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A laboratory gravity gradiometry experiment used two vertically separated clouds of chemical atoms in vacuum, launched into free fall. Shared laser beams applied three pulses ($\pi /2,\pi ,\pi /2$), splitting and recombining matter waves. Balanced photodiodes measured probe-light absorption to infer internal-state populations; scanning the last pulse's wave phase produced an interference fringe.^[ref]

The difference of fringe wave phases obeys $\delta \varphi ={\mathbf{k}}_{\mathrm{eff}}\cdot \Delta \mathbf{g}{\tau }^2$, with effective wave vector ${\mathbf{k}}_{\mathrm{eff}}$ and pulse spacing $\tau$: shared reference motion cancels, leaving differential gravitational acceleration. Eight lead bricks, each approximately $12.5\mathrm{kg}$, were placed symmetrically around the lower chamber, about $0.2\mathrm{m}$ from the fountain apex. The measured signal was $(8.1\pm 2.1)\times 10^{-9}g$, versus a calculated $8.2\times 10^{-9}g$, where $g$ denotes terrestrial gravitational acceleration.^[ref]

Controls deliberately shook the shared reference reflector and compared actively isolated and unisolated operation. Tilting tests exposed sensitivity to Coriolis acceleration and centrifugal acceleration. Thus shared translation is rejectable, while rotation and finite measurement uncertainty remain limitations. The brick signal detects spatially varying gravity; it does not establish a precision value of the gravitational constant.^[ref]

For a smooth horizontal circular hoop of radius $a$, rotated at constant angular velocity $\omega$ about a vertical axis through a rim point, measure $\theta$ at that pivot from the diameter toward the bead. In rotating Cartesian coordinate system coordinates, $\mathbf{r}=a(1+\cos 2\theta ,\sin 2\theta ,0)$. The tangential component of the centrifugal acceleration gives the displayed equation; the Coriolis acceleration is normal to the hoop. The farthest point $\theta =0$ is stable for $\omega \ne 0$, with small oscillation angular frequency $|\omega |$. The pivot itself is unstable. This geometry differs from a vertical rotating hoop and has no tangential gravitational term.

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.

For a locally eastward aircraft velocity at northern latitude $\lambda$, the horizontal Coriolis acceleration is southward with magnitude $2\Omega v\sin \lambda$. Banking to direct part of the lift northward cancels this acceleration. If vertical lift balances weight, the force ratio gives the displayed angle. Including vertical Coriolis acceleration replaces $g$ by $g-2\Omega v\cos \lambda$. This is a local rotational correction, not a full navigation model of motion along a curved latitude circle.

At colatitude $\theta$, ignore terms quadratic in Earth angular speed and use uniform vertical gravity. A vertically launched particle has zeroth-order height $z=vt-g{t}^2/2$. Its eastward Coriolis acceleration is $-2\omega \sin \theta \dot{z}$, so its horizontal velocity is $-2\omega \sin \theta z$. Integrating over the flight time $2v/g$ gives the westward displacement displayed above. Descent reverses the horizontal acceleration but does not cancel the accumulated displacement.

Use local axes radial, tangential in the orbital plane, and perpendicular to that plane around a circular Kepler orbit with ${\Omega }^2=GM/R^3$. The gravitational tidal tensor is ${\Omega }^2\mathrm{diag}(2,-1,-1)$. A uniformly corotating frame adds the gradient of centrifugal acceleration, ${\Omega }^2\mathrm{diag}(1,1,0)$, giving the displayed effective tensor for stationary offsets. Thus the radial differential acceleration is $3GMa/R^3$. These are accelerations, or forces per unit test mass; an actual differential force includes that mass. Moving offsets also experience Coriolis acceleration.

For constant density ${\rho }_0$ and body force ${\Omega }^2(3x-2qz){\mathbf{e}}_x$, dotting the incompressible flow momentum equation with ${\rho }_0\mathbf{u}$ gives the kinetic energy flux $(K+P)\mathbf{u}$ and source ${\rho }_0{\Omega }^2(3x-2qz)u_x$, where $K={\rho }_0|\mathbf{u}{|}^2/2$. Coriolis acceleration does no work. Absorbing ${\Phi }_t=-3{\Omega }^2{x}^2/2$ gives the displayed balance. For nonzero $q$, the remaining force has curl $-2{\Omega }^{2}q{\mathbf{e}}_y$, so it cannot be incorporated into a scalar mechanical potential. The model can exchange energy with its maintained vertical-shear background; at $q=0$ ordinary Jacobi energy in a shearing sheet conservation is recovered under zero boundary flux.

For constant mass density ${\rho }_0$, a frame with constant angular velocity $\mathbf{\Omega }$ absorbs centrifugal acceleration into reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. Indeed $\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})=-\nabla |\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. Transforming the material acceleration in the Euler equations gives the displayed equation. The Coriolis acceleration remains velocity dependent and cannot be absorbed into a scalar potential. A conservative physical body force can be included by adding its potential to $\Pi$.

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 $m{u}^2/2$.

For ideal magnetohydrodynamics at constant mass density $\rho$, a shearing sheet with velocity $-q\Omega x{\mathbf{e}}_y+{\mathbf{v}}_h(z,t)$ and magnetic field ${\mathbf{B}}_h(z,t)+B_z{\mathbf{e}}_z$ obeysHorizontal invariance and vanishing vertical velocity eliminate nonlinear advection. These equations retain Coriolis acceleration, orbital shear and magnetic tension. The vertical pressure adjusts to balance gravity and magnetic pressure.

In a uniform-density incompressible flow inside a smooth rigid sphere rotating at angular speed $\Omega$, a normal mode with phase $e^{i\sigma t+im\varphi }$ has intrinsic frequency $s=\sigma +m\Omega$. For $s\ne 0,\pm 2\Omega$, eliminating velocity from the linearized Euler equations gives the displayed pressure equation, with $W=p^{\prime }/\rho$. Impermeability gives $R{W}_R+(2m\Omega /s)W+(1-4{\Omega }^2/s^2)z{W}_z=0$ on the sphere. There is no condition $W=0$: the rigid wall supplies variable normal pressure. These inertial waves are restored by Coriolis acceleration, and their pressure equation is hyperbolic for $|s|<2|\Omega |$.

For constant isothermal sound speed, time-independent shearing-sheet tidal potential ${\Phi }_t$, and inviscid flow, define $E=\Sigma u^2/2+c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})+\Sigma {\Phi }_t$. The barotropic energy density and kinetic energy balances combine to give ${\partial }_{t}E+\nabla \cdot [(E+P)\mathbf{u}]=0$. Coriolis acceleration does no work. Integral conservation requires vanishing net boundary energy flux.

The time-independent particle Lagrangian in a shearing sheet has conserved rotating-frame energyThe velocity-linear Coriolis acceleration term cancels from this expression. Up to the reference-orbit constant, it is the second-order expansion of inertial specific orbital energy minus ${\Omega }_0$ times inertial specific angular momentum. Its negative radial tidal term allows inelastic collisions to lower the total energy while increasing the radial extent of a ring.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

The material-boundary and vorticity conditions give $\alpha =S/[r(r-1)]$ and $\beta =Sr/(r-1)$. A pressureless particle subject to linear drag, Coriolis acceleration and tidal gravity has the drag-polynomial stability for all positive stopping rates with $K=2\Omega (2\Omega -S)$, $L=\Omega (\alpha +\beta -S)$ and $P=\alpha \beta$. In Keplerian shear, $K/{\Omega }^2=1$, $L/{\Omega }^2=3(r+1)/[2r(r-1)]$ and $P/{\Omega }^2=9/[4(r-1{)}^2]$. The exact criterion is $r\ge 3$, including the boundary $K=L>P$ at $r=3$. Attraction is local to the centre under the prescribed interior velocity field; a trajectory leaving the patch cannot be followed using that field alone.

Expand a unit-mass particle's Lagrangian about an axisymmetric circular orbit of radius $r_0$ and frequency ${\Omega }_0$, using $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Circular-orbit balance removes the linear radial term; total-time-derivative invariance of a Lagrangian removes $r_0{\Omega }_0\dot{y}$. To second order,Here $q$ is the orbital shear parameter and ${\Omega }_z$ is the vertical epicyclic frequency. The Euler-Lagrange equations retain the local Coriolis acceleration and tidal gravity.

Let $\hat{\mathbf{u}}$ and $\hat{p}$ be the complex plane wave amplitudes and set $k_x=k_{\perp }$, $k_z=k_{\parallel }$. The linearized Euler equations in the rotating frame areHere $p$ is pressure per unit mass density, including the centrifugal potential. The last line is incompressibility, not an additional equation of state.

Taking the curl eliminates pressure. For constant $\mathbf{\Omega }$ and a divergence-free vector field, $\nabla \times (2\mathbf{u}\times \mathbf{\Omega })=2(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$. Thus the inertial-wave vorticity equation givesApply $\mathbf{k}\times$ once more and use $\mathbf{k}\times (\mathbf{k}\times \hat{\mathbf{u}})=-k^2\hat{\mathbf{u}}$. For a nonzero oscillatory amplitude,so the inertial-wave dispersion relation isThe restoring influence is the Coriolis acceleration. In particular $|\omega |\le 2|\Omega |$. For $k_{\parallel }=0$, the projected linear dynamics have zero frequency; these are steady balanced disturbances rather than the nonzero-frequency branch used in the division above. The spatially uniform $k=0$ mode must be treated separately from this dispersion relation.

Let $L_{\perp }=k_{\perp }^{-1}$ and let $U$ be the common size of horizontal and vertical velocity perturbations, as found in part b. The linear inertial wave time and nonlinear advection time areThe critical balance in a rapidly rotating fluid is thereforeThe Rossby number is of order $ϵ$. Horizontal geostrophic flow balances the leading pressure gradient against Coriolis acceleration, soThese are dimensionally comparable statements, rather than comparisons of a dimensional frequency with a dimensional velocity. Equivalently, use $L_{\perp }$ and $(2\Omega {)}^{-1}$ as units. ThenThus rotation stays $O(1)$ in these units, while the motions evolve on the slower $O({ϵ}^{-1})$ time. Vertical advection $u_{\parallel }{\partial }_z$ is smaller than horizontal advection by $ϵ$.

Write ${\mathbf{u}}_{\perp }={\mathbf{u}}_{\perp }^{(0)}+{\mathbf{u}}_{\perp }^{(1)}+\cdots$, with ${\mathbf{u}}_{\perp }^{(1)}=O(ϵU)$, and $p=p_0+p_1+\cdots$. At leading horizontal order,Define the stream function $\Phi =p_0/(2\Omega )$, including its horizontally uniform part in this convention. ThenThe next-order incompressibility condition is not zero horizontal divergence:This term has size $ϵk_{\perp }U$ and is essential: its product with the large rotation rate is a leading contribution to the vertical vorticity equation.

To see that contribution explicitly, write $D_{\perp }={\partial }_t+{\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }$. At the first horizontal inertial order,Taking its vertical curl eliminates $p_1$. The two-dimensional advection identity and the two divergence orders giveEquivalently the ageostrophic flow satisfieswhose divergence is $-D_{\perp }\zeta /(2\Omega )$. This explicitly verifies that the next-order divergence has the required sign.

The vertical momentum equation, at the same slow order, isFinally ${\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }f=\{\Phi ,f\}$, where the canonical two-dimensional Poisson bracket is $\{f,g\}=f_x{g}_y-f_y{g}_x$. The reduced equations for an anisotropic rotating fluid areThe omitted vertical-advection, ageostrophic-advection and relative-vorticity stretching terms are smaller by $O(ϵ)$ in the equations retained here. Linearizing these equations recovers ${\omega }^2=4{\Omega }^2{k}_{\parallel }^2/k_{\perp }^2$, the anisotropic limit of part a.

For a uniform axial magnetic field, the anisotropic Shear Alfvén wave is predominantly horizontally polarized, whereas the pseudo-Alfvén wave is predominantly vertically polarized. Without rotation both have characteristic frequency ${\omega }_A\sim |k_{\parallel }|v_A$. Pressure projection makes the effective Coriolis acceleration couple these two transverse wave polarization states at the inertial-wave rateThus the useful comparison is not $\Omega$ alone, nor anisotropy alone, butIn this limit rotational coupling of shear and pseudo-Alfvén waves is negligible on an Alfvén propagation time, so the two familiar linear polarizations are decoupled to leading order. The factor $k_{\parallel }$ cancels: simply making both disturbances more anisotropic does not by itself remove their relative rotational coupling.

For a nonlinear reduction one must also make the rotational mixing slow compared with the nonlinear evolution being retained. Under Alfvénic critical balance, $|k_{\parallel }|v_A\sim k_{\perp }U$, the same inequality gives ${\omega }_I\ll k_{\perp }U$. Rapid rotation relative to the horizontal nonlinear rate can therefore coexist with leading Alfvénic decoupling in the windowHigh electrical conductivity makes magnetic diffusion slow, and the corresponding inviscid approximation is also assumed. If $2\Omega \gtrsim k_{\perp }{v}_A$, rotational and magnetic restoring effects are comparable and the modes are mixed magneto-Coriolis disturbances; in the rotation-dominated limit they separate into fast inertial-like and slow magnetostrophic branches, rather than into independently evolving shear and pseudo-Alfvén polarizations.

At finite amplitude, independence is generally one-way in the usual anisotropic ideal magnetohydrodynamics reduction: the Alfvénic subsystem has no leading feedback from pseudo-Alfvénic perturbations, but it can still advect them. The small rotational-coupling condition removes an additional two-way coupling, rather than all nonlinear advection.

This is an asymptotic finite-timescale statement. The two nonrotating Alfvén polarizations are degenerate, so even a weak nonzero rotational coupling can rotate their superposition over the much longer time ${\omega }_I^{-1}$. Exact decoupling for arbitrarily long times does not follow from a small ratio alone. The strictly $k_{\parallel }=0$ nonpropagating limit is separate.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.