112
DYNAMICAL OCEANOGRAPHY
At the interface (χ =0)the relative velocity becomes
ˆ
u
0 − u
0 =
α
2
(T + T ∧ e 3 ),
(5.67)
and has rotated 45 ◦ with respect to T. For the Ekman volume transport per unit
length (in m 2 s −1 ), a similar expression can be derived as for the bottom Ekman
layer. The result is
M E∗ =
T ∗ ∧ e 3
ρ 0 f 0
,
(5.68)
and hence the Ekman mass transport is always perpendicular and to the right of
the wind stress (in the northern hemisphere).
◮
Example 5.3: Physics of the Ekman spiral
As we have seen in Fig. 5.7b and Fig. 5.9b, the velocity vector turns in the
Ekman boundary layers. To explain this in more detail we again consider Example
5.2 where a simple case of the bottom boundary layer was presented. For the
geostrophic flow far away from the boundary layer, we have from a c
∗ = −2Ω ∧ v ∗
and from geostrophic equilibrium
a
c
∗ =
⎛
⎝
f 0 v ∗
−f 0 u ∗
0
⎞
⎠ ;
1
ρ 0
∇ ∗ p ∗ =
⎛
⎝
f 0 v ∗
−f 0 u ∗
0
⎞
⎠ ,
where f 0 =2Ωor in dimensionless form (with the flow in example 5.2)
∇p = a
c =
⎛
⎝
v 0
−u 0
0
⎞
⎠ =
⎛
⎝
0
−1
0
⎞
⎠ .
In Fig. 5.10a, −∇p and the Coriolis acceleration a c are sketched. The pressure
decreases in the meridional direction; without the rotation of the Earth fluid would
flow from high to low pressure. Through the Coriolis acceleration the flow is deflected to the right and in a final steady state, the pressure gradient exactly balances
the Coriolis acceleration. In the boundary layer, the effect of friction becomes important and the velocity decreases, but the pressure gradient is the same as the
one outside the boundary layer. Consider the situation in the boundary layer in
Fig. 5.10b, where the velocity has decreased and hence the Coriolis acceleration
also decreases. The velocity vector must turn counterclockwise such that the resulting frictional acceleration and the Coriolis acceleration are able to balance the
pressure gradient.
DYNAMICAL OCEANOGRAPHY
At the interface (χ =0)the relative velocity becomes
ˆ
u
0 − u
0 =
α
2
(T + T ∧ e 3 ),
(5.67)
and has rotated 45 ◦ with respect to T. For the Ekman volume transport per unit
length (in m 2 s −1 ), a similar expression can be derived as for the bottom Ekman
layer. The result is
M E∗ =
T ∗ ∧ e 3
ρ 0 f 0
,
(5.68)
and hence the Ekman mass transport is always perpendicular and to the right of
the wind stress (in the northern hemisphere).
◮
Example 5.3: Physics of the Ekman spiral
As we have seen in Fig. 5.7b and Fig. 5.9b, the velocity vector turns in the
Ekman boundary layers. To explain this in more detail we again consider Example
5.2 where a simple case of the bottom boundary layer was presented. For the
geostrophic flow far away from the boundary layer, we have from a c
∗ = −2Ω ∧ v ∗
and from geostrophic equilibrium
a
c
∗ =
⎛
⎝
f 0 v ∗
−f 0 u ∗
0
⎞
⎠ ;
1
ρ 0
∇ ∗ p ∗ =
⎛
⎝
f 0 v ∗
−f 0 u ∗
0
⎞
⎠ ,
where f 0 =2Ωor in dimensionless form (with the flow in example 5.2)
∇p = a
c =
⎛
⎝
v 0
−u 0
0
⎞
⎠ =
⎛
⎝
0
−1
0
⎞
⎠ .
In Fig. 5.10a, −∇p and the Coriolis acceleration a c are sketched. The pressure
decreases in the meridional direction; without the rotation of the Earth fluid would
flow from high to low pressure. Through the Coriolis acceleration the flow is deflected to the right and in a final steady state, the pressure gradient exactly balances
the Coriolis acceleration. In the boundary layer, the effect of friction becomes important and the velocity decreases, but the pressure gradient is the same as the
one outside the boundary layer. Consider the situation in the boundary layer in
Fig. 5.10b, where the velocity has decreased and hence the Coriolis acceleration
also decreases. The velocity vector must turn counterclockwise such that the resulting frictional acceleration and the Coriolis acceleration are able to balance the
pressure gradient.