Find the question you need
Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Name the current question
Chapters 1–2. Separate velocity, shear and the model’s coordinates.
Build the calculation
Chapters 3–5. Derive thermal wind and test a density-section example.
Recover what is missing
Chapters 6–7. Add a reference velocity and examine transport.
Check the evidence
Chapters 8–10. Match observations to assumptions, limits and common questions.
Practise and continue
Chapters 11–12. Transfer the calculation and choose connected guides.
Full chapter index · Practice and worked checks · Physical Oceanography subject root
A density section can tell you how an ocean current changes with depth without telling you the current’s absolute direction. The upper water may move more eastward than the lower water while both layers still move westward.
That distinction is the central question here. We derive a simple geostrophic-shear relationship, work an original fictional section and add the missing reference information. Start with Physical Oceanography for the wider subject map. These calculations describe a teaching model, not an observed current or an operational forecast.
Chapter index
Name the current question · Chapters 1–2
1. Ask whether you need velocity or shear
2. Fix the coordinates and approximation
Build the calculation · Chapters 3–5
3. Derive the vertical shear
4. Calculate a fictional density section
5. Interpret the difference before naming a direction
Recover what is missing · Chapters 6–7
6. Add three possible reference velocities
7. See why the reference matters to transport
Check the evidence · Chapters 8–10
8. Choose observations that supply the missing information
9. Keep the limits with the answer
10. Questions that clarify the inference
Practise and continue · Chapters 11–12
11. Practise with changed gradients and references
12. Continue through ocean evidence
CHAPTER 1 OF 12 · Name the current question
1. Ask whether you need velocity or shear
Velocity describes motion relative to a stated frame. Vertical shear describes how that velocity changes with vertical position. A velocity of 0.10 m/s eastward and a difference of 0.10 m/s between two depths are therefore different statements.
Imagine two walkers. One is moving 1 m/s faster eastward than the other. That difference alone does not tell you whether either walker is travelling east or west. Adding the same velocity to both preserves their difference.
An ocean density section creates a related inference problem. Under suitable balances, horizontal density differences constrain vertical differences in geostrophic velocity. A separate observation or justified constraint is needed to fix a reference velocity.
CHAPTER 2 OF 12 · Name the current question
2. Fix the coordinates and approximation
Use x eastward, y northward and z upward. Let u be eastward geostrophic velocity, so a negative u means westward motion. Let f be the Coriolis parameter, positive in this Northern Hemisphere example, and g gravitational acceleration.
Adopt a Boussinesq model: use constant reference density ρ0 in the horizontal momentum balance while retaining the density field ρ in hydrostatic pressure. The density variations are small relative to ρ0. Assume a smooth field, hydrostatic balance and a suitable large-scale geostrophic approximation.
This is not a model for every ocean motion. Rapid acceleration, strong friction, small scales and the equatorial limit can make neglected terms important. The Naval Postgraduate School’s physical-oceanography notes explain these balances and the reference-level problem.
The northward geostrophic momentum balance is fu = −(1/ρ0) ∂p/∂y. Hydrostatic balance is ∂p/∂z = −ρg. These relate pressure gradients to current and to the weight of the water column respectively.
Differentiate the geostrophic equation with respect to z. With constant f and ρ0, interchange the derivatives of the smooth pressure field and substitute the hydrostatic relationship. This gives f ∂u/∂z = (g/ρ0) ∂ρ/∂y, or:
∂u/∂z = [g/(ρ0f)] ∂ρ/∂y.
For these coordinates, a positive northward density gradient gives a positive upward change in eastward velocity when f is positive. Reversing the horizontal gradient or the sign of f reverses the shear. The relationship is often called thermal wind, but in seawater the relevant density structure can involve salinity as well as temperature.
Suppose water at the northern end of a 100 km section is 0.50 kg/m³ denser than water at the southern end at each height in the modelled layer. Treat that horizontal gradient as uniform through a 200 m vertical interval, from z = −200 m to z = 0.
| Model input or result | Value | Meaning |
|---|---|---|
| Reference density ρ0 | 1,025 kg/m³ | Stipulated Boussinesq constant |
| Coriolis parameter f | 0.0001 s⁻¹ | Positive for this fictional case |
| Gravity g | 9.81 m/s² | Stipulated constant |
| Northward density gradient | 0.50 / 100,000 = 0.000005 kg/m⁴ | Density difference divided by northward distance |
| Vertical shear ∂u/∂z | About 0.00047854 s⁻¹ | Eastward velocity change per upward metre |
| Upper-minus-lower velocity | About 0.095707 m/s | Shear multiplied by 200 m |
The calculation is 9.81 × 0.000005 / (1,025 × 0.0001), followed by multiplication by 200. Units reduce first to s⁻¹, then to m/s. A common error is entering 100 rather than 100,000 for a distance stated in kilometres.
CHAPTER 5 OF 12 · Build the calculation
5. Interpret the difference before naming a direction
The result is uupper − ulower ≈ 0.095707 m/s. It does not yet say uupper = 0.095707 m/s. That second statement adds the assumption ulower = 0.
Integrating a vertical derivative introduces a reference value. For uniform shear s in this layer, u(z) = uref + s(z − zref), with zref = −200 m. The density-derived calculation supplies s; it does not supply uref.
Adding a depth-uniform velocity offset leaves the shear unchanged. Within this local model, the corresponding depth-independent pressure-gradient contribution is also not fixed by the vertical derivative. A complete ocean circulation must satisfy its wider constraints, but the density section alone has not specified them.
CHAPTER 6 OF 12 · Recover what is missing
6. Add three possible reference velocities
Keep the same density field and shear. Compare three choices for the lower reference velocity.
| Lower reference velocity | Upper velocity | Upper-water direction |
|---|---|---|
| 0 m/s | +0.095707 m/s | Eastward |
| +0.050 m/s | +0.145707 m/s | Eastward |
| −0.150 m/s | −0.054293 m/s | Westward |
In the third case, the upper layer is moving more eastward relative to the lower layer, yet both velocities are westward. A positive velocity difference is compatible with a negative absolute velocity.
Calling a deep level a “level of no motion” is an assumption about its velocity. Depth alone does not establish it. NOAA’s ocean-transport methodology explicitly distinguishes a relative geostrophic contribution from a reference-level contribution and an ageostrophic component.
CHAPTER 7 OF 12 · Recover what is missing
7. See why the reference matters to transport
Suppose the fictional profile is uniform along a 20 km wide section and linear through its 200 m depth. The section area is 4,000,000 m². With uref = 0, the mean velocity is half the upper value, about 0.047854 m/s, giving about 191,415 m³/s eastward.
Now add −0.150 m/s everywhere. The transport changes by area × offset = 4,000,000 × (−0.150) = −600,000 m³/s. The resulting net is about −408,585 m³/s, westward, even though the density-derived shear is unchanged.
This section is normal to the eastward velocity component. A differently oriented section needs the appropriate normal component. Real mass and energy budgets also require boundaries, other flow components and storage terms; this small model does not establish a basin-wide transport.
CHAPTER 8 OF 12 · Check the evidence
8. Choose observations that supply the missing information
A suitably interpreted current measurement at a reference depth can constrain the velocity offset. Sea-surface height and appropriate pressure information can also contribute to an absolute geostrophic estimate. Each route has its own reference definitions and uncertainties.
An instantaneous current-meter reading is not automatically a pure geostrophic reference. Tidal motion, wind-driven flow, other ageostrophic contributions and mismatched sampling times can affect the comparison. Match the averaging period, location and physical quantity before adding measurements to the density-derived profile.
Temperature alone may be insufficient to establish the density contrast. Salinity and pressure definitions matter, and a precision seawater calculation needs appropriate thermodynamic variables. Our density difference is supplied directly as a teaching input; it is not inferred from a temperature colour map.
The derivation concerns a geostrophic component under hydrostatic and Boussinesq assumptions. It does not assert that every measured current obeys that balance. Near the equator, division by a small f signals a model limitation rather than a trustworthy prediction of an enormous velocity.
A uniform gradient through the layer was another substantive assumption. If ∂ρ/∂y changes with z, integrate the varying shear over depth instead of multiplying one value by the total interval. A narrow layer of strong shear and a weakly sheared broad layer need not share the same profile as a constant-gradient model.
Finally, check the coordinate convention. A derivation using depth positive downward has different intermediate signs. Translate the coordinates before comparing formulas; do not fix an apparent disagreement by changing a sign without understanding its origin.
Is thermal wind an additional current caused only by temperature?
Here it names a relationship between density structure and vertical geostrophic shear. It is not an extra velocity to add indiscriminately to every observed flow, and seawater density can depend on salinity as well as temperature.
Does more eastward motion with height mean eastward surface flow?
No. The reference-velocity table shows a profile that becomes less westward upward while remaining westward throughout. The difference and the absolute direction are separate results.
Can a deeper reference always be assumed stationary?
No. A no-motion choice requires justification appropriate to the region and question. A deeper measurement may still contain substantial flow.
CHAPTER 11 OF 12 · Practise and continue
11. Practise with changed gradients and references
Case A: halve the supplied horizontal density difference to 0.25 kg/m³ while retaining the distance, f and vertical interval. Shear and upper-minus-lower velocity halve, giving about 0.047854 m/s. With a reference velocity of −0.080 m/s, the upper velocity is about −0.032146 m/s.
Case B: restore the original gradient but use a 100 m interval. The difference is again about 0.047854 m/s. Explain why this equality with Case A comes from two different changes in the model.
Case C: reverse f while retaining the coordinate definitions and density gradient. The upper-minus-lower difference reverses sign. This is a mathematical comparison under stipulated balances; it is not permission to transfer a real current section unchanged between hemispheres without checking its setting.
Mechanics and observing systems
Return to Physical Oceanography and Fluid Dynamics for pressure, momentum and approximation choice. Oceanographic Engineering connects scientific questions to dependable observations.
Distinguish another current component
Ekman transport develops the ideal wind-driven, layer-integrated response. It helps explain why a total current measurement may contain more than the geostrophic component calculated here.
Transfer the reasoning
The Gold Standard of Education and How X Works connect this example to wider learning. Whenever evidence gives a difference or a derivative, identify the reference needed before claiming an absolute quantity.
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Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.