6Fluid dynamics on a rotating frame
IB Fluid Dynamics
6.1 Equations of motion in a rotating frame
The Lagrangian (particle) acceleration in a rotating frame of reference is given
by
Du
Dt
+ 2Ω × u + Ω × (Ω × x),
as you might recall from IA Dynamics and Relativity. So we have the equation
of motion
ρ
∂u
∂t
+ u · ∇u + 2Ω × u
= −∇p − ρω × (Ω × x) + gρ.
This is complicated, so we want to simplify this. We will first compare the
centrifugal force compared to gravity. We have
|Ω| =
2π
1 day
s
−1
≈ 2π × 10
−5
s
−1
.
The largest scales are 10 × 10
4
km. Compared to gravity g, we have
|Ω × (Ω × x)|
|g|
≤
(2π)
2
× 10
−10
. × 10
7
10
≈ 4 × 10
−3
.
So the centrifugal term is tiny compared to gravity, and we will ignore it.
Alternatively, we can show that
Ω ×
(
ω × x
) can be given by a scalar potential,
and we can incorporate it into the potential term, but we will not do that.
Next, we want to get rid of the non-linear terms. We consider motions for
which
|u · ∇u| |2Ω ×u|.
The scales of these two terms are U
2
/L and ΩU respectively. So we need
R
o
=
U
ΩL
1.
This is known as the Rossby number. In our atmosphere, we have
U ∼ 10 m s
−1
and L ∼ 1 × 10
3
km. So we get
R
o
=
10
10
6
· 10
−4
≈ 0.1.
So we can ignore the non-linear terms. Thus we get
Proposition (Euler’s equation in a rotating frame).
∂u
∂t
+ 2Ω × u = −
1
ρ
∇p + g.
Definition
(Coriolis parameter/planetary vorticity)
.
We conventionally write
2Ω = f , and we call this the Coriolis parameter or the planetary vorticity.
Note that since we take the cross product of
f
with
u
, only the component
perpendicular to the velocity matters. Assuming that fluid flows along the
surface of the earth, we only need the component of
f
normal to the surface,
namely
f = 2Ω sin θ,
where θ is the angle from the equator.