Closing Remarks

We close by repeating something we said in Chapter 1:

A vector is an objective entity.

In elementary mathematics we learn to think of a vector as an ordered triple of components. There is, of course, no harm in this if we keep in mind a certain Cartesian frame. But if we fix those components then in any other frame the vector is determined uniquely. Absolutely uniquely! So a vector is something objective, but as soon as we specify its components in one frame we can find them in any other frame by the use of certain rules.

We emphasize this because the situation is exactly the same with ten-sors. A tensor is an objective entity, and fixing its components relative to one frame, we determine the tensor uniquely — even though its components relative to other frames will in general be different.

2.9 Problems

2.1 Find the dual basis to ei.

(a) e₁= 2i₁+ i₂− i₃, e₂= 2i₂+ 3i₃, e₃= i₁+ i₃;

(b) e₁= i₁+ 3i₂+ 2i₃, e₂= 2i₁− 3i2+ 2i₃, e₃= 3i₁+ 2i₂+ 3i₃; (c) e₁= i₁+ i₂, e₂= i₁− i2, e₃= 3i₃;

(d) e₁= cos φi₁+ sin φi₂, e₂=− sin φi1+ cos φi₂, e₃= i₃. 2.2 Let

˜

e₁=−2i₁+ 3i₂+ 2i₃, e₁= 2i₁+ i₂− i₃,

˜

e₂=−2i₁+ 2i₂+ i₃, e₂= 2i₂+ 3i₃,

˜

e₃=−i₁+ i₂+ i₃, e₃= i₁+ i₃.

Find the matrix A^j_i of transformation from the basis ˜ei to the basis ej.

2.3 Let

˜

e₁= i₁+ 2i₂, e₁= i₁− 6i3,

˜e₂=−i2− i3, e₂=−3i1− 4i2+ 4i₃,

˜e₃=−i1+ 2i₂− 2i3, e₃= i₁+ i₂+ i₃. Find the matrix of transformation of the basis ˜ei to ej. 2.4 Find

(a) a_jδ^jk, (b) aia^jδⁱ_j, (c) δⁱ_i, (d) δijδ^jk, (e) δijδ^ji, (f) δ^j_iδ_j^kδⁱ_k.

2.5 Show that ijk^ijl = 2δ^l_k. 2.6 Show that _ijk^ijk= 6.

2.7 Find (a) ijkδ^jk, (b) ijk^mkjδ_mⁱ , (c) ijkδ^k_mδ^j_n, (d) ijkaⁱa^j, (e) ijk|^ijk|, (f) ijk^imnδ^j_m. 2.8 Find (a× b) × c.

2.9 Show that (a× b) · a = 0.

2.10 Show that a· (b × c)d = (a · d)b × c + (b · d)c × a + (c · d)a × b.

2.11 Show that (e× a) × e = a if |e| = 1 and e · a = 0.

2.12 Let ekbe a basis ofR³, let x = x^kek, and suppose h₁, h₂, h₃are fixed positive numbers. Show that hk|x^k| is a norm in R³.

Tensors

3.1 Dyadic Quantities and Tensors

We have met sets of quantities like g^ij or gij. Such a table of 3× 3 = 9 coefficients could be considered as a vector in a nine-dimensional space, but we must reject this idea for an important reason: if we change the frame vectors and calculate the relations between the new and old components, the results differ in form from those that apply to vector components. The components of the metric tensor transform according to certain rules, how-ever, and it is found that these transformation rules also apply to various quantities encountered in physical science. We indicated in Chapter 1 that these quantities, represented by 3×3 matrices, form a class of objects known as second-order tensors. Our plan is to present the relevant theory in a way that parallels the vector presentation of Chapter 2.

We begin to realize this program with the introduction of the dyad (or tensor product ) of two vectors a and b, denoted a⊗ b. We assume that the tensor product satisfies many usual properties of a product:

(λa)⊗ b = a ⊗ (λb) = λ(a ⊗ b), (a + b)⊗ c = a ⊗ c + b ⊗ c,

a⊗ (b + c) = a ⊗ b + a ⊗ c, (3.1) where λ is an arbitrary real number. However, the tensor product is not symmetric: if a is not proportional to b then a⊗ b = b ⊗ a. From now on, we shall write out the dyad without the⊗ symbol: ab = a ⊗ b.

Let us once again consider the space of three-dimensional vectors with the frame ei. Using the expansion of the vectors in the basis vectors and the properties (3.1), we represent the dyad ab as

ab = aⁱe_ib^je_j= aⁱb^je_ie_j.

29

This introduces exactly nine different dyads e_ie_j. We now consider a linear space whose basis is this set of nine dyads and call it the space of second-order tensors (or tensors of second-order two). The numerical coefficients of the dyads are called the components of the tensor. Thus an element of this space, a tensor A, has the representation

A = a^ijeiej.

To maintain the property of objectivity of the elements of this space, we require that upon transformation of the frame the components of A trans-form correspondingly. Note that we have introduced superscript indices for the components of A. This was done in keeping with the development of Chapter 2.

In preparation for the next section let us introduce the dot product of a dyad ab by a vector c:

ab· c = (b · c)a. (3.2)

So the result is a vector co-oriented with a. Analogously we can introduce the dot product from the left:

c· ab = (c · a)b. (3.3)

Exercise 3.1. (a) A dyad of the form ee, where e is a unit vector, is sometimes called a projection dyad. Explain. (b) Write down matrices for the dyads i₁i₁, i₂i₂, and i₃i₁.

3.2 Tensors From an Operator Viewpoint

An alternative to viewing a second-order tensor as a weighted sum of dyads is to view the tensor as an operator. From this standpoint a tensor A is considered to map a vector x into a vector y according to the equation

y = A· x.

Conversely, a given linear relation between x and y will define the operator A uniquely. Thus if we have A· x = B · x for all x, then we have A = B.

Let us show that the components are really uniquely defined in any basis by the equality y = A· x. The tensor A is represented by the expression a^ije_ie_j in some basis e_i. It is clear that the operation A·x is linear in x, so we define A uniquely if we specify its action on all three vectors of a basis.

Taking x = e^k, the corresponding y is

A· x = a^ije_ie_j· e^k= a^ike_i.

Dot multiplying this by e^l we get

a^lk = e^l· A · e^k.

In this way we can find the components of a tensor A in any basis:

a^ij= eⁱ· A · e^j, aⁱ_·j= eⁱ· A · ej, a_ij= e_i· A · ej, a_i^·j= e_i· A · e^j.

Note that in “mixed components” we position the indices in such a way that their association with the various dyads remains clear.

Analyzing the above reasoning, we can find that we have proved the quotient law for tensors of order two. If y is a given vector and there is a linear transformation from x to y for an arbitrary vector x, then the linear transformation is a tensor and we can write y = A· x. This statement is sometimes useful in establishing the tensorial character of a set of scalar quantities (i.e., the components of A).

We may also define common algebraic operations from the operator viewpoint. Given tensors A and B, the sum is the tensor A + B uniquely defined by the requirement that

(A + B)· x = A · x + B · x

for all x. If c is a scalar, cA is defined by the requirement that (cA)· x = c(A · x)

for all x. In particular, any product of the form 0A gives a zero tensor denoted 0. The dot product A· B is regarded as the composition of the operators B and A:

(A· B) · x ≡ A · (B · x).

The dot product y· A, called pre-multiplication of A by a vector y, is defined by the requirement that

(y· A) · x = y · (A · x) for all vectors x.

A simple but important tensor is the unit tensor denoted by E and defined by the requirement that for any x

E· x = x · E = x. (3.4)

It is evident that in any Cartesian frame i_i we must have E =

3 i=1

iiii. (3.5)

In any frame we have

E = eⁱei= eje^j (3.6) for the mixed components. Consequently, the raising and lowering of indices gives

E = g_ijeⁱe^j = g^ije_ie_j (3.7) in non-mixed components. We see that the role of the unit tensor belongs to the metric tensor! Throughout our discussion of second-order tensors we shall emphasize the close analogy between tensor theory and matrix theory.

Equations (3.5) and (3.6) show that the matrix representation of E in either Cartesian or mixed components is the 3× 3 identity matrix

⎛

⎝1 0 0 0 1 0 0 0 1

⎞

⎠ .

This does not hold for the non-mixed components of (3.7).

Exercise 3.2. Use (3.4) along with (2.4) and (2.5) to show that the various components of E are given by

e^ij = g^ij, eⁱ_·j= δⁱ_j, eij = gij, e_j^·i= δⁱ_j. Hence establish (3.6).

Our consideration of A as an operator leads us to introduce the notion of an inverse tensor : if

A· A⁻¹= E,

then A⁻¹ is called the inverse of A. The inverse of a tensor is also a tensor.

An important special case occurs when the matrix of the tensor is a diagonal matrix. If in a Cartesian frame ii we have

A =

3 i=1

λiiiii

then the corresponding matrix representation is

to which there corresponds the matrix

⎛

This means that B = A⁻¹. Correspondingly,

⎛

Exercise 3.3. Establish the formula

(A· B)⁻¹= B⁻¹· A⁻¹ for invertible tensors A, B.

A second-order tensor A is singular if A· x = 0 for some x = 0.

Hence A is nonsingular if A· x = 0 only when x = 0. Recall that a matrix A is said to be nonsingular if and only if det A= 0. The connection between the uses of this terminology in the two areas is as follows. If we take a mixed representation of the tensor A, the equation A· x = 0 yields a set of simultaneous equations in the components of x; these equations have a nontrivial solution if and only if the determinant of the coefficient matrix (i.e., the matrix representing A) is zero. Moreover, taking any other representation of A and a dual representation of the vector, we again arrive the same conclusion regarding the determinant. This brings the use of the term “singular” to the tensor A. By definition the determinant of a second-order tensor A, denoted det A, is the determinant of the matrix of its mixed components:

det A =|ai^·j| = |a^k_{· m}| = 1

g|ast| = g|a^pq|.

The first equality is the definition as stated above; the rest are left for the reader to establish. Various other formulas such as

det A = 1

6ijk^mnpa_m^{· i}a_n^{· j}a_p^·k can be established for the determinant.

We close this section with an important remark. We can derive all de-sired properties of a tensor, and perform actions with the tensor, in any co-ordinate frame. Convenience will often dictate the use of Cartesian frames.

But if we obtain an equation or expression through the use of a Cartesian frame and can subsequently represent this result in non-coordinate form, then we have provided rigorous justification of the latter. As we have said before, tensors are objective entities and ultimately all results pertaining to them must be frame independent.

3.3 Dyadic Components Under Transformation

The standpoint for deriving the transformation rules is that in any basis a tensor is the same element of some space, and only (3.1) and the rules we derived for vectors can govern the rules for transforming the components of a tensor. Let us begin with the transformation of the components when we go to the reciprocal basis. We set

a_ijeⁱe^j= a^ije_ie_j

and take dot products as in (3.2) and (3.3):

ek· aijeⁱe^j· em= ek· a^ijeiej· em. This gives

aij(ek· eⁱ)(e^j· em) = a^ij(ek· ei)(ej· em), hence

akm = a^ijgkigjm.

We see that the components of the metric tensor are encountered in this transformation.

Now we can construct the formulas for transforming the tensor compo-nents when the change of basis takes the general form

ei= A^j_i˜ej. From

˜

a^ij˜e_ie˜_j = a^kme_ke_m= a^kmA^p_k˜e_pA^q_me˜_q we obtain

˜

a^ij = a^kmAⁱ_kA^j_m. (3.8) Similarly, the inverse transformation

˜

e_i= ˜A^j_iej

leads to

a^ij = ˜a^kmA˜ⁱ_kA˜^j_m. (3.9) Equations (3.8) and (3.9) together imply that

A^k_jA˜ⁱ_k= δⁱ_j. Various expressions for A,

A = a^ijeiej = akle^ke^l= a^·j_i eⁱej = a^k_{· l}eke^l

= ˜a^ij˜eie˜j = ˜akl˜e^k˜e^l= ˜a^{· j}_i ˜eⁱ˜ej= ˜a^k_{· l}˜ek˜e^l, lead to other transformation formulas such as

˜

aij = ˜A^k_iA˜^l_jakl, aij = A^k_iA^l_j˜akl, and

˜

a^{· j}_i = ˜A^k_iA^j_la_k^{· l}, ˜aⁱ_{· j}= Aⁱ_kA˜^l_ja^k_{· l}.

Remembering the terminology of § 2.5, we see why the aij are called the covariant components of A while the a^ij are called the contravariant com-ponents. The components aⁱ_·j and a^·j_i are called mixed components.

Quick summary We have

A = ˜a^ij˜e_i˜e_j = ˜a_kl˜e^k˜e^l= ˜a^{· j}_i ˜eⁱ˜e_j= ˜a^k_{· l}˜e_ke˜^l

= a^ije_ie_j = a_kle^ke^l= a_i^·jeⁱe_j= a^k_{· l}e_ke^l where

˜

a^ij= Aⁱ_kA^j_la^kl, a^ij = ˜Aⁱ_kA˜^j_la˜^kl,

˜

a_ij= ˜A^k_iA˜^l_ja_kl, a_ij = A^k_iA^l_ja˜_kl,

˜

aⁱ_{· j}= Aⁱ_kA˜^l_ja^k_{· l}, aⁱ_·j = ˜Aⁱ_kA^l_ja˜^k_{· l},

˜

a^{· j}_i = ˜A^k_iA^j_la_k^{· l}, a^·j_i = A^k_iA˜^j_la˜^{· l}_k. Exercise 3.4. (a) Express the transformation law

˜b^ij = Aⁱ_kA^j_mb^km

in matrix notation. (b) Repeat for a transformation law of the form

˜bij = ˜A^k_iA˜^m_j bkm.

Exercise 3.5. Our A^j_i values give the transformation from one basis to another; they define a transformation of the space that is an operator, and hence a second-order tensor. Write out the tensor for which the A^j_i are components. Repeat for the inverse transformation.

3.4 More Dyadic Operations

The dot product of two dyads ab and cd is defined by ab· cd = (b · c)ad.

The result is again a dyad, with a coefficient b· c. Extensions of this and the formulas (3.2) and (3.3) to operations with sums of dyads and vectors (using (3.1) and the vectorial rules) gives us a number of rules which the dot product obeys. Let A and B be dyads, a and b be vectors, and λ and µ be any real numbers. Then

A· (λa + µb) = λA · a + µA · b, (λA + µB)· a = λA · a + µB · a.

Similar identities hold for dot products taken in the opposite orders. These results show that linearity may be assumed in working with these opera-tions.

Now let us pursue the close analogy between the dot product and matrix multiplication. We begin with the simple case of a Cartesian frame. We take a dyad A and a vector b and express these relative to a basis i_k:

A = a^kmikim, b = b^jij.

(We have inserted the summation symbol because j stands in the upper position twice and the summation convention would not apply.) Written out, this is the system of three equations

c¹= a¹¹b¹+ a¹²b²+ a¹³b³,

Here we have a matrix equation of the form

c = Ab (3.11)

where c and b are column vectors and A is a 3× 3 matrix. The analogy between the dot product and matrix multiplication is evident from (3.10) and (3.11). This analogy extends beyond the confines of Cartesian frames.

Let us write, for example,

A = a^kme_ke_m, b = b_je^j.

This time c = A· b gives

c^kek = a^kmekem· bje^j = a^kmekδ_m^j bj, hence

c^k= a^kjb_j. The corresponding matrix equation is, of course,

⎛

With suitable understanding we could still write this as (3.11). Note what happens when we express both the dyad and the vector in terms of covariant components:

A = akme^ke^m, b = bje^j. We obtain

c_k = a_kmg^mjb_j

and the metric tensor appears. The corresponding matrix form is

⎛

Because the metric tensor can raise an index on a vector component, we may also write these equations in the forms

ck= akmb^m

Let us examine the dot product between two dyads. There are various possibilities for the components of the dyad

C = A· B,

depending on how we choose to express A and B. If we use all contravariant components and write

A = a^kme_ke_m, B = b^kme_ke_m,

then

C = c^kne_ke_n where

c^kn= a^kmg_mjb^jn. Similarly, the use of all covariant components as in

A = akme^ke^m, B = bkme^ke^m, leads to

C = ckne^keⁿ where

ckn= akmg^mjbjn. Mixed components appear when we express

A = a^kmekem, B = bkme^ke^m. Then

C = A· B = a^kmekem· bjne^jeⁿ= a^kmδ^j_mbjnekeⁿ = a^kjbjnekeⁿ. Defining

c^k_{· n}= a^kjbjn

we have

C = c^k_{· n}ekeⁿ. We leave other possibilities to the reader as

Exercise 3.6. (a) Discuss how the formulation c_k^{· n} = akjb^jn arises.

(b) Show how all the forms above correspond to matrix multiplication.

(c) What happens if mixed components are used on the right-hand sides to express A and B?

Another useful operation that can be performed between tensors is dou-ble dot multiplication. If ab and cd are dyads, we define

ab·· cd = (b · c)(a · d).

That is, we first dot multiply the near standing vectors, then the remaining vectors, and thereby obtain a scalar as the result.

Exercise 3.7. (a) Calculate A· · E if A is a tensor of order two. How does this relate to the trace of the matrix that represents A in mixed components? (b) Let A and B be tensors of order two. Write down several different component forms for the quantity A·· B.

Yet another operation is the scalar product of two second-order tensors A and B, denoted by A• B. This represents a natural extension of the operation

ab• cd = (a · c)(b · d) between two dyads ab and cd.

3.5 Properties of Second-Order Tensors

Now we would like to consider in more detail those tensors that occur most frequently in applications: tensors of order two. First we recall that such a tensor is represented in dyadic form as

A = a^ijeiej.

Those who work in the applied sciences are probably more accustomed to the matrix representation

⎛

⎝a¹¹ a¹² a¹³ a²¹ a²² a²³ a³¹ a³² a³³

⎞

⎠ . (3.12)

When we use the matrix form (3.12) the dyadic basis of the tensor remains implicit. Of course when we use a unique, say Cartesian, frame for the space of vectors, then it does not matter whether we show the dyads. The correspondence between the dyadic and matrix representations suggests that we can introduce many familiar ideas from the theory of matrices.

The tensor transpose

Let us begin with the notion of transposition. For a matrix A = (a^ij) the transposed matrix A^T is

A^T = (a^ji).

Similarly we introduce the transpose operation for the tensor A:

A^T = a^jie_ie_j. (3.13)

This operation yields a new tensor, in each representation of which the corresponding indices appear in reverse order:

A^T = a^jie_ie_j= a_jieⁱe^j= a^j_·ieⁱe_j = a_j^·ie_ie^j.

A useful relation for any second-order tensor A and any vector x is

A· x = x · A^T. (3.14)

This follows when we write x = x^ke_k and use (3.13) to see that A^T = a^ijejei.

Equation (3.14) can be used to define the transpose. Also note that (A^T)^T = A

for any second-order tensor A.

Exercise 3.8. Let A and B be tensors of order two. Demonstrate that A• B = A ·· B^T = A^T ·· B.

Exercise 3.9. Let A be a second-order tensor. Find A·· A^T. Demonstrate that A·· A^T = 0 if and only if A = 0.

Exercise 3.10. (a) Show that if A and B are tensors of order two, then (A· B)^T = B^T · A^T.

(b) Let a and b be vectors and C be a tensor of order two. Show that a· C^T · b = b · C · a.

(c) Show that if A is a nonsingular tensor of order two, then the components of the tensor B = A⁻¹ are given by the formulas

b_i^·j= 1

2 det Aikl^jmna_m^{· k}a_n^{· l}. (d) Verify the following relations:

det A⁻¹= (det A)⁻¹, (A· B)⁻¹= B⁻¹· A⁻¹, (A^T)⁻¹= (A⁻¹)^T, (A⁻¹)⁻¹= A.

Tensors raised to powers

By analogy with matrix algebra we may raise a tensor to a positive integer power:

A²= A· A, A³= A· A², A⁴= A· A³,

and so on. Note that A^k still represents a linear operator. Negative integer powers are defined by raising A⁻¹ to positive integer powers:

A⁻²= A⁻¹· A⁻¹, A⁻³= A⁻²· A⁻¹, A⁻⁴ = A⁻³· A⁻¹, and so on. These operations can be used to construct functions of tensors using Taylor expansions of elementary functions. For example,

e^x= 1 + x 1!+x²

2! +x³ 3! +· · · . By this we can introduce the exponential of the tensor A:

e^A= E +A 1!+A²

2! +A³ 3! +· · · .

The issue of convergence of such series is approached in a manner similar to the absolute convergence of usual series, but with use of a norm of the tensor A (see § 3.12). Note that e^A represents a linear operator. We can introduce other functions similarly. This technique is used in the study of nonlinear elasticity, for example.

Symmetric and antisymmetric tensors

Among the class of all second-order tensors, an important role is played by the symmetric tensors. These include the strain and stress tensors of the theory of elasticity. The tensor of inertia is symmetric, as is the metric tensor. All these satisfy the relation

A = A^T. It follows from (3.14) that

A· x = x · A (3.15)

and

(A· x) · y = x · (A · y) (3.16) if A is symmetric. The reader will recall that the unit tensor E satisfies a relation of the form (3.15). A tensor A is said to be antisymmetric if

A =−A^T.

Exercise 3.11. Give the matrix forms corresponding to the cases of sym-metric and antisymsym-metric tensors. How many components can be indepen-dently specified for a symmetric tensor? For an antisymmetric tensor?

Both symmetric and antisymmetric tensors arise naturally in the phys-ical sciences. Their significance is also shown by the following

Theorem 3.1. Any second-order tensor can be decomposed as a sum of symmetric and antisymmetric tensors:

A = B + C where B = B^T and C =−C^T.

Proof. Take

B = 1 2

A + A^T

, C = 1

2

A− A^T ,

and check all the statements.

The dyad ab can be decomposed into symmetric and antisymmetric parts as

ab = 1

2(ab + ba) +1

2(ab− ba) for example.

Exercise 3.12. Show that if A is symmetric and B is antisymmetric then A·· B = 0.

Exercise 3.13. Demonstrate that the quadratic form x· A · x does not change if the second-order tensor A is replaced by its symmetric part.

Given an antisymmetric tensor C = c^iji_ii_jin a Cartesian frame, we can construct a vector

ω = ω^kik

according to the formulas

ω¹= c³², ω²= c¹³, ω³= c²¹. It is easy to verify directly that

C· x = ω × x, x· C = x × ω,

where x is an arbitrary vector. These formulas are written in non-coordinate form so they hold in any frame. The reader can derive the formulas forω, which is called the conjugate vector, for an arbitrary frame.

The cross-products of a tensor A = a^ijeiej and a vector x are defined by the formulas

A× x = a^ijei(ej× x), x× A = a^ij(x× ei)ej. Exercise 3.14. Show that C = E× ω = ω × E.

3.6 Eigenvalues and Eigenvectors of a Second-Order Sym-metric Tensor

We now consider the question of which basis yields a tensor of simplest form. As the analogous question in matrix theory relates to eigenvalues and eigenvectors, we extend these notions to tensors. The pair

(λ, x) (x= 0) is called an eigenpair if the equality

A· x = λx (3.17)

holds. Hence x is an eigenvector of A if A operates on x to give a vector proportional to x. Equation (3.17) may also be written in the form

(A− λE) · x = 0.

Exercise 3.15. Find the eigenpairs of the dyad ab. Now try to position an eigenvector on the left: x· ab = λx. You should find that this differs from the previous eigenvector, so it makes sense to introduce left and right eigenvectors. In the case of a symmetric tensor they coincide.

The eigenvalues of a second-order tensor A are found as solutions of the characteristic equation for A, which is derived as follows. In components (3.17) becomes

a^ijeiej· x^kek= λxⁱei

or

a^ijgjkx^kei= λx^kδ_kⁱei. Writing this as

(aⁱ_·k− λδkⁱ)x^k= 0,

we have a system of three simultaneous equations in the three variables x^k. A nontrivial solution exists if and only if the determinant of the coefficient matrix vanishes:

we have a system of three simultaneous equations in the three variables x^k. A nontrivial solution exists if and only if the determinant of the coefficient matrix vanishes:

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