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Let the rotating frame have origin acceleration ${\mathbf{A}}_0$ and angular velocity $\mathbf{\Omega }(t)$ relative to an inertial frame. For a vector referred to the moving axes, the transport rule isApply it twice to position. With rotating-frame velocity $\mathbf{v}$ and acceleration ${\mathbf{a}}_R$,Thus Newton's equation expressed in the rotating frame isThe added terms are translational, Coriolis, Euler and centrifugal fictitious forces. They reflect the frame's acceleration, rather than additional interactions with nearby objects.

The centrifugal force points outward from the rotation axis and, for constant $\mathbf{\Omega }$, has potential $-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. The Coriolis force is perpendicular to velocity, so does no work on the particle measured in that rotating frame; it changes direction rather than speed directly. The Euler force is present when the frame's rotation rate or direction changes.

On Earth, gravity and centrifugal acceleration combine into effective gravity. For slow horizontal motion at latitude $\lambda$, the vertical component of Earth's rotation gives Coriolis parameter $f=2\Omega \sin \lambda$: an initially northward motion deflects eastward in the northern hemisphere. This supplies a useful local explanation of rotating flows and Foucault precession. A bucket of rotating liquid illustrates the centrifugal potential: its steady free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$, giving $z={\Omega }^2{r}^2/(2g)+\mathrm{constant}$. These equations require a declared frame; adding a centrifugal force while retaining inertial-frame acceleration would count the same kinematic effect twice.

A spatially uniform external gravitational acceleration moves a body's centre of mass without stretching it. Tidal forces arise from the variation of that acceleration across the body. Let a distant mass $M$ lie at distance $D$ along unit vector $\mathbf{n}$, and let $|\mathbf{r}|\ll D$ denote position relative to the body's centre. Subtract the centre's acceleration. To first order,The tidal tensor is $T_{ij}=GM(3n_i{n}_j-{\delta }_{ij})/D^3$. It stretches along the line to the source with eigenvalue $2GM/D^3$ and compresses in the two perpendicular directions with eigenvalues $-GM/D^3$. Its trace vanishes outside the source, consistently with Laplace's equation for the gravitational potential.

After removal of the constant and uniform-acceleration terms, the tidal potential isAn ideal equilibrium ocean on a sphere of radius $R$ follows an equipotential of the body's gravity plus this perturbation. For surface gravity $g$, the leading displacement isup to a constant fixing total water volume. The quadrupolar form produces bulges on both the near and far sides; the far-side bulge follows from differential acceleration, without postulating a repulsive gravitational force.

The Moon and Sun both raise terrestrial tides, with relative strength set mainly by $M/D^3$, rather than by the total gravitational pull $M/D^2$. Their contributions reinforce near new and full Moon to produce spring tides, and partly cancel near quadrature to produce neap tides. Earth's rotation relative to the tidal pattern gives a strong semidiurnal component in the simplest geometry. Real tide heights and timing depend on ocean depth, coastlines, basin resonances, Coriolis acceleration, friction and the declination of the forcing bodies. Tidal dissipation produces phase lags and torques, exchanging rotational and orbital angular momentum and gradually altering the spin and orbit. The equilibrium bulge is therefore a useful starting model, not an exact local tide forecast.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

Integrating coaxial cylindrical shells givesRolling without slipping has spin speed $u/a$, so $K=m{u}^2/2+I{u}^2/(2a^2)=3m{u}^2/4$.

For the rolling-cylinder loop contact threshold, set $\ell =R-a$, the radius of the centre's circular path, and let $\theta$ run from the loop bottom. The centre starts $h$ above its lowest centre position. Its height gain from the bottom is $\ell (1-\cos \theta )$. Energy conservation givesAt the top, contact requires $N=m(u_t^2/\ell -g)\ge 0$, hence $u_t^2\ge g\ell$. Inserting the energy relation proves the limiting conditionThe same condition keeps $N\ge 0$ elsewhere on the circle, sinceis smallest at the top.

For the centre's acceleration, the inward normal component is $u^2/\ell$ and the tangential component is $-(2g/3)\sin \theta$. If $A=h/\ell -1\ge 7/4$ and $x=\cos \theta$,This convex quadratic has its largest endpoint value at $x=1$. The maximum on the circular loop is at the bottom, directed vertically upward. For the limiting release height,The approach track's curvature is unspecified, so a global bound for an arbitrary approach shape cannot be proved from the sketch alone; the result above concerns the specified circular part and the usual smooth approach interpretation.

The finite-friction qualification of a marginal rolling loop also matters: the contact threshold presumes the stated no-slip idealization. At equality, near the top, required tangential friction vanishes linearly with angular distance from the top but $N$ vanishes quadratically. Thus any finite coefficient of static friction requires a release height above this limiting bound to maintain rolling everywhere; no coefficient is provided. This does not change the requested ideal contact criterion.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.

The bob experiences weight $Mg$ vertically downward and a rod force towards the pivot. The tangential component of weight is $-Mg\sin \theta$, soThe rod force supplies the radial acceleration and contributes no tangential torque about the pivot.

To the same small-angle order, writing $\chi ={\omega }_{0}t+\varphi$,Thus $K+U=MgL{\theta }_0^2/2$ is constant. Put $b=\dot{\theta }(0)$, a signed angular velocity. ThenIf the printed angular speed $\beta$ is intended as the signed rate, set $b=\beta$. If it means magnitude only, $b=\pm \beta$ gives two possible phases; the amplitude is unique. At zero amplitude the phase is arbitrary.

For a finite-radius spherical pendulum bob, the PDF dimension $L$ runs from pivot to centre of mass. The parallel axis theorem gives $I_P=M{L}^2+2M{a}^2/5$. The restoring torque remains $-MgL\theta$, henceWith $a=[3M/(4\pi \rho ){]}^{1/3}=0.02605\mathrm{m}$,The finite bob increases the inertia without changing the gravitational lever arm.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

For equal mean mass densities, $M_a/M_E=(r/R_E{)}^3$. Thus the asteroid's surface gravitational acceleration is $g_a=g_{E}r/R_E$ and its escape speed isAn explicit estimate of jumping ability is an Earth jump height $H=0.50\mathrm{m}$, giving $v_J=\sqrt{2g_{E}H}\simeq 3.1\mathrm{m}{\mathrm{s}}^{-1}$. ConsequentlyThis is an order-of-magnitude estimate, scaling with assumed take-off speed; no unique human jump speed is supplied. A spherical nonrotating asteroid and negligible atmosphere are assumed.

A gravitational field is force per unit test mass, $\mathbf{g}=\mathbf{F}/m$. Newton's law of universal gravitation gives the field of a point mass M:The potential is potential energy per unit mass, with zero at infinity for an isolated mass. Fields and potentials superpose. The potential difference fixes the work per unit mass, and a conservative gravitational field has zero circulation around a closed path.

For a continuous mass distribution, the gravitational Gauss law is $\oint \mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$. A spherical body therefore has the same exterior field as a point mass at its centre. Inside a uniform sphere of density $\rho$,The field is continuous at its surface and vanishes at its centre. An empty spherical shell has zero interior field but a nonzero constant interior potential. Near a small region of Earth's surface one may approximate $\mathbf{g}$ as uniform and the potential change as $g\Delta h$. Farther away its spatial gradient produces tidal relative accelerations. Escape to infinity from radius R requires speed $\sqrt{2GM/R}$ when rotation and other bodies are neglected.

Take $q=\ddot{x}$ positive when the hanging mass descends and the other mass moves uphill. For the sliding block, Newton's second law gives $mg-T=mq$ and $T-Mg\sin \theta =Mq$. ThereforeFor the pulley coupled to a rolling disc, tension and weight act through its centre; the contact normal is perpendicular to the slope. Static friction provides the torque required for rolling without slipping. Define signed $F$ down the slope, so $F>0$ for uphill acceleration.

The rotational equation and rolling constraint give $Fa=Iq/a$, hence $F=Mq/2$. The translational equations become $T-Mg\sin \theta -F=Mq$ and $mg-T=mq$, givingThus the threshold remains $m>M\sin \theta$: the moment of inertia increases the effective mass but does not change the sign of the driving force. For downhill acceleration $F$ reverses. The normal force is $Mg\cos \theta$; static friction does no work at the stationary contact point and sufficient roughness is assumed.

The Lorentz force determines the motion of a charged particle:At nonrelativistic speeds, $\mathbf{p}=m\mathbf{v}$. The electric field does work at rate $q\mathbf{E}\cdot \mathbf{v}$, while the magnetic part of the Lorentz force is perpendicular to velocity and does no work. A uniform electric field gives constant acceleration $q\mathbf{E}/m$; with a transverse initial velocity, the trajectory is a parabola.

In a uniform magnetic field, the parallel velocity is constant, while the perpendicular component undergoes circular motion. The angular cyclotron frequency, orbit radius and helical pitch areThe sign of $q$ determines the sense of rotation. For perpendicular uniform electric fields and magnetic fields, the guiding centre has the electric drift velocityIndeed, $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$, so the remaining motion is cyclotron rotation about that drifting centre, with possible motion along the magnetic field. The drift is independent of electric charge sign and mass. At relativistic energies, retain $\mathbf{p}=\Gamma m\mathbf{v}$ and the corresponding relativistic orbit radius rather than assuming constant nonrelativistic acceleration.

Expand the cloud's potential energy about a stable equilibrium: $U(z)=U(0)+U^{\prime }(0)z+U^{\prime \prime }(0)z^2/2+\cdots$. Equilibrium gives $U^{\prime }(0)=0$, stability gives $U^{\prime \prime }(0)=k>0$, and reflection symmetry also removes odd terms. For small displacements, $U-U(0)\simeq k{z}^2/2$ and $F=-dU/dz\simeq -kz$.

Including the electric and damping forces, Newton's second law is $m\ddot{z}=qE(t)-b\dot{z}-kz$. Dividing by $m$ gives the Lorentz oscillator modelThe specified convention uses $2\gamma$ as the damping coefficient per unit mass.

For the steady response, represent the field by $E_0{e}^{i\omega t}$ and the displacement by $\tilde{z}{e}^{i\omega t}$, taking real parts for physical quantities. Substitution givesDefine the nonnegative amplitude $z_0=|\tilde{z}|$ and $\delta =atan2(2\gamma \omega ,{\omega }_0^2-{\omega }^2)$. ThenFor positive $q$, $\varphi =-\delta$. For electrons $q<0$, the phase of a negatively charged driven oscillator is $\varphi =\pi -\delta$: the low-frequency displacement is opposite to the field, rather than in phase with it. A signed amplitude could absorb this extra $\pi$, but the phase convention must then be stated. The induced dipole $p=qz$ has phase $-\delta$ for either charge sign. Transient free oscillations have been omitted because the question asks for steady state.

For the electron cloud, $q/m=-e/m_e$ andThe visible range corresponds to $\omega =2\pi c/\lambda$ between $2.35\times 10^{15}$ and $4.71\times 10^{15}\mathrm{rad}{\mathrm{s}}^{-1}$, below this resonance. The resonant wavelength is about $180\mathrm{nm}$, in the ultraviolet.

The acceleration amplitude is ${\omega }^2{z}_0$. At fixed incident field amplitude, the scattered intensity is therefore proportional toWell below resonance and outside strong damping corrections, $z_0$ is nearly frequency-independent and this becomes the Rayleigh scattering wavelength dependence $I_{\mathrm{sc}}\propto {\omega }^4\propto {\lambda }^{-4}$. The shortest visible wavelengths are consequently scattered much more strongly into directions away from the direct solar beam than the longest ones, producing a blue sky. The simple fourth-power approximation gives a factor 16 between 400 and 800 nm at equal incident intensity. Retaining the undamped denominator with the given resonance raises that factor to about 22.7, since the upper end is not extremely far below resonance. The exact ratio needs the unspecified damping and incident solar spectrum; the stronger scattering of the blue end is the robust conclusion here.

The model describes frequency-dependent scattering, not the detailed perceived color after solar-spectrum weighting and human visual response. It also explains why treating the electrons as free at visible frequencies would miss the bound-charge fourth-power limit.

A charge $q$ moving nonrelativistically in prescribed fields obeys the Lorentz force lawThe magnetic part is perpendicular to velocity, so $d(m{v}^2/2)/dt=q\mathbf{E}\cdot \mathbf{v}$: the electric field can change kinetic energy, while the magnetic field alone changes direction without doing work.

In a uniform electric field with no magnetic field, acceleration is constant: $\mathbf{v}={\mathbf{v}}_0+(q\mathbf{E}/m)t$ and $\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+q\mathbf{E}{t}^2/(2m)$. The trajectory is a parabola when there is initial motion perpendicular to the field, and a line when motion is purely along it. Reversing the charge reverses the electric acceleration.

In a uniform magnetic field with no electric field, resolve velocity into parallel and perpendicular components. The parallel component is constant; the perpendicular component rotates at the cyclotron frequencyThe resulting path is a helix, or a circle if $v_{\parallel }=0$, with pitch $2\pi v_{\parallel }/{\omega }_c$. Charge sign determines its sense of rotation. These relations underlie magnetic bending and mass/charge analysis. At relativistic energies, use momentum $\mathbf{p}=\gamma m\mathbf{v}$; at fixed energy the cyclotron frequency becomes $|q|B/(\gamma m)$.

For uniform crossed fields $\mathbf{E}\perp \mathbf{B}$, the velocity ${\mathbf{v}}_D=\mathbf{E}\times \mathbf{B}/B^2$ satisfies $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$. The orbit is a gyromotion about a guiding center drifting withindependent of mass and charge sign. A velocity selector transmits particles with the appropriate direction and speed $E/B$ without deflection. If the electric field has a component along the magnetic field, that component accelerates the parallel motion instead. Nonuniform fields can add other drifts; the simple helix and uniform crossed-field expressions require the stated uniform-field assumptions.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

The inclined parcel has tension $T$ up the slope, weight $mg$ vertically downward, normal reaction $N=mg\cos \theta$ and friction $\mu mg\cos \theta$ down the slope. The hanging mass has weight downward and tension upward. The ideal string makes the two acceleration magnitudes equal.

The velocity slope is $a=2.4/1.5=1.60\mathrm{m}{\mathrm{s}}^{-2}$, up the track for the parcel and downward for the hanging mass. The free-body diagrams show gravity itself as one vector; its slope-parallel and normal components are used in the equations rather than counted as additional forces.

For a prograde equatorial synchronous circular orbit, equate gravitational acceleration and centripetal acceleration. With planetary radius $R_M$, $GM=g_s{R}_M^2$, and rotation period $P$,The tabulated $g_s$ is a gravitational acceleration; its printed unit must be read as $\mathrm{m}{\mathrm{s}}^{-2}$, rather than $\mathrm{m}{\mathrm{s}}^{-1}$. Using $P=88642\mathrm{s}$ givesThe orbital altitude is measured from the surface, rather than the centre.

At the visible horizon, the line of sight is tangent to the spherical surface. The limiting central angle ${\theta }_h$ satisfies $\cos {\theta }_h=R_M/r_o$. The visible spherical cap has area $2\pi R_M^2(1-\cos {\theta }_h)$, givingAbout $41.7\%$ of the whole surface is geometrically visible, with ${\theta }_h\simeq 80.4^{\circ }$. This is a surface-area fraction, not an apparent-disc area ratio. Useful imaging near the horizon is further limited by foreshortening and the long atmospheric path.

Use a shallow, homogeneous active upper layer of mean thickness $h_0$ and mass density ${\rho }_1$, over a much deeper, effectively stationary layer with mass density ${\rho }_2>{\rho }_1$. The Boussinesq approximation keeps the small density contrast only in buoyancy, giving reduced gravity $g^{\prime }=g({\rho }_2-{\rho }_1)/{\rho }_0$. The hydrostatic approximation and shallow-water approximation make horizontal velocity nearly depth independent and turn the interface slope into the horizontal force $-g^{\prime }\nabla h$.

Linearize about rest and uniform $h_0$, with $|h-h_0|\ll h_0$, neglecting products of perturbations and convective acceleration. The f-plane approximation sets the Coriolis parameter to constant $f_0$. Neglect wind forcing, viscosity, entrainment and bottom drag. The two horizontal momentum equations then balance local acceleration, the Coriolis force and the pressure gradient. The signs $-f_{0}v$ and $+f_{0}u$ follow from eastward and northward coordinates.

The thickness equation is the linearized conservation of mass law: horizontal divergence decreases $h$, while horizontal convergence increases it. Here $h$ is active-layer thickness, not total ocean depth, and $g^{\prime }$ rather than $g$ gives the restoring force because the motion principally displaces the thermocline.

Assume a stable, lossless elastic body with the usual symmetry of the elasticity tensor and boundary conditions that eliminate the surface terms in integration by parts, such as a traction-free outer surface. Its elastic stiffness operator is then self-adjoint. When density varies spatially, the precise eigenproblem is generalized:Here $\alpha$ abbreviates the pair of mode labels. Equivalently, ${\rho }^{-1}\hat{H}$ is a self-adjoint operator in the mass inner product. Stability makes ${\omega }_{\alpha }^2$ real and nonnegative. Modes at different eigenvalues are orthogonal, and an orthonormal basis can be selected within each degenerate eigenspace. Completeness permits expansion of a suitable displacement field in these mass-weighted elastic normal modes.

Insert the expansion and project onto ${\mathbf{u}}_{\alpha }^{\ast }$. Mass normalization diagonalizes both the acceleration term and the restoring term. For a symmetric stress glut tensor supported away from the boundary,The symmetry of the stress glut tensor removes the antisymmetric part of the displacement gradient, leaving the strain. The result is one independent forced oscillator per mode, with the stated excitation integral. If the modes instead have mass norm $N_{\alpha }$, that integral is divided by $N_{\alpha }$. A source touching the boundary requires the corresponding surface forcing to be retained rather than silently dropping it.

Rigid translations and rotations can be zero-frequency modes of a free body. Their strain vanishes, so an internal symmetric seismic moment tensor does not excite them through this coupling. Attenuation, radiative boundaries or other non-self-adjoint effects require modifications of the simple conservative expansion.

Write the four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$ and four-acceleration $A^{\mu }=d{U}^{\mu }/d\tau$. Since $U_{\mu }{A}^{\mu }=0$, spatial orthogonality $\mathbf{U}\cdot \mathbf{A}=0$ implies $U^0{A}^0=0$. Thus $A^0=cd\gamma /d\tau =0$ and the speed is constant. If “acceleration” is read as the laboratory three-acceleration instead, $\gamma \mathbf{v}\cdot d\mathbf{v}/dt=0$ immediately gives the same conclusion.

For four-acceleration in uniform circular motion, the speed is $r|\omega |<c$, so $\gamma =(1-r^2{\omega }^2/c^2{)}^{-1/2}$ is constant. Differentiating the four-velocity using $d/d\tau =\gamma d/dt$ givesThe proper acceleration is $\sqrt{-A_{\mu }{A}^{\mu }}={\gamma }^{2}r{\omega }^2$. The vanishing time component reflects constant energy; the spatial components describe the changing momentum direction.

tidal forces measure differences in gravitational acceleration across an extended region. Neighboring freely falling particles each have zero four-acceleration, but their separation need not accelerate uniformly. Let $u^{\mu }$ be the tangent to a reference timelike geodesic and ${\xi }^{\mu }$ its infinitesimal separation from a neighboring one, chosen from an affinely parametrized geodesic congruence. Using the Riemann curvature tensor convention specified in Solution 3,Changing the overall curvature convention changes this displayed sign. The covariant derivatives compare vectors by parallel transport, avoiding a coordinate-dependent subtraction of accelerations.

In the Newtonian limit, expand the force field at two nearby points:For $\Phi =-GM/r$, a local orthonormal frame aligned with the radius givesThere is radial stretching and transverse compression; the trace vanishes outside the mass. Two freely falling test masses and a measurement of their changing separation implement gravity gradiometry. Their common acceleration drops out, while a relative acceleration proportional to separation remains. A local inertial frame removes Christoffel symbols at a point, but cannot remove nonzero curvature over a finite region.