GFD – 2 Spring 2010 P.B. Rhines Problem set 2 out: 14 April back: 21 April 1. • Write on one page a summary of the most important ideas so far in GFD-2. Include some of the simple equations/formulas but mostly text. Do this without using notes/text. • Now on a 2d page rewrite it using your notes/text. Page 1 will not be graded..it’s for feedback. • Include a paragraph suggesting how the class lectures could be improved: [e.g., slow down, use more student-at-board sessions, use scalar equations rather than vectors, don’t change reference frames so much, key more closely to Vallis text sections, talk about small scales as well as general circulation, do more lab demos, use more observations…] The feedback has been useful…PR 2. Sketch the f/h contours (‘barotropic PV contours’) for an wide circular mountain in the middle of a mid-latitude ocean basin. •What parameter (based on the mountain’s height and width, ocean depth, β etc.) determines whether free flow will tend to go over the mountain or around it? To stay on an f/h=constant contour your path deviates from a latitude line by an amount such that Δ(f/h) =0 = Δf/h –(f/h2)Δh; since Δf = βΔy we have Δy = fΔh/βh. If the width of the mountain is L, the parameter (βL/f)(h/Δh) determines whether the flow goes over ( if it >>1) or around (if it <<1). Note β/f ~ 1/a where a is the Earth’s radius so the important parameter is (L/a)(h/Δh). •What if this is an f-plane (flat Earth, β=0): what parameter determines whether quasi-geostrophic flow will go over or around the mountain. Here there is no β effect to allow the flow slip easily over the mountain. The ‘f/h’ contours are always blocked by it. But, we know about spin up by vortex stretching where the PV = (f+ζ)/h is conserved following the fluid. If the mountain is so small that the vortex stretching (actually squashing) produces a velocity change ΔU << the mean flow U, then the fluid will go ‘over’. We estimate, with ΔPV = 0 => Δζ/h ~ -fΔh/h2. Now, Δζ ~ ΔU/L, where L is the mountain diameter (because that is the horiz. length scale of the vorticity change). So, ΔU/hL ~ f(Δh/h2). So,

ΔU/U ~ fL/U (Δh/h) ~ Ro-1 (Δh/h) << 1 for the flow to go over the mountain. Ro is the Rossby number. This non-dimensional parameter must be small for the fluid to pass easily over the mountain; otherwise the induced vorticity sweeps the flow around the periphery of the mountain. 3 Geostrophic balance dominates the horizontal momentum equations (MOM) in large-scale, slowly varying flows (Ro = U/fL << 1, fT >> 1). Hydrostatic pressure balance dominates the vertical MOM typically for thin aspect ratio, when (H/L)2 <<1 (H is the vertical length scale of the flow, which may or may not equal the ocean depth). Because constant-pressure (isobaric) surfaces are nearly horizontal, pressure, p, is often used as a vertical coordinate rather than z. Classical hydrographers and meteorologists both do this. Define α = 1/ρ, the ‘specific volume’, and δ the ‘specific volume anomaly’,

•Show that α dp0

p

∫ = − gdzz(η)

z( p)

∫ where z(η) is the position of the sea surface. Using

hydrostatic vertical MOM balance, dp = −ρgdz => dp / ρ = −gρdz / ρ =∫∫ − gdz =∫ [Φ]

•If we define δ =α(S,T,p) - α(S=35.0psu, T=0, p) as the ‘specific volume anomaly’ show that the geostrophic balance can thus be written as

fk̂ × [u(0) − u(p)] = ∇(D ')

where D ' = δdp0

p

∫ is called ‘dynamic height anomaly’. k̂ is a vertical unit vector and

f=2Ωsin(θ), θ is latitude. Consider one scalar component of the thermal wind equation:

z

0

∫ fuzdz =z

0

∫ (g / ρ)ρy dz = (g / ρ)ρyp(z )

0

∫ (−dp / ρg)

= − (ρy / ρ2

p(z )

0

∫ )dp = ∂∂y

dp / ρp

0

∫ =∂∂y

α dpp

0

∫ .

Now since the reference value of specific volume, α(35,0,p) , depends only on pressure p, its integral with respect to p does not vary in space: we can replace α by its changing value δ in the above integral. Notice that D’ = -Φ’, the ‘geopotential anomaly’, tying this with the fundamental idea of the Earth’s geopotential. Why is the term ‘height’ appropriate (make a sketch). {There is a subtle issue here: the above height gradient above is taken along a constant pressure surface: you can show the equivalence to our more familiar horizontal pressure gradient taken along a constant-z level surface.}

•Show how the transport (the integral of the horizontal velocity over this depth range) can

be written in terms of the vertical integral of pδdp0

p

∫ Using the result just above, the total

transport (the vertical integral of the velocity) from z=-H to z=0 is the double integral

p= p(z=−H )

p=0

∫ δ (p ')dp 'dpp '= p(z=−H )

p '= p

∫ . Recall that integration by parts allows you to write

for example (a b)x = axb + abx yielding: axbdx = − abx dx + [ab]∫∫ ;

now choose dx => dp, ax = 1 and b= δdp∫

=>p= p(z=−H )

p=0

∫ δdp 'dpp '= p(z=−H )

p '= p

∫ =− pδdp + [pp(z=−H )

0

∫ δdp ]p(z=−H )0

p= z(−H )

0

∫ = pδdp0

p(z=−H )

∫

since the term in brackets vanishes.

• At sea we also use p as a vertical coordinate in CTD profiling; what is the equivalance between meters and decibars ? ( A decibar is a pressure unit of 1/10 bar, where 1 bar = 105 Pascal (close to average atmospheric pressure; the Pascal is the SI unit for pressure or stress, with units force/area = Newton/m2= kg m-1sec-2 . It sounds like a joke to say that atmospheric pressure is about 1000 millibars. ) 1 decibar = 10-1 Bar = 104 Pascal ≡ 104 Newton m-2; the weight of 1m high column of water (1 m2 in horizontal area)≅103 kg x 9.8 m sec-2 = 0.98x104 N m-2 = 0.98 decibar. However, as one of you noticed, seawater’s density varies with depth (both because of T,S and P). The in situ density ρ changes from roughly 1026 kg m-3 to roughly 1040 kg m-3 from surface to 5 km depth (varies with T and S). So the relationship between m and db shifts slightly. It matters if you lowering a ctd and measuring pressure at the ctd, while approaching close to the bottom which you know in meters of depth, say from a topographic chart. Hopefully you will have a pinger on your ctd which can act as a bottom finder by bouncing sound off the bottom and measuring travel time. • Attached is a map of dynamic height anomaly (~ his caption is ‘acceleration potential’) for the surface of the South Pacific, relative to a depth of 1000db x 2000 dbx. The contours of Φ’ are in units of m2sec-2 from J.Reid, Prog. Oceanography 1986, and his ‘adjusted steric height’ which tries to show the surface circulation, by accounting for the unknown reference velocity u(p=1000 db) using ocean tracer information. The subtropical gyre and circumpolar current appear stronger in the ‘adjusted’ map; what does this imply about the velocity field at 1000db? It says that there is a deep velocity in approximately the same direction as the surface flow, and roughly 1/3 as strong. See Reid’s Fig 22 below for his estimate of the total flow at 1000db (which should be the difference between his first two figs.)

4. (math techniques question). A mass-spring oscillator driven by a force sinusoidal in time obeys Ytt + K

2Y = Asinω0t with initial conditions Y=0 at t=0; Yt=0 at t=0. Solve this initial value problem, paying attention to the resonance that occurs when ω0 is near K. Textbooks often say that this equation has solutions of the form Bsin ω0t when ω0 ≠ K and C t sin ω0t when ω0 = K. Your solution should clarify this peculiar statement. Done in class…homogenous + particular; free + forced.