2.8. Vector Dot and Cross Products

The dot product of two vectors u and v is defined in (2.2). It can also be written u·v =  cos θ, where u and υ are the vectors’ magnitudes and θ is the angle between the vectors (see Exercises 2.14 and 2.15). The dot product is therefore the magnitude of one vector times the component of the other in the direction of the first. The dot product u·v is equal to the sum of the diagonal terms of the tensor uivj and is zero when u and v are orthogonal.
The cross product between two vectors u and v is defined as the vector w whose magnitude is sin θ where θ is the angle between u and v, and whose direction is perpendicular to the plane of u and v such that u, v, and w form a right-handed system. In this case, u × v = v × u. Furthermore, unit vectors in right-handed coordinate systems obey the cyclic rule e1 × e2 = e3. These requirements are sufficient to determine:

u×v=(u2v3u3v2)e1+(u3v1u1v3)e2+(u1v2u2v1)e3,

image (2.20)

(see Exercise 2.16). Equation (2.20) can be written as the determinant of a matrix:

u×v=det[e1e2e3u1u2u3v1v2v3].

image

In indicial notation, the k-component of u × v can be written as:

(u×v)·ek=i=13j=13εijkuivjεijkuivj=εkijuivj.

As a check, for k = 1 the nonzero terms in the double sum in (2.21) result from i = 2, j = 3, and from i = 3, j = 2. This follows from the definition (2.18) that the permutation symbol is zero if any two indices are equal. Thus, (2.21) gives:

(u×v)·e1=εij1uivj=ε231u2v3+ε321u3v2=u2v3u3v2,

image

which agrees with (2.20). Note that the fourth form of (2.21) is obtained from the second by moving the index k two places to the left; see the remark following (2.18).
Example 2.8
Use the definition of the cross and dot products to evaluate (a×b)·cimage in index notation and determine its value when a = e1, b = e2, and c = e3.
Solution
Use (2.21) to find: a×b=εijkaibjimage, which has free index is k. Form the dot product using (2.2): (a×b)·c=[εijkaibj]·c=εijkaibjckimage. When a = e1, b = e2, and c = e3, then a1 = b2 = c3 = 1; all other components of the three unit vectors are zero. Thus, the triple sum implied by εijkaibjckimage reduces to one term, so that (e1×e2)·e3=ε123a1b2c3=1image. Note that the placement of parentheses in the original expression is important; a×(b·c)=εijkaiblclimage is a second-order tensor since b·c=blclimage is a scalar and this leaves two free indices (j and k).

2.9. Gradient, Divergence, and Curl

The vector-differentiation operator “del”1 is defined symbolically by:

=e1x1+e2x2+e3x3=i=13eixieixi.

image (2.22)

When operating on a scalar function of position ϕ, it generates the vector:

ϕ=i=13eiϕxieiϕxi,

image

whose i-component is ϕ/xiimage. The vector ∇ϕ is called the gradient of ϕ, and ∇ϕ is perpendicular to surfaces defined by ϕ = constant. In addition, it specifies the magnitude and direction of the maximum spatial rate of change of ϕ (Figure 2.7). The spatial rate of change of ϕ in any other direction n is given by:
image
Figure 2.7 An illustration of the gradient, ∇ϕ, of a scalar function ϕ. The curves of constant ϕ and ∇ϕ are perpendicular, and the spatial derivative of ϕ in the direction n is given by n·∇ϕ. The most rapid change in ϕ is found when n and ∇ϕ are parallel.

ϕ/n=ϕ·n.

image

In Cartesian coordinates, the divergence of a vector field u is defined as the scalar

·u=u1x1+u2x2+u3x3=i=13uixiuixi.

So far, we have defined the operations of the gradient of a scalar and the divergence of a vector. We can, however, generalize these operations. For example, the divergence, ·Timage, of a second-order tensor T can be defined as the vector whose j-component is:

(·T)·ej=j=13TijxiTijxi.

image

The divergence operation decreases the order of the tensor by one. In contrast, the gradient operation increases the order of a tensor by one, changing a zero-order tensor to a first-order tensor, and a first-order tensor to a second-order tensor, i.e., ∂ui/∂xj.
The curl of a vector field u is defined as the vector ∇×u, whose i-component can be written as:

(×u)·ei=j=13k=13εijkukxjεijkukxj,

using (2.21) and (2.22). The three components of the vector ∇×u can easily be found from the right-hand side of (2.24). For the i = 1 component, the nonzero terms in the double sum result from j = 2, k = 3, and from j = 3, k = 2. The three components of ∇×u are finally found as:

(×u)·e1=u3x2u2x3,(×u)·e2=u1x3u3x1,and(×u)·e3=u2x1u1x2.

image (2.25)

A vector field u is called solenoidal or divergence free if ∇·u = 0, and irrotational or curl free if ∇ × u = 0. The word solenoidal refers to the fact that the divergence of the magnetic induction is always zero because of the absence of magnetic monopoles. The reason for the word irrotational is made clear in Chapter 3.
Example 2.9
If a is a positive constant and b is a constant vector, determine the divergence and the curl of a vector field that diverges from the origin of coordinates, u = ax, and a vector field indicative of solid body rotation about a fixed axis, u = b × x.
Solution

·u=ax1x1+ax2x2+ax3x3=a+a+a=3a,(×u)·e1=ax3x2ax2x3=0,(×u)·e2=ax1x3ax3x1=0,and(×u)·e3=ax2x1ax1x2=0.

image

 
Thus, u = ax has a constant nonzero divergence and is irrotational. Using u = (b2x3 – b3x2)e1 + (b3x1 – b1x3)e2+ (b1x2 – b2x1)e3 in (2.23) and (2.25) produces:

·u=(b2x3b3x2)x1+(b3x1b1x3)x2+(b1x2b2x1)x3=0,(×u)·e1=(b1x2b2x1)x2(b3x1b1x3)x3=2b1,(×u)·e2=(b2x3b3x2)x3(b1x2b2x1)x1=2b2,and(×u)·e3=(b3x1b1x3)x1(b2x3b3x2)x2=2b3.

image

 
Thus, u = b × x is divergence free and rotational.

2.10. Symmetric and Antisymmetric Tensors

A tensor B is called symmetric in the indices i and j if the components do not change when i and j are interchanged, that is, if Bij = Bji. Thus, the matrix of a symmetric second-order tensor is made up of only six distinct components (the three on the diagonal where i = j, and the three above or below the diagonal where ij). On the other hand, a tensor is called antisymmetric if Bij = –Bji. An antisymmetric tensor's diagonal components are each zero, and it has only three distinct components (the three above or below the diagonal). Any tensor can be represented as the sum of symmetric and antisymmetric tensors. For if we write:

Bij=12(Bij+Bji)+12(BijBji)=Sij+Aij,

image

then the operation of interchanging i and j does not change the first term, but changes the sign of the second term. Therefore, (Bij + Bji)/2 ≡ Sij is called the symmetric part of Bij, and (Bij Bji)/2 ≡ Aij is called the antisymmetric part of Bij.
Every vector can be associated with an antisymmetric tensor, and vice versa. For example, we can associate the vector ω having components ωi, with an antisymmetric tensor:

R=[0ω3ω2ω30ω1ω2ω10].

The two are related via:

Rij=k=13εijkωkεijkωk,andωk=i=13j=1312εijkRij12εijkRij.

As a check, (2.27) gives R11 = 0 and R12 = ε123ω3 = ω3, in agreement with (2.26). (In Chapter 3, R is recognized as the rotation tensor corresponding to the vorticity vector ω.)
A commonly occurring operation is the doubly contracted product, P, of a symmetric tensor T and another tensor B:

P=k=13l=13TklBklTklBkl=Tkl(Skl+Akl)=TklSkl+TklAkl=TijSij+TijAij,

image (2.28)

where S and A are the symmetric and antisymmetric parts of B. The final equality follows from the index-summation convention; sums are completed over both k and l, so these indices can be replaced by any two distinct indices. Exchanging the indices of A in the final term of (2.28) produces P = TijSijimageTijAjiimage, but this can also be written P=TjiSjiTjiAjiimage because Sij and Tijimage are symmetric. Now, replace the index j by k and the index i by l to find:

P=TklSklTklAkl.

image (2.29)

This relationship and the fourth part of the extended equality in (2.28) require that TijAij=TklAkl=0image, and:

TijBij=TijSij=12Tij(Bij+Bji).

image

Thus, the doubly contracted product of a symmetric tensor T with any tensor B equals T doubly contracted with the symmetric part of B, and the doubly contracted product of a symmetric tensor and an antisymmetric tensor is zero. The latter result is analogous to the fact that the definite integral over an even (symmetric) interval of the product of a symmetric and an antisymmetric function is zero.
Solution

R:R=RijRij=(εijkωk)(εijlωl)=(εkij)(+εjil)ωkωl=(δkiδilδklδii)ωkωl.

image

 
Inside the last parentheses, δkiδil = δkl and δii = 3, so:

R:R=(δkl3δkl)ωkωl=2δklωkωl=2ωk2=2(ω12+ω22+ω32)=2|ω|2.

image

 

2.11. Eigenvalues and Eigenvectors of a Symmetric Tensor

The reader is assumed to be familiar with the concepts of eigenvalues and eigenvectors of a matrix, so only a brief review of the main results is provided here. Suppose τ is a symmetric tensor with real elements. Then the following facts can be proved:
1. There are three real eigenvalues λk (k = 1, 2, 3), which may or may not all be distinct. (Here, the superscript k is not an exponent, and λk does not denote the kth-component of a vector.) These eigenvalues (λ1, λ2, and λ3) are the roots or solutions of the third-degree polynomial:

det|τijλδij|=0.

image

2. The three eigenvectors bk corresponding to distinct eigenvalues λk are mutually orthogonal. These eigenvectors define the directions of the principal axes of τ. Each b is found by solving three algebraic equations:

(τijλδij)bj=0

image

    (i = 1, 2, or 3), where the superscript k on λ and b has been omitted for clarity because there is no sum over k.
3. If the coordinate system is rotated so that its unit vectors coincide with the eigenvectors, then τ is diagonal with elements λk in this rotated coordinate system:

τ=[λ1000λ2000λ3].

image

 
Solution
For the given velocity profile u1(x2), it is evident that S11 = S22 = 0, and 2S12 = 2S21 = du1/dx2 = 2Γ. The strain rate tensor in the original coordinate system is therefore:

S=[0ΓΓ0].

image

 
The eigenvalues are determined from:

det|Sijλδij|=det|λΓΓλ|=λ2Γ2=0,

image

 
which has solutions λ1 = Γ and λ2 = Γ. The first eigenvector b1 is given by:

[0ΓΓ0][b11b21]=λ1[b11b21],

image

 
which has solution b11=b21=1/2image, when b1 is normalized to have magnitude unity. The second eigenvector is similarly found so that:

b1=[1/21/2],andb2=[1/21/2].

image

 
These eigenvectors are shown in Figure 2.8. The direction cosine matrix of the original and the rotated coordinate system is therefore:

C=[1/21/21/21/2],

image

 
which represents rotation of the coordinate system by 45°. Using the transformation rule (2.12), the components of S in the rotated system are found as follows:

S12=Ci1Cj2Sij=C11C22S12+C21C12S21=1212Γ1212Γ=0,S21=0,S11=Ci1Cj1Sij=C11C21S12+C21C11S21=Γ,andS22=Ci2Cj2Sij=C12C22S12+C22C12S21=Γ.

image

 
(Instead of using (2.12), all the components of S in the rotated system can be found by carrying out the matrix product CT · S · C.) The matrix of S in the rotated frame is therefore:

S=[Γ00Γ].

image

 
image
Figure 2.8 Original coordinate system Ox1x2 and the rotated coordinate system Ox1x2image having unit vectors that coincide with the eigenvectors of the strain-rate tensor in Example 2.4. Here the strain rate is determined from a unidirectional flow having only cross-stream variation, and the angle of rotation is determined to be 45°.
The foregoing matrix contains only diagonal terms. For positive Γ, it will be shown in the next chapter that it represents a linear stretching at a rate Γ along one principal axis, and a linear compression at a rate Γ along the other; the shear strains are zero in the principal-axis coordinate system of the strain rate tensor.

2.12. Gauss’ Theorem

This very useful theorem relates volume and surface integrals. Let V be a volume bounded by a closed surface A. Consider an infinitesimal surface element dA having outward unit normal n with components ni (Figure 2.9), and let Q(x) be a scalar, vector, or tensor field of any order. Gauss’ theorem states that:

VQxidV=AniQdA.

image (2.30)

The most common form of Gauss’ theorem is when Q is a vector, in which case the theorem is:

Vi=13QixidVVQixidV=Ai=13niQidAAniQidA,orV·QdV=An·QdA,

image

which is commonly called the divergence theorem. In words, the theorem states that the volume integral of the divergence of Q is equal to the surface integral of the outflux of Q.
Alternatively, (2.30) defines a generalized field derivative, denoted by Dimage, of Q when considered in its limiting form for a vanishingly small volume:

DQ=limV01VAniQdA.

image (2.31)

Interestingly, this form is readily specialized to the gradient, divergence, and curl of any scalar, vector, or tensor Q. Moreover, by regarding (2.31) as a definition, the recipes for the computation of vector field derivatives may be obtained in any coordinate system. As stated, (2.31) defines the gradient of a tensor Q of any order. For a tensor of order one or higher, the divergence and curl are defined by including a dot (scalar) product or a cross (vector) product, respectively, under the integral:

·Q=limV01VAn·QdA,and×Q=limV01VAn×QdA.

image (2.32, 2.33)

Example 2.12
Obtain the recipe for the divergence of a vector Q(x) in Cartesian coordinates from the integral definition (2.32).
Solution
 

([n·Q]EADH+[n·Q]FBCG)dA=([e1·Q(x)+e1·Δx12Q(x)x1+···]+[e1·Q(x)+e1·Δx12Q(x)x1+···])Δx2Δx3=(e1·Q(x)x1+...)Δx1Δx2Δx3.

image

 
Similarly for the other two directions:

([n·Q]ABCD+[n·Q]EFGH)dA=(e2·Q(x)x2+···)Δx1Δx2Δx3,([n·Q]ABFE+[n·Q]DCGH)dA=(e3·Q(x)x3+···)Δx1Δx2Δx3.

image

 
Assembling the contributions from all six faces (or all three directions) to evaluate (2.32) produces:

·Q=limV01VAn·QdA=limΔx10Δx20Δx301Δx1Δx2Δx3(e1·Q(x)x1+e2·Q(x)x2+e3·Q(x)x3+...)Δx1Δx2Δx3,

image

 
and when the limit is taken, the expected Cartesian-coordinate form of the divergence emerges:

·Q=e1·Q(x)x1+e2·Q(x)x2+e3·Q(x)x3.

image

 

2.13. Stokes’ Theorem

Stokes’ theorem relates the integral over an open surface A to the line integral around the surface’s bounding curve C. Here, unlike Gauss’ theorem, the inside and outside of A are not well defined so an arbitrary choice must be made for the direction of the outward normal n (here it always originates on the outside of A). Once this choice is made, the unit tangent vector to C, t, points in the counterclockwise direction when looking at the outside of A. The final unit vector, p, is perpendicular the curve C and tangent to the surface, so it is perpendicular to n and t. Together the three unit vectors form a right-handed system: t × n = p (see Figure 2.10). For this geometry, Stokes’ theorem states:

A(×u)·ndA=Cu·tds,

image (2.34)

where s is the arc length of the closed curve C. This theorem signifies that the surface integral of the curl of a vector field u is equal to the line integral of u along the bounding curve of the surface. In fluid mechanics, the right side of (2.34) is called the circulation of u about C. In addition, (2.34) can be used to define the curl of a vector through the limit of the circulation about an infinitesimal surface as:

n·(×u)=limA01ACu·tds.

image (2.35)

The advantage of integral definitions of field derivatives is that such definitions do not depend on the coordinate system.
Example 2.13
Obtain the recipe for the curl of a vector u(x) in Cartesian coordinates from the integral definition given by (2.35).
Solution
This recipe is obtained by considering rectangular contours in three perpendicular planes intersecting at the point (x, y, z) as shown in Figure 2.11. First, consider the elemental rectangle lying in a plane defined by x = const. The central point in this plane is (x, y, z) and the rectangle’s area is ΔyΔz. It may be shown by careful integration of a Taylor expansion of the integrand that the integral along each line segment may be represented by the product of the integrand at the center of the segment and the length of the segment with attention paid to the direction of integration ds. Thus we obtain:

(×u)·ex=limΔy0limΔz0{1ΔyΔz[uz(x,y+Δy2,z)uz(x,yΔy2,z)]Δz+1ΔyΔz[uy(x,y,zΔz2)uy(x,y,z+Δz2)]Δy}.

image

 
image
Figure 2.11 Drawing for Example 2.11. The point (x,y,z) is located at the center of the cruciform. The arrows show the direction of ds on the edge of the rectangular contour that lies in the x = const. plane. The area inside this contour is ΔyΔz.
Taking the limits produces:

(×u)·ex=uzyuyz.

image

 
Similarly, integrating around the elemental rectangles in the other two planes leads to:

(×u)·ey=uxzuzxand(×u)·ez=uyxuxy.

image

 

Exercises

2.1. For three spatial dimensions, rewrite the following expressions in index notation and evaluate or simplify them using the values or parameters given, and the definitions of δij and εijk wherever possible. In b) through e), x is the position vector, with components xi.
a. b·cimage where b = (1, 4, 17) and c = (–4, –3, 1).
b. (u·)ximage where u a vector with components ui.
c. ϕimage, where ϕ=h·ximage and h is a constant vector with components hi.
d. ×uimage, where u = Ω × x and Ω is a constant vector with components Ωi.
e. C·ximage, where

C={123012001}

image

2.3. For two three-dimensional vectors with Cartesian components ai and bi, prove the Cauchy-Schwartz inequality: (aibi)2 ≤ (ai)2(bi)2.
2.4. For two three-dimensional vectors with Cartesian components ai and bi, prove the triangle inequality: |a|+|b||a+b|image.
2.8. Show that the condition for the vectors a, b, and c to be coplanar is εijkaibjck = 0.
2.9. Prove the following relationships: δijδij = 3, εpqrεpqr = 6, and εpqiεpqj = 2δij.
2.10. Show that C·CT = CT·C =δ, where C is the direction cosine matrix and δ is the matrix of the Kronecker delta. Any matrix obeying such a relationship is called an orthogonal matrix because it represents transformation of one set of orthogonal axes into another.
2.11. Show that for a second-order tensor A, the following quantities are invariant under a rotation of axes:

I1=Aii,I2=det|A11A12A21A22|+det|A22A23A32A33|+det|A11A13A31A33|,andI3=det(Aij).

image

    [Hint: Use the result of Exercise 2.10 and the transformation rule (2.12) to show that I1Aii=Aii=I1image. Then show that AijAji and AijAjkAki are also invariants. In fact, all contracted scalars of the form AijAjk ċċċ Ami are invariants. Finally, verify that I2=12[I12AijAji]image, and I3=13[AijAjkAkiI1AijAji+I2Aii]image. Because the right-hand sides are invariant, so are I2 and I3.]
2.12. If u and v are vectors, show that the products uiυj obey the transformation rule (2.12) in any number of spatial dimensions, and therefore represent a second-order tensor.
2.13. Show that δij is an isotropic tensor. That is, show that δij = δij under rotation of the coordinate system. [Hint: Use the transformation rule (2.12) and the results of Exercise 2.10.]
2.14. If u and v are arbitrary vectors resolved in three-dimensional Cartesian coordinates, use the definition of vector magnitude, |a|2=a·aimage, and the Pythagorean theorem to show that u·v = 0 when u and v are perpendicular.
2.15. If u and v are vectors with magnitudes u and υ, use the finding of Exercise 2.14 to show that u·v = cos θ where θ is the angle between u and v.
2.16. Determine the components of the vector w in three-dimensional Cartesian coordinates when w is defined by: u·w = 0, v·w = 0, and w·w = u2υ2 sin2 θ, where u and v are known vectors with components ui and υi and magnitudes u and υ, respectively, and θ is the angle between u and v. Choose the sign(s) of the components of w so that w = e3 when u = e1 and v = e2.
2.17. If a is a positive constant and b is a constant vector, determine the divergence and the curl of u = ax/x3 and u = b×(x/x2) where x=x12+x22+x32xixiimage is the length of x.
2.18. Obtain the recipe for the gradient of a scalar function in cylindrical polar coordinates from the integral definition (2.32).
2.20. Obtain the recipe for the divergence of a vector function in spherical polar coordinates from the integral definition (2.32).
2.21. Use the vector integral theorems to prove that ·(×u)=0image for any twice-differentiable vector function u regardless of the coordinate system.
2.22. Use Stokes’ theorem to prove that ×(ϕ)=0image for any single-valued twice-differentiable scalar ϕ regardless of the coordinate system.