A Introduction to Cartesian Tensors

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1 A Introduction to Cartesian Tensors In this text book a certain knowledge of tensors has been assumed. We restrict ourselves to Cartesian tensors, since all equations in fluid mechanics can in principle be developed in Cartesian coordinate systems. The most important elements of Cartesian tensors are summarized in this chapter; otherwise the literature should be consulted. A.1 Summation Convention When dealing with quantities in index notation we make use of Einstein s summation convention, which states that all indices which appear twice within an expression are to be summed. In R 3 the summation indices run from 1 to 3: P = F i u i = t i = τ ji n j = x = x i e i = 3 F i u i, i=1 3 τ ji n j, j=1 3 x i e i. Indices which appear twice are called dummy indices. Since they vanish after carrying out the summation, they may be arbitrarily named: i=1 F i u i = F k u k = F j u j, x i e i = x l e l = x m e m. As well as the dummy indices, single indices can also appear in equations. These free indices must be identical in all terms of the same equation: t i = τ ji n j, e i = a ij g j, a ij = b ik c kj + d ijl n l.

2 472 A Introduction to Cartesian Tensors Otherwise they may be arbitrarily named: t m = τ jm n j, t k = τ mk n m. In order to by unambiguous, the summation convention requires that an index appears no more than twice within an expression. A forbidden expression would be t i = a ij b ij n j (wrong!), but the following would be allowed t i = pδ ij n j +2ηe ij n j. A.2 Cartesian Tensors A tensor consists of tensor components and basis vectors.thenumberof linearly independent basis vectors gives the dimension of the tensor space. In three dimensional space R 3, from which, in what follows, we shall always start from, there are three linearly independent vectors, which along with three linear factors are in the position to determine a point in space uniquely. Such a set of three vectors which span a (not necessarily orthogonal) coordinate system can be used as a set of basis vectors. If these basis vectors are functions of position, the coordinate system which they span is called a curvilinear coordinate system. (Think for example of polar coordinates where the direction of the basis vectors is a function of the polar angle.) As basis vectors we choose fixed, orthogonal unit vectors, which we denote by e i (i =1, 2, 3). The coordinate system spanned by these is the Cartesian coordinate system with the coordinate axes x i (i =1, 2, 3). We differentiate between tensors of different orders. Tensors of order zero are scalars. Since a scalar is completely independent of the choice of coordinate system, no basis vector is needed to describe it. Tensors of order one are vectors. The example of the position vector, x = 3 x i e i = x i e i, i=1 (A.1) shows that each component of a tensor of order one appears along with one basis vector. Tensors of order two (dyadics) can be thought of as being formed from two vectors a and b multiplied together, so that each term a i e i of the vector a is multiplied with each term b j e j of the vector b: T = a b = 3 i=1 j=1 3 a i b j e i e j = a i b j e i e j. (1.2)

3 A.2 Cartesian Tensors 473 This product is called the dyadic product, and is not to be confused with the inner product a b (whose result is a scalar), or the outer product a b (whose result is a vector). Since the dyadic product is not commutative, the basis vectors e i e j in (1.2) may not be interchanged, since a i b j e j e i would correspond to the tensor b a. If we denote the components of the tensor T with t ij in (1.2) we obtain T = t ij e i e j. (1.3) Therefore to every component of a second order tensor there belong two basis vectors e i and e j.inr 3 nine of these basis vector pairs form the so called basis of the tensor. Completely analogously tensors of any order may be formed: the dyadic product of a tensor of order n and one of order m forms a tensor of order (m + n). The basis of an nth order tensor in R 3 consists of 3 n products each of n basis vectors. Since the basis vectors for Cartesian tensors (unit vectors e i )areconstant, it suffices to give the components of a tensor if a Cartesian coordinate system has already been layed down. Therefore, for a vector x it is enough to state the components x i (i =1, 2, 3), and a second order tensor T is fully described by its components t ij (i, j =1, 2, 3). Therefore, if we talk about the tensor t ij, we shall tacitly mean the tensor given in (1.3). The notation in which the mathematical relations between tensors are expressed solely by their components is the Cartesian index notation. Because we assume fixed and orthonormal basis vectors e i, Cartesian index notation is only valid for Cartesian coordinate systems. It is possible to develop this to general curvilinear coordinate systems, but we refer for this to the more advanced literature. The components of tensors up to the second order may be written in the form of matrices, so for example T ˆ= t 11 t 12 t 13 t 21 t 22 t 23. (1.4) t 31 t 32 t 33 Note however that not every matrix is a tensor. In order to derive some rules we shall digress from the pure index notation and carry the basis vectors along, using a mixed notation. First we shall deal with the inner product (scalar product): a b =(a i e i ) (b j e j )=a i b j ( e i e j ). (1.5)

4 474 A Introduction to Cartesian Tensors Because of the orthogonality of the unit vectors, the product e i e j is different from zero only if i = j. If we expand (1.5) we can easily convince ourselves that it is enough to carry out the summation a b = a i b i = a j b j. (1.6) Clearly within a summation, the product e i e j will cause the index on one of the two vector components to be exchanged. We can summarize all possible products e i e j into a second order tensor: { 1 for i = j δ ij = e i e j = (1.7) 0 for i j This tensor is called the Kronecker delta, or because of its properties stated above, the exchange symbol. Multiplying a tensor with the Kronecker delta brings about an exchange of index in this tensor: a ij δ jk = a ik, (1.8) a i b j δ ij = a i b i = a j b j. (1.9) Applying the Kronecker delta in (1.5) therefore furnishes the inner product in Cartesian index notation a b = a i b j δ ij = a i b i. (1.10) We now consider the outer product (vector product) of two vectors: c = a b =(a i e i ) (b j e j )=a i b j ( e i e j ). (1.11) Now the outer product of two orthogonal unit vectors is zero if i = j, since this is outer product of parallel vectors. If i j, the outer product of the two unit vectors is the third unit vector, possibly with negative sign. It easily follows that the relation e i e j = ɛ ijk e k (1.12) holds if we define ɛ ijk as a third order tensor having the following properties: +1 if ijk is an even permutation (i.e. 123, 231, 312) ɛ ijk = 1 if ijk is an odd permutation (i.e. 321, 213, 132). (1.13) 0 if at least two indices are equal We call ɛ ijk the epsilon tensor or the permutation symbol. Inserting (1.12) into (1.11) leads to c = a i b j ɛ ijk e k. (1.14) We read off the components of c from this equation as c k = ɛ ijk a i b j, (1.15) where we have used the fact that the order of the factors is arbitrary; we are dealing with components, that is, just numbers.

5 A.2 Cartesian Tensors 475 We shall now examine the behavior of a tensor if we move from a Cartesian coordinate system with basis vectors e i to another with basis vectors e i. The dashed coordinate system arises from rotating (and possibly also from translating) the original coordinate system. If we are dealing with a zeroth order tensor, that is a scalar, it is clear that the value of this scalar (e.g. the density of a fluid particle) cannot depend of the coordinate system. The same holds for tensors of all orders. A tensor can only have a physical meaning if it is independent of the choice of coordinate system. This is clear in the example of the position vector of a point. If x and x denote the same arrow (Fig. A.1) in the dashed and the undashed coordinate systems, then x = x, (1.16) that is, x i e i = x i e i. (1.17) To decompose the vector x into its components relative to the dashed coordinate system, we form the scalar product with e j and obtain x i e i e j = x i e i e j. (1.18) The scalar product of the unit vectors in the same (dashed) coordinate system e i e j, using (1.7), furnishes just δ ij. The scalar product of the unit vectors of the dashed and undashed coordinate systems forms the matrix a ij = e i e j (A.19a) or a ij =cos( x i,x j ). (A.19b) We call the matrix a ij the rotation matrix. It is not associated with a basis and therefore is not a tensor. Inserting (A.19a) into (1.18) leads to the desired transformation law for the components of a vector: x j = a ijx i. (A.20) Fig. A.1. Rotation of the coordinate system

6 476 A Introduction to Cartesian Tensors If we take the scalar product of (1.17) with e j we decompose the vector x into its components relative to the undashed system and thus we obtain the inverse x j = a ji x i. (A.21) The transformation and its inverse may look formally the same, but we note that in (A.20) we sum over the first index and in (A.21) over the second. Knowing the transformation law for the components we can easily derive that for the basis vectors. To do this we relabel the dummy indices on the right-hand side of (1.17) as j so that we can insert (A.21). We obtain the equation x i e i = x i a ji e j, (A.22) from which, using the fact that x i is arbitrary (independent variable), we can read off the transformation as e i = a ji e j. In order to be able to compare this with the components (A.20), we relabel the index i as j (and vice versa), and therefore write e j = a ij e i. (A.23) We see that for Cartesian coordinate systems both the components and the basis vectors of a tensor obey the same transformation laws. Thus we take the inverse directly from (A.21) as e j = a ji e i, (A.24) where we could also have obtained this formally be inserting (A.20) into (1.17). Before we consider the transformation laws for tensors of a higher order we shall take note of one well known property of the rotation matrix. To do this we exchange the indices in the transformation (A.20) (e.g.: x i = a kix k ), insert this into (A.21) and thus obtain x j = a ji a ki x k. (A.25) Since the vector components are independent variables we can read off the following identity from (A.25) a ji a ki = δ jk, (A.26a) which reads AA T = I (A.26b) in matrix notation. Since AA 1 = I is the equation which determines the inverse of A, we conclude from (A.26a) that the transpose of the rotation matrix is equal to its inverse (orthogonal matrix). The transformation for the components of a tensor of arbitrary order results from the transformations for the unit vectors (A.23) and (A.24). For clarity we shall restrict ourselves to a second order tensor whose basis we

7 A.2 Cartesian Tensors 477 express in terms of the basis of the dashed coordinate system using the transformation (A.24) as T = t ij e i e j = t ij a ik a jl e k e l. (A.27) Because of T = T = t kl e k e l we can read off the components in the rotated system directly from (A.27) as t kl = a ik a jl t ij. (A.28) If in T we replace the basis vectors using (A.23), we obtain t kl = a ki a lj t ij. (A.29) The same procedure is carried out for tensors of any order. The transformation behavior of tensor components is characteristic of them and therefore is used as the definition of a tensor. If we drop the basis vectors and use pure Cartesian index notation, the transformation behavior is the only criterion by which we can decide if a given expression is a tensor. Let us take an example: we shall examine whether the gradient of a scalar function is a scalar of order one. The equation u = Φ reads in index notation u i = Φ, (A.30) x i or in the rotated coordinate system u j = Φ x. (A.31) j If u is a first order tensor, using the transformation (A.20) should transform (A.30) into (A.31) u Φ j = a ij u i = a ij, (A.32) x i or using the chain rule, u j = a Φ ij x k x k x i. (A.33) By x k = a jkx j we have x k x j = a jk, (A.34) x i x i and since x j and x i are independent variables for i j, wewrite x j x i = δ ij, (A.35) so that we replace (A.34) with x k x i = a ik. (A.36)

8 478 A Introduction to Cartesian Tensors We should note that the result of (A.35) is the Kronecker delta and it is therefore a second order tensor, and should not be confused with (A.36), whose result is the rotation matrix and is therefore not a tensor. If we insert (A.36) into (A.33), we obtain u j = a Φ ija ik x k which, because of (A.26a), is identical to u j = δ jk Φ x k = Φ x j, (A.37). (A.38) This result corresponds to (A.31), and so the gradient of a scalar function is a second order tensor. The gradient of a tensor of the nth order comes from forming the dyadic product with the Nabla operator and is therefore a tensor of the (n +1)th degree. An important example of this in fluid mechanics is the velocity gradient: ( ) u = e i (u j e j )= u j e i e j. (A.39) x i x i This is a second order tensor with the components u ˆ= t ij = u j. (A.40) x i The coordinate with respect to which we differentiate is given by the first index of t ij (the row index in matrix representation) and the component of u is determined by the second index (the column index). In index notation we usually write the velocity gradient as u i / x j, that is in matrix representation as the transpose of (A.40). Although the matrix representation is not needed in index notation, in going from matrix equations to index notation (or vice versa), we should be aware of the sequence of indices determined by (A.39). The divergence of the velocity vector (or of another first order tensor) reads u i / x i in index notation, and formally corresponds with the scalar product of the Nabla operator with the vector u. Thus symbolically the divergence reads u or else div u. The result is a scalar. In general, the divergence of an nthordertensorisan(n 1)th order tensor. Therefore the divergence of a scalar is not defined. An important quantity in fluid mechanics is the divergence of the stress tensor τ ji / x j, which is a vector. Every second order tensor can be decomposed into a symmetric and an antisymmetric part. From the identity t ij = 1 2 (t ij + t ji )+ 1 2 (t ij t ji ) (A.41)

9 A.2 Cartesian Tensors 479 we obtain the symmetric tensor c ij = 1 2 (t ij + t ji ), (A.42) and the antisymmetric tensor b ij = 1 2 (t ij t ji ). (A.43) We can see that the symmetric part satisfies c ij = c ji and the antisymmetric part satisfies b ij = b ji. It follows immediately for the antisymmetric tensor that its diagonal elements (where i = j) mustbezero.whileasymmetric tensor has six independent components, an antisymmetric tensor is fully described by three components: 0 b 12 b 13 [b ij ]= b 12 0 b 23. b 13 b 23 0 (A.44) In this connection we wish to refer to an important property of the ɛ tensor. To do this we multiply the decomposition of a second order tensor with the ɛ tensor: p k = ɛ ijk t ij = ɛ ijk c ij + ɛ ijk b ij, (A.45) where c ij and b ij are again the symmetric and antisymmetric parts respectively of t ij. We rewrite this equation as follows: p k = 1 2 (ɛ ijkc ij + ɛ ijk c ji )+ 1 2 (ɛ ijkb ij ɛ ijk b ji ), (A.46) which is allowable because of the properties of c ij and b ij. We now exchange the dummy indices in the second expression in brackets: p k = 1 2 (ɛ ijkc ij + ɛ jik c ij )+ 1 2 (ɛ ijkb ij ɛ jik b ij ). (A.47) From the definition of the ɛ tensor (1.13) it follows that ɛ ijk = ɛ jik,sothat the first bracket vanishes. We obtain the equation p k = ɛ ijk b ij, whichwritteninmatrixformreads p 1 p 2 = b 23 b 32 b 31 b 13 =2 b 23 b 13. p 3 b 12 b 21 b 12 (A.48a) (A.48b)

10 480 A Introduction to Cartesian Tensors Applying the ɛ tensor to an arbitrary second order tensor using (A.45) therefore leads to the three independent components of the antisymmetric part of the tensor (compare (A.48b) with (A.44)). From this we conclude that application of the ɛ tensor to a symmetric tensor furnishes the null vector: ɛ ijk c ij =0, if c ij = c ji. (A.49) Here follow four identities of the ɛ tensor, given without proof: ɛ ikm ɛ jln =det δ ij δ il δ in δ kj δ kl δ kn. δ mj δ ml δ mn (A.50) Contraction by multiplication with δ mn (setting m = n) leadsto [ ] δij δ ɛ ikn ɛ jln =det il. (A.51) δ kj δ kl Contracting again by multiplying with δ kl furnishes ɛ ikn ɛ jkn =2δ ij, (A.52) and finally for i = j ɛ ikn ɛ ikn =2δ ii =6. (A.53) Table A.1 contains a summary of the most important rules of calculation in vector and index notation. Table A.1. Operation Symbolic Notation Cartesian Index Notation Scalar product c = a b c = δ ija ib j = a ib i c = a T c k = δ ija it jk = a it ik Vector product c = a b c i = ɛ ijk a jb k Dyadic product T = a b t ij = a ib j Gradient of a scalar field c =grada = a c i = a x i Gradient of a vector field T =grad a = a t ij = aj x i Divergence of a vector field c =div a = a c = ai x i Divergence of a tensor field c =divt = T c i = tji x j a k Curl of a vector field c =curl a = a c i = ɛ ijk x j Laplace operator on a scalar c = ϕ = ϕ c = 2 ϕ x i x i

11 B Curvilinear Coordinates In applications it is often useful to use curvilinear coordinates. In order to derive the component equation for curvilinear coordinates we can start from general tensor calculus, which is valid in all coordinate systems. However, if we restrict ourselves to curvilinear but orthogonal coordinates, we can move relatively easily from the corresponding equations in symbolic notation to the desired component equations. Since it is orthogonal coordinate systems which are needed in almost all applications, we shall indeed restrict ourselves to these. We consider the curvilinear orthogonal coordinates q 1, q 2, q 3,whichcan be calculated from the Cartesian coordinates x 1, x 2,andx 3 : q 1 = q 1 (x 1,x 2,x 3 ), q 2 = q 2 (x 1,x 2,x 3 ), q 3 = q 3 (x 1,x 2,x 3 ), or in short: q i = q i (x j ). We assume that (B.1) has a unique inverse: x i = x i (q j ) (B.1) (B.2a) or x = x(q j ). (B.2b) If q 2 and q 3 are kept constant, the vector x = x(q 1 ) describes a curve in space which is the coordinate curve q 1. x/ q 1 is the tangent vector to this curve. The corresponding unit vector in the direction of increasing q 1 reads: e 1 = x/ q 1 x/ q 1. (B.3) If we set x/ q 1 = b 1,weseethat x q 1 = e 1 b 1, (B.4)

12 482 B Curvilinear Coordinates andinthesameway x = e 2 b 2, q 2 x = e 3 b 3, q 3 with b 2 = x/ q 2 and b 3 = x/ q 3. Because of x = x(q j ) it follows that (B.5) (B.6) d x = x dq 1 + x dq 2 + x dq 3 = b 1 dq 1 e 1 + b 2 dq 2 e 2 + b 3 dq 3 e 3, (B.7) q 1 q 2 q 3 and, since the basis vectors are orthogonal to each other, the square of the line element is d x d x = b 2 1 dq1 2 + b 2 2 dq2 2 + b 2 3 dq3 2. (B.8) For the volume element dv (Fig. B.1) we have dv = b 1 dq 1 e 1 ( b 2 dq 2 e 2 b 3 dq 3 e 3 )=b 1 b 2 b 3 dq 1 dq 2 dq 3. (B.9) The q 1 surface element of the volume element dv (i.e. the surface element normal to the q 1 direction) is then ds 1 = b 2 dq 2 e 2 b 3 dq 3 e 3 = b 2 b 3 dq 2 dq 3. (B.10) In a similar manner we find for the remaining surface elements ds 2 = b 3 b 1 dq 3 dq 1, ds 3 = b 1 b 2 dq 1 dq 2. (B.11) (B.12) Fig. B.1. Volume element in the curvilinear orthogonal coordinate system

13 B Curvilinear Coordinates 483 The continuity equation, Cauchy s equation of motion and the entropy equation read symbolically: ϱ + u ϱ + ϱ u =0, t ϱ D u Dt = ϱ k + T, and [ ] s ϱt + u s = Φ + (λ T ). t In Cauchy s equation we write the material derivative in the form (1.78), as this is more useful for getting the equations in curvilinear coordinates: ϱ [ ] u t u ( u)+ ( u2 /2) = ϱ k + T. (B.13) Now in order to reach the component form of these equation, the Nabla operations, and (gradient, divergence and curl) are given in curvilinear coordinates. The components of the vector Φ are: q 1 : ( Φ) 1 = 1 Φ, b 1 q 1 q 2 : ( Φ) 2 = 1 Φ, b 2 q 2 q 3 : ( Φ) 3 = 1 Φ. b 3 q 3 and (B.14) If u 1, u 2 and u 3 are the components of the vector u in the direction of increasing q 1, q 2 and q 3,wehave: 1 u = b 1 b 2 b 3 [ q 1 (b 2 b 3 u 1 )+ q 2 (b 3 b 1 u 2 )+ ] (b 1 b 2 u 3 ) q 3. (B.15) Since the basis vectors are orthonormal, the Laplace operator Δ = = 2 can be easily calculated, if, in (B.15) we identify the components of u with the components of : { [ ] 1 b2 b 3 Δ = + b 1 b 2 b 3 q 1 b 1 q 1 q 2 [ b3 b 1 b 2 q 2 ] + q 3 [ ]} b1 b 2. b 3 q 3 (B.16)

14 484 B Curvilinear Coordinates u has the components q 1 : ( u) 1 = 1 [ (b 3 u 3 ) ] (b 2 u 2 ), b 2 b 3 q 2 q 3 q 2 : ( u) 2 = 1 [ (b 1 u 1 ) ] (b 3 u 3 ), (B.17) b 3 b 1 q 3 q 1 q 3 : ( u) 3 = 1 [ (b 2 u 2 ) ] (b 1 u 1 ). b 1 b 2 q 1 q 2 The components of the divergence of the stress tensor are: [ 1 q 1 : ( T) 1 = (b 2 b 3 τ 11 )+ (b 3 b 1 τ 21 )+ ] (b 1 b 2 τ 31 ) + b 1 b 2 b 3 q 1 q 2 q 3 + τ 21 b 1 + τ 31 b 1 τ 22 b 2 τ 33 b 3, b 1 b 2 q 2 b 1 b 3 q 3 b 1 b 2 q 1 b 1 b 3 q 1 1 q 2 : ( T) 2 = b 1 b 2 b 3 1 q 3 : ( T) 3 = b 1 b 2 b 3 [ q 1 (b 2 b 3 τ 12 )+ q 2 (b 3 b 1 τ 22 )+ + τ 32 b 2 b 3 b 2 q 3 + τ 12 b 2 b 1 b 2 q 1 τ 33 b 2 b 3 b 3 q 2 τ 11 b 2 b 1 b 1 q 2, [ q 1 (b 2 b 3 τ 13 )+ q 2 (b 3 b 1 τ 23 )+ + τ 13 b 3 b 1 b 3 q 1 + τ 23 b 3 b 2 b 3 q 2 τ 11 b 3 b 1 b 1 q 3 τ 22 b 3 b 2 b 2 q 3. ] (b 1 b 2 τ 32 ) + q 3 ] (b 1 b 2 τ 33 ) + q 3 (B.18) Here for example the stress component τ 13 is the component in the direction of increasing q 3 which acts on the surface whose normal is in the direction of increasing q 1. The Cauchy-Poisson law in symbolic form holds for the components of the stress : T =( p + λ u) I +2ηE. The components of the rate of deformation tensor are given by e 11 = 1 u 1 + u 2 b 1 + u 3 b 1, b 1 q 1 b 1 b 2 q 2 b 3 b 1 q 3 e 22 = 1 u 2 + u 3 b 2 + u 1 b 2, b 2 q 2 b 2 b 3 q 3 b 1 b 2 q 1 e 33 = 1 u 3 + u 1 b 3 + u 2 b 3, b 3 q 3 b 3 b 1 q 1 b 2 b 3 q 2 2 e 32 = b 3 b 2 (u 3 /b 3 ) q 2 + b 2 b 3 (u 2 /b 2 ) q 3 =2e 23,

15 B Curvilinear Coordinates e 13 = b 1 (u 1 /b 1 ) + b 3 (u 3 /b 3 ) =2e 31, and b 3 q 3 b 1 q 1 2 e 21 = b 2 (u 2 /b 2 ) + b 1 (u 1 /b 1 ) =2e 12. (B.19) b 1 q 1 b 2 q 2 As an example of how to apply this we consider spherical coordinates r, ϑ, ϕ with the velocity components u r, u ϑ, u ϕ. The relation between Cartesian and spherical coordinates is given by the transformation (cf. Fig. B.4) x = r cos ϑ, y = r sin ϑ cos ϕ, z = r sin ϑ sin ϕ. (B.20) The x axis is the polar axis and ϑ is the polar angle. With q 1 = r, q 2 = ϑ, and q 3 = ϕ (B.21) it follows that b 1 = { cos 2 ϑ +sin 2 ϑ (sin 2 ϕ +cos 2 ϕ) } 1/2 =1, b 2 = { r 2 sin 2 ϑ + r 2 cos 2 ϑ (cos 2 ϕ +sin 2 ϕ) } 1/2 = r, b 3 = { r 2 sin 2 ϑ (sin 2 ϕ +cos 2 ϕ) } 1/2 = r sin ϑ. (B.22) The line element reads d x =dr e r + r dϑ e ϑ + r sin ϑ dϕ e ϕ, (B.23) and the volume element is dv = r 2 sin ϑ dr dϑ dϕ. (B.24) Forthesurfaceelementsweobtain ds r = r 2 sin ϑ dϑ dϕ, ds ϑ = r sin ϑ dr dϕ, ds ϕ = r dr dϑ. (B.25) The components of grad Φ = Φ are r : ( Φ) r = Φ r, ϑ : ( Φ) ϑ = 1 Φ r ϑ, ϕ : 1 ( Φ) ϕ = r sin ϑ Φ ϕ. (B.26)

16 486 B Curvilinear Coordinates For div u = u it follows that [ u =(r 2 sin ϑ) 1 r (r2 sin ϑu r )+ ϑ (r sin ϑu ϑ)+ ] ϕ (ru ϕ). The components of curl u = u are [ r : ( u) r =(r 2 sin ϑ) 1 ϑ (r sin ϑu ϕ) ] ϕ (ru ϑ) [ ϑ : ( u) ϑ =(r sin ϑ) 1 ϕ (u r) ] r (r sin ϑu ϕ), ϕ : ( u) ϕ = r 1 [ r (ru ϑ) ϑ (u r) ]., (B.27) (B.28) We now wish to calculate the rth component of the Navier-Stokes equations. To do this we require the rth component of u ( u) and of T : { u ( u)} r = 1 [ r u ϑ r (ru ϑ) ] ϑ (u r) [ 1 r sin ϑ u ϕ ϕ (u r) ] r (r sin ϑu ϕ), (B.29) [ 1 ( T) r = r 2 sin ϑ r (r2 sin ϑτ rr )+ ϑ (r sin ϑτ ϑr)+ ] ϕ (rτ ϕr) 1 r (τ ϑϑ + τ ϕϕ ), (B.30) where, from (3.1b) for incompressible flow: τ rr = p +2ηe rr, τ ϑϑ = p +2ηe ϑϑ, τ ϕϕ = p +2ηe ϕϕ, τ ϑr =2ηe ϑr, τ ϕr =2ηe ϕr, and τ ϕϑ =2ηe ϕϑ. (B.31) The components of the rate of deformation tensor are e rr = u r / r, e ϑϑ = 1 r { u ϑ/ ϑ + u r }, e ϕϕ = 1 r sin ϑ ( u ϕ/ ϕ)+ 1 r (u r + u ϑ cot ϑ),

17 B Curvilinear Coordinates e ϕϑ =2e ϑϕ =sinϑ [ ] 1 ϑ r sin ϑ u ϕ + 1 [ ] 1 sin ϑ ϕ r u ϑ, 1 2 e rϕ =2e ϕr = r sin ϑ u r/ ϕ + r sin ϑ [ ] 1 r r sin ϑ u ϕ, and 2 e ϑr =2e rϑ = r [ ] 1 r r u ϑ + 1 r u r/ ϑ. (B.32) By inserting these equations into Cauchy s equation, we obtain the rth component of the Navier-Stokes equations for incompressible flow ϱ + u [ ϕ ur r sin ϑ = ϱk r + 1 r 2 sin ϑ { u r t u [ ϑ (ruϑ ) u ] r + r r ϑ ϕ (r sin ϑu ϕ) r { r ] [ r 2 sin ϑη (u ϑ/r) +sinϑη u r ϑ r ϑ (u 2 r + u2 ϑ + u2 ϕ ) } = r [ { r 2 sin ϑ p + η u r r + η u r r ] + ϕ +r 2 sin ϑη ( ] } 1 r r sin ϑ u ϕ) + p r 2 η r 2 } ] + [ η u r sin ϑ ϕ + [ ] uϑ ϑ + u r + + p r 2 η u ϕ r 2 sin ϑ ϕ 2 η ( ) r 2 u r + u ϑ cot ϑ. (B.33) All terms containing p together result in p/ r. In spherical coordinates the Laplace operator reads Δ = 1 r 2 r [ r 2 r ] 1 + r 2 sin ϑ [ ( sin ϑ ) + 1 ϑ ϑ sin ϑ 2 ] ϕ 2. (B.34) We see that the doubly underlined terms can be written together as the differential operator ηδu r. For the singly underlined terms we can write η r { 1 r 2 sin ϑ [ r (r2 sin ϑu r )+ ϑ (r sin ϑu ϑ)+ ]} ϕ (ru ϕ) we can convince ourselves of this by differentiating it out. The expression in curly brackets is, by (B.27) equal to u, and in incompressible flow is zero. ;

18 488 B Curvilinear Coordinates Fig. B.2. Cartesian coordinates If we carry out all the differentiation on the left-hand side we find { u r ϱ t + u u r r r + 1 r u u r ϑ ϑ + 1 r sin ϑ u u r ϕ ϕ u2 ϑ + } u2 ϕ = r = ϱk r p {Δu r + η r 2r [ 2 u r + u ϑ ϑ + u ϑ cot ϑ + 1 sin ϑ u ] } ϕ (B.35) ϕ as the rth component of the Navier-Stokes equations. The remaining components are obtained in the same manner. We shall now summarize the results for Cartesian, cylindrical and spherical coordinates. B.1 Cartesian Coordinates a) Unit vectors: e x, e y, e z b) Position vector x : x = x e x + y e y + z e z c) Velocity vector u : u = u e x + v e y + w e z d) Line element: d x =dx e x +dy e y +dz e z

19 B.1 Cartesian Coordinates 489 e) Surface elements: ds x =dydz ds y =dxdz ds z =dxdy f) Volume element: dv =dx dy dz g) Gradient of the scalar Φ : h) Laplace operator on the scalar Φ : i) Divergence of the vector u : j) Curl of the vector u : [ w curl u = u = y v ] z grad Φ = Φ = Φ x e x + Φ y e y + Φ z e z k) Laplace operator on the vector u : ΔΦ = Φ = 2 Φ x Φ y Φ z 2 div u = u = u x + v y + w z [ u e x + z w ] [ v e y + x x u ] e z y Δ u = u = Δu e x + Δv e y + Δw e z l) Divergence of the stress tensor T : div T = T =( τ xx / x + τ yx / y + τ zx / z) e x + +( τ xy / x + τ yy / y + τ zy / z) e y + +( τ xz / x + τ yz / y + τ zz / z) e z

20 490 B Curvilinear Coordinates m) Rate of deformation tensor E : e xx = u/ x e yy = v/ y e zz = w/ z 2 e xy =2e yx = u/ y + v/ x 2 e xz =2e zx = u/ z + w/ x 2 e yz =2e zy = v/ z + w/ y n) Continuity equation: ϱ t + x (ϱu)+ y (ϱv)+ z (ϱw)=0 o) Navier-Stokes equations (with ϱ, η =const): x : ϱ ( u/ t + u u/ x+ v u/ y+ w u/ z) y : ϱ ( v/ t + u v/ x+ v v/ y+ w v/ z) = ϱk x p/ x + ηδu = ϱk y p/ y + ηδv z : ϱ ( w/ t + u w/ x+ v w/ y+ w w/ z)=ϱk z p/ z + ηδw B.2 Cylindrical Coordinates a) Unit vectors: e r =+cosϕ e x +sinϕ e y e ϕ = sin ϕ e x +cosϕ e y e z = e z b) Position vector x : x = r e r + z e z c) Velocity vector u : u = u r e r + u ϕ e ϕ + u z e z d) Line element: d x =dr e r + r dϕ e ϕ +dz e z

21 B.2 Cylindrical Coordinates 491 Fig. B.3. Cylindrical Coordinates e) Surface elements: ds r = r dϕ dz ds ϕ =drdz ds z = r dr dϕ f) Volume element: dv = r dr dϕ dz g) Gradient of the scalar Φ : grad Φ = Φ = Φ r e r + 1 Φ r ϕ e ϕ + Φ z e z h) Laplace operator on the scalar Φ : ΔΦ = Φ = 2 Φ r Φ r r Φ r 2 ϕ Φ z 2 i) Divergence of the vector u : div u = u = 1 r { (ur r) r + u ϕ ϕ + (u } z r) z

22 492 B Curvilinear Coordinates j) Curl of the vector u : { 1 u z curl u = u = r ϕ u } ϕ z + 1 { (uϕ r) r r u r ϕ { ur e r + } e z z u z r } e ϕ + k) Laplace operator on the vector u : Δ u = u = {Δu r 1r [ 2 u r +2 u ]} ϕ e r + ϕ + {Δu ϕ 1r [ 2 u ϕ 2 u ]} r e ϕ + Δu z e z ϕ l) Divergence of the stress tensor T : { 1 (τ rr r) div T = T = + 1 r r r { 1 (τ rϕ r) r r r + m) Rate of deformation tensor E : n) Continuity equation: ϱ t + 1 r τ ϕr ϕ + τ zr z τ } ϕϕ r } τ ϕϕ ϕ { 1 (τ rz r) + 1 τ ϕz r r r ϕ + τ zz z e rr = u r r e ϕϕ = 1 u ϕ r ϕ + 1 r u r e zz = u z z 2 e rϕ =2e ϕr = r (r 1 u ϕ ) + 1 r r 2 e rz =2e zr = u r z + u z r 2 e ϕz =2e zϕ = 1 u z r ϕ + u ϕ z r (ϱu r r)+ 1 r + τ zϕ z + τ rϕ r } e z u r ϕ ϕ (ϱu ϕ)+ z (ϱu z)=0 e r + e ϕ +

23 o) Navier-Stokes equations (with ϱ, η =const): r : ϕ : z : { ur t + u u r r [ u r u ϕ {Δu r 1r [ 2 u r +2 u ϕ ϕ ϱ r + u u r z z + 1 r = ϱk r p r + η { uϕ u ϕ ϱ + u r t r + u u ϕ z z + 1 r p ϕ + η B.3 Spherical Coordinates 493 ]} ϕ u2 ϕ = ]} [ u ϕ u ϕ {Δu ϕ 1r [ 2 u ϕ 2 u r ϕ } = ϱk ϕ 1 r { uz ϱ t + u u z r r + u u z z z + 1 r u ϕ ]} ϕ + u r u ϕ = ]} u z ϕ = ϱk z p z + ηδu z B.3 Spherical Coordinates a) Unit vectors: e r =cosϑ e x +sinϑ cos ϕ e y +sinϑ sin ϕ e z e ϑ = sin ϑ e x +cosϑ cos ϕ e y +cosϑ sin ϕ e z e ϕ = sin ϕ e y +cosϕ e z b) Position vector x : x = r e r Fig. B.4. Spherical coordinates

24 494 B Curvilinear Coordinates c) Velocity vector u : d) Line element: u = u r e r + u ϑ e ϑ + u ϕ e ϕ d x =dr e r + r dϑ e ϑ + r sin ϑ dϕ e ϕ e) Surface elements: ds r = r 2 sin ϑ dϑ dϕ ds ϑ = r sin ϑ dr dϕ ds ϕ = r dr dϑ f) Volume element: dv = r 2 sin ϑ dr dϑ dϕ g) Gradient of the scalar Φ : grad Φ = Φ = Φ r e r + 1 Φ r ϑ e ϑ + 1 Φ r sin ϑ ϕ e ϕ h) Laplace operator on the scalar Φ : ΔΦ = Φ = 1 [ r 2 r 2 Φ ] 1 + r r r 2 sin ϑ [ sin ϑ Φ ] + ϑ ϑ i) Divergence of the vector u : { 1 (r 2 sin ϑu r ) div u = u = r 2 + (r sin ϑu ϑ) sin ϑ r ϑ j) Curl of the vector u : { 1 (r sin ϑuϕ ) curl u = r 2 (ru } ϑ) e r + sin ϑ ϑ ϕ { 1 ur + r sin ϑ ϕ (r sin ϑu } ϕ) e ϑ + r + 1 { (ruϑ ) u } r e ϕ r r ϑ k) Laplace operator on the vector u : Δ u = {Δu r 2r [ 2 u r + u ϑ ϑ + u ϑ cot ϑ + 1 sin ϑ { + Δu ϑ + 2 u r r 2 ϑ 1 r 2 sin 2 ϑ { + Δu ϕ 1 r 2 sin 2 ϑ ]} u ϕ ϕ ]} [ u ϑ +2cosϑ u ϕ ϕ [ u ϕ 2sinϑ u r ϕ 2cosϑ u ϑ ϕ 1 2 Φ r 2 sin 2 ϑ ϕ 2 + (ru } ϕ) ϕ e r + e ϑ + ]} e ϕ

25 l) Divergence of the stress tensor T : T = { 1 [ (r 2 sin ϑτ rr ) r 2 sin ϑ r τ } ϑϑ + τ ϕϕ e r + r { [ 1 (r 2 sin ϑτ rϑ ) + r 2 sin ϑ r + τ } rϑ τ ϕϕ cot ϑ e ϑ + r { [ 1 (r 2 sin ϑτ rϕ ) + r 2 sin ϑ r + τ rϕ + τ ϑϕ cot ϑ r } e ϕ m) Rate of deformation tensor E : B.3 Spherical Coordinates (r sin ϑτ ϑr) ϑ + (r sin ϑτ ϑϑ) ϑ + (r sin ϑτ ϑϕ) ϑ e rr = u r r e ϑϑ = 1 u ϑ r ϑ + 1 r u r 1 u ϕ e ϕϕ = r sin ϑ ϕ + 1 r (u r + u ϑ cot ϑ) 2 e ϕϑ =2e ϑϕ =sinϑ [ ] 1 ϑ r sin ϑ u ϕ + 1 sin ϑ 2 e rϕ =2e ϕr = 1 u r r sin ϑ ϕ + r sin ϑ [ r 2 e ϑr =2e rϑ = r [ ] 1 r r u ϑ + 1 u r r ϑ + (rτ ] ϕr) ϕ + (rτ ] ϕϑ) + ϕ + (rτ ] ϕϕ) + ϕ [ 1 ϕ 1 r sin ϑ u ϕ r u ϑ ] n) Continuity equation: [ ϱ t + 1 r 2 sin ϑ r (r2 sin ϑϱu r )+ ϑ (r sin ϑϱu ϑ)+ ] ϕ (rϱu ϕ) =0 o) Navier-Stokes equations (with ϱ, η =const): r : ϱ { u r t + u u r r r + 1 r u ϑ = ϱk r p r + η u r ϑ + 1 r sin ϑ u ϕ {Δu r 2r [ 2 u r + u ϑ ϑ + u ϑ cot ϑ + 1 sin ϑ u r ϕ u2 ϑ + u2 ϕ r ] } = ]} u ϕ ϕ

26 496 B Curvilinear Coordinates { u ϑ ϑ : ϱ t + u u ϑ r r + 1 r u ϑ p ϑ + η = ϱk ϑ 1 r { uϕ u ϕ ϕ : ϱ + u r t = ϱk ϕ 1 r sin ϑ u ϑ ϑ + 1 r sin ϑ u u ϑ ϕ { Δu ϑ + 2 r 2 u r ϑ 1 r 2 sin 2 ϑ r + 1 r u u ϕ ϑ ϑ + 1 r sin ϑ u ϕ { p ϕ + η 1 Δu ϕ r 2 sin 2 ϑ ϕ + u } r u ϑ u 2 ϕ cot ϑ = r [ u ϑ +2cosϑ u ]} ϕ ϕ u ϕ ϕ + u ϕu r + u ϑ u ϕ cot ϑ r [ u ϕ 2cosϑ u ϑ ϕ 2sinϑ u r ϕ } = ]}

27 C Tables and Diagrams for Compressible Flow Table C.1 Pressure, density, temperature and area ratio as dependent on the Mach number for calorically perfect gas (γ =1.4) Subsonic M p/p t ϱ/ϱ t T/T t a/a t A /A

28 498 C Tables and Diagrams for Compressible Flow M p/p t ϱ/ϱ t T/T t a/a t A /A

29 C Tables and Diagrams for Compressible Flow 499 M p/p t ϱ/ϱ t T/T t a/a t A /A

30 500 C Tables and Diagrams for Compressible Flow Supersonic M p/p t ϱ/ϱ t T/T t a/a t A /A

31 C Tables and Diagrams for Compressible Flow 501 M p/p t ϱ/ϱ t T/T t a/a t A /A

32 502 C Tables and Diagrams for Compressible Flow M p/p t ϱ/ϱ t T/T t a/a t A /A

33 C Tables and Diagrams for Compressible Flow 503 M p/p t ϱ/ϱ t T/T t a/a t A /A

34 504 C Tables and Diagrams for Compressible Flow M p/p t ϱ/ϱ t T/T t a/a t A /A

35 C Tables and Diagrams for Compressible Flow 505 Table C.2 Pressure, density, temperature, total pressure and Mach number M 2 behind a normal shock as dependent on the Mach number M 1 in front of the shock for calorically perfect gas (γ =1.4). M 1 p 2 /p 1 ϱ 2 /ϱ 1 T 2 /T 1 p t2 /p t1 M

36 506 C Tables and Diagrams for Compressible Flow M 1 p 2 /p 1 ϱ 2 /ϱ 1 T 2 /T 1 p t2 /p t1 M