Chapter 6

Conservation of : and Elastic

The description of the of fluids plays a fundamental role in applied mathematics. The basic model is a system of partial di↵erential of evolution type.

6.1 Equations of Motion

Fluids consists of ; thus, on a microscopic level, a fluid is a discrete material. To obtain a useful approximation, we will describe fluid motion on the macroscopic level by taking into account that act on a parcel of fluid, which we assume to be a collection of suciently many molecules of the fluid so that the continuity assumption is valid. More precisely, we will assume the fluid to be a continuous medium contained in three-dimensional R3 such that every parcel of the fluid, no how small in compar- ison with the whole body of fluid, can be viewed as a continuous material. Mathematically, a parcel of fluid (at every of ) is a bounded open subset A of the fluid that is assumed to be in an open set whose closure is the region containing the fluid. To ensure correctness of theR math- ematical operations that follow, we assume that each parcel A has a C1 boundary. At every given moment of time, a of fluid is identified with a point in . R Let ⇢ = ⇢(, t)denotethedensityandu = u(x, t)thevelocityofthefluid at x and time t R. The motion of the fluid is modeled by 2R 2 257 258 Fluids di↵erential equations for ⇢ and u determined by the fluid’s internal material properties, its container, and the external forces acting on the fluid. The components of u with respect to the usual coordinates of three- dimensional space are denoted by (u1,u2,u3). We assume that the u is twice continuously di↵erentiable and satisfies other properties that are necessary for the correctness of the mathematical operations used in the following discussion. We will make remarks about the additional properties of u as needed. The changing position of the particle of fluid with initial position x 0 2R at time t =0isgivenbythecurvet (x0,t)in that is the solution of the initial value problem (IVP) 7! R

⇠˙ = u(⇠,t),⇠(0) = x0 (6.1) (see Appendix A.3 for a theorem on existence of solutions of ordinary dif- ferential equations). When is viewed as a function of two variables : R3 R R3 it is called the flow of u. Note that the notation (A, t), where ⇥ ! A is a subset of R3,isthesetofallpointsinR3 obtained by solving to time t the di↵erential with initial condition ⇠(0) = x0 for each x0 in A. For each such initial point, the value of the solution at time t (a point in R3) belongs to the set (A, t). 3 Let x1, x2,andx3 denote the Cartesian coordinates in R and e1, e2, e3 the usual unit direction vectors. Using this notation, the vector u is u = u1e1 + u2e2 + u3e3. The in Cartesian coordinates (also called nabla or del) is @ @ @ := e1 + e2 + e3 r @x1 @x2 @x3 or, equivalently, @ @x1 := @ . r 0 @x2 1 @ B @x3 C @ A This operator applied to a function f : R3 R gives its gradient in Cartesian coordinates ! @f @x1 f = 0 @f 1 . r @x2 @f B C B @x3 C @ A Fluids 259

Applied to a vector field u (with the notation u,where denotes the usual inner product in ), the gradientr· operator gives· the in Cartesian coordinates @u @u @u u = 1 + 2 + 3 . r· @x1 @x2 @x3 The expression (u )u,oftenwrittenu u,isthevector ·r ·r u @u1 + u @u1 + u @u1 1 @x1 2 @x2 3 @x3 @u2 @u2 @u2 u1 + u2 + u3 . 0 @x1 @x2 @x3 1 u @u3 + u @u3 + u @u3 1 @x1 2 @x2 3 @x3 @ A AparcelA of fluid identified at time zero is moved to (A, t) at time t. Reynolds’s transport theorem states that d ⇢(x, t) dx = ⇢ (x, t)+div(⇢u)(x, t) dx (6.2) dt t Z(A,t) Z(A,t) (see A.11). By conservation of , the rate of change of the total mass in A does not change as the parcel is transported by the flow; therefore, the left-hand side of equation (6.2) vanishes. Since A may be taken arbitrarily small (for example, a ball with arbitrarily small radius), it follows that

⇢t +div(⇢u)=0 (6.3) or, equivalently, ⇢ + (⇢u)=0. (6.4) t r· Equation (6.4) (or (6.3)) is called the equation of continuity; it states that the mass of a fluid is conserved by the fluid motion. Equation (6.2) is a general statement of the rate of change of total mass that holds as long as u is an arbitrary (smooth) vector field with flow . Adi↵erentialequationforthevelocityfieldu is obtained from the equa- tion for the conservation of momentum using Newton’s second law of motion. Note that a fluid has mass. It might also be charged. Thus, a fluid is sub- jected to body forces, which by definition are forces that act per unit of mass or per unit of charge. The most important body is the gravitational force, which acts on every fluid simply because fluids have mass. Unlike the motion of or rigid bodies, fluids (by definition) have internal 260 Fluids

(force per area) that is caused by the of the fluid on itself. Stress is modeled by a function that assigns a vector in R3 to each pair consisting of a point (x, t) in space-time and a unit-length (space) vector ⌘ in R3 at this point. The value of the stress function at this pair is called the stress vector at the point x in the direction of the outer normal ⌘ at time t on imag- inary surfaces in space that contain this point and have this outer normal at time t. The stress vector has units of force per area. Using conservation of momentum and , it is possible to prove that the stress function at each point in space-time is a symmetric linear transformation of space (see [17]). Let us simply incorporate these facts about the stress as assumptions. Alinearandsymmetrictransformationonvectorsmayberepresented by a (diagonalizable) in the usual Cartesian coordinates. Thus, for each point (x, t)inspace-time,(x, t)maybeviewedasasymmetricmatrix, which is thus defined by six numbers (the elements on and above the main diagonal of the matrix). Total stress over the current position of parcel A at time t is given by

TS := (x, t)⌘(x, t) d , S Z@(A,t) where ⌘ is the outer unit normal on the boundary of (A, t)andd is the element of surface area. Using the body force b per unit of mass, theS total body force on (A, t)is

TB := ⇢(x, t)b(x, t) d , V Z(A,t) where d is the element of volume. TheV total momentum of (A, t)is

⇢(x, t)u(x, t) d . V Z(A,t) By Newton’s second law of motion (the time rate of change of momen- tum on a body is equal to the sum of the forces acting on the body), the mathematical expression for the conservation of momentum is d ⇢(x, t)u(x, t) d = (x, t)⌘(x) d + ⇢(x, t)b(x, t) d . dt V S V Z(A,t) Z@(A,t) Z(A,t) (6.5) Fluids 261

The region (A, t) in space is moving with time. Using the equation of continuity (6.3), the transport theorem (A.11), and some algebra, it follows that the left-hand side of the momentum balance (6.5) is given by

d ⇢(x, t)u(x, t) d = ⇢(x, t)(u +(u )u)(x, t) d , (6.6) dt V t ·r V Z(A,t) Z(A,t) where ut denotes the partial of u with respect to t. The expression ut +(u )u that appears in the right-hand side of equa- tion (6.6) is the material· derivativer of the velocity field u,whichisoften Du denoted by (x, t); its definition takes into account the motion of the fluid Dt particles with time:

d u((t, x ),t)=u ((t, x ),t)+Du((t, x ),t)˙(t, x ) dt 0 t 0 0 0 = ut((t, x0),t)+Du((t, x0),t)u((t, x0),t) =(u +(u )u)((t, x ),t) t ·r 0 Du = ((t, x ),t). Dt 0

The total stress in the direction of an arbitrary constant vector field v in R3 is given by ((x, t)⌘(x)) vd . · S Z@(A,t) Using the of the stress , we have

(⌘) v =(v) ⌘. · · This fact together with the implies

((x, t)⌘(x)) vd = div((x, t)v) d . · S V Z@(A,t) Z(A,t) By an easy calculation in components with represented as a matrix and v a constant vector, we have the identity

(v)=( )v. (6.7) r r· 262 Fluids

By our notation, div(v)= (v). Thus, we may view (the divergence of ), as a linear operation onr vectors. In Cartesian coordinates,r· this operator is given by @11 + @12 + @13 @x1 @x2 @x3 = @21 + @22 + @23 . r· 0 @x1 @x2 @x3 1 @31 + @32 + @33 @x1 @x2 @x3 Using these facts, the total stress@ is A

TS = (x, t)⌘(x) d = (x, t) d . (6.8) S r· V Z@(A,t) Z(A,t) By substitution of formulas (6.6) and (6.8) into the conservation of mo- mentum equation (6.5), we have the equivalent expression Du ⇢ ⇢b d =0 (6.9) Dt r· V Z(A,t) for conservation of momentum. This equation holds for every parcel of fluid. Under the assumption that the integrand is continuous (which might be astrongassumptioninsomecircumstances),thedi↵erentialequationfor conservation of momentum is Du ⇢ = + ⇢b;(6.10) Dt r· or, equivalently, Cauchy’s equation

⇢(u +(u )u)= + ⇢b. (6.11) t ·r r· There is a third : the conservation of . It is re- quired, for example, in case temperature changes are important (see, for example, [43]). But, for simplicity, we will treat only situations where this law can reasonably be ignored. Equation (6.11) holds for arbitrary elastic media, not just fluids. This model would be exact if internal stress could be described exactly by a sym- that was derived without additional assumptions from elec- tromagnetism. While a fundamental representation of internal stress might be possible in principle, it would likely be so complex as to be useless for making predictions. Instead, an approximation—based on physical princi- ples and physical intuition—called a constitutive law for the type of fluid Fluids 263

(or elastic material) being modeled is used to define the stress tensor as a function of the other state variables. The choice of constitutive stress law is the most important ingredient in a viable model. Since constitutive laws are approximations, predictions derived from corresponding models must be validated by physical . Perhaps the simplest model for internal stress arises from the constitutive assumption that the stress is the same in all directions (no shear stress). In this case, there is a function p on space-time, called the , such that the stress tensor is given by

(x, t)= p(x, t)I, (6.12) where I denotes the identity transformation on space. The minus sign is taken because the stress on the boundary of a fluid parcel is ⌘,where⌘ is the outer unit normal on the boundary of the parcel and a positive pressure (from outside the parcel) is assumed to act in the direction of the inner normal. Indeed, pressure is correctly defined to be the normal component of force per area. A fluid that satisfies constitutive law (6.12) is called an ideal fluid. Using formula (6.7), the divergence of the stress for an ideal fluid is

= p, r· r and the corresponding

⇢(u +(u )u)= p + ⇢b, t ·r r ⇢ + (⇢u)=0. (6.13) t r· System (6.13) is four equations, counting the three components of the vector equation for conservation of momentum, for five unknowns: the three components of the velocity u,thedensity⇢ and the pressure p. One more equation is needed to close the system. The missing ingredient is conser- vation of energy. A physically realistic approach to 1 2 requires a digression into thermodynamics. While the 2 ⇢ u is a quantity from classical , models of internal energy requirek k a deeper analysis that is not completely understood. Thus, the most general form of the equations of fluid remain somewhat controversial. For- tunately, for some practical applications, internal energy considerations can be bypassed or modeled with relatively simple constitutive laws. Another approach to closing the system is to make a simplifying assumption on the 264 Fluids nature of the fluid. Two standard (closely related) possibilities are the as- sumptions that the fluid is incompressible (that is, the flow preserves volume) or that the of the fluid is constant. The incompressibility assumption is equivalent to assuming the velocity field is divergence free: div u =0.Inthiscase,usingtheidentity

(⇢u)= ⇢ u + ⇢ u, r· r · r the (five scalar) equation of motion

⇢(u +(u )u)= p + ⇢b, t ·r r ⇢ + ⇢ u =0, (6.14) t r · u =0. (6.15) r· are called Euler’s equations for an ideal fluid. Amorerestrictiveassumptionisconstantdensity;itleadstothe(four) equations

⇢(u +(u )u)= p + ⇢b, t ·r r u =0 (6.16) r· for the four unknowns for the components of u and the pressure. Because our fluid is confined to a region of space, boundary conditions must be imposed. Physical experiments showR that the correct boundary condition is u 0onthesolidboundariesof .Thisiscalledtheno-slip boundary condition.⌘ To demonstrate this factR yourself, consider cleaning a metal plate by using a hose to spray it with ; for example, try cleaning a dirty automobile. As the pressure of the water increases, the size of the particles of dirt that can be removed decreases. But, it is very dicult to remove all the dirt by spraying alone. This can be checked by polishing with a clean cloth. In fact, the velocity of the spray decreases rapidly in the boundary layer near the plate. Dirt particles with suciently small diameter are not subjected to flow that are high enough to dislodge them. Euler’s equations are not well-posed with the no-slip boundary condition; that is, under this boundary condition, the equations of motion do not have unique solutions depending continuously on the initial position of the fluid. The mathematically correct boundary condition for Euler’s equations is that the fluid does not penetrate the boundary; or, equivalently, the fluid velocity is everywhere tangent to the boundary. The no-slip condition is allowed, Fluids 265 but not required. While this fact implies that Euler’s equations cannot be the correct model for physical fluids, Euler’s equations give experimentally verifiable predictions as long as measurements are taken far from the fluid’s boundary. Indeed, this observation is fundamental in many applications (for example, flow over an airplane wing) where there is a thin layer of fluid near the boundary that must obey a more realistic stress constitutive law in order to satisfy the no-slip boundary condition; but, away from this layer, the motion of the fluid is well-approximated by Euler’s equations. To obtain a more realistic model, the of the fluid must be taken into account. The modeling process requires several assumptions, which are di↵erent for di↵erent types of fluids. For fluids similar to water and air, called Newtonian fluids, the main assumptions are the isotropy of the fluid (that is, the fluid is the same in all directions) and a linear relation between stress and velocity. For some fluids, for example blood, accurate models require nonlinear stress-velocity relations. The modeling process to obtain the stress-velocity relation for Newtonian fluids is described in books on fluid mechanics (see, for example, [17, 43]). A discussion and derivation of this stress-strain relation in the context of the equation of motion for elastic media is presented in Section 6.13. The same underlying ideas lead to stress for fluids. The stresses are all given by pressure (due to molecular motion). This part of the stress is modeled by the Eulerian fluid where = pI. Stresses in moving fluids are determined by constitutive laws, usually formulated as linear relations between stress and strain; that is, Hooke’s law. Strains are relative changes in length. They are measured by the (infinitesimal) changes in positions of material points of the fluid, which is modeled by detecting the stretching and compression of vectors (representing directions) at points in the fluid. Strain, the relative change in length, is dimensionless. Let p be a point in the fluid and v aunitvectoratp.Themotionofp in the moving fluid starting at some fixed time T with velocity field u is given by t (p, t), where 7! @ (p, t)=u((p, t),t)(6.17) @t and (p, T )=0.Thetemporalvariablet measures time starting at time T . The of p after time t is given by (p, t):=(p, t) p.Let↵ be acurveatp parameterized by s with tangent vector v;thatis,s ↵(s)is 7! acurvesuchthat↵(0) = p and ↵0(0) = v.Thefluidflowmoveseachpoint, hence the curve tangent to v.Theinfinitesimaldistortionofthecurveisthe 266 Fluids directional derivative d (↵(s),t) = D(p, t)v, ds s=0 where D denotes the derivative of the transformation p (p, t)forfixedt. 7! The length of ↵ starting at p is

s `(s):= ↵0(⌧) d⌧, | | Z0 and the length along the distorted curve is

s L(s):= D(↵(⌧),t)↵0(⌧) d⌧. | | Z0 The di↵erence of the squares of these lengths is easily computed using the Taylor approximation to be

L2(s) `2(s)=(D(p, t)v 2 v 2)s2 + O(s3) | | | | =( D(p, t)v, D(p, t)v v, v )s2 + O(s3) h ih i = (D(p, t)T D(p, t) I)v, v s2 + O(s3). h i To take into account the size of the displacement, the linear transformation D(p, t)T D(p, t) I is recast in the form D(p, t)T D(p, t) I =(Du(p)+I)T (Du(p)+I) I = Du(p)T Du(p)+Du(p)T + Du(p).

The Lagrange-Green (strain) tensor E is defined by

1 E(p, t)(v, w)= (D(p, t)T D(p, t) I)v, w 2h i 1 = (D(p, t)T D(p, t)+D(p, t)T + D(p, t))v, w , 2h i

1 where the factor 2 is inserted to agree with the usual definition of this tensor (which is used in Section 6.13) and w is a vector at p. A displacement (or ) is called small if the product D(p, t)T D(p, t)maybesafely neglected. From a mathematical perspective, the same result is achieved by Fluids 267 simply declaring the linear strain tensor " to be the linear approximation of the Lagrange-Green tensor; that is, the linear strain tensor is given by 1 "(p, t)(v, w)= (D(p, t)T + D(p, t))v, w . (6.18) 2h i The spatial derivative of the fluid motion is determined by di↵erentiation of the di↵erential equation (6.17) in the direction v:

@ D(p, t)v = Du((p, t),t)D(p, t)v, @t with D(p, T )=I. By an algebraic manipulation, this di↵erential equation can be recast into the form @ ((D(p, t) I)v = Du((p, t),t)(D(p, t) I)v + Du((p, t),t)v. @t At t = T , the time rate-of-change of the di↵erential of displacement at p in the direction v is @ (D(p, t)v = Du(p, T )v. @t t=T The choice of T is arbitrary. Thus, the time rate-of-change of the linear strain tensor is given by 1 "˙(p, t)(v, w)= (Du(p, t)T + Du(p, t))v, w , (6.19) 2h i where u is the fluid velocity. It has units of inverse time. Stress has units of force per area. Hooke’s law (stress is proportional to strain) suggests the dynamic stress must be some constant µ,whichhasunitsofmassperlength per time, " ˙ so that = pI +2µ"˙. The factor 2 is put in to remove the factor 1/2intheLagrange-Greentensor. Our construction of the stress tensor is almost correct! The only problem is the interpretation of Hooke’s law. The strain is a matrix; simple propor- tionality is too strong. Hooke’s law is more general: it states that stress is linearly related to strain. More generally, there is some matrix valued linear transformation of matrices K so that

= pI + K"˙. 268 Fluids

There is no reason to take K =2µI.Onphysicalgrounds,K applied to "˙ is isotropic and symmetric. Isotropic means that K is under . Using these hypotheses, it is possible to prove that K has to be a special type of linear transformation. In fact, there are only two µ and and = pI + K"˙ = pI +2µ" + div uI. For reasons that arise from thermodynamics, the viscosities are related by = 2/3µ.Thefinalformoftheconstitutiverelationisthestresstensor 2 = pI + K"˙ = pI +2µ" µ div uI. (6.20) 3

Using ij =0ifi = j and ij =1ifi = j,thecomponentsof" are given by 6 @ui @uj 2 ij = pij + µ + µ( u)ij. @xj @xi 3 r· The function p is the pressure, µ is called the viscosity, and 1/2(@ui/@xj + @uj/@xi)isthe(i, j)-component of the linear strain tensor. With this choice of ,theboundaryvalueproblem µ ⇢(u +(u )u)= p + µu + ( u)+⇢b, t ·r r 3 r r· ⇢ + (⇢u)=0, (6.21) t r· u =0on@ , R consisting of Cauchy’s equation and the —which are to- gether called the Navier-Stokes equations for a Newtonian fluid—is well- posed with the no-slip boundary condition. Here u denotes the Laplacian applied component-wise to the vector u;thatis,

2 2 2 @ u1 @ u1 @ u1 2 + 2 + 2 @x1 @x2 @x3 2 2 2 @ u2 @ u2 @ u2 u = 2 + 2 + 2 . 0 @x1 @x2 @x3 1 2 2 2 @ u3 @ u3 @ u3 @x2 + @x2 + @x2 B 1 2 3 C @ A In some realistic situations (for example, the flow of water at constant temperature) it is reasonable to make two further assumptions: the viscosity µ and density ⇢ are constant. Using the equation of continuity, the con- stant density assumption implies that the vector field u is divergence free. Fluids 269

This is equivalent to the assumption that the fluid is incompressible (see Ap- pendix A.11). Under these assumptions, the Navier-Stokes equations (6.21) simplify to

⇢(u +(u )u)= p + µu + ⇢b, t ·r r u =0, (6.22) r· u =0on@ . R

Exercise 6.1. (a) Prove that if u is a suciently smooth vector field, then

u =( u). r· r· Conclude that if u divergence free, then u =0. (b) Use part (a) to conclude r· that for incompressible flow with body force b satisfying b =0(such a body force r· is called conservative), the pressure is given by

p = ⇢ ((u )u). r· ·r Exercise 6.2. Consider incompressible flow with no body force. Let be a har- monic function ( =0) on three-dimensional space and f an arbitrary real func- tion of one real variable. (a) Show that the velocity field u(x, t)=f(t) solves r the Navier-Stokes equations, ignoring the boundary condition. Find the exact ex- pression for the pressure. (b) Is it possible to find nontrivial solutions as in part (a) on a bounded domain such that the velocity field vanishes at the boundary? DIscuss.

Exercise 6.3. Show that it is not necessary to assume the density is constant (everywhere) to obtain equations (6.22) from equations (6.21). More precisely, prove that it suces to have D⇢/Dt =0(that is, the density is constant along fluid particle paths).

Exercise 6.4. Define

1,x0;  (x)= 1 3x2 +2x3, 0

(a) Show that is continuously: di↵erentiable and write out the formula for 0. (b) Recall the formula (the continuity equation) ⇢t +div(⇢u)=0. Suppose that u is the constant vector field (2, 0, 0) and the density is distributed in 270 Fluids space so that the density is the same on each plane that is orthogonal to the x-axis. This means ⇢(x, y, z, t) does not depend on y or z; it can be viewed as a function of x and t only. Using these simplifications, consider the initial value problem

⇢t +div(⇢u)=0,⇢(x, 0) = (x).

Determine the value of ⇢(3, 5/4). Hint: Look for a traveling solution. Exercise 6.5. The total kinetic energy of a fluid with density ⇢ and velocity u over the parcel A moving with the fluid via the flow is 1 KE := ⇢(x, t)u(x, t) u(x, t) d . 2 · V Z(A,t) Use the transport theorem and a computation in coordinates to show that dKE = ⇢u (u +(u ˙ )u) d . dt · t r V Z(A,t) 6.2 Scaling: The Reynolds Number and

For simplicity, the Navier-Stokes equations (6.22) for incompressible flow is considered in this section. By introducing a characteristic length L and a characteristic velocity V for the flow in question together with the corre- sponding natural time-scale ⌧ = L/V ,thestatevariablesarerendereddi- mensionless via the assignment X = x/L, s = t/⌧, U(X, s)=u(x, t)/V , and P (X, s)=p(x, t)/(V 2⇢). Using the kinematic viscosity ⌫ := µ/⇢, the Reynolds (dimensionless) number Re := LV/⌫, and the Froude num- ber Fr := U/ L b ,whichdependsonthemagnitudeofthebodyforce,the system of equations| | for the velocity and pressure can be rescaled to the di- p mensionless form of the Navier–Stokes equations for an incompressible fluid

1 1 b Us +(U )U = U P + , ·r Re r Fr2 b | | U =0, r· U =0 in@ ⇤ (6.23) R where ⇤ is the image of the region under the change of coordinates. The existenceR of this scaling is important:R If two flows have the same Reynolds Fluids 271 and Froude numbers, then the flows have the same dynamics. For example, flow around a scaled model of an airplane in a tunnel might be tested at the same Reynolds and Froude numbers expected for the airplane under certain flight conditions. Perhaps the same Reynolds number can be obtained for testing by increasing the velocity in the wind tunnel to compensate for the smaller length scale of the scaled model. In principle, the behavior of the flow around the scaled model is the same as for the full-sized aircraft.

6.3 The Zero Viscosity Limit

The scaled Euler’s equation in a region ⇤ is given by R U +(U )U = P + B, s ·r r U =0, r· U ⌘ =0 on@ ⇤, (6.24) · R where ⌘ is the outward unit normal vector field on @ ⇤,canbeviewedasan idealization of the scaled Navier–Stokes equations (6.23)R for a fluid with zero viscosity. The no-slip boundary condition for the Navier–Stokes equations is replaced by the condition that there is no fluid passing through the boundary. A naive expectation is that the limit of a family of solutions of the Navier– Stokes equations, as the Reynolds number increases without bound, is a solution of Euler’s equations. After all, the term U/Re would seem to approach zero as Re . Note, however, the possibility that the second of the velocity!1 field are unbounded in the limit and the di↵erent boundary conditions for the Navier-Stokes and Euler equations. For these and other reasons, the limiting behavior of the Navier-Stokes equations for large values of the Reynolds number is not yet completely understood. The necessity of di↵erent boundary conditions in passing from the Navier- Stokes to the Euler equations is the starting point for one of the most im- portant aspects of fluid dynamics, which was introduced by L. Prandtl in 1904 (see [63, 66]), called boundary layer theory. The fundamental idea is that for flows of interest in , for instance, the viscous e↵ects are only important in a thin layer near the boundary where the no-slip con- dition must hold; away from the boundary, the flow is well-approximated by Euler’s equations. Boundary layer theory will be discussed more fully in Section 6.12.3. 272 Fluids

The possibility of large second derivatives of the velocity field is important in another area of fluid dynamics: the study of turbulence, a subject that is beyond the scope of this book.

6.4 The Low Reynolds’s Number Limit

In the zero viscosity limit, the pressure is scaled by P = p/(V 2⇢). Physically, the pressure is comparable to the momentum per unit volume. For flows with low Reynolds’s numbers, the pressure is comparable to the forces due to viscosity. For this reason, the correct scaling in this regime is X = x/L, s = t/⌧, U(X, s)=u(x, t)/V ,andP (X, s)=p(x, t)L/(µV ), which leads to the equation L 1 1 U + U U = P + U. V s ·r Rer Re After multiplying by the Reynolds number and passing to the limit as Re 0, we obtain the well-posed !

P =U, r U =0, r· U =0 on@ ⇤. (6.25) R This system, usually called Stokes’s equations, is a useful approximation in many di↵erent situations where the velocity is small, the viscosity is large, or the body size is small.

6.5 Flow in a Pipe

As an example of the solution of a fluid flow problem, let us consider perhaps the most basic example of the subject: a special case of flow in a round pipe.1 Consider cylindrical coordinates r, ✓,andz where the z-axis is the axis of symmetry of a round pipe with radius a.Moreprecisely,weconsiderthe coordinate transformation

x1 = r cos ✓, x2 = r sin ✓, x3 = z.

1The general nature of flow in a round pipe is not completely understood (see, for example, B. Eckhardt (2008), Turbulence transition in pipe flow: some open questions. Nonlinearity. 21, T1–T11.) Fluids 273

The basis vector fields for cylindrical coordinates are defined in terms of the usual basis of Euclidean space by

e := (cos ✓, sin ✓, 0),e:= ( sin ✓, cos ✓, 0),e:= (0, 0, 1). r ✓ z We denote the coordinates of a vector field F on the cylinder (r, ✓, z):r {  a with respect to the basis vector fields er, e✓,andez by Fr, F✓,andFz respectively.} The use of subscripts to denote coordinates in this section must not be confused with partial derivatives of F . It is natural to expect that there are some flow regimes for which the velocity field has its only nonzero component in the axial direction of the pipe; that is, the velocity field has the form

u(r, ✓, z, t)=(0, 0,uz(r, ✓, z, t)), (6.26) where the components of this vector field are taken with respect to the basis vector fields er, e✓, ez and uz denotes the z component of the field (not the with respect to z). We will express the Euler and the Navier–Stokes equations in cylindrical coordinates. For a function f and a vector field F = Frer + F✓e✓ + Fzez on Euclidean space, the basic operators are given in cylindrical coordinates by @f 1 @f @f f = e + e + e , r @r r r @✓ ✓ @z z 1 @ 1 @F @F F = (rF )+ ✓ + z , (6.27) r· r @r r r @✓ @z 1 @ @f 1 @2f @2f f = r + + r @r @r r2 @✓2 @z2 (see Exercise 6.7). To obtain the incompressible Navier–Stokes equations in cylindrical coordinates, consider the unknown velocity field u = urer +u✓e✓ + uzez.WritethisvectorfieldintheusualCartesiancomponentsbyusing the definitions of the direction fields given above, insert the result into the Navier–Stokes equations, and then compute the space derivatives using the operators given in display (6.27). After multiplication of the vector consisting of the first two of the resulting component equations (that is, the equations in the directions ex and ey) by the matrix cos ✓ sin ✓ , sin ✓ cos ✓ ✓ ◆ 274 Fluids we obtain the equivalent system @u 1 1 1 @u @p r +(u )u u2 = u (u +2 ✓ ) , @t ·r r r ✓ Re r r2 r @✓ @r @u 1 1 1 @u 1 @p ✓ +(u )u + u u = u (u 2 r ) , @t ·r ✓ r r ✓ Re ✓ r2 ✓ @✓ r @✓ @u 1 @p z +(u )u = u , @t ·r z Re z @z u =0, (6.28) r· where the operators and are represented in cylindrical coordinates as in formulas (6.27). r The Euler equations in cylindrical coordinates for the fluid motion in the pipe are obtained from system (6.28) by deleting the terms that are divided by the Reynolds number. If the velocity field u has the form given in equa- tion (6.26), then u automatically satisfies the appropriate boundary condition for the incompressible Euler equation; that is, the Neumann boundary con- @u dition @n =0,wheren is a unit normal on the cylinder that models the wall of the pipe. Thus, the Euler equations for the (scaled) velocity and pressure fields u and p reduce to the system @p @p @u @u @p @u =0, =0, z + u z = , z =0. @r @✓ @t z @z @z @z The first two equations imply that p is a function of z and t only; the second two equations imply that @u @p z = . (6.29) @t @z After di↵erentiation of equation (6.29) with respect to z,itfollowsthat @2p/@z2 = 0. Therefore, p = ↵ + z for some functions ↵ and that depend t only on t. Using equation (6.29), uz = v(r) 0 (s) ds for an arbitrary choice of initial velocity v that may depend on r,butnot✓ or z.Thegeneral R solution, for the class of velocities that have zero first and second components is t u(x, y, z, t)=v(r) (s) ds),p(x, y, z, t)=↵(t)+(t)z. Z0 All such solutions, the initial velocity depends only on r.Thenopen- etration boundary condition is always satisfied by the assumption that the Fluids 275

first two components of u vanish. The no slip boundary condition can also be satisfied by taking v(a)=0.Ofcourse,sofarourpipehasinfinitelength. A physically more realistic requires some pressure di↵erential to push the flow. Suppose for example the pressure is constant with value p0 at z =0 and p1 at z =1,andp0 >p1.Themodelshouldthenpredictanonzeroflow velocity with positive uz component. In this case, the pressure must be p = p +(p p )z 0 1 0 and the third component of velocity is u = v(r)+(p p )t. z 0 1 The flow moves in the expected direction, but the velocity increases without bound as t grows to infinity. Thus, the model predicts an unrealistic flow for an arbitrary choice of initial velocity v. Note that for arbitrary ↵,thepairoffunctionsuz =0andp = ↵(t) is a solution of the Euler model with appropriate boundary conidtion. At first sight it might seem that this simple example shows that solutions of Euler’s equations with zero initial velocity and the no penetration boundary condition does not have unique solutions. But, this is not the case: the non uniqueness of pressure is allowed. The reason is simple: the gradient of the pressure appears in the equations of motion. Thus, the pressure is never unique in a fluid flow problem; it is required to be unique up to the addition of a constant function of time. There are several cases for Euler flow with the no penetration boundary condition. For =0,thepressuredoesnotdependonthepositioninthe pipe and the fluid velocity field is constant with respect to z (along the flow direction in the pipe). This is called plug flow. Because of its mathematical simplicity, plug flow is often used as a model. For example, plug flow is often used to model flow in tubular reactors studied in chemical . What about Navier–Stokes flow? By considering the same pipe, the same coordinate system, and the same hypothesis about the direction of the velocity field, the Navier–Stokes equa- tions reduce to @p @p @u @u 1 @p @u =0, =0, z + u z = u , z =0, @r @✓ @t z @z Re z @z @z with the no-slip boundary condition at the wall of the pipe given by u (a, ✓, z, t) 0. z ⌘ 276 Fluids

This apparently simple system of fluid equations is dicult to solve. But, we can obtain a solution under two additional assumptions: The velocity field is in steady state and it is symmetric with respect to rotations about the central axis of the pipe. In other words, the partial derivatives of u with respect to t and ✓ vanish. With these assumptions and taking into account that @uz/@z =0,itsucestosolvethesingleequation 1 1 @ @u @p r z = . Re r @r @r @z Because @p/@r =0and@p/@✓⇣ =0, @p/@z⌘ depends only on z and the left- hand side of the last equation depends only on r.Thus,thefunctionson both sides of the equation must have the same constant value, say . It follows that p = ↵ + z and d(ru (r)) z0 =( Re)r dr with the initial condition uz(a)=0.Thegeneralsolutionofthisordinary di↵erential equation has a free parameter because there is only one initial condition given for a second-order di↵erential equation. The term with the free parameter contains the factor ln r,whichblowsupasr approaches zero (that is, at the center of the pipe). With the free parameter set to zero, the solution 1 u (r)= Re (r2 a2). z 4 is continuous in the pipe and physically realistic. This steady state veloc- ity field, called Poiseuille flow, predicted from the Navier-Stokes model is parabolic with respect to the radial coordinate with the fastest flow at the center of the pipe and flow velocity zero at the pipe wall. Poiseuille flow is a close approximation to physical flow in a pipe for small Reynolds num- bers. As the Reynolds’s number is increased, a critical value is reached at which the radial symmetry hypothesis used to obtain the Poiseuille flow is violated. Above this critical value, the steady state physical flow as measured in experiments and the velocity field given by the Navier-Stokes model be- come more complex as the Reynolds’s number is increased. The flow regime changes from laminar flow to turbulent flow. The causes of the onset of tur- bulent flow, the transition from laminar to turbulent flow, and the nature of fully turbulent flow are not fully understood; they are important unsolved problems in physics and applied mathematics. Fluids 277

Exercise 6.6. Consider Poiseuille flow in a section of length L of an infinite round pipe with radius a. Suppose that the pressure is p in at the inlet of the section and the flow at the center of the pipe is v in. Determine the pressure at the outlet. What happens in the limit as the Reynolds number grows without bound? Compare with the prediction of plug flow.

Exercise 6.7. (a) Write a detailed derivation of the gradient, divergence, and Laplacian in cylindrical coordinates (see display (6.27)). There are several methods that can be employed. One method begins by defining f c(r, ✓, z)=f(r cos ✓, r sin ✓, z) so that f(x, y, z)=f c( x2 + y2, arctan(y/x),z) and continues by di↵erentiating the latter equality with respect to x, y,andz. (b) Derive the expressions for the p gradient, divergence, and Laplacian in spherical coordinates.

Exercise 6.8. [Hagen-Poiseuille law] Return to unscaled variables and reconsider steady-state Poiseuille flow in a round pipe of radius a.. WIth the z coordinate in the direction of the pipe axis, the equations of motion are

px =0,py =0,pz = µ(vxx + vyy),vz =0 with the no slip boundary condition at the pipe wall. (a) Write out the full phys- ically realistic general solution for the pressure and and velocity. (b) Suppose the pressure at z =0is p0 and the pressure at some z = `>0 is p1 with p0 >p1.The pressure drop on this section of pipe is p p . Determine the relation between 0 1 the pressure drop and the volumetric flow rate in the pipe (volume per time of fluid passing a given cross section of the pipe). This relation is called the Hagen- Poiseuille law. (c) Determine the average velocity through a cross section.

Exercise 6.9. Determine the profile for Stokes flow in a round pipe of radius a under the assumption that the flow is symmetrical with respect to rotations around the central axis of the pipe.

Exercise 6.10. Consider flow over a flat infinite surface that is tilted ✓ radians with respect to the horizontal. Ignore the banks of the river, assume that the flow velocity is all in the downstream direction, and the body force driving the flow is . (a) Show that the velocity at z units above the plate (river bottom) is u = 1 ⇢gz(2h z)sin✓. (b) The Mississippi River drops 2.5 inches per mile. 2µ Suppose its average depth is 30 feet. Compute the surface velocity of the river? Does the velocity computed using these assumptions agree with observation? Which if any of the assumptions is not realistic? Discuss. 278 Fluids 6.6 Bernoulli’s Form of Euler’s Equations

6.6.1 Isentropic Flow In this section we will assume that our fluid is incompressible. In particular, we will assume the density is constant.Moregenerally,thesameresultsare valid for isentropic flow; that is, there is some function q called the enthalpy, such that 1 grad q = grad p. ⇢ In case the density is constant, q := p/⇢. Using the vector identity 1 grad(u u)=u curl u +(u )u, 2 · ⇥ ·r we may rewrite Euler’s equation of motion (6.13) in the form 1 p u u curl u = b +grad( (u u) ). t ⇥ 2 · ⇢ The definition 1 p ↵ := u 2 , 2| | ⇢ is used to obtain Bernoulli’s form of Euler’s equations

u = u curl u +grad↵ + b, t ⇥ div u =0, u ⌘ =0 in@ . (6.30) · R 6.6.2 Potential Flow An important specialization of Bernoulli’s form of the incompressible Euler’s equations, called potential flow, is the case where the velocity field u is the gradient of a potential so that u =grad and the body force is the gradient of a potential B so that b =gradB.Bysubstitutionintosystem(6.30),the identity curl(grad u)=0,andsomerearrangement,theequationsofmotion can be recast into the form @ 1 p grad( + grad 2 + B)=0, =0. (6.31) @t 2| | ⇢ Fluids 279

As a result, we see immediately that there is a number C,constantwith respect to the space variable, such that @ 1 p + grad 2 + B = C. (6.32) @t 2| | ⇢ In particular, if u is a steady state velocity field, then p 1 + u 2 B = C. (6.33) ⇢ 2| | This is Bernoulli’s law. In case the body force is gravity, Bernoulli’s law states that (at a height h above a reference surface) 1 p + ⇢ u 2 + g⇢h = C (6.34) 2 | | An important consequence of Bernoulli’s law is Bernoulli’s principle: At constant height, if the velocity of an incompressible fluid increases, then the pressure decreases. For example, flow over an airplane wing (which is more curved on top) produces because the velocity of air relative to the wing is greater for flow over the top of the wing compared to flow over its bottom. Exactly why the flow is faster over the top of an airplane wing can be explained, but not simply (see, for example, [43]).

Exercise 6.11. Consider an open tank containing water with a round hole in the bottom of the tank. Show that the velocity of the fluid in the drain is (approxi- mately) proportional to the square root of the depth of the fluid in the tank. More precisely, this velocity is approximately p2gh, where h is the depth of the water in the tank and g is the due to gravity. What assumptions are required to make the velocity exactly p2gh.

Potential Flow in Two In view of the second equation of system (6.31), the potential is a harmonic function. Therefore, by considering a hypothetical flow on a two-dimensional plane with Cartesian coordinates (x, y)andvelocityfieldu =(˙x, y˙), the potential is locally the real part of a holomorphic function, say f = + i . Moreover, the pair , satisfies the Cauchy–Riemann equations @ @ @ @ = , = . @x @y @y @x 280 Fluids

Thus, the assumption that u =grad implies the fluid are solutions of an ordinary di↵erential equation that can be viewed in two di↵erent ways: as the gradient system @ @ x˙ = , y˙ = ;(6.35) @x @y or the Hamiltonian system

@ @ x˙ = , y˙ = . (6.36) @y @x

The function , a Hamiltonian function for system (6.36), is called the . The orbits of system (6.36), called stream lines, all lie on level sets of .Becausethestreamlinesarealsoorbitsofthegradientsys- tem (6.35), there are no periodic fluid motions in a region where the function is defined. It should be clear that function theory can be used to study planar po- tential flow. For example, if is a harmonic function defined in a simply connected region of the complex plane such that the boundary of the region is a level set of ,then is the imaginary part of a holomorphic function defined in the region, and therefore is the stream function of a steady state flow. This fact can be used to find steady state solutions of Euler’s equations. As an example, let us consider plug flow in a round pipe with radius a and notice that every planar slice containing the axis of the pipe is invariant under the flow. Thus, it seems reasonable to consider the two-dimensional flow on the strip := (x, y):0

1 (Q (w). In particular, stream lines of map to stream lines of w 1 7! (Q (w)). And, this flow is a solution of Euler’s equation in the domain . R For example, let us note that w := Q(z)=pz has a holomorphic branch defined on the strip such that this holomorphic function maps into the region in the firstS quadrant of the complex plane bounded aboveS by the R 1 2 hyperbola (,⌧):⌧ = a .Infact,Q (w)=w so that { } x = 2 ⌧ 2,y=2⌧. The velocity field of the new “flow at a corner” is

(2c, 2c⌧). The corresponding pressure is found from Bernoulli’s equation (6.33). In fact, there is a constant p1 such that

p = p 2c2(2 + ⌧ 2). (6.37) 1 The stream lines for the flow at a corner are all hyperbolas. The flow near a wall is essentially plug flow. In fact, if we consider, for example, the velocity field on a vertical line orthogonal to the -axis, say the line with equation = ,thenthevelocityfieldnearthewall,where⌧ 0, 0 ⇡ is closely approximated by the constant vector field (2c0, 0). In other words, the velocity profile is nearly linear.

Exercise 6.12. Consider the plug flow vector field u =(c, 0) defined in a hori- zontal strip in the upper half plane of width 2a.Findthepushforwardofu into 1 2 the first quadrant with respect to the map Q(z)=pz with inverse Q (w)=w . Is this vector field a steady state solution of Euler’s equations at the corner for an incompressible fluid? Explain.

6.7 Equations of Motion 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 dynamics and particle mechanics. 282 Fluids

6.7.1 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. Consider a rectangular inertial coordinate system, with coordinates (⇠,⌘,⇣), and the motion of a particle whose position vector in this coordinate system is R.Supposethereisasecond(moving)rectangularcoordinatesystemwith rectangular coordinates (x, y, z)andthesameoriginastheinertialcoordi- nates. Let Q denote the position of the particle with respect to the moving coordinates. The symbols Q and R refer to the same point in space; but, Q = R when Q is given as the triple of numbers that specify its coordinates in6 the moving coordinate system and R is given as the triple of numbers that specify its coordinates in the inertial coordinate system. This can be a confusing distinction. Suppose that a Q = b 0 c 1 with respect to the moving coordinates.@ ThisA means

Q = aex + bey + cez, where ex, ey,andez are the unit basis vectors (with respect to the usual inner product in the inertial coordinate frame) in the positive directions on the coordinate axes of the rectangular moving coordinate system. These basis vectors have coordinate representations in the 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 Thus, the coordinate@ representationA @ of Q Ain the inertial@ coordinatesA 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 @ A @ A @ A Fluids 283 are the inertial coordinates of the point R.Morecompactly,wewillwrite R = AQ, where A is the matrix with components aij.Inthegeneralcase,wherethe two coordinate systems might not share a common origin, the origin of the moving frame has a position vector in the inertial frame. Also there is a notion of parallel transport in Euclidean space. The inertial frame may be moved parallel to itself until its origin coincides with the origin of the moving frame. As before, there is a matrix A that changes the coordinates of the particle in the moving frame to the coordinates of this point in the transported inertial frame. Thus, the coordinates with respect to the original inertial frame are obtained by adding ;thatis, R = + AQ. (6.38)

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 6.13). Here, A gives the inertial coordinates of the position hivector Q and its inverse gives the moving coordinates of a point expressed in the coordinates of the inertial frame. Because A is orthogonal, AAT = AT A = I. By Newton’s second law, the equation of motion of the particle with position R moving in the inertial frame is mR¨ = F, (6.39) where m is the mass of the particle and F is the (vector) sum of the forces acting on the particle. Our goal is to express the equation of motion for our particle in the moving coordinate system. A convenient method is to first describe the motion using the position, velocity, and acceleration of the origin of the moving coordinate system in inertial coordinates and the inertial representations r := AQ of the particle’s position Q, v := AQ˙ of its velocity Q˙ ,anda := AQ¨ of its acceleration. In view of equation (6.38), the velocity of the particle can be expressed in the form R˙ =˙ + AQ˙ + AQ,˙ =˙ + AA˙ T (AQ)+v =˙ + AA˙ T r + v. (6.40) 284 Fluids

The transformation ⌦:= AA˙ T has a special property: it is skew-symmetric; that is, ⌦T = ⌦. This fact follows simply by di↵erentiation of the identity AAT = I with respect to the temporal parameter. It follows that

AA˙ T + AA˙ T =0; (6.41) therefore, ⌦T =(AA˙ T )T = AA˙ T = AA˙ T = ⌦. By using the definition of skew-symmetry, the matrix representation of ⌦ must have the form 0 ! ! 3 2 ! 0 ! . (6.42) 0 3 1 1 ! ! 0 2 1 The action of ⌦on a vector@W (that is, ⌦WA)canalsobeexpressedusing the vector cross product (with respect to the usual orientation of space) by

⌦W = ! W, (6.43) ⇥ where ! is the vector with components (!1,!2,!3). Thus, the velocity may be expressed in either of the forms

R˙ =˙ +⌦r + v, R˙ =˙ + ! r + v. (6.44) ⇥ Using the first equation in display (6.44) for the velocity of our particle, we find its acceleration to be

R¨ =¨ +˙v + ⌦˙ r +⌦r.˙ (6.45)

Since, by definition, r = AQ and v = AQ˙ , we have that

r˙ = AA˙ T (AQ)+AQ˙ =⌦r + v, (6.46) and

v˙ = AA˙ T (AQ˙ )+AQ¨ =⌦v + a. (6.47) Fluids 285

These formulas forr ˙ andv ˙ are substituted into equation (6.45) to obtain the acceleration in the form R¨ =¨ + a +2⌦v + ⌦˙ r +⌦2r;(6.48) or, equivalently, R¨ =¨ + a +2! v +˙! r + ! (! r). (6.49) ⇥ ⇥ ⇥ ⇥ The equation of motion (6.39) may be recast in the forms ma = m(¨ +2⌦v + ⌦˙ r +⌦2r)+F (6.50) or ma = m(¨ +2! v +˙! r + ! (! r)) + F, (6.51) ⇥ ⇥ ⇥ ⇥ where the vectors a, v,andr give the acceleration, velocity and position of the particle, moving under the influence of the force F ,withrespectto an observer in the moving coordinate system. Here, the acceleration and the state vectors r and v are expressed in the inertial coordinates. Their coordinates in the moving frame are simply obtained from multiplication on the left by the orthogonal matrix AT .Forexample,usingequation(6.50), the equation of motion in the inertial coordinates is mAQ¨ = m(¨ +2⌦AQ˙ + ⌦˙ AQ +⌦2AQ)+F ;(6.52) and, in the moving coordinates it is mQ¨ = m(AT ¨ +2AT ⌦AQ˙ + AT ⌦˙ AQ + AT ⌦2AQ)+AT F. (6.53) The last equation can also be written in the form mQ¨ = m(AT ¨ +2Q˙ + ˙ Q +2Q)+AT F, (6.54) where := AT ⌦A;thatis,and⌦arenamesforthesamelineartransfor- mation expressed in di↵erent bases. Since the equation of motion (6.50) (or (6.54)) is valid for the non-inertial observer, this observer treats the terms on the right-hand side of the equation of motion as additional forces. The term ¨ is the acceleration of the origin of the non-inertial frame, ⌦˙ r =˙! r is the 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. 286 Fluids

Exercise 6.13. Prove that the coordinate transformation in display (6.38) is orthogonal.

6.7.2 Rotation To consider pure rotation, imagine a rotating disk whose motion is measured with respect to a rectangular rotating coordinate system (with coordinates (Q1,Q2,Q3)), 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 (⇠,⌘,⇣) has the same 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, which does not 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,whichdeterminesthecoordinate transformation R = AQ from the moving frame to the inertial frame, takes the form BT B,where R cos ✓ sin ✓ 0 := sin ✓ cos ✓ 0 (6.55) 0 1 R 001 @ A is the rotation matrix (in inertial coordinates) for the motion about the Q3- axis in the moving frame. By a direct computation, it is easy to check that

0 ✓˙ 0 ⌦=AA˙ T = ✓˙ 00 , (6.56) 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, ✓˙). Fluids 287

For rotations in three-dimensional space, we have the useful identity

AA˙ T = AT A.˙ (6.57)

Thus, ⌦= AT A˙.

6.8 Fluid Motion in Rotating Coordinates

The equation of motion (6.53) 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 momentum and mass) for a fluid in a gravitational field

Du ⇢ = p + µu + ⇢g Dt r D⇢ = ⇢ u, (6.58) Dt r· where it is important to note that the derivatives 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) and four equations (if we view the conservation of momentum as three scalar equations). To close the system we need one addi- tional scalar equation, for example, an that relates density and pressure 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.