The Atmosphere, the Ocean, and Environmental Change

GG 140 - Lecture 21 - Ocean Currents

Chapter 1: Review of Exam 2 [00:00:00]

Professor Ron Smith: I’m going to go over the exam first thing here. OK so problem one, I show you a top view of a Focault pendulum. So imagine you’re looking down at this thing. And the bob is swinging back and forth on this line. So just kind of looking down like this. I asked you to sketch the forces on the bob and explain how the track will rotate if the pendulum is in the Southern Hemisphere. So in the Southern Hemisphere, Coriolis force acts to the left of the motion vector.

So when that bob is moving upwards on this blackboard, the Coriolis force is going to be in that direction. And when it’s moving in the other direction, the Coriolis force is going to be like that. So the Coriolis force shifts depending on the direction the bob is moving.

So what is that going to do to the plane in which the bob is swinging? Well this part when it’s swinging that way it’s going to deflect it a little bit like that. And then when it’s swinging back and this force acts on it, it’s going to deflect it a little bit like that. And then that’s going to repeat that way and then that way and then that way. So basically the plane in which the bob is rotating is going to rotate in the counterclockwise direction.

So if you came back several hours later and it started out here, now it’s going to be oscillating like that. And, of course, if we were in the Northern Hemisphere, it would be the opposite. It would rotate in the clockwise direction.

Questions on that? Yes.

Student: What does CCW mean?

Professor Ron Smith: Counterclockwise. Is that something the TA put on there?

Student: Yeah.

Professor Ron Smith: Yeah.

Student: I said it was to the left.

Professor Ron Smith: Yeah. See left is a little bit ambiguous because what does left mean here? I don’t know.

OK, question two. Explain how cool winds and a gust wind are created by a thunderstorm. So the basic idea there is you get rain coming out of the base of the cloud. You know that below the cloud, the relative humidity is less than 100%. So as soon as that rain falls below cloud base, it’s going to begin to evaporate. Now maybe only a small fraction will evaporate, maybe all of it will evaporate.

I remember when I used to live in Colorado, you’d see–in the summertime you’d see thunderstorms, deep convective storms, with the rain coming out of the base, and it would all evaporate. None of it reached the surface of the Earth. When you see that, by the way, when it doesn’t reach the surface here, it’s called virga.

But in any case, whether it all evaporates or just some of it, it takes heat to evaporate water, and so—and that heat comes out of the air. So there’s a cooling that takes place, an evaporative cooling that takes place just below the cloud base, and that takes air and makes it cooler and more dense. So it then falls out of the sky until it hits the surface, and then spreads out. And the front of that is called the gust front.

So the basic idea here is that evaporation of raindrops below the cloud base cools the air, makes it sink and spread out. So that’s the idea behind–if you’ve experienced this in the summertime, you hear a thunderstorm in the distance, and then for a few minutes later the wind begins to pick up. It’ll be a cool wind, typically, and blowing from the direction of the thunderstorm. That’s this. That’s that cool air coming down from the base of the cloud.

Questions on that?

Question three. I give you 45 degrees north. There’s a pressure gradient, 0.002 Pascals per meter, with pressure increasing towards the west. So on a map view, north, south, east, west, the question states that the pressure is greater towards the west because it’s high pressure over here relative to low pressure over there. The lines of constant pressure, if you drew them in, the isobars would be oriented north-south like that.

An air parcel sitting in there is going to feel a pressure gradient acting from high to low. So that’s the pressure gradient force. And if we can assume that the air is in geostrophic balance, the Coriolis force must be equal and opposite to that.

We’re in the Northern Hemisphere, 45 degrees north latitude. So if the Coriolis force is like that, the wind must be such that the Coriolis force is at right angles to the wind, and to the right of the wind. So there’s only one possible solution for the wind, and that would be that vector. So that the Coriolis force is to the right of the wind motion. So, well I’m getting ahead of the story, this is part B. But basically, the velocity direction would be towards the south. You could state that as a northerly wind if you wanted, but the direction will be towards the south.

Now, the magnitude of the geostrophic wind is given by the pressure gradient divided by 2ρΩsinϕ. So let’s put in some numbers on that. It’s 0.002, 2–the density of air at sea level is about 1.2 kilograms per cubic meter. The rotation rate of the Earth is 7.27 times 10^-7. And the sine of 45 degrees is about 0.71. So that turns out to be 16.2 meters per second.

I would recommend–you notice I left off the units, but I don’t recommend that. If you put the units for everything, Pascals per meter, kilograms per cubic meter, seconds to the minus 1, you can work it out and you should get units of speed meters per second. And if you don’t, maybe you’ve left out something. Maybe you’ve got the units wrong or you’ve left out a factor or something like that. So it’s always good to check the units on that.

For part C, again, what you’ve assumed is geostrophic balance, which says that the Coriolis force and the pressure gradient force are equal and opposite. That is the definition of geostrophic balance.

Yeah.

Student: Is the rotation rate 10^-7 or 10^-5?[INAUDIBLE]

Professor Ron Smith: 10 to the negative–sorry, this is wrong. 10^-5. Thank you.

If you ever–I mean I gave you that constant on the equation sheet, but if you ever need it and can’t remember it, just take 2 pi, which is the number of radians in a circle, and divide it by the length of a day expressed in seconds. And you’ll get that number, 7.27 times 10 to the–so it’s just the rotation rate of the Earth basically, expressed in radians per second.

Four. Explain why the sky appears blue, but a cloud appears white under similar illumination from the Sun. So the idea is we have our observer. We have a cloud here. And the Sun is illuminating the cloud, but is also seeing radiation that’s been scattered–I don’t have to draw another beam there. Some light is scattered out of this beam coming directly to the observer, and that’s blue light. But this is white light.

So the difference has to do with the size of the particles that are doing the scattering. In the case of the sky, the particles are molecules. And the condition that the wavelength is much, much greater than the diameter of the particle is met, and that puts us into the Rayleigh scattering regime.

In that case, short wavelengths are scattered much more strongly than longer wavelengths. Remember, the part of the spectrum we can see with our eye, the visible part of the spectrum has red, green and blue. Blue being the shorter. Red being the longer wavelength. So in Rayleigh scattering where short wavelengths are scattered more strongly, that’ll be the blue light that’s scattered more than the green and the red. So this light scattered out of the Sun’s beam to your eye by molecules is going to be dominated by the blue light.

For the cloud itself, the particles are much larger. They’re the cloud droplets. They fall into the category that the wavelength is the same order of magnitude as the cloud droplet. Which puts us into the Mie scattering range where all wavelengths are scattered equally. So whatever color was illuminating this, you’d have the same color coming out. So if it’s white light illuminating the cloud, and the Sun’s radiation is usually composed of roughly equal mixtures of blue, green and red, so that would appear white to our eye. Then the light scattered by the cloud is also going to be white.

By the way, if you have a setting Sun illuminating the cloud where some of the blue light has already been scattered out, it may be that the cloud is illuminated by red light. In that case, the cloud would appear red as well. But the point here is that under Mie scattering, it scatters equally whatever is illuminating the cloud. So it’ll keep the color the same as it scatters radiation from the cloud.

Questions there?

Question five. Why are hurricanes not found over the sea near the equator? Well right near the equator within four or five degrees of the equator, there’s not sufficient Coriolis force. In order to form a hurricane, as the air moves in and begins to lift upwards, there’s got to be a Coriolis force to give it a sense of spin. And if you don’t have a sufficient Coriolis force, then you can’t really form a hurricane.

Part B, in the tropical south Atlantic. So let’s assume you’re south of the Equator enough to have a Coriolis force, but it turns out in the Atlantic Ocean, there’s a cold current that comes up from the southern ocean that drops the ocean temperature below that threshold value–it’s 27, 28 degrees Celsius, and therefore, you don’t have the ocean warmth to create a hurricane.

Questions there?

Question six I thought was straightforward. Basically is why water drops form on the outside of a cool glass of water? So you’ve got cold water or cool water in there at some temperature. And you want to know when beads of water would form on the outside. Well, of course the water is not coming from inside the glass through the glass, that water is coming from the atmosphere around. The only role of the water in the glass is to control the temperature of the outer surface of the glass.

And the key condition is that the temperature of that glass, if it’s less than the dew point of the atmosphere around it, then you will get condensation. The idea is that the glass will remove heat from the air adjacent to it, drop the saturation vapor pressure down, and if it can drop it low enough, it can drop the temperature down below the dew point, then you’ll bring water out of the vapor state and condense it on the outside of the glass. So I was looking for a clear explanation of that.

Question seven. How do the raindrops form that we find falling from a tall cumulonimbus cloud? In the summertime you get tall cumulus clouds with heavy rain out the bottom. The fact that they’re tall means that the top part of it is certainly going to be at a temperature lower than zero degrees Celsius. So there’s going to be super cooled water up here. And that’s going to be the key for generating precipitation.

So using the ice phase mechanism then you can convert these to snowflakes. They’ll fall, when they fall below the zero degree line, they’ll melt and form raindrops and then they’ll fall to Earth. So what you had to say there was to mention, at least mention, the ice phase mechanism because that’s how the hydrometer would be formed. And then you had to describe how it would melt on the way down to form a raindrop.

Question 8. I imagine on the Earth some air moving northward with a temperature of 20 degrees C, and some air moving southward with a degree of—with a temperature of 10 degrees C. And I asked you to compute how much heat is being transported northward.

The idea there is that the heat transported is given by the rate–I’ll put a dot over it to indicate it’s a rate–the rate at which you’re moving mass forward times and heat capacity of the air times–well, if it’s a two stream difference and you want to compute the net amount of heat, then you’d use the temperature differences between the two streams. That’s the way I did the problem.

So my answer for this was 10¹¹ kilograms per second for the mass flow rate, and that was given. The heat capacity of air is 1,004, and that has units of Joules per kilogram per degree Kelvin. And then the temperature difference between the two streams, the ΔT was 10 degrees. And look how the units are going to work there. This is going to be in Kelvin or Celsius, in either case it’ll cancel. The kilograms will cancel, and you get something with Joules per second, which is a watt, by the way.

So I ended up with–what did I get–1.5x10¹⁷–oh sorry. Let’s see, about 1.004x10¹⁵ watts, which could be expressed as about 1.004 petawatts. Peta meaning the shortcut for 10¹⁵. So it’s approximately 1 times 10¹⁵ watts.

Now, there was some confusion because while I described these two air masses moving northward and southward, in the end I said how much heat is transported northward. What I meant to say, what I hoped you would interpret that is the NET amount transported northwards. And that’s what I’ve computed here. It may be, though, that some of you computed just the amount of heat transported northward by the northward moving current. Maybe that could be supported given by the way I wrote the question.

But what I intended was the net heat given the fact that there’s both a northward and a southward block of air moving. So look at the grading and see what the TAs did about that.

Questions there?

We’re getting close to the end here.

Question nine says that on a rainy day a centimeter of rain falls on a 10,000 square kilometer area. Estimate the total latent heat. So you imagine some big cloud system raining out, putting a layer of water on the surface of the Earth. You just have to compute the volume of that layer. Then from the volume multiplied by the density of fresh water to get the mass of that layer. And then knowing the latent heat of condensation, you know how much heat had to be released in order to condense that much vapor to a liquid.

So from the mass you just multiply that by the latent heat of condensation. So the dimensions of this give the volume, the mass–the density of water, which is 1,000 kilograms per cubic meter gives you the mass. And then the latent heat of condensation gives you the rest. And the answer is about 1.5x10¹⁷ Joules.

Student: 2.5.

Professor Ron Smith: Sorry, 2.5. Thank you. That 2.5 comes from the latent heat. Thank you.

The last problem then was to explain the reason for the rainy season in these two cities. Jerusalem lies here. And Asuncion, Paraguay, except it shifted in longitude occurs somewhere here.

Now, Jerusalem has a–it’s in the Northern Hemisphere. Wettest month is January, and notice from the temperatures, that’s also the cooler month. So Jerusalem gets a wintertime rainy season, and that is clearly due to the southward shift of the polar front. So what’s happened is that during the winter season, this polar front has moved southward and is giving you frontal cyclones that come through that latitude and give you the rain.

In Asuncion, Paraguay, Southern Hemisphere, the wettest month is December. In the Southern Hemisphere that’s summertime. Notice also from the temperature that that’s the warmer month. So you know immediately that is a summertime precipitation maximum. And that guarantees that it’s going to be a southward shift of the ITCZ, giving you precipitation there. So the ITCZ in the Southern Hemisphere summer moves across the Equator into the Southern Hemisphere and brings precipitation to Asuncion.

So that’s going to be convective rain, not frontal rain, and it’s going to be in the summer wet season.

Any questions on that? The average grade on that exam was a bit lower than the first one. It was 74%. And if you got maybe 10 degrees—10 degrees—10 points lower than that, maybe drop me a note, come see me, we’ll talk about how to improve things in the future. But for some reason, the grades were a little bit lower on this than the other one. And take that into account when you’re–don’t be overly critical of yourself, because apparently your classmates found that to be a tough exam as well.

We’re going to continue where we left off. Question, yes?

Student: I was wondering what the average score corresponds to in grade?

Professor Ron Smith: I don’t do that calculation until the end of the course. I keep them just as numerical scores until the end of the course. You can judge that for yourself, perhaps, based on the average grade. But that’s not a calculation that I do until I get all the scores at the end.

Chapter 2: Atmospheric Forcing of the Ocean: Wind Stress [00:23:54]

So we’re talking about atmospheric forcing of the ocean, and I showed this slide last time. I’ll just go back over it. Basically the ocean is driven by the atmosphere above it in these three ways. Heat added, and taken away, fresh water added and taken away, and the wind stress. We can quantify this, and recent investigators have tried to do this by producing maps, in this case, of the heat flux in and out of the ocean, or the water flux in or out of the ocean, or the wind stress applied to the ocean surface. So we know a bit about this forcing.

And last time I derived some formulas for this, how a layer of depth D, which will feel the direct effect of this forcing, would respond to fresh water, heat, or wind stress. I won’t go back over those derivations, but for the wind stress, which is I think the most difficult of the three, the concept involved this Ekman layer. When the wind blows over the ocean and puts a frictional stress on it, the first few tens of meters or hundreds of meters responds to that directly, but in an odd way.

It moves off–well, it has a spiral with depth. I’m not going to talk about that. But if you average that out, the net of all that motion is directly to the right of the wind stress. That’s this big arrow here. That’s the direction of Ekman transport, exactly at right angles to the wind stress that’s being applied.

So if you had a wind, say, from the southwest to the northeast, giving you wind stress in that direction–that’s the black arrow–and you gave it a few hours to come into this new state of balance called the Ekman force balance, that force would be balanced by the Coriolis force, and if that’s going to be the Coriolis force, then the Ekman transport must be in that direction. So the CF is at right angles to the net Ekman transport.

Now, this is not identical with the geostrophic balance. Remember, geostrophic balance was a balance between Coriolis force and pressure gradient force. I have no pressure gradient force in this problem. This instead is a balance between a frictional stress being applied to the ocean, and the Coriolis force. But it’s very real. I mean you can measure this easily in the ocean. If the wind is blowing briskly from one direction you can measure the surface water moving off in the right direction.

I’ll mention this in a week or so. The same thing applies to icebergs. When the wind blows on an iceberg, it moves at right angles to that force. The old sea captains would make notes of that in their log. Winds are from the south today, but the icebergs are moving off to the east. What’s going on there? Well, you guys know, it has to do with this kind of a balance including the Coriolis force.

So again, not the same as geostrophic balance, but they both involve the Coriolis force in an important way. These are the formulas that I derived last time. If you added an amount of heat per unit area Q over A, you can compute how the temperature of some depth D will change. If you add a layer of fresh water to the ocean of thickness little d, and then you mix that in to some layer of depth D, there’s how the salinity will change. If you put a wind stress τ on the ocean, and that is mixed down to depth d, that’s what the speed will be of that Ekman flow that goes off at right angles. Yeah.

Student: In the Ekman wind stress, is the rho–in the denominator,is that the density of water or air?

Professor Ron Smith: Correct. Correct. Water. I left off the subscript, but that should be water. when you’re computing tau, however, I gave you a formula for τ. It was ρ u squared. That was the density of air, and the speed of the wind. So in the formula for τ that I gave you, it’s the density of air that appears there. But down to the bottom here it’s the density of sea water.

Student: So they don’t divide out?

Professor Ron Smith: No because there a factor of a thousand different.

Student: OK.

Professor Ron Smith: The density of air at sea level is about 1.2 kilograms per cubic meter. The density of sea water’s about 1,000 kilograms per cubic meter. Yeah.

Student: Just real quick again, the little d versus big D, the big D is the depth and–

Professor Ron Smith: D–in all cases, D is this layer that is feeling the direct influence of these forcings. And it’s hard to predict exactly what that’ll be. It’ll depend on how much turbulence there is, how deep things are going to get mixed. But it’s something very small compared to the total ocean depth. It may be only a few tens of meters if the winds are weak, and there’s not much turbulence. It could be 100 meters or 200 meters if the winds are very strong and you’re mixing the ocean very vigorously. So I can’t give you a fixed value for d, it’ll depend on conditions. Good questions there.

Chapter 3: Thermohaline Currents [00:29:38]

So if those are our three forcings, then what do they do to the ocean? Well the first two, the heat and the fresh water input, will modify the temperature and the salinity. The temperature and salinity will in turn change the density of the seawater. And then gravity will act on those density differences to produce currents down in the ocean. And those currents are called thermohaline currents. The name reminds you is they have to do with heat and salt. But they’re acting through the role of heat and salt in controlling seawater density.

Typically these currents are slow and deep. As I will show you in the next few minutes, they are responsible for the water mass distributions in the ocean, for the conveyor belt circulation, and for example, also for estuary circulation. So these are very important circulations in the ocean, even though they tend to be rather slow in the speed of their motions. These are the thermohaline currents. So let’s take a look at the Atlantic Ocean along this section of 25 west longitude. Here is a temperature section given where this goes north to south, sea level down to 6,000 meters. And the black is the terrain.

Now, the first thing that strikes you about this image perhaps is the steepness of this terrain. Oh my goodness, are there really these sharp spikes and cliffs in the ocean? And the answer is please remember that when you make a diagram like this, there is a lot of vertical exaggeration. For example, that distance is 5 kilometers or 6 kilometers, whereas this distance is something like 15,000 kilometers.

So there is really an enormous vertical exaggeration. Where I could squish this down or stretch this out, these supposedly steep slopes would become very, very gradual. Probably only 1% or 2% slopes. So don’t be fooled by this. It’s only the vertical exaggeration that gives you that apparent steepness.

Now, for the temperature itself, you can’t read the labels very well, but some of these are slightly negative, temperature of minus 0.02—sorry 0.2 or 0.4. If I’ve got a temperature of minus 0.4 Celsius, is that super cooled water? Is that super cooled water? How many think yes? Show of hands. How many think no? The danger of putting no up, I’m going to ask you why. Why don’t you think that’s super cooled water?

Student: Is it because it has salt in it?

Professor Ron Smith: Yes. So it’s not super cooled water because the salt actually depresses the sea level by a degree or two. So in fact, you can have a temperature down to maybe minus 1 or 1 1/2 degrees Celsius, and still be above the freezing point because the salt has suppressed the freezing point slightly. So this is not super cooled water. It’s not about to freeze if you introduce some kind of a freezing nucleus to it or something.

I want you to notice this band of blue coming down here. We’re going to identify that as Antarctic Bottom Water. It’s very cold water that’s formed where the ice sheets from Antarctica float out over the ocean and cool the ocean down. And that water becomes so cold and so dense that it falls to the bottom of the ocean and then spreads northward. In this diagram it reaches the Equator.

Depending–in some other sections, it actually gets a little bit north of the Equator. This is amazing. So a water mass formed under the ice shelves of Antarctica falls to the bottom of the ocean, and then under its own weight slowly moves, sinking and spreading, actually in some cases getting north of the Equator. That’s Antarctic Bottom Water, AABW. One of the important water masses in the Atlantic Ocean is the AABW.

There is a layer of warm water floating right on the top in the tropical and equatorial regions. And there are some other water masses up in here that will be a little bit easier to identify when we look at the salinity field. So let me look at the salinity field here. Atlantic salinity, and 34.7, 34.9, and so on.

The Antarctic Bottom Water you can maybe see it there, but it’s not as easily seen in the temperature field. But look at this thing. There’s a tongue. Apparently, a water mass is being formed here in the Antarctic Ocean–so sorry, in the southern Ocean. Not close to the shores of Antarctica, but in the Southern Ocean. It sinks and then it spreads northwards as a tongue, again, reaching the Equator. Actually going a little bit north of the Equator. That’s called the Antarctic Intermediate Water, AAIW.

There’s also a water mass here. It’s formed in the North Atlantic. It sinks and it spreads southward getting south of the Equator. That’s the North Atlantic Deep Water, NADW, North Atlantic Deep Water.

So what’s going on in all of these cases? At the surface of the ocean where the atmosphere and the ocean touch, the atmosphere is imprinting a certain temperature, salinity characteristic to the water. That determines the water’s density. That water will then sink down to its appropriate density level and spread out. That was a case for the Antarctic Bottom Water, the Antarctic Intermediate Water, and the north Atlantic Deep Water.

Notice, their roots all go back to the surface of the ocean, because that’s where they can get their properties. They get the temperature from the heat fluxes, they get their salinity from the fresh water fluxes, and then they sink and they spread out and they’re found thousands of kilometers away from where they were formed, and moving very slowly there under the direction of these currents.

Oxygen doesn’t play any role in how these water masses move, but they provide a good tracer. Here’s dissolved oxygen, and you find the lowest values here. And notice this curious structure. Right at the ocean surface, of course, you find the highest values of oxygen. Well, the oxygen is being taken in from the atmosphere, mixed down a couple of hundred meters, and then following that water mass–here it is following the north Atlantic Deep Water, here it is following the Antarctic Intermediate Water right there.

But surprisingly, the lowest oxygen is in the tropics, just below the–not touching the atmosphere, but just below it. So there’s not a very good communication or mixing between the surface of the ocean and this water or it wouldn’t have that depleted oxygen. The reason why there is not strong communication there is because it’s very stable. There’s a thermocline there, just like an inversion in the atmosphere, prevents vertical mixing. So you get this strongly depleted oxygen water here, so close to its oxygen source, but yet cut off by this stable inversion preventing vertical mixing. So we can–

Yeah, question.

Student: Is the oxygen depleted there because organisms use it up?

Professor Ron Smith: Organisms use–as soon as the water leaves–I should have said this, thanks Julia–as soon as the water leaves the surface it begins to lose oxygen because animals use it up, and respiration takes place to use up the oxygen. So the older the water is, the less oxygen it has in it.

Student: And so the fact that the oxygen can mix more deeply in other regions of the world, does that mean that it won’t be used up as–

Professor Ron Smith: No, it might mean that, but it might just mean that that water has been down there a longer period of time.

So the cartoon then of the Atlantic Ocean, the same north-south section I’ve been telling you about, south to north, looks like this. You’ve got this warm surface water heated by the Sun kind of floating, it’s quite shallow. You’ve got the Antarctic Intermediate Water coming in here, you’ve got the North Atlantic Deep Water, and then you’ve got the Antarctic Bottom Water. Eventually they lose their characteristics. They mix in with their environments, but for at least hundreds, maybe thousands of kilometers, during their transport, they retain enough of their property that we can identify them by where their source region was.

So this is the concept of the water mass. In oceanography we have this concept of the water mass. It’s a mass of water obtaining its temperature, salinity properties at the air-sea boundary, and then sinking and moving slowly, and for years retaining that property until eventually it mixes in with its environment. And these motions are driven by these slow thermohaline-driven currents.

Questions on this?

So this is the concept of the water mass in oceanography. It’s a bit like the concept of the air mass in meteorology where on a day like this, cool, brisk, I can say maybe this is a Canadian air mass. It got its dryness, its cool temperatures in Canada, now it’s moved down to me in New England, but I know from its properties where it came from. Whereas a couple days from now if it’s really moist and humid and warm, likely that’s Caribbean air coming up from the Caribbean transported to my location, retaining some of the properties it acquired when it was down in the lower latitudes. So the concepts are roughly similar to each other.

Now, I wanted to show you a famous thermohaline current called the conveyor belt circulation. I wanted to show this diagram again–you’ve seen it before–to remind you that the Atlantic Ocean is slightly saltier than the Pacific, and slightly cooler than the Pacific, mostly just saltier. But as a result, the Atlantic Ocean–look at the density numbers–the Atlantic Ocean water is on average slightly denser than the Pacific Ocean.

So imagine you’ve got these two oceans, OK, they’re separated by North and South America, but they have different densities. If there’s any leak between the two of them, that’s going to try to equilibrate by setting up a circulation between them, and that exactly is what we think happens to give this large scale conveyor belt circulation. Notice that in the Pacific Ocean, the water is less dense. It seems to rise and then flow into the Atlantic. The Atlantic Ocean being somewhat denser, this water sinks, and then flows on the bottom of the ocean back into the Pacific. So there’s kind of an overturning circulation.

Now OK, it has to go through a very circuitous route because they’re not directly connected. They can’t make their way across this barrier or across this barrier. But they do share the Southern Ocean in common. So they can communicate and close this circulation by running the currents down through the southern Ocean. That’s the conveyor belt circulation. Another example of a slow thermohaline circulation.

A third example is the estuary circulation. An estuary is defined as a semi-enclosed body of water with some connection to the ocean, and with fresh water inputs. Long Island Sound would be a good example of that. There are fresh water inputs–the Connecticut River, the Quinnipiac River that we studied, the Housatonic River. And there is a connection to the ocean that goes out through the race here so that you can mix ocean water in and out.

What we find is that while the ocean salinity is typically 35 parts per thousand, the salinity in Long Island Sound’s only about 20 parts per thousand. I think in your field trip you found 18, but generally that’s pretty close to this. So what’s happening is it’s a place where mixing occurs between fresh water coming off the land and ocean water mixing in from the ocean.

So remember, salty water is denser. Fresher water is less dense. So what happens is that the denser, fully salt ocean water slides in along the bottom, has fresh water added to it, making it less dense so it rises, and then it floats out on the top. So right through here there’s going to be a two-layer circulation. It’s illustrated up here with the fresh water being added, and that slightly reduced salinity water flowing out at the top, and the full ocean water flowing in at the bottom.

So you get a two-layer circulation right at the entrance region to an estuary, driven by this salinity difference. That would be called a thermohaline circulation. Does everybody get that? What’s driving it?

So if I turned off these rivers in some magical way, and let the mixing continue, after a few months, this would come up to 35 parts per thousand. It’s the fresh water inputs that’s keeping this lower than the 35. And it’s that two-layer circulation that’s providing the mixing between the estuary and the open ocean.

Questions there? Yeah.

Student: So it’s still called thermohaline, even though in this case it’s just driven by salinity?

Professor Ron Smith: True. Because you still got the haline part of it. So it would still be called thermohaline.

The Mediterranean Sea has one of these, but it works just in the opposite direction. Remember, the Mediterranean Sea is, especially in summertime, a very dry climate. It doesn’t rain there, and there’s an excess of evaporation over precipitation. Even when you include the rivers that flow into it, there’s an excess of evaporation over precipitation. So it becomes saltier than the ocean itself.

And there is a connection. It’s the Straits of Gibraltar. That little narrow connection between Spain and Morocco that allows some exchange to occur. Just remember now, in this case, the Mediterranean water, while it’s fairly warm, the salt outweighs that. It’s a denser water mass.

So here’s a cross section through the Straits of Gibraltar, and here’s the Mediterranean Sea in the ocean. So that dense, salty water flows out on the bottom, because it’s denser, and the ocean water at 35 parts per thousand flows in on the top. A two-layer circulation driven by the salt difference. Just remember, it’s opposite in a sense to the estuary that I showed you just before this.

This has a place in the history books. In both World Wars, the German U-boat captains knew about this, and the British had set up a sonar listening station at Gibraltar to listen for the engines of U-boats trying to get in and out of the Mediterranean Sea. But the U-boats captains realized they could come out to this place, drop down about 100 meters, turn off their engines, and then just drift silently past the British into the Mediterranean.

Then they could start their engines, sink a few ships, and when they wanted to come back home, they would come over here, sink a bit deeper, turn off their engines, and this current would take them back out into the Atlantic past the British sonar station. So this was known at the time, and it was used in a military fashion.

Questions on that?

Chapter 4: Wind Driven Currents [00:48:03]

Now, we’ve got to get to the wind driven currents. We won’t get very far on it today, but I wanted to alert you to the complexity of this subject. We’ve got to think of a chain of events that occurs as to how the wind generates the major ocean currents. And there are four bullets to this. The wind stress acts on the ocean as the frictional stress. That generates an Ekman layer, which moves water off at right angles to the wind stress.

That Ekman layer air flow then piles up water in certain parts of the world ocean, giving a little bit of elevation variation. We call that ocean topography. It’s not large, it’s only a few centimeters or tens of centimeters, but it’s very important. And then that ocean topography, through the hydrostatic law, sets up horizontal pressure gradients in the ocean. If the water’s heaped up here and a little lower here, you’re going to have higher pressure under the heaped up water, lower pressure under the lower water.

So you’re going to have a horizontal pressure gradient. If you have a horizontal pressure gradient, I think you know what’s going to happen. You’re going to set up a geostrophic current. A few hours later, you’re going to have a geostrophically balanced current, and that’s going to be an ocean current this time.

So that’s what we’re going to be looking for. So the winds over the ocean, for example, in the Atlantic Ocean, the winds in the tropical regions are from east to west. The Ekman flux will be towards the north. Up in mid-latitudes, the winds are westerlies. The Ekman flux is toward the south. The Ekman flow is going to pile up water in the middle of the Atlantic. And then with that high pressure, you’re going to generate an anti-cyclone in the way of ocean currents around that going in this same direction.

And for years and years, the theoretical oceanographers knew that that little variation in sea level had to be there. But finally about 15 years ago–well, more than that now–we were able to measure this for the first time by flying a satellite in orbit and having a laser point down from the satellite, bounce it off the ocean surface, time how long it takes to get that signal down and back up to the satellite. We were able to measure ocean height for the first time within a few centimeters of accuracy.

Then for the first time, we were able to actually make a map of this tiny little elevation in the Earth’s surface. And here it is. It’s called Ocean Dynamic Topography. It’s given in units of centimeters. And for example, when you see red here, the water is piled up about that much. So you wouldn’t ever notice this if you were sailing across the ocean. Remember, the ocean’s five kilometers deep. What does 40 centimeters mean? But this is relative to the geoid, relative to a constant level surface on the ocean. And it’s that piling up then that gives rise to a pressure gradient that then causes the major ocean currents in the ocean.

We’re out of time, but this will be the theme on Monday. We’ll talk about how these currents get moving.

[end of transcript]

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