4. Conservation of momentum, energy, and vorticity on a rotating sphereCopyright 2014, David A. RandallIntroductionThis chapter is a quick review of basic equations and theoretical results to be used later. We systematically explore the conservation of momentum, vorticity, and energy in its various forms. We also briefly introduce some approximations that are useful when considering large-scale motions in the Earth’s atmosphere. In the first part of the chapter, we write the equations as they apply at a point, without approximations other than the continuum approximation, and show molecular fluxes rather than turbulent fluxes. Various useful approximations are discussed in the middle of the chapter, and used in the latter part of the chapter. Conservation of momentum on a rotating sphereThe length of a sidereal day is 86,164 s, so the Earth rotates about its axis with an angular velocity of 2π86,164 s( )≅ 7.292 × 10−5 s−1. This angular velocity can be represented by a vector, Ω, pointing towards the celestial North Pole, as shown in Fig. 4.1. Consider a coordinate system that is rotating with the Earth. As shown in the QuickStudy on the equation of motion, Newton's statement of momentum conservation, as applied in the rotating coordinate system, is DVDt= −2Ω × V − Ω × Ω × r( )− ∇φa−α∇p −α∇ ⋅F.(1)Here r is a position vector extending from the center of the Earth to a particle of air whose position is generally changing with time. The gravitational potential is φa. The pressure-gradient term is −α∇p, where α is the specific volume, and p is the pressure. The quantity F is the stress tensor associated with molecular viscosity. The dimensions of F are density times velocity squared, e.g., the units could be (kg m-3) (m s-1)2 = kg m-1 s-2. Note that ∇ ⋅ F is a vector.! Revised Thursday, January 23, 2014! 1An Introduction to the Global Circulation of the AtmosphereThe term −2Ω × V represents the Coriolis acceleration, whose direction is perpendicular to V, and the term −Ω × Ω × r( ) represents the centripetal acceleration. You should be able to show thatVe= Ω × r = Ωr cosϕ( )eλ,(2)where eλ is a unit vector pointing east, ϕ is latitude, and Ve is the velocity (as seen in the inertial frame) that a particle at radius r and latitude ϕ experiences due to the Earth's rotation (refer to Fig. 4.1). The total velocity of the particle is V + Ve. With this notation, we find that−Ω × Ω × r( )= Ω2r cosϕ( )eλ× eΩ= Ω2re,(3)where re is the vector shown in Fig. 4.1, and eΩ is a unit vector pointing toward the celestial north pole. The centripetal acceleration points outward, in the direction of re, which is perpendicular to the axis of the Earth’s rotation. It can be shown thatFigure 4.1: Sketch defining notation used in the text.!!!!!!"""!"##! Revised Thursday, January 23, 2014! 2An Introduction to the Global Circulation of the AtmosphereΩ2re= ∇12Ω × r2⎡⎣⎢⎤⎦⎥.(4)According to (4), the centripetal acceleration can be regarded as the gradient of a potential, called the “centrifugal potential.” The “apparent” gravity,g , due to the combined effects of true gravity and the centripetal acceleration, can be defined as g = ga+ Ω2re,(5)where ga≡ −∇φa, and using (4) we see that the potential of g isφ=φa−12Ω × r2=φa−12Ωr cosϕ( )2.(6)so that g ≡ −∇φ. We refer to φas the “geopotential.” The horizontal pressure-gradient force vanishes when the pressure is uniform along isosurfaces of the geopotential. Because of the centripetal acceleration (and also because of spatial inhomogeneities of true gravity, due to inhomogeneities of the Earth’s mass), these isosurfaces are not spherical. The centripetal acceleration causes them to bulge outward in low latitudes, and pull inward near the poles; the resulting shape of a geopotential isosurface is an “oblate spheroid.”As is well known, the Earth’s gravitational potential, φa, decreases as one over the square of the distance from the center of the Earth. In addition, φa varies with latitude and longitude, at a fixed distance from the center of the Earth, due to inhomogeneities in the distribution of mass within the solid Earth and the oceans. As a result, a surface of constant φa is not quite spherical. Moreover, as can be seen from (4) and (6), the centripetal acceleration also varies geographically and with distance from the center of the Earth. As a result, surfaces of constant φ are only approximately spherical. In particular, the centripetal acceleration causes these surfaces to bulge outward at low latitudes, so that their shapes are well approximated by “oblate spheroids.” To the extent that we want to consider geographical variations of φ, the oblateness of the Earth’s surface should also be taken into account. For most purposes, however, g ≅ ga= −gk,(7)! Revised Thursday, January 23, 2014! 3An Introduction to the Global Circulation of the Atmospherebecause the centripetal acceleration is small compared to ga. Here k is a unit vector pointing upward, away from the center of the Earth. When we use (6) with a spatially constant value of g, as is conventional, we must also approximate the shape of the Earth as a sphere, with minor topographical bumps.Using (6) we can now write the equation of motion (1) asDVDt= −2Ω × V − ∇φ−α∇p −α∇ ⋅F.(8)Another useful form of this equation is∂V∂t+ 2Ω + ∇ × V( )⎡⎣⎤⎦× V + ∇12V ⋅ V⎛⎝⎜⎞⎠⎟= −∇φ−α∇p −α∇ ⋅F.(9)To obtain (9) from (8), we have used the vector identityV ⋅∇( )V = ∇ × V( )× V + ∇12V ⋅V⎛⎝⎜⎞⎠⎟.(10)We will have occasion to use both (8) and (9). Now consider spherical coordinates, λ,ϕ, r( ). The unit vectors in the λ,ϕ, r( ) coordinates are eλ, eϕ, and er, respectively. As shown in the QuickStudy on “Vectors and Coordinate Systems,” the vector operators that will be used in this book, i.e., the gradient, divergence, curl, and Laplacian, can be expressed in spherical coordinates as follows:∇A =1r cosϕ∂A∂λ,1r∂A∂ϕ,∂A∂r⎛⎝⎜⎞⎠⎟,(11)∇ ⋅ H =1r cosϕ∂Hλ∂λ+1r cosϕ∂∂ϕHϕcosϕ( )+1r2∂∂rHrr2( ),(12)∇ × H