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An inclined plane is a flat surface tilted through an angle $\theta$ above horizontal. A ramp exchanges a smaller lifting force for a longer travel distance; for a released body, it reduces the downhill component of gravitational weight and makes the resulting acceleration easier to measure. Its simplest dynamical model is a nonrotating block on a stationary, rigid surface, with constant mass $m>0$, uniform gravitational acceleration of magnitude $g>0$, and $0\le \theta <\pi /2$. Assume continued contact, negligible drag force, no tipping, and no applied pull. The only forces are gravitational weight, the normal force, and any tangential friction.

The central distinction is between a block resting on the inclined plane and one already sliding. Static friction supplies the force needed to maintain rest, up to its bound; kinetic friction has a prescribed magnitude in the ideal constant-coefficient model. These laws are empirical approximations whose range depends on the contacting surfaces.^[ref]

Inertia is a body's persistence in its state of rest or constant velocity when the net external force vanishes in an inertial frame. It is not an extra force that pushes a moving body forward. Newton's first law describes this persistence; for a point mass, Newton's second law quantifies how a nonzero net external force changes its momentum. At constant mass, ${\mathbf{F}}_{\mathrm{net}}=m\mathbf{a}$, so a larger mass gives a smaller acceleration for the same net force.

A moving cart does not need a forward net force to maintain its velocity. Ordinary carts slow because friction and drag force produce an opposing net external force; reducing those interactions reveals Newton's first law more clearly. The air-track test of inertial motion separates this persistence from the initial push. In overdamped particle dynamics, the inertial acceleration term is small compared with drag force over the time scales considered; that approximation does not mean that inertia has disappeared.

A gas syringe makes the distinction between a constant-temperature path and equal-temperature endpoints observable. The Ideal Gas Law Apparatus (TD-8596A), Instruction Manual 012-08420B by PASCO scientific specifies a TD-8596A gas syringe with a mechanical stop, an internal thermistor, an attached pressure sensor and data acquisition. Record the stop volume with the coupling disconnected, then set $40\mathrm{mL}$ and seal the coupling. Record pressure and temperature, compress quickly to the stop near $20\mathrm{mL}$, and hold until both readouts settle. The manual gives roughly $30\mathrm{s}$ for settling and about $0.5\mathrm{s}$ for the thermistor response. Releasing the plunger gives a second transient and a final volume to record.^[ref]

The manual describes a pressure overshoot and temperature rise during rapid compression, followed by cooling at the held volume; the settled pressure remains higher than initially. This illustrates approximately adiabatic compression followed by heat rejection, rather than an isothermal compression throughout. Use the settled initial and compressed readings to test Boyle's law, correcting both syringe readings by the unmarked tubing volume $V_d$:Use absolute pressure, not gauge pressure. A transient comparison of $pV/T$ must pair simultaneous pressure and temperature readings: the sensor lag can separate their peaks.^[ref]

For an additional test of the path, repeat with small compression steps, wait for stable temperature and pressure at every step, and repeat in reverse. Keep the sealed amount of substance, connecting tubing and ambient temperature fixed; prevent hand warming of the barrel and check for leaks by watching for continued pressure drift at held volume after thermal equilibrium. These are proposed controls, rather than reported measurements. Agreement of equilibrated points with Boyle's law supports the fixed-temperature ideal gas law approximation, but does not prove that the rapid segments between them were isothermal processes. A single local temperature sensor can miss temperature gradients; plunger friction can produce direction-dependent readings. The pressure and volume data estimate reversible work only insofar as the path approaches a reversible thermodynamic process; they do not directly measure heat. After motion stops, cooling at fixed volume has zero boundary work and $Q=\Delta U<0$, even though the complete equal-endpoint-temperature ideal gas run has $\Delta U=0$.

An isothermal process changes a thermodynamic system while keeping its temperature constant. It need not keep its pressure, volume or internal energy constant, and it does not generally have zero heat transfer. An adiabatic process specifies zero heat transfer instead. For example, a slowly expanding ideal gas can absorb heat from a heat bath and deliver the same amount of work while its temperature stays fixed.

For a fixed amount of substance and composition, internal energy of an ideal gas depends only on temperature. With $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work delivered to the surroundings, the first law of thermodynamics therefore givesThis equality does not determine either transfer until the path is specified. Isothermal ideal-gas work gives the reversible value; an insulated free expansion has a different value even when its equilibrium endpoints have the same temperature. The closed system convention here excludes changes in bulk kinetic energy and external potential energy.

A conducting cylinder in a large, stirred heat bath approximates an isothermal process when the piston moves slowly enough for thermal conduction to keep up. A quasistatic process keeps the gas close to internal thermodynamic equilibrium; a reversible thermodynamic process additionally requires negligible friction and vanishing driving differences in pressure and temperature. Exact equality of the gas and heat bath temperatures gives no finite heat transfer rate through a finite thermal resistance. The reversible description is a limit of arbitrarily small differences and arbitrarily slow motion, illustrated by the finite-rate isothermal piston model.^[ref]

For a real gas, internal energy at constant temperature can change with volume because molecular interaction potential energy changes. In one smooth thermodynamic phase of a fixed-composition simple compressible system,Thus an isothermal process generally has $Q=\Delta U+W_{\mathrm{by}}$, with no reason to set $\Delta U=0$. The right-hand side vanishes for the ideal gas law, but not generally for another equation of state.^[ref]

Researcher associated with experimental rock friction and Byerlee's law.

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

An ideal string of negligible mass and fixed length, transmitting tension when taut. In an ideal guide with no friction, the tension magnitude is the same on both sides: otherwise the force balance on an arbitrarily short massless segment would fail. Its fixed length gives a mechanical constraint relating the endpoint velocities. The two endpoint tension powers sum to zero when the guide is stationary; the string can transfer energy between connected particles without adding net energy.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

The component of a contact force perpendicular to the local contact plane.

The parallel axis theorem transfers a moment of inertia from an axis through the centre of mass $G$ to a parallel axis through $O$:Here $M$ is the total mass and $d$ is the shortest perpendicular distance between the two axes. Both moments of inertia concern the same mass distribution and the same axis direction. The axis for $I_G$ must pass through the centre of mass; $I_O$ may refer to an axis outside the body. No uniform-density or planar-shape assumption is needed. Within any family of parallel axes, the centre of mass axis minimizes the moment of inertia.^[ref]

A short proof isolates why the centre of mass matters. Choose parallel axes perpendicular to an $xy$ plane, put $G$ at the origin, and place $O$ at $(d,0)$. Thenbecause $\int xdm=0$ in centre of mass coordinates. Individual mass elements can move closer to the new axis, but their weighted cross terms cancel. For arbitrary parallel axes $A$ and $B$, use $I_B-I_A=M(d_B^2-d_A^2)$, where each distance is measured from the parallel centre of mass axis; adding $M$ times the squared separation of $A$ and $B$ is generally incorrect. A displacement along the axis has $d=0$. Changing the axis direction instead requires the inertia tensor or, in its applicable planar setting, the perpendicular axis theorem.

For a uniform disc, mass $M$ and radius $R$, the central perpendicular moment of inertia is $I_G=M{R}^2/2$. Shifting that axis to the rim gives $d=R$ and $I_O=3M{R}^2/2$, three times $I_G$. In the computed $M=2\mathrm{kg}$, $R=0.30\mathrm{m}$ example, $I_G=0.090\mathrm{kg}{\mathrm{m}}^2$, $M{d}^2=0.180\mathrm{kg}{\mathrm{m}}^2$ and $I_O=0.270\mathrm{kg}{\mathrm{m}}^2$. At angular speed $10rad/s$ the rim-pivoted disc has rotational kinetic energy $13.5\mathrm{J}$. Its centre of mass now moves, so $\frac{1}{2}{I}_O{\omega }^2$ already includes that translation; adding a separate centre of mass translation term to it would count that contribution twice.

The physical pendulum measurement described under moment of inertia provides a practical check: infer $I_O=Mgh(P/2\pi {)}^2$ at several pivot positions, then subtract $M{h}^2$ to obtain $I_G$. Keep all pivot axes parallel and measure $h$ to the centre of mass of the entire moving assembly. The same assembly should yield the same $I_G$ within measurement precision. If a smartphone is attached, its mass and its own moment of inertia belong in both the geometric prediction and the centre of mass calculation. The published bar-and-phone study explicitly adds both contributions using the parallel axis theorem.^[ref]

For a careful replication, release small swings without a push, repeat timings at several small amplitudes, and check that the extrapolated oscillation period is stable as amplitude decreases. Time many complete oscillations and repeat the run; two crossings of a photogate per cycle must not be mistaken for two cycles. Inspect decay of amplitude and compare early and late timings to detect significant pivot friction or air drag. Mark the bar and phone mount to detect relative slip, tighten the attachment, and check that the pivot support does not rock or migrate between holes. Repeat with an independent photogate or video readout to check the phone's timing. These are proposed controls for the replication, not additional results reported by the authors. The ideal model excludes appreciable damping, large swings and flexing; subtraction of $M{h}^2$ also magnifies relative errors when $I_G$ is much smaller than $I_O$.

The oscillation period need not increase monotonically with pivot distance: the shift raises the moment of inertia but also increases the gravitational restoring torque. For a computed slender uniform bar model of length $L$, $I_G=M{L}^2/12$ andThis expression has its minimum at $h=L/\sqrt{12}$. With $L=1\mathrm{m}$, $h=0.50\mathrm{m}$ and stipulated $g=9.81m/{\mathrm{s}}^2$, the prediction is $P\simeq 1.638\mathrm{s}$; it is not a recorded timing or a prediction for the bar with a phone attached. The competing terms explain why a timing experiment can reveal a mass distribution rather than merely its weight.

On releasing the tensile load, the extended chain first retracts towards less aligned conformations because they have much greater configurational multiplicity. Viscous friction and internal relaxation determine the time course; stored free energy is dissipated as heat. An unconstrained ideal chain does not shrink to a point: its mean vector is zero but its root mean square end-to-end length is $b\sqrt{N}\simeq 2.37\mathrm{nm}$ for the supplied ideal-chain parameters.

A real polypeptide also has sequence-dependent interactions. If the original native structure is the stable state in the prevailing solvent and the molecule remains intact and capable of folding, it can subsequently refold, gaining favourable intramolecular and solvent free energy. The folded state is not selected by chain entropy alone, and formation of native contacts need not be instantaneous. Kinetic traps, proline isomerization, protein aggregation, misfolding, surface attachments or chemical damage may prevent complete recovery or produce hysteresis. Thus reversible refolding is a physically motivated possibility, not guaranteed simply by removal of force. This distinction connects the recoil of a stretched polymer to the thermodynamics discussed in Question 6(c).

In a frame rotating with angular velocity $\mathbf{\Omega }$, a particle of relative velocity $\mathbf{v}$ experiences the Coriolis forceFor nearly horizontal atmospheric motion, the relevant vertical rotation component is $\Omega \sin \ell$, with latitude $\ell$. In the northern hemisphere the horizontal deflection is to the right of the motion. A low-pressure center produces an inward pressure-gradient force; an anticlockwise circulation has an outward Coriolis force to oppose it. A high-pressure center instead supports clockwise circulation, with inward Coriolis force opposing the outward pressure-gradient force.

The sense reverses in the southern hemisphere. There, anticlockwise circulation is anticyclonic and corresponds to high atmospheric pressure. HenceThis is the large-scale rotating-fluid interpretation; friction, centrifugal acceleration and departures from ideal balance modify actual wind profiles. NASA's description independently identifies the southern, counterclockwise Great Red Spot as a high-pressure system.^[ref]

Treat the marble as a frictionless point particle; no size, rolling constraint or friction law is supplied. Neglect its vertical kinetic energy as requested, but retain the gravitational potential energy $mgz=m{\omega }^2(x_1^2+y_1^2)/2$.

For completeness keep the general constant cup rotation rate $\nu =\Omega +\varphi$. In cup components the center's velocity is $\mathbf{w}=R\Omega (\sin \varphi t,\cos \varphi t)$. With $\mathsf{J}(x,y)=(-y,x)$, the horizontal velocity norm is $|\dot{\mathbf{q}}+\nu \mathsf{J}\mathbf{q}+\mathbf{w}|$. ThusIn the nonrotating translating case $\varphi =-\Omega$, this reduces toThe time-only constant $m{R}^2{\Omega }^2/2$ may be dropped. Integrating the cross-velocity term by parts gives an equivalent LagrangianIt differs by a total derivative and a time-only term, so gives the same trajectories but different expressions for the canonical momenta. The small-vertical-velocity approximation is typically consistent at small surface slope ${\omega }^2|\mathbf{q}|/g$; it need not remain valid during unlimited resonant growth.

For dry friction, the tangential contact force opposes sliding and has limiting magnitude ${\mu }_f{N}_c$, where $N_c\ge 0$ is the compressive normal force and ${\mu }_f=\tan \varphi$. Sticking requires $|F|\le {\mu }_f{N}_c$; impending sliding reaches equality.

The cohesionless Mohr-Coulomb failure criterion applies the same local condition on internal planes: yield begins when some plane has $|{\tau }_{\mathrm{shear}}|=N_c\tan \varphi$. In the tension-positive Cauchy convention used here, $N_c=-N$ for a compressive plane, soA cohesive material would add a positive intercept, but the later formulas in this paper are explicitly the zero-cohesion case. The model does not supply a tensile strength law.

The Mohr circle has center $C=({\tau }_{xx}+{\tau }_{yy})/2$ on the normal-stress axis and radius $R=\sqrt{({\tau }_{xx}-{\tau }_{yy}{)}^2/4+{\tau }_{xy}^2}$. In the compressive regime $C<0$, yield occurs when the circle first touches either friction line $F=\pm (-N)\tan \varphi$. The perpendicular distance of its center from either line is $|C|\sin \varphi$. Tangency therefore gives $R^2=C^2{\sin }^2\varphi$, or $C^2-R^2=C^2{\cos }^2\varphi$. But $C^2-R^2={\tau }_{xx}{\tau }_{yy}-{\tau }_{xy}^2$. ThusThe compressive sign restriction remains part of this equivalence: squaring alone would also produce algebraic branches that do not represent cohesionless compressive friction.

Insert the two constitutive stresses into the radial component of the Cauchy momentum equation. After collecting terms, $(\lambda +2\mu )(u_r^{\prime \prime }+u_r^{\prime }/r-u_r/r^2)=0$, so $u_r=Ar+B/r$. Its stresses areThe far pressure gives $A=-p_{\infty }/[2(\lambda +\mu )]$. At the tunnel wall, compressive bracing imposes ${\tau }_{rr}(a)=-P$, so $B=(P-p_{\infty })a^2/(2\mu )$. HenceWith no shear and ${\tau }_{\theta \theta }=k{\tau }_{rr}$, divide the yield identity by the nonzero ${\tau }_{rr}^2$ to obtain $4k=(1+k{)}^2{\cos }^2\varphi$. Solving its quadratic givesTo locate first yield, the stress-circle center is the constant $-p_{\infty }$, while its radius is $|P-p_{\infty }|a^2/r^2$. The radius is maximal at $r=a$, so that is where it first meets the friction envelope as the bracing departs from the hydrostatic value. At the wall ${\tau }_{rr}=-P$ and ${\tau }_{\theta \theta }=P-2p_{\infty }$; hence $k=(2p_{\infty }-P)/P$ andFor lowering bracing from $P=p_{\infty }$, the hoop stress becomes more compressive, so $k=k_{+}>1$ and $\boxed{P_{\mathrm{crit}}=p_{\infty }(1-\sin \varphi )}$. Increasing bracing instead selects $k_{-}$ and reaches the other threshold $p_{\infty }(1+\sin \varphi )$. Stating the loading direction removes any ambiguity between the two reciprocal stress ratios. These conclusions describe first yield of the elastic field, not the later plastic-zone solution.

Take $x$ eastward, $y$ northward and $z$ upward. In a rotating reference frame, the horizontal Cauchy momentum equation isHere ${\mathbf{F}}_h$ represents other accelerations, such as friction, and $f$ is the Coriolis parameter. When the Rossby number $U/(|f|L)$ and the ratio $1/(|f|\tau )$ are small, and friction is weak, both advective and local acceleration can be neglected. The Coriolis force then balances the pressure gradient force: this is geostrophic balance, with $fv=p_x/\rho$ and $fu=-p_y/\rho$. The geostrophic wind runs along atmospheric isobars, with low pressure on its left in the northern hemisphere.

A rapid change of pressure need not bring the corresponding geostrophic wind into existence at once. An initially weak wind therefore leaves an inward pressure gradient force around the new low. For a smoothed anomaly of horizontal scale $L=5\times 10^5\mathrm{m}$, geostrophic balance would requireFor the illustrative midlatitude value $|f|=10^{-4}{\mathrm{s}}^{-1}$, the implied Rossby number is already about $0.67$, so curved-flow acceleration matters. With a small tropical value $|f|=10^{-5}{\mathrm{s}}^{-1}$, the inferred speed is $333\mathrm{m}{\mathrm{s}}^{-1}$ and the Rossby number about $67$: the assumed geostrophic balance is untenable. Geostrophic adjustment redistributes the anomaly and emits internal gravity waves, leaving a weaker balanced component.

For a target speed $U_{\ast }=10\mathrm{m}{\mathrm{s}}^{-1}$ on the same scale, the required scale estimate isThus $6\mathrm{hPa}$, or $0.30$ of the initial magnitude, is the midlatitude estimate; at the illustrative tropical $|f|=10^{-5}{\mathrm{s}}^{-1}$ it is $0.6\mathrm{hPa}$, or $0.030$ of the initial magnitude. The paper gives no latitude, so there is no unique numerical reduction. At the equator itself one must use equatorial dynamics rather than divide by $f=0$. Even the tropical $10\mathrm{m}{\mathrm{s}}^{-1}$ estimate has Rossby number $2$, so it describes the scale permitted by the stated balance, rather than a controlled tropical geostrophic balance.

There are two qualifications to “unstable”. A literal top-hat pressure anomaly has a discontinuous edge, where no finite classical geostrophic wind can balance its distributional pressure gradient. The $500\mathrm{km}$ estimate assumes a smoothed edge; a thinner edge requires stronger winds. More generally a large low is not, by its amplitude alone, proved to have an exponentially growing instability. For a concrete counterexample to the claim that such an amplitude cannot be stationary, set $p\prime (s)=-P\exp (-s^2/a^2)$, with cylindrical radius $s$, constant $\rho$ and $f>0$. The cyclonic gradient windis real and satisfies $v_{\theta }^2/s+f{v}_{\theta }=p_s/\rho$ for every $P>0$, with a regular limit at the centre. This is a stationary axisymmetric flow; no claim about its perturbation stability is needed. Here the physically justified conclusion is that a rapidly imposed anomaly without its balancing wind is unbalanced.

Small tropical Coriolis parameter allows only weak horizontal pressure gradients to coexist with ordinary large-scale winds. Hydrostatic balance connects the vertical variation of these pressure gradients to horizontal temperature gradients; equivalently the approximate thermal wind relation gives $f{u}_z\simeq -(g/T)T_y$. For comparable vertical wind shear, small $|f|$ means smaller $|T_y|$. Internal gravity waves can spread heating-induced buoyancy anomalies efficiently before rotation arrests the adjustment. This motivates the weak temperature gradient approximation for the tropical free troposphere; it does not require the land surface or atmospheric boundary layer to have uniform temperature.

The equations describe the linear reduced gravity shallow water equations for a uniform upper layer over a deep, resting lower layer, or equivalently one separated internal vertical mode. The layer is shallow compared with its horizontal scales, permitting hydrostatic balance, and its horizontal velocity is independent of depth. The Boussinesq approximation retains the small density contrast through $g\prime =g({\rho }_2-{\rho }_1)/{\rho }_0>0$. We neglect friction, external forcing, background currents, and quadratic perturbation terms. On an equatorial beta plane, $f=\beta y$ with $\beta >0$, and the equatorial Kelvin wave branch imposes $v=0$. The meridional Cauchy momentum equation then reduces exactly within this linear model to a balance between the Coriolis force and pressure gradient force.

Strictly, the printed $h$ must denote the thickness perturbation $\eta$, with total thickness $h_0+\eta$ and $|\eta |\ll h_0$. A signed oscillation around zero cannot represent the total positive layer thickness. Define $c=\sqrt{g\prime h_0}$. For a nonzero zonal wavenumber $k$, the zonal Cauchy momentum equation and conservation of mass giveOn the eastward branch $\omega =ck$, the wave polarization is $u=(c/h_0)\eta$. Meridional geostrophic balance givesThe branch $\omega =-ck$ instead grows as $\exp (+\beta y^2/(2c))$ and is excluded by decay away from the equator. Matching the paper's Gaussian function therefore givesThe usual equatorial deformation radius is $R_{\mathrm{eq}}=\sqrt{c/\beta }$, with the envelope written $e^{-y^2/(2R_{\mathrm{eq}}^2)}$. Thus the printed exponent uses $L_{\mathrm{eq}}=\sqrt{2}{R}_{\mathrm{eq}}$. Calling that width a deformation radius is a convention; using the standard radius without the factor $1/2$ would fail the printed meridional equation.

At an instant, alternating positive and negative thickness anomalies have their largest magnitudes on the equator and decay symmetrically away from it. Eastward arrows occupy the positive anomalies, westward arrows the negative anomalies; the velocity vanishes at the nodes and weakens away from the equator. The whole pattern travels eastward at speed $c$. Meridionally, $fu=-g\prime {\eta }_y$ is geostrophic balance. For a positive anomaly north of the equator, the northward pressure gradient force balances the southward Coriolis force; both directions reverse south of the equator. Zonally, $fv=0$ and $u_t=-g\prime {\eta }_x$: the zonal pressure gradient force accelerates the flow, so it is not in zonal geostrophic balance. On the equator the meridional terms both vanish smoothly.