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For uniform angular velocity $\omega$, potential $\varphi$, and additional forces doing no work in the rotating frame,The Coriolis force does no work; the centrifugal potential accounts for the remaining apparent force. This energy is conserved precisely when the potential has no explicit time dependence in the rotating coordinates. An inertial time-independent conservative force need not meet that condition unless its potential is also invariant under the frame's rotation. The balance prevents applying a rotating-frame conservation law to a nonaxisymmetric stationary inertial potential.

For a smooth plane at inclination $\theta$ rotating about a vertical axis at constant angular velocity $\omega$, choose rotating coordinates $\xi$ horizontally and $\eta$ uphill, with their cross product the upward normal vector. A sliding particle under gravity satisfiesThe normal reaction is $N=m(g\cos \theta -2\omega \dot{\xi }\sin \theta +{\omega }^2\eta \sin \theta \cos \theta )$ while contact is maintained. The Coriolis force does no work in the rotating frame, giving the conserved specific energyThe last term is the centrifugal potential; this quantity need not equal the inertial mechanical energy.

Use pressure per unit constant density, $\pi$, with the gravitational potential and centrifugal potential absorbed into it. Strict geostrophic balance is the exact balancewith material acceleration omitted, for a homogeneous incompressible flow. Writing $f=2\Omega \ne 0$, its components are $fv={\pi }_x$, $fu=-{\pi }_y$ and ${\pi }_z=0$. Differentiating the first two equations with respect to $z$ gives $u_z=v_z=0$. Alternatively, the vector calculus identityshows, on taking the curl of the balance, that $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus $w_z=0$ too, proving the Taylor–Proudman theorem:The physical pressure still includes its background hydrostatic pressure; it is the reduced pressure $\pi$ that is independent of $z$. Between horizontal impermeable boundaries, the vertically constant $w$ is zero.

For boundaries $z=z_b(x,y)$ and $z=z_t(x,y)$, impermeability condition gives $w_b={\mathbf{u}}_H\cdot {\nabla }_H{z}_b$ and $w_t={\mathbf{u}}_H\cdot {\nabla }_H{z}_t$. A nearly columnar flow crossing contours of $H=z_t-z_b$ must therefore have $w_t-w_b\simeq {\mathbf{u}}_H\cdot {\nabla }_{H}H\ne 0$. Exact columnarity must give way to small corrections from ageostrophic flow. The leading horizontal geostrophic flow can remain nearly independent of $z$, and the leading reduced pressure has ${\pi }_z=0$, but $w_z$ is now nonzero and exact three-dimensional strict geostrophic balance no longer holds. For small slopes, the dominant vertical vorticity equation iswhere $q=v_x-u_y$ is relative vorticity. Equivalently, $D[(q+f)/H]/Dt\simeq 0$. Crossing geostrophic contours stretches or squashes columns: increasing $H$ produces cyclonic relative vorticity, decreasing $H$ produces anticyclonic relative vorticity. The full vorticity equation also contains tilting terms, neglected at this columnar order.

For the circular shallowing, fluid arriving from upstream has potential vorticity $f/h$. Its height is $h(1-ϵ)$ inside, so conservation givesThis uses the stated column-stretching description and initially open streamlines. To obtain the leading disturbance velocity field, use the conventional streamfunction definition $u=-{\Psi }_y$, $v={\Psi }_x$, and write $\Psi =-Uy+\psi (r)$. Then $q={\nabla }_H^2\psi =r^{-1}(r{\psi }_r{)}_r$. Regularity at $r=0$ gives ${\psi }_r=-fϵr/2$ inside. Outside, ${\psi }_r=B/r$; matching tangential velocity at $r=a$ fixes $B=-fϵa^2/2$. Choose the irrelevant additive constant so thatBoth $\psi$ and ${\psi }_r$ match at $a$, so there is no extra vorticity sheet on the rim. This is the circular topographic anticyclone. The radial expression specifies the disturbance; the complete streamfunction is $\Psi =-Uy+\psi$, since a radial function alone cannot represent the required uniform far-field current. Reversing the streamfunction sign convention reverses the displayed proportionality constant, but leaves the physical velocity unchanged.

Differentiation gives the full relative velocity field in the rotating frame:away from the narrow transition across the change of depth. The disturbance turns clockwise for $fϵ>0$ and decays as $1/r$. The horizontal field is the leading small-$ϵ$ columnar inversion: the exact variable-depth transport across a sharp rim needs higher-order corrections, rather than an exactly two-dimensional velocity continuous across a literal vertical step.

Take $U>0$ and $ϵ>0$. The greatest opposing zonal disturbance speed is $Ca$, attained at $(0,-a)$. If $Ca<U$, then $u>0$ everywhere and every particle advances in $x$, precluding closed streamlines. If $Ca>U$, the interior stagnation point $(0,-U/C)$ is a center, since the interior contours are circles about that point. The exterior stagnation point $(0,-C{a}^2/U)$ is a saddle: linearization there has real eigenvalues of opposite sign. Closed contours around the center are bounded by a separatrix through the saddle. The two points meet at the rim at onset, giving the closed-streamline threshold for a circular topographic anticyclone:At this threshold $U/(fa)=ϵ/2\ll 1$, consistent with small Rossby number. Equality is the onset, not a finite-area closed cell. Beyond onset, upstream data alone no longer determine the potential vorticity of trapped fluid; the threshold follows from the continuation of the supplied open-flow model.

The variable $\psi$ is a rotating-flow pressure potential including the conservative body-force and centrifugal potentials. For constant mass density,up to an additive function of time, where $\Phi$ is any imposed conservative potential. The last term is the centrifugal potential; its negative gradient is the centrifugal acceleration. The Coriolis force remains explicit in the equation because it is velocity-dependent and cannot be absorbed into a scalar position potential.

The further constraint is incompressible flow:For the specified shear, ${\mathbf{u}}_0\cdot \nabla {\mathbf{u}}_0=0$ and ${\nabla }^2{\mathbf{u}}_0=0$. Its Coriolis acceleration is $4\Omega Ax{\mathbf{e}}_x$, so the corresponding total basic pressure potential can be chosen as ${\psi }_0=-2\Omega A{x}^2$. This supplies the force balance required for the basic state used below.

The Time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

In the rotating-hoop bead problem, use a signed angle in the hoop plane from the downward vertical. The gravitational potential energy is $-mga\cos \theta$, the rotating-frame speed is $a\dot{\theta }$, and the distance from the rotation axis is $a\sin \theta$. The centrifugal potential givesGravity is stationary in these rotating coordinates, and the frictionless constraint does no work. Projecting the rotating equation along the hoop, or differentiating its effective potential in the one-dimensional Euler-Lagrange equation, givesIf ${\omega }^2>g/a$, the two off-axis positions areTheir unsigned angle to the downward vertical is the same; they lie on opposite sides of the hoop. The second derivative of the effective potential there is $m{\omega }^2{a}^2{\sin }^2{\theta }_0>0$. Equivalently, writing $\theta ={\theta }_0+\delta$ gives to first orderThe same holds at $-{\theta }_0$. Thus both equilibria are stable, with small-oscillation angular frequency $|\omega |\sin {\theta }_0$.

In the inertial frame the steady convective acceleration is $-R{\Omega }^2{\mathbf{e}}_R$. For a barotropic fluid, $\nabla p/\rho =\nabla h$, where $h$ is the barotropic enthalpy function. The equilibrium momentum equation therefore giveson each connected fluid region. Uniform rotation is essential: it makes the centrifugal acceleration derivable from the centrifugal potential $-{\Omega }^2{R}^2/2$.

The Cowling approximation sets the gravitational-potential perturbation to zero while retaining background gravity. The barotropic pressure force linearizes as $\delta (\nabla p/\rho )=\nabla \delta h$, with $\delta h=p^{\prime }/\rho =c_s^2{\rho }^{\prime }/\rho$. Denote its mode amplitude by $W$. The advective time derivative on a scalar mode is ${\partial }_t+\Omega {\partial }_{\varphi }$, so the stated positive-frequency exponential produces $i\sigma$ with $\sigma =\omega +m\Omega$. Linearizing the radial centrifugal term produces $-2\Omega v_{\varphi }$, while linearizing the azimuthal convective term produces $2\Omega v_R$. ThusThese are the pressure-gradient and Coriolis acceleration terms in the corotating-frame form. Independently, linearizing continuity equation givesThe equilibrium mass density is axisymmetric, so no additional azimuthal background-density derivative appears.

For $\sigma \ne 0$ and $D=4{\Omega }^2-{\sigma }^2\ne 0$, invert the horizontal momentum system:Inserting these into the continuity equation, dividing by $i\sigma$ and multiplying by $D$ givesThe two terms proportional to $\rho W_R$ cancel; this cancellation leaves precisely the background-density derivative in the last term. Since $-D/{\sigma }^2=1-4{\Omega }^2/{\sigma }^2$, this is exactlyThe singular frequency cases require the original velocity equations rather than this inversion.

In the low-frequency anelastic approximation for a rotating barotropic star, neglect the left side. Spherical background mass density satisfies ${\rho }_R/R={\rho }_z/z$, understood by its smooth limiting form on the equatorial plane. For $W=z{R}^m$,The remaining equation reduces toWriting $q=2\Omega /\sigma$, the bracket factors as $(m+1-q)(q+1)$. The regular desired branch is thereforeThe other algebraic factor corresponds to $\sigma =-2\Omega$, where the eliminated horizontal system is singular; it is not classified by this pressure-equation inversion. A constant-density background makes the bulk factor identically zero but does not invalidate the displayed solution.

For the regular branch, an explicit velocity check is particularly informative. Up to a common mode normalization,It satisfies $R{v}_R+z{v}_z=0$ and $\nabla \cdot \mathbf{v}=0$. Thus $\nabla \cdot (\rho \mathbf{v})=0$ for every spherical mass density profile, directly verifying the anelastic continuity condition. The motion is tangential to spherical shells and is the sectoral inertial mode of a slowly rotating barotropic star. For single-valued azimuthal modes, $m$ is a positive integer. With the source's exponential convention,For $m\ge 2$ the pattern is retrograde relative to the star but prograde in the inertial frame. This is a leading slow-rotation anelastic mode, not an exact solution of the compressible equations whose left-hand term was discarded.

Let $\Psi$ be the gravitational potential, with ${\nabla }^2\Psi =4\pi G\rho$, and let ${\varpi }^2=x^2+y^2$ measure distance from the rotation axis. Uniform solid-body rotation contributes the centrifugal potential $-{\Omega }^2{\varpi }^2/2$. In the corotating frame, static force balance therefore gives the uniformly rotating stellar hydrostatic equilibrium equationsThe final coefficient follows because ${\nabla }^2(x^2+y^2)=4$.

A tangent displacement along a regular connected equipotential surface has zero dot product with $\nabla \varphi$, and hence with $\nabla P$. Thus $P=P(\varphi )$ locally. Taking the curl of the hydrostatic equilibrium equation gives $\nabla \rho \times \nabla \varphi =0$, so $\rho =\rho (\varphi )$ on such surfaces as well. The Poisson equation then shows that ${\nabla }^2\varphi$ is constant on each equipotential surface.

The magnitude $g=|\nabla \varphi |$ need not be constant. For instance $\varphi =A(x^2+y^2)+B{z}^2$, with positive $A\ne B$, has constant Laplacian but different pole and equator gradient magnitudes on an ellipsoidal equipotential. In a rotating star, flattening and centrifugal acceleration likewise make the effective gravity latitude dependent.

With uniform composition, an equation of state determines $T$ from $P$ and $\rho$; for the supplied ideal gas plus radiation pressure law its right side increases monotonically with $T$. Hence $T=T(\varphi )$, and the radiative conductivity $\chi (\rho ,T)$ is also constant on an equipotential surface. Put $f(\varphi )=-\chi dT/d\varphi$. Then $\mathbf{F}=f(\varphi )\nabla \varphi$. On the stellar surface, below mass shedding, the outward normal is $\nabla \varphi /g$, soThis is the radiative gravity-darkening law. The effective temperature measures emergent flux; it is not the same quantity as the interior material temperature that was constant on an equipotential.

A rapidly rotating star is oblate, with reduced effective gravity and cooler effective temperature around the equator. The poles are hottest. The original sketch uses a Roche model of a uniformly rotating star only to illustrate the shape and surface-flux variation:

The obstruction to purely local radiative equilibrium is visible by taking the divergence:On an equipotential surface, $\rho ϵ$, $f$, $f^{\prime }$ and the second term are constant, but $g^2$ generally varies. Unless $f^{\prime }=0$ or another special circumstance removes the mismatch, $\nabla \cdot \mathbf{F}=\rho ϵ$ cannot hold at every latitude. This is the radiative-equilibrium obstruction in a rotating barotropic star; heat advection by stellar meridional circulation resolves the missing energy transport in the leading hydrostatic treatment.

For the steady circulation, mass conservation gives $\nabla \cdot (\rho \mathbf{v})=0$. The specific enthalpy identity and hydrostatic equilibrium giveConsequentlyBoth $h$ and $\varphi$ are constant on the bounding equipotential surface. By the divergence theorem and steady continuity equation,Integrating the given thermal energy equation therefore gives equipotential luminosity conservation with stellar circulation:Individual circulation streams can cross the surface; only their net steady mass flux has to vanish.

Since $f$ is constant on that surface, integrating the Poisson equation givesThus $dT/d\varphi =-L/[\chi (4\pi Gm-2{\Omega }^{2}V)]$, while $dP/d\varphi =-\rho$. Dividing these derivatives and inserting $\chi =4ac{T}^3/(3\kappa \rho )$ yields the radiative gradient in a uniformly rotating star:The bracket must be positive for the regular outward-gravity configuration being used.

For an interior equipotential, both mass factors must be the enclosed mass $m$. The printed final bracket uses $M$, but its derivation gives $m$; at the outer surface $m=M$ and the printed bracket is recovered. For a direct check, a uniform-density model has $m=\rho V$, so the true correction ${\Omega }^{2}V/(2\pi Gm)={\Omega }^2/(2\pi G\rho )$ is independent of the chosen enclosed volume. Replacing $m$ by a fixed total mass would incorrectly make it volume dependent.

For any vector $\mathbf{a}$, differentiating its components and the rotating basis gives the rotating-frame derivative formulaApply it twice to the position $\mathbf{x}$, with constant angular velocity $\mathbf{\omega }$. The inertial acceleration iswhere dots denote rotating-frame derivatives. Newton's second law therefore becomesThe last terms are the Coriolis force and the force corresponding to centrifugal acceleration.

For the particle on a uniformly rotating inclined plane, choose orthonormal vectors ${\mathbf{e}}_1$ along the horizontal $\xi$ axis and ${\mathbf{e}}_2$ up the slope, and set ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, the upward normal vector. Then ${\mathbf{e}}_z=\sin \theta {\mathbf{e}}_2+\cos \theta {\mathbf{e}}_3$ andHere $N$ is the normal reaction. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame Coriolis acceleration and centrifugal acceleration are respectivelyThe normal rotating-frame acceleration is zero, so the normal reaction is not generally $mg\cos \theta$; it satisfiesProjecting the same equation onto the two tangential axes gives exactly

Multiply the equations by $\dot{\xi }$ and $\dot{\eta }$ and add. The Coriolis force terms cancel, as expected because this force does no work relative to the rotating frame. The normal reaction also does no work along the plane. The conserved quantity is rotating-frame kinetic energy plus gravitational and centrifugal potential energy:This need not be the conserved inertial energy, since the externally driven rotating plane can exchange energy with the particle.

Let $\mathbf{v}$ denote the velocity relative to the uniformly rotating frame, and write $r_{\perp }^2=x^2+y^2$. For rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, the Euler equations for an inviscid fluid in that frame areThe negative quadratic term is the centrifugal potential, whose negative gradient is the outward centrifugal acceleration per unit mass. In steady flow, dotting the equation with $\mathbf{v}$ eliminates the Coriolis force, since $\mathbf{v}\cdot (\mathbf{\Omega }\times \mathbf{v})=0$. The remaining acceleration term is $\mathbf{v}\cdot \nabla (|\mathbf{v}{|}^2/2)$. ThusBy part (b), the steady rotating-frame Bernoulli integral isSteadiness here is measured in the rotating frame, including the body-force potential used in that frame.

For nonzero $\omega$, the equilibrium points satisfy $\sin 2\theta =0$. The equilibrium point $\theta =0$ is stable: it puts the bead at the point farthest from the rotation axis. Writing $\theta =\eta$ small in the tangential ordinary differential equation givesThus the small oscillation is simple harmonic motion, with angular frequency $|\omega |$ and ordinary frequency $|\omega |/(2\pi )$.

The endpoints $\theta =\pm \pi /2$ represent the same rim point $O$. A perturbation there instead satisfies $\ddot{\eta }=+{\omega }^2\eta$ and is unstable. Equivalently, the centrifugal potential along the hoop is $-\frac{1}{2}{\omega }^2|\mathbf{r}{|}^2=-2a^2{\omega }^2{\cos }^2\theta$, whose minimum is at zero and maximum at $O$. If $\omega =0$, every bead position with zero relative velocity is an equilibrium point, but perturbing that velocity produces unrestored drift; there is no nonzero restoring frequency.

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ is the sum of the two gravitational potentials and the centrifugal potential. The circular binary's Kepler third law gives ${\Omega }^2=G(M_1+M_2)/a^3=GM/a^3$, hence

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic balance becomes $\nabla (h+\Psi )=0$, or $h+\Psi =\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are equipotential surfaces of $\Psi$.

This approximation represents the gravitational potential by a central point mass while retaining the centrifugal potential. A surface through polar radius $R_p$ satisfies $R_p/r+q(r/R_p{)}^2{\sin }^2\theta /2=1$, where $q={\Omega }^2{R}_p^3/(GM)$. The inner closed surface is regular below $q=8/27$; at the limiting value its equatorial radius is $3R_p/2$ and equatorial effective gravity vanishes. The approximation is useful for illustrating oblateness and gravity darkening, rather than modeling the density distribution of every rotating star.

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.

Uniform solid-body rotation adds the centrifugal potential $-{\Omega }^2{\varpi }^2/2$ to the gravitational potential. In the corotating frame, hydrostatic equilibrium becomes $\nabla P=-\rho \nabla \varphi$. The Poisson equation gives ${\nabla }^2\varphi =4\pi G\rho -2{\Omega }^2$. On regular connected equipotential surfaces, $P$ is constant; taking the curl shows that $\rho$ is constant too. Uniform composition and an invertible equation of state then make $T$ constant on each such surface. These statements do not require $|\nabla \varphi |$ to be constant there.