Mathematical derivation of the Coriolis force

The Equations of Motion in a Rotating Coordinate System

Chapter 3

Since the earth is rotating about its axis and since it is

convenient to adopt a frame of reference fixed in the earth, we need to study the equations of motion in a rotating

coordinate system.

Before proceeding to the formal derivation, we consider briefly two concepts which arise therein:

Effective gravity and Coriolis force

g* g g*

g

effective gravity g on a spherical earth

Ω²R Ω²R

R R

effective gravity on an earth with a slight equatorial bulge g is everywhere normal to the earth’s surface

Effective Gravity

Effective Gravity

If the earth were a perfect sphere and not rotating, the only gravitational component g* would be radial.

Because the earth has a bulge and is rotating, the effective

gravitational force g is the vector sum of the normal gravity to the mass distribution g*, together with a centrifugal force

Ω²R, and this has no tangential component at the earth’s surface.

g = g * + Ω ²R

A line at rest in an inertial system A line that rotates with

the roundabout

Apparent trajectory of the ball in a rotating

coordinate system

Ω

The Coriolis force

Mathematical derivation of the Coriolis force

Representation of an arbitrary vector A(t)

i k

j

Ω

i ´

j ´ k ´

A(t) = A₁(t)i + A₂(t)j + A₃(t)k

rotating

reference system

A(t) = A_1´(t)i ´ + A_2´ (t)j ´ + A_3´ (t)k´

inertial

reference system

The derivative of A(t) with respect to time d

dt

dA dt

dA dt

dA dt

aA

i j k

= ¹ + ² + ³

the subscript “a” denotes the derivative in an inertial reference frame

In the rotating frame of reference

( )

( )

a 1

1

1

1

1

d dA d

A ...

dt dt dt

dA A ...

dt

d A ...

dt

′

′

′

′ ′ ′

= + +

′ ′ ′

= + ∧ +

⎡ ⎤ ′

= ⎢⎣ + ∧⎥⎦

A i

i

i i

i + Ω

Ω

The time derivative of an arbitrary vector A(t)

Ω

a a + da

θ

b

Unit vector perpendicular to Ω and a

d dt

a a b

a

=

= ∧

| ||Ω|sin Ω

θ

Let a be any vector rotating with angular velocity Ω

Position vector r(t)

O

The relative velocity in a rotating frame of reference is

u = d = ^′ i′ + ^′ j′ + ^′ k′ dt

dr dt

dr dt

dr dt

r _{1 2 3}

and u_a = +u Ω ∧ r

Want to calculate u_a = d dt

ar

the absolute velocity of an air parcel

This air parcel starts relative to the earth with a poleward velocity V.

It begins relative to space with an additional eastwards velocity component ΩR_e .

V

Ω

u_a = +u Ω ∧ r

Example

Earth’s radius ΩR_e

We need to calculate the absolute acceleration if we wish to apply Newton's second law

d dt

au_a

Measurements on the earth give only the relative velocity u and therefore the relative acceleration

d dt

u

With A = u_a

d dt

d dt

a a a

a

u u

= + ∧Ω u

u_a = +u Ω ∧ r

d dt

d dt

aua = u + 2Ω ∧ + ∧u Ω (Ω ∧ r)

d dt

d dt

aua = u + 2Ω ∧ + ∧u Ω (Ω ∧ r)

absolute acceleration

relative acceleration

Coriolis-

acceleration Centripetal acceleration

Ω Ω Ω Ω Ω

∧ ∧ = ∧ ∧

= −

( ) ( )

| |

r R

R

2

R

r r

Ω Ω

Ω Position vector r is split up

r R

Ω

| |Ω r = ( .r Ω Ω ) + R

The Centripetal acceleration is directed inwards towards the axis of rotation and has magnitude |Ω|²R.

Centripetal acceleration

Newton’s second law in a rotating frame of reference

ρ d dt

aua

= F In an inertial frame:

density force per

unit mass In a rotating frame:

ρ[d ( )]

dt

u + 2Ω ∧ +u Ω ∧ Ω ∧ r = F

ρ d ρΩ ρΩ dt

u = F − 2 ∧ − u ∧ ( Ω ∧ r )

Alternative form

Coriolis force Centrifugal force

Let the total force F = g* + F´ be split up

ρ d ρΩ ρΩ

dt

u = ′ +F g * −2 ∧ −u ∧ (Ω ∧ r)

Ω ∧ (Ω ∧ r) = −|Ω|² R

With g = g * + Ω ²R

ρ d ρ ρΩ

dt

u = ′ +F g − 2 ∧ u

The centrifugal force associated with the earth’s rotation no longer appears explicitly in the equation; it is contained in the effective gravity.

When frictional forces can be neglected, F’ is the pressure gradient force

′ = −∇

F p _T per unit volume

ρ d ρ ρΩ

dt

u = ′ +F g − 2 ∧ u

d

dtu p_T

= − ∇1 + − ∧

ρ g 2Ω u ^per

unit mass

This is the Euler equation in a rotating frame of reference.

total pressure

u

−2Ω ∧ u

the Coriolis force acts normal to the

rotation vector and normal to the velocity.

is directly proportional to the magnitude of u and Ω.

Note: the Coriolis force does no work because u (⋅ 2Ω ∧ u) ≡ 0 Ω

The Coriolis force does no work

Define p_T = p₀( )z + p where

p₀(z) and ρ₀(z) are the reference pressure and density fields p is the perturbation pressure

Important: p₀(z) and ρ₀(z) are not uniquely defined Euler’s equation becomes

dp

dz⁰ = − ρg ₀

D

Dtu p

u g

+ ∧ = − ∇ + ⎡ −

⎣⎢ ⎤

⎦⎥

2Ω 1 ₀

ρ

ρ ρ ρ

the buoyancy force

Perturbation pressure, buoyancy force

g = (0, 0, −g)

But the total force is uniquely defined.

Important: the perturbation pressure gradient

and buoyancy force are not uniquely defined.

1 0

p ⎛ ρ − ρ ⎞

− ∇ + ⎜ρ g ⎝ ρ ⎟⎠

⎛ ρ − ρ0 ⎞

⎜ ρ ⎟

⎝ ⎠

g

− ∇1 ρ p

Indeed 1 ⁰ 1 _T

p ⎛ ρ − ρ ⎞ p

− ∇ +ρ g ⎜⎝ ρ ⎟⎠ = − ∇ρ + g

A mathematical demonstration

D 1 p ' b ˆ

Dt = − ∇ + ρ

u k

Momentum equation Continuity equation

∇ ⋅ =u 0

The divergence of the momentum equation gives:

( ) ( )

2

p ' ⎡ b ˆ ⎤

∇ = − ∇⋅ ρ ⋅∇ −∇⋅ ρ ⎣ u u k ⎦

This is a diagnostic equation!

Newton’s 2nd law

ρD w ρ

D t

p z

= − ∂ g

∂ −

mass × acceleration = force

buoyancy form

Dw 1 p'

Dt z b

= − ∂ + ρ ∂

p po z p o z

= +

= +

( ) ' ( ) '

ρ ρ ρ

Put

Then

where d p o d z

g o

= − ρ

where b g ⎛ ρ−ρo ⎞

= − ⎜ ⎟

⎝ ρ ⎠

buoyancy force is NOT unique

b g ⎛ ρ− ρ o ⎞

=− ⎜ ⎟

⎝ ρ ⎠

it depends on choice of reference density ρ_o(z)

but 1 p 1 p'

g b

z z

∂ ∂

− − = − +

ρ ∂ ρ ∂ ^{is unique}

Buoyancy force in a hurricane

ρ

( ) z

ρ

( ) z

U(z) z

Τ + ΔΤ

θ = constant

T

Initiation of a thunderstorm

θ = constant

tropopause negative buoyancy

outflow

inflow

LCL LFC

original heated air

negative buoyancy positive

buoyancy

Some questions

» How does the flow evolve after the original thermal has reached the upper troposphere?

» What drives the updraught at low levels?

– Observation in severe thunderstorms: the updraught at cloud base is negatively buoyant!

– Answer: - the perturbation pressure gradient

outflow

inflow

original heated air

negative buoyancy

LFC LCL

positive buoyancy

HI HI

HI HI

LO

p'

Scale analysis

» Assume a homogeneous fluid ρ = constant.

» Euler’s equation becomes:

D

Dtu p

+ 2Ω ∧ = − ∇u 1 ρ

U

L U P

L

2

2Ω Δ

scales: ρ

D Dt U L

U

U

L Ro u

u

/ ~ /

2 2 2

2

Ω ∧ = =

Ω Ω

Then

Rossby number

2Ω = 10⁻⁴ s⁻¹ L = 10⁶ m L

U = 10 ms⁻¹

R o U

= L =

× =

− −

2

10

10 ₄ 10₆ 10 ¹ Ω

Extratropical cyclone

U = 50 m/sec

2Ω = 5 X 10⁻⁵ L = 5 x 10⁵ m

R o U

= L =

× ₋ × × =

2

5 0

5 1 0 ₅ 5 1 0 ₅ 2 Ω

Tropical cyclone

L

L = 10 - 100 m U = 10 ms⁻¹ 2Ω = 10⁻⁴s⁻¹

R o U

= L =

× = →

− − −

2

1 0

1 0 ₄ 1 0 1 0 0 1 0 ³ 1 0 ⁴

Ω ( , )

Dust devil

L

L = 100 m U = 50 ms⁻¹

2Ω = 10⁻⁴s⁻¹

R o = 5 × 1 0³

Waterspout

L = 10 m U = 200 m s⁻¹ 2Ω = 10⁻⁴s⁻¹ Ro = ×2 10⁵

Aeroplane wing

Flow system L U m s⁻¹ Ro

Ocean circulation 10³ - 5 × 10³ km 1 (or less) 10⁻² - 10⁻³ Extra-tropical cyclone 10³ km 1-10 10⁻² - 10⁻¹ Tropical cyclone 500 km 50 (or >) 1

Tornado 100 m 100 10⁴

Dust devil 10-100 m 10 10³ - 10⁴

Cumulonimbus cloud 1 km 10 10²

Aerodynamic 1-10 m 1-100 10³ - 10⁶

Bath tub vortex 1 m 10^-1 10³

The Rossby number

(i) Large scale meteorological and oceanic flows are strongly constrained by rotation (Ro << 1), except possibly in

equatorial regions.

(ii)Tropical cyclones are always cyclonic and appear to

derive their rotation from the background rotation of the earth. They never occur within 5 deg. of the equator

where the normal component of the earth's rotation is small.

(iii) Most tornadoes are cyclonic, but why?

(iv) Dust devils do not have a preferred sense of rotation as expected.

(v) In aerodynamic flows, and in the bath (!), the effect of the earth's rotation may be ignored.

Summary

» Many of the flows we shall consider have horizontal dimensions which are small compared with the earth's radius.

» In studying these, it is both legitimate and a great

simplification to assume that the earth is locally flat and to use a rectangular coordinate system with z pointing vertically upwards.

» Holton (§2.3, pp31-35) shows the precise circumstances under which such an approximation is valid.

» In general, the use of spherical coordinates merely refines the theory, but does not lead to a deeper understanding of the phenomena.

Coordinate systems and the earth's sphericity

Φ Ω

i

j k

Äquator

equator

φ Ω

j i

k

Take rectangular coordinates fixed relative to the earth and centred at a point on the surface at latitude.

Ω

φ φ Ω

| Ω | s in φ k

| Ω | c o s φ j

f-plane or β-plane

| | c o s | | s in

= φ + φ

Ω Ω j Ω k

2 Ω ∧ u = | Ω | c o s φ ∧j u + | Ω | s i n φ k ∧ u

2

2 2

2 2 Ω ∧ =

− +

−

⎡

⎣

⎢⎢

⎢⎢

⎤

⎦

⎥⎥

⎥⎥ u

Ω Ω

Ω Ω

v w

u u

sin cos

sin cos

φ φ

φ φ In component form

I will show that for middle latitude, synoptic-scale weather systems such as extra-tropical cyclones, the terms involving cos φ may be neglected.

2Ω ∧ =u 2 | Ω | cos φ ∧ +j u 2 | Ω | sin φ ∧k u

The important term for large-scale motions

To a good approximation

2 Ω ∧ u = 2 | Ω | s i n φ k ∧ u = f ∧ u f = 2 | Ω | s i n φ f = f k

Coriolis parameter

» Much of the significant weather in middle latitudes is associated with extra-tropical cyclones, or depressions.

» We shall base our scaling on such systems.

» Let L, H, T, U, W, P and R be scales for the horizontal size, vertical extent, time, |u_h|, w, perturbation pressure, and density in an extra-tropical cyclone, say at 45° latitude, where f (= 2Ω sin φ ) and 2Ω cos φ are both of order 10⁻⁴.

U ms W ms

L m km H m km

T L U s day P Pa mb

R kg m

= =

= =

= =

=

− − −

−

10 10

10 10 10 10

10 1 10 10

1

1 2 1

6 3 4

5 3

3

; ;

( ); ( );

/ ~ (~ ); ( )

.

and δ

Scale analysis of the equations for middle latitude synoptic systems

horizontal momentum equations

Du

Dt v w p

x Dv

Dt u p

y

− + = −

+ = −

2 2 1

2 1

Ω Ω

Ω

sin cos

sin

φ φ

ρ

∂

∂

φ ρ

∂

∂

scales U²/L 2 ΩU sin φ 2ΩW cos φ δP/ρL

10⁻⁴ 10⁻³ 10⁻⁶ 10⁻³

orders

D

Dtu^h f _{h h}p

k u

+ ∧ = − ∇1 ρ

vertical momentum equations Dw

Dt u p

z^T g

− 2 = − 1 −

Ω cosφ

ρ

∂

∂

scales UW/L 2ΩU cosφ δP_T/ρΗ g orders

the atmosphere is strongly hydrostatic on the synoptic scale.

negligible

7 3

10⁻ 10⁻ 10 10

But are the disturbances themselves hydrostatic?

Question: when we subtract the reference pressure p₀ from p_T, is it still legitimate to neglect Dw/Dt etc.?

vertical momentum equations

Dw 1 p

2 u cos b

Dt z

− Ω φ = − ∂ + ρ ∂

scales UW/L 2ΩU cosφ δP/ρΗ∗ gδT/T₀

orders

still negligible

7 3 1 1

10

10

≥ 10

10

H* = height scale for a perturbation pressure difference δp of 10 mb assume that H* ≤ H

assume that δT ≈ 3^oK/300 km

Dw 1 p

2 u cos b

Dt z

− Ω φ = − ∂ + ρ ∂

orders ^{10−7 10−3 ≥ 10−1 10−1}

0 1 p b

z

= − ∂ + ρ ∂

In synoptic-scale disturbances, the perturbations are in close hydrostatic balance

Remember: it is small departures from this equation which drive the weak vertical motion in systems of this scale.

In summary

The hydrostatic approximation permits enormous simplifications in dynamical studies of large-scale

motions in the atmosphere and oceans.

The hydrostatic approximation

End of

Chapter 3