sign in sign up
<<
Home , Equations of motion

330 4.7 of in Moving Coordi- nate Systems

Moving coordinate systems are discussed in this section in the context of general mechanical systems. The theory is applied to the equations of fluid and .

4.7.1 Moving Coordinate Systems An inertial coordinate system is a coordinate system in which Newton’s laws of motion are valid; in particular, a free particle moves along a (straight) line. Imagine an inertial coordinate system in Euclidean , with rectangu- lar coordinates (⇠,⌘,⇣) and the motion of a particle whose vector in this coordinate system is R. Suppose there is a second (moving) rectangular coordinate system (with coordinates (x, y, z)) such that, after the transla- tion in space that moves its origin to the origin of the inertial coordinate system, its translated frame (ordered basis) of coordinate unit direction vec- tors [ex,ey,ez] can be rigidly rotated in space to coincide with the inertial frame [e⇠,e⌘,e⇣ ]. In other words, assume that the moving frame is given at each instant of by a and of the inertial frame. Translation in space, which more properly should be called parallel trans- lation or parallel transport, is given simply by vector addition. For example, let denote the vector from the origin of the moving frame to the origin of theV inertial frame. The vector Q := R ,wheretheadditioniswith respect to the inertial vector space, representsV the parallel translation of R along the straight line joining the origins to the position of the origin of the moving frame. Likewise, the inertial frame [e⇠,e⌘,e⇣ ], is parallel transported to frame [e1,e2,e3]. The vector connecting the origin of the moving frame to the point Q in space is expressed in the coordinates of the moving frame by

a Q = b . 0 c 1 @ A In other words,

Q = aex + bey + cez. Fluids 331

The basis vectors for the moving frame have coordinate representations in the parallel transported inertial coordinates:

a11 a12 a13 e = a ,e= a ,e= a . x 0 21 1 y 0 22 1 x 0 23 1 a31 a32 a33 @ A @ A @ A Thus, the coordinate representation of Q in the parallel transported inertial coordinates is a11 a12 a13 a a + b a + c a . 0 21 1 0 22 1 0 23 1 a31 a32 a33 Or, equivalently, @ A @ A @ A ↵ a11 a12 a13 a = a a a b 0 1 0 21 22 23 1 0 1 a31 a32 a33 c are the coordinates of R@inA the parallel@ transportedA @ A inertial frame. In a more compact form, R = + AQ, (4.78) V where A is the with components aij.Thismatrixchangestheparallel transported inertial frame [e1,e2,e3]tothemovingframe;thatis,

Ae1 = ex, Ae2 = ey, Ae3 = ez. The addition in (4.78) is with respect to the moving frame. 1 T The matrix A is an orthogonal transformation; that is, A = A with respect to the usual inner product, which will be denoted by angled brackets (see Exercise 4.22). Because A is orthogonal, hi AAT = AT A = I.

Using this fact, R = A(AT + Q). (4.79) V In other words, the coordinates of R in the parallel transported inertial frame are the components of the vector AT + Q. By Newton’s second law, the equationV of motion of the particle with position R moving in the inertial frame is

mR¨ = F, (4.80) 332 Fluids where m is the of the particle and F is the (vector) sum of the acting on the particle. Our goal is to express the equation of motion of the particle in the moving coordinate system. To do this, the quantities r := AQ, v := AQ,˙ a := AQ¨ will be employed to simplify the calculations used to obtain the transformed equation of motion. They are the position, , and of the particle in the parallel transported inertial frame. In view of equation (4.78), the velocity of the particle can be expressed in the form R˙ = ˙ + AQ˙ + AQ,˙ V = ˙ + AA˙ T (AQ)+v V = ˙ + AA˙ T r + v. (4.81) V To make further progress, note that the transformation ⌦:= AA˙ T has a special property: it is skew-symmetric; that is, ⌦T = ⌦. This fact is easily proved by di↵erentiation with respect to time in the identity AAT = I.In fact, AA˙ T + AA˙ T =0; (4.82) therefore, ⌦T =(AA˙ T )T = AA˙ T = AA˙ T = ⌦. By using the definition of skew-, the matrix representation of ⌦ must have the form 0 ! ! 3 2 ! 0 ! . (4.83) 0 3 1 1 ! ! 0 2 1 The of ⌦on a vector@W given by ⌦WAcan also be represented using the vector cross product (with respect to the usual orientation of space) and the vector ! := (!1,!2,!3)by ⌦W = ! W, (4.84) ⇥ Thus, the velocity of the particle may be expressed in two ways: R˙ = ˙ +⌦r + v, V R˙ = ˙ + ! r + v. (4.85) V ⇥ Fluids 333

Using the first equation in display (4.85) for the velocity of the particle, its acceleration is

R¨ = ¨ +˙v + ⌦˙ r +⌦r.˙ (4.86) V Since r = AQ and v = AQ˙ ,theirtimederivativesare

r˙ = AA˙ T (AQ)+AQ˙ =⌦r + v, (4.87) and

v˙ = AA˙ T (AQ˙ )+AQ¨ =⌦v + a. (4.88)

These formulas forr ˙ andv ˙ are substituted into equation (4.86) to obtain the particle’s acceleration in the form

R¨ = ¨ + a +2⌦v + ⌦˙ r +⌦2r;(4.89) V or, equivalently,

R¨ = ¨ + a +2! v +˙! r + ! (! r). (4.90) V ⇥ ⇥ ⇥ ⇥ The equation of motion (4.80) may be recast in the forms

ma = m( ¨ +2⌦v + ⌦˙ r +⌦2r)+F (4.91) V V or ma = m( ¨ +2! v +˙! r + ! (! r)) + F , (4.92) V ⇥ ⇥ ⇥ ⇥ V where the vectors a, v,andr give the acceleration, velocity and position of the particle, moving under the influence of the F ,withrespecttoan observer in the moving coordinate system but referred to the frame [e1,e2,e3]. These equations in the frame [ex,ey,ez]areobtainedbymultiplicationonthe left by the orthogonal matrix AT .Forexample,usingequation(4.91),the equation of motion is given by

mAQ¨ = m( ¨ +2⌦AQ˙ + ⌦˙ AQ +⌦2AQ)+F ;(4.93) V V and, in the moving coordinate frame [ex,ey,ez], it is mQ¨ = m(AT ¨ +2AT ⌦AQ˙ + AT ⌦˙ AQ + AT ⌦2AQ)+AT (F ). (4.94) V V 334 Fluids

The last equation can also be written in the form

mQ¨ = m(AT ¨ +2Q˙ + ˙ Q +2Q)+AT (F ), (4.95) V V where := AT ⌦A;thatis,and⌦arenamesforthesamelineartransfor- mation expressed in di↵erent bases. Since the equation of motion (4.91) (or (4.95)) is a valid expression of Newton’s equation of motion for the particle on the influence of the force F from the point of view of a non-inertial observer sitting at the moving origin and measuring with respect to the translated inertial coordinates, this observer treats the terms on the right-hand side of the equation of motion as additional forces. The vector ¨ is the acceleration of the origin of the non- inertial frame, ⌦˙ r =˙! r is theV acceleration due to the rotation of the moving frame, 2m⌦v =2m!⇥ v is the , and m⌦2r = m! (! r) is the .⇥ Of course, from Newtonian mechanics, F is⇥ the⇥ only force acting on the particle. The remaining fictitious forces are artifacts of reference to a non-inertial coordinate system.

Exercise 4.22. Prove that the coordinate transformation A in display (4.78) is orthogonal.

4.7.2 Pure Rotation To consider pure rotation, imagine a rotating disk whose motion is measured with respect to a rectangular rotating coordinate system (with coordinates (x, y, z)), whose third coordinate-axis coincides with the axis of rotation of the disk, whose origin is at the intersection of the disk with this axis, and whose first two coordinates are fixed in the disk. In addition, suppose without loss of generality, that an inertial frame with coordinates (⇠,⌘,⇣)hasthesame origin. Of course, we could choose the inertial coordinates so that the ⇣-axis corresponds with the axis of rotation; but, we will not make this assumption to illustrate an important feature of three-dimensional space: Each rotation is determined by a vector (specifying the axis of rotation) and the rotation about this direction. Thus, under our assumption that there is a fixed axis of rotation, there is an orthogonal transformation B,whichdoesnot depend on time, taking the in the direction of the ⇣-axis of the inertial frame to one of the two unit vectors in the direction of the axis of rotation. The matrix A,whichdeterminesthecoordinatetransformation Fluids 335

R = AQ from the moving frame to the inertial frame, takes the form BT B, where R cos ✓ sin ✓ 0 := sin ✓ cos ✓ 0 (4.96) 0 1 R 001 @ A is the rotation matrix (in inertial coordinates) for the motion about the z-axis in the moving frame. By a direct computation, it is easy to check that

0 ✓˙ 0 ⌦=AA˙ T = ✓˙ 00 , (4.97) 0 1 000 @ A which is exactly the in matrix form; that is, the action of this matrix on a vector W is W, ⇥ where is the vector with components (0, 0, ✓˙). For in three-dimensional space, we have the useful identity

AA˙ T = AT A.˙ (4.98)

Thus, ⌦= AT A˙. For pure rotation where the origins of the inertial and rotating coordinates coincide, the equation of motion (4.93) reduces to

mAQ¨ = m(2⌦AQ˙ + ⌦˙ AQ +⌦2AQ)+F ;(4.99) and, in case the is constant, to

mAQ¨ = m(2⌦AQ˙ +⌦2AQ)+F. (4.100) In this case, only the centrifugal and Coriolis forces appear.

Exercise 4.23. Revisit Exercise ?? and derive the equation of motion referred to a coordinate system rotating with the hoop using the abstract equation of motion (4.100). 336 Fluids 4.8 Motion in Rotating Coordinates

The equation of motion (4.94) of a particle in a moving coordinate system is valid for a particle of fluid. We will obtain the transformation of the Eulerian equations of motion for a fluid to a moving coordinate system. Let us consider the equations of motion (conservation of and mass) for a fluid in a gravitational field

Du ⇢ = p + µu + ⇢g Dt r D⇢ = ⇢ u, (4.101) Dt r· where it is important to note that the on the left-hand sides are the material derivatives of the velocity and the density respectively; that is, Du @u = + u u, Dt @t ·r D⇢ @⇢ = + u ⇢. Dt @t ·r Also, as a remark, note that the equations of motion do not form a closed sys- tem; there are three state variables u, p,and⇢ and only two equations. More precisely, there are five state variables (if we include as separate variables the three components of u)andfourequations(ifweviewtheconservationof momentum as three equations). To close the system we need one addi- tional scalar equation, for example, an equation of state that relates density and or the statement that the density is constant. For simplicity, let us consider a moving coordinate system whose origin is fixed at the origin of the inertial coordinates and the transformation from the moving coordinates to the inertial coordinates given by

R = AQ.

The previous analysis of the transformation of the equation of motion for a particle will be repeated almost verbatim except that, for fluid motion, the state variables are functions of both space and time. The density is a scalar field; that is, at each point of space and time, this field assigns a real number. It is clear that this number is invariant under a arbitrary change of coordinates. We will show that the material Fluids 337 of a scalar field is invariant under our (time-dependent orthogonal) change of coordinates. How does a scalar field change coordinates? This is easy, we simply compose the field with the change of coordinates. Consider a (perhaps time-dependent) scalar field given in the inertial coordinates by f(R, t)and in the moving coordinates by F (Q, t). The coordinate representations are related by f(R, t)=F (AT R, t). (4.102) We will show that the material derivatives are similarly related; that is, d d d f(R, t)= F (AT R, t)= F (Q, t). (4.103) dt dt dt Indeed, using the identity (4.82), we have that d f(R, t)=DF(AT R, t)(A˙ T R + AT R˙ )+F (AT R, t) dt t T T T T = DF(A R, t)(A˙ R + A (AQ˙ + AQ˙ )) + Ft(A R, t) T T T T = DF(A R, t)(A˙ AQ + Q˙ + A AQ˙ )) + Ft(A R, t)

= DF(Q, t)Q˙ + Ft(Q, t) d = F (Q, t). (4.104) dt By our result for scalar fields, we see that the left-hand side of the equation for the conservation of mass is invariant under our coordinate transformation. Of course, this means that the right-hand side is also invariant. To see this fact more directly, let us first note how a (perhaps time-dependent) vector field changes coordinates. The general principle is simple from a geometric point of view. Let X be a vector field on Rk and let F be a transformation from Rk to Rk given by R = F (Q, t)(whereourchoiceofsymbolsreflects the application to the fluid equations). We wish to find the vector field Y corresponding to X via this change of coordinates. To do so, pick (q, ⌧)and r such that r = F (q, ⌧)anda(smooth)curve : J Rk,whereJ is an ! open interval in R containing ⌧,suchthat(⌧)=q and d ˙ (⌧):= (s) = X(q, ⌧). ds s=⌧ For example, we can take (s)=q+(s ⌧)X(q, ⌧). The change of coordinates defines a new curve s F ((s),s) 7! 338 Fluids whose tangent vector at s = ⌧ is Y (r, ⌧ ); that is, d Y (r, ⌧ ):= F ((s),s) . ds s=⌧ By the chain rule, we also have @F Y (r, ⌧ )=DF(q, ⌧)X(q, ⌧)+ (q, ⌧). (4.105) @t For our special change of coordinates R = F (Q, t)=AQ and equa- tion (4.105) takes the form

Y (r, ⌧ )=AX(AT r, ⌧ )+AA˙ T r. (4.106)

Using as before ⌦:= AA˙ T ,wealsohavethegeneralrelation

Y (r, ⌧ )=AX(q, ⌧)+⌦Aq, (4.107) which for the fluid velocity field is

u(r, ⌧ )=AU(q, ⌧)+⌦Aq. (4.108)

The right-hand side of the conservation of mass (the second equation in display (4.101)) is ⇢ u,wherethederivativeisthedivergenceofthe vector field u.Intheinertialcoordinates,thedivergencecanalsobeviewedr· as the trace of the derivative of u.Thetraceisaninvariantofalinear transformation (it can be defined as the sum of the eigenvalues counting multiplicities). The trace is also give by the formula

k tr(B)= Be ,e , (4.109) h i ii i=1 X where B is a linear transformation, e1,e2,...,en is an orthonormal basis and the angled brackets denote the{ inner product.} Since the trace is an invariant, it follows that

1 tr(B)=tr(CBC ) whenever C is an invertible matrix. Let us also note an easy consequence of (4.109): the trace of a skew-symmetric matrix is zero. Fluids 339

Using equation (4.107), it follows that

DY (r, ⌧ )=ADX(AT r, ⌧ )AT +⌦. (4.110)

Hence, Y (r, ⌧ )= X(q, ⌧), (4.111) r· r· where, for example, Y denotes the divergence of the vector field Y . By applying formular· (4.111) to the fluid velocity field u,wehaveproved that the conservation of mass in the moving coordinate system is given by d% = % U, (4.112) dt r· where ⇢(R, t)=%(Q, t). Let us determine the change of coordinates for the of avectorfield.Tothisend,supposethat

R˙ = u(R, t); that is, u is the velocity field of the (time-dependent) position R. Using equation (4.108), ⌦:=AA˙ T , and Q˙ = U(Q, t), the of the u(R, t)is d d u(R, t)=AU˙ (Q, t)+A U(Q, t)+⌦AQ˙ +⌦AQ˙ + ⌦˙ AQ dt dt d = A U(Q, t)+2⌦AU +⌦2AQ + ⌦˙ AQ. (4.113) dt Here, for instance, the time derivative of u(R, t)mustnotbeconfusedwith the @u(R, t)/@t;thetimederivativeisthederivativeofthe function t u(R(t),t). Also, note that new terms are introduced due to the time-dependent7! change of coordinates. To complete the change of coordinates for the conservation of momentum, we must determine the gradient of the pressure and the Laplacian in the moving coordinate system. For a scalar field, for example the pressure field, our change of coordinates is expressed by the relation (4.102). By di↵erentiation, we have that

Df(R, t)=DF(Q, t)AT ; 340 Fluids therefore, f(R, t)=(DF(Q, t)AT )T = A F (Q, t). (4.114) r r In particular, p(R, t)=A P (Q, t). (4.115) r r The Laplacian u is the vector field whose components are the Laplacians of the components of u. We will show that if the moving coordinates are given by R = AQ and u(R, t)=AU(Q, t)+A⌦Q,then

u = AU. (4.116)

Let us suppose that the vector u has components ui.Eachui is a scalar field. We will compute ui in the moving coordinates. Using the relation between u and U and the usual basis e ,e ,e ,wehavethat { 1 2 3} T T T ui(R, t)=ei u(R, t)=ei AU(Q, t)+ei ⌦AQ. (4.117)

According to the general formula (4.114), which expresses how the gradient changes coordinates, it follows that

T T ui(R, t)=A (ei AU(Q, t)+ei ⌦AQ) r rT T T = A(ei ADU(Q, t)+ei ⌦A) T T T T = A(DU(Q, t) A ei + A ⌦ ei) = A(DU(Q, t)T AT e + AT ⌦T e AT ⌦AQ)+⌦AQ. (4.118) i i According to the general formula (4.111), which expresses how the divergence of a field changes coordinates, it follows that

u (R, t)= u (R, t) i r·r i = (DU(Q, t)T AT e + AT ⌦T e AT ⌦AQ) r· i i = (DU(Q, t)T AT e )(4.119) r· i T T because A ⌦ ei is a constant with respect to the space variables and the divergence of AT ⌦AQ vanishes. The desired result follows from the formula

eT AU(Q, t)= (DU(Q, t)T AT e ). (4.120) i r· i To prove it, we will simply compute both sides of equation (4.120) using the components of U and A.LetA have components aij and the vector Fluids 341

U have components Ui.TheLaplacianofU is a vector with components Ui,``, where the subscripts following the comma denote partial derivatives with respect to the corresponding variables Q1, Q2 and Q3,andsummation on repeated indices in a single term is assumed. Using this notation and T the (Einstein) summation convention, AU = ajkUk,`` and ei AU(Q, t) is the vector with components aikUk,``.Fortheright-handside,thevector T T DU A ei has components aikUk,`.Itsderivativeisamatrixwithcomponents aikUk,`m and trace aikUk,``,asrequired. We have proved that the conservation of momentum equation (4.101) in the moving coordinates (referred to the inertial frame) is given by dU %A + A P = µAU + %g %(2⌦AU +⌦2AQ + ⌦˙ AQ)(4.121) dt r and referred to the moving frame is dU % + P = µU + %AT g %(2AT ⌦AU + AT ⌦2AQ + AT ⌦˙ AQ)(4.122) dt r or, with such that := AT ⌦A and W = W as in the representa- tion (4.84), ⇥ dU % + P = µU + %AT g %(2 U + ( Q)+ ˙ Q). (4.123) dt r ⇥ ⇥ ⇥ ⇥ The identity ˙ :=AT ⌦˙ A (4.124) requires a proof that is left to the reader.

4.8.1 Water Draining in Sinks Versus Hurricanes Hurricanes always rotate counterclockwise. Large scale low pressure systems in the Southern Hemisphere always rotate clockwise. Water (usually) drains counterclockwise in a bathtub or sink in the Northern Hemisphere and clock- wise in the Southern Hemisphere. The model equations for fluid dynamics will be used to discuss these phenomena. For fluid motion on the surface of the earth, as viewed by an observer rotating with the earth, the apparent motion is governed by equation (4.123) and the corresponding equation of continuity (4.112). By custom, we take our (right-hand rectangular) inertial coordinate system (⇠,⌘,⇣)withthepositive 342 Fluids

⇣-axis passing through the North Pole and the origin at the center of the earth. The rotation is counterclockwise (west to east) looking down from above the North Pole. We will consider a (right-hand rectangular) rotating coordinate system (x, y, z) fixed to the earth whose origin is at the center of the earth and whose positive ⇣-axis passes through the North Pole. The orthogonal transformation A from the rotating to the inertial coordinates is

cos ✓ sin ✓ 0 A = sin ✓ cos ✓ 0 . (4.125) 0 1 001 @ A As in equation (4.97),

0 ✓˙ 0 ⌦=AA˙ T = ✓˙ 00 (4.126) 0 1 000 @ A and, by simple calculations using the identity (4.98),

0 :=AT ⌦A =⌦, = 0 . (4.127) 0 1 ✓˙ @ A Let us assume that the angular velocity of the earth is

2⇡ 1 5 1 ✓˙ := hr 7.3 10 sec . (4.128) 24 ⇡ ⇥ The equation of motion (4.123) applied to the earth takes the form

d 1 µ U + P = U + g 2 U ( Q). (4.129) dt %r % ⇥ ⇥ ⇥

To determine the e↵ect of the Coriolis acceleration 2 U for air moving with velocity U,whichismeasuredbytherotatingobserverrelativetowhat ⇥ is perceived as the stationary surface of the earth, imagine a position Q in the Northern Hemisphere on the surface of the earth and note that the vector at Q can be decomposed into a vector tangent to the earth at Q and a vector normal to the earth at this point. In reality, the velocity U is a vector in space that may not be tangent to the earth, but for a large scale weather Fluids 343 system (like a hurricane) we may measure velocity as if it were tangent to the surface of the earth. The formation of a hurricane (in an idealized scenario) starts when there is alargelowpressureairmasssurroundedbyrelativelyhigh-pressureair.The pressure di↵erence causes high pressure air to move toward the low pressure region (see Exercise 4.27). Imagine a circle parallel to the earth whose center is at the center of the low pressure region. In an ideal situation, the velocity vectors of the air flow at each point on the circle point toward its center. Recall that the cross product ↵ of two vectors ↵ and has length ↵ sin ',where' is the angle between⇥ the vectors and it points in the direction,| || | say n,suchthat(↵,, n)isaright-handsystemofvectors;or, equivalently, the matrix [↵,, n]partitionedbycolumnshaspositivedeter- minant. In this case, we say that the basis ↵,, n is positively oriented. The cross product of a vector that is tangent{ to} the surface of the earth and the velocity field U is a vector that is normal to the earth; it does not a↵ect rotation. On the other hand, the cross product of U and a vector normal to the earth is tangent to the earth. In particular, consider the normal component of taken in the direction of the outer normal on the surface of the earth and the velocity vector U,whichpointstowardthecenterofthe circle with center at the center of the low pressure air. By the right-hand rule, the cross product of these vectors is tangent to the circle and points in the clockwise direction on the circle viewed from above. The Coriolis acceleration is twice the negative of this cross product. Hence, the Coriolis acceleration is counterclockwise in the Northern Hemisphere. In the Southern Hemisphere, the analysis is the same except that the normal component of points toward the center of the earth instead of away from the center of the earth. If follows that the Coriolis acceleration is clockwise. Exactly the same analysis that shows the Coriolis acceleration is coun- terclockwise in the Northern Hemisphere applies to water moving toward adrain.Byobservation,hurricanesalwaysrotatecounterclockwiseinthe Northern Hemisphere but water does not always drain in this direction. What is the di↵erence? It must be that the size of the Coriolis e↵ect is di↵erent. How can we determine the size of this e↵ect? The determination of relative size is not obvious in fluid mechanics. The usual comparison is via scaling as in the determination of the Reynolds num- ber in the scaled fluid model equations (4.35) (see [18]). We will consider a length scale L (a characteristic length) and a velocity scale V (a characteris- tic velocity) together with the natural induced time-scale T := L/V . Notice 344 Fluids that in equation (4.129), all the terms have units of acceleration L/T 2. Us- ing our scaling, the fluid acceleration dU/dt can be viewed as having units L/T 2 = V 2/L.Itsrelativesizeisviewedasthesquareofthecharacteristic velocity divided by the characteristic length. The relative size of the Coriolis acceleration is twice the characteristic velocity the angular velocity. By custom, the relative size is measured by the (dimensionless) Rossby number (see [35, 49]) V Rossby number := fluid acceleration/Coriolis acceleration = . 2 L | |(4.130) The Coriolis acceleration is important whenever this number is smaller than one. For a hurricane, we may take as a characteristic length the diameter of the low pressure region, say L = 1000kilometers and a characteristic velocity of V = 100kilometers/hour. Using the angular velocity of the earth given in display (4.128), the corresponding Rossby number is 2400 0.19. 4000⇡ ⇡ Thus, the Coriolis force is important. For water draining in a sink, we may take the diameter of the drain (say 5 centimeters) as a characteristic length. Let us leave the characteristic velocity V (measured in centimeters/second) unspecified for the . The Rossby number is

V 105 218V. (4⇡) 7.3 5 ⇡ ⇥ ⇥ Hence, the Coriolis force is not important unless the fluid velocity is on the 2 order of 0.5 10 centimeters/second, which seems small for water draining in a sink except⇥ possibly for still water as it just begins to drain. Does this cause water draining in a sink to rotate counterclockwise (see [?,shapiro]? Our discussion so far leaves out consideration of the centrifugal accelera- tion ( Q). This term is given in components by ⇥ ⇥

Q1 ( Q)= ✓˙2 Q . (4.131) 0 2 1 ⇥ ⇥ 0 @ A Fluids 345

˙2 2 2 In other words, this vector has magnitude ✓ Q1 + Q2;itpointstowardthe axis of rotation and is parallel to the equatorial plane. The magnitude is p largest at the equator and vanishes at the poles. Because of its direction, the centrifugal acceleration is naturally added to the gravitational acceleration thus defining an e↵ective gravitational force acting on a fluid at the surface of the earth. The e↵ective does not point toward the center of mass of the earth; it also has a nonzero component that points toward the poles. In practice, this term is often neglected because it is small relative to the gravitational acceleration (see Exercise 4.29).

Exercise 4.24. Suppose a particle moves radially outward with constant velocity on a disk that is stationary with respect to an inertial coordinate system. Deter- mine the motion of the particle with respect to a rotating coordinate system that is rotating with angular velocity #. Assume that the first two coordinate axes of the rotating frame rotate in the plane of the disk and its third axis is perpendicular to the disk at the center of the disk. The inertial frame has the same configuration but is not rotating.

4.8.2 A Counterintuitive Result: The Proudman-Taylor Theorem Imagine a cylindrical tank partially filled with water that is being rotated rapidly at a constant angular velocity about its axis of symmetry. The Rossby number, for a given characteristic fluid velocity and characteristic length, will approach zero as the rotation increases without bound. Thus, it is (physically) reasonable to ignore the inertial acceleration given by the material derivative of the velocity of the fluid. The resulting equations of motion are called the geostrophic equations. Likewise, because the of water is small, it seems reasonable to ignore the Laplacian term in the equations of motion. By doing so, we are tacitly expecting that solutions with small non-zero viscosity will be near the corresponding solutions with zero viscosity. While this expectation may be true in special cases, results in this setting—called singular —where the highest-order deriva- tives are multiplied by a small parameter, are mathematically challenging. Undaunted, we will also assume that the density of the fluid remains con- stant and that the fluid is incompressible (that is, U = 0). With all these assumptions and in view of equation (4.129), ther geostrophic· equations for 346 Fluids inviscid incompressible flow are 1 P = g 2 U ( Q), (4.132) %r ⇥ ⇥ ⇥ U =0. (4.133) r· The Proudman-Taylor theorem states that every (smooth) solution of the inviscid incompressible geostrophic equations is constant in the direction of the rotation axis; that is, @U i =0 (4.134) @Q3 for i =1, 2, 3. To prove this theorem, apply the curl ( )tobothsidesofthe equation of motion. It is not dicult to simply writer⇥ out all the terms in components and compute. But, it is perhaps more elegant (and certainly more instructive) to use some results from vector analysis; in particular, for vector fields X and Y ,

( X)=curlgradX =0, r⇥ r (X Y )=( Y )X ( X)Y +(Y )X (X )Y r⇥ ⇥ r· r· ·r ·r =(divY )X (div X)Y + Y,grad X X, grad Y . (4.135) h ih i Using the first identity in display (4.135) and the constant density, it follows that the curl of the pressure term vanishes. The gravity term is constant; hence, it is clearly curl free. The vector is constant and U is divergence free; therefore, by an application of the second identity in display (4.135), we have that curl(2 U)= 2( )U. (4.136) ⇥ ·r By Exercise 4.30, 1 ( Q)= ( Q 2). (4.137) ⇥ ⇥ r 2| ⇥ | Thus, using again the first identity in display (4.135), this term is curl free. In summary, after applying the curl operator, we have proved that the velocity field must satisfy the equation

( )U =0. (4.138) ·r Fluids 347

The vector (as in equation (4.127)) is given by the transpose of the vector (0, 0, ✓˙); thus, in component form, equation (4.138) states exactly the desired result (4.134). This completes the proof. Returning to our rotating cylindrical water tank, let us suppose that the hypotheses of the Proudman-Taylor theorem are valid. At the surface of the water, we expect that U3 = 0. By the theorem, U3 =0everywhere and the fluid velocity U =(U1,U2, 0) does not depend on Q3;inotherwords, the flow is the same on every plane perpendicular to the axis of rotation. This is a remarkable claim. The mathematics is rigorous, but the result is based on several assumptions. Is the result true for real fluids? The answer is yes. G.I. Taylor performed several experiments to verify this result. In his first experiments (see [81]), small amounts of dye were injected into a small volume of a rotating cylinder filled with water. The dye was drawn out into two-dimensional sheets that were perpendicular to the bottom of the cylinder. In other words, the flow marked by the dye was the same in the vertical direction. In a second, more elaborate experiment, Taylor put a small solid cylindrical object on the bottom of a rotating tank and demonstrated by injecting dye into the flow that there is a stagnant cylindrical column that lay directly above the submerged cylinder. The two-dimensional flow on every horizontal plane knows that cylinder is there! A beautiful description of this experiment is in Taylor’s original paper [82]: ...the only possible two dimensional motion satisfying the re- quired conditions is one in which a cylinder of fluid moves as if fixed to the body. The boundary of such a cylinder would act as a solid body, and the liquid outside would behave as though a solid cylindrical body were being moved through it. No fluid would cross this boundary, and the liquid inside it would, in general, be at rest realive [to] the solid body. This idea appears fantastic, but the experiments now to be described show that the true motion does, in fact, approximate to this curious type.

Exercise 4.25. Prove the identity (4.124).

Exercise 4.26. Prove the identities (4.135).

Exercise 4.27. Prove that fluids move from high-pressure regions toward low- pressure regions.