Etudes et évaluation de processus océaniques par des hiérarchies ...

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5.8. GEOSTROPHIC EQUILIBRIUM 29
∂ t v + fu + g∂ y η = 0 (5.42)
∂ t η + H(∂ x u + ∂ y v) = 0 (5.43)
+boundary conditions.
Important: When approaching the equator f tends to zero, so rotation no longer dominates
and most of the considerations following are not applicable. Equatorial dynamics is different!
In equations (5.41) and (5.42) we have neglected the viscous term which can be safely done
as ν(sea water)≈ 10 −6 m 2 /s.
Exercise 18: What is the Rossby number of the basin wide circulation in the North Atlantic
when u = 10 −1 m/s? What is the Rossby number of a Gulf Stream eddy when u = 1m/s and
the radius R = 30km ?
tel-00545911, version 1 - 13 Dec 2010
5.8 Geostrophic Equilibrium
Large-scale ocean currents usually change on a time scale much larger than f −1 and are thus
often well approximated by the stationary versions of eqs. (5.41) – (5.43) which are,
fv = g∂ x η (5.44)
−fu = g∂ y η (5.45)
this is called the geostrophic equilibrium. For a flow in geostrophic equilibrium all variables can
be expressed in terms of the free surface, you can easily calculate that the vorticity is given
by ζ = (g/f)∇ 2 η. Note that to every function η(x,y) there is a unique flow in geostrophic
equilibrium associated to it. In the case with no rotation (f = 0) the stationary solutions of
the linearised equations have η = 0 and ∂ x u + ∂ y v = 0. In the case without rotation η(x,y)
does not determine the flow.
Exercise 19: What happened to the stationary version of equation (5.43) ?
Exercise 20: Across the Gulf Stream, which is about 100km wide there is a height difference
of approx. 1m. What is the corresponding geostrophic speed of the Gulf Stream.
Exercise 21: In a sea surface height (SSH) map, how can you distinguish cyclones from
anti-cyclones? What happens on the southern hemisphere?
The function Ψ = (gH/f)η is called the geostrophic transport stream-function as ∂ x Ψ = Hv
and ∂ y Ψ = −Hu which means that: (i) isolines of Ψ, and also of η, are stream-lines of the
geostrophic velocity field, and (ii) Ψ(B) − Ψ(A) is the transport that passes between points
A and B. In oceanography transport is usually measured in Sverdrup (1Sv =10 6 m 3 /s), which
corresponds to a cube of water of side length 100 meters passing in 1 second.
When the stationarity assumption is not made the eqs. (5.41) – (5.43) can be used to derive
an equation for η only where u and v can be derived from η. This leads to:
Exercise 22: derive eqs. (5.46) – (5.48).
∂ t
[
∂tt η + f 2 η − gH∇ 2 η ] = 0 (5.46)
∂ tt u + f 2 u = −g(∂ tx η + f∂ y η) (5.47)
∂ tt v + f 2 v = −g(∂ ty η − f∂ x η) (5.48)
Exercise 23: show that the geostrophic equilibrium is a solution of eqs. (5.46) – (5.48).
Exercise 24: show that the only stationary solution of eqs. (5.46) – (5.48) is geostrophic
equilibrium.