Process Solutions

Process Solutions Jay McCreary Summer School on: Dynamics of the North Indian Ocean National Institute of Oceanography Dona Paula, Goa June 17 – July 29, 2010

Introduction • Interior ocean dynamics • Coastal ocean dynamics • Equatorial ocean dynamics

Interior ocean dynamics

Response to a switched on zonal wind When β = 0 (f-plane), easterly winds force northward Ekman drift (northern hemisphere), which spins up two counter-rotating, geostrophic gyres, a process known as Ekman pumping. When β≠ 0, Rossby waves extend the response west of the wind region, and adjust the response to a state of Sverdrup balance. –

Interior-ocean equations Equations for the un, vn, and pn for a single baroclinic mode are Even this linear set of equations, though, is difficult to solve. A useful (and still popular) simplification is to drop the acceleration and damping terms from the momentum equations. In addition, the horizontal viscosity terms are assumed small dropped in the interior ocean, and are only retained to represent western boundary currents. It is often useful to view this set of equations as corresponding to a linear, 1½-layer system, in which diffusion results from an entrainment velocity w1 rather than by a damping term, –A/cn2pn.

Solving the complete set of equations for a single one in vn, and, for convenience, dropping the subscript n, setting τy = 0, and F = τx/H, gives (1) whereas solving the approximate setfor the interior ocean gives (2) Equation (2) is valid provided the first three terms of (1) are small compared to the fourth. That will be true provided that and similarly that

Free waves A useful way of understanding the impact of the approximations is to see how they distort the free waves of the system. Typically, one proceeds by “pretending” that f and βare both constant. Then, solutions have the form of plane waves, In the HIG Notes, σ→ –σ. providing dispersion relations for the waves. The resulting dispersion relation for the complete system is whereas that for the approximate one is

Free waves To plot σ(k,l), for the moment consider a slice along l = 0. When σ is large, the kβ term is small, and can be neglected to first order. The resulting curve describes gravity waves. σ/f σ/f σ/f When σ is small, the σ2/c2 term is small, yielding the Rossby-wave curve. NOTE: The limits on the axes are not accurate. For example, the gravity-wave curve should bottom out near σ/f = 1. k/α k/α k/α When l ≠ 0, the curves extend along the l/α axis, to form circular “bowls.”

Free waves How does the interior-ocean approximation distort the curves? It eliminates gravity waves and the Rossby curve becomes a straight line (non-dispersive). Moreover, σ is independent of ℓ, so that the Rossby wave curve is not a bowl in k-ℓ space, but a plane. σ/f σ/f When are the waves accurately simulated in the interior model? When kR = k/α << 1, ℓR << 1, and σ/f << 1. k/α k/α

Solution forced by switched-on τx Let the ocean be unbounded and forced by a switched-on wind patch with the separable form Neglecting damping, the v-equation we would like to solve is (1) Note that y-derivatives are absent from (1), a major simplification. A simple way to solve this equation is to use Laplace and Fourier transformsin t and x, which replaces derivatives with the numbers, As a result, differential equation (1) reduces to an algebraic one, which can be easily solved.

Solution forced by switched-on τx The equation for double-transformed v is then where ^ and ~ over the v indicate a Laplace and Fourier transformed variable. Solving for transformed v gives To invert the Laplace transforms, we use the inverse Laplace transforms to get (2)

Solution forced by switched-on τx To invert the Fourier transforms, in (2), we use to obtain the solution The u field can be found by integrating v according to where the integral is from +∞ because no information can appear west of the forcing since Rossby waves propagate westward.

Solution forced by switched-on τx (a) (b) (c) Term (a) – (b) advances westward as a Rossby wave, leaving a Sverdrup flow (c) behind. Note that (a) and (c) cancel when t 0, so that the initial circulation is just Ekman flow, –(b).

Solution forced by switched-on τx (3) Although the v field of the Rossby wave separates from v of the forcing region, the u field does not because of the x-integral. It might seem from the form of (3) that u extends farther west than the front, but one can show that the integral vanishes there. –

β-plane Interior-ocean dynamics The initial response is the same as on the f-plane. Subsequently, the radiation of Rossby waves adjusts the circulation to Sverdrup balance. f-plane –

Coastal ocean dynamics

Forcing by a band of alongshore wind τy All the solutions discussed in this part of my talk are forced by a band of alongshore winds of the form, Since this wind field is x-independent, it has no curl. Therefore, the response is entirely driven at the coast by onshore/offshore Ekman drift. The time dependence is either switched-on or periodic Y(y)

Response to switched-on τy f-plane In a 2-dimensional model(x, h), alongshore winds drive upwelling and coastal currents only locally, in the region of the wind. The offshore decay scale is the Rossby radius of deformation. In a 3-d model (x, y, h) with β = 0, in addition to local upwelling by we, coastal Kelvin waves extend the response north of the forcing region. The pycnocline tilts in the latitude band of the wind, creating a pressure force to balance the wind stress. 1½-layer model f-plane

Response to switched-on τy β-plane When β≠ 0, Rossby waves carry the coastal response offshore, leaving behind a state of rest in which py balances τy everywhere. A fundamental question about eastern-boundary currents, then, is: Why do they exist at all?

Response to switched-on τy There is upwelling in the band of wind forcing. There is a surface current in the direction of the wind, and a subsurface CUC. McCreary (1981) obtained a steady-state, coastal solution to the LCS modelwith damping. The model allows offshore propagation of Rossby waves. A steady coastal circulation remains, however, because the offshore propagation of Rossby waves is damped by vertical diffusion.

Response to switched-on and periodic τy In a solution to an OGCM forced by switched-on, steady winds (left panels), coastal Kelvin waves radiate poleward and Rossby waves radiate offshore, leaving behind a steady-state coastal circulation. In response to a periodic wind (right panels), the adjustment never ceases, and the coastal currents exhibit upward phase propagation. Does all the current propagate offshore, or does some remain attached to the coast? σ = 0 σ = 2π/200 days

Coastal-ocean equations Equations for the un, vn, and pn for a single baroclinic mode are which are difficult to solve. A useful simplification for coastal oceans is to drop the acceleration and damping terms from the un equation, neglect horizontal mixing, and ignore forcing by τx (although the latter two are not necessary). In this way, the alongshore flow is in geostrophic balance.

Solving the complete set of equations for single one in vn, and, for convenience, dropping the subscript n, setting τx = 0, and G = τy/H, gives (1) whereas solving the approximate setfor coastsgives (2) As for the interior equations, As before, y-derivatives are absent from (2), a major simplification. Equation (2) will be valid provided that the second and third terms of (1) are small compared to the fourth. That will be true provided that

Free waves If we look for plane-wave [exp(ikx + ily – iσt)] solutions to (2), the resulting dispersion relation is (3) Equation (3) is quadratic in k, and has the solutions Note that the roots are either real or complex depending on size of the last term under the radical, which defines a critical latitude, Poleward of ycrsolutions are coastally trapped (β-plane Kelvin waves) whereas equatorward of ycr they radiate offshore (Rossby waves).

Free waves How does the interior-ocean approximation distort the curves? It eliminates gravity waves and the Rossby curve has the correct shape for ℓ = 0. But, σ is independent of ℓ, so that the Rossby wave curve is not a bowl in k-l space, but a curved surface. σ/f σ/f When are the waves accurately simulated in the interior model? When ℓR << 1, and σ/f <<1. k/α k/α

Solution forced by periodic τy Neglecting damping terms, an equation in p alone is It is useful to split the total solution (q) into interior (q') and coastal (q") pieces. The interior piece (forced response) is x-independent, and so is simply The coastal piece (free-wave solution) is where we choose k1, rather than k2, because it either describes waves with westward group velocity (long-wavelength Rossby waves) or that decay to the west(eastern-boundary Kelvin waves).

Solution forced by periodic τy To connect the interior and coastal solutions, we choose P so that that there is no flow at the coast, To solve for P, it is useful to define the quantity in which case, Define Go = τy/H. Then, the solution for total p is (4)

Solution forced by periodic τy The solution has interesting limits when y ∞ and y 0. In the first limit, so that a β-plane Kelvin wave with an amplitude in curly brackets. In the second limit, so that a long-wavelength Rossby wave propagating westward at speed cr.

Equatorial ocean dynamics

Spin-up of an inviscid, baroclinic mode d (1 month) In response to forcing by a patch of easterly wind, Kelvin and Rossby waves radiate from the forcing region, reflect from basin boundaries, and eventually adjust the system to a state of Sverdrup balance. Rossby wave Kelvin wave d (6 months) Equatorial jet If the wind oscillates at the annual (or semiannual) period, these adjustments continue indefinitely. Coastal Kelvin waves continuously radiate from the equator around the perimeter of the basin, followed by the propagation of Rossby waves into the basin interior. d (1 year) Reflected Rossby-wave packet d (5 years) Sverdrup flow

Steady, linear response with damping With diffusion: When the LCS model includes diffusion, realistic steady flows can be produced near the equator. EUC McCreary (1981) Oscillating wind: Suppose the wind oscillates at the semiannual or annual cycle. Then, the equatorial currents oscillate, penetrate into the deep ocean along beams, and exhibit upward phase propagation, similar to the coastal circulation.

Equatorial-ocean equations Equations for un, vn, and pn for a single baroclinic mode are (1) Because f vanishes at the equator, no terms can be dropped that allow for mathematically simple solutions near the equator. A useful assumption, though, is to set f = βy, known as the equatorial β-plane approximation. As a result, one can look for solutions as expansions in Hermite functions.

Equatorial gravity and Rossby waves We look for free-wave solutions to (1) of the form, ℓ(y)exp(ikx– iσt), without damping (A = 0), and, for convenience, we drop the subscript n. The resulting v equation is (2) Because f varies so much near the equator, we cannot assume ℓ(y) = exp(iℓy). It is useful to introduce the non-dimensional variable, for y, and to rewrite (2) as (3) The operator in parentheses is the Hermite operator. Its eigenfunctions are the solutions to NOTE: In the HIG Notes, ℓ → m. Where ℓ = 0, 1, 2, …, and they are referred to as Hermite functions.

Equatorial gravity and Rossby waves The figure plots the first six Hermite functions ℓ (ℓ = 0–5). The scaling factor, LR = (cn/β), the equatorial Rossby radius of deformation for baroclinic mode n. (For n = 1, LR is roughly 250 km.) Note that the ℓ are less equatorially trapped(extend farther off the equator) as ℓ increases. Note also that they alternate between being symmetric and antisymmetricabout the equator. Ascani (2002) Fedorov and Brown (2007)

Equatorial gravity and Rossby waves The solutions to (3) can be represented as expansions in Hermite functions (4) where vℓ is a wave amplitude. Each term in expansion (4) is an individual equatorial wave. Inserting (4) into (3) gives which provides the dispersion relation for equatorial, Rossby and gravity waves.

Equatorial Kelvin wave The equatorial Kelvin wave has v = 0, and so was missed in the preceding solutions. To find it, set v = A = 0 in (1), and look for a free-wave solution of the form (y)exp(ikx– iσt). With these restrictions, equations (1) reduce to The first and third equations imply and the second then gives (5)

Equatorial Kelvin wave The solution to (5) is The solution that grows exponentially in y, which corresponds to the root, k = –σ/c, is physically unrealistic in an unbounded basin and must be discarded. Therefore, the only possible wave is (6) which describes the structure and dispersion relation for the equatorial Kelvin wave. In (6), I have used the property that and redefined the arbitrary constant amplitude to be Po = π¼P'o.

Free waves For each ℓ, there is a gravity wave(large σ) and a Rossby wave (small σ). The plot indicates the waves for ℓ = 1, 2, and 3. There is a new type of wave, the mixed Rossby-gravity (Yanai) wave, which joins the Rossby (gravity) wave curves for large negative (positive) values of k. σ/σo 1 Another additional wave is the equatorial Kelvin wave. 3 k/αo

Free waves As for the coastal model, there are two wavenumbers, kℓ1,2, associated with each m value. The wavenumberskℓ1 (kℓ2) describe waves with eastward (westward) group velocity or decay. Also as for the coastal model, the wavenumbers are real for small σ (Rossby waves) and become complex as σ increases. Eventually, they become real again for even larger σ (gravity waves). σ/σo The region of complex roots for ℓ = 1 waves is indicated by the shading.. Such waves exist only along boundaries, where they superpose to generate β-plane coastal Kelvin waves. 1 3 k/αo

Equatorial gravity and Rossby waves The vℓ,uℓ, and pℓ fields for equatorially trapped Rossby and gravity waves are whereVℓ is a constant amplitude, j = 1,2, and

Equatorial gravity and Rossby waves σ/σo The u, v, and p fields for a Yanai wave when cn = 250 cm/s and P = 30 days. For this P, σ/σo = .36 and λ = 7.3º. 1 3 k/αo Courtesy of Francois Ascani

Equatorial gravity and Rossby waves σ/σo The u, v, and p fields for a Yanai wave when cn = 250 cm/s and P = 360 days. For this P, σ/σo = .03 and λ = 0.64º. 1 3 k/αo Courtesy of Francois Ascani

Equatorial gravity and Rossby waves σ/σo The u, v, and p fields for an ℓ = 1 Rossby wave when cn = 250 cm/s and P = 360 days. For this P, σ/σo = .03 and λ = 240º. 1 3 k/αo Courtesy of Francois Ascani

Equatorial gravity and Rossby waves σ/σo The u, v, and p fields for an ℓ = 2 Rossby wave when cn = 250 cm/s and P = 360 days. For this P, σ/σo = .03 and λ = 140º. 1 3 k/αo Courtesy of Francois Ascani

Free waves The first equatorially trapped waves to be discovered were gravity-wave resonanceswith periods of O(10 days) (Wunsch and Gill, 1976; Deep-Sea Res.). There are no publications that explore the possibility of Rossby-wave resonances. The equatorial Kelvin wave was discovered after it was predicted to be dynamically important in El Nino (Knox and Halpern, 1982, JMR). σ/σo The mixed Rossby-gravity (Yanai) wave was first observed in the atmosphere by Yanai. In the ocean, it was (probably) first detected in the Indian Ocean by Tsai et al. (1992)using altimeter data. 1 3 k/αo Who first detected an equatorial Rossby wave?

Free waves The first equatorially trapped waves to be discovered were gravity-wave resonanceswith periods of O(10 days) (Wunsch and Gill, 1976; Deep-Sea Res.). There are no publications that explore the possibility of Rossby-wave resonances. The equatorial Kelvin wave was discovered after it was predicted to be dynamically important in El Nino (Knox and Halpern, 1982, JMR). σ/σo The mixed Rossby-gravity (Yanai) wave was first observed in the atmosphere by Yanai. In the ocean, it was (probably) first detected in the Indian Ocean by Reverdin and Luyten (1986)using altimeter data. Tom Farrar 2010 1 3 k/αo Who first detected an equatorial Rossby wave?

Yoshida Jet Kozo Yoshida wrote down the first solution for an x-independent equatorial current driven by zonal winds. The (more complete) theoretical solution developed somewhat later (Dennis Moore) has come to be called the “Yoshida Jet” (Jim O’Brien). The basic dynamics of the Yoshida Jet can be understood from the zonal-momentum equation. Neglecting the pressure-gradient and mixing terms, in the zonal momentum equation gives Offshore, Ekman balance (fvn = τx/Hn) holds, whereas at the equator un continues to accelerate (unt = τx/Hn). The switch from one dynamical regime to the other occurs at y ≈ α0–½ = (β/cn) –½ .

Bounded Yoshida Jet Zonal flows along the equator don’t continue to accelerate in reality or models. Why not? If only the damping term (A/cn2)unis included in the zonal momentume equation, the flow stops accelerating, but it is unrealistically fast(the unit is km/s) and extends to the bottom. If the pressure-gradient term pnxis then included, the flow field has both a realistic amplitude and structure.

Bounded Yoshida Jet d (1 month) Rossby wave Kelvin wave d (6 months) Equatorial jet In response to forcing by a patch of easterly wind, a Yoshida Jet initially develops in the forcing region. Subsequently, Kelvin and Rossby waves radiate from the forcing region, generating an eastward jet along the equator, both east and west of the forcing region: the bounded Yoshida Jet.

Eastern-boundary reflections Suppose the ocean is forced by a patch of oscillating zonal wind confined to the interior ocean. It generates an equatorial Kelvin wave, that radiates to the eastern boundary of the basin. There can be no zonal flow through the boundary. How does the system adjust to prevent this flow? Dennis Moore showed that a packet of equatorial waves with the zonal velocity field, (7) are generated at the eastern boundary. In (7), the wavenumbers kℓ1 correspond to waves with westward group velocity or decay.

Process Solutions