Evaluación ética

Notice that the definition of a tensor does not mention components of the vectors. A tensor must be a rule [that] gives the same real number independently of the reference frame in which the vectors’ components are calculated.

Bernard Schutz, §3.2

3.1 Exercises

3.2 Prove that the set of all one-forms is a vector space.

Solution: While most instructors will be satisfied with a much more succinct answer, we give a detailed response to help clarify the concepts. One-forms were introduced in Schutz Eqs. (3.6a,b):

˜s = ˜p + ˜q,

˜r = α ˜p,



Schutz Eq. (3.6a) (3.1)

where ˜s and ˜r above must obey the following for all arguments A:

˜s( A)= ˜p( A)+ ˜q( A),

˜r( A)= α ˜p( A).



Schutz Eq. (3.6b) (3.2) To prove that the set of all one-forms is a vector space, we must show that this set meets axioms (1) and (2) given in Schutz Appendix A, p. 374. Axiom (1) states that

“[The set] V is an abelian group with operation + (A + B = B + A ∈ V ) and identity 0 (A+ 0 = A).” For a brief account of group theory see (e.g. Schutz,1980).

Very briefly, a group is a set of objects that can be combined with a binary operation, here “+” but it can be very general, to form new elements that are also in the set.

There are four axioms of group theory that must be met: (i) The closure property, A+ B ∈ V , is the one we just mentioned that the result of the binary operation (here “+”) between two elements results in an element of the set, (ii) Associativity property, A+ (B + C) = (A + B) + C, (iii) There must be an identity element, i.e.

an element that does not alter any elements of the set, A+ 0 = A for all A ∈ V , (iv) For every element A ∈ V there must be an inverse element B that combines to produce the identity element, A+ B = 0 for any A ∈ V . So applying this to a vector space, Axiom (1) tells us that the binary operation is vector addition and the identity 47

element is the zero vector 0. Some books will emphasize that 0 in this context is a vector by writing 0. We are asked to apply this to one-forms.

The sum of two one-forms must also be a one-form (so closure property (i) met) which is satisfied byeqn.(3.1),˜s = ˜p+ ˜q. We require that the order of summation does not matter (this is the “Abelian” property), which is satisfied byeqn.(3.2) because a one-form acting on a vector evaluates to a real and the sum of two reals does not depend upon the order. Similarly the property (ii) of associativity is also met. We also require a zero (there must be an identity element (iii)). The zero one-form gives zero for any vector. So say ˜q is the zero one-form. Then assumingeqn.(3.1) and by eqn.(3.2)

˜s( A)= ˜p( A)+ ˜q( A)= ˜p( A)+ 0 = ˜p( A), (3.3) so ˜p + 0 = ˜p and we have a zero. Finally setting α = −1 ineqns.(3.1) and (3.2) we see that for each ˜p we can always construct the inverse element ˜r = −1 · ˜p that sums to give the zero element, ˜p + ˜r = ˜p − 1 · ˜p = 0. So the set of one-forms with addition as defined ineqns.(3.1,3.2) satisfies Axiom (1).

Axiom (2) of Appendix A requires that multiplication of an element of the vector space by a real number gives another element of the vector space, with four requirements:

(i) a(A+ B) = aA + aB, (ii) (a+ b)A = aA + bA,

(iii) (ab)A= a(bA), (iv) 1(A)= A, (3.4)

with A, B elements of the vector space, a, b reals. Although it is not made explicit in eqns.(3.1) and (3.2), it was clear from the context that α ∈ R. Byeqns.(3.1and3.2) it is immediately clear that multiplication of a one-form by a real scalar meets all the requirements of Axiom 2. For instance Axiom 2(i) requires

α(˜p + ˜q) = α ˜p + α ˜q. (3.5)

On the LHS ofeqn. (3.5) we have a one-form, say ˜s = α( ˜p + ˜q), where ˜s is the one-form such that

˜s( V )= α · ( ˜p + ˜q)( V )= α ·

˜p( V )+ ˜q( V )



usedeqn.(3.2)

= α ˜p( V )+ α ˜q( V ). because ˜p( V )and ˜q( V )are reals (3.6) On the RHS ofeqn.(3.5) we have a one-form, say˜s = α ˜p + α ˜q, such that

˜s ( V )= (α ˜p + α ˜q)( V )= α ˜p( V )+ α ˜q( V ) usedeqn.(3.2)

= ˜s( V ). usedeqn.(3.6) (3.7)

So because ˜s and ˜s are one-forms that evaluate to the same real number when they operate on the same vector V, we conclude ˜s = ˜s andeqn. (3.5) holds, i.e.

multiplication by a scalar α is distributive over addition of one-forms. The other three properties in Axiom 2 follow similarly.

49 Exercises

3.4 Given the following vectors inO:

A→

O (2, 1, 1, 0), B→

O (1, 2, 0, 0), C →

O (0, 0, 1, 1), D →

O (−3, 2, 0, 0), (b) find components of ˜p if

˜p( A)= 1, ˜p( B)= −1, ˜p( C)= −1, ˜p( D)= 0.

Solution: Using the expression

˜p( A)= pαA^α Schutz Eq. (3.8) (3.8)

we can write a linear system in the four unknown components:

⎛

Note the matrix is written with the rows given by the vectors (so that one-forms appeared as columns) but this choice was arbitrary. This was solved in the accompanying Maple^TMworksheet, giving ˜p →_O (−2, −3, 15, −23)/8.

(d) determine whether the one-forms ˜p, ˜q, ˜r, ˜s are linearly independent if

˜q( A)= 0, ˜q( B)= 0, ˜q( C)= 1, ˜q( D)= −1,

˜r( A)= 2, ˜r( B)= 0, ˜r( C)= 0, ˜r( D)= 0,

˜s( A)= −1, ˜s( B)= −1, ˜s( C)= 0, ˜s( D)= 0. (3.10)

Solution: Given the values of the four one-forms, ˜p, ˜q, ˜r, ˜s applied to the four known vectors A, B, C, Dwe can, in principle, find all components of all four one-forms, repeating the procedure we did in Exercise 3.4(b). And then one could write a matrix M where the columns of M are taken from the one-form components. If the determinant of M is zero the one-forms are linearly dependent. But that is a lot of work.

There is a simpler way to test for linear dependence. If the one-forms are linearly dependent, then there are nontrivial real numbers a, b, c, d such that

a˜p + b ˜q + c˜r + d ˜s = ˜t = 0

But then

The latter can only be true if the determinant is zero, but



so the one-forms must be linearly independent.

3.5 Justify steps from Schutz Eq. (3.10a) to Eq. (3.10d), where

A^¯αp_¯α= (Λ^¯α_βA^β)(Λ^μ_¯αp_μ), Schutz Eq. (3.10a)

= (Λ^μ_¯αΛ^¯α_β) (A^βp_μ), Schutz Eq. (3.10b)

= δ^μ_βA^βp_μ, Schutz Eq. (3.10c)

= A^βp_β. Schutz Eq. (3.10d) (3.15)

Solution: We start with the respective Lorentz transformations of the vector and one-form, and then use the inverse property of the two Lorentz transformations. Summing over the dummy index results in the desired expression. In particular,

A^¯αp_¯α= (Λ^¯α_βA^β)(Λ^μ_¯αp_μ), by Schutz Eqs. (2.7) and (3.9) respectively

= (Λ^μ_¯αΛ^¯α_β) (A^βp_μ), just rearranged the terms

= δ^μ_β(A^βpμ), byeqn.(3.23)

= A^βp_β. sum over μ, use properties of the Kronecker delta

51 Exercises

3.6 Consider the basis{eα} of a frame O and a basis {˜λ⁰, ˜λ¹, ˜λ², ˜λ³} for the space of one-forms, with

˜λ⁰→

O (1, 1, 0, 0), ˜λ¹→

O (1,−1, 0, 0),

˜λ²→

O (0, 0, 1,−1), ˜λ³→

O (0, 0, 1, 1). (3.16) Note that{˜λ^β} is not the basis dual to {eα}.

(a) Show that ˜p  ˜p(eα)˜λ^αfor arbitrary ˜p.

Solution: Applying a one-form to a basis vector results in the corresponding compo-nent of the one-form,

pα = ˜p(eα), Schutz Eq. (3.7) (3.17)

but this component “belongs to” the basis one-form dual to the basis vector eα. So using these components with a different basis such as ˜λ^α leads to a different one-form. That is, ˜p = pα˜ω^α, when and only when ˜ω^α is dual toeα so thateqn.(3.91) applies. Indeed some texts use the same symbol for both the basis vectors and their corresponding dual basis one-forms, (e.g. Hobson et al., 2006, Eq. (3.2)), which emphasizes this correspondence but de-emphasizes the distinction between one-forms and vectors.

More formally, consider an arbitrary one-form ˜p and vector A.

˜p(eα)˜λ^α( A)= pα˜λ^α( A)= pα˜λ^α(A^βeβ)= pαA^β ˜λ^α(eβ)

= pαA^α iff ˜λ^α(eβ)= δ^αβ. (3.18) But it is clear that ˜λ^α(eβ) δ^α_βby inspection of the given basis. For example, δ⁰₁= 0 but

˜λ⁰(e1)=

1 1 0 0

⎛

⎜⎜

⎝ 0 1 0 0

⎞

⎟⎟

⎠ = 1 · 0 + 1 · 1 + 0 · 0 + 0 · 0 = 1. usedeqn.(3.8)

3.7 Prove that the basis one-forms transform under a change of basis as follows:

˜ω^¯α = Λ^¯α_β˜ω^β. Schutz Eq. (3.13) (3.19)

Solution: We were told just before Schutz Eq. (3.13) that its derivation is analogous to the corresponding relation for basis vectors, see derivation of Schutz Eq. (2.13) in

§2.2. Imagine that ˜p is an arbitrary one-form. Let ˜ω^αand ˜ω^¯αbe the sets of basis one-forms in framesO and O respectively. ˜p can be expressed in terms of either basis set

and we want this to be the same geometrical object,

˜p = pα˜ω^α = p_¯α˜ω^¯α.

Note the analogy with Schutz Eq. (2.12), which specifies that vectors are frame independent. It is important to realize that in general p0  p_¯0. Part of the subtlety arises here because the notation, while very standard, is actually misleading. In fact a few textbooks (e.g. Hobson et al.,2006; Weinberg, 1972) use a different symbol for the components of ˜p in the O frame, namely ¯pα. Here we will continue with the standard notation used by Schutz and most others

p_α˜ω^α = p¯α˜ω^¯α same geometrical object

= Λ^β_¯αp_β˜ω^¯α. usedeqn.(3.87) (3.20) At this point we may relabel dummy indices and replace β with α,

pα˜ω^α = Λ^α_¯αpα˜ω^¯α. relabeled dummy index (3.21) Because the equality above must hold for arbitrary ˜p it is clear that we require

˜ω^α = Λ^α_¯α ˜ω^¯α.

If you are uncomfortable with this last step, see SP3.5below.

3.10 (a) Given a frameO whose coordinates are {x^α}, show that

∂x^α/∂x^β = δ^α_β.

Solution: Let’s use an asterix to denote an arbitrary but fixed index, like α∗. When α∗  β∗, then x^α∗and x^β∗ are independent variables and of course ∂x^α∗/∂x^β∗ = 0. But when α∗ = β∗ then x^α∗ = x^β∗ are the very same variable and of course

∂x^α∗/∂x^α∗= 1. This completes the proof, which is valid in any coordinate system.

To be more concrete, consider pseudo-Cartesian coordinates wherein x⁰= t, x¹= x, x²= y, x³= z. When α  β we have terms like, say,

∂x⁰

∂x¹ = ∂t

∂x = 0,

because t and x are independent variables. But when α= β we have terms like, say,

∂x³

∂x³ =∂z

∂z = 1.

3.10 (b) For any two frames, we have:

∂x^β

∂x^¯α = Λ^β_¯α. Schutz Eq. (3.18) (3.22)

53 Exercises

Show that (a) and the chain rule imply

Λ^β_¯αΛ^¯α_μ= δ^βμ. (3.23) This is the inverse property again.

Solution: We start with the result from Exercise 3.10(a), and apply a Lorentz transformation as follows:

∂

∂x^μx^β = δ^βμ from Exercise 3.10(a)

∂

∂x^μ

 Λ^β_¯αx^¯α

= δ^βμ sub coordinate transform

Λ^β_¯α ∂x^¯α

∂x^μ = δ^β_μ transform is a constant Λ^β_¯αΛ^¯α_μ= δ^β_μ. from Schutz Eq. (3.18)

Eqn. (3.22) reveals that the Lorentz transformations are a type of coordinate transformation with the special property that they apply globally. Later in the text, when one studies GR, the idea of so-called general covariance will be of funda-mental importance and general coordinate transformations x^α(x^¯α)will be considered wherein ∂x^α/∂x^¯αis not constant throughout spacetime; however the inverse property eqn.(3.23) will still apply locally.

3.12 Let S be the two-dimensional plane x = 0 in three-dimensional Euclidean space. Let

˜n  0 be a normal one-form to S.

(a) Show that if V is a vector that is not tangent to S, then˜n( V ) 0.

Solution: Applying the general rule for the contraction of V and ˜n to this three-dimensional Euclidean space, we have

˜n( V )= nxV^x+ nyV^y+ nzV^z. usedeqn.(3.8) (3.24) A normal one-form must produce zero when contracted with any vector tangent to the surface. It follows that for the x= 0 surface in 3D space,

˜n →O (nx, 0, 0), (3.25)

where any nx  0 will give us a non-zero, normal one-form. So the contraction eqn.(3.24) reduces to

˜n( V )= nxV^x.

Because V is not tangent to surface S it must have a component in theex-direction, V^x 0. Thus the contraction is non-zero.

(c) Show that any normal to S is a multiple of ˜n.

Solution: Any normal to the surface S must have ny= 0 and nz= 0, as we argued in Exercise 3.12(a) in derivingeqn.(3.25). To be non-zero it requires nx 0. So any

˜n →O a(nx, 0, 0), (3.26)

where a 0 and a ∈ R will serve also as a non-zero, normal one-form.

3.13 Prove, by geometric or algebraic arguments, that the gradient of f , denoted by the tensor ˜df , is normal to surfaces of constant f .

Solution: Consider an arbitrary point p = (t, x, y, z) where f (t, x, y, z) = fp. Now imagine taking an infinitesimal step (t, x, y, z) such that the change in value of f ,

f = ∂f

∂tt+∂f

∂xx+∂f

∂yy+∂f

∂zz= 0. (3.27)

This ensures we do not leave the surface of constant f . So a tangent vector to the surface of constant f is obtained from an arbitrary multiple of such a step:

A→

O a(t, x, y, z), (3.28)

where a∈ R and a  0. The gradient one-form applied to such a tangent vector is

˜df ( A)= a

∂f

∂tt+∂f

∂xx+∂f

∂yy+∂f

∂zz



used Schutz Eq. (3.15)

= 0. usedeqn.(3.27) (3.29)

3.15 Supply the reasoning leading from

f= fαβ˜ω^αβ Schutz Eqs. (3.23) to

˜ω^αβ(eμ,eν)= δ^αμδ^β_ν. Schutz Eq. (3.24)

Solution: This exercise provides a good opportunity to deepen one’s understand of tensor algebra. We proceed step-by-step:

f= fαβ˜ω^αβ, what we mean by a basis fμν = f(eμ,eν), what we mean by components

f_μν = fαβ ˜ω^αβ(eμ,eν), sub first line into second line (3.30)

˜ω^αβ(eμ,eν)= δ^αμδ^β_ν. solving above (3.31)

55 Exercises

The last step deserves a few words of explanation. While fμν = fαβδ^α_μδ^β_ν is obviously a solution, how do we know this is the solution? Here is one way to convince yourself:

f_μν = fαβA^αβ_μν, re-writeeqn.(3.30) with A^αβ_μνas unknown

f_{σ γ}δ^σ_μδ^γ_ν = fαβA^αβ_μν, re-write LHS

f_αβδ^α_μδ^β_ν = fαβA^αβ_μν, relabel dummy indices, see SP3.5 0= fαβ

δ^α_μδ^β_ν− A^αβμν

. rearrange

But fαβ is arbitrary so the only way this last line above can always be true is for δ^α_μδ^β_ν = A^αβμν.

3.17 (a) Suppose that h is a₀

2

tensor with the property that, for any two vectors Aand B(where B 0),

h( , A)= α h( , B),

where α is a number that may depend on Aand B. Show that there exist one-forms ˜p and ˜q such that

h= ˜p ⊗ ˜q.

Solution: In general,

h( C, A)= hγβ C^γA^β.

Treat C as an arbitrary vector. We were given that for arbitrary vectors Aand B, but with B 0

h( , A)= αh( , B). (3.32)

Suppose C^γ →

O (1, 0, 0, 0). Then

h( C, A)= αh( C, B)

h_0βA^β = αh0μB^μ (3.33)

˜q( A)= α ˜q( B). (3.34)

The LHS and RHS ofeqn.(3.33) have the form of a one-form contracted with vectors Aand Brespectively, so we wrote that explicitly ineqn.(3.34), defining qβ ≡ h0β. So far there is no restriction on ˜q; we simply choose

α= ˜q( A)

˜q( B), (3.35)

and we note the stipulation that B 0.

Now suppose C^γ →

O (1, 1, 0, 0). Then h( C, A)= αh( C, B)

h_0βA^β + h1βA^β = α(h0μB^μ+ h1μB^μ) (3.36) h_1βA^β = α(h1μB^μ). subtractedeqn.(3.33) (3.37) Now α is no longer a free variable, being set byeqn.(3.33) or equivalentlyeqn.(3.35).

Botheqns.(3.33) and (3.37) can be satisfied if

a h_0β = h1β (3.38)

for an arbitrary a since theneqn.(3.37) is simplyeqn.(3.33) multiplied on both sides by a. And this must be the unique solution as is obvious from considering the case where the vectors have only a single component, e.g. A →

O (0, 1, 0, 0) and B → (1, 0, 0, 0). For theneqns.(3.33) and (3.37) reduce to simple algebraic equations, e.g.O

h₀₁ = αh00

h11 = αh10

⇒h₀₁ h11 = h₀₀

h10 = a⁻¹. (3.39)

Repeating the argument above for different C, we find a different constant for each first index of h. In particular, with C^γ →

O (1, 0, 1, 0) we conclude that bh0β = h2β, for some constant b. With C^γ →

O (1, 0, 0, 1) we find ch0β= h3β. That is, if the tensor h is written as a matrix, the rows are arbitrary scalar constants p_μ= (1, a, b, c) of the first row (h0ν)≡ qν. And so

h_μβ= pμq_β (3.40)

and so the tensor h has the form of an outer product,

h= ˜p ⊗ ˜q. (3.41)

tensor provides a map of vectors to vectors and one-forms to one-forms.

Solution: You can think of a₁

1

tensor as a machine with two input slots, a slot for one-forms and another for vectors. Inserting a one-form and a vector into their respective slots produces a real number. So once you’ve filled the one-form slot, this machine takes one more input, a vector, to produce a real. But a machine that takes a vector as input to produce a real is, by definition, a one-form. Now let’s go back to the₁

1

 tensor, the machine with two empty slots. We’ve just seen that the act of inserting the one-form converts the machine to one-form; so we can say a₁

1

tensor is a map from one-forms to one-forms. By a similar argument, a₁

1

tensor is also a map from vectors to vectors.

57 Exercises

It’s also satisfying to see how this plays out using the mathematical symbols. Write T acting on the vector alone:

T( ;v) = T( ; v^βeβ) usedeqn.(2.4)

= T^νμv^β eν⊗ ˜ω^μ( ;eβ) used Schutz Eq. (3.61)

= T^νμv^β ˜ω^μ(eβ)eν applied basis to argument

= T^ν_μv^βδ^μ_β eν usedeqn.(3.91)

= T^ν_μv^μ

eν. summed over β (3.42)

And now we are done because the RHS of eqn.(3.42) is the product of four reals

T^ν_μv^μ

, i.e. one for each value of ν, times the four basis vectors,eν. This is a vector, cf.eqn.(2.4).

And similarly we can write T acting on the one-form alone:

T(˜p; ) = T(pα˜ω^α; ) used Schutz Eq. (3.11)

= T^ν_μp_αeν⊗ ˜ω^μ(˜ω^α; ) used Schutz Eq. (3.61)

= T^νμp_α ˜ω^μδ^α_ν usedeqn.(3.91)

= T^ν_μpν

˜ω^μ. summed over α (3.43)

Again we are done here because the RHS is clearly a one-form, i.e. it is the product of four reals

T^ν_μp_ν

, one for each value of μ, and the set of four one-form bases ˜ω^μ. So a₁

1

tensor is a map from vectors to vectors in the sense that when you contract it with a vector the result is a vector. Likewise a₁

1

tensor is a map from one-forms to one-forms because contracting it with a one-form results in a one-form.

3.19 (b) Derive the equation for the dot product of two one-forms:

˜p · ˜q = −p0q0+ p1q1+ p2q2+ p3q3. Schutz Eq. (3.53) (3.44)

Solution: To derive the formula for the inner product of one-forms in terms of components,eqn.(3.44), we start with the definition:

˜p · ˜q = 1

2[( ˜p + ˜q)²− ˜p²− ˜q²]. Schutz Eq. (3.52) (3.45) This involves only addition and squares of one-forms. The square of a one-form has been defined in component form ineqn.(3.83). The addition of one-forms was defined abstractly ineqns.(3.1,3.2), and in SP3.3we show that this implies that we just add the components; if˜s = ˜p + ˜q then sα = pα+ qα. So the square of the sum will be:

(˜p + ˜q)²= ˜s²= η^αβsαsβ, byeqn.(3.83)

= η^αβ(p_α+ qα)(p_β+ qβ), component-wise addition

= η^αβ(pαpβ+ qαqβ+ pαqβ+ pβqα). components are just reals (3.46)

Substitutingeqn.(3.46) intoeqn.(3.45) and using{η^αβ} from Schutz Eq. (3.44):

˜p · ˜q = 1

2[( ˜p + ˜q)²− ˜p²− ˜q²]

= 1

2[η^αβ(p_αp_β+ qαq_β+ pαq_β+ pβq_α)− η^αβp_αp_β− η^αβq_αq_β],

= 1

2[η^αβ(p_αq_β+ pβq_α)] cancelled terms

= η^αβp_αq_β η^αβsymmetric

= −p0q₀+ p1q₁+ p2q₂+ p3q₃, (3.47)

which iseqn.(3.44).

3.20 In Euclidean three-space in Cartesian coordinates, we do not normally distinguish between vectors and one-forms, because their components transform identically.

Prove this in two steps. Suppose we are in Euclidean three-space in Cartesian coordinates.

(a) Show that

A^¯α = Λ^¯α_βA^β and

P_¯β= Λ^α_¯βP_α (3.48)

are the same transformation if the matrix (Λ^¯α_β)is equal to the transpose of its inverse. Such a matrix is said to be orthogonal.

Hint: Solution available in (Schutz,1985, Appendix B).

(b) The metric of such a space has components{δij, i, j = 1, . . . , 3}. Prove that a transformation from one Cartesian coordinate system to another must obey

δ_{¯i ¯j}= Λ^k_¯iΛ^l_¯jδ_kl (3.49)

and that this implies (Λ^k_¯i)is an orthogonal matrix. See Exercise 3.32 for the analog of this in SR.

Solution: All we are given is that the metric tensor for Cartesian three-space is {δij, i, j = 1, 2, 3}. The metric tensor is used in forming the inner product of vectors, which we know must be frame invariant. So write the inner product between two three-space vectors in two different frames as

59 Exercises

δ_{¯i ¯j}A^¯iB^¯j= A· B= δklA^kB^l

= δkl(A^¯iΛ^k_¯i)(B^¯jΛ^l_¯j), (3.50) and so upon “cancelling” the A^¯iB^¯jon either side (allowed because these are arbitrary vectors) and rearranging we see that

δ_{¯i ¯j}= Λ^k_¯iΛ^l_¯jδkl, (3.51)

as required. If you were uncomfortable with cancelling the A^¯iB^¯jon either side, see SP3.5.

Now to show that this implies an orthogonal transformation matrix, sum over l:

δ_{¯i ¯j}= Λ^k_¯iΛ^l_¯jδ_kl, fromeqn.(3.51)

= Λ^k_¯iΛ^k_¯j. after summing over l (3.52) The RHS of eqn.(3.52) is the product of a matrix by its transpose, and for this to equal the identity matrix (i.e. the LHS), we require the matrix to be orthogonal.

And now you see clearly why we never learned about one-forms when working in Cartesians coordinates in Euclidean space, and why we called the gradient of a scalar field a vector.

3.23 (a) Prove that the set of all_M

N

tensors for fixed M and N forms a vector space.

(You must define addition of such tensors and their multiplication by numbers.) Solution: This is like Exercise 3.2, but now we need to define what we mean by the addition of two_M

N

tensors and the multiplication of an_M

N

tensor by a scalar. So we are guided byeqns.(3.5) and (3.6) above. That is, we note that_M

N

tensors produce real numbers that can be added like real numbers, so the generalization ofeqns.(3.1) and (3.2) is trivial. The tensor S where

S= P + Q (3.53)

is defined to be that which gives the sum of the two values obtained by applying the input to P and Q. That is,

S(˜a¹,˜a², . . . ,˜a^M; b₁, b₂, . . . , b_N)=

P(˜a¹,˜a², . . . ,˜a^M; b1, b2, . . . , bN)+ Q(˜a¹,˜a², . . . ,˜a^M; b1, b2, . . . , bN), (3.54) where the notation (˜a¹,˜a², . . . ,˜a^M; b1, b2, . . . , bN)was used to represent the M one-form inputs and N vector inputs. The choice of one-one-forms first, we will see later, gives the basis in the order Schutz gave in Exercise 3.23(b). We have followed the convention that superscript integers are used as indices of different one-forms. That is, ˜a¹ and ˜a² are two different one-forms, not components of the same one-form.

Similarly, subscripts are used to denote different vectors. Some authors will use a different notation to avoid this ambiguity; Carroll (2004) uses parentheses around the indices, like˜a⁽¹⁾, to distinguish them from components.

In analogy with eqn. (3.2) we can define multiplication of an _M

N

 tensor by a scalar α

R= αP (3.55)

to be the tensor that, for a given input, gives just α times the real number produced by supplying the input to P:

R(˜a¹,˜a², . . . ,˜a^M; b1, b2, . . . , b_N)= αP(˜a¹,˜a², . . . ,˜a^M; b1, b2, . . . , b_N). (3.56) The set of _M

N

 tensors for fixed M and N forms a vector space by the same argument as given for Exercise 3.2. What do we mean by the zero_M

N

tensor? This is the tensor that gives zero for any input,

0(˜a¹,˜a², . . . ,˜a^M; b₁, b₂, . . . , b_N)= 0.

The set of_M

N

tensors, with addition defined byeqns. (3.53) and (3.54) and scalar multiplication defined byeqns.(3.55) and (3.56) then meets axiom (1) in Appendix A:

_M

N

 tensors form an abelian group with the operation of addition. Similarly the requirements of Axiom (2) in Appendix A are clearly met.

3.23 (b) Prove that a basis for the vector space formed from the set of all_M

N

tensors for fixed M and N is the set:

{eα⊗ eβ⊗ · · · ⊗ eγ ⊗ ˜ω^μ⊗ ˜ω^ν⊗ · · · ⊗ ˜ω^λ} (3.57) with M vectors labeled with α . . . γ and N one-forms labeled μ . . . λ.

Solution: This is a nice question because it forces us to think about what we mean by a basis. The answer is a straightforward generalization of the argument for the basis of the₀

2

tensors starting after Schutz Eq. (3.22) and ending with Eq. (3.26) of §3.4.

The notation is cumbersome because one needs to refer to M superscripts and N subscripts where M and N are arbitrary positive integers. In defining the basis eqn. (3.57) above Schutz has used a series of Greek letters like α . . . γ . Here we put subscript indices on the Greek letters α1, α2, . . . αM to be explicit about how many there are. Remember that each Greek letter index can take on four values, e.g.

α₁∈ {0, 1, 2, 3} corresponding to the four dimensions.

As in Schutz Eq. (3.23) we write the_M

N

tensor as a sum of components times the basis that we seek:

R= R^{α1,α2,...,αM}_{β1,β2,...,βN} ˜ω_{α1,α2,...,αM}^{β1,β2,...,βN}. (3.58) And furthermore, the components correspond to the real values produced by applying the tensor to arguments that are the basis one-forms and basis vectors. So,

61 Exercises

R^{α1,α2,...,αM}_β

1,β₂,...,β_N = R( ˜ω^α1,˜ω^α2, . . . ,˜ω^αM;eβ1,eβ2, . . . ,eβN), (3.59) which is the generalization of the formula given between Schutz Eq. (3.23) and Eq. (3.24). Now, we simply substitute the tensoreqn.(3.58) intoeqn.(3.59) to obtain:

R^μ1,...,μM_ν

1,...,ν_N =R^α1,...,αM _β1,...,βN˜ωα₁,...,α_M ^β1,...,βN(˜ω^μ1, . . . ,˜ω^μM;eν₁, . . . ,eν_N).

(3.60) This implies the analogue to Schutz Eq. (3.24),

˜ω_α1,...,αM^β1,...,βN(˜ω^μ1, . . . ,˜ω^μM;eν1, . . . ,eνN)= δ_α^μ₁¹. . . δ_α^μM

M δ^β_ν¹

1. . . δ^β_ν^N

N. (3.61) Usingeqn.(3.91) we identify

δ^β_ν¹₁ = ˜ω^β1(eν₁), δ^β_ν²₂ = ˜ω^β2(eν₂), . . . δ^β_ν^N

N = ˜ω^βN(eν_N). (3.62) Based upon the dualism between vectors and one-forms, we identify:

δ_α^μ1

1 = eα1(˜ω^μ1), δ_α^μ2

2 = eα2(˜ω^μ2), . . . δ_α^μM

M = eαM(˜ω^μM).

So focusing on just the tensor, i.e. dropping the arguments, we are left with the basis that is the analogue to Schutz Eq. (3.25),

So focusing on just the tensor, i.e. dropping the arguments, we are left with the basis that is the analogue to Schutz Eq. (3.25),

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