Circulation and Vorticity, the General Circulation, Thermal Wind, the Indian Monsoon and Tropical Cyclones

This chapter answers the second half of the dynamics list of Section XE7.1 Atmospheric Science of XE7 Atmospheric and Oceanic Sciences, one of the nine optional sections of the Engineering Sciences (XE) paper: “Circulations and Vorticity, General Circulation of the Atmosphere. Broad Features of Indian Monsoons; Fundamentals of Tropical Cyclones.” It teaches circulation, Kelvin's theorem in outline and relative and absolute vorticity, the Hadley, Ferrel and polar cells with the jet streams and the thermal-wind relation, the broad features of the Indian monsoon and the fundamentals of tropical cyclones. The numericals are vorticity from a wind field, conservation of absolute vorticity, the angular-momentum argument for the subtropical jet, thermal wind and the gradient-wind speed in a tropical cyclone. Constants used: Ω = 7.292×10⁻⁵ s⁻¹, Earth radius a = 6.371×10⁶ m, g = 9.81 m/s²; other values are stated in each question.

1. Circulation and vorticity; Kelvin's theorem

Circulation around a closed curve is the line integral of the velocity along it, C = ∮ V·dl. By Stokes' theorem C = ∬ (∇×V)·dA, so the circulation equals the vertical vorticity summed over the enclosed area. For horizontal flow the vertical component is the relative vorticity ζ = ∂v/∂x − ∂u/∂y: positive for anticlockwise (cyclonic in the Northern Hemisphere) rotation. A flow in solid-body rotation at angular speed ω has ζ = 2ω; pure shear u(y) has ζ = −∂u/∂y, so westerly wind that weakens to the north has positive vorticity.

The Earth itself rotates, so a fluid at rest relative to it has the planetary vorticity f (twice the local vertical component of Ω). The absolute vorticity is η = ζ + f. Per Kelvin's theorem (in outline): for an inviscid, barotropic fluid (density a function of pressure alone) the absolute circulation around a closed material curve is conserved, DC_a/Dt = 0. For a horizontal loop of area A, C_a = (ζ + f)A. In a baroclinic fluid, where isobaric and isopycnic surfaces intersect, solenoids generate circulation at the rate −∮ α dp, which is how a sea breeze starts (land and sea at different temperatures); friction also changes circulation.

Consequence (barotropic, non-divergent flow): η = ζ + f is conserved (more generally D(ζ + f)/Dt = −(ζ + f)∇_h·V, so divergence reduces and convergence raises vorticity, the way a skater spins faster on pulling in the arms). Worked example. Air with ζ = 0 at 30°N moves to 45°N with no divergence: ζ = f₃₀ − f₄₅ = 7.29×10⁻⁵ − 10.31×10⁻⁵ = −3.02×10⁻⁵ s⁻¹. Poleward motion adds planetary vorticity, so relative vorticity becomes anticyclonic. This exchange between ζ and f is what makes the large-scale flow meander as Rossby waves.

⚠️ Vorticity is not the same as curved streamlines
A flow with straight parallel streamlines but wind speed changing across the flow (shear) has vorticity, and a curved flow with speed falling as 1/r (an irrotational vortex) has none outside its core. Evaluate ζ from derivatives of u and v, never from the shape of the streamlines. Also, ζ + f is conserved, not ζ alone.

2. The general circulation: cells, surface winds and jet streams

Differential heating (the equator absorbs more than it emits, the poles less) drives a meridional overturning, shaped by Earth's rotation into three cells in each hemisphere.

The three-cell picture, each hemisphere
CellExtentCirculation and surface wind
Hadley (thermally direct)Equator to about 30°Air rises at the ITCZ, flows poleward aloft, sinks in the subtropical highs, returns equatorward at the surface as the trade winds (north-easterly in the Northern Hemisphere, south-easterly in the Southern)
Ferrel (thermally indirect)About 30° to 60°Surface westerlies; driven by eddies (cyclones and anticyclones), not by direct heating
Polar (thermally direct)About 60° to the poleCold air sinks at the pole, flows equatorward as polar easterlies, rises near 60°

The Intertropical Convergence Zone (ITCZ) is the belt where the two trade-wind systems converge, with deep cumulonimbus, heavy rain and weak surface winds (the doldrums). It follows the thermal equator and shifts toward the summer hemisphere; over South Asia it moves to about 20 to 25°N in summer. The subtropical highs near 30° are sinking, dry regions (the great deserts). Air moving equatorward in the Hadley cell is turned west by Coriolis, hence the easterly trades.

Jet streams are narrow, fast westerly currents near the tropopause. The subtropical jet (near 30°, about 12 km) lies at the poleward edge of the Hadley cell; the polar-front jet (50 to 60°, about 9 to 10 km) lies above the sharp temperature contrast at the polar front. Typical core speeds are 30 to 60 m/s, with peaks beyond 100 m/s in winter. The subtropical jet has a simple explanation in angular momentum: air at rest at the equator has absolute angular momentum M = Ωa² (cos φ = 1). If it moves poleward with M conserved, M = (Ωa cos φ + u) a cos φ gives u = Ωa sin²φ / cos φ. At 30°, u = 7.292×10⁻⁵ × 6.371×10⁶ × 0.25/0.866 = 134 m/s. Real jets are much weaker (about 50 to 60 m/s) because eddies and friction remove momentum, but the argument shows why the poleward-moving air is accelerated to the east.

3. The thermal-wind relation

Combine geostrophic balance with the hydrostatic equation and differentiate with height: the vertical shear of the geostrophic wind is proportional to the horizontal temperature gradient. In height coordinates ∂u_g/∂z = −(g/(fT)) ∂T/∂y and ∂v_g/∂z = (g/(fT)) ∂T/∂x. In vector form ∂V_g/∂z = (g/(fT)) k × ∇T. The thermal wind is the difference between the geostrophic winds at two levels; it blows parallel to the isotherms with the cold air on its left in the Northern Hemisphere (on its right in the Southern).

Worked example. Temperature falls northward at 1.0×10⁻⁵ K/m (10 K per 1000 km) in a layer with mean T = 260 K at a latitude where f = 1.0×10⁻⁴ s⁻¹. Then ∂u_g/∂z = −(9.81 / (1.0×10⁻⁴ × 260)) × (−1.0×10⁻⁵) = 3.77×10⁻³ s⁻¹, i.e. 3.8 m/s per km. Between 1 km and 9 km, Δu = 3.77×10⁻³ × 8000 = 30.2 m/s: westerlies strengthen with height. That is why the westerly jets sit at the top of the troposphere above the strongest temperature gradient, and why a warm-cored anticyclone aloft in the tropical summer with warm air to the north reverses the shear to give upper easterlies.

⚠️ Thermal wind is a shear, not a wind at a level
The thermal wind is the vector difference V_g(upper) − V_g(lower) and describes how the wind changes with height. It does not say what the wind is at any one level; a southerly surface geostrophic wind plus a westerly thermal wind gives a south-westerly aloft. Also the relation requires geostrophic and hydrostatic balance, so it fails near the equator.

4. Broad features of the Indian monsoon

The monsoon is a seasonal reversal of the low-level wind, driven by the contrast between land and ocean heating and by the seasonal migration of the ITCZ. In summer the Indian land mass and Tibetan Plateau heat strongly, a heat low forms over north-west India, and the cross-equatorial south-westerly flow carries moisture from the Indian Ocean: the south-west monsoon of June to September, which brings roughly three-quarters of India's annual rainfall. In winter the flow reverses to the dry north-easterly; the north-east monsoon brings rain to Tamil Nadu in October to December.

  • Onset. The south-west monsoon normally reaches Kerala around 1 June, advances north in surges, and covers the whole country around the first half of July (the normal date is about 8 July). Withdrawal begins from north-west India in September.
  • Monsoon trough. An elongated low-pressure axis from the heat low over north-west India and Pakistan across the Gangetic plain to the head of the Bay of Bengal. Rain is heaviest to its south. Bay of Bengal lows and depressions form on it and move west-north-west, giving much of the rain over central India.
  • Low-level jet (Somali or Findlater jet). A narrow cross-equatorial south-westerly jet with its core near 1 to 1.5 km (about 850 hPa) over the Arabian Sea, with speeds of 15 m/s or more. It supplies the moisture and is Coriolis-turned from south-easterly trades of the Southern Hemisphere as they cross the equator. Aloft, the Tibetan anticyclone and the Tropical Easterly Jet (about 150 hPa) lie over southern India.
  • Active and break spells. In an active spell the trough lies in its normal position and rain is heavy over central India. In a break spell (several days) the trough moves north to the Himalayan foothills: rain is heavy there and in the north-east but scanty over most of the country.

ENSO and IOD (outline). El Niño (warm eastern Pacific) shifts the Walker circulation and generally weakens the Indian monsoon, with drought risk; La Niña generally favours it. The relation is not exact and has weakened in some decades. The Indian Ocean Dipole, the west-minus-east sea-surface temperature difference of the Indian Ocean, favours the monsoon when positive (warm west) and can offset an El Niño.

5. Fundamentals of tropical cyclones

Formation conditions (all needed together): sea-surface temperature of about 26.5 °C or more through a depth of about 50 m, which supplies heat and moisture; a latitude beyond about 5° from the equator, so that f is large enough to give the vortex rotation; weak vertical wind shear, so that the warm core is not ventilated away; a moist, conditionally unstable troposphere; and a pre-existing disturbance with low-level cyclonic vorticity and convergence to start with. Hence they do not form on the equator, nor over cold water or land. In the North Indian Ocean the seasons are April to June and October to December, with the Bay of Bengal more active than the Arabian Sea.

Structure. Low-level air spirals inward in the boundary layer, picks up heat and moisture from the sea (friction lets it cross isobars), and rises in the eyewall, a ring of towering cumulonimbus at radius of tens of kilometres, where the strongest winds and heaviest rain occur. Within it lies the eye (diameter about 20 to 60 km), where air subsides, warms and dries: clear, light winds and the lowest pressure. Outside are spiral rainbands. At the top of the storm (10 to 15 km) air flows outward in an anticyclonic sense. The storm has a warm core, and winds fall off above the surface. Over land or cold water, the heat and moisture source is cut and friction rises, so it weakens.

Intensity and balance. Intensity is measured by the central pressure and the maximum sustained wind. India Meteorological Department grades North Indian Ocean systems by maximum sustained wind: depression, deep depression, cyclonic storm (34 knots and above), severe, very severe and extremely severe cyclonic storms, and super cyclonic storm (120 knots and above). Away from the core the wind is in gradient-wind balance, V²/r + fV = (1/ρ)∂p/∂r. Near the eyewall, V²/r ≫ fV so the balance is nearly cyclostrophic, V ≈ √((r/ρ)∂p/∂r). Example: at r = 50 km, f = 5.0×10⁻⁵ s⁻¹, ρ = 1.1 kg/m³ and ∂p/∂r = 0.10 Pa/m (10 hPa per 10 km), (1/ρ)∂p/∂r = 0.0909 m/s², V² + 2.5V − 4545 = 0 and V = 66 m/s; the cyclostrophic estimate is 67 m/s.

🎯 Latent heat is the fuel
Evaporation at the warm sea surface stores energy in vapour; condensation in the eyewall releases it as L_v per kilogram, warming the core. The warm core lowers the surface pressure, which steepens the pressure gradient, which strengthens the winds and the evaporation: a positive feedback that stops when the storm moves over land or cold water, or sheared flow, or the supply of energy is cut. Latent heat release, not the Coriolis force, powers the storm; Coriolis only sets the rotation.

Key takeaways

  • Circulation C = ∮V·dl equals vorticity integrated over the area. Absolute vorticity is η = ζ + f; for barotropic, non-divergent flow it is conserved (Kelvin's theorem in outline), so poleward motion makes relative vorticity anticyclonic and divergence reduces it.
  • The general circulation has Hadley (direct, 0 to 30°), Ferrel (indirect, 30 to 60°) and polar cells; trade winds blow from the north-east in the Northern Hemisphere, the ITCZ is the convergence belt, and the subtropical highs are dry sinking regions.
  • Jet streams are westerly maxima near the tropopause, at about 30° (subtropical) and 50 to 60° (polar front). Angular-momentum conservation from the equator would give 134 m/s at 30°; the actual jet is about 50 to 60 m/s because of eddies and friction.
  • Thermal wind: ∂V_g/∂z = (g/(fT)) k × ∇T, a vertical shear tied to the horizontal temperature gradient; in the Northern Hemisphere it blows with the cold air to its left, and westerlies strengthen with height where the air is cold to the north.
  • The Indian summer monsoon: onset near 1 June over Kerala, monsoon trough, Somali low-level jet, active and break spells (trough at the foothills in a break); El Niño tends to weaken it, and a positive IOD can offset this.
  • Tropical cyclones need warm sea (about 26.5 °C or more), a latitude beyond about 5°, weak shear and a seed disturbance. They have a warm core, an eye and an eyewall; the wind is in gradient balance, nearly cyclostrophic near the eyewall, and latent heat powers them.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Kelvin's circulation theorem states that the absolute circulation around a closed material curve is conserved for a fluid that is

    1. viscous and baroclinic, because friction and density gradients balance each other
    2. compressible, with the Mach number greater than one everywhere
    3. incompressible and at rest relative to the Earth at all times
    4. inviscid and barotropic, with density a function of pressure alone
    Show answer

    Answer: D — inviscid and barotropic, with density a function of pressure alone

    DC_a/Dt = −∮ α dp + friction, so the circulation is conserved only when the solenoidal term vanishes (barotropic) and friction is absent (inviscid). A baroclinic fluid generates circulation, which is how a sea breeze starts.

  2. A sea breeze circulation develops in the afternoon mainly because

    1. the Coriolis force deflects the air at the coast and so drives it from the sea toward the land
    2. conservation of absolute vorticity forces the air over the sea to move toward the heated land surface
    3. land and sea air have different temperatures, so pressure and density surfaces intersect and generate circulation
    4. the sea surface is warmer than the land in the afternoon and heats the air above it more strongly
    Show answer

    Answer: C — land and sea air have different temperatures, so pressure and density surfaces intersect and generate circulation

    Differential heating creates a baroclinic state in which the solenoid term −∮ α dp is non-zero, so circulation is generated: surface flow from cool sea to warm land. In the afternoon the land is warmer than the sea, so the claim that the sea is warmer is reversed, and Coriolis plays only a minor role at such small scales.

  3. In the Northern Hemisphere, the trade winds near the surface in the Hadley cell blow from the

    1. north-east toward the equator
    2. south-west toward the pole
    3. north-west toward the equator
    4. south-east toward the pole
    Show answer

    Answer: A — north-east toward the equator

    Surface air in the Hadley cell moves equatorward, and Coriolis turns it to the right in the Northern Hemisphere, to a north-easterly wind (blowing from the north-east). South-westerlies are the westerlies of mid-latitudes.

  4. The Ferrel cell between about 30° and 60° is called thermally indirect because

    1. it has no vertical motion at all and only moves air horizontally between the two other cells
    2. it is driven by the Coriolis force alone, which supplies all the energy of the motion
    3. it is found only over the oceans, where the heating of the air by the surface is indirect
    4. warm air sinks and cold air rises in it, so it is driven by eddy transports rather than by direct heating
    Show answer

    Answer: D — warm air sinks and cold air rises in it, so it is driven by eddy transports rather than by direct heating

    Warm air sinks and colder air rises in the Ferrel cell, the reverse of a heat engine, so it is maintained by momentum and heat transport by cyclones and anticyclones. The Coriolis force does no work and cannot supply energy.

  5. In the Northern Hemisphere troposphere, westerly winds increase with height where the air is

    1. warmer to the north than to the south, by the thermal-wind relation
    2. colder to the north than to the south, by the thermal-wind relation
    3. equally warm everywhere, since the vertical shear depends on pressure only
    4. very humid, since moisture accelerates the wind
    Show answer

    Answer: B — colder to the north than to the south, by the thermal-wind relation

    ∂u_g/∂z = −(g/(fT)) ∂T/∂y, so ∂u_g/∂z > 0 when ∂T/∂y < 0: the air is colder toward the pole. Warm air to the north would give westerlies that weaken with height and eventually easterlies, as in the tropical summer upper troposphere.

  6. During a break spell of the Indian south-west monsoon

    1. the trough shifts to the Himalayan foothills, with heavy rain there and weak rain over most of India
    2. the monsoon trough moves to the southern tip of the peninsula, with heavy rain over Kerala alone
    3. the monsoon withdraws permanently from the whole country within a few days of the start of the spell
    4. the low-level jet strengthens and the trough lies over central India with the heaviest rain
    Show answer

    Answer: A — the trough shifts to the Himalayan foothills, with heavy rain there and weak rain over most of India

    In a break the trough moves north to the foothills of the Himalaya, so central and peninsular India go dry while the foothills and the north-east get heavy rain. The statement with the trough over central India and the strong jet describes an active spell. A break lasts days, after which the monsoon revives.

  7. Which combination is generally associated with a weaker-than-normal Indian summer monsoon?

    1. La Niña together with a positive Indian Ocean Dipole, with a warm western Indian Ocean
    2. A cooler-than-normal Arabian Sea together with a stronger-than-normal low-level jet
    3. An earlier-than-normal northward shift of the ITCZ to about 25°N during April
    4. El Niño, which shifts the Walker circulation and favours subsidence over India
    Show answer

    Answer: D — El Niño, which shifts the Walker circulation and favours subsidence over India

    El Niño is associated with a shift of the ascending branch of the Walker circulation eastward and enhanced subsidence over the Indian region, a risk factor for a weak monsoon (a statistical tendency, not a rule). La Niña and a positive IOD favour the monsoon.

  8. Tropical cyclones do not form within about 5° of the equator mainly because

    1. the sea surface there is always too cold to supply the heat and moisture needed
    2. the surface pressure there is always higher than elsewhere, so no low can form
    3. there is no water vapour in the equatorial atmosphere to release latent heat on condensing
    4. the Coriolis parameter is too small to give the converging air the rotation needed to form a vortex
    Show answer

    Answer: D — the Coriolis parameter is too small to give the converging air the rotation needed to form a vortex

    Near the equator f ≈ 0, so converging air gains little cyclonic spin and the pressure-gradient force cannot be balanced in a rotating vortex. The equatorial sea is very warm and moist, so temperature and humidity are not the reason.

  9. A barotropic, non-divergent parcel at rest relative to the Earth (ζ = 0) at 20°N moves to 40°N. Its relative vorticity on arrival is

    1. positive, equal to f₄₀ − f₂₀, because the parcel gains planetary vorticity on moving poleward
    2. zero, because relative vorticity is conserved when there is no divergence or friction
    3. negative, equal to f₂₀ − f₄₀, because absolute vorticity is conserved while f increases
    4. negative, equal to −f₄₀, because the parcel loses all of its planetary vorticity on arrival
    Show answer

    Answer: C — negative, equal to f₂₀ − f₄₀, because absolute vorticity is conserved while f increases

    η = ζ + f is conserved: 0 + f₂₀ = ζ + f₄₀, so ζ = f₂₀ − f₄₀ < 0 because f grows with latitude. Relative vorticity is not conserved alone, and the parcel keeps its absolute, not its planetary, vorticity.

  10. Select ALL correct statements about the general circulation.

    1. The ITCZ is a belt of convergence and deep convection that follows the thermal equator seasonally
    2. The subtropical highs near 30° are regions of descending air, giving dry conditions
    3. The polar cell is thermally indirect, with warm air sinking at the pole
    4. Surface winds in the mid-latitude Ferrel cell are mainly easterly
    Show answer

    Answer: A — The ITCZ is a belt of convergence and deep convection that follows the thermal equator seasonally; B — The subtropical highs near 30° are regions of descending air, giving dry conditions

    The ITCZ and the sinking subtropical highs are standard features. The polar cell is thermally direct, with cold air sinking at the pole, and the mid-latitude surface winds are westerly.

  11. Select ALL correct statements about the Indian summer monsoon.

    1. The low-level (Somali) jet carries moisture across the Arabian Sea in a south-westerly flow
    2. The monsoon trough extends from the heat low over north-west India to the head of the Bay of Bengal
    3. The normal onset date over Kerala is around 1 July
    4. A positive Indian Ocean Dipole always guarantees a strong monsoon whatever the ENSO state
    Show answer

    Answer: A — The low-level (Somali) jet carries moisture across the Arabian Sea in a south-westerly flow; B — The monsoon trough extends from the heat low over north-west India to the head of the Bay of Bengal

    The Somali jet and the monsoon trough are standard features. The normal onset over Kerala is about 1 June, not 1 July, and the IOD is only a tendency, not a guarantee.

  12. Select ALL correct statements about a mature tropical cyclone.

    1. The eye is a region of subsiding, warm, relatively dry air with light winds
    2. The strongest winds and heaviest rain occur in the eyewall
    3. The storm is powered mainly by the Coriolis force doing work on the air
    4. Near the eyewall the wind balance is nearly cyclostrophic, V²/r ≈ (1/ρ)∂p/∂r
    Show answer

    Answer: A — The eye is a region of subsiding, warm, relatively dry air with light winds; B — The strongest winds and heaviest rain occur in the eyewall; D — Near the eyewall the wind balance is nearly cyclostrophic, V²/r ≈ (1/ρ)∂p/∂r

    Subsidence warms and dries the eye, the eyewall holds the peak wind and rain, and near it V²/r ≫ fV, so the balance is nearly cyclostrophic. The energy comes from latent heat release; the Coriolis force does no work.

  13. A westerly wind decreases from u = 12 m/s at y = 0 to u = 4 m/s at y = 400 km to the north, with v = 0 throughout. Find the relative vorticity ζ = ∂v/∂x − ∂u/∂y, in units of 10⁻⁵ s⁻¹, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 2.0

    ∂u/∂y = (4 − 12)/(4×10⁵ m) = −2×10⁻⁵ s⁻¹, so ζ = −(−2×10⁻⁵) = +2.0×10⁻⁵ s⁻¹ (cyclonic).

  14. Air with relative vorticity ζ = 0 at 30°N moves to 45°N with absolute vorticity conserved (barotropic, non-divergent flow). Take Ω = 7.292×10⁻⁵ s⁻¹. Find its relative vorticity at 45°N, in units of 10⁻⁵ s⁻¹, to two decimal places (include the sign).

    Numerical answer — type the value.

    Show answer

    Answer: -3.02

    f₃₀ = 2Ω sin 30° = 7.292×10⁻⁵; f₄₅ = 2Ω sin 45° = 10.312×10⁻⁵. ζ = f₃₀ − f₄₅ = −3.02×10⁻⁵ s⁻¹.

  15. Air at rest relative to the Earth at the equator moves poleward conserving absolute angular momentum, so that u = Ωa sin²φ / cos φ. With Ω = 7.292×10⁻⁵ s⁻¹ and a = 6.371×10⁶ m, find u in m/s at φ = 30°, rounded to the nearest integer.

    Numerical answer — type the value.

    Show answer

    Answer: 134

    Ωa = 7.292×10⁻⁵ × 6.371×10⁶ = 464.6 m/s. sin²30° / cos 30° = 0.25/0.8660 = 0.2887. u = 464.6 × 0.2887 = 134.1 m/s.

  16. In a layer with mean temperature T = 260 K, temperature falls northward at 1.0×10⁻⁵ K/m at a latitude where f = 1.0×10⁻⁴ s⁻¹. Taking g = 9.81 m/s² and the thermal-wind relation ∂u_g/∂z = −(g/(fT)) ∂T/∂y, find the increase of the geostrophic westerly wind between heights 1 km and 9 km, in m/s, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 30.2

    ∂u_g/∂z = −(9.81/(1.0×10⁻⁴ × 260)) × (−1.0×10⁻⁵) = 3.773×10⁻³ s⁻¹. Over 8000 m, Δu = 3.773×10⁻³ × 8000 = 30.2 m/s.

  17. In a tropical cyclone at radius r = 40 km, f = 5.0×10⁻⁵ s⁻¹ and the pressure-gradient force per unit mass is (1/ρ)∂p/∂r = 0.080 m/s² toward the centre. Solving the gradient-wind equation V²/r + fV = (1/ρ)∂p/∂r for the positive root, find V in m/s, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 55.6

    fr = 2.0 m/s and (1/ρ)(∂p/∂r) r = 0.080 × 4×10⁴ = 3200. V² + 2.0V − 3200 = 0, so V = (−2 + √(4 + 12800))/2 = (−2 + 113.15)/2 = 55.6 m/s.