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

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26 CHAPTER 5. DYNAMICS OF THE OCEAN
5.5 Rotation
When considering the motion of the ocean, at time scales larger than a day, the rotation of the
earth is of paramount importance. Newton’s laws of motion only apply when measurements are
done with respect to an inertial frame, that is a frame without acceleration and thus without
rotation. Adding to all measurements (and to boundary conditions) the rotation of the earth
would be very involved (the tangential speed is around 400m/s and the speed of the ocean
typically around 0.1m/s), one should then also have “rotating boundaries”, that is the rotation
would only explicitly appear in the boundary conditions, which then would be very involved. It
is thus a necessity to derive Newton’s laws of motion for a frame rotating with the earth, called
geocentric frame, to make the problem of geophysical fluid dynamics treatable by calculation.
5.6 The Coriolis Force
tel-00545911, version 1 - 13 Dec 2010
❑
y r
✻
y f
P
α = Ωt
x r
✯
✲
x f
Figure 5.3: A moving point P observed by a fix and a rotating coordinate system
Let us start with considering a movement of a point P that is observed by two observers,
one in an inertial frame (subscript . f ) and one in a frame (subscript . r ) that is rotating with
angular velocity Ω. The coordinates at every time t transform following:
(
xf
y f
)
=
(
xr cos(Ωt) − y r sin(Ωt)
x r sin(Ωt) + y r cos(Ωt)
In a inertial (non-rotating) frame Newtons laws of motion are given by:
∂ tt
(
xf
y f
)
=
( F
x
f
F y f
)
)
(5.29)
(5.30)
Where F . . are forces per mass, to simplify notation. So, in an inertial frame if the forces vanish
the acceleration vanishes too. How can we describe such kind of motion in a rotating frame.
Combining eqs. (5.29) and (5.30), performing the derivations and neglecting the forces in
eq. 5.30, we obtain:
∂ tt
(
xf
y f
)
=
This is only satisfied if:
(
(∂tt x r − 2Ω∂ t y r − Ω 2 x r ) cos(Ωt) − (∂ tt y r + 2Ω∂ t x r − Ω 2 y r ) sin(Ωt)
(∂ tt x r − 2Ω∂ t y r − Ω 2 x r ) sin(Ωt) + (∂ tt y r + 2Ω∂ t x r − Ω 2 y r ) cos(Ωt)
)
= 0. (5.31)
∂ tt x r − 2Ω∂ t y r − Ω 2 x r = 0 and (5.32)
∂ tt y r + 2Ω∂ t x r − Ω 2 y r = 0. (5.33)