Consider a western boundary current, perhaps the one you used last time. We will use the ECCO dataset, which I will preprocess for you to make it easy at fox-kemper.com/data

Choose one latitude, and make a section

- Confirm that the Sverdrup relation holds for the depth-averaged velocity, away from the western boundary current.
- Calculate the velocity shear across the latitude dv/dz.
- Calculate the result for dv/dz from the thermal wind relation, based on the potential density anomaly of the section. f dv/dz = - g/(rho_0) drhoanom/dx (Note: there is a small difference between d/dx of density and potential density called the thermobaric effect. This is neglected in this model.)
- How well do the two dv/dz estimates agree?
- Is the assumption of a level of no motion a good one? That is, is there any depth where v=0 at all longitudes?
- Take a look at the pressure anomaly field. Can you see the impact of the surface height displacements? What about the interior density variations? Does f v = dp/dx /rho_0 = dphihydid/dx make sense? Do you see a level of equal pressure (i.e., a level where phihyd is constant, so v=0, so level of no motion)?
Overall Objective: How can you understand aspects of the western boundary current: is it driven by surface height anomalies (eta), or thermal wind (rhoanom), or pressure (phihydid which has both eta and rhoanom effects), or the return flow of Sverdrup? What makes the most sense to you?

NOTES on rhoanom and phihydid******

Rhoanom is the *potential* density deviation from the background density (rhoanom=potlrho-rho_0). You can expect flow to generally be directed along surfaces of constant rhoanom. You can also understand the values at any depth as corresponding to one another, i.e., rhoanom=60kg/m^3 at 1000m would be denser if adiabatically rearranged to 500m than a parcel already at 500m with rhoanom=40kg/m^3. You can also expect horizontal gradients of rhoanom to correspond to roughtly to thermal wind dv/dz, except for thermobaric effects which should be small. The units are kg/m^3. Generally, you should expect rhoanom to increase with depth.phihydid is the perturbation pressure away from hydrostatic (in class we called it p'), which includes both the effects of surface height elevation and the accumulation with depth due to density by the hydrostatic relation on the perturbation pressure (rho-rho_0). Thus, by looking at phihydid, you don't have to mess with calculating the accumulating pressures. The hydrostatic equation relevant for phihydid is:

d/dz(phihydid)= - (rho-rho_0) g/(rho_0)

Thus, while phihydid is proportional to pressure, it has units of m^2/s^2, so multiply by rho_0 to get to Pascals if you like. Perhaps best of all is: phihydid*rho_0/10^4 which is the pressure in decibars, roughly equivalent to sea surface height displacements. Thus, at the surface you expect

phihydid*rho_0/10^4 \approx eta