Lesson 5
Boussinesq approximation
Conservation of Momentum
Derivation of the Equations of Motion
(EOM)
Boussinesq Approximation
Recall for an incompressible
fluid:
Assume density variations are small and
therefore can compare local variations of
the density field to uniform background
density ρ′ << 1
Navier-Stokes equations
Must be able to identify terms and coordinate
systems
∇⋅v = 0
ρ0
Conservation of Momentum
Newton’s Second Law
Relates the rate of change of absolute
momentum following the motion in an
inertial reference frame to the sum of the
forces acting on the fluid
p = mv
dv dp
∑ F = ma = m =
dt dt
Valid for nearly all circumstances in the ocean
and lower atmosphere
Conservation of Momentum
Use this to derive our equations of motion
Let’s re-examine the net forces acting on a
fluid parcel
Navier-Stokes Equations
This information helps to construct equations
that calculate the net force - and therefore
acceleration - acting on a parcel
Sum the forces acting in each direction and
equate it to the acceleration of the parcel
1
Primary Forces
Lesson 5
Vertical Components
Where are we going…
Vector form of EOM
Gravity
Centrifugal force
PGF
Coriolis
Spherical form of EOM
Shallow
Cartesian (rectangular)
Horizontal Componets
PGF
Viscous
Coriolis
Primitive
Equations
Equations of Motion: Goal
Du
1 ∂P
=−
+ υ∇ 2 u + (2Ωv sin φ − 2Ωw cos φ )
Dt
ρ ∂x
Dv
1 ∂P
=−
+ υ∇ 2v − (2Ωusin φ )
Dt
ρ ∂y
EOM: Derivation
Now, to deal with the rotating frame of
reference....
Inertial (absolute)
frame
A = Axiˆ + Ay ˆj + Az kˆ
Dw
1 ∂P
=−
+ υ∇ 2 w + (2Ωucos φ ) − g
Dt
ρ ∂z
f-plane
β-plane
Non-inertial (rotating) frame
A = Ax ′ˆi ′ + Ay ′ ˆj ′ + Az′kˆ ′
2
EOM: Derivation
EOM: Derivation
Da A Dr A =
+ Ω× A
Dt
Dt
DaU a DU r
=
+ 2Ω × U r − Ω 2 r
Dt
Dt
Relationship between inertial and rotating
frames of reference:
Relationship between absolute velocity
and velocity relative to Earth
Acceleration following the motion in an
inertial frame is equal to the acceleration
following the motion in a rotating frame
plus the coriolis acceleration and
centripetal acceleration
EOM: Vector
EOM: Spherical Coordinates
Ua = Ur + Ω × r
Rewrite to get N2L relative to a rotating
coordinate frame:
Vector EOM
1
dU r
= −2Ω × U r − ∇P + g + υ∇ 2U r
dt
ρ
1
2
3
4
5
Assume the Earth’s departure from a
sphere is negligible
Expand vector form of EOM so that the surface
of the Earth corresponds to a coordinate surface
Coordinate axes (λ, φ, z):
λ: longitude (−π ≤ λ ≤ π )
π
π
φ: latitude (− 2 ≤ φ ≤ 2 )
z: height above Earth’s surface
a: radius of the Earth
Unit vectors to describe motion
3
EOM: Spherical Coordinates
EOM: Spherical Coordinates
Relative velocity vector becomes:
V = u iˆ + v ˆj + w kˆ
Where:
dλ
u ≡ acos φ
dt
dφ
v≡a
dt
dz
w≡
dt
EOM: Derivation
Determine the magnitude and direction
Zonal (East-West) distance
Meridional (North-South) distance
Vertical distance
dx = acos φdλ
dy = adφ
dz = da
EOM: Derivation
Diˆ
∂iˆ
u
=u =
(sin φˆj − cos φkˆ )
Dt
∂x acos φ
Dˆj
∂ˆj
∂ˆj
utan φ ˆ v ˆ
=u +v =−
i− k
Dt
∂x
∂y
a
a
Determine the magnitude and direction
4
EOM: Derivation
Dkˆ
∂kˆ
∂kˆ u
v
= u + v = iˆ + ˆj
Dt
∂x
∂y a
a
Determine the magnitude and direction
EOM: Derivation
EOM: Spherical
Substitute into relative velocity equation:
DV
Du uv tan φ uw ˆ Dv u2 tan φ vw ˆ Dw u 2 + v 2 ˆ
=(
−
+
)i + (
+
+
)j +(
−
)k
Dt
Dt
a
a
Dt
a
a
Dt
a
Describes the spherical expansion of the
acceleration following the motion
Recall, our vector EOM had our force
terms…now we have to expand the force terms
to get the complete EOM in spherical
coordinates
Where are we going…
Du uv tan φ uw
1 ∂P
−
+
=−
+ 2Ωv sin φ − 2Ωw cos φ + Fx
Dt
a
a
ρ ∂x
Dv u 2 tan φ vw
1 ∂P
+
+
=−
− 2Ωusin φ + Fy
Dt
a
a
ρ ∂y
Dw u 2 + v 2
1 ∂P
−
=−
− g + 2Ωucos φ + Fz
Dt
a
ρ ∂z
Vector form of EOM
Spherical form of EOM
Shallow
Cartesian (rectangular)
Lesson 6
Primitive
Equations
f-plane
β-plane
5