Currents and Wind-driven Circulation: Inertial and Geostrophic Flow, Air-Sea Fluxes, Ekman and Sverdrup Transports

This chapter continues the ocean physics of the Engineering Sciences (XE) paper, Section XE7.2 of XE7 Atmospheric and Oceanic Sciences. It answers the middle of the heading: “Inertial Currents; Geostrophic Motion; Air-Sea Surface Fluxes; Wind-driven Circulation, Ekman and Sverdrup Transports”. It teaches inertial circles, geostrophic velocity from sea-surface slope and dynamic height, the bulk formulae for wind stress and heat fluxes, the Ekman layer, Ekman transport and pumping, and Sverdrup transport with the idea of western intensification. The numericals are the inertial period, geostrophic speed, bulk latent heat flux, Ekman transport and Sverdrup transport; every constant is given in the stem.

1. Inertial currents

Suppose the wind stops and no pressure gradient acts. A water parcel moving with speed V is then acted on by the Coriolis force alone, f V, which is perpendicular to its velocity and does no work. The parcel moves in a circle: the inertial circle, of radius r = V/f, travelled in the inertial period T = 2π/f = 2π/(2Ω sin φ) = 11.97 h / sin φ. Here f = 2Ω sin φ is the Coriolis parameter and Ω = 7.292 × 10⁻⁵ s⁻¹ is the rotation rate of the Earth. The motion is clockwise in the northern hemisphere and anticlockwise in the southern, and the period is about 24 h at 30° latitude, 17 h at 45° and 12 h at the pole. There are no inertial oscillations at the equator, where f = 0.

Worked example. At 45° N with V = 0.30 m s⁻¹: f = 2 × 7.292 × 10⁻⁵ × 0.7071 = 1.031 × 10⁻⁴ s⁻¹, so r = 0.30/1.031 × 10⁻⁴ = 2.9 km and T = 2π/f = 6.09 × 10⁴ s = 16.9 h. Inertial currents are excited when a storm passes and appear as clockwise-rotating current vectors in mooring records, with the frequency of f and a slight upward shift of frequency with the background vorticity.

⚠️ Trap: inertial period
T = 2π/f uses f = 2Ω sin φ, so T is 12 h/sin φ, not 24 h/sin φ. The 24 h figure is the period of the Foucault pendulum's plane rotation, T = 24 h/sin φ, which is twice the inertial period.

2. Geostrophic motion, sea-surface slope and dynamic height

For large-scale, steady flow far from boundaries the horizontal momentum balance is between the Coriolis force and the pressure gradient (small Rossby number). Geostrophic balance: f v_g = (1/ρ) ∂p/∂x and f u_g = −(1/ρ) ∂p/∂y. Flow follows isobars, with high pressure on the right in the northern hemisphere (left in the southern). Using the hydrostatic relation, the pressure gradient at the surface is set by the slope of the sea surface η: u_g = −(g/f) ∂η/∂y, v_g = (g/f) ∂η/∂x. A rise of 1 m over 100 km at f = 10⁻⁴ s⁻¹ gives (9.81/10⁻⁴) × 10⁻⁵ = 0.98 m s⁻¹; sea-surface height from satellite altimetry is therefore a direct measure of the surface geostrophic current.

Below the surface the pressure gradient also depends on the density field. The dynamic height (geopotential anomaly) Φ = ∫ δ dp, with δ the specific volume anomaly, is computed from a CTD (conductivity-temperature-depth) cast. Between two stations a distance L apart, the velocity at the upper level relative to a deeper level of no motion is v = (Φ₁ − Φ₂)/(f L), with Φ in m² s⁻². Differentiating the geostrophic and hydrostatic balances in z gives the thermal wind relation ∂v_g/∂z = −(g/(ρ₀ f)) ∂ρ/∂x: horizontal density gradients mean the current changes with depth. The unknown is the reference level; classically it is taken where the current is assumed to vanish (the level of no motion).

⚠️ Trap: direction and f
Geostrophy fails near the equator (f → 0, so the formula gives infinite speed) and for small-scale, fast flow where the Rossby number U/(fL) is not small. Always check the hemisphere: the same slope gives opposite flow directions on either side of the equator.

3. Air-sea surface fluxes

The atmosphere drives the ocean through momentum and heat exchange across the surface. Using bulk formulae with 10 m wind speed U, air density ρₐ, and empirical transfer coefficients, positive upward (ocean loses): wind stress τ = ρₐ C_D U², sensible heat Q_H = ρₐ c_pa C_H U (T_s − T_a), latent heat Q_E = ρₐ L_e C_E U (q_s − q_a), where q is specific humidity and L_e the latent heat of vaporisation. Typical values: ρₐ = 1.2 kg m⁻³, C_D ≈ 1.3 × 10⁻³, C_H ≈ 1.0 × 10⁻³, C_E ≈ 1.2 × 10⁻³, c_pa = 1004 J kg⁻¹ K⁻¹, L_e = 2.5 × 10⁶ J kg⁻¹. All three are proportional to U (stress to U²) and the transfer coefficients vary with wind speed and stability, so exam stems give them.

The radiative terms are the absorbed shortwave (1 − A) Q_sw, with albedo A, and the net longwave loss, which uses the Stefan-Boltzmann law σ ε T⁴ with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ for the sea surface (ε ≈ 0.97) less the downwelling atmospheric longwave. The net surface heat flux into the ocean is Q_net = (1 − A) Q_sw − Q_LW − Q_H − Q_E. In the tropics Q_E dominates the losses. Worked example. U = 8 m s⁻¹, C_D = 1.3 × 10⁻³: τ = 1.2 × 1.3 × 10⁻³ × 64 = 0.10 N m⁻². With T_s − T_a = 2 K: Q_H = 1.2 × 1004 × 10⁻³ × 8 × 2 = 19 W m⁻². With q_s − q_a = 4 g kg⁻¹: Q_E = 1.2 × 2.5 × 10⁶ × 1.2 × 10⁻³ × 8 × 0.004 = 115 W m⁻². The Bowen ratio Q_H/Q_E ≈ 0.17.

⚠️ Trap: units of humidity
Specific humidity is usually quoted in g kg⁻¹; convert to kg kg⁻¹ (divide by 1000) before using Q_E. Doubling the wind speed doubles Q_H and Q_E but quadruples the stress.

4. Wind-driven circulation: the Ekman layer and Ekman transport

Near the surface the wind stress is balanced by Coriolis force and friction (vertical eddy viscosity A_z), with the pressure gradient neglected: −f v = (1/ρ) ∂τₓ/∂z, f u = (1/ρ) ∂τ_y/∂z. The solution is the Ekman spiral: the surface current is 45° to the right of the wind in the northern hemisphere (left in the southern), and with depth the current decays and turns clockwise (NH) over the Ekman depth D_E = π √(2A_z/f), typically tens of metres. Integrated over the layer the friction terms cancel and the result does not depend on A_z: Ekman volume transport per unit width: (Uₑ, Vₑ) = (τ_y/(ρ f), −τₓ/(ρ f)), which is 90° to the right of the wind (NH) and equals |τ|/(ρ f).

Worked example. τ = 0.10 N m⁻² westward (x negative) at f = 1.0 × 10⁻⁴ s⁻¹, ρ = 1025 kg m⁻³: |M| = 0.10/(1025 × 1.0 × 10⁻⁴) = 0.98 m² s⁻¹, directed to the north (right of a westward wind). Across a 1000 km front this is 9.8 × 10⁵ m³ s⁻¹ ≈ 0.98 Sv (1 Sv = 10⁶ m³ s⁻¹). Coastal upwelling: along an eastern-boundary coast of the northern hemisphere (California, Peru-Chile in the south by mirror), an equatorward wind drives offshore Ekman transport, and cold nutrient-rich water rises to replace it. At the equator f → 0 and the Ekman transport grows large, so equatorial winds drive divergence and upwelling on both sides.

5. Ekman pumping

If the wind stress varies in space, Ekman transport has a divergence, and continuity forces a vertical velocity at the base of the Ekman layer: w_E = (1/ρ) [∂/∂x (τ_y/f) − ∂/∂y (τₓ/f)] ≈ (1/(ρ f)) k·(∇ × τ) where f is treated as constant. A positive wind-stress curl (cyclonic) gives upwelling (suction, w_E > 0) and a negative curl (anticyclonic) gives downwelling (pumping, w_E < 0) in the northern hemisphere. The signs reverse in the southern hemisphere because f < 0. Subtropical gyres, under the westerlies and trades, sit under negative curl and have convergence and downwelling, which depresses the thermocline and creates the high sea level at the gyre centre.

Worked example. The stress changes by 0.10 N m⁻² across 1000 km, so curl τ = 1.0 × 10⁻⁷ N m⁻³. At f = 1.0 × 10⁻⁴ s⁻¹: w_E = 1.0 × 10⁻⁷/(1025 × 1.0 × 10⁻⁴) = 9.8 × 10⁻⁷ m s⁻¹ ≈ 31 m per year (1 yr = 3.156 × 10⁷ s). This is tiny compared with horizontal currents, yet it is what drives the basin-scale flow.

6. Sverdrup transport and western intensification

Below the Ekman layer, in the interior, the depth-integrated flow is geostrophic. Taking the curl of the depth-integrated geostrophic and Ekman balance (friction neglected) gives the Sverdrup relation: β V = (1/ρ₀) k·(∇ × τ), i.e. V = (1/(ρ₀ β)) k·(∇ × τ), where V is the depth-integrated meridional volume transport per unit zonal width (m² s⁻¹), β = ∂f/∂y = 2Ω cos φ/R ≈ 2.0 × 10⁻¹¹ m⁻¹ s⁻¹ at 30°. Physically, wind-stress curl injects vorticity, and the interior balances it by moving water meridionally so that f changes. Negative curl (subtropical gyre) gives equatorward interior flow; positive curl (subpolar gyre) gives poleward flow. Total transport across a basin of width W is V × W.

Worked example. Over a basin 4000 km wide curl τ = −1.5 × 10⁻⁷ N m⁻³ and β = 2.0 × 10⁻¹¹: V = −1.5 × 10⁻⁷/(1025 × 2.0 × 10⁻¹¹) = −7.3 m² s⁻¹ (equatorward), total = 7.3 × 4 × 10⁶ = 29 Sv. The Sverdrup interior flow cannot close by itself: equatorward flow across the whole interior must be returned poleward somewhere. A narrow, fast western boundary current (Gulf Stream, Kuroshio, Agulhas, Brazil and East Australian currents) closes it. Westward intensification arises because β increases the vorticity change needed for a parcel to go north, and only on the western side can friction (Stommel) or lateral viscosity (Munk) supply the vorticity balance in a narrow band; on an eastern boundary the same balance would need a boundary layer wider than the interior itself. Sverdrup balance does not hold in the boundary layers, near the equator, or where friction matters.

⚠️ Trap: what Sverdrup transport includes
V is the transport of the whole water column (Ekman layer included) and is a volume transport per unit width. Multiply by the basin width for Sv. It gives the meridional component only; the zonal component follows from continuity.

Key takeaways

  • Inertial period T = 2π/f = 11.97 h/sin φ and radius V/f; the motion is clockwise in the northern hemisphere.
  • Geostrophic flow keeps high pressure on the right in the northern hemisphere: v_g = (g/f) ∂η/∂x, and dynamic height gives the velocity relative to a reference level.
  • Bulk fluxes: τ = ρₐ C_D U², Q_H and Q_E linear in U; convert humidity to kg kg⁻¹ and watch the sign convention.
  • Ekman transport = |τ|/(ρ f), 90° to the right of the wind in the northern hemisphere; surface current is 45° to the right.
  • Ekman pumping w_E = curl τ/(ρ f): positive curl gives upwelling, negative curl downwelling in the northern hemisphere.
  • Sverdrup: V = curl τ/(ρ₀ β) in the interior; the western boundary current closes the circulation.

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. In the northern hemisphere, geostrophic flow around a region of high sea-surface elevation is

    1. clockwise, keeping the high elevation on its right
    2. anticlockwise, keeping the high elevation on its right
    3. directly down the slope, away from the high elevation
    4. clockwise, keeping the high elevation on its left
    Show answer

    Answer: A — clockwise, keeping the high elevation on its right

    With high pressure on the right of the flow in the northern hemisphere, going around a high means clockwise circulation. Down-slope flow would be ageostrophic, and keeping the high on the left is the southern-hemisphere rule.

  2. A steady wind blows towards the east over the open northern-hemisphere ocean. Ignoring other forces, the depth-integrated Ekman transport is directed

    1. towards the north, 90° to the left of the wind
    2. towards the south, 90° to the right of the wind
    3. towards the east, along the wind
    4. towards the south-east, 45° to the right of the wind
    Show answer

    Answer: B — towards the south, 90° to the right of the wind

    The transport is 90° to the right of the wind in the northern hemisphere, so eastward wind gives southward transport. The 45° direction belongs to the surface current only, not the integrated transport.

  3. A positive (cyclonic) wind-stress curl over the open northern-hemisphere ocean produces

    1. Ekman convergence, downwelling and a deepening thermocline
    2. Ekman convergence, piling water up and raising the sea surface
    3. Ekman divergence with upwelling and a shoaling thermocline
    4. no vertical motion, only a change in Ekman depth
    Show answer

    Answer: C — Ekman divergence with upwelling and a shoaling thermocline

    Positive curl gives divergent Ekman transport, so water rises from below, w_E > 0, and the thermocline shoals. Downwelling and a raised surface go with negative curl.

  4. Why is the Gulf Stream narrow and fast on the western side of the North Atlantic while the interior return flow is slow and broad?

    1. The equatorward interior flow must be returned, and friction closes the β-effect vorticity balance only in a narrow western layer
    2. The Earth rotates eastward, so the ocean water is dragged back against the western coasts of every basin
    3. The wind stress is always strongest right at the western boundary of the basin
    4. The Coriolis parameter is larger at the western boundary of every basin
    Show answer

    Answer: A — The equatorward interior flow must be returned, and friction closes the β-effect vorticity balance only in a narrow western layer

    Mass conservation needs a poleward return flow, and only on the western side can friction supply the required vorticity in a narrow layer. The wind need not peak there, and f depends on latitude, not on which side of the basin.

  5. In the bulk formulae, when the 10 m wind speed doubles with all coefficients and temperature and humidity differences unchanged, the wind stress and the latent heat flux change by factors of

    1. factors 2 and 2
    2. factors 4 and 2
    3. factors 4 and 4
    4. factors 2 and 4
    Show answer

    Answer: B — factors 4 and 2

    τ = ρₐ C_D U² scales as U², while Q_E = ρₐ L_e C_E U Δq is linear in U. Treating both as quadratic ignores that heat fluxes use one factor of U.

  6. Two CTD stations 100 km apart have dynamic heights of 20.00 m² s⁻² and 19.00 m² s⁻² at the surface relative to a deep level of no motion. At f = 1.0 × 10⁻⁴ s⁻¹, the surface speed relative to that level is

    1. 0.10 m s⁻¹
    2. 1.0 m s⁻¹
    3. 0.010 m s⁻¹
    4. 10 m s⁻¹
    Show answer

    Answer: A — 0.10 m s⁻¹

    v = ΔΦ/(f L) = 1.00/(1.0 × 10⁻⁴ × 1.0 × 10⁵) = 0.10 m s⁻¹. Forgetting to convert 100 km to metres gives 100 times too large a speed.

  7. A westerly wind of constant stress blows along 40° N and 40° S in two different oceans, with the same magnitude of stress, density and |f|. Compared with the northern case, the Ekman transport in the southern case has

    1. the same magnitude, directed 90° to the left of the wind (equatorward)
    2. the same magnitude and direction, since only the wind matters
    3. half the magnitude, directed to the right of the wind
    4. the same magnitude, directed 90° to the right of the wind (poleward)
    Show answer

    Answer: A — the same magnitude, directed 90° to the left of the wind (equatorward)

    In the southern hemisphere f is negative, so the transport turns 90° to the left of the wind; for a westerly that is towards the equator (north). The magnitude |τ|/(ρ|f|) is unchanged, so the two hemispheres differ in direction only.

  8. Which statement about the thermal wind relation ∂v_g/∂z = −(g/(ρ₀ f)) ∂ρ/∂x is correct?

    1. It holds only if the density is uniform everywhere in the horizontal plane and the flow is steady
    2. A horizontal density gradient means the geostrophic current varies with depth
    3. It gives the absolute surface current directly, with no reference level needed
    4. It follows from the Ekman balance, not from geostrophy and hydrostasy together
    Show answer

    Answer: B — A horizontal density gradient means the geostrophic current varies with depth

    Combining geostrophic and hydrostatic balances shows that tilted density surfaces (∂ρ/∂x ≠ 0) imply vertical shear. It gives only the shear, so a reference velocity is still needed. It follows from geostrophy plus hydrostasy, not from the Ekman balance.

  9. Over a warm tropical ocean the surface (SST) is 28 °C and the air temperature is 27 °C, with the air at about 80 % relative humidity, so the surface-to-air humidity difference is a few g kg⁻¹. Which term of the net surface heat flux is expected to dominate the heat loss, and why?

    1. Sensible heat, because water has a large heat capacity and warms the air above it
    2. Net longwave radiation, since 28 °C water emits strongly like a blackbody and the sky returns little
    3. Conduction of heat downward into the deep ocean through the thermocline
    4. Latent heat: the humidity gap is large while the temperature gap is small
    Show answer

    Answer: D — Latent heat: the humidity gap is large while the temperature gap is small

    With T_s − T_a only 1 K, Q_H is small, but q_s − q_a is a few g kg⁻¹, so evaporation removes far more heat. Net longwave loss is tens of W m⁻² and is typically smaller than Q_E in the tropics.

  10. Which of the following statements about Ekman dynamics are correct?

    1. In the northern hemisphere the surface current is 45° to the right of the wind
    2. The Ekman volume transport per unit width is |τ|/(ρ f) and does not depend on the eddy viscosity
    3. Equatorward wind along an eastern-boundary coast of the northern hemisphere drives coastal upwelling
    4. The Ekman transport is parallel to the wind
    Show answer

    Answer: A — In the northern hemisphere the surface current is 45° to the right of the wind; B — The Ekman volume transport per unit width is |τ|/(ρ f) and does not depend on the eddy viscosity; C — Equatorward wind along an eastern-boundary coast of the northern hemisphere drives coastal upwelling

    The first three are standard results of the Ekman model. The transport is perpendicular, not parallel, to the wind.

  11. Which of the following are true of the Sverdrup balance βV = (1/ρ₀) k·(∇ × τ)?

    1. It relates the depth-integrated meridional transport to the wind-stress curl in the ocean interior
    2. Under negative curl in the northern hemisphere the interior flow is equatorward
    3. It requires a return flow in a western boundary layer, where the balance itself does not hold
    4. It applies at the equator, where β is largest and f is zero
    Show answer

    Answer: A — It relates the depth-integrated meridional transport to the wind-stress curl in the ocean interior; B — Under negative curl in the northern hemisphere the interior flow is equatorward; C — It requires a return flow in a western boundary layer, where the balance itself does not hold

    The first three are correct. The relation is a geostrophic, frictionless interior result and fails at the equator, where geostrophy is invalid; β is largest there but the balance is not used.

  12. Which of the following statements about inertial and geostrophic motion are correct?

    1. Inertial motion in the northern hemisphere is clockwise with period 2π/f
    2. The inertial radius equals V/f
    3. A geostrophic current has a nonzero speed only if the sea surface or the pressure surfaces are tilted
    4. The inertial period is shortest at the equator
    Show answer

    Answer: A — Inertial motion in the northern hemisphere is clockwise with period 2π/f; B — The inertial radius equals V/f; C — A geostrophic current has a nonzero speed only if the sea surface or the pressure surfaces are tilted

    The first three are correct. The inertial period 11.97 h/sin φ is shortest at the poles and infinite at the equator.

  13. Take Ω = 7.292 × 10⁻⁵ s⁻¹. The inertial period of the water at latitude 30° N is ____ hours (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 23.9

    f = 2Ω sin 30° = 7.292 × 10⁻⁵ s⁻¹. T = 2π/f = 6.2832/7.292 × 10⁻⁵ = 8.617 × 10⁴ s = 23.9 h. Using Ω sin φ instead of 2Ω sin φ doubles it.

  14. The sea surface rises by 0.80 m over a horizontal distance of 200 km. With g = 9.81 m s⁻² and f = 1.0 × 10⁻⁴ s⁻¹, the surface geostrophic speed is ____ m s⁻¹ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.39

    v = (g/f)(Δη/L) = (9.81/1.0 × 10⁻⁴)(0.80/2.0 × 10⁵) = 9.81 × 10⁴ × 4.0 × 10⁻⁶ = 0.392, i.e. 0.39 m s⁻¹.

  15. A uniform alongshore wind stress of 0.080 N m⁻² acts on the open ocean at a latitude where f = 1.2 × 10⁻⁴ s⁻¹. Take ρ = 1025 kg m⁻³. The Ekman volume transport across a 500 km length of line perpendicular to the transport direction is ____ × 10⁵ m³ s⁻¹ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 3.25

    Per unit width |M| = τ/(ρ f) = 0.080/(1025 × 1.2 × 10⁻⁴) = 0.6504 m² s⁻¹. Times 5.0 × 10⁵ m gives 3.252 × 10⁵ m³ s⁻¹, i.e. 3.25.

  16. Use ρ₀ = 1025 kg m⁻³ and β = 2.0 × 10⁻¹¹ m⁻¹ s⁻¹. Over a subtropical basin 4000 km wide the wind-stress curl is −1.5 × 10⁻⁷ N m⁻³. The magnitude of the total Sverdrup transport across the basin is ____ Sv (one decimal place; 1 Sv = 10⁶ m³ s⁻¹).

    Numerical answer — type the value.

    Show answer

    Answer: 29.3

    |V| = 1.5 × 10⁻⁷/(1025 × 2.0 × 10⁻¹¹) = 1.5 × 10⁻⁷/2.05 × 10⁻⁸ = 7.317 m² s⁻¹. Times 4.0 × 10⁶ m = 2.927 × 10⁷ m³ s⁻¹ = 29.3 Sv.

  17. Using the bulk formula Q_E = ρₐ L_e C_E U (q_s − q_a) with ρₐ = 1.2 kg m⁻³, L_e = 2.5 × 10⁶ J kg⁻¹, C_E = 1.2 × 10⁻³, U = 10 m s⁻¹ and q_s − q_a = 3.0 g kg⁻¹, the latent heat flux is ____ W m⁻² (integer).

    Numerical answer — type the value.

    Show answer

    Answer: 108

    Convert q: 3.0 g kg⁻¹ = 0.0030. Q_E = 1.2 × 2.5 × 10⁶ × 1.2 × 10⁻³ × 10 × 0.0030 = 3600 × 10 × 0.0030 = 108 W m⁻². Leaving q in g kg⁻¹ gives 1000 times more.