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For a conventional Northern-Hemisphere anticyclone let $G=-p_r/\rho >0$. Gradient wind balance gives $V^2/r-fV+G=0$, with $V_{\pm }=fr[1\pm \sqrt{1-4G/(f^{2}r)}]/2$. The minus branch approaches the geostrophic wind at weak forcing; the plus branch is anomalous. Reality requires $G\le f^{2}r/4$, so a smooth central pressure maximum must have a radial pressure force vanishing at least linearly toward its centre.

The geostrophic wind is the horizontal flow for which the pressure gradient force balances the Coriolis force. It runs along atmospheric isobars and has speed proportional to their pressure gradient, at fixed density and nonzero Coriolis parameter. Curvature, friction and small spatial scales can invalidate that approximation.

A surface of constant pressure. Its intersection with a fixed-height horizontal plane is an atmospheric isobar. An isobaric surface is generally not a geopotential surface: the horizontal geopotential variation on it determines the geostrophic wind.

In the rotating terrestrial frame, horizontal pressure gradient force and Coriolis force balance in geostrophic balance. Gravity establishes the approximately vertical hydrostatic equilibrium; it is not an additional horizontal force in this balance. With eastward $x$, northward $y$, upward unit vector $\hat{\mathbf{z}}$ and Coriolis parameter $f=2\Omega \sin \varphi$,The geostrophic wind runs along atmospheric isobars, with lower pressure on its left in the Northern Hemisphere.

Interpreting the stated south-to-north gradient as $p_y>0$, $p_y=200\mathrm{Pa}/(2.00\times 10^5\mathrm{m})=10^{-3}\mathrm{Pa}{\mathrm{m}}^{-1}$, while $p_x=0$. Using the external nominal Earth's rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$ gives the meridional-pressure-gradient geostrophic windBoth zonal velocities are westward; their signs reverse if the supplied gradient is interpreted as a northward decrease. ^[ref]

The validity criteria for geostrophic balance require horizontal acceleration and friction to be small compared with the balancing forces: Rossby number $U/(|f|L)\ll 1$ and slow evolution compared with $1/|f|$. The approximation fails near the equator, on sufficiently small or rapidly changing scales, in strongly curved flow needing appreciable centripetal acceleration, and in the frictional surface layer. Neglecting friction near the ground is the stated idealization.

For an illustrative geostrophic wind with linear drag, use acceleration $-r\mathbf{u}$ with $r>0$. The steady horizontal momentum equation givesDrag reduces the wind and hence its Coriolis acceleration; the wind crosses isobars toward lower pressure. Here it points southwest. The pressure-gradient acceleration points south, Coriolis northwest and drag northeast; the three forces sum to zero. The cross-isobar angle satisfies $\tan \alpha =r/f$ for $f>0$. A turbulent boundary layer has a vertically varying stress rather than necessarily this constant linear drag.

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.

Take $x$ eastward, $y$ northward and $z$ upward, and let $u$ be the eastward geostrophic wind. With Coriolis parameter $f$ independent of $z$, geostrophic balance, hydrostatic balance, and the ideal gas law giveHere $R_d$ is the specific gas constant for dry air. Taking a vertical derivative of the first equation and interchanging mixed partial derivatives givesIt matters that $\rho$ is not constant in the pressure gradient force. Substitutingcancels the two terms proportional to $gfu/(R_{d}T)$ and yields the height-coordinate thermal wind relationIf $|u{T}_z|\ll |T{u}_z|$, this reduces toThe discarded term has relative scale $H|T_z|/T$ when $u_z\sim u/H$. Taking $H=10\mathrm{km}$, $T=250\mathrm{K}$ and a atmospheric lapse rate of $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ gives about $0.26$. Thus this height-coordinate formula is a useful estimate, with potentially tens-of-percent compressibility-related correction through the vertical temperature gradient; it is not an exact consequence of the two balances. If the temperature gradient vanishes vertically, the displayed approximation is exact within those balances.

The hydrostatic approximation is very good on the broad, shallow atmospheric scales considered. The geostrophic balance is normally justified by a small Rossby number and fails as a general approximation near the equator. At $30^{\circ }\mathrm{N}$, $f=7.292\times 10^{-5}{\mathrm{s}}^{-1}$. With the ideal $134\mathrm{m}{\mathrm{s}}^{-1}$ wind and a $1000$--$2000\mathrm{km}$ horizontal scale, $\mathrm{Ro}=U/(fL)\simeq 0.9$--$1.8$: quantitative accuracy is limited for this especially strong ideal jet. Small vertical aspect ratio alone does not ensure geostrophic balance, although a straight, steady zonal flow can satisfy it even when this scale estimate of the Rossby number is not small.

For the requested estimate, suppose the upper jet stream is $H=10\mathrm{km}$ above a weak surface wind, has temperature $T\simeq 250\mathrm{K}$, and spans a meridional width $L$. Integrating the approximate thermal wind relation in scale form givesThe northward temperature gradient is negative because the eastward wind strengthens upward. A $1000$--$2000\mathrm{km}$ jet width gives roughly $25$--$50\mathrm{K}$. No unique numerical difference follows without specifying the jet width, height and representative temperature; these are explicitly estimated quantities.

A horizontal pressure gradient force accelerates air from higher toward lower pressure. In a rotating frame the Coriolis force deflects the flow. Away from strong friction, at large spatial scales and small Rossby number, a slowly changing flow can reach geostrophic balance:where $f=2\Omega \sin \phi$ is the Coriolis parameter. The geostrophic wind runs along atmospheric isobars, with speed $|{\nabla }_{h}p|/(\rho |f|)$. Closely spaced atmospheric isobars therefore indicate stronger geostrophic wind when density and latitude are comparable. In the northern hemisphere lower pressure lies to the left of this wind, producing counterclockwise circulation around a low; the sense reverses in the southern hemisphere.

A curved flow also has centripetal acceleration. For cyclonic circular flow in the northern hemisphere, the gradient wind relation isNear the ground, friction reduces wind speed and weakens its Coriolis contribution, so air crosses atmospheric isobars toward lower pressure. The simple geostrophic formula fails near the equator, where $f$ becomes small, and where rapid changes or tight curvature make neglected acceleration important. Vertically, large-scale motion is often approximated by hydrostatic balance, $dp/dz=-\rho g$.

Bernoulli's principle can relate speed changes to pressure changes along an appropriate inviscid streamline, but its simple dynamic-pressure relation is not a replacement for the rotating force balance governing a broad atmospheric pressure map.

Signed radial balance. Let $r$ increase outward from the vortex centre and $U$ be signed azimuthal velocity, positive anticlockwise. At $45^{\circ }$N, $f=2{\Omega }_{\oplus }\sin 45^{\circ }>0$. The positive speed is $V=|U|$, $f$ is the Coriolis parameter, the density is $\rho$, and $p_r=\partial p/\partial r$. In circular motion the inward centripetal acceleration is $-U^2/r$, while the outward radial acceleration from the Coriolis force is $fU$. The pressure gradient acceleration is $-p_r/\rho$. HenceThe term $U^2/r$ is an acceleration, not an additional physical force in the rotating-frame momentum equation. It can be moved to the force-balance side and depicted as outward centrifugal force in a frame following the circular parcel.

For a cyclone, $U=+V$ and $p_r/\rho =V^2/r+fV>0$: pressure increases outward and the centre is a low. Its inward pressure force balances the outward Coriolis and centrifugal terms. For a conventional anticyclone, $U=-V$ and $G=-p_r/\rho >0$; the balance isThe inward Coriolis force balances the outward pressure and centrifugal terms, and pressure has a maximum at the centre.

Anticyclonic branches and near-centre regularity. The anticyclonic gradient-wind branches follow from the quadratic balance. Solving givesThe minus root is the normal branch connected continuously to geostrophic flow for small forcing. The plus root is an anomalous high-speed branch and must not be discarded algebraically, even though it often fails the stability conditions relevant to large-scale anticyclones. Reality requiresA finite smooth central pressure therefore has zero radial gradient at the origin, with $-p_r/\rho$ no larger than order $r$. For $p=p_c-\rho C{r}^2/2+\cdots$, the limiting curvature obeys $C\le f^2/4$. A pressure maximum with a finite nonzero limiting outward pressure force cannot support the required circular anticyclonic balance arbitrarily close to its centre.

For positive $\mathrm{Ro}=V/(fr)$ and geostrophic magnitudes,The cyclonic gradient wind is subgeostrophic and the normal anticyclonic wind supergeostrophic. On the latter branch $0\le \mathrm{Ro}\le 1/2$. At a point on the eastern side of a cyclone, geostrophic and actual winds point north but the actual arrow is shorter; the ageostrophic difference points south. On the eastern side of a normal anticyclone both geostrophic and actual winds point south, with the actual arrow longer and its ageostrophic addition also south.

At lower latitude, the same $V$ and radius give a larger Rossby number and larger differences from geostrophic wind. The contour topology remains a central low or a conventional central high, but for fixed $V,r$ the required radial pressure gradients change with $f$: the cyclonic gradient is $\rho (V^2/r+fV)$ and the anticyclonic outward pressure force is $fV-V^2/r$. The latter conventional branch cannot be maintained once the reduced Coriolis contribution is too weak. Actual spatial gradients, rather than an arbitrary spacing of contour values, determine how tightly drawn contours should be.

The high-centre conclusion is conditional on the conventional anticyclonic branch. Gradient balance alone also allows a sufficiently rapid clockwise circulation about a low: if $V>fr$, then $p_r/\rho =V^2/r-fV>0$. For example, $V=2fr$ gives $p_r/\rho =2f^{2}r>0$. Such a configuration is outside the usual geostrophic anticyclone regime. Tornado-scale vortices have very large Rossby number, so the dominant balance is nearly cyclostrophic, $p_r/\rho \simeq V^2/r>0$. Their centres are low pressure irrespective of the sign of rotation; low pressure supplies the inward acceleration needed to bend the fast flow.

Approximate the surface layer as locally steady with small horizontal advective acceleration. Write $J\mathbf{u}=\mathbf{k}\times \mathbf{u}$ and $\mathbf{P}={\rho }^{-1}{\nabla }_{h}p$. Since $J^2=-I$, the drag-modified force balance can be inverted exactly:The geostrophic wind is ${\mathbf{u}}_g=J\mathbf{P}/f$, so the ageostrophic flow isIts cross-isobar component is $-\alpha {\nabla }_{h}p/[\rho (f^2+{\alpha }^2)]$, always towards lower pressure, in either hemisphere. Thus the surface wind spirals into a cyclone and out of an anticyclone. In the northern hemisphere, the nearly tangential circulation is counterclockwise around the low and clockwise around the high; those rotations reverse in the southern hemisphere. The original force diagrams below show pressure, drag and Coriolis force closing to zero at the east side of each system.

Surface convergence into a low requires upward motion out of the friction layer by the continuity equation; divergence from a high requires subsidence. In unsaturated adiabatic motion the rising parcel expands and cools, while descending air compresses and warms. The relevant warming or cooling relative to its surroundings depends on the environmental stratification, and latent heating can modify the ascending-air result.

For otherwise unsupported weather systems this is quasi-geostrophic Ekman spin-down: drag dissipates kinetic energy, convergence tends to fill a low and divergence tends to erode a high. In a simple column with upward velocity at its base and negligible vertical velocity aloft, cyclonic vorticity decreases because ${\partial }_{z}w<0$; the opposite sign weakens anticyclonic vorticity. Cooling beneath a fixed upper-level pressure also tends to raise the surface pressure through the hydrostatic column, while warming tends to lower it. Real baroclinic development, upper-level mass transport and condensation can sustain or deepen a cyclone despite surface friction, so temperature changes alone do not prove a universal decay rate.