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 Effective Gravity g is everywhere normal to the earth’s surface Ω 2 R 2 R R Ω R g* g g* g effective gravity g effective gravity on an earth on a spherical earth with a slight equatorial bulge 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. gg*R= + Ω 2 The Coriolis force A line that rotates with the roundabout Ω A line at rest in an inertial system Apparent trajectory of the ball in a rotating coordinate system Mathematical derivation of the Coriolis force Representation of an arbitrary vector A(t) Ω A(t) = A (t)i + A (t)j + A (t)k k 1 2 3 k ´ A(t) = A1´(t)i ´ + A2´ (t)j ´ + A3´ (t)k´ inertial j reference system rotating j ´ reference system i ´ i The time derivative of an arbitrary vector A(t) The derivative of A(t) with respect to time d A dA dA dA a =+ijk12 + 3 dt dt dt dt the subscript “a” denotes the derivative in an inertial reference frame ddAdAi′ ′ In the rotating frame a1=++i′′A ... of reference dt dt1 dt dA′ = ii′′′1 +∧+A...()Ω dt 1 ⎡⎤d ′ =+∧Ω ()A...1i+′ ⎣⎦⎢⎥dt Let a be any vector rotating with angular velocity Ω Ω a + da a θ b da = ||||sinabΩ θ dt =∧Ω a Unit vector perpendicular to Ω and a Position vector r(t) dar O Want to calculate ua = dt the absolute velocity of an air parcel The relative velocity in a rotating frame of reference is dr dr dr dr uijk==12′′′ + ′ + 3′ ′ dt dt dt dt and uua = + Ω ∧ r Example Ω uua =+Ω ∧ r V ΩRe Earth’s radius This air parcel starts relative to the earth with a poleward velocity V. It begins relative to space with an additional eastwards velocity component ΩRe . We need to calculate the absolute acceleration if we wish to apply Newton's second law daua dt Measurements on the earth give only the relative velocity u and therefore the relative acceleration du dt uua = + Ω ∧ r With A = u a d uud aa=+∧ a Ω u dt dt a d uud aa=+∧+∧∧2ΩΩΩur() dt dt absolute d uud aa=+∧+∧∧2ΩΩΩur() acceleration dt dt relative Coriolis- Centripetal acceleration acceleration acceleration Centripetal acceleration Ω Ω Position vector r is split up |Ω| RR rr=+(.ΩΩ ) R ΩΩ∧∧=()rR Ω∧ (Ω ∧ )r Ω r 2 r =−||Ω R Ω The Centripetal acceleration is directed inwards towards the axis of rotation and has magnitude |Ω|2R. Newton’s second law in a rotating frame of reference d u In an inertial frame: ρ aa= F dt density force per unit mass In a rotating frame: du ρ[()]+∧+∧∧=2ΩΩΩurF dt Alternative form du ρρΩρΩ=−Fu2() ∧− ∧Ω ∧ r dt Coriolis force Centrifugal force Let the total force F = g* + F´ be split up du ρρΩρΩ= Fg*′ +−2() ∧−∧∧ uΩ r dt Ω ∧ ()||Ω ∧ rR= − Ω 2 With gg*R= + Ω 2 du ρρρΩ= Fg′ +−2 ∧ u dt 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 total pressure F ′ = −∇ p T per unit volume du ρρρΩ= Fg′ +−2 ∧ u dt du 1 =− ∇p +gu −2Ω ∧ per dt ρ T unit mass This is the Euler equation in a rotating frame of reference. The Coriolis force does no work the Coriolis force acts normal to the Ω rotation vector and normal to the velocity. u is directly proportional to the magnitude of u and Ω. −2Ω ∧ u Note: the Coriolis force does no work because u(⋅ 2Ω ∧ u)≡ 0 Perturbation pressure, buoyancy force dp0 Define ppzpT = 0 ()+ where =−g ρ dz 0 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 g = (0, 0, −g) Euler’s equation becomes Du 1 ⎡ρρ− 0 ⎤ +∧=−∇+2Ω ugp ⎢ ⎥ Dt ρ ⎣ ρ ⎦ the buoyancy force 1 Important: the perturbation pressure gradient −∇p ρ ⎛⎞ρ −ρ0 and buoyancy forceg ⎜⎟ are not uniquely defined. ⎝⎠ρ 1 ⎛⎞ρ −ρ0 But the total force −∇+ p g ⎜⎟ is uniquely defined. ρρ⎝⎠ 11⎛⎞ρ−ρ0 Indeed −∇+ppgg⎜⎟ =−∇+T ρρρ⎝⎠ A mathematical demonstration D1u =− ∇p 'b + kˆ ∇ ⋅u = 0 Dt ρ Momentum equation Continuity equation The divergence of the momentum equation gives: 2 ∇p' =−⎡ ∇⋅ρ⋅∇uu −∇⋅ρ b kˆ ⎤ ⎣ ()()⎦ This is a diagnostic equation! Newton’s 2nd law mass × acceleration = force ρDw=−∂p − gρ Dt ∂z buoyancy form p = p ()z + p ' o dp Put where o =−g ρ ρ=+ρρ()z ' o o dz Dw 1∂ p' ⎛⎞ρ−ρo Then =− +b where bg=− ⎜⎟ Dtρ∂ z ⎝⎠ρ buoyancy force is NOT unique ⎛⎞ρ−ρ bg=− ⎜⎟o ⎝⎠ρ it depends on choice of reference density ρo(z) 1p∂∂ 1p' but −−=−+gbis unique ρ∂zz ρ∂ Buoyancy force in a hurricane ρ ()z o ρ ()z o Initiation of a thunderstorm z θ = constant Τ + ΔΤ T U(z) negative buoyancy tropopause outflow original heated air θ = constant positive buoyancy LFC LCL inflow negative 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 positive buoyancy HI outflow p' original heated air LFC LO LCL HI inflow HI HI negative buoyancy Scale analysis » Assume a homogeneous fluid ρ = constant. » Euler’s equation becomes: Du 1 +∧=−∇2Ω u p Dt ρ U 2 ΔP scales: 2ΩU L ρL DDtu / UL2 / U Then ~ ==Ro 222Ω ∧ u ΩΩU L Rossby number Extratropical cyclone 2Ω = 10−4 s−1 L = 106 m L U = 10 ms−1 U 10 Ro == = 10−1 2Ω L 10−46× 10 Tropical cyclone U = 50 m/sec 2Ω = 5 X 10−5 L = 5 x 105 m U 50 Ro == = 2 2 Ω L 510× −55× 510× Dust devil L L = 10 - 100 m U = 10 ms−1 2Ω = 10−4s−1 U 10 Ro == =→10−−34 10 2Ω L 10−4 × (, 10 100 ) Waterspout L = 100 m U = 50 ms−1 2Ω = 10−4s−1 Ro = 510× 3 L Aeroplane wing 5 L = 10 m U = 200 m s−1 2Ω = 10−4s−1 Ro = 210× The Rossby number 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 Summary (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. Coordinate systems and the earth's sphericity » 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. Take rectangular coordinates fixed relative to the earth and centred at a point on the surface at latitude. Ω k k j j i i Φφ equatorÄquator φ Ω Ω ||sinΩ φ k ||cosΩ φ j f-plane or β-plane φ ΩΩ=φ+φ||cos||sinj Ω k 2||cos||sinΩ ∧=u Ω φ∧+j u Ω φ∧ku In component form ⎡−+22ΩΩvwsinφφ cos ⎤ ⎢ ⎥ 2Ω ∧=u ⎢ 2 Ω u sin φ ⎥ ⎢ ⎥ ⎣⎢ − 2 Ω u cosφ ⎦⎥ 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||cosΩ φ∧+ju 2||sinΩ φ∧ku The important term for large-scale motions To a good approximation 22Ω ∧ ukufu= ||sinΩ φ ∧ = ∧ f = 2 | Ω | sin φ fk= f Coriolis parameter Scale analysis of the equations for middle latitude synoptic systems » Much of the significant weather in middle latitudes is associated with extra-tropical cyclones, or depressions.