We staled earlier thai seawaier densities used in the calculation of gcostrophic velocities are computed from data on temperature, salinity and depth. Of course, it is not depth per se that affects the density of a sample of sea water hut the pressure it is under, which is related to the depih by the hydrostatic equation for the purpose of making geostrophic calculations, it is usual to interconven hydrostatic pressure and depth by using the following approximate relationship:

pressure = lO^xdepth (3.14)

where pressure is in hewtons per square metre (N m"-') or pascals (I Pa -I N m'~). and depth is in metres

Can you see where the factor of Hi4 comes from?

Relationship 3.14 is simply the hydrostatic equation p - p.?-. in which it has been assumed that $ is exucth 10 m s~: t instead oí" M.S tn s"J| and that the density of seawater is a constant 1 x i (J1 kg m ' (rather than ranging between ) .025 and 1.029 x lO'kg m" 'l. This approximation is very useful for "back-of-envelope* calculations involving pressure at depth in the ocean, as we shall now demonstrate.

One possible source of error in geostrophic calculations is the variable pressure of the atmosphere. An oceanographic section may be of the order of 100 km across and it is quite likely that there will be a difference in atmospheric pressure between (he two hydrographic stations,

Would >uch a difference in atmospheric pressure be taken into account by Equation 3.13?

No. it would not. in deducing Equation 3.13 wc used the /m/msutic equation t3.8). which gives pressures resulting from the weight of overlying vrrj/f r only. Nevertheless, latera! variations in atmospheric pressure would contribute to any slope in ihe sea-surface und hence to the slopes of isobars at depth. As the ultimate purpose of Equation 3.13 is to obtain geostrophic current velocities from i sobarte slopes, it would be useful to know the ex ten) to which variations in atmospheric pressure affect these slopes.

The pressure resulting from a standard atmosphere is I bar. or 10* Pa. Given relationship 3.14 above, what depth of seawater would give rise to the same pressure?

The answer must be 10710J - 10m If a standard atmosphere is equivalent to Him of seawater. a difference in atmospheric pressure of a lew millibars -which is what we might expect over distances of the order of 100 km - would he equivalent to a few centimetres of seawater. Thus, variations in atmospheric pressure typically give rise to isobane slopes ol a few centimetres in 100 km or about I in 107. and hence contribute between about I and lOSir ol' the total horizontal pressure gradient (cf. end of Section 3.3.2)

Would geostrophic current How resulting from spatial variations in atmospheric pressure change with depth?

No. it would not. because it would be a slope current * and as such would be in the 'bamtropie' component of the flow (cf. Figure 3.18(b)). Thus, although the effects of variations in atmospheric pressure cannot be completely ignored, in theory they could easily be corrected for. However, weather systems may travel several hundred kilometres per day. so differences in atmospheric pressure between two stations are likely to fluctuate over the period in which measurements are made. In practice, therefore, the effect of atmospheric pressure on geostrophic current flow is difficult to take account of, and in most situations is considered sufficiently small to be ignored.

We deduced above that 10 m of seawater is equivalent to a pressure of 1 bar, i.e. that 1 m of seawater is equivalent to 1 decibar (dbar). For many purposes this is a very useful approximation. However, like all fluids, seawater is compressible, and for greater depths quite significant errors result from converting pressures in decibars directly to depths in metres. For example, a pressure of 11 240 dbar has been recorded at the bottom of the Marianas Trench, but soundings of the area indicate that the maximum depth is about 10880 m.

You have seen how. to determine geostrophic current velocities, we need to quantify departures of isobaric surfaces from the horizontal. But what does 'horizontal' really mean? The simplest answer is that a horizontal surface is any surface at right angles to a vertical plumb-line. i.e. a plumb-line hanging so that it is parallel to the direction in which the force of gravity acts. The reason for this apparently perverse approach is that the upper layers of the solid Earth are neither level nor uniformly dense, so that a surface over which gravitational potential energy is constant is not smooth but has a topography, with bumps and dips on a horizontal scale of tens to thousands of kilometres and with a relief of up to 200 m. If there were no currents, the sea-surface would be coincident with an equipotential surface. The particular equipotential surface that corresponds to the sea-surface of a hypothetical motionless ocean is known as the marine geoid.

In the context of geostrophic current flow, the important aspect of a 'horizontal' or equipotential surface is that the potential energy of a parcel of water moving over such a surface remains constant. If a parcel of water moves from one equipotential surface to another, it gains or loses potential energy, and the amount of potential energy gained or lost depends on the vertical distance moved and the value of g at the location in question.

Now imagine a situation in which a steady wind has been blowing sufficiently long for the slopes of isobaric and isopycnic surfaces to have adjusted so that there is a situation of geostrophic equilibrium. If the wind speed now increases up to another steady value, more energy is supplied to the upper ocean; the speed of the current increases and the slopes of isobaric and isopycnic surfaces become steeper - i.e. they depart even further from the horizontal, the position of least potential energy. In other words, the ocean gains potential energy as well as kinetic energy. You might like to think of this as being analogous to motorcyclists riding around a 'wall of death". As the motorcyclists travel faster and faster, so their circuits move higher and higher up the 'wall', thus increasing the potential energy of riders and machines.

Because of the relationship between isobaric slope and potential energy, departures of isobaric surfaces from the horizontal are often discussed in terms of 'dynamic height', i.e. vertical distances are quantified in terms of changes in potential energy (or 'work') rather than simply distance. The units of work that have been adopted for this purpose are known as 'dynamic metres' because they are numerically very close to actual metres. How closely dynamic metres and geometric metres correspond depends on the local value for g: if g is 9.80 m s-2. 1 dynamic metre is equivalent to 1.02 geometric metres.

Figure 3.21 The mean dynamic topography of the sea-surface relative to 1500 dbar. The contours are labelled in dynamic metres and values range from 4.4 to 6.4, corresponding to a relief of about 2 m. (Note: Close to certain land masses, the contours suggest flow 'coming out of the coastline; this is effectively an artefact arising from assumptions made in generating the contours.) This computed map should be compared with the Frontispiece, which shows the dynamic topography of the sea-surface in December 2000, as determined by satellite-borne radar altimeters.

You have effectively encountered dynamic height already, in Question 3.7(b). When you calculated the value of /zB - /¡a as 0.3 m, you could equally well have deduced that the difference in the dynamic height of isobar p\ between stations A and B was about 0.3 dynamic metres. Such variations in dynamic height are described as dynamic topography. Because of the way geostrophic calculations are made (cf. Question 3.7), dynamic heights are always given relative to some depth or pressure at which it is assumed, for the purposes of the calculations, that the isobaric surface is horizontal. In Question 3.7 and Figure 3.16(b), the depth was zo and the pressure po.

It is possible to determine the dynamic topography of any isobaric surface relative to another. The simplest situation to imagine is that in which the isobaric surface under consideration is the uppermost one - the sea-surface. Figure 3.21 shows the mean dynamic topography of the sea-surface relative to the 1500 dbar isobaric surface, for the world ocean. Clearly, maps like Figure 3.21 may be used to determine geostrophic current flow in the upper ocean.

1 low does the directum of geostrophic currcni flow re lute lo the contours of d>namic height?

The current flows at right angles to the slope of the isobaric surface, and therefore flows along the contours of dynamic height. The direction of flow will be such that the isobaric surface slopes up towards the right in the Northern Hemisphere and up towards the left in the Southern Hemisphere.

Bearing in mind Kqualinn _v I I (the gradient equation), can you suggest how the contours of dynamic height reflect the geostrophic current velocity 1

Figure 3.22 Topographie map of the mean sea-surface (i.e. the marine geoid), as determined using a satellite-borne radar altimeter. The mean sea-surface topography reflects the topography of the sea-floor rather than geostrophic current flow, as the effect of the latter is about two orders of magnitude smaller, even in regions of strong current flow.

Note that uncertainties in the shape of the geoid are the biggest source of possible error in estimates of dynamic topography determined by satellite altimetry (e.g. as in the Frontispiece).

The greater the geostrophic current velocity, the greater the isobaric slope, and - by analogy with ordinary topographic contours - the closer together the contours of dynamic height.

With these points in mind, try Question 3.8. QUESTION 3.8

(a) What is meant by 'the dynamic topography of the sea-surface relative to 1500 dbar"?

<b> Identify on Figure 3.21 the regions corresponding to flow in (i) the Gull Stream, and (ii) the Antarctic Circumpolar Current, referring to Figure 3.1 if necessary.

(c) Where is the fastest part of the Antarctic Circumpolar Current, according to Figure 3.21?

<d) At 120' F and about 65c S there is a closed contour corresponding to 4.4 dynamic metres. In the region of this contour, does ihe sea-surtace form a 'hill* or a depression? Given (he direction of current flow, is this what you would expect?

Figure 3.22 is a topographic map of the mean sea-surface as obtained by satellite altimetry. The mean sea-surface is a very close approximation to the marine geoid because even in regions where there are strong and relatively steady currents (e.g. the Gulf Stream or the Antarctic Circumpolar Current) the contribution to the sea-surface topography from current flow is only about one-hundredth of that resulting from variations in the underlying solid crust (cf. Figure 3.21). In a sense, the topographic surface in Figure 3.22 is the 'horizontal' surface, departures from which are shown by the topographic surface in Figure 3.21.

Figure 3.22 is a topographic map of the mean sea-surface as obtained by satellite altimetry. The mean sea-surface is a very close approximation to the marine geoid because even in regions where there are strong and relatively steady currents (e.g. the Gulf Stream or the Antarctic Circumpolar Current) the contribution to the sea-surface topography from current flow is only about one-hundredth of that resulting from variations in the underlying solid crust (cf. Figure 3.21). In a sense, the topographic surface in Figure 3.22 is the 'horizontal' surface, departures from which are shown by the topographic surface in Figure 3.21.

Was this article helpful?

## Post a comment