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Home , Coriolis force, Eddy (fluid dynamics), Equations of motion, Force, Sliding (motion)

CHAPTER FIVE

Equations of

The that describe the currents, , , , and other forms of motion in the are of a nonlinear nature for which there are no complete, exact, analytical . The best we can do is to with partial solutions, many of which provide excellent insight into the at work. In this chapter, we will examine the indi- vidual terms in the complete . In Chapter 6, we will examine a series of simplified equations that demonstrate important aspects of ocean circulation. In Chap- ters 9 and 10, we will use simplified forms of these same equations in our discussion of waves. Any quantitative discussion of forces and requires a . The system most commonly used in is the rectilinear, Cartesian system in which the is assumed to be flat. A spherical coordinate system would be more realistic, but it is also more complicated. The Cartesian system is adequate for most problems in physi- cal oceanography. The usual convention is to assume a plane in which the x axis points east, the y axis points north, and the z axis is up; more precisely, the z axis is in the direc- tion opposite the gravitational vector. Doing so makes the horizontal xy plane an equal surface (see ": Equal Potential Surfaces" later in this chapter for defini- tion). The corresponding components are u, v, and w. Although meteorologists and oceanographers may agree on the coordinate system in Figure 5.1, they use different conventions for describing and currents. A north cur- rent is a flowing toward the north; a north is a wind blowing from the north. The convention is confusing, but there is little likelihood that it will be changed. To mini- mize the confusion, this text refers to northerly winds and northward currents. 's law states that the the of a is pro- portional to the sum of the forces acting on the particle:

du_lvF dt Equations of Motion 81

FIVE

-x:-u:West x:u:East :ion

Figure 5.1 The Cartesian, flat earth coordinate system used in this text. ' fluid exact, In discussing fluid motion, the relationship is usually written which indi- (5.2) series :hap- where it is now understood that the forces are per unit volume, since Eq. (5.2) follows on of from Eq. (5.1): t. The vhich (5.3) – V c, but thysi- As written, Eqs. (5.1) and (5.2) apply to the components of the forces acting in the )oints east–west or x direction Similar equations can be written for the components acting lirec- along the other two axes: equal efini- —dt 10 2, r x m in dv 1 (5.4) I cur- dt p Y torth. dw i tnini- v F dt p z

: pro- There are four important forces acting on a fluid particle in the ocean: gravity, pres- sure , , and . In a generalized way, Eq. (5.4) may be written

(5.1) x = gravity + gradient + Coriolis + friction (5.5) 82 Equations of Motion

The mathematical expression for the forces of gravity, pressure, and Coriolis may be expressed simply. The various forms of the frictional forces are less easy to express in a precise manner, and they are considerably more difficult to measure in the ocean. The two problems are not unrelated. Note that by choosing a coordinate system such that the z axis is along the direction of gravity, there will be no gravitational force in either the x or y direction.

Acceleration Before looking at the various force terms, it is necessary to examine the acceleration of . Newton's second law is usually introduced in terms of particle (a block down an or the movement of billiard balls). As written, Eq. (5.4) applies to the motion of a particle. In (or ), there are two kinds of acceleration for which there are operational definitions, as the following example demonstrates. Consider the motion in a channel of constant depth but narrowing width (Fig- ure 5.2). The volume of water entering the channel is constant over , and, in the absence of turbulence, so is the rate of flow along the channel. A current meter suspended

A (a)

Velocity at B

Particle Velocity

— — Velocity at A

Time (b)

Figure 5.2 For steady flow within a channel, the water must flow faster as the channel narrows. Cur- rent meters at points A and B record a constant velocity and zero acceleration. (a) Because the chan- nel narrows, the velocity measured at point B is higher than at point A. (b) However, a particle moving along the channel is accelerated as it moves from A to B. The local acceleration is zero; the average particle acceleration between A and B is not.

Equations of Motion 83

at any point within the channel would measure a constant velocity (i.e., zero acceleration). However, if one could tag a water particle (perhaps by using a floating cork) and record its velocity as it traverses the channel, its would increase as a constant volume of water is forced to flow through a narrower channel. In this example, the local acceleration is zero; the particle acceleration is not. In many problems, it is desirable to write Eq. (5.4) in terms of the local acceleration rather than the particle acceleration. The two are related in the following way: particle acceleration = local acceleration + acceleration terms du du du du du dt dt+u—dx+v—dy+w—dz dv dv dv dv dv (5.6) dt dx dy dz dw aw aw dw dw dt dt u-d-x -"Ty +w-Tz or, in general, DE_ d a a d (5.7) Dt dt — Tt +17x +11—±}11dY dz a where we will follow the convention of a number of texts and use

D = d Dr dt to emphasize the distinguishing characteristics of acceleration of fluids. The motion of a particle is called Lagrangian motion. The flow past a point is called Eulerian motion. The derivation of Eq. (5.6) is given in Box 5.1.

Box 5.1 Acceleration The velocity of a fluid is not only a of time but also of : u = f(x,y,z,t) By the chain rule of differentiation, du au duds au dy au dz dt at ax di dy di az di (5.1') du du du du =—at+u—ax+v—dy+W-Tz

For emphasis, the total differential is often written Du du du du du Dr dt dt dx dy dz Note that D/Dt is the particle acceleration, and alat is the local acceleration. Thus, Eq. (5.1') can be written 84 Equations of Motion

Du du Dt dt = + uux + vu y + wux where we again adopt the notation that du ux —ax Similarly, Dv dv v, + uv x + vvy + wv, Dt dt Dw dw = wt +uwx+vwy+wwx Dt dt In vector notation, Du — = ut +(V V)u Dt Dv — = v, + (V • V)v Dt Dw — = v t + (V V)w Dt or, in general, DV– —+ (V. V)V (5.2') Dt

Pressure Gradient Of the various terms in Eq. (5.5), perhaps the is the easiest to visualize. A particle will move from high pressure to low pressure, and the acceleration is simply proportional to the pressure gradient. A mechanical analog is a ball on a frictionless inclined plane. The ball rolls down the plane (from high to low pressure), and the acceler- ation of the ball is proportional to the inclination of the plane (pressure gradient). In math- ematical terms, Eq. (5.5) now becomes (see Box 5.2 for derivation)

Du 1 — = ---dP + other forces Dt P ax Dv 1 dp — = — — — + other forces (5.8) Dt P aY Dw 1 — = — — —dP + other forces Dt P az Pressure arise in a variety of ways. One of the simplest is by a sloping water surface. Imagine a container with an ideal fluid (constant density, incompressible, and without ) whose density is pa and that in some manner it is possible to have the water surface slope as in Figure 5.3 without causing any other motion. Remembering Equations of Motion 85

1 2

Pi = Pagz P2 =Pa g (z +ha)

Figure 5.3 The slope of the surface creates a horizontal pressure gradient throughout the entire fluid. The pressure gradient is proportional to the slope of the sea surface.

that the pressure at any point in a motionless fluid is simply the of the fluid above—that is, the hydrostatic pressure, Eq. (1.2)—

Pi = PagZ (5.9) (5.2') P2 = Pa g( Z + AZ) The resulting pressure gradient term is

1 dp 1 p2 - Pa & Pa AX ualize. AZ ;imply = g — (5.10) AX onless ;celer- = gix math- where ix is the slope of the fluid surface in the x direction. It can be easily shown in a homogeneous fluid that the horizontal pressure gradient is identical everywhere within the fluid; the result is the same regardless of the length Z chosen in Figure 5.3. Thus, if there were no other forces acting, Eq. (5.8) says that the entire fluid in Figure 5.3 would be uniformly accelerated toward the lower pressure. (5.8) Box 5.2 Pressure Gradient

Consider a cube of fluid of density p with sides dr, dy, Liz, and let this element of fluid be in a channel where the pressure increases from left to right (i.e.,p 2 > p 1 (Figure 5.1'). Remembering oping that a pressure force is pressure times the cross-sectional , with the force vector acting sible, to the cross section, the force on the two sides of the cube would be have = PI AYAZ, F2 = P2AYAz ,ering 86 Equations of Motion

Figure 5.1'

Let the pressure p2 be slightly larger than pi:

P2 = P1 + AP The mass of the fluid element is simply the density times the volume:

m = piltdytiz Equating the acceleration of the cubic mass to the pressure forces,

Du = 1 4p Dt p By letting the cube become very small, one approaches the differential form

Du = 1 ap = 1 Dt p az pPx Note that F2 was given a negative sign because the force was directed in the —x direction. The meaning of the negative sign in the final is simply that the particle is accelerated from high pressure toward low pressure, which is in the direction opposite to the pressure gradient. A similar derivation can be done in the other two directions. Combined, these give

Dv 1 (5:3') Dt Dw= _— 1 Dt p PZ In vector notation, these become

(5.4') Equations of Motion 87

Coriolis Force The is the most difficult of the four forces to comprehend because physical intuition is of little avail. Most of us have some qualitative ideas of what to expect from the forces of gravity, pressure, and friction; but there is little in our experience to indicate what happens to a particle under the influence of the Coriolis force. The first thing to understand about the Coriolis force is that it is not a true force at all; rather, it is a device for compensating for the fact that the particle which is being acceler- ated by the forces of gravity, pressure, and friction is being accelerated on a rotating earth. However, the and we make of forces and particle movement use a reference system that is fixed. The reference system, and hence our observations, ignore the earth's . For most problems, that is an adequate assumption; the forces and are sufficiently large that the effect of the earth's rotation can be ignored. There are exceptions. Two examples will indicate the nature of the problem. The earth, with a of about 6400 km, rotates once every 24 h. It has tangential , as indicated in Figure 5.4. Let us assume that a constant-level balloon is set in motion at 40° N with a southward velocity of 1 m/s, and that no horizontal forces act on it. (There are vertical and gravity forces which are responsible for maintaining the balloon at constant level, but no horizontal forces.) Remember that the 1 nVs southward velocity is measured relative to the earth. In terms of a coordinate system that allows for earth rota- tion, the particle also has an eastward velocity at 40° N of 356 in/s. To an observer fixed in space, the true velocity would be almost due east, an eastward component of 356 m/s and a southward component of 1 m/s. According to Newton's first law, a particle in motion will continue to move at a con- stant velocity in the absence of any force. With a southward speed of 1 m/s, it will take about 30 h for the balloon to pass an earth observer at 39° N. The earth's tangential veloc- ity at 39° N is about 362 m/s. To the space observer nothing would be amiss, but to the earth observer it would appear that the balloon was not only going south at 1 m/s but, in the absence of any horizontal forces, it had somehow picked up a westward component of

Tangential Path of Particle Apparent Path (m/s) on Nonrotating of Particle on Earth Rotating Earth

Figure 5.4 Because of the change in the tangential speed of the earth with , a particle mov- ing toward the appears to an earth observer to be accelerated to the west. 88 Equations of Motion

velocity of about 6 m/s, the difference between 362 and 356 m/s. The "force" that must be applied to account for the apparent westward acceleration is the Coriolis force. For a second example, consider a suspended at the North Pole and free to swing in any direction. Assume that at 12 noon it is set in motion such that it is swinging along the 900 E/90° W longitude axis (Figure 5.5). In the absence of other forces, it will continue to swing in the same direction as the earth rotates under it. To an observer looking down on the North Pole, the earth rotates counterclockwise 15°/h with respect to the pen- dulum, which is not rotating. To an observer on earth standing near the pole, it is not the earth that appears to rotate but the pendulum, which rotates clockwise 15°/h. In 6 h, the pendulum would be rotating along the 180° E—W axis, some 90° from where it began. In 12 h, the pendulum will have rotated such that it is again swinging along the 90° E/90° W axis. A similar result occurs at the South Pole, but for an observer in space looking down on the South Pole, the earth appears to be rotating clockwise with respect to the pendulum. Thus to an observer on the earth standing near the pole, the pendulum will appear to rotate counterclockwise. Imagine now a similar pendulum swinging along the east—west axis at the equator. As the earth rotates under it, the pendulum will continue to swing along the east—west axis. It can be shown that the time required for the pendulum to rotate 180° so that it is swinging in the original plane is 12h T= (5.11) sin/ where ot) is latitude. At 90° latitude, the period is 12 h; at the equator, the period is infinity. Such a pendulum is called a , after Jean Bernard Foucault, who dem- onstrated such a pendulum in in 1851. As the two preceding examples demonstrate, there is a class of problems for which it is necessary to allow for the effect of the earth's rotation. We have a choice. We can use the center of the earth as our fixed and add the tangential velocity of the earth to all of our calculations. This would obviate the need for an artificial force but

180° North Pole Time Zero One Hour Later

Figure 5.5 A pendulum at the North Pole, free to swing in any vertical plane, will continue to swing in its original plane. However, to an observer on the earth, who is rotating with the earth, it appears that the pendulum is rotating clockwise at the rate of 153/h. Equations of Motion 89

t be would vastly complicate most routine calculations. The alternative is to continue to make our calculations on the assumption that the earth is not rotating, but add an artificial force e to (the Coriolis force) to ensure that the results are correct when the earth's rotation cannot ;ing be ignored. Oceanography and have adopted the latter convention. will In our coordinate system, this force is given with sufficient accuracy for most ocean- :ing ographic problems by writing Eq. (5.5) as u 1 the 11- = -- —(913 + vf + other forces Dt P ax the (5.12) 12 Dv 1 dp xis. — = --- – uf + other forces Dt P aY )wn am. where f 2,0 sine and Q is the of the earth, 21r/24 h (more precisely tate 2it/86,164 s, which is the length of the sidereal day) or 7.29 x 10-5 s-1 and e is latitude. A complete derivation of the Coriolis force is best done with vector algebra (see Box tor. 5.3 for such a derivation). However, some insight in understanding at least one component lest of the Coriolis force can be gained by considering the centrifugal acceleration of a particle t is on the surface of the earth, which is

U2 2 = s2 q (5.13) 11) q where U= ,2,q (Figure 5.6). If the particle is moving in an eastward direction u, the centrif- ugal acceleration is h it the the U + u (eastward) but

Figure 5.6 Eastward velocity of a particle relative to the surface of the earth increases the centrifu- ing gal acceleration of the particle relative to the earth. As the expanded diagram shows, this to Irs the can be divided into two components, one of which is in the xy plane of the earth's surface. 90 Equations of Motion

2 (u + u2 2(fu u2 U = + + =122q +212u+ — (5.14) q q q Since U is at least a hundred times greater than any at speed u, the last term is small enough to be ignored. The second term is the Coriolis force.

Box 5.3 Coriolis Force The derivation of the Coriolis term can best be done with vector algebra. With the center of the earth as the origin of the coordinate system, a point on the surface of the earth is given by R= ix + jy + Key to this derivation is choosing a coordinate system relative to a fixed point on the earth. The coordinates i, j, k are east, north, and up with respect to a point on the surface of the earth (Figure 5.2'). However, as viewed from outside the earth, this coordinate system is not fixed in space but is rotating. As the earth rotates, the coordinate system rotates with it. Therefore, taking the of R with respect to time yields two sets of terms: dR . —c-rt = (ixt Iczt) + (At + Yir zkt) (5.5') where the subscript refers to the derivative with respect to time.

Figure 5.2' The first set of terms is the movement of R with respect to the fixed coordinate system, the one that rotates with the earth. The second set of terms is the movement of the coordinate system itself as the earth rotates about its axis. We shall call this first set of terms the usual velocity (V), since this is the velocity we are all familiar with when we ignore, as we almost always do, the rotation of the earth. We will adopt the convention that a dot over a term means the rate of change (velocity) with respect to our fixed coordinate system.

The second set of terms is the movement of the coordinate system. It is the movement of a fixed point on the earth relative to the origin. In vector terms, it is the of the vector radius and the angular velocity of the earth. xi t + yjt + zkt = xR Equations of Motion 91

Thus the movement of a point on the surface of the earth relative to a fixed coordinate system, ;14) whose origin is the center of the earth, is of two kinds: movement relative to that fixed coordinate system and the movement of the coordinate system itself as the earth rotates. I 11 :erm dR =12+nxR (5.6') dt The next step is to take the second derivative of R with time. The most straightforward way is to take the derivative of Eq. (5.5') and proceed to separate out the resulting 12 terms. A more elegant approach is to note that Eq. (5.6') defines an d . —= +azX (5.7') dt rth. Then trth in + x)(R + x ing 11) = ( d2R 2 = R IF xR + 2fix + Si x (0, x R) 5') dt Since we assume that the angular rotation of the earth is constant, the second term on the right is zero.

d2R =it+ 211,x1i + SZ x (St xR) (5.8') dt2 The term on the left is the acceleration of a point on the surface of the earth relative to a coordi- nate system whose origin is the center of the earth. This is the true velocity of a particle with respect to a coordinate system whose fixed origin is the center of the earth. For those of us who prefer our coordinate system to be based on a fixed on the surface of the earth, the accel- eration is composed of three terms. The first is the acceleration relative to the fixed coordinate system. We might call this the usual acceleration, since this is what we mean by acceleration when we ignore the earth's rotation. = dV dt The second and third terms on the right are the acceleration of the coordinate system. The third term is generally folded into the value of gravity and will be ignored for the . The second he term is the Coriolis acceleration. The final step is to translate the Coriolis terms to the fixed coordinate system on the gurface 0, of the earth. Using the usual notation of i,x is east, j, y is north and k,w is up he R = ix + jy + kz ge R = iu + jv + kw

= jS2 cost, + kS2sint9. îa :or where .0 is latitude 92 Equations of Motion

i j 211x R =2 0 S2 cos 2.) S2 sin 2) (5.9')

= 2i(wS2 cos t) —142 sin 6) + 2j(uS2 sin t.) —0) + 2k(0 — uf2 cos t)) The i and j terms are the horizontal components of the Coriolis acceleration. The k term is in the vertical direction of gravity and does not concern us, but for those who attempt to measure gravity from a moving platform such as a ship, it is the EiitvOs correction. We can now substitute back into Eq. (5.8% d2R dV , ..., --Ezazxn dt 2 dt (5.10') d2R . i( du _ dv dw fv + 2w flcos6)+ j(—+ fu)+k( —2uS2cos /)) dt2 dt dt dt where f = 2C2sim). In the absence of any external forces, the left hand side of Eq (5.10') is zero and we can write dV =— 211, x V dt

du — = fv — w212cos 6 (5.11') dt

dv fu dt which is in the form of Eq. (5.12). By transferring the Coriolis terms in Eq. (5.10') to the opposite side of the equation, they become the Coriolis force, not the Coriolis acceleration. The vertical velocity terms are usually dropped because the average vertical velocities in the ocean are one to several orders of magnitude less than the horizontal velocities. Of course, this term cannot be ignored in problems of launches or , where the vertical velocity may be of the same order or larger than the horizontal terms. Returning finally to the last term in Eq. (5.8'), it can be shown in a similar manner that 2 2 x x R) = IQ q sin 6 – ki2 q cost) (5.12') where q = R cos 6 and is the of a point on the earth from the earth's axis (Figure 5.2'). Note that these terms are a function of position only and are independent of velocity. They are important in determining the of the earth (see "Gravity: Equal Potential Sur- faces" in this chapter).

As can be seen in Figure 5.6, the direction of the Coriolis force can be resolved into two components, one normal to the plane of the earth and the other parallel to the plane of the earth. The latter has the value 2S2u sin6 and is the horizontal component of the Corio- lis force that applies to east–west motion. Note that if the particle were moving westward, the same absolute value would apply, but the second term on the right in Eq. (5.14) would Equations of Motion 93 be negative. The vector in Figure 5.6b would be pointed toward the earth's axis, and the component in the plane of the earth would be pointing north. The vertical component of the Coriolis force can be ignored in almost all oceano- graphic applications. Unless otherwise stated, any future reference to the Coriolis force in this text means the horizontal component only. We will also adopt the convention of plac- ing the Coriolis term on the right side of the equation of motion and treating it as force, rather than putting it on the left side of the equation and thinking of it as an acceleration. An analysis of Eq. (5.12) indicates the following:

1. The Coriolis force is proportional to the velocity of the particle relative to the earth; if there is no velocity, there is no Coriolis force. 2. The Coriolis force increases with increasing latitude; it is a maximum at the North and South Poles, but with opposite sign, and is zero at the equator. 3. The Coriolis force always acts at right to the direction of motion. In the , it acts to the right (for an observer looking in the direction of motion); in the (where the sine of the latitude is nega- tive), the Coriolis force acts to the left.

In a system in which the Coriolis force is important, physical intuition is of little help in predicting what will occur. rotate and are accelerated normal to their direction of movement. An interesting example is to consider what happens to a ball that rolls down a frictionless inclined plane with a slope i. In the absence of a Coriolis term, the governing equation is simply du . (5.15) =-81 and, assuming that the ball starts from a resting position at the top of the incline, the veloc- ity attune t is

u = —git (5.16) and the distance traveled is

1 X = — — git2 (5.17) 2 E However, if the Coriolis force is added, the equation becomes

u 61- = -gi + fy dt (5.18) ,dv -fu dt the of which is 94 Equations of Motion

x = --7(1 — cos ft) . (5.19) g Y = + --2- (ft — sin ft)

The path described by the ball is indicated in Figure 5.7. The ball starts down the inclined plane, but as soon as it begins to move, the Coriolis force begins to accelerate it to the right, normal to the incline. As the ball picks up speed, the Coriolis effect becomes larger and the curvature becomes noticeable. Eventually, the ball is running normal to the incline. However, the Coriolis force continues to accelerate the ball to the right, and now the ball starts to run back up the incline. As it does, it slows down. Under the assumption of no frictional losses, the ball will continue to curve up the inclined plane, continuously losing speed until it reaches the top. At that point, the velocity is zero (and so, therefore, is the Coriolis force). The ball now begins once more to roll down the inclined plane, and the process is repeated.

Particle Path Is a Cycloid

Figure 5.7 By adding a Coriolis acceleration to the equation governing a particle sliding down an incline plane, the particle follows the path of a cycloid.

Before the reader dashes off to undertake this experiment with a large piece of ply- wood as the incline, he or she should pause and plug a few numbers into Eq. (5.19). The ball must roll for 5 min to see a curvature of 1 part in 100. Even with an incline as small as 0.1%, the ball will have traveled nearly 500 m in this time, and for the ball to reach the bottom and come back up to the top would require an inclined plane somewhat larger than the United States. To do the experiment on an inclined plane of only a few square miles would require an inclined plane whose slope was measured in parts per million. One should also note that we have adopted a "flat earth" coordinate system, which adds an additional complexity to erecting an inclined plane with a slope of a few parts per million. Equations of Motion 95

Gravity: Equal Potential Surfaces With the assumed coordinate system, gravity acts along the z axis. Although gravity varies slightly from place to place, the change is insignificant for nearly any problem in . Surface gravity changes about 0.5% (9.78 rn/s 2 at the equator and 9.83 m/s2 at the poles). The decrease in gravitational potential is related to the spinning earth. The first term on the right in Eq. (5.14) is centrifugal acceleration, which varies between zero at the poles and 0.034 m/s2 at the equator. (See also the discussion of gravity in the derivation of the Coriolis force.) The remainder of the 0.05 m/s 2 difference between the poles and the equator is related to the fact that the radius of the earth at the equator is about 22 km larger than the polar radius, a circumstance that can be explained in large part by the expected equilibrium shape of our rotating earth. If the earth were of uniform density, gravity would decrease linearly with depth; but because the earth's density increases with depth, gravity actually increases with depth through the crust. The change is small:

g(z) = go +.023 x 10-4 Z in/s2 (5.20) where the depth is measured in meters. Even in the bottom of the deepest trench, the value of gravity is only 0.25% larger than at the surface. Gravity measurements at sea are often made from a moving ship. An instrument that measures acceleration cannot distinguish one type of acceleration from another. Short- period accelerations, such as the roll and heave of the ship, can be averaged out, but the cen- trifugal acceleration due to the ship's east—west movement cannot. In Eq. (5.14) and Figure 5.6, the term Wu was divided into horizontal and vertical components. The horizontal component was the Coriolis force; the vertical component, 212u cos 6, points in the direc- tion of the gravitational vector and is the Etitvtis correction, which must be applied to all observations of gravity if made from a moving platform (see also the discussion in Box 5.3 on the derivation of the Coriolis force). A ship with an eastward velocity of 10 knots would have an ElltvOs correction of at least 50 milligals (1 gal = 0.01 m/s 2) in midlatitudes. Since we have adopted a right-handed coordinate system, the z axis is up, and grav- ity force's acceleration should have a negative value since it acts toward the center of the earth. Similarly, all ocean depths should have negative values, and these should become increasingly negative as the depth increases. The convention is ignored in this section and throughout most of this text. When possible, ocean depths are treated as positive values: The reader, however, is warned to be wary of this problem if attempting analytical solu- tions of equations similar to those discussed in later chapters of this book. An equal potential surface is one normal to the gravitational vector. There is no change in the gravitational potential of a particle as it moves along an equal poten- tial surface. To a very good approximation, the mean sea surface of the ocean (after one averages out the waves and tides) is an equal potential surface. Its departure from such a surface is mostly the result of the forces that determine the mean ocean surface currents. The sea surface slopes necessary to maintain even the strongest of ocean currents are of the order of 10-5 or less (see "Geostrophic Flow" in Chapter 6), and nowhere in any ocean does the mean sea surface depart by as much as 2 m from a common equal potential surface. 96 Equations of Motion

However, the mean ocean surface is not flat and smooth. A radar altimeter in an earth-circling satellite shows the ocean skin as a highly wrinkled, complex surface, with bumps, troughs, and other departures from the mean as large as tens of meters and slopes as large as 10-3, a hundred times larger than those sea surface slopes that drive the stron- gest ocean currents. The reason for this apparent paradox is that the earth itself is not homogeneous, and because of this the direction of the gravitational vector changes much more rapidly than would be the case otherwise. Figure 5.8 is a simple example of what occurs. Because the density of the is greater than water, the gravitational vector is deflected toward the greater mass of the seamount. Similarly, the gravitational vector is deflected away from a deep trench filled with . As a consequence, the mean sea surface of the as observed from space bears a strikingly qualitative resemblance to the underlying bottom topography: bumps over and troughs over trenches.

Equalpotentlal Sea Surface

Bottom

Figure 5.8 The direction of the gravitation vector is determined by the mass distribution of the earth. Since the mass of the earth is not uniformly distributed, there are small differences in the directio''n of the gravity vector. As a consequence, the average slope of the sea surface (which, to a high degree of approximation, resembles an equal potential surface) reflects the slope of the bottom topography.

Friction The fourth and final force to be discussed is friction. Frictional forces (1) allow the wind blowing over the surface to establish waves and currents, (2) provide an important mecha- nism for the exchange of between adjacent parcels of water, and (3) ulti- mately drain the kinetic energy from the ocean and it to heat energy. The processes by which this is done are reasonably well understood. The details are not. The energy flow due to friction is unidirectional, from kinetic energy to heat energy. The water in a vigorously stirred barrel will slow down and eventually come to rest after Equations of Motion 97

the stirring has stopped. By analogy, if one could turn off the wind, the , and the tidal forces of the and sun so that new energy was not imparted to the ocean, the ocean would eventually "run down" and the ocean currents and waves would grow smaller and slowly come to rest, the kinetic energy slowly being transformed to heat energy. However, the total kinetic energy of the ocean is the equivalent of less than a 0.01°C increase in ocean . Friction causes the ocean currents to deaccelerate. One expression of ocean friction is to add a frictional term proportional to velocity:

friction (x)= —Ju friction (y)= —Jv (5.21) friction (z) = —Jw Eq. (5.21) says that the larger the currents and kinetic energy, the larger the friction term. This formulation has the virtue of simplicity and for this reason has found applicability, but it contributes little to the understanding of the physics involved in frictional forces. As a start to understanding the physics, let us consider molecular viscosity. The ideal fluids that one studies in elementary physics have no viscosity. They are infinitely slippery. The wind in an ideal blowing over an ideal ocean would have no effect. One would slide by the other, causing no currents or waves. A viscous wind blowing over the surface of a viscous ocean will set the surface water in motion. Because the water is vis- cous, the frictional applied to the water will be transmitted downward. If the wind stops, the water will begin to slow down, and eventually the movement will stop as the effect of water viscosity acts to transfer the kinetic energy to heat energy. The molecular viscosity of water is known: dynamic viscosity Ai a- 1 x 10-3 kg/m s kinematic viscosity v -a 1 x le m2is where

v=

I t Knowing the viscosity, the rate of transfer can be calculated:

di) = - (5.22) az The process of molecular transfer of kinetic energy and momentum is similar to the processes discussed in Chapter 4 for molecular transfer of heat and salt. As with molecular of heat and salt, the calculated rate of transfer of momentum and kinetic energy is much slower than that observed in the ocean. For example, assume there was a way to maintain a surface current of 0.5 m/s at the surface skin of the ocean. How long would it take to transfer kinetic energy downward from the surface by molecular processes alone? Figure 5.9 shows the velocity distribution at the end of one day, 10 days, and a year. Even the most casual observational evidence suggests that the effect of such a surface current would be felt much deeper much sooner.

1 98 Equations of Motion

The problem is not the molecular theory of viscosity, but rather that such calcula- tions ignore the turbulence that stirs the water. Just as it was demonstrated in Chapter 4 that turbulent stirring can enhance , so can turbulence more rapidly transfer and intermix parcels of water with different kinetic and thus enhance the transfer by molecular viscosity of kinetic energy from one parcel of water to another. As with the diffusion of heat and salt, the final transfer is by molecular processes, but turbu- lent stirring of the water allows the process to go more rapidly than it would otherwise.

Eddy Viscosity. The same arguments used in Chapter 4 to relate diffusion to molecu- lar diffusion have been used to relate eddy viscosity to molecular viscosity. Eddies or swirls of moving fluid are analogous to molecules, and mixing lengths are equivalent to molecular mean free path lengths. As with eddy diffusion, a coefficient of eddy viscosity can be substituted which is many times larger than the coefficient of molecular viscosity. Again it must be emphasized, as in Chapter 4, that there is a problem of scale when such a substitution is made, whether or not one accepts the validity of the analogy. For example, the vertical mixing resulting from swirls of water a few centimeters in , with a charac- teristic mixing length of 1 m, would be much less than that resulting from larger eddies 200 km across, which break off from the . Thus a wide range of eddy viscosity values can be found in the oceanographic literature.

Speed (m/s)

0.1 0.2 0.3 0.4 0.5

Figure 5.9 The horizontal velocity generated by a 0.5 mis surface current after 1 day, 10 days, and 1 year, assuming that energy is transferred downward by molecular viscosity alone.

Equations of Motion 99

If eddy viscosity is considered an exact physical analog to molecular viscosity, the frictional terms in the equations of motion can be given explicit form: d2,4 (d 2u d2u) friction (x) = Ah TxT + 4- Az dz2

A (d2v d2v) A d2v friction = -cixT + 4-‘z (9z2 (5.23)

(92 w .921 (92w friction (z) Ah( dx2 ay2 + Az az2

where the subscripts h and z denote horizontal and vertical eddy viscosity coefficients and the units are those of kinematic viscosity (m2/s). Characteristic values are scale dependent and are of the order of

Ah = 102-105 m2/s

Az = 10-4-10-2 m2/s

Wind Stress. Frictional forces not only dissipate kinetic energy within the ocean, but they also allow for the transfer of momentum and kinetic energy from the atmosphere to the ocean. As noted earlier, the wind in a viscous atmosphere blowing on the surface of a vis- cous ocean creates waves and currents. If one assumes that the energy transfer into the interior of the fluid is accomplished by an eddy viscosity process, as described earlier, the transfer process can be written as x 49214 = pAz dz2 (5.24) ar .92v = PAz

where '1", and -ty are the x and y components of in units of force per unit area. The absolute magnitude of the wind stress term is thought to be proportional to the square of the wind velocity (see "Wind Stress: " in Chapter 6). A derivation of Eqs. (5.23) and (5.24) is given in Box 5.4. Making the frictional dissipation term proportional to velocity, as in Eq. (5.21), is helpful in picturing in a semiquantitative way how kinetic energy is dissipated in the ocean. Arguing by analogy that one can scale up from molecular viscosity to eddy viscos- ity and substitute eddy viscosity coefficients for a molecular viscosity coefficient, as in Eq. (5.23), has proven useful in examining a variety of oceanographic problems. However, as useful as such an analogy is, it is unsettling to be forced to work with what might be thought of as a constant of nature (like the velocity of light or the latent heat of evapora- tion) that actually varies by a factor of 10 to 1000, depending on the circumstances to which it is applied.

100 Equations of Motion

Box 5.4 Friction Shearing Stresses (Molecular and Eddy Viscosity) A viscous fluid is subject to shearing stresses. In the example discussed in connection with Figure 5.9, the surface skin of the ocean was made to move at a constant velocity. Momentum was trans- ferred downward into the interior of the fluid by random molecular motion. This is the process of molecular viscosity: du rxz = — aZ where Txz is the shearing stress . The first subscript refers to the direction of the stress; the second subscript refers to the plane in which the stress acts. The subscript z is normal to the xy plane; At is the molecular viscosity in units of kg/m-s. Shearing stress has the same units as pres- sure. The force applied on the surface of the fluid is force per unit surface area. There are three elements of the stress tensor acting in each of three directions. In the x direction, they are du xz = p-- du xy ay= - (5.13) du rxx = 11—ax Consider the force in the x direction that these stresses apply to a fluid element (Figure 5.3'). The force applied to the cube is the difference between the shearing force in the upper and lower faces.

Figure 5.3'

In terms of the acceleration of the mass of the cube, the x-directed stress forces are du du m—dt = (PArAY4z)—dt = xz2 (A t4Y) xZI (A X AY) du 1 ( r xz, xzi ) _ 1 41- dt p AZ P Az In the limit, as the cube becomes small, this reduces to du _1 d r xz _ l ar du) dtp dz p -d-ze az du 1 a2u = —27 dt P By a similar argument, the other stress terms can be found, and -

Equations of Motion 101

du 1 d2u 1 d2u l d2u — = P —p —zTa 2 + igure dt P P dx2 a2v rans- dv 1 d2v l d2 v ss of • P dz P dY (5.14') dw 1 d2 w 1 d2 w 1 d2w =11—p —dz F +11---i-+11— 2 dt P dY P dx —d =-11 V2V = W2V ; the dt p e xy xes- where v is the kinematic viscosity, which is of the order of le m 2/s. On a scale of meters, the hree largest values of velocity shear observed in the ocean are of the order of 5 x 10 -2/s. Thus it is unlikely that molecular viscosity plays an important role in the equations of motion in oceanography. The ocean is fully turbulent. Swirls and eddies of from tens of centimeters to tens of kilometers are thought to occur and move at random within the ocean. The randomness, however, has strong dimensional constraints. Eddies that move horizontally along surfaces, more 13') or less, of constant are both larger and move more easily than those that move verti- cally and must work against the buoyancy forces. One can imagine that these eddies are transfer- ring momentum, both vertically and horizontally, in a manner analogous to that performed by random molecular motion in the case of molecular viscosity. It is then possible to define an eddy ;ure viscosity coefficient or Austausch coefficient (A) analogous to the molecular kinematic viscosity and coefficient. By analogous arguments, one can derive the eddy viscosity version of Eq. (5.14): du _ d ( A dis)_, ±(A 24A du) — Tcz "z az r ay y ay ax x (5.15') dv d (A ?±) (I( d_v.) _±( .(2!_.) -cTz dz ay C'Y ay ) -F axVix ax) Frictional forces are only important in horizontal components of the equations of motion. In Eq. (5.15') it is not assumed that A is constant. However, such an assumption is made in most cases when these terms are used. Another assumption is that .3' A, = Ay = Az where the subscript refers to horizontal eddy viscosity, and thatA h >> A.

du d2u A (3214 a2U -dt = ith -dy2 (5.16') d2vd2v d2v —dv = A —+ A (—+—) dt Z az2 h ax2 dy2

Reynolds Stresses. The most satisfactory formulation for the frictional transfer of kinetic energy is in terms of Reynolds stresses. Unfortunately, it is still difficult to make with the accuracy necessary to apply such formulations. The key to measur- ing Reynolds stresses is the ability to separate the mean velocity from its turbulent compo- nent. This is difficult to do with horizontal currents; it continues to be almost impossible with flow in the z direction, when both the mean velocity and the turbulent component are so small.

102 Equations of Motion

To derive the Reynolds stress terms, it is necessary to subtract the mean from the instantaneous velocity. That which remains is the turbulent component. It is further neces- sary to make the separation in terms of the three components u, v, and w:

(5.25)

By suitable averaging, one can separate the velocity components in the equation of motion, as well as in the equation of continuity Eq. (4.10), into terms that carry the mean velocity components, Ft, i, and Tv- , and the turbulent components, u', V, and w'. The tur- bulent components of velocity become the frictional components, where the denotes a time-averaged mean of the product of the turbulent components:

friction (x) = --d u'u' )— dx dy dz

friction (y)= --(v'u')- (v'v' )- (5.26) dx)0Y dz d friction (z) = -- (w'u')- —(w'V)- dx dy dz For a derivation of Eq. (5.26), see Box 5.5. Note the similarities between Eqs. (5.26) and (5.23). There is a formal relationship between the Reynolds stress terms derived here and the eddy frictional terms derived ear- lier. For the x component, it is X X u'u = —A — u'v' = —Ax ax —Y dy u'w' = —Az —az (5.27) — X v'u' = — X x v'v' = -AY — A, v'w' = —Az — dZ c..j. As the turbulent velocity components increase, one would expect the coefficient of eddy viscosity to increase. Because the vertical turbulent velocity values are so much smaller than the horizontal turbulent velocity values, one would expect from this simple scale analysis alone that A, < < Ah.

Box 5.5 Reynolds Stress Terms The key to deriving the Reynolds stress terms is understanding the averaging process. Any velocity component can be averaged over time: T U = —judt (5.17') T o where u is the instantaneous velocity component and 17 the average value of that component as observed over time T. The instantaneous velocity can be written in terms of its average value and fluctuating component u

Equations of Motion 103

he u=a+te It follows that the average value of u', averaged over the same period, must be zero. 1 =- f u'dt=0 (5.18') T 0 5) Consider next the averaging of the product of separate components u and v: uv = (17 + u')(i + v')= t717 + +17u' + u'v' of Now take the average of each of the four terms over the same time interval T: an T T - 1 r 1 __ 1 _.' ir- uv = — uvdt — uv dt + — f uv' dt+— v-14' dt u'v dt ; a T 0 T 0 T 0 T 0 T 0

uv = =uv + uv + vu + u v Since ii and g are constant over the interval T and since LT' =17' = 0 uv = -u-v+u'v' (5.19') 6) Note that although the average value of the fluctuating velocity component of u' and v' is zero, the average value of the product of these fluctuating components need not be zero. To apply this averaging process to the equation of motion, it is first necessary to manipulate it a bit. Ignoring friction, the x component of the equation of motion, Eq. (5.12), can be written put + puux + pvuy + pwuz = + pfv (5.20') The equation of continuity, Eq. (4.6), is ip I- p, +(pu)x+(pv)y+(pw),= 0, Multiplying the continuity equation by u, up, + u(pu)x + u(pv)y + u(pw)z = 0

7) Add the preceding equation to the x component of the equation of motion: [pu, + up,]+[u(pu)x+ puux]+[u(pv)y + pvuy]+[u(pw),+ pwuzi= -px + pfv which can be written, assuming p is a constant, (up), +(puu)x +(puv)y +(puw)z = -px + pfv Next substitute li + u', and so on, for the instantaneous velocity components and then average over the same time interval T: [(ii + ulp]t +[p(Ti + ulx+

+[p(u-u+ u'u') ]+[p(u-v +)]+[p(u-w+ u'w') ]= -p +sx The averaging bars have been dropped over terms assumed constant over the time interval. Rearranging terms, 104 Equations of Motion

(i p)f +(piTiii)y+(piWi)z= –px + pf –(pu'u') –(pu'v') –(pu'w') The preceding equation is identical in form with Eq. (5.20') except that average rather than instantaneous velocity components are used, and there are three additional terms representing the cross products of the fluctuating components. By reversing the manipulation done to arrive at Eq. (5.20'), one can write the x component of the equation of motion as

+ Wit; +i71.7y + Cliaz = px fi x –(u'v') –(u'w')

The last three terms are called the Reynolds stress terms. A similar derivation can be made for the y component. Where the averaging process implicit with the Reynolds stress consideration is understood, the averaging bars can be dropped, and we can write

u, + uux + vuy +wu z=_!p + fv–(u'ul x – (u'v') y –(u'w'), 1 (5.21') v,+ uvx +vvy +wvz = py – fu–(v'ul x –(v'v') y –(v'w') z

Equations of Motion In summary, there are four forces that balance the acceleration term: (1) the pressure-gra- dient force, (2) gravity, (3) friction, and (4) the Coriolis force; the latter is in some sense a in that it is required because we assume a nonrotating coordinate system when in fact all our observations are made on a rotating earth. We have further defined our coordinate system so that the gravity vector points in the vertical direction, which elimi- nates the gravity term from the xy plane. There are several choices for the analytical expressions for the frictional terms. Examples are given in Eqs. (5.21), (5.23), (5.24), and (5.26). In Eq. (5.28) we write the horizontal components of the equation of motion in both partial differential and vector form using generalized friction terms X 9; 01) . du du v dudy+ wc-5--duz---p7- "ox+fv+XP

av dv av dv 1 ap —at +u—av +v .5;--Fw —az = --p —ay - fu+Y (5.28)

1 —3V +(V. V)V = Vp - 2flxV+g+43

In future chapters, we will use different formulations of the friction terms depending on the process being examined. In many cases, the frictional terms are so much smaller than the other terms in the equation of motion that they can be ignored. It is important to write these equations down and note the individual terms, not because there will be any attempt in this text to consider Eq. (5.28) in any detail, but because the relationships that are discussed in succeeding chapters should be considered against Eq. (5.28). These relationships are all approximations in that certain terms are Equations of Motion 105

ignored. The reader should always be aware of what those terms are and of the justification for ignoring them.

Vorticity A number of oceanography problems can best be illuminated if the equations of motion are written in terms of a balance in vorticity. The vorticity of a particle is a measure of its about its axis. Vorticity is proportional to the of a particle. By convention, clockwise spin is negative vorticity; counterclockwise spin is positive vortic- ity. The forces that impart spin (or vorticity) are sometimes referred to as . The ver- tical component of vorticity, with which we are concerned, is a measure of the spin around the vertical axis. It is given by

dv du ax dy (5.29)

It is possible to write a series of equations similar to Eq. (5.28) showing the balance of torques and rate of change in vorticity. For many problems that oceanographers treat, it is more convenient to think in terms of the balance of vorticity than of the balance of linear forces, as given in Eq. (5.28). However, there is nothing esoteric about vorticity balance. Any problem that can be solved in terms of vorticity can, in principle, be solved in terms of Eq. (5.28). The vorticity balance equation equivalent of Eq. (5.28) is Eq. (5.30). For a derivation of the vorticity equation, see Box 5.6.

du 4 + + at,) dX 737 u -d- v4 + w (5.30) x &+v1+(4+f(--x dy- dz dY d— + Ty ) ax dY Note that the pressure-gradient term disappears when the equations of motion are written in terms of a balance in vorticity. Also note the term (df/ay), the change in the Cori- olis parameter with latitude. This term is the key to understanding why currents such as the Gulf Stream are narrow and swift (see "The General Circulation: Western Boundary Currents," in Chapter 6). The oceanographic convention is to refer to this term as beta, 0=-. (ray. There are several commonly used vorticity terms in geophysical fluid They include

relative vorticity : (5.31) absolute vorticity : The absolute vorticity divided by Z, a layer of water with that vorticity, is called the

potential vorticity : (+f) (5.32)

Most importantly, it can be shown that in the absence of friction, the potential vortic- ity must be conserved (see Box 5.6 for derivation). Thus, as a layer of water moves from 106 Equations of Motion

one latitude to another, there must be a change in either the layer depth or in the relative vorticity. If the thickness of the layer of water changes, it must be accompanied by either a change in the latitude or a change in the relative voracity. Similarly, if the water changes latitude, it must be accompanied by either a change in the layer thickness or a change in the relative vorticity (Figure 5.10).

a.

Column Thickening \ No Change In Latitude

b.

Column Moving Northward 1 No Change in Thickness

Figure 5.10 If the potential vorticity of a water layer is to be conserved, then the relative vorticity must change if (a) the latitude remains the same but the water layer thickness changes, or (b) the layer thickness remains the same but latitude changes.

Box 5.6 Vorticity Voracity can be defined as the cross product of velocity: V xV = i(wy —v )+ — wx )+k(vx —uy ) (5.22') When V xV = 0 , the motion is said to be irrotational. The two horizontal components of the equation of motion, Eq. (5.28), are often written in the form of a single vorticity equation in terms of the vertical vorticity, which is defined as 4 = avkx- away. We begin with the horizontal components of the equation of motion. Equations of Motion 107

(5.23')

Differentiate the y component with respect to x and the x component with respect to y, and assume that the differentials with respect to density (p) and vertical velocity (w) are small enough to be neglected. Note that af/ax is zero but that afidy is not. Subtract the x component equation from the y component and collect terms. The result is

u4x +1v4z vfy + (4 + f)(u. + vy) =x—Xy (5.24') Since

+373 (.1= vfy

(see Box 5.1) we can write

2,- (4+ f)+(g+ lux +11y ) = -Xy (5.25')

The terms 4 and 4 + f are defined as relative voracity and absolute voracity, respectively. Consider a layer of constant density (p) and thickness (Z). First integrate the equation of continuity over a layer of thickness Z:

f(ux +vy )dz=-,1 wzdz

(ux + vy)Z = -}z (-1 )d z = - ddZt

Then substitute into Eq. (5.25') D ,__1 clZ = _ _ -1)7 (+ f z kc dt AY which in the absence of friction can be rewritten D ( 4+ f = 0 (5.26') Dt Z The term (4+ f)/ z is the potential vorticity.