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If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

Let $e_i^{\prime }(t)$ be a Cartesian basis fixed in the rotating frame. A vector $A=A_i^{\prime }{e}_i^{\prime }$ then satisfiesbecause a derivative of a body-fixed basis vector is ${\dot{e}}_i^{\prime }=\omega \times e_i^{\prime }$. Applying this transport formula twice to $r$ gives, for constant $\omega$,The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_{\varphi },e_z)$ with $e_z$ upward andProjection of $m\ddot{r}=-mg{e}_z+N$ along the wire tangent $\cos \theta e_r+\sin \theta e_z$ eliminates the smooth-wire normal force $N$ and gives

Multiply the three equations by $\dot{x},\dot{y},\dot{z}$ and add. The Coriolis acceleration does no work because its two terms cancel, leavingEquivalently, the radiation-modified Jacobi constant is

In a frame with unit angular velocity, transforming the acceleration introduces the Coriolis acceleration and centrifugal acceleration. Radiation reduces the attraction of $M_1$ by the factor $1-{\beta }_1$, soWith the effective potentialits Cartesian components are exactlyThis is the photogravitational restricted three-body problem.

For the Keplerian shearing sheet velocity $\mathbf{U}=-(3/2)\Omega x{\mathbf{e}}_y$, its material acceleration vanishes because $\mathbf{U}\cdot \nabla \mathbf{U}=0$. The $x$ component of the Coriolis acceleration is $-3{\Omega }^{2}x$, which is cancelled by $-\nabla {\Phi }_t=+3{\Omega }^{2}x{\mathbf{e}}_x$ with the signs in the stated equation; constant $P_0$ supplies no force. The flow is also incompressible.

Let $\mathbf{u}=\mathbf{U}+{\mathbf{u}}^{\prime }$ and $h^{\prime }=P^{\prime }/\rho$. Axisymmetry removes $\mathbf{U}\cdot \nabla {\mathbf{u}}^{\prime }$, while ${\mathbf{u}}^{\prime }\cdot \nabla \mathbf{U}=-(3/2)\Omega u_x^{\prime }{\mathbf{e}}_y$. Keeping first-order terms givesDifferentiate the momentum equations in time and use incompressibility to eliminate $h^{\prime }$. The radial velocity obeysFor a plane wave proportional to $e^{i(k_{x}x+k_{z}z-\omega t)}$, this gives the inertial wave dispersion relation

When $k_x=0$ and $k_z\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $h^{\prime }=0$. The remaining motion has $\omega =\pm \Omega$ and satisfies $u_y^{\prime }=\mp i{u}_x^{\prime }/2$: each horizontal layer executes an epicyclic motion, with the phase varying vertically but no pressure or vertical-velocity perturbation.

The term $-2\mathbf{\Omega }\times \dot{\mathbf{x}}$ is the Coriolis acceleration, and $-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})$ is the centrifugal acceleration. With $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$, the Cartesian components are

Dot the momentum equation with $\mathbf{u}$. The Coriolis acceleration does no work, while the buoyancy-variable equation giveswhich cancels the buoyancy work $-N^2\theta u_z$. ThusUsing incompressibility,Therefore

For ${\mathbf{u}}_0=-3\Omega x{\mathbf{e}}_y/2$, advection vanishes and the Coriolis acceleration exactly cancels $3{\Omega }^{2}x{\mathbf{e}}_x$, so constant $P_0$ and ${\theta }_0=0$ complete the equilibrium.

Write perturbations $(\mathbf{v},p,\vartheta )e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ with $\mathbf{k}=(k_x,0,k_z)$. Incompressibility gives $\mathbf{k}\cdot \mathbf{v}=0$. It also makes the quadratic terms $\mathbf{v}\cdot \nabla \mathbf{v}=i(\mathbf{k}\cdot \mathbf{v})\mathbf{v}$ and $\mathbf{v}\cdot \nabla \vartheta$ vanish exactly; the background-advection terms vanish because $k_y=0$. The amplitude equations areEliminating the amplitudes yields the inertia-gravity wave dispersion relation

For $N^2=0$, $\omega =\pm \Omega k_z/k$, so the group velocity isIt is perpendicular to $\mathbf{k}$ and hence to the phase velocity. An inertial wave packet transports energy along beams lying in its phase surfaces.

If $N^2>0$ and $k_x/k_z\to 0$, incompressibility suppresses vertical motion and $\omega \to \Omega$: horizontal epicyclic motion and rotation dominate. If $k_z/k_x\to 0$, radial motion is suppressed and $\omega \to N$: vertical buoyancy oscillations dominate. Intermediate ratios give hybrid inertia-gravity waves.

If $N^2<0$, exponential growth occurs exactly whenOnly modes with sufficiently large radial wavenumber permit enough vertical displacement for unstable buoyancy to overcome rotational restoration; the other orientations remain stabilized by the Coriolis force.

The first two equations are horizontal momentum balances: local acceleration plus the Coriolis acceleration equals the pressure-gradient force. The third is linearized mass conservation: convergence raises the free surface, represented by $\varphi$, while divergence lowers it. The constant $c$ is the long-wave gravity-wave speed.

For a Keplerian shearing sheet, the shear rate is $S=3\Omega /2$ and the shearing-sheet tidal potential isA uniform steady solution isThe Coriolis acceleration of this linear shear flow balances the radial tidal acceleration. The only nonzero background component of the viscous stress tensor that matters is $T_{xy}=-\nu ({\Sigma }_0){\Sigma }_{0}S$, which is spatially constant. Hence $\nabla \cdot \mathbf{T}=0$: a local uniform patch has no stress gradient or torque divergence to drive an accretion flow.

For the axisymmetric perturbation, defineLinearization about the uniform dust density and drifting equilibrium givesThe factors $2{\Omega }_K$ and ${\Omega }_K/2$ are the Coriolis acceleration and Keplerian-shear couplings. Applying ${\mathcal{D}}_{\tau }({\mathcal{D}}_{\tau }^2+{\Omega }_K^2)$ to the continuity equation and using the three momentum equations eliminates ${\mathbf{u}}^{\prime }$. All remaining forcing terms are proportional to $e^{-i\omega t}$, sowhere $F$ is independent of time.

Let $\delta ={\Sigma }^{\prime }/{\Sigma }_0$ and define the derivative following the background Keplerian shear byThe linearized continuity and momentum equations areThe coefficient $\Omega /2$ combines the background shear with the Coriolis acceleration.

At latitude ${\theta }_0$, resolve the planetary angular velocity asFor velocity $(u,v,w)$, the Coriolis acceleration isThe traditional approximation drops the terms involving the horizontal rotation component $\Omega \cos {\theta }_0$. The retained horizontal force is thereforeIts direct velocity-scale requirement isFor an incompressible flow, $W/U=O(H/L)$, so a sufficient condition istogether with the small aspect ratio that makes hydrostatic pressure the leading vertical momentum balance. The approximation consequently becomes delicate near the equator.

The beta plane is the local Taylor expansionwhereand $a$ is the planetary radius. It requires a local Cartesian region,and $H/a\ll 1$. Away from the equator, treating the variation as a perturbation of an f-plane also requires $\beta L/|f_0|\ll 1$; near the equator one instead retains the linear term as the leading Coriolis parameter.

Assume an inviscid homogeneous ocean, no horizontal pressure gradient, and no wind stress. Horizontal uniformity then removes advective acceleration, and the parcel equations on a beta plane arewith $u=\dot{x}$ and $v=\dot{y}$.

Since $\dot{u}=f(y)\dot{y}$, integration from the stated initial data givesThis is conservation of the parcel's zonal canonical momentum: its zonal speed records the meridionally accumulated Coriolis acceleration. Multiplying the two momentum equations by $u$ and $v$ and adding givesThus kinetic energy and speed are constant:Eliminating $\dot{x}$ gives the required ordinary differential equation

For $f_0,\beta >0$, the equator is at $y_e=-f_0/\beta$. ThereThe parcel must encounter a meridional turning point before reaching the equator if it is to remain strictly in the Northern Hemisphere. The energy relation therefore requiresor $\beta V/f_0^2<1/2$. Equality is the limiting trajectory that reaches the equator with zero meridional speed.

For ${\mathbf{u}}_0=-\frac{3}{2}\Omega x{\mathbf{e}}_y$, advection and viscosity vanish, while its Coriolis acceleration cancels $-\nabla {\Phi }_t$; constant pressure and $\theta =0$ complete the equilibrium. For axisymmetric perturbations,

Linearize the inviscid momentum equation in the uniformly rotating frame. The Coriolis acceleration is $2\mathbf{\Omega }\times \delta \mathbf{u}$, and the equilibrium pressure gradient cancels gravity. For perturbations proportional to $e^{i{k}_{x}x-i\omega t}$, the horizontal components areThe vertical component is

Linearizing mass conservation,givesFinally, linearizing the adiabatic pressure equation givesThese are the stated five equations. Self-gravity contributes no perturbation because it is neglected, and the equilibrium centrifugal term has already been absorbed or omitted.

For the incompressible velocity $u_x=\alpha y$, $u_y=-\beta x$ in a shearing sheet, advection produces $-\alpha \beta (x,y)$ and the Coriolis acceleration produces $(2\Omega \beta x,2\Omega \alpha y)$. Including the radial tide $2\Omega Sx$ gives $p/\rho =(A{x}^2+B{y}^2)/2+C$. A material ellipse has $\beta =r^2\alpha$. A vortex patch is immersed in surrounding fluid, so this pressure need not vanish on its boundary; imposing that condition would describe a different free-boundary problem.

A dilute dust component can be approximated as a pressureless fluid with velocity $\mathbf{v}$ and surface density of a disk ${\Sigma }_d$. In a gas velocity $\mathbf{u}$, linear drag force gives acceleration $-(\mathbf{v}-\mathbf{u})/t_s$ for positive aerodynamic stopping time $t_s$. Its equations include Coriolis acceleration, the shearing-sheet tidal potential, and ${\partial }_t{\Sigma }_d+\nabla \cdot ({\Sigma }_d\mathbf{v})=0$. Neglecting gas backreaction is justified only when the prescribed-gas approximation is appropriate. Strong orbit crossing or appreciable dust velocity dispersion can invalidate the single-valued pressureless-fluid description.

The Rossby number compares inertial acceleration with Coriolis acceleration. For speed $U$, horizontal length $L$, and Coriolis parameter $f$, it is $\mathrm{Ro}=U/(|f|L)$. Small Rossby number supports geostrophic balance and the quasi-geostrophic approximation.

A rotating fluid is described in a rotating reference frame, where the Coriolis acceleration and centrifugal acceleration supplement the physical forces.