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Unformatted text preview: M.Sc. in Computational Science Fundamentals of Atmospheric Modelling
´ Peter Lynch, Met Eireann
Mathematical Computation Laboratory (Opp. Room 30) Dept. of Maths. Physics, UCD, Belﬁeld. January–April, 2004. Lecture 8 Linear Wave Solutions 2 Introduction
In this lecture we will investigate the wave solutions of the linearized Shallow Water Equations (SWE). The equations are linearized by the perturbation method: • The dependent variables are separated into mean and perturbation components • The mean state is assumed to be known and constant in time • The perturbations are assumed to be of small amplitude, so that quadratic and higher terms in them can be ignored • The system then becomes a linear system in the perturbation variables. The Basic State
Recall from previous work that the Shallow Water Equations are du ∂Φ − fv + =0 dt ∂x dv ∂Φ + fu + =0 dt ∂y dΦ ∂ u ∂v +Φ + =0 dt ∂x ∂y The total time derivative is: ∂ ∂ ∂ ∂ d = +V· = +u +v . dt ∂t ∂t ∂x ∂y The independent variables are • x: Zonal (eastward) coordinate • y: • t:
3 Meridional (northward) coordinate Time.
4 The dependent variables are • u: • v: • Φ: Zonal (westtoeast) component of the wind Meridional (southtonorth) component of the wind Geopotential, Φ = gh where h is depth/height. The Perturbation Equations
Consider a small perturbation of the basic state ¯ u = u + u (x, y, t) v= v (x, y, t) ¯ y ) + Φ (x, y, t) Φ = Φ( We assume that u  ¯ u, v  ¯ u, Φ  ¯ Φ Thus, squares and higher powers in the perturbations or primed quantities can be neglected. The equations for the primed quantities can then be linearized. This is essential for the progress of our analysis. In general, linear equations may be solved analytically, whereas nonlinear equations must be treated numerically.
5 6 The Basic State
We consider a basic state of constant zonal (eastwest) ﬂow ¯ u, independent of time and of the space coordinates, and a ¯ corresponding basic geopotential ﬁeld Φ The shallow water equations for this state reduce to ¯ ∂Φ ¯ fu + = 0; v = 0, ∂y That is, the basic state is in geostrophic balance. Substituting the assumed form of the solution into the SWE, The perturbation equations take the following form: ∂u ∂u ¯ +u ∂t ∂x − fv + ∂Φ =0 ∂x Divergence and Vorticity
It is straightforward to derive divergence and vorticity equations for the perturbation variables. Recall the deﬁnitions of divergence and vorticity: δ= ∂v ∂u + ∂x ∂y ; ζ= ∂v ∂u − ∂x ∂y . ∂v ∂Φ ∂v ¯ +u + fu + =0 ∂t ∂x ∂y ∂Φ ∂Φ ¯ ∂ u + ∂v = 0 ¯ +u +Φ ∂t ∂x ∂x ∂y ¯ In the special case of vanishing mean zonal ﬂow (u = 0): ∂u ∂Φ − fv + =0 ∂t ∂x ∂v ∂Φ + fu + =0 ∂t ∂y ∂Φ ¯ ∂ u + ∂v = 0 +Φ ∂t ∂x ∂y These are called the Laplace Tidal Equations.
7 Adding the xderivative of the u equation to the y derivative of the v equation gives an equation for the divergence: ∂δ ∂δ ¯ + u − f ζ + βu + 2Φ = 0 . ∂t ∂x Subtracting the xderivative of the v equation from the y derivative of the u equation gives an equation for the vorticity: ∂ζ ∂ζ ¯ + u + f δ + βv = 0 . ∂t ∂x
8 The vorticity equation may also be written in the form: d (ζ + f ) + f δ = 0 . dt This equation expresses the Conservation of absolute vorticity. General Remarks on Wave Motion
To have wave motion it is necessary to have some restoring mechanism, so that a particle which is disturbed from an equilibrium position will be induced to return there. Examples of restoring forces are the elastic force in springs, gravity acting on a pendulum, electrostatic and electromagnetic forces in oscillating circuits, negative feedbacks in biological systems, etc. We will consider some restoring forces important in the atmosphere. Exercise:
Fill in the missing steps in the derivation of the vorticity and divergence equations. Note: from now on, the primes on u , etc., are omitted. 9 10 Compressibility: If a ﬂuid is compressible, the increase in pressure of a parcel of ﬂuid which is compressed will tend to make it expand again. This eﬀect can result in compression or sound waves propagating through the ﬂuid. In the present context (for the Shallow Water Equations) we have assumed that the ﬂuid is incompressible. Thus, sound waves have been eliminated as possible solutions. Gravity: If the ﬂuid surface is not horizontal, there will be pressure forces due to diﬀering masses of ﬂuid above diﬀerent points, i.e., there will be a pressure gradient from deep to shallow parts of the ﬂuid. This will tend to push ﬂuid away from places where the surface is elevated towards regions where it is depressed. This tendency to even out disturbances in the height ﬁeld can lead to gravity waves. The betaeﬀect: The variation in the planetary vorticity f , which we have denoted by β , combined with the conservation of absolute vorticity, amounts to a restoring force and results in the Rossby waves which we will study below.
11 Longitudinal v. Transverse waves: We have eliminated the longitudinal waves by the assumption of incompressibility. Transverse waves may be primarily vertical or horizontal. However, there is something of a paradox: How can pure transverse waves (i.e., waves having no velocity component in the direction of travel of the waves) exist in an inviscid ﬂuid? Since there is no shearing force, there seems to be no way of communicating the wave motion orthogonally to the ﬂow. Yet such waves do exist in the atmosphere! (The argument of their nonexistence is used in seismology to deduce the liquid nature of the earth’s core). Exercise:
Try to resolve this paradox. Note: A gadankenexperiment may be productive even if it does not yield the result that is sought.
12 Pure Types of Wave Solution
I: Vertical Transverse Solutions: Gravity Waves ¯ Let us assume that the mean ﬂow u is zero, and consider motions in an eastwest plane. Moreover, we neglect the eﬀect of rotation. Thus ∂ f ≡ 0; v ≡ 0; ≡0 ∂y The perturbation equations reduce to ∂u ∂ Φ + =0 ∂t ∂x ∂ Φ ¯ ∂u +Φ =0 ∂t ∂x We can immediately eliminate u and derive an equation for Φ: ∂ 2Φ ¯ ∂ 2Φ −Φ 2 =0 ∂t2 ∂x (Exercise: Show that the same equation governs u.)
13 This is the familiar wave equation, usually appearing as ∂ 2Φ 1 ∂ 2Φ =2 2 ∂x2 c ∂t and it has solutions of the general form Φ = Φ1(x − ct) + Φ2(x + ct) . The wave phase speed c is given by √ ¯ c=± Φ Exercise: Repeat the analysis with a mean ﬂow u, and show √¯ ¯ ¯ that the wave speed is now given by c = u ± Φ. Exercise: Take the scaleheight of the atmosphere, H ≈ 10 km as the mean depth. Calculate the gravitywave speed in this case. Exercise: If the duckpond in St. Stephen’s Green has depth h = 1 m, at what speed do the waves on the pond travel? Check by observation!
14 To examine the structure of the gravity wave solutions, let us consider a single sinusoidal component: u u0 = exp[ik (x − ct)] . Φ Φ0 Note: It is analytically convenient to consider a complex exponential solution, but the physical solution is the real part of this expression. The diﬀerential operators now have the following expression: ∂ ∂ ∼ ik ; ∼ −ikc ∂x ∂t and the wave equations for u and Φ become u u0 ik(x−ct) ¯ = −[k 2(c2 − Φ)] e = 0. Φ Φ0 √ ¯ Clearly, this implies c = ± Φ. ∂t2 ¯ −Φ ∂x2 ∂2 ∂2 The divergence and vorticity reduce to δ = (ux + vy ) = iku ζ = (vx − uy ) ≡ 0 (recall that ∂/∂y ≡ 0.) Note: Pure gravity waves are irrotational, with vanishing vorticity. The momentum equation gives the relationship between u and Φ: ∂u ∂ Φ Φ + =0 ⇒ −ikcu + ikφ = 0 ⇒ u= . ∂t ∂x c Thus, for c > 0, i.e., eastwardmoving waves, the perturbation wind u and geopotential (or height) are positive and negative together: Where the ﬂuid is elevated, the movement is eastward. 15 16 Recall that we consider only the real part of the complex expressions for u and Φ to be physically signiﬁcant. Thus, the physical solutions for geopotential, zonal wind and divergence are Φ = {Φ0 exp[ik (x − ct)]} = Φ0 cos[k (x − ct)] u = {u0 exp[ik (x − ct)]} = (Φ0/c) cos[k (x − ct)] For c > 0, they are in phase. The divergence is δ = {iku0 exp[ik (x − ct)]} = (−k Φ0/c)sin[k (x − ct)] It is 90◦ out of phase with u and Φ. Thus, maxima and minima in the divergence are a quarter wavelength west of peaks and troughs in the height ﬁeld. That is, maximum divergence is west of a peak, and maximum convergence to the east. Therefore, the mass ﬂux will cause the peak to move eastward (see Figure).
17 The essential mechanism for the propagation of gravity waves is the change of pressure due to varying weight of ﬂuid above a given horizontal surface.
18 Exercises: (1) Repeat the above description of the relative phases of the Φ, u and δ ﬁelds in the case where c < 0, i.e., for westward wave propagation. (2) Derive the vertical velocity, and combine it with the zonal velocity u to describe the particle trajectories. II: Horizontal Transverse Solutions: Rossby waves These solutions are of central importance in meteorology and oceanography. They are also found in other rotating ‘ﬂuid’ systems, for example, galaxies. We will use the vorticity and divergence equations, in which the β term occurs explicitly. We consider the simplest case: • We ignore all variations with latitude (i.e., in the y direction) except for the β term. • We seek solutions for which the motion is purely horizontal • We will also assume that the zonal velocity is constant, ¯ u, and that the perturbation zonal velocity u vanishes identically • This implies purely transverse wave motion (i.e., no component of perturbation velocity in the direction of travel of the wavefronts) 19 20 We have u ≡ 0; ∂ ≡ 0; ∂y w ≡ 0; The divergence and vorticity become δ = (ux + vy ) ≡ 0 ; ζ = (vx − uy ) = vx . The vorticity equation becomes ∂ζ ∂ζ ¯ + u + βv = 0 , ∂t ∂x This may be written in terms of the meridional velocity v : Recall the perturbation divergence and vorticity equations: ∂δ ∂δ ¯ + u − f ζ + βu + 2Φ = 0 . ∂t ∂x ∂ζ ∂ζ ¯ + u + f δ + βv = 0 . ∂t ∂x The divergence equation reduces to a diagnostic relationship: f ζ = 2Φ . This equivalent to geostrophic balance: f v = Φx ⇒ f vx = Φxx .
21 Now let us assume a sinusoidal variation of v with time and the x coordinate: v = v0 exp[ik (x − ct)] . Substituting this into the vorticity equation, we get ¯ k 2 (c − u ) + β v = 0 which can hold only if the quantity in square brackets vanishes. This gives us an expression for the phase speed: ∂ 2v ∂ ∂v ¯ + u 2 + βv = 0 . ∂t ∂x ∂x ¯ c=u− β . k2
22 The dispersion relationship ¯ c=u− β . k2 is the celebrated Rossby wave formula. The wavenumber k and wavelength L are related by k = 2π/L. Thus the phase speed can also be written ¯ c=u− βL2 . 4π 2 The geostrophic balance gives a relationship between v and Φ: f v = Φx = ik Φ ; Φ = −(if /k )v So, assuming v is real and considering the real parts of the solution, we get v = v0 cos[k (x − ct)] Φ = (f v0/k ) sin[k (x − ct)]
23 Schematic ﬁgure of Rossby wave structure. Top: Vertical section. Bottom: Horizontal section.
24 Predominantly Zonal Flow
25 Meandering Pattern Develops
26 Strongly Meridional Regime
27 Cutoﬀ Low. Return to zonal Regime
28 Properties of Rossby Waves
Rossby waves always move westward relative to the ﬂow. They are the prototype of the large scale wavelike disturbances in the atmosphere. The motion is horizontal and there is geostrophic balance between the pressure or height disturbance and the wind ﬁeld. The divergence vanishes identically under the assumptions we have made. In the absence of rotation, these wave solutions do not exist; they owe their existence to the eﬀects of rotation and their westward movement to the latitudinal gradient of the vertical component of the planetary vorticity f , i.e., the beta eﬀect. The dynamics of the planetary waves were ﬁrst elucidated by the Swedish meteorologist CarlGustav Rossby, around 1940, using the principle of conservation of absolute vorticity.
29 Summary of Properties of Rossby Waves • The solutions derive from the principle of conservation of absolute vorticity • They owe their existence to the rotation Ω of the Earth • The dynamics were ﬁrst elucidated by CarlGustav Rossby ¯ • They waves always move westward relative to u • The motion is purely horizontal • The meridional wind is in geostrophic balance • The divergence vanishes identically 30 Exercises
Exercise: Consider the conservation absolute vorticity d (ζ + f ) = 0 . dt ¯ Linearize this equation about a constant mean ﬂow u (don’t forget the beta term!) and assume a wavelike solution with dependency proportional to exp[ik (x−ct)]. Deduce the Rossby wave formula β βL2 ¯ ¯ c=u− 2 =u− 2 . k 4π Exercise: Find the phase speed of a Rossby wave, assuming that β = 10−11 m−1 s−1, and that the wavelength is 3,000 km, Exercise: Show that the vorticity equation ∂ ∂v ∂ 2v ¯ + u 2 + βv = 0 . ∂t ∂x ∂x is a hyperbolic partial diﬀerential equation (PDE).
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