CHAPTER FIVE - Hunter College · Doing so makes the horizontal xy plane an equal potential surface ... the total differential is often written ... the hydrostatic pressure,

CHAPTER FIVE

Equations of Motion

The equations that describe the currents, waves, tides, turbulence, and other forms of fluidmotion in the ocean are of a nonlinear nature for which there are no complete, exact,analytical solutions. The best we can do is to work with partial solutions, many of whichprovide excellent insight into the forces at work. In this chapter, we will examine the indi-vidual terms in the complete equations of motion. In Chapter 6, we will examine a seriesof 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 ofwaves.

Any quantitative discussion of forces and motions requires a coordinate system. Thesystem most commonly used in oceanography is the rectilinear, Cartesian system in whichthe earth is assumed to be flat. A spherical coordinate system would be more realistic, butit 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 pointseast, 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 equalpotential surface (see "Gravity: Equal Potential Surfaces" later in this chapter for defini-tion). The corresponding velocity components are u, v, and w.

Although meteorologists and oceanographers may agree on the coordinate system inFigure 5.1, they use different conventions for describing winds and currents. A north cur-rent is a current flowing toward the north; a north wind 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.

Newton's second law states that the mass times the acceleration of a particle is pro-portional to the sum of the forces acting on the particle:

du_lvF

dt

Figure 5.1 The Cartesian, flat earth coordinate system used in this text.

In discussing fluid motion, the relationship is usually written

(5.3)

Equations of Motion 81

FIVE

:ion

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

' fluidexact,whichindi-

series:hap-on of

t. Thevhichc, butthysi-)ointslirec-equal

efini-

m inI cur-torth.tnini-

: pro-

(5.1)

(5.2)

where it is now understood that the forces are per unit volume, since Eq. (5.2) followsfrom Eq. (5.1):

– V

As written, Eqs. (5.1) and (5.2) apply to the components of the forces acting in theeast–west or x direction Similar equations can be written for the force components actingalong the other two axes:

—dt 10 2, r x

dv 1(5.4)

dt p Y

dw iv Fdt p z

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

density x particle acceleration = gravity + pressure gradient + Coriolis + friction (5.5)

82 Equations of Motion

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

Acceleration

Before looking at the various force terms, it is necessary to examine the acceleration offluids. Newton's second law is usually introduced in terms of particle mechanics (a blocksliding down an inclined plane or the movement of billiard balls). As written, Eq. (5.4)applies to the motion of a particle. In continuum mechanics (or fluid mechanics), there aretwo kinds of acceleration for which there are operational definitions, as the followingexample demonstrates.

Consider the water motion in a channel of constant depth but narrowing width (Fig-ure 5.2). The volume of water entering the channel is constant over time, and, in theabsence 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 particlemoving along the channel is accelerated as it moves from A to B. The local acceleration is zero; theaverage particle acceleration between A and B is not.


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 itsvelocity as it traverses the channel, its speed would increase as a constant volume of wateris forced to flow through a narrower channel. In this example, the local acceleration iszero; the particle acceleration is not.

In many problems, it is desirable to write Eq. (5.4) in terms of the local accelerationrather than the particle acceleration. The two are related in the following way:

particle acceleration = local acceleration + field acceleration terms

du du du du dudt dt+u—dx+v—dy+w—dz

dv dv dv dv dv

dt dx dy dzdw aw aw dw dwdt dt u-d-x -"Ty +w-Tz

or, in general,

D d a a dE_Dt dt — Tt +17x +11—±}11 adY dz

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 aparticle is called Lagrangian motion. The flow past a point is called Eulerian motion. Thederivation of Eq. (5.6) is given in Box 5.1.

(5.6)

(5.7)

Box 5.1 Acceleration

The velocity of a fluid is not only a function of time but also of space:

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

du du du du=—at+u—ax+v—dy+W-Tz

For emphasis, the total differential is often written

Du du du du duDr dt dt dx dy dz

Note that D/Dt is the particle acceleration, and alat is the local acceleration. Thus, Eq. (5.1') canbe written

(5.1')


Du duDt dt = + uu x + vu y + wux

where we again adopt the notation that

duu —x ax

Similarly,

Dv dvDt

v, + uv x + vvy + wv,dt

Dw dwDt dt

= w t +uwx+vwy+wwx

In vector notation,

Du— = u t +(V V)uDtDv— = v, + (V • V)vDtDw— = v t + (V V)wDt

or, in general,

DV– —+ (V . V)V

Dt

Of the various terms in Eq. (5.5), perhaps the pressure gradient is the easiest to visualize.A particle will move from high pressure to low pressure, and the acceleration is simplyproportional to the pressure gradient. A mechanical analog is a ball on a frictionlessinclined 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 forcesDt P az

Pressure gradients arise in a variety of ways. One of the simplest is by a slopingwater surface. Imagine a container with an ideal fluid (constant density, incompressible,and without viscosity) whose density is pa and that in some manner it is possible to havethe water surface slope as in Figure 5.3 without causing any other motion. Remembering

(5.2')

Pressure Gradient


1

2

Pi = Pagz

P2 = Pa g (z +ha)

(5.2')

ualize.;implyonless;celer-math-

(5.8)

opingsible,have

,ering

Figure 5.3 The slope of the sea surface creates a horizontal pressure gradient throughout the entirefluid. 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 weight of the fluidabove—that is, the hydrostatic pressure, Eq. (1.2)—

Pi = PagZ

P2 = Pag(Z + AZ)

The resulting pressure gradient term is

1 dp 1 p2 -

Pa & Pa AX

AZ= g —

AX

= gix

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 Zchosen in Figure 5.3. Thus, if there were no other forces acting, Eq. (5.8) says that theentire fluid in Figure 5.3 would be uniformly accelerated toward the lower pressure.

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 achannel where the pressure increases from left to right (i.e.,p 2 > p 1 (Figure 5.1'). Rememberingthat a pressure force is pressure times the cross-sectional area, with the force vector acting normalto the cross section, the force on the two sides of the cube would be

= PI AYAZ, F2 = P2AYAz

(5.9)

(5.10)

Dv 1DtDw 1= _—Dt p PZ

In vector notation, these become

(5:3')

(5.4')


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 4pDt p

By letting the cube become very small, one approaches the differential form

Du = 1 ap = 1Dt p az pPx

Note that F2 was given a negative sign because the force was directed in the —x direction. Themeaning of the negative sign in the final equation is simply that the particle is accelerated fromhigh pressure toward low pressure, which is in the direction opposite to the pressure gradient. Asimilar derivation can be done in the other two directions. Combined, these give


Coriolis Force

The Coriolis force is the most difficult of the four forces to comprehend because physicalintuition is of little avail. Most of us have some qualitative ideas of what to expect from theforces of gravity, pressure, and friction; but there is little in our experience to indicate whathappens 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 observations and measurements we make of forces and particle movementuse a reference system that is fixed. The reference system, and hence our observations,ignore the earth's rotation. For most problems, that is an adequate assumption; the forcesand accelerations 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. Theearth, with a radius of about 6400 km, rotates once every 24 h. It has tangential velocities,as indicated in Figure 5.4. Let us assume that a constant-level balloon is set in motion at40° N with a southward velocity of 1 m/s, and that no horizontal forces act on it. (There arevertical buoyancy and gravity forces which are responsible for maintaining the balloon atconstant level, but no horizontal forces.) Remember that the 1 nVs southward velocity ismeasured 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 inspace, the true velocity would be almost due east, an eastward component of 356 m/s anda 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 takeabout 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 theearth observer it would appear that the balloon was not only going south at 1 m/s but, inthe absence of any horizontal forces, it had somehow picked up a westward component of

TangentialSpeeds (m/s)

Path of Particleon NonrotatingEarth

Apparent Pathof Particle onRotating Earth

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


velocity of about 6 m/s, the difference between 362 and 356 m/s. The "force" that must beapplied to account for the apparent westward acceleration is the Coriolis force.

For a second example, consider a pendulum suspended at the North Pole and free toswing in any direction. Assume that at 12 noon it is set in motion such that it is swingingalong the 900 E/90° W longitude axis (Figure 5.5). In the absence of other forces, it willcontinue to swing in the same direction as the earth rotates under it. To an observer lookingdown 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 theearth that appears to rotate but the pendulum, which rotates clockwise 15°/h. In 6 h, thependulum would be rotating along the 180° E—W axis, some 90° from where it began. In 12h, 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 downon 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 rotatecounterclockwise.

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—westaxis. It can be shown that the time required for the pendulum to rotate 180° so that it isswinging in the original plane is

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 Foucault pendulum, after Jean Bernard Foucault, who dem-onstrated such a pendulum in Paris in 1851.

As the two preceding examples demonstrate, there is a class of problems for which itis necessary to allow for the effect of the earth's rotation. We have a choice. We can use thecenter of the earth as our fixed frame of reference and add the tangential velocity of theearth to all of our calculations. This would obviate the need for an artificial force but

180°

Time Zero One Hour Later

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

T= 12h sin/

(5.11)

North Pole


t be

e to;ingwill:ing

thethe12

xis.)wnam.tate

tor.lestt is

11)

h itthethebut

would vastly complicate most routine calculations. The alternative is to continue to makeour calculations on the assumption that the earth is not rotating, but add an artificial force(the Coriolis force) to ensure that the results are correct when the earth's rotation cannotbe ignored. Oceanography and meteorology have adopted the latter convention.

In our coordinate system, this force is given with sufficient accuracy for most ocean-ographic problems by writing Eq. (5.5) as

11-u

= --1

—(913 + vf + other forcesDt P axDv 1 dp— = --- – uf + other forcesDt P aY

where f 2,0 sine and Q is the angular velocity of the earth, 21r/24 h (more precisely2it/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 Box5.3 for such a derivation). However, some insight in understanding at least one componentof the Coriolis force can be gained by considering the centrifugal acceleration of a particleon the surface of the earth, which is

U 22= s2 qq

where U= ,�2,q (Figure 5.6). If the particle is moving in an eastward direction u, the centrif-ugal acceleration is

U + u (eastward)

(5.12)

(5.13)

ingIrs

Figure 5.6 Eastward velocity of a particle relative to the surface of the earth increases the centrifu-gal acceleration of the particle relative to the earth. As the expanded diagram shows, this addition tothe centrifugal force can be divided into two components, one of which is in the xy plane of theearth's surface.


U(u + =

u2 2(fu u22

+ + =122 q +212u+ —q q q

Since U is at least a hundred times greater than any ocean current at speed u, the last termis 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 theearth 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 inspace but is rotating. As the earth rotates, the coordinate system rotates with it. Therefore, takingthe derivative 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, theone that rotates with the earth. The second set of terms is the movement of the coordinate systemitself 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, therotation 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 afixed point on the earth relative to the origin. In vector terms, it is the cross product of the vectorradius and the angular velocity of the earth.

xi t + yjt + zkt = xR

(5.14)


Thus the movement of a point on the surface of the earth relative to a fixed coordinate system,whose origin is the center of the earth, is of two kinds: movement relative to that fixed coordinatesystem and the movement of the coordinate system itself as the earth rotates.

dR =12+nxR (5.6')

dtThe 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 moreelegant approach is to note that Eq. (5.6') defines an operator

d .—= +azX (5.7')dt

Then

11) = ( + x)(R + x

d2R2 = R IF xR + 2fix + Si x (0, x R)

dtSince we assume that the angular rotation of the earth is constant, the second term on the right iszero.

dt2The 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 withrespect to a coordinate system whose fixed origin is the center of the earth. For those of us whoprefer our coordinate system to be based on a fixed position on the surface of the earth, the accel-eration is composed of three terms. The first is the acceleration relative to the fixed coordinatesystem. We might call this the usual acceleration, since this is what we mean by accelerationwhen we ignore the earth's rotation.

= dVdt

The second and third terms on the right are the acceleration of the coordinate system. The thirdterm is generally folded into the value of gravity and will be ignored for the moment. The secondterm is the Coriolis acceleration.

The final step is to translate the Coriolis terms to the fixed coordinate system on the gurfaceof the earth. Using the usual notation of i,x is east, j, y is north and k,w is up

R = ix + jy + kz

R = iu + jv + kw

= jS2 cost, + kS2sint9.

where .0 is latitude

;14)

:erm

rth.trth

ining

5')

he

0,hege

îa:or

I 11

d2R =it+ 211,x1i + SZ x (St xR) (5.8')

211x R =2

i j

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 thevertical direction of gravity and does not concern us, but for those who attempt to measure gravityfrom 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

d2R . i( du _ fv + 2w flcos6)+ j(—+ fu)+k( dw

—2uS2cos /))dt2 dt dt dt

dv

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 6dt

dv fudt

which is in the form of Eq. (5.12). By transferring the Coriolis terms in Eq. (5.10') to the oppositeside of the equation, they become the Coriolis force, not the Coriolis acceleration. The verticalvelocity terms are usually dropped because the average vertical velocities in the ocean are one toseveral orders of magnitude less than the horizontal velocities. Of course, this term cannot beignored in problems of rocket launches or ballistics, where the vertical velocity may be of thesame 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

x x R) = IQ2 q sin 6 – ki2 2 q cost) (5.12')

where q = R cos 6 and is the distance 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 areimportant in determining the gravitational field 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 intotwo components, one normal to the plane of the earth and the other parallel to the plane ofthe 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

(5.10')

(5.11')



be negative. The vector in Figure 5.6b would be pointed toward the earth's axis, and thecomponent 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 inthis 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; ifthere is no velocity, there is no Coriolis force.

2. The Coriolis force increases with increasing latitude; it is a maximum at the North andSouth Poles, but with opposite sign, and is zero at the equator.

3. The Coriolis force always acts at right angles to the direction of motion. In theNorthern Hemisphere, it acts to the right (for an observer looking in the directionof motion); in the Southern Hemisphere (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 littlehelp in predicting what will occur. Pendulums rotate and particles are accelerated normalto their direction of movement. An interesting example is to consider what happens to aball that rolls down a frictionless inclined plane with a slope i. In the absence of a Coriolisterm, the governing equation is simply

du .=-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

X = — —1 git22

(5.15)

(5.17)

(5.18)

E

However, if the Coriolis force is added, the equation becomes

61-u

= -gi + fydtdv, -fudt

the solution of which is

Particle Path Isa Cycloid


x = --7(1 — cos ft) .

gY = + --2- (ft — sin ft)

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

(5.19)

Figure 5.7 By adding a Coriolis acceleration to the equation governing a particle sliding down anincline 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). Theball must roll for 5 min to see a curvature of 1 part in 100. Even with an incline as small as0.1%, the ball will have traveled nearly 500 m in this time, and for the ball to reach thebottom and come back up to the top would require an inclined plane somewhat larger thanthe United States. To do the experiment on an inclined plane of only a few square mileswould require an inclined plane whose slope was measured in parts per million. Oneshould also note that we have adopted a "flat earth" coordinate system, which adds anadditional complexity to erecting an inclined plane with a slope of a few parts per million.


Gravity: Equal Potential Surfaces

With the assumed coordinate system, gravity acts along the z axis. Although gravity variesslightly from place to place, the change is insignificant for nearly any problem in physicaloceanography. Surface gravity changes about 0.5% (9.78 rn/s 2 at the equator and 9.83 m/s2at the poles). The decrease in gravitational potential is related to the spinning earth. The firstterm on the right in Eq. (5.14) is centrifugal acceleration, which varies between zero at thepoles and 0.034 m/s2 at the equator. (See also the discussion of gravity in the derivation ofthe Coriolis force.) The remainder of the 0.05 m/s 2 difference between the poles and theequator is related to the fact that the radius of the earth at the equator is about 22 km largerthan the polar radius, a circumstance that can be explained in large part by the expectedequilibrium shape of our rotating earth.

If the earth were of uniform density, gravity would decrease linearly with depth; butbecause the earth's density increases with depth, gravity actually increases with depththrough 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 valueof gravity is only 0.25% larger than at the surface.

Gravity measurements at sea are often made from a moving ship. An instrument thatmeasures 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 Figure5.6, the term Wu was divided into horizontal and vertical components. The horizontalcomponent 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 allobservations of gravity if made from a moving platform (see also the discussion in Box 5.3on the derivation of the Coriolis force). A ship with an eastward velocity of 10 knots wouldhave 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 theearth. Similarly, all ocean depths should have negative values, and these should becomeincreasingly negative as the depth increases. The convention is ignored in this section andthroughout 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 nochange in the gravitational potential energy 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 oneaverages out the waves and tides) is an equal potential surface. Its departure from such asurface 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 ofthe order of 10-5 or less (see "Geostrophic Flow" in Chapter 6), and nowhere in any oceandoes the mean sea surface depart by as much as 2 m from a common equal potentialsurface.

Bottom


However, the mean ocean surface is not flat and smooth. A radar altimeter in anearth-circling satellite shows the ocean skin as a highly wrinkled, complex surface, withbumps, troughs, and other departures from the mean as large as tens of meters and slopesas 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 nothomogeneous, and because of this the direction of the gravitational vector changes muchmore rapidly than would be the case otherwise. Figure 5.8 is a simple example of whatoccurs. Because the density of the seamount is greater than water, the gravitational vectoris deflected toward the greater mass of the seamount. Similarly, the gravitational vector isdeflected away from a deep trench filled with seawater. As a consequence, the mean seasurface of the oceans as observed from space bears a strikingly qualitative resemblance tothe underlying bottom topography: bumps over seamounts and troughs over trenches.

Equalpotentlal Sea Surface

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 ofthe gravity vector. As a consequence, the average slope of the sea surface (which, to a high degree ofapproximation, 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 windblowing over the surface to establish waves and currents, (2) provide an important mecha-nism for the exchange of kinetic energy between adjacent parcels of water, and (3) ulti-mately drain the kinetic energy from the ocean and turn it to heat energy. The processes bywhich 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


the stirring has stopped. By analogy, if one could turn off the wind, the sun, and the tidalforces of the moon and sun so that new energy was not imparted to the ocean, the oceanwould eventually "run down" and the ocean currents and waves would grow smaller andslowly 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 inocean temperature.

Friction causes the ocean currents to deaccelerate. One expression of ocean frictionis 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 idealfluids that one studies in elementary physics have no viscosity. They are infinitely slippery.The wind in an ideal atmosphere blowing over an ideal ocean would have no effect. Onewould slide by the other, causing no currents or waves. A viscous wind blowing over thesurface of a viscous ocean will set the surface water in motion. Because the water is vis-cous, the frictional stress applied to the water will be transmitted downward. If the windstops, the water will begin to slow down, and eventually the movement will stop as theeffect 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 m2iswhere

v=

Knowing the viscosity, the rate of momentum transfer can be calculated:

di)= - (5.22)

az

The process of molecular transfer of kinetic energy and momentum is similar to theprocesses discussed in Chapter 4 for molecular transfer of heat and salt. As with moleculardiffusion of heat and salt, the calculated rate of transfer of momentum and kinetic energyis much slower than that observed in the ocean. For example, assume there was a way tomaintain a surface current of 0.5 m/s at the surface skin of the ocean. How long would ittake 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. Eventhe most casual observational evidence suggests that the effect of such a surface currentwould be felt much deeper much sooner.

I t

1

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

0.1 0.2 0.3 0.4 0.5


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 4that turbulent stirring can enhance molecular diffusion, so can turbulence more rapidlytransfer and intermix parcels of water with different kinetic energies and thus enhance thetransfer by molecular viscosity of kinetic energy from one parcel of water to another. Aswith 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 eddy diffusion to molecu-lar diffusion have been used to relate eddy viscosity to molecular viscosity. Eddies orswirls of moving fluid are analogous to molecules, and mixing lengths are equivalent tomolecular mean free path lengths. As with eddy diffusion, a coefficient of eddy viscositycan 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 asubstitution 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 size, with a charac-teristic mixing length of 1 m, would be much less than that resulting from larger eddies200 km across, which break off from the Gulf Stream. Thus a wide range of eddy viscosityvalues can be found in the oceanographic literature.

Speed (m/s)

If eddy viscosity is considered an exact physical analog to molecular viscosity, thefrictional terms in the equations of motion can be given explicit form:

friction


friction

A (d2v d2v) A d2v= -cixT + 4-‘z (9z2

(9 2w .921 (92w(z) Ah( dx 2 ay2 + Az az2

(d 2u d2u) d2,4friction (x) = Ah TxT + 4- Az dz2

(5.23)

where the subscripts h and z denote horizontal and vertical eddy viscosity coefficients andthe units are those of kinematic viscosity (m2/s). Characteristic values are scale dependentand 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 theyalso allow for the transfer of momentum and kinetic energy from the atmosphere to theocean. 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 theinterior of the fluid is accomplished by an eddy viscosity process, as described earlier, thetransfer process can be written as

x 49214= pAz dz2

ar .92v

= PAz

where '1", and -ty are the x and y components of wind stress in units of force per unit area.The absolute magnitude of the wind stress term is thought to be proportional to the squareof the wind velocity (see "Wind Stress: Ekman Transport" in Chapter 6). A derivation ofEqs. (5.23) and (5.24) is given in Box 5.4.

Making the frictional dissipation term proportional to velocity, as in Eq. (5.21), ishelpful in picturing in a semiquantitative way how kinetic energy is dissipated in theocean. 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, asuseful as such an analogy is, it is unsettling to be forced to work with what might bethought 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 towhich it is applied.

(5.24)


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 Figure5.9, 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 ofmolecular viscosity:

durxz = —

aZ

where Txz is the shearing stress tensor. The first subscript refers to the direction of the stress; thesecond subscript refers to the plane in which the stress acts. The subscript z is normal to the xyplane; 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 threeelements of the stress tensor acting in each of three directions. In the x direction, they are

du= p--xz

du

= - (5.13)xyay

durxx = 11—ax

Consider the force in the x direction that these stresses apply to a fluid element (Figure5.3'). The force applied to the cube is the difference between the shearing force in the upper andlower faces.

Figure 5.3'

In terms of the acceleration of the mass of the cube, the x-directed stress forces are

du dum—dt = (PArAY4z)—dt = xz2 (A t4Y) xZI (A X AY)

du 1 (r xz, xz i ) _ 1 41-

dt p AZ P AzIn 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

dt = —27P

By a similar argument, the other stress terms can be found, and

du 1 d2 u 1 d2 u l d2u—dt = P —p —Taz 2 +P P dx2

a2vdv 1 d2 v l d2 v

• P dz P dYdw 1 d2 w 1 d2 w 1 d2w

dt

-

=11—p —Fdz +11---i-+11— 2P dY P dx

—d =-11 V 2V = W2Vdt p

where v is the kinematic viscosity, which is of the order of le m 2/s. On a scale of meters, thelargest values of velocity shear observed in the ocean are of the order of 5 x 10 -2/s. Thus it isunlikely that molecular viscosity plays an important role in the equations of motion inoceanography.

The ocean is fully turbulent. Swirls and eddies of dimensions from tens of centimeters totens 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, moreor less, of constant geopotential 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 byrandom molecular motion in the case of molecular viscosity. It is then possible to define an eddyviscosity coefficient or Austausch coefficient (A) analogous to the molecular kinematic viscositycoefficient. By analogous arguments, one can derive the eddy viscosity version of Eq. (5.14):

— cTz "z az r ay y ay ax xdu _ d ( A dis)_, ±(A24A du)

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 mostcases when these terms are used. Another assumption is that

A, = Ay = Az

where the subscript refers to horizontal eddy viscosity, and thatA h >> A.

du d2u A (3214 a2U

- = ith -dy2dt

dv d2vd2v d2v— = A —+ Ah (—+—)Z az2 ax2 dy2dt

(5.14')

(5.15')

(5.16')


Reynolds Stresses. The most satisfactory formulation for the frictional transfer of kineticenergy is in terms of Reynolds stresses. Unfortunately, it is still difficult to make oceanobservations 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 impossiblewith flow in the z direction, when both the mean velocity and the turbulent component areso small.

igurerans-ss of

; thee xyxes-hree

13')

;ureand

.3'


To derive the Reynolds stress terms, it is necessary to subtract the mean from theinstantaneous 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:

By suitable averaging, one can separate the velocity components in the equation ofmotion, as well as in the equation of continuity Eq. (4.10), into terms that carry the meanvelocity 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 bar denotes atime-averaged mean of the product of the turbulent components:

friction (x) = --d u'u' )—dx dy dz

friction (y)= --(v'u')- (v'v' )-dx)

(5.26)0Y dz

friction (z) = --d

(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 Xu'v' = —A — u'w' = —A —

Y dy z az

v'v' = -A —X X

v'w' = —Az —Y A,dZ

c..j.

u'u = —A —x ax

—v'u' = —x

(5.27)

As the turbulent velocity components increase, one would expect the coefficient of eddyviscosity to increase. Because the vertical turbulent velocity values are so much smallerthan the horizontal turbulent velocity values, one would expect from this simple scaleanalysis alone that A, < < Ah.

Box 5.5 Reynolds Stress Terms

The key to deriving the Reynolds stress terms is understanding the averaging process. Any scalarvelocity 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 asobserved over time T. The instantaneous velocity can be written in terms of its average value andfluctuating component u

(5.25)


u=a+teIt follows that the average value of u', averaged over the same period, must be zero.

=-1 f u'dt=0T 0

Consider next the averaging of the product of separate components u and v:

uv = (17 + u')(i + v')= t717 + +17u' + u'v'Now take the average of each of the four terms over the same time interval T:

T T- 1 r 1 __ 1 _.'uv = — uvdt — uv dt + — f uv' dt+— v-14' dt u'v dt

T 0T0T0T0T0

=uv = uv + uv + vu + u vSince ii and g are constant over the interval T and since LT' =17' = 0

uv = -u-v+u'v' (5.19')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 manipulateit 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

p, +(pu)x+(pv)y+(pw),= 0,

Multiplying the continuity equation by u,

up, + u(pu) x + u(pv)y + u(pw) z = 0

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),+ pwu z i= -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 averageover the same time interval T:

[(ii + ulp] t +[p(Ti + ulx+

he

5)

ofanir-; a

6)

ipI-

7)

+[p(u-u+ u'u') ]+[p(u-v +)]+[p(u-w+ u'w') ]= -p x +s

The averaging bars have been dropped over terms assumed constant over the time interval.Rearranging terms,

(5.18')


(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 thaninstantaneous velocity components are used, and there are three additional terms representing thecross 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 they component. Where the averaging process implicit with the Reynolds stress consideration isunderstood, the averaging bars can be dropped, and we can write

u, + uux + vuy +wuz =_!p + fv–(u'ul x – (u'v') y –(u'w'),

v,+ uvx +vvy +wvz =1

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 afictitious force in that it is required because we assume a nonrotating coordinate systemwhen in fact all our observations are made on a rotating earth. We have further defined ourcoordinate 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 analyticalexpressions 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 bothpartial differential and vector form using generalized friction terms X 9; 01) .

du du du + du - "Pv dy wc-5--z---p7ox+fv+X

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

—3V +(V. V)V = Vp - 2flxV+g+431

In future chapters, we will use different formulations of the friction terms depending onthe process being examined. In many cases, the frictional terms are so much smaller thanthe 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, notbecause there will be any attempt in this text to consider Eq. (5.28) in any detail, butbecause the relationships that are discussed in succeeding chapters should be consideredagainst Eq. (5.28). These relationships are all approximations in that certain terms are

(5.21')


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

Vorticity

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

dv duax dy

It is possible to write a series of equations similar to Eq. (5.28) showing the balanceof torques and rate of change in vorticity. For many problems that oceanographers treat, itis more convenient to think in terms of the balance of vorticity than of the balance of linearforces, 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 termsof Eq. (5.28). The vorticity balance equation equivalent of Eq. (5.28) is Eq. (5.30). For aderivation of the vorticity equation, see Box 5.6.

dXt,)4 + + 4 &+v1+(4+f(--d—

du a

x + Ty ) ax dYv + w737 u -d-x

dy- dz dY

Note that the pressure-gradient term disappears when the equations of motion arewritten 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 asthe Gulf Stream are narrow and swift (see "The General Circulation: Western BoundaryCurrents," 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 dynamics

They include

(5.29)

(5.30)

relative vorticity :

absolute vorticity :(5.31)

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

a.

b.


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

Column Thickening \No Change In Latitude

1 Column Moving Northward

No Change in Thickness

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

Box 5.6 Vorticity

Voracity can be defined as the cross product of velocity:

V xV = i(w y —v )+ — wx )+k(vx —uy )

When V xV = 0 , the motion is said to be irrotational. The two horizontal components ofthe equation of motion, Eq. (5.28), are often written in the form of a single vorticity equation interms of the vertical vorticity, which is defined as 4 = avkx- away. We begin with the horizontalcomponents of the equation of motion.

(5.22')


(5.23')

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

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

Since

+337 (.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

Dt Z

The term (4+ f)/ z is the potential vorticity.

(5.24')

(5.26')

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CHAPTER FIVE - Hunter College · Doing so makes the horizontal xy plane an equal potential surface ... the total differential is often written ... the hydrostatic pressure,