ATOC 5051: Introduction to Physical HW #4: Given Oct. 21; Due Oct 30

1. [15pts] Figure 1 shows the longitude - time plot of the Depth of 20°C isotherm Anomaly (D20A) along the Pacific equator from Jan 2004-Jan 2005. D20 represents the depth of the central in the equatorial Pacific, which deepens by as much as 30m in January 2004 and to a lesser degree, Jan 2005. Note that positive D20A represents thermocline ! deepening and negative D20A, thermocline shoaling. D20 anomaly (m) (a) From Figure 1, estimate the period and propagation speed, and calculate the wavelength. Provide discussions on how you obtain these estimates. (6pts)

The period of the wave, which is the time it spans the two phase lines, is T=~80days. Take 1°~110km. The time it takes the wave to cross the basin, with Width=100°=100x110000m=11x106m, is ~50days. Therefore, cp=Width/(50*86400)=2.5(m/s). Based on the definition, wavelength 170E 90W L=T*cp=80*86400*2.5=17280(km).

(b) From Figure 1, can you infer whether they are Figure 1. D20A along the equatorial symmetric or anti-symmetric , why? (3pts) Pacific from Jan 2004-Jan 2005. From Fig. 1, I infer that they’re symmetric waves Units are meters (m). The two solid because D20A, which mirrors level anomaly, lines represent the phase lines of obtains large amplitude on the equator. For anti- two waves, which also represent the symmetric waves, D20 and are ~0 at the propagating fronts of deepened equator. D20.

(c) What type of waves are they? Provide the reasons Figure 1. D20A along the for why you think they are this type of waves and not otherequatorial types. (6pts) Pacific from Jan They are equatorial Kelvin waves, because equatorial2004 Kelvin-Jan waves 2005. are Units symmetric are meters about the equator and propagate eastward. They can’t be IGWs because(m). The their two periods solid are lines 80days, too long for IGWs to exist. They can’t be mixed Rossby-Gravityrepresent (MRG) waves the becausephase linesMRG ofwaves two are anti-symmetric about the equator. They can’t be Rossbywaves, waves which because also longrepresent Rossby thewaves propagate westward (both cp and cg). Note that Short Rossbypropagating waves are fronts too short of comparing deepened with such as a long, observed wavelength. D20.

2. [15pts] Figure 2 shows the longitude-time plot of sea surface height anomaly (SSHA) from TOPEX/Poseidon satellite observations at 21oN of the North Pacific from 1993 to 1995.

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(a) Toward which direction are the SSHAs propagating? (2pts) The SSHAs propagate toward the west.

(b) The two dark solid lines in Figure 2 are phase lines. Estimate the period T, wavelength L, and phase speed Cp of the waves. (6pts) From the phase lines of Fig 2, we obtain the following estimates: Period T=600 days (the time spans the two adjacent peaks of the low SSHAs at a specific longitude); Wavelength: L= 45°=45 x 110km x cos(21°)=4621km (the distance between the two phase lines at a specific time); Cp=-0.095 m/s.

(c) Based on the wave properties you obtain from (a) and (b), what waves are they? (2 pts) They are long Rossby waves.

Write down the relation and phase speed equations of this type of wave. (2pts) βc 2 βc 2 The dispersion relation is: ω = − k , and phase speed c = − . f 2 p f 2

(d) Are they gravity waves (1pt)? Why (2pts)? No. They cannot be gravity waves, because their period is 600days (almost two years), whereas gravity € 2π € wave’s longest period is Tmax = , which is only in the order of few days. f 21N

Figure 2. Longitude- € time plot of Sea Surface Height Anomalies (SSHAs) at 21N of the Pacific Ocean.

2 3. [20pts] Figure 3 shows the time evolution of currents (arrows) and layer thickness variations (color 1 contours) from the 1 - layer reduced gravity model, with equatorial beta-plane approximation and 2 harmonic horizontal mixing. The baroclinic mode speed is c=2.74m/s. These signals represent the biweekly oscillations excited by the meridional winds, which oscillate at 14-day period (7 days northward and 7 days southward) in the central equatorial ocean basin. These waves are frequently observed in €the equatorial Indian, Pacific and Atlantic .

(a) From Figure 3, estimate the wave length and . (4pts) Wavelength is the distance from crest to crest or trough to trough. From the solution on day 30 (bottom-right panel of Fig. 3), L=23°=2530km; From day 20 to day 30, the ‘front’ of zonal current u propagates from ~53° to 60°. So, Cg=(7x110000)/864000=0.9(m/s).

(b) Are they symmetric or antisymmetric waves, why? (3pts) They are anti-symmetric waves because based on the definition, their zonal currents and depth of thermocline are anti-symmetric about the equator, and the meridional currents are symmetric. (c) What type of wave are they? (1pt) Discuss why it cannot be other types of equatorial waves. (6pts) It is Yanai (mixed Rossby gravity) wave. This is because its energy propagates eastward, and its period is 14days, and it is antisymmetric about the equator. It cannot be because the period is too short for propagating Rossby wave to exist. It cannot be Kelvin wave because Kelvin wave is symmetric about the equator. It cannot be IGW because the minimum frequency of the IGW is βc(2n +1) when n=1 (the first meriodinal mode). 2 × 7.29 ×10−5 × cos0 × 2.74 × 3 ω = βc3 = =1.37x10-5s-1. min 6370000 € 2π 5 So the longest period for the IGW to exist is: Tmax = = 4.58 ×10 s=5.3 days. At 14-day ω min period, IGWs do not exist. €

(d) Write down the dispersion relation of this wave (1pt), and calculate its wavelength at biweekly period (2pts). You may check whether it€ is close to your estimate from (b).

Yanai wave’s dispersion relation is: 2π For T=14days, ω = = 5.19 ×10−6 s. T ∂f ∂(2Ωsinϕ) 2Ωcosφ 2 × 7.29 ×10−5 β = = = = = 2.28 ×10−11 ∂y ∂(Rϕ) R 6370000 2.288 ×10−11 5.19 ×10−6 k = −€ + = (−4.408 +1.894) ×10−6 5.19 ×10−6 2.74 2π 6.28 € L = = ×106 = 2500000(m) = 2500km k 2.514 This is close to the estimate of 2530km. € [NOTE: you can calculate cg from the dispersion relation.

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3 ∂ω 1 c = = = 0.89m /s. This is close to the estimated 0.9m/s g k β 1 ∂ + ω 2 c above.]

(e) After the wave arrives at the eastern boundary, what will happen to its energy? (3pts) € The energy will go poleward as coastal Kelvin wave with damping on its way of propagation. There won’t be any wave signal to be reflected westward to the ocean interior, because IGW does not exist at this period, and Rossby wave does not exist at T=14days.

Figure 3. Currents and layer thickness from the 1.5-layer model: X-axis: longitude (degree); y-axis: latitude (degree).

4. [35pts] Figure 4 shows the idealized Pacific basin. Assume the ocean is inviscid and the first baroclinic mode speed is c=2.8m/s. The basin width (east-west) is 16,000km. Useful values are:

4 2Ωcosφ 1degree=110km, β = where φ is latitude and R=6370km is the earth's radius and Ω=7.29x10-5 R s-1. [Note: For simplicity, I only considered the case with northern and southern boundaries of the rectangular basin representing open ocean (and there are no solid boundaries there).] € € €

Figure 4. Idealized tropical-midlatitude Pacific Ocean forced by periodic wind in the central subtropical south Pacific basin.

(i) [20pts] Assume the central equatorial basin (5°S-5°N) is forced by zonal wind oscillating at a 20- day period. (a) Will persons at A, B, C and D see sea level signals? (4pts) B and C will see sea level signals, but D and A won’t.

(b) If yes, through what processes? If not, why? (6pts) At the 20-day period, propagating Rossby waves won’t be excited. So, only KWs are excited by the symmetric winds. MRG waves won’t be excited, because symmetric winds won’t excite anti-symmetric MRG waves. IGWs won’t exist because their periods are shorter than 20days. B will see sea level signals through equatorial KWs; C will see sea level signals by coastal KWs when EQ KWs arrive at the eastern boundary; D and A won’t see sea level signals because RWs don’t exist at 20-day period.

(c) Draw a schematic diagram showing the possible “processes” that can affect A, B, C and D, respectively, if they can affect these locations. (4pts)

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(d) Approximately after how many days of the winds are switched on will A, B, C and D observe the sea level signals, if they can? Please make explicit statements if a location can’t be affected (6pts) T1=16,000000/2.8=2857142.86s=33 (days). So it takes 33days for B to see sea level signals. Then, sea level signal propagates to 30N at C along the eastern boundary via coastal Kelvin wave: T2=(30x111000)/2.8=1189285.7s=~14 (days). So, it takes approximately 33+14=47(days) for C to see sea level signals.

(ii) Assume the equatorial basin (5°S-5°N) is forced by the same zonal winds but oscillation at a 5- year period. (15pts) (a) Will persons at A, B, C and D see sea level signals? If yes, through what physical processes? (10pts) Yes, A, B, C and D will all see sea level signals. At such a long period, large-scale symmetric winds will excite long symmetric equatorial Rossby waves and equatorial Kelvin waves. (1) Firstly, A will see a sea level signal via equatorial long Rossby wave, which is directly forced by the zonal wind in the central equatorial basin; (2) Secondly, Rossby wave reflected from the eastern boundary, after equatorial Kelvin wave impinges on the eastern boundary, can also propagate to the western boundary and affect location A; (3) Thirdly, After the equatorial Kelvin wave impinges on the eastern boundary, part of the energy will propagate along the eastern boundary northward to higher latitude, and subsequently radiate westward as Rossby wave. Upon arriving at the western boundary, part of the energy will propagate southward as coastal Kelvin wave and then affect location A, while part of the energy can be reflected as short Rossby wave and then it is killed by the weak mixing. (2) B will see sea level signals through equatorial Kelvin wave. (3) C will see sea level signal through coastal Kelvin wave. (4) D will see sea level signal through midlatitude Rossby wave, which radiates westward from the eastern boundary near latitude C. (b) How long will it take for point D to be affected, if this location can be affected? (5pts) First, equatorial Kelvin wave propagates from the central equatorial basin to B: T1=16,000000/2.8=2857142.86s=33 (days); Then, sea level signal propagates to 30N along the eastern boundary via coastal Kelvin wave:

6 T2=(30x111000)/2.8=1189285.7s=13.76 (days); Subsequently, sea level signal propagates westward as Rossby wave.

βc 2 2Ωcosφ 2.82 cr = − 2 = − × 2 2 × f7.29 ×10−5R× cos30(2oΩsinφ) 2.82 = − × 6370000 (2 × 7.29 ×10−5 × sin30o ) =0.029(m/s) € T3=(16000000)/0.029=6385.7(days)≈17.5(yrs) It takes T=T1+T2+T3≈6432 (days)≈17.6(yrs) for D to see the sea level signal. € 5. [15pts] (i) Assume the warmer and saltier MOW€ with density is on top of the relatively fresher and colder NADW with density €, and is slightly€ smaller than initially. After sometime, what phenomenon may occur, and why? (3ps) After some time, stratification will become unstable because of “double diffusion” and salt fingering may occur. Temperature diffuses 100 times faster than salinity. After certain time, the original MOW becomes cold but salinity changes little. This will increase its density and thus becomes heavier than the NADW, resulting in salt fingering.

(ii) During northern summer, the observed eastern equatorial Pacific cold tongue often shows “cusplike” features (Figure 5).

Figure 5. Observed SST in the equatorial Pacific. The color bar shows SST magnitude.

(a) Based on what you have learnt from this course so far, what are the most likely causes of the cusplike features (2pts)? Provide the reason for your answers. (4pts) Barotropic instability and baroclinic instability. In the equatorial Pacific Ocean, strong shears

exist between NEC, NECC and SEC. The criterion for the barotropical instability β −Uyy changes sign somewhere in these background currents. So, the strong sheared regions favor barotropic instability. On the other hand, strong North-South SST gradients on the south and north edges of the cold tongue indicates meridional gradients of density, which favor baroclinic instability. €

Where are the energies from for these cusplike features to grow? (2pts) Barotropic instability extracts kinetic energy from the background flow, and baroclinic instability extracts potential energy from the background state.

(b) Figure 6 shows an amazing picture of cusplike cloud in the atmosphere. What is the most likely cause for the cusplike features? (2pts)

7 Kelvin-Helmholtz instability. What is the necessary condition for this cusplike cloud to occur, and where is the energy from? (2pts)

Figure 6.

Richardson number less than 0.25 in theory, but approximately Ri < 0.5 in real ocean. Vertical shear of currents in the ocean (wind in the atmosphere) favors Kelvin- Helmholtz instability. The shear, however, has to act against the buoyancy in a stably stratified ocean. The stronger the current shear and the weaker the stratification, the more likely the K-H instability will occur. It extracts kinetic energy from the background flow.

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