rather than merely noticing equal magnitudes and opposite directions.

For a book resting on a table, the table's upward normal force and Earth's downward gravitational weight both act on the book. They can balance in the book's free-body diagram, but they are not a third-law interaction pair. The partner of the table's normal force on the book is the book's downward contact force on the table. The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. Their equality does not require equal accelerations: Earth's much greater mass makes its response correspondingly smaller for a given net force.

Similarly, friction on a sliding block from a surface is paired with friction on that surface from the block. For the collection containing only the block, that friction is an external force; including the contacting surface makes both members internal forces. Changing the collection changes the bookkeeping of momentum, not the physical interaction.

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

A free nonrelativistic charge driven by a weak plane electromagnetic wave has peak acceleration $|q|E_0/m$. Its time-averaged Larmor formula power is $q^4{E}_0^2/(12\pi {ϵ}_0{m}^2{c}^3)$. Divide by incident light intensity ${ϵ}_{0}c{E}_0^2/2$ to obtain the displayed cross section, independent of frequency. For an electron, it is $8\pi r_e^2/3$. The approximation assumes negligible Compton recoil kinematics and small driven velocity compared with the speed of light.

The instantaneous acceleration contribution from rocket thrust $T$ acting on remaining rocket mass $M$ is $a_T=T/M$. Gravity and other external forces contribute separately. A constant rocket thrust therefore produces an increasing thrust acceleration as rocket propellant is consumed.

Disruption of a self-gravitating body when differential gravitational acceleration across it exceeds its binding acceleration. For a body of radius $R$, mass $m$ at distance $r$ from a much more massive body $M$, the competing scales are $GMR/r^3$ and $Gm/R^2$; their equality gives $r\sim R(M/m{)}^{1/3}$, up to geometry and internal-structure factors. The tidal tensor gives the directional differential acceleration. A tidal disruption event near a black hole is a specific astronomical application.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

The acceleration tidal tensor is $T_{jk}={\partial }_k{g}_j=-{\partial }_j{\partial }_k\Phi$ for gravitational acceleration $\mathbf{g}=-\nabla \Phi$. It describes differential acceleration $\Delta \mathbf{g}\simeq T\Delta \mathbf{r}$. Some authors instead call the positive Hessian matrix ${\partial }_j{\partial }_k\Phi$ the tidal tensor, reversing this sign convention. For a point mass and the positive potential convention $\mathbf{g}=\nabla \psi$, the tensor is $Gm(3\mathbf{R}{\mathbf{R}}^{\mathsf{T}}/R^5-1/R^3)$.

The tidal tensor is the Jacobian matrix of the Newtonian gravitational field, rather than the gravitational acceleration itself. Its entries have units ${\mathrm{s}}^{-2}$: multiplying by a displacement produces an acceleration. For a smooth Newtonian gravitational potential, the tidal tensor is a symmetric matrix, since mixed partial derivatives commute. A spatially uniform gravitational acceleration has zero tidal tensor, however large that acceleration is.

A timelike geodesic has timelike tangent and vanishing covariant acceleration in an affine parameter. Choosing proper time gives $g(\dot{\gamma },\dot{\gamma })=-1$.

Physically, an ideal massive test particle in free fall follows a timelike geodesic when nongravitational forces and appreciable self-gravity can be neglected. The normalization above uses $c=1$; in SI units it is $g(u,u)=-c^2$. Zero covariant acceleration does not mean constant coordinate velocity: a dropped object accelerates relative to the ground. Conversely, a person standing on the ground is pushed away from free fall by the supporting surface. A photon follows a null geodesic, whose parameter cannot be proper time. See Tong's geodesics in spacetime.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

The 1972 paper Around-the-World Atomic Clocks: Observed Relativistic Time Gains by J. C. Hafele and others reports four cesium-beam atomic clocks flown around the world in October 1971, eastward and westward. Their readings were compared with the U.S. Naval Observatory reference time scale before and after travel. An electronic time-interval counter resolved differences to the nearest nanosecond; intercomparisons among the flying clocks continued during travel. The final ensemble results were $-59\pm 10\mathrm{ns}$ eastward and $+273\pm 7\mathrm{ns}$ westward; the quoted spreads are standard deviations. These are corrected clock offsets, not raw readings or aircraft travel times. Four clocks allowed identification of individual rate changes through patterns in their mutual differences. The authors also report magnetic shielding and laboratory orientation checks. Their analysis could not detect a disturbance common to every flying clock, and the small ensemble limits statistical conclusions. This is a concrete reunited-clock experiment with apparatus, readout and drift controls.^[ref]

The companion paper Around-the-World Atomic Clocks: Predicted Relativistic Time Gains by J. C. Hafele and others computes $-40\pm 23\mathrm{ns}$ eastward and $+275\pm 21\mathrm{ns}$ westward from the recorded flight paths. These are predictions, with flight-data uncertainty estimates. Both time dilation from motion and gravitational time dilation matter: aircraft clocks at altitude experience a different gravitational potential, while the rotating Earth means a ground clock is not the inertial home clock of our simple model. Eastward and westward motion change the aircraft's velocity relative to an approximately nonrotating inertial frame, not merely its speed relative to the ground. The measured agreement therefore tests combined special relativity and general relativity; the westward gain does not contradict the maximum-proper time result for fixed endpoints in Minkowski spacetime.^[ref]

A complementary microscopic test is Measurements of relativistic time dilatation for positive and negative muons in a circular orbit by J. Bailey and others. The authors measured muon decay lifetimes in a storage ring at Lorentz factor $\gamma =29.33$: $64.419\pm 0.058\mu \mathrm{s}$ for positive muons and $64.368\pm 0.029\mu \mathrm{s}$ for negative muons. The reported positive-muon lifetime agrees with the independently measured lifetime at rest and special relativity to a fractional error of $2\times 10^{-3}$ at 95% confidence. This is a decay-lifetime readout under continuous centripetal acceleration, rather than a reunion of portable clocks. It supports the clock hypothesis for this physical process and regime; the inspected abstract does not establish universal immunity of every clock mechanism to acceleration. ^[ref]

An instantaneous reversal is an idealization. A real traveller changes velocity over a finite time, follows a smooth world line, and has finite proper acceleration. Special relativity still calculates the elapsed proper time using the home inertial frame and the same integral; general relativity is not needed merely because the traveller accelerates in otherwise flat spacetime. The clock hypothesis identifies an ideal clock's reading with this geometric proper time.

One explicit smooth trip between the same $O$ and $R$ starts and ends at rest in the home inertial frame. ChooseIts velocity reverses continuously at $T/2$, and its proper acceleration is bounded because $v_{\max }<c$ and $dv/dt$ is finite. With $v_{\max }=0.6c$ and $T=10\mathrm{yr}$, numerical integration givesThis comparison fixes the reunion spacetime events and maximum speed, not the turnaround distance: the smooth trip reaches $v_{\max }T/\pi \simeq 1.910$ light-years, whereas the ideal trip reaches $3$ light-years. The smooth traveller spends less time at high speed and therefore accumulates more proper time than the ideal traveller, while still accumulating less than the home clock.

For a sufficiently short smooth reversal replacing the corner, the proper time approaches the piecewise inertial result. A reversal interval of home duration $\delta t$ contributes between zero and $\delta t$ to the clock reading; shrinking that interval cannot create a finite extra loss concentrated at the corner. The finite difference survives in the long coasting legs. Keeping the same reversal manoeuvre while lengthening those legs changes the age difference. Conversely, two travellers following mirrored paths with the same speed history accumulate equal proper times, even though both accelerate. Thus acceleration distinguishes paths and enables reunion here, but an amount of acceleration alone does not determine the age difference.

The changing simultaneity assignment also becomes continuous. At a traveller spacetime event $(t,x(t))$ with instantaneous velocity $v(t)$, the momentarily comoving inertial frame assigns home time $t_H=t-v(t)x(t)/c^2$. Differentiating givesDuring the reversal at positive $x$, the last term advances the assigned home time rapidly. It describes the traveller's changing relativity of simultaneity, not a locally measured alteration of the home clock's ticking.