1. Atmospheric & Oceanic Processes Lecture 7: Taylor Columns and Large-Scale Eddies in Atmosphere and Ocean

2. Dye distributions from Lab experiment
Dye distributions from Lab experiment *Lecture01: on the left we see a pattern from dyes (colored red and green) stirred into a nonrotating fluid in which the turbulence is three- dimensional; on the right we see dye patterns obtained in a rotating fluid in which the turbulence occurs in planes perpendicular to the rotation axis and is thus two-dimensional.

3. Dye-dispersion in non-rotating fluid (WWWLink)
In the non-rotating tank the dye disperses much as we might intuitively expect - have a look at the pictures and movie loop below. Note that the we can also see a side view on the left-hand side of the images below obtained using a mirror sloped at 45 degrees.
Dye-dispersion in non-rotating fluid (WWWLink)
In the rotating body of water, by contrast, something glorious happens - have a look at the images and the movie loop below. We see beautiful vertical streaks of dye falling vertically; the vertical streaks become drawn out by horizontal fluid motion in to vertical `curtains' which wrap around one-another.  The vertical columns - called `Taylor Columns' after G.I. Taylor who discovered them - are a result of the rigidity imparted to the fluid by the rotation of the tank. The water moves around in columns which are aligned parallel to the rotation vector. Since the rotation vector is directed upward, the columns are vertical. Thus we see that rotating fluids are not really like fluids at all! Note that the movie is recorded in the frame of reference of the tank - i.e. by a camera mounted above the rotating table, rotating at exactly the same speed.
Dye-dispersion in rotating fluid (WWWLink)

4. We previously (Lecture05) derived the equation of motion on a rotating earth:
Du/Dt = -2Ω×u - p/ - () + Ffric, where  = gz - Ω2.r2/2.
For barotropic fluid:  = Function of “p” only. So dp/ = dG(p), i.e. G = ∫dp/; then:
Du/Dt = -2Ω×u - (G(p)+) + (u/z)/z
Compare the magnitudes of various terms in the above equation of motion and derive the appropriate parameters Rossby#  = U/(fL) and Ekman# Ev = 2/(fD2), where U, L, D and f (=2|Ω|) are scales of velocity, horizontal length, vertical length and rotation rate, respectively.
For small Rossby & Ekman numbers, then:
2Ω×u + (G(p)+) = 0
Taking curl (i.e. ×; noting that ×(a×b) = a.b - b.a + b.a – a.b)
Ω. (u) = 0  u/z = 0.
This is the Taylor-Proudman Theorem: Slow, steady, frictionless flow of a barotropic fluid—the velocity u, both horizontal and vertical components, cannot vary in the direction of the rotation vector Ω. In other words the flow is two-dimensional.
Homework #7.1

5. The Taylor-Proudman theorem states that slow, steady, frictionless flow of a barotropic, incompressible fluid is two-dimensional and does not vary in the direction of the rotation vector Ω.

6. The T-P theorem demands that vertical columns of fluid move along contours of constant
fluid depth because, they cannot be stretched in the direction of Ω. Thus fluid columns act as
if they were rigid columns and move along contours of constant fluid depth. Horizontal flow is thus deflected as if the obstacle extended through the whole depth of the fluid.

7. GFD7_taylor (WWWLink)
Paper dots on the surface of the fluid shown in the experiment. The dots move around, but not over, a submerged obstacle.

8. . z x y

10. GFD8_gravity_rotation (WWWLink)

11. y z
PE’s (per unit volume) before and after the exchange are:
PEbef = 1gz1 + 2gz PEaft = 2gz1 + 1gz2
The change in PE is then: PE = PEaft – PEbef = g {(2-1)z1 – (2-1)z2}.
Therefore, PE = -g (2-1) (z2-z1) = o (b2-b1) (z2-z1), (*)
where b = -g/o. But, the density gradients of the surrounding environment are:
M2 = b/y and N2 = b/z.
Therefore: b2 – b1  M2.y + N2.z = N2y.[(M2/N2) + (z/y)], (**)
where y = (y2-y1), and z = (z2-z1).
Clearly, z/y = s = slope of path of parcel-exchange. (***)
What is M2/N2 ? Consider the red line where , i.e. b = constant, called an isopycnal.
Then on this line, b = 0 = b/z.z + b/y.y = N2.z + M2.y
Therefore (z/y) = sb = -M2/N2 is the slope of the isopycnal of the surrounding.
Then, (**) gives: b2 – b1 = N2y.[- sb + s], and
(*) w/(***) gives: PE = o N2 (-sb + s)s y2 .
If there is to be an instability, PE < 0. Since sb > 0, we then have: sb > s > 0 for instability
The maximum release of PE, i.e. Max|PE|, can be calculated, it occurs when s = sb/2.
This kind of instability is called “baroclinic instability”, it is common in atmosphere & ocean. The resulting exchange of fluid reduces horizontal density difference but stratify it vertically.

12. Homework #6.5
Show by looking at the y-momentum equation that du/dz is positive so that the surface current at day 3 is eastward.
Day 3
Day 10

14. The temperature, T, on the 500-mbar surface at 12 GMT on June 21, 2003
The temperature, T, on the 500-mbar surface at 12 GMT on June 21, The contour interval is 2◦C. A region of pronounced temperature contrast separates warm air (pink) from cold air (blue). The coldest temperatures over the pole get as low as −32◦C.

15. A cross section of zonal wind, u (color-scale, green indicating away from us and brown toward us, and thin contours every 5ms−1), and temperature, T (thick contours every 5◦C ), through the atmosphere at 80◦ W, extending from 20◦ N to 70◦ N, on June 21, 2003, at 12 GMT, as marked in previous slide. Note that ∂u/∂p < 0 in regions where ∂T/∂y < 0 and vice versa.