Noncommutative differential geometry with

Noncommutative differential geometry with application to physics

By Roa Mohammed Makki A thesis submitted to Swansea University in fulfilment of requirement for the Degree of Doctor of Philosophy of Science Department of Mathematics Swansea University Wales, United Kingdom. January, 2015

Abstract The object of this thesis is to study noncommutative differential geometry working on the κ-Minkowski space. A central metric is introduced to this calculus allowing the functions to commute with the Minkowski metric to first order in the deformation parameter λ, and this condition forces a unique deformed calculus. But this property is not associative to O(λ2 ) and we use an explicit associator to fix this and find an associated star structure. However, this associator does not satisfy the pentagon identity to O(λ4 ). The Schrödinger equation comes with an O(λ) correction, which can be interpreted as a critical mass. We could relate it to the Planck mass and find a numerical value of λ. The noncommutative massless Klein-Gordon equation has a possible correction to its propagation speed. Numerical evidence is given that this effect is real, rather than just a phase velocity. For the given numerical value of λ, this effect may be measurable. In addition, electromagnetism is studied using this calculus, and at the Dirac equation We also show that theory of bar categories allows a star operation on the finite difference calculus on the lattice, which may have applications to numerical analysis.

1

Declaration This work has not previously been accepted in substance for any degree and is not being concurrently submitted in candidature for any degree. Signed ...................................................................... (candidate) Date ........................................................................ STATEMENT 1 This thesis is the result of my own investigations, except where otherwise stated. Where correction services have been used, the extent and nature of the correction is clearly marked in a footnote(s). Other sources are acknowledged by footnotes giving explicit references. A bibliography is appended. Signed ..................................................................... (candidate) Date ........................................................................ STATEMENT 2 I hereby give consent for my thesis, if accepted, to be available for photocopying and for inter-library loan, and for the title and summary to be made available to outside organisations. Signed ..................................................................... (candidate) Date ........................................................................ 2

Contents Abstract

1

Declaration

2

Dedication

7

Acknowledgement

8

1 Introduction

9

2 Preliminaries on Algebra and Geometry

12

2.1

Riemannian Geometry . . . . . . . . . . . . . . . . . . . . . . 12

2.2

Star Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.3

Noncommutative differential calculi . . . . . . . . . . . . . . . 16

2.4

Modules and covariant derivatives . . . . . . . . . . . . . . . . 18

2.5

Monoidal Categories . . . . . . . . . . . . . . . . . . . . . . . 21

2.6

Bar categories . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.7

The Hopf algebra q-deformed SL2 . . . . . . . . . . . . . . . . 26

3 Preliminaries on Physics 3.1

29

Background to special relativity . . . . . . . . . . . . . . . . . 29 3.1.1

Minkowski space . . . . . . . . . . . . . . . . . . . . . 30

3

3.1.2 3.2

3.3

Lorentz transformation . . . . . . . . . . . . . . . . . . 30

Quantum theory . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.2.1

The uncertainty principle

. . . . . . . . . . . . . . . . 31

3.2.2

Schrödinger equation . . . . . . . . . . . . . . . . . . . 32

3.2.3

Klein-Gordon equation . . . . . . . . . . . . . . . . . . 33

3.2.4

Spinors and the Dirac equation . . . . . . . . . . . . . 35

Gauge theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4 The Majid-Ruegg space time

39

4.1

The κ-Poincaré algebra . . . . . . . . . . . . . . . . . . . . . . 39

4.2

κ-Minkowski space . . . . . . . . . . . . . . . . . . . . . . . . 41

4.3

A new differential calculus . . . . . . . . . . . . . . . . . . . . 42 4.3.1

A central metric . . . . . . . . . . . . . . . . . . . . . . 42

4.3.2

Constructing the calculus . . . . . . . . . . . . . . . . 45

4.3.3

Nonassociativity at O(λ2 ) . . . . . . . . . . . . . . . . 49

4.4

Conjugate Modules . . . . . . . . . . . . . . . . . . . . . . . . 57

4.5

Levi-Civita and covariant derivative . . . . . . . . . . . . . . . 62 4.5.1

Differentiation and Normal Orders . . . . . . . . . . . 64

4.5.2

Motion Along a Constant Vector Field . . . . . . . . . 68

5 The Schr¨ odinger equation and propagation speeds

70

5.1

The Schrödinger equation and effective mass . . . . . . . . . . 70

5.2

Plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.3

Verifying the change in propagation speed for the massless KG equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6 Electromagnetism

88

6.1

Curvature at O(λ) . . . . . . . . . . . . . . . . . . . . . . . . 88

6.2

Finding the current in the noncommutative case . . . . . . . . 90 4

6.3

Electromagnetic plane waves . . . . . . . . . . . . . . . . . . . 96

7 Monopole solutions of the Dirac equation

7.1

7.0.1

The Lorentz group . . . . . . . . . . . . . . . . . . . . 98

7.0.2

SL2 (C) acting on spinors . . . . . . . . . . . . . . . . . 101

7.0.3

Solving Dirac equation for 4-spinors . . . . . . . . . . . 106

7.0.4

Solving Dirac equation for 8-spinors . . . . . . . . . . . 110

Trying to find a noncommutative monopole . . . . . . . . . . . 112 7.1.1

Simplifying the Dirac equation . . . . . . . . . . . . . . 112

7.1.2

Solving the zero current equations. . . . . . . . . . . . 120

7.1.3

Solving the Dirac equation with zero current . . . . . . 123

8 Noncommutative geometry on a Cubic Lattice 8.1

8.2

8.3

98

127

Numerical Partial Derivatives and Non-commutative Geometry 128 8.1.1

Finite difference calculus on R . . . . . . . . . . . . . . 128

8.1.2

The problem with star operation for a lattice . . . . . . 132

A Bar Category . . . . . . . . . . . . . . . . . . . . . . . . . . 133 8.2.1

A star calculus in a bar category . . . . . . . . . . . . 133

8.2.2

A category of graded modules . . . . . . . . . . . . . . 133

8.2.3

The monoidal category C . . . . . . . . . . . . . . . . . 135

A Conjugate on C . . . . . . . . . . . . . . . . . . . . . . . . . 136

9 Conclusion and possible further work

143

Bibliography

145

Appendix A Solution to 4-spinor Dirac equation

153

Appendix B Finding a noncommutative monopole I

162

5

Appendix C Finding a noncommutative monopole II

6

191

Dedication This work is dedicated with love and affection, to my mother Jalilah and the soul of my father Mohammed. To her, who carried in her heart the honest passion and raised her palms in pray for me with sincerity. The one who devoted her life giving generously not asking for anything to rise and educate me, hoping I reach my goals. To him, the price worthy teacher, who taught me to challenge in the destiny of learning. Whom I consider the symbol of love, passion and warmth. The one who used to say ” every daughter is her fathers admirer ”, I hope you are proud today.

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Acknowledgement First of all I thank Almighty Allah for guiding me through my life. I praise Him, I adore Him. I wish to acknowledge the effort and guidance of my diligent supervisor Dr Edwin Beggs for his total dedication, kindness, time and patience in guiding me through this thesis. No amount of words is enough to appreciate your help. Thank you so much for your countless efforts to make this work a success. I also wish to thank Dr Jeff Giansiracusa . I acknowledge and appreciate the support of King Abdullah bin Abdulaziz al-Saud and King Salman bin Abdulaziz al-Saud throughout my scholarship. I also thank Um Al-Qura University in Makkah and the Saudi Cultural Bureau in United Kingdom and Ireland for their financial support. I can not express how much I am grateful to my beloving husband Mustafa Makki and my adorable daughter Seba for their moral support and endless love. And never the least, to my brother and my sisters, Suhair, Sahar, Suzan, Afaf, Mohannad and Maram, I am truly thankful . Thank you family and friends for all the prayers and good wishes.

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Chapter 1 Introduction The noncommutativity of coordinates was introduced in quantum theory, but not really viewed as ”geometry”. C ∗ -algebras were introduced by Von Neumann [59], and these became an idea of functions on a ”noncommutative topological space”. Noncommutative differential geometry goes back to the work of Connes on Dirac operators [24] and Woronowicz on differential calculi on quantum groups [70]. It is now sometimes supposed that noncommutative differential geometry is the first approximation to quantum gravity. Some effect has been made to look for noncommutative effect in the large scale structure of space time, for example variations in the speed of light [3][4][5]. In the thesis I will continue in this project using κ-Minkowski or the Majid-Ruegg model. There is a 4D differential calculus [60], and a 5D calculus [64] which is related to the coaction of the κ-Poincaré algebra. However, we use a different calculus: In chapter 4, it is imposed that the metric is central and the functions commute with the Minkowski metric to first order in the deformation parameter, which gives a unique deformed calculus, to first order. This is an extremely useful property but it is quite easy to see that 9

it is not associative at O(λ2 ). This can be fixed by introducing an explicit associator, but in turn that associator does not satisfy the pentagon identity at O(λ4 ). The choice of the associator is not unique. In the presence of a non-trivial associator, we use the idea of a bar category to define a conjugate structure. We consider differential operators using the calculus we constructed, and look at the effect on the Schrödinger equation and propagation speeds in chapter 5. Our calculations indicate a factor of

1 2

difference in the time delay

from that calculated in [5]. In addition we look at electromagnetism using this calculus, and at the Dirac equation. There is a classical monopole to the Dirac equation, which is well studied from several points of view. We carry out on explicit construction using two hemispheres and a gauge equivalence, in such a manner as we might hope to extend it to the noncommutative case. We use a double spinor field construction to ensure that we also have a solution to the current equation. We attempt to deform this to first order in λ- however we do not succeed in solving this, and it remains an open problem. In chapter 8 we consider the very different problem where the calculus is well known, but there was a problem with the star structure. The differential calculus on the group Z2 gives a finite difference calculus, which has been proposed as a model for exact numerical analysis [36][37][46]. However there is no standard way to obtain a star calculus. We show that there is a star differential calculus, using the ideas of bar categories. Both the non-standard κ-Minkowski calculus and the deformed conjugate structure for the lattice calculus use non-trivial category theory, and associators and bar categories. Note that some of this work has been written up in the form of a joint

10

paper and placed in the archive as arXiv:1306.4518v2 [14]

11

Chapter 2 Preliminaries on Algebra and Geometry 2.1

Riemannian Geometry

We use [20] as our main reference on Riemannian geometry, quoting in places. An inner product on a vector space, hi : V ⊗ V → R , Obeys hv, wi = hw, vi , hv, w + ui = hv, wi + hv, ui , hv, λwi = λhv, wi , and is given in terms of a basis e1 , .., en of V by hei , ej i = gij , where (gij ) is a symmetric matrix. 12

λ∈R.

In differential geometry for a manifold M , a Riemannian metric is an inner product (which is assumed non-degenerate) on the tangent or cotangent bundle. We set, in terms of coordinates x1 , ..., xn , hdxi , dxj i = g ij , where (g ij ) is a symmetric invertible matrix of elements of C ∞ (M ), whose inverse is (gij ). We can write the metric in terms of a square of distance ds2 =

X

gij dxi dxj .

Although Riemannian geometry is usually given with a positive definite metric, e.g. ds2 = dx2 + dy 2 + dz 2 , many results are valid on indefinite metric, like Minkowski space, ds2 = dt2 − dx2 − dy 2 − dz 2 . A connection on the tangent bundle of a manifold M is a linear map: ∇ : Vect(M) ⊗R Vect(M) → Vect(M) . where we write ∇(v i

∂ ⊗ w) = v i ∇i w , and ∇i satisfies the Leibniz rule, for ∂xi

a function ∇i (f w) =

∂f w + f ∇i (w) . ∂xi

We often use the notation ∇i (v j

∂ ∂ ) = v j ;i j j ∂x ∂x

We say that ∇ preserves the metric if ∂ j k (v w gjk ) = (v j ;i wk + v j wk ;i )gjk . i ∂x 13

(2.1)

Define the Christoffel symbols by ∇i (

∂ ∂ ) = Γkij k , j ∂x ∂x

and the connection ∇ is torsion free if Γkji = −Γkij . Then the usual Levi-Civita connection is defined by the following: Theorem 2.1.1. For a Riemannian manifold, there is a unique torsion free connection preserving the metric in (2.1). This is called the Levi-Civita connection. The formula for the Christoffel symbol is 1 ∂gsi ∂gij ∂gjs Γkij = g ks − + . 2 ∂xj ∂xs ∂xi As we are mostly concerned with forms, we note that the Levi-Civita derivative of a form is ∇∂i dxj = −Γjik dxk . In more generality, a vector bundle associates a vector space V to every point of the manifold. This is done in a locally trivial manner, i.e., there is a cover by open sets Ui so that for every Ui the vector bundle is Ui × V . The vector space operations are defined over every point of M . The tangent and cotangent bundle are examples of vector bundles. A section of a vector bundle is a map s from the manifold M to the vector bundle, so that every s(m) is in the vector space at the point m ∈ M . Locally, given the open sets Ui above, we can write s : Ui −→ Ui × V given by s(m) = (m, t(m)) for t : Ui −→ V . We can add sections by (t0 + t)(m) = t0 (m) + t(m) , and multiply by λ in the field (usually R or C) by (λs)(m) = λs(m) . 14

Given a function f on M taking values in the field, we can multiply a section by f , giving (f.s)(m) = f (m)s(m) . This makes the sections of the bundle into a module over the functions of M.

2.2

Star Algebras

The theory of C ∗ -algebras goes back to the work of Murray and Von Neumann [59], and was extended by many others. We shall mostly be concerned by the algebraic theory of star algebras rather than C ∗ -algebras, but it is useful to set down the definition. Definition 2.2.1. (Normed vector space, quoted from [20]) A norm in a vector space V over C (or R), is a positive real valued function kak such that kak ≥ 0 unless a = 0, kαak = |α|.kak and kabk ≤ kakkbk for a, b ∈ V, α ∈ C. We quote the definition of a ?-algebra from [50]. For the field of complex numbers C, the notion of a ?-algebra means an algebra A equipped with antilinear map ( )∗ : A → A obeying ?2 = id and for all a, b ∈ A and λ ∈ C, (a + b)∗ = a∗ + b∗

,

(ab)∗ = b∗ a∗

(λa)∗ = λa∗

,

(a∗ )∗ = a .

A C ∗ -algebra is a complete normed algebra A with the star operation such that a 7→ a∗ for a in the algebra with the star map having the following properties [59] kabk ≤ kakkbk , ka∗ ak = kakka∗ k = kaa∗ k , 15

kak = ka∗ k . An example of a C ∗ -algebra is Mn (C) which acts on the Hilbert space Cn by matrix multiplication and uses the operator norm k·k. Another example is the continuous complex valued functions on a compact Hausdorff topological space, with the supremum norm. Example 2.2.1. (Noncommutative Torus)[34] In the theory of C ∗ -algebra, for any irrational number θ, the noncommutative torus Tθ2 is a noncommutative C ∗ -algebra generated by two unitary elements U1 and U2 with the relation U2 U1 = e2 π i θ U1 U2 . The noncommutative torus with irrational θ is also known as irrational rotation algebra.

2.3

Noncommutative differential calculi

To do differential geometry, we either need vector fields or differential forms [21]. The definitions of noncommutative differential geometry are usually given in terms of forms, as these make up an algebra with the wedge product. Definition 2.3.1. A differential structure on an algebra A is a differential graded algebra (Ωn A, ∧, d) for n ≥ 0. This means, 1) ∧ : Ωn A ⊗ Ωm A → Ωn+m A an associative multiplication. 2) d : Ωn A → Ωn+1 A obeys d2 = 0. 3) d(τ ∧ ω) = dτ ∧ ω + (−1)|τ | τ ∧ dω where τ ∈ Ω|τ | A 4) Ω0 A = A and ∧ is the usual multiplication on A.

16

It is usual to add for a calculus that A generates the calculus, i.e., that Ω1 A = A.dA and that Ωn+1 A = Ωn A ∧ Ω1 A for all n ≥ 1, but this is not part of the standard definition of a D.G.A. Example 2.3.1. [25] A differential structure on the noncommutative torus might be e1 = U1−1 dU1

,

e2 = U2−1 dU2 ,

where the ei are central, i.e. a.ei = ei .a for all a ∈ Tθ2 , and e1 ∧ e1 = e2 ∧ e2 = 0 ,

e1 ∧ e2 = −e2 ∧ e1 = 0 .

Form this we can calculate d(e1 ). First note d(1) = d(U1 U1−1 ) = 0 , so dU1 .U1−1 + U1 d(U1−1 ) = 0 , so d(U1−1 ) = −U1−1 .dU1 .U1−1 . Now d(e1 ) = d(U1−1 ) ∧ dU1 + U1−1 .d2 U1 = −U1−1 dU1 ∧ U1−1 dU1 = −e1 ∧ e1 = 0 . The 1-forms Ω1 A consist of sums of elements of the form a.db where a, b ∈ A. There is a star operation on Ω1 A given by (a.db)∗ = db∗ .a∗ , and this gives a bimodule map ? : Ω1 A → Ω1 A by ?(η) = η ∗ . A 1-form ξ is Hermitian if ξ ∗ = ξ. The star operation extends to 2-forms Ω2 A by (dξ)∗ = d(ξ ∗ ) and (ξ ∧ η)∗ = −η ∗ ∧ ξ ∗ for ξ, η ∈ Ω1 A. 17

2.4

Modules and covariant derivatives

The connection between vector bundles and the algebra of functions on a space is well known, but we use [31][32][58] as references for covariant derivatives on modules. Let A be the algebra of functions on a topological space, then the sections of any vector bundle form a module. For example, the collection of wind speeds on earth at a given time forms a section of the tangent bundle to the sphere of the earth, (i.e., a vector field). We can multiply such vector fields v(x) by a function f (x) by using pointwise multiplication, i.e., f (x) v(x) for x ∈ S 2 . We also have pointwise addition v(x) + w(x) for two vector fields. In the noncommutative setting, we consider modules in place of bundles. Definition 2.4.1. (Left Module) Let A be an algebra, a vector space E is a left A-module if and only if there exists function λ : A × E → E such that for all a1 , a2 ∈ A and e1 , e2 ∈ E, 1) λ(a1 + a2 , e1 ) = λ(a1 , e1 ) + λ(a2 , e1 ) , λ(a1 , e1 + e2 ) = λ(a1 , e1 ) + λ(a1 , e2 ), 2) λ(a1 a2 , e1 ) = λ(a1 , λ(a2 , e1 )) , 3) λ(1A , e1 ) = e1 , 4) For z in the field, λ(z a1 , e1 ) = z λ(a1 , e1 ) = λ(a1 , z e1 ) . Example 2.4.1. C2 is a  a  c

left M2 (C) module with left action,     b x ax + by   =   . d y cx + dy

We can equivalently define a right module using µ : E × A → E. Definition 2.4.2. (Bimodule) If A and B are two algebras, then an A − Bbimodule is a vector space E such that 1) E is a left A-module and a right B-module. , 18

2) For all a ∈ A, b ∈ B and e ∈ E, (ae)b = a(eb). An A − A-bimodule is also known as an A-bimodule. Example 2.4.2. M2 (C) is an M2 (C) bimodule, by matrix multiplication similarly to the previous example. Take two vector bundles over a manifold M , for example Ω1 M and Ω1 M . We have the relation, for sections s, t of the cotangent bundle s(m)f (m) ⊗ t(m) = s(m) ⊗ f (m)t(m) , for every function f on M , so we need s.f ⊗ t = s ⊗ f.t. We use the notation F ⊗A E for the tensor product of a right A-module F and a left A-module E, which is F ⊗ E with the additional relation f.a ⊗ e = f ⊗ a.e , for all e ∈ E, f ∈ F and a ∈ A. Note that F ⊗E is much bigger than F ⊗A E. Definition 2.4.3. A map ∇ : E → Ω1 A ⊗A E given a left A-module E, is called a left A-covariant derivative and it satisfies the condition ∇(a.e) = da ⊗ e + a.∇e , for all e ∈ E and a ∈ A. This is related to the usual coordinate idea of covariant derivative by ∇v = dxi ⊗ ∇i v . In classical geometry, we do not have to be concerned about the difference between left and right multiplication, as they are the same. However, in noncommutative geometry we do. The Leibniz rule given in Definition 2.4.3 19

is for left multiplication. We could try to impose the same rule for right multiplication, ∇(e · a) = ∇(e) · a + e ⊗ da . but there would be almost no examples of both applying. The following idea is due to several authors [31][32][58], but we quote the following two definitions from the later paper [8][48]. Definition 2.4.4. A pair (∇, σ) is a bimodule covariant derivative on an A-bimodule E, where ∇ : E → Ω1 A ⊗A E is a left A-covariant derivative, and σ : E ⊗A Ω1 A → Ω1 A ⊗A E is a bimodule map called the generalised ‘braiding’ obeying ∇(e.a) = ∇(e).a + σ(e ⊗ da) . Note that it is convenient to call σ the generalised‘braiding’ as it reverses order, but only in some cases does it actually satisfy the braid relations. Now we define the covariant derivative on tensor products of bimodules. In fact this is the justification for why Definition 2.4.4 is useful, and can be found in [17]. Definition 2.4.5. [17] Given (∇E , σE ) a bimodule covariant derivative on the bimodule E and ∇F a left covariant derivative on the left module F , there is a left A-covariant derivative on E ⊗A F given by ∇E ⊗ idF + (σE ⊗ idF )(idE ⊗ ∇F ) . Further if F is also an A-bimodule with a bimodule covariant derivative (∇F , σF ), then there is a compatible braiding on E ⊗A F given by σE ⊗ F = (σE ⊗ id)(id ⊗ σF ) .

20

2.5

Monoidal Categories

We need a category in which we can take products of objects, in a similar manner to taking tensor products of vector spaces. A monoidal or tensor category is a category C equipped with a functor ⊗ : C × C → C which is associative up to a map [43][47] ΦX,Y,Z : (X ⊗ Y ) ⊗ Z → X ⊗ (Y ⊗ Z) ,

∀X, Y, Z ∈ C ,

obeying the pentagon identity. That is [67]

((X ⊗ Y ) ⊗ Z) ⊗ W

ΦX,Y,Z ⊗ idW

(X ⊗ (Y ⊗ Z)) ⊗ W ΦX,Y ⊗Z,W

ΦX⊗Y,Z,W (X ⊗ Y ) ⊗ (Z ⊗ W )

X ⊗ ((Y ⊗ Z) ⊗ W ) ΦX,Y,Z⊗W idX ⊗ ΦY,Z,W X ⊗ (Y ⊗ (Z ⊗ W ))

Fig. 1 It also admits a unit object 1 and natural equivalences between the functors. i.e. there should be given lX : X → X ⊗ 1 and rX : X ⊗ 1 → X obeying the triangle condition.

21

(X ⊗ 1) ⊗ Y

Φ

X ⊗ (1 ⊗ Y ) id ⊗ r

l ⊗ id X ⊗Y

Fig. 2 Example 2.5.1. Finite dimensional vector space over C. Here the unit object is 1 = C, with the usual tensor product. The map Φ is the identity, Φ((x ⊗ y) ⊗ z) = x ⊗ (y ⊗ z). Example 2.5.2. For a unital algebra A, the category of bimodules A MA forms a monoidal category. The tensor product is ⊗A , the unit element 1 is A. The map l : X → X ⊗A A is given by l(x) = x ⊗ 1A , where 1A ∈ A is the identity in A. Φ is the identity. Definition 2.5.1. [50] A braided monoidal category (C, ⊗, Ψ) is a monoidal category (C, ⊗) which is commutative in the sense that there is a natural equivalence between the two functors ⊗, ⊗op : C × C → C ΨX,Y : X ⊗ Y → Y ⊗ X obeying the hexagon condition,

22

,

∀X, Y ∈ C ,

X ⊗ (Y ⊗ Z) Φ−1 id ⊗ Ψ X ⊗ (Z ⊗ Y )

(X ⊗ Y ) ⊗ Z

Φ−1

Ψ

(X ⊗ Z) ⊗ Y

Z ⊗ (X ⊗ Y ) Ψ ⊗ id Φ−1 (Z ⊗ X) ⊗ Y

Fig. 3 Note that vector spaces over C is a braided monoidal category, we just reverse order, Ψ(v ⊗ w) = w ⊗ v . However, A MA is not in general braided. We can check that order reversal as above is not a bimodule map. Ψ(v ⊗ w.a) = w.a ⊗ v = w ⊗ a.v 6= Ψ(v ⊗ w).a = w ⊗ v.a , as a.v 6= v.a in general.

2.6

Bar categories

In noncommutative geometry it can be quite complicated to deal with modules of star algebras, as ? reverses order. Possibly the best way is to introduce 23

the idea of a conjugate bimodule E of an A-bimodule E where A is a star algebra. E consists of elements e for e ∈ E, with the algebra action, a · e = e · a∗

,

e · a = a∗ · e .

This idea was abstracted to a general monoidal category in [11], and we quote the definition of a bar category from there. Definition 2.6.1. [11] A bar category is a tensor category (C, ⊗, 1C , l, r, Φ) together with the following data, 1) A functor bar : C → C (written as X 7→ X). 2) A natural equivalence bb between the identity and the bar ◦ bar functors on C. 3) An invertible morphism ? : 1C → 1C . 4) A natural equivalence Υ (capital upsilon) between bar ◦ ⊗ and ⊗ ◦ (bar × bar) ◦ flip from C × C to C. These are required to obey the rules: a) The following compositions are rX¯ and lX¯ respectively: l

ΥX,1

?−1 ⊗id

r

Υ1

id⊗?−1

X X −→ X ⊗ 1C −→C 1C ⊗ X −→ 1C ⊗ X , ,X

X C X −→ 1C ⊗ X −→ X ⊗ 1C −→ X ⊗ 1C .

b) The following condition is satisfied by Υ and the associator Φ: ΦZ,Y ,X (ΥY,Z ⊗ id)ΥX,Y ⊗Z ΦX,Y,Z

= (id ⊗ ΥX,Y )ΥX⊗Y,Z .

c) ?? = bb1C : 1C → 1C . d) bbX = bbX¯ : X 7→ X . For modules over the star algebra A, we have 1) 1C = A . 24

2) ? : A → A sends a 7→ a∗ . 3) Υ(e ⊗ f ) = f ⊗ e . 4) bb(e) = e . Example 2.6.1. (Vector spaces over C.) If V is a complex vector space, take V = V as a set, with elements denoted v ∈ V for v ∈ V to avoid confusion. We have addition and multiplication v+w =v+w

,

λ v = λ∗ v .

for v, w ∈ V and λ ∈ C (using λ∗ for conjugate). Now we can make a single treatment of linear and conjugate linear maps. For example an inner product on V which is linear in the first component and conjugate linear in the second becomes a linear map for V ⊗ V to C. Then hv ⊗ λ wi = hv ⊗ λ∗ wi = λ∗ hv ⊗ wi , hλ v ⊗ wi = λhv ⊗ wi . The maps in the definitions are given by Υ:V ⊗W → W ⊗V , Υ(v ⊗ w) = w ⊗ v , bb : V → V

,

bb(v) = v .

The Φ is trivial, i.e. Φ((a ⊗ b) ⊗ c) = a ⊗ (b ⊗ c) . Example 2.6.2. [11](Bimodule over a Star algebra.) For a ?-algebra A, Let E be an A-bimodule. We take E as a vector space as in Example 2.6.1. and with A-actions a.e = e.a∗ and e.a = a∗ .e. The 25

formulae for Φ, Υ and bb are as before. We use ⊗A for bimodules. We need to check that Υ and bb are bimodule maps, and that Υ is well defined, i.e, Υ(v ⊗ a w) = a w ⊗ v = w a∗ ⊗ v = w ⊗ a∗ v = w ⊗ v a∗∗ = w ⊗ v a = Υ(v a ⊗ w) . and, for example the right action Υ((v ⊗ w).a) = Υ(a∗ .(v ⊗ w)) = Υ(a∗ v ⊗ w) = wa∗ v = w ⊗ v.a = Υ(v ⊗ w).a

bb(v.a) = v.a = a∗ v = va∗∗ = va = bb(v).a (The left action is similar).

2.7

The Hopf algebra q-deformed SL2.

In [70], Woronowicz described a deformation of the algebra of functions on the Lie group SL2 , with a parameter q, where q = 1 gives the classical case. We use mainly [53] as a reference We quote the relations for Cq [SL2 ] in the conventions of [50], ba = qab,

db = qbd,

cb = bc,

ad − da = q(1 − q −2 )bc ,

ca = qac,

dc = qcd.

ad − q −1 bc = 1.

Ma=a⊗a+b⊗c ,

M b = a ⊗ b + b ⊗ d,

Mc=c⊗a+d⊗c ,

M d = c ⊗ b + d ⊗ d.

(a) = (d) = 1 , S(a) = d,

S(b) = −qb,

(b) = (c) = 0. S(c) = −q −1 c,

26

S(d) = a.

We can also add a star structure, to give the Hopf star algebra Cq [SU2 ], as follows a∗ = d,

d∗ = a,

c∗ = −qb,

b∗ = −q −1 c.

On the Hopf algebra Cq [SU2 ] we take the 3-d calculus of [70], with basis e− = ddb − qbdd,

e+ = q −1 adc − q −2 cda,

e0 = dda − qbdc

left-invariant 1-forms, while the right module relations and exterior derivative are given by, again quoting the relations from [53] :         −1 2 −2 a b qa q b a b q a q b =  e± , e0  =  e0 e±  c d qc q −1 d c d q 2 c q −2 d da = ae0 +qbe+ ,

db = ae− −q −2 be0 ,

de0 = q 3 e+ ∧ e− ,

dc = ce0 +qde+ ,

de± = ∓q ±2 [2; q −2 ]e± ∧ e0 ,

q 2 e+ ∧ e− + e− ∧ e+ = 0,

dd = ce− −q −2 de0 .

(e± )2 = (e0 )2 = 0

e0 ∧ e± + q ±4 e± ∧ e0 = 0

where [n; q] = (1 − q n )/(1 − q) denotes a q-integer. Take a Hopf algebra C[t, t−1 ] with ∆t = t ⊗ t and St = t−1 . This is just functions on the circle algebra, where t is the inclusion function from S 1 to C, taking S 1 to be unit norm complex numbers. We can specify a Z grading on Cq [SL2 ] by: deg(a) = deg(c) = 1 ,

deg(b) = deg(d) = −1.

and this gives a coaction on a monomial by: MR (w) = w ⊗ tdeg(w) . This gives a left coaction: MR : Cq [SU2 ] → Cq [SU2 ] ⊗ C[t, t−1 ], 27

and the coinvariants for this is the simplest or standard noncommutative sphere (they were classified by Podle´s [61]) and is the degree 0 subalgebra, according to the given grading. The grading extends to the differential calculus by deg(e± ) = ±2 ,

deg(e0 ) = 0.

and the 1-forms on Cq [S 2 ] are given by the degree 0 forms in e+ and e− . Thus a2 e− and bde+ are 1-forms on Cq [SU2 ], but e0 or abe− are not.

28

Chapter 3 Preliminaries on Physics 3.1

Background to special relativity

Special relativity was based on the failure to find a drift velocity of the earth with respect to a hypothetical ”aether” in the Michelson-Morley experiment [35], and that Maxwell’s equations were not consistent with Galilean relativity. (This gives a simple rule for moving frames of reference, x0 = x − vt, t0 = t). Special relativity assures that the velocity of light is constant for all observers in constant motion (not accelerated), in fact that all the laws of physics are the same for such observers [63]. The special theory of relativity was published by Einstein in 1905[69].

29

3.1.1

Minkowski space

We take this account from [63]. A Minkowski space is an R4 with coordinates (t, x, y, z) and inner product given by the equation     t t0        0 x x  2 0 0 0 0    h   ,  0 i = c tt − xx − yy − zz . y  y      z z0

(3.1)

For a small vector (dt, dx, dy, dz) in space time, we use the length ds2 = c2 dt2 − (dx2 + dy 2 + dz 2 ).

(3.2)

where c is the speed of light [63]. We write coordinates xµ = (t, x, y, z)

(3.3)

and then the metric can be written as the inner product of (dt, dx, dy, dz) with itself, i.e., ds2 = gµν dxµ dxν ,

(3.4)

where 

gµν



2

c 0 0 0      0 −1 0 0 , =    0 0 −1 0    0 0 0 −1

(3.5)

In particle physics, it is quite common to set c = 1.

3.1.2

Lorentz transformation

As in section 3.1.1, we use [63] as our main reference. Lorentz transformations are linear maps T : R4 −→ R4 which keep the inner product the same, i.e., hT v, T wi = hv, wi 30

,

v, w ∈ R4

(3.6)

One example is rotations in xyz space leaving t the same. In addition we get Lorentz boosts, which in the x-direction is given by vx ) c2 y0 = y

t0 = γ(t −

,

x0 = γ(x − v t)

,

z0 = z .

(3.7)

1 where v is the relative velocity and γ = p , and this can be written 1 − β2 as a matrix [63]      γv 0 0 0 γ − t ct      c2  0    x  −vγ x γ 0 0  =   (3.8)  0    y   0    y 0 1 0      z0 z 0 0 0 1

3.2

Quantum theory

The basic reference for this section is [63][19] but other sources will be given as appropriate for historical reasons.

3.2.1

The uncertainty principle

In [41] Heisenberg discussed the problem of measuring both the position and momentum of a particle.

Fig. 4 31

This is caused by a combination of light carrying momentum and by the diffraction of light when it enters the camera. When the light is of high frequency the first factor is dominant, and for low frequencies the second factor is the problem. If we call the uncertainty in position 4x and the uncertainty in momentum 4p then we get the inequality 4p 4 x ≥

~ 2

(3.9)

where ~ is the Planck constant [19].

3.2.2

Schr¨ odinger equation

Quantum mechanics can be modelled mathematically by linear operators on a Hilbert space of functions on space, say R, with inner product Z ∞ hf, gi = f ∗ g dx . −∞

∂ and we represent the ∂t potential by a real function V which is time independent. The momentum ∂ is p = −i~ , so we have the commutator [65], ∂x In one dimension, the total energy is denoted by i~

∂ ∂ (x f ) + i~x f ∂x ∂x ∂f ∂f = −i~f − i~x + i~x ∂x ∂x = −i~f .

[p, x]f = −i~

This means that we get an uncertainty on swapping the order of measuring momentum and position consistent with (3.9). p2 Now the equation E = +V (Total energy=Kinetic+Potential energy) be2m comes the Schrödinger’s equation when applied to a complex valued function Ψ, ∂Ψ −~2 ∂ 2 Ψ = +V Ψ , i~ ∂t 2m ∂x2 32

(3.10)

or in more space dimensions, i~

∂Ψ −~2 2 = ∇ Ψ + ΨV ∂t 2m

(3.11)

For a fixed time t, we interpret Ψ∗ Ψ as a probability density for the location of the quantum particle. We need, Z ∞

Ψ∗ Ψdx = 1

(3.12)

−∞

but Ψ is a function of time by (3.10), so if we have (3.12) at time t = 0, we need it to be true for t = 1, ∂ ∂t

Z

∞

Ψ∗ Ψdx = 0

(3.13)

−∞

This is true if the potential is real. The Schrödinger equation is for a particle (or several particles) moving in a potential. It can be used to study energy levels in the Hydrogen atom [18]. The atom can absorb or emit radiation in amounts which correspond to differences of energy levels [56]. For more particles (e.g. 2 electrons for Helium) it is much more difficult to get exact solutions. Also high speed electrons give relativistic corrections, though this is not a vital factor for the Hydrogen atom [35].

3.2.3

Klein-Gordon equation

Our main reference for this is Ryder’s book [63]. This is an attempt to make a relativistic Schrödinger equation. The differential operators are defined as ∂µ =

∂ ∂ ∂ ∂ ∂ = (∂0 , ∂1 , ∂2 , ∂3 ) = ( , , , ) µ ∂x ∂t ∂x ∂y ∂z ∂ = ( , ∇), ∂t

(3.14)

giving the Lorentz invariant second-order differential operator ∂2 = ∂ ∂µ = 2 , −∇2 , ∂t µ

33

(3.15)

called the d’Alembertian operator. The energy-momentum 4-vector of a particle is pµ = (

E , Px , Py , Pz ), c2

(3.16)

giving the equation for a single particle, p2 = p µ pµ =

E2 − p.p = m2 c2 . c2

(3.17)

1 . Note that m is somefor p = γ m v and E = γ m c2 , where γ = r v 2 1−( ) c times called the rest mass of the particle, and does not depend on velocity. The Klein-Gordon equation is a wave equation for a particle with no spin - a scalar particle. We take φ to be complex valued, giving zero spin. We get the Klein-Gordon equation by substituting differential operators for E and p in (3.17), in the fashion standard in quantum theory [63] E −→ i~

∂ , ∂t

p −→ −i~∇

(3.18)

i~ ∂ 2 X ∂ ) − (−i~ i )2 = m2 c2 c ∂t ∂x i X ∂2 ~2 ∂ 2 = m2 c2 − 2 2 + ~2 c ∂t ∂(xi )2

(

Applying to φ gives (m2 c2 + ~2 )φ = 0

(3.19)

If we put m = 0 in (3.19), this gives the usual wave equation with propagation speed c, 2 ∂ 2φ 2 ∂ φ = c . ∂t2 ∂(xi )2

If we put φ = ei(wt+α1 x1 ) in the massive Klein-Gordon equation we get w2 =

m 2 c4 + c2 α12 , ~2 34

(3.20)

so we seem to have a change in velocity |v| = | −

ω |>c. α1

However this is only the phase velocity for one Fourier component - no information can be transmitted at speed greater than c (see [15] ,[66]) . This is why, in section 5.3, we will be careful to look at the propagation speed, also called the group velocity.

3.2.4

Spinors and the Dirac equation

Our main reference for this is Ryder’s book [63]. The Dirac equation was invented to be a first order differential equation which implied the KleinGordon equation. Take the simplest case i γ µ ∂µ ψ = −m ψ apply this twice i γ ν ∂ν i γ µ ∂µ ψ = −m i γ ν ∂ν ψ −γ ν γ µ ∂ν ∂µ ψ = m2 ψ we want this to be ψ = −

m2 c2 φ, ~2

We need γ ν γ µ ∂ν ∂µ ψ = ψ For c = 1, this requires γ0 γ0 = 1 ,

γ i γ i = −1 ,

35

i 6= 0

and γa γb + γa γb = 0 ,

a 6= b , 0 ≤ a, b ≤ 3 .

We see that the γ a can not commute in general. We take them to be matrices, giving the Dirac equation. Then ψ is a column vector, which is called a spinor. The Dirac equation is a first order wave equation, for spin- 12 particles, with relativistic invariance. The particles are represented by 4D complex column vectors, known as bispinors [27][68]. The massless Dirac Equation is given by [63] i(∂µ − i q Aµ )γ µ ψ = 0 .

(3.21)

Here we have used a covariant derivative ∂µ − i q Aµ , replacing the partial derivative earlier. As we shall see in the next section, the gauge potential Aµ in the covariant derivative gives an electromagnetic field. The smallest size of matrices satisfying the required equations is 4 by 4, and we get the standard form of the γ µ matrices,    1 0 0 0 0       0 1 0 0 0  , γ1 =  γ0 =     0 0 −1 0  0    0 0 0 −1 −1    0 0 0 −i 0       0 0 0 i 0  , γ3 =  γ2 =     −1 0 i 0 0    −i 0 0 0 0

 0 0 1   0 1 0   −1 0 0  0 0 0  0 1 0   0 0 −1   0 0 0  1 0 0

(3.22)

There is a current carried by the field in the Dirac equation, given by: ja = q ψ γ a ψ ,

for

a = 0, 1, 2, 3 where ψ = ψ ∗ γ 0 ,

(3.23)

which is conserved as in [63]: ∂µ jµ = (∂µ ψ)γ µ ψ + ψγ µ (∂µ ψ) = (imψ)ψ + ψ(−imψ) = 0 . 36

(3.24)

This satisfies the following equation, linking the electromagnetic fields with the current [63]: ∂µ F µν = jν .

3.3

(3.25)

Gauge theory

U1 gauge theory is about covariant derivatives on sections of line bundles, and classically gives Maxwell’s equations of electrodynamics. Each copy of C at a point of the manifold can be multiplied by a circle valued function on the manifold. This gives a gauge transformation, as described in [63] . The unitary group U1 is just the circle of unit norm complex numbers with multiplication as the operation. As we are only dealing with the space-time R4 we shall assume that the bundle is trivial, which is a fancy way of saying that we just deal with derivatives of functions, i.e. we set E = A in the language of section 2.4 . We shall only deal with Maxwell’s theory, there will be no spinor fields. We add a gauge potential i Aµ to the covariant derivative to get Oµ = ∂µ − i q Aµ . Here the real potential Aµ is related to the electric and magnetic fields by [63] Fµν = where

 F µν

0

∂Aν ∂Aµ − , µ ∂x ∂xν −E 1 −E 2 −E 3

(3.26) 

   1 3 2  E 0 −B B   . =  2  3 1 E B 0 −B    E 3 −B 2 B 1 0

(3.27)

Here, the E i is the electric field vector and B i is the magnetic field vector. To see what a change of gauge is, just consider a very simple equation (the

37

idea is the same for more complicated gauge equations), which is (3.21), and make a change of φ by a circle valued complex function u, φ0 = uφ ,

u−1

∂ 0 ∂ φ = (uφ) ∂xµ ∂xµ ∂u ∂φ = φ+u µ µ ∂x ∂x 0 ∂φ −1 ∂φ −1 ∂u u = u φ + ∂xµ ∂xµ ∂xµ 0 ∂φ ∂u ∂φ u−1 ( µ − i q Aµ φ0 ) = u−1 µ φ + µ − i q Aµ φ ∂x ∂x ∂x −1 ∂u φ = u µ ∂x ∂φ0 −1 ∂u 0 = 0 − (i q A + u )φ µ ∂xµ 0 ∂xµ ∂φ −1 ∂u − (i q A + u )φ0 = 0 . (3.28) µ ∂xµ ∂xµ

so φ0 obeys the same equation, but with a different potential. This is a gauge transformation. In a gauge theory, such gauge transformations are taken to ∂u be symmetries of the theory. Note that u−1 µ is imaginary. ∂x More generally Gauge theory can be done with other, nonabelian, Lie algebras, but we will not be concerned with this.

38

Chapter 4 The Majid-Ruegg space time 4.1

The κ-Poincar´ e algebra

In R2 with the Euclidean metric, isometries are given by translations and rotations about the origin, and any combinations of these (we need to include a reflection if we do not preserve right-left symmetry). For Minkowski space (see section 3.1.1) the isometries are made of translations and Lorentz transformations - this gives the Poincaré group. Now look at R2 in more detail - Minkowski space is similar. Tv = Translation by a vector V and MA = Multiply by a matrix A. For R2 , A will just be a rotation (or reflection) matrix.  Rθ = 

cos θ − sin θ sin θ

cos θ

  .

(4.1)

Any isometry of R2 is of the form Tv MA for some v and A. The group operation is Tv MA Tw MB (u) = Tv+Aw MAB

(4.2)

The translations are a subgroup of the isometries. Also multiplying by matrices are subgroup of the isometries. But they do not commute - we do not get 39

a direct (or cross) product - that would give (Tv MA )(Tw MB ) = Tv+w MAB . But from (4.2) we see that the direct product rule is modified by an action of the matrices on the translations, to give a semi-direct product. We see that the Poincaré group is also a semi direct product of the translations and the Lorentz transformations. In the noncommutative case, in [55] Majid and Ruegg point out that there is a similar product structure on the deformed κ-Poincaré algebra, and that this can be used to identify the translations in the κ-Poincaré algebra. However in this case the translations do not have trivial coproduct, and they arrive at the relations of κ-Minkowski space. The κ-Poincaré algebra Pκ , ¯i ; κ (antiHermitian generators of translations Pµ , rotations Mi and boosts N real; i, j, k = 1, 2, 3; µ, ν = 0, 1, 2, 3) is defined by: [Pµ , Pν ] = 0,

[Mi , Mj ] = ijk Mk ,

[Mi , Pj ] = ijk Pk ,

[Mi , P0 ] = 0,

¯j ] = ijk N ¯k , [Mi , N ¯i , P0 ] = Pi , [N

¯i , Pj ] = δij κ sinh [N

(4.3) (4.4) (4.5)

P0 , κ

¯i , N ¯j ] = −ijk (Mk cosh P0 − 1 Pk P~ M ~ ). [N κ 4κ2

(4.6) (4.7)

The coproducts are given by, ∆P0 = P0 ⊗ 1 + 1 ⊗ P0 ,

∆Mi = Mi ⊗ 1 + 1 ⊗ Mi , P0

P0

∆Pi = Pi ⊗ e 2κ + e− 2κ ⊗ Pi , P0

P0

¯i = N ¯i ⊗ e 2κ + e− 2κ ⊗ N ¯i + ∆N

P0 P0 ijk (Pj ⊗ Mk e 2κ + e− 2κ Mj ⊗ Pk ). 2κ

(4.8) (4.9) (4.10)

The M, N generators correspond to Lorentz transformations, and the Pi to translations. (4.9) leads to the noncommutativity of κ-Minkowski. The 4-dimensional differential calculus was obtained by Oeckl [60], but this is 40

not directly related to κ-Poincaré. There is a 5-dimensional calculus which is directly related to the calculus on κ-Poincaré by a construction done by Sitarz [64]. Having said this, we will do something different, we shall look for a metric which is central, and that shall require finding a different calculus.

4.2

κ-Minkowski space

In space time we have the coordinates x0 = t, x1 = x, x2 = y, x3 = z. The Minkowski metric is given by c2 dt ⊗ dt − dx ⊗ dx − dy ⊗ dy − dz ⊗ dz, where c P is the velocity of light. This is written as ab dxa ⊗ dxb . The Majid-Ruegg or κ-deformed space-time is given by making this noncommutative [55]. This involves an imaginary λ. We have the following commutator relations, where we use the convention that i, j, k take values 1, 2, 3 only, [xi , xj ] = 0 ,

[xi , t] = λxi .

(4.11)

In [33] this κ-Minkowski space algebra is studied as a topological algebra, but we only need the algebraic structure for what we shall do. We can calculate commutators of functions to first order in λ as follows, [x2 , y] = x(xy − yx) + (xy − yx)x = x[x, y] + [x, y]x. n+1 x , y = xn+1 y − yxn+1 = x(xn y − yxn ) + (xy − yx)xn = x[xn , y] + [x, y]xn . by induction, we can prove that j n 0 (x ) , x = n λ (xj )n . 41

(4.12)

(x0 )a (x1 )b (x2 )c (x3 )d , x0

= [(x0 )a , x0 ](x1 )b (x2 )c (x3 )d +(x0 )a [(x1 )b , x0 ](x2 )c (x3 )d +(x0 )a (x1 )b [(x2 )c , x0 ](x3 )d +(x0 )a (x1 )b (x2 )c [(x3 )d , x0 ] = λ(b + c + d)(x0 )a (x1 )b (x2 )c (x3 )d (. 4.13)

where as for i 6= 0 0 a 1 b 2 c 3 d i (x ) (x ) (x ) (x ) , x = −λ a (x0 )(a−1) xi (x1 )b (x2 )c (x3 )d .

(4.14)

so if we split any function into a sum of products α(x0 ) β(x1 x2 x3 ) for i 6= 0, then to first order in λ ∂α α(x0 ), x0 = 0 , α(x0 ), xi = −λ xi ∂x0 X ∂β 1 2 3 i β(x1 x2 x3 ), x0 = λ , β(x x x ), x xi =0 ∂xi ∂β 0 x β(x1 x2 x3 ), (x0 )2 = 2λ xi ∂xi ∂α X i ∂β [β, α] = λ x ∂x0 ∂xi

4.3 4.3.1

(4.15) (4.16) (4.17) (4.18)

A new differential calculus A central metric

The κ-Minkowski space is normally given either a 4D differential calculus [60] or a 5D invariant differential calculus [64]. However, we do something completely different, we deform the differential calculus according to the principle that the metric is central to first order in λ, i.e., that it commutes with all functions. Why do we want a central metric? the answer is that without that it is not possible to contract indices in a consistent fashion. We can look on the metric in two ways: 42

1) As a bilinear map from Vect ⊗ Vect to the algebra, (ignoring conjugates), 2) or as an element of Ω1 A ⊗A Ω1 A, (again ignoring conjugates) . Classically, we have for the inner product of vectors, hv, wi = gab v a wb , g = gab dxa ⊗ dxb . We can think of the link between these formulae as the following, V W

create g Ω1

Ω1 evaluate

Fig. 5 so the inner product can be thought of as creating a copy of g, and then evaluating the vector fields and 1-forms as duals. Now suppose that we have a 3-index tensor given by a product of 3 vector fields v ⊗ w ⊗ u, and use g to contract the indices of v and w. V

W

U

create g

ev : Vect ⊗A Ω1 A → A Fig. 6

43

In terms of indices, v ⊗ w ⊗ u → v a wb gab u, but if g is not central this is not a legal operation in Vect ⊗A Vect ⊗A Vect, i.e. it does not give a well defined answer. In Vect ⊗A Vect ⊗A Vect, for a ∈ A, v ⊗ wa ⊗ u = v ⊗ w ⊗ au , and introducing g gives v ⊗ wa ⊗ g ⊗ u = v ⊗ w ⊗ g ⊗ au , so we get v ⊗ w ⊗ ag ⊗ u = v ⊗ w ⊗ ga ⊗ u . so we need, for all v, w, the evaluation of the first 3 parts of the following must be zero, v ⊗ w ⊗ (a g − g a) ⊗ u = 0 . Now if a g 6= g a we expect a non-zero result for some v, w. Thus, unless we are very careful just what tensor product are used (i.e. check ⊗ not ⊗A ) we will possibly find inconsistent results in applying inner products in the middle of tensors. It corresponds to the property that the inner product h i : Vect ⊗A Vect → A is a bimodule map. To check this, supposing g is central, V ⊗W a

V ⊗W

g

ga =

Fig. 7 as v ⊗ w a ⊗ g = v ⊗ w ⊗ a g = v ⊗ w ⊗ g a . 44

4.3.2

Constructing the calculus

In [10], it was shown that the Minkowski metric was not central in the usual κ-Poincaré differential calculus, and that in fact there is no central metric possible for that calculus (though if we restrict to 2-dimensions it is possible). Here we shall take the Minkowski metric, and insist that it is central, and use that to find an alternative calculus. Of course this calculus is no longer invariant to the κ-Poincaré algebra, but that need not to be a problem for the physics. There are reasons why κ-Minkowski algebra is interesting which are entirely independent of κ-Poincaré. The κ-Minkowski algebra is rotationally invariant, and in the universe we may expect the Lorentz symmetry to be broken by the microwave background radiation. We shall assume the following form of all commutators of functions and forms, where Λslp is a numerical multiple of λ, [dxl , xp ] = Λslp dxs .

(4.19)

This is our main assumption, and under it we shall be able to find all the parameters Λslp . There are constraints on Λslp given by the commutation relations (4.11) d(xi xj − xj xi ) = 0 again, note that this is only to first order in λ. Now, as d is a derivation, dxi xj − xj dxi + xi dxj − dxj xi = 0 .

(4.20)

i j dx , x = [dxj , xi ] .

(4.21)

so for i, j ∈ {1, 2, 3}, Λsij = Λsji . 45

(4.22)

Also from (4.11), d(xi x0 − x0 xi ) = λ dxi dxi x0 − x0 dxi + xi dx0 − dx0 xi = λ dxi i 0 dx , x − [dx0 , xi ] = λ dxi

(4.23)

Λsi0 dxs − Λs0i dxs = λ dxi   λ s=i =  0 s= 6 i

i.e., Λsi0 − Λs0i

(4.24)

applying d to (4.19) gives the usual relations for the higher order calculus dxi ∧ dxj + dxj ∧ dxi = 0.

(4.25)

Now apply the commutation relations to the classical Minkowski metric. Why could we not consider an O(λ) correction to the metric g + λg 0 ? This would not make any difference to our calculations, as for an algebra element, [g + λg 0 , a] = [g, a] + λ[g 0 , a] , and then λ[g 0 , a] is O(λ2 ). So an O(λ) correction to the classical metric makes no difference to O(λ) to the commutator. Proposition 4.3.1. For the metric to be central, to first order in λ we must have all commutations relations [dxa , xb ] = 0, except for (i = 1, 2, 3) i i dx , x = −c2 λ dx0

,

0 i dx , x = −λ dxi .

(4.26)

Proof. We use the result, in Ω1 A ⊗A Ω1 A, that for ξ, η ∈ Ω1 A and a ∈ A [ξ ⊗ η, a] = ξ ⊗ [η, a] + [ξ, a] ⊗ η

46

(4.27)

to write, hX

a

b

ab dx ⊗ dx , x

g

i

=

X

ab ([dxa , xg ] ⊗ dxb + dxa ⊗ [dxb , dxg ])

X

ab (Λsag dxs ⊗ dxb + dxa ⊗ Λtbg dxt ) X X = sb Λasg dxa ⊗ dxb + as Λbsg dxa ⊗ dxb X = (sb Λasg + as Λbsg )dxa ⊗ dxb (4.28) =

For all a, b, g and when summed over s, we have, X

(sb Λasg + as Λbsg ) = 0

(4.29)

If we put a = b = g = 0, in (4.29), we get, Λ000 = 0

(4.30)

for a = 0, b = i 6= 0, g = 0 in (4.29), −Λ0i0 + c2 Λi00 = 0

(4.31)

But for a = b = 0, g = i 6= 0 in (4.29) we get, Λ00i = 0

(4.32)

Now from (4.24) with s = 0 and by using (4.31), (4.32) we get, Λ0i0 = Λ00i = c2 Λi00 = 0

(4.33)

i.e., all Λabg with two of a, b, g being zero must vanish. Now, if we put a = i 6= 0, b = j 6= 0, g = 0 in (4.29), we get, −Λij0 − Λji0 = 0

(4.34)

and for a = 0, b = j 6= 0, g = i 6= 0 in (4.29), we get, −Λ0ji + c2 Λj0i = 0 47

(4.35)

in equation (4.23), if we put s = j 6= i , we get, Λj0i = Λji0

(4.36)

Λ0ij = Λ0ji

(4.37)

Λ0ji = c2 Λj0i = c2 Λji0 = −c2 Λij0 = −c2 Λi0j = −Λ0ij

(4.38)

and in (4.22) for s = 0, we get,

then we have:

but combining this with (4.37) gives, Λ0ij = 0 ,

i 6= j

(4.39)

From (4.34), if i = j we get, Λii0 = 0

(4.40)

and from (4.35), (4.36) and (4.24) for i = j we get, Λ0ii = Λi0i = −c2 λ

(4.41)

i.e., Λabg with one of a, b, g being zero will vanish unless we have two indices equal i, then Λ0ii = Λi0i = −c2 λ , Λii0 = 0. For the non-zero case, if we put a = j 6= 0, b = k 6= 0, g = i 6= 0 in (4.29) we get, −Λjki − Λkji = 0

(4.42)

then using (4.22) and (4.33) we get, Λkij = Λkji = −Λjki = −Λjik = Λijk = Λikj = −Λkij i.e., it is antisymmetric, then all Λkij = 0.

48

(4.43)

For higher forms, we note that the commutation relations in Proposition 4.3.1 differentiate to give d(dxi xi − xi dxi ) = −c2 λ d(dx0 ) , d(dx0 xi − xi dx0 ) = −λ d(dxi ) . which gives −2 dxi ∧ dxi = 0

dxo ∧ dxi + dxi ∧ dx0 = 0 .

and

Thus we get the usual antisymmetry of the dxi in the noncommutative calculus to O(λ).

4.3.3

Nonassociativity at O(λ2 )

Every associative algebra obeys the Jacobian identity for commutators [a, b] = ab − ba, 0 = J(A, B, C) = [[A, B], C] + [[B, C], A] + [[C, A], B]

(4.44)

We shall use this equation with one variable taking values in Ω1 and the other two taking values in the algebra Ω0 , and if the differential calculus was associative, we would have J(A, B, C) = 0. However if we calculate the Jacobi identity for the calculus in Proposition 4.2.1., we find the following non-zero result, [x0 , [dx0 , xi ]] + [dx0 , [xi , x0 ]] + [xi , [x0 , dx0 ]] = −λ2 dxi .

(4.45)

We could stop here and say that the algebra was not associative, however we should make three points: The first is that, since the parameter λ giving noncommutativity is very small (much searching has still not found a definite value), λ2 is so small as to be totally outside any current possible measurement. However the error from treating a non-central metric as though it were 49

central is O(λ)-much larger than any error caused by assuming associativity when it is not true, which is O(λ2 ). The second is that nonassoiciativity is now an accepted part of some parts of physics [22][51]. It is known that any semiclassical deformation of classical calculus on a curved space must violate associativity at the order of the deformation parameter squared [13][40]. The third is that we can correct the nonassoiciativity to make the error O(λ4 ), by using an associator which is a machine for altering the order of bracketing on tensor products, ΦA,B,C : (A ⊗ B) ⊗ C → A ⊗ (B ⊗ C). A product µ is associative with respect to the associator if, where id is the identity µ µ(a ⊗ b) ⊗ c = µ(id ⊗ µ)Φ (a ⊗ b) ⊗ c

(4.46)

In ordinary vector spaces, associative algebras etc. Φ is trivial, all it does is to change the position of the brackets. However we need a more complicated form here to order λ2 , Φ (a ⊗ b) ⊗ c = a ⊗ (b ⊗ c) + λ2 φ (a ⊗ b) ⊗ c

(4.47)

where φ : (A ⊗ B) ⊗ C −→ A ⊗ (B ⊗ C) . The multiplication µ is associative with respect to Φ so we write, µ(id ⊗ µ)Φ(a ⊗ b) ⊗ c

= µ(id ⊗ µ)

2

a ⊗ (b ⊗ c) + λ φ (a ⊗ b) ⊗ c = a . (b . c) + λ2 µ(id ⊗ µ)φ (a ⊗ b) ⊗ c (ab)c − a(bc) = λ2 µ(id ⊗ µ)φ (a ⊗ b) ⊗ c (4.48)

we can rewrite J(a, b, c) in (4.44) to be J(a, b, c) =

(ab)c − a(bc) − (ba)c − b(ac) + (bc)a − b(ca) + (ca)b − c(ab) − (ac)b − a(cb) − (cb)a − c(ba)

we ignore terms higher than λ2 , we can take the multiplication of φ to be the usual multiplication. If we use the shorthand φ(abc) = µ(id⊗µ)φ (a⊗b)⊗c 50

then the Jacobian identity for φ is given by,   φ(abc) − φ(bac) + φ(bca)  J(a, b, c) = λ2  +φ(cab) − φ(acb) − φ(cba)

(4.49)

Now Φ must satisfy the Pentagon identity [67]

ΦA,B,C ⊗ idD

((A ⊗ B) ⊗ C) ⊗ D

(A ⊗ (B ⊗ C)) ⊗ D ΦA,B⊗C,D

ΦA⊗B,C,D (A ⊗ B) ⊗ (C ⊗ D)

A ⊗ ((B ⊗ C) ⊗ D) ΦA,B,C⊗D idA ⊗ ΦB,C,D A ⊗ (B ⊗ (C ⊗ D))

Fig. 8

Using the formula (4.47), to order λ2 this becomes, for a, b, c, d elements of A, B, C, D , φ(a, b, c)⊗d+φ(a, b⊗c, d)+a⊗φ(b, c, d) = φ(a⊗b, c, d)+φ(a, b, c⊗d) (4.50) we can calculate, for i = 1, 2, 3 J(dxi , xe , xf ) = c2 λ2 (δei dxf − δf i dxe )

(4.51)

J(dx0 , xe , xf ) = λ2 (t(f ) s(e) dxe − t(e) s(f ) dxf )

(4.52)

If we use   0 f = 1, 2, 3 t(f ) =  1 f =0

  1 e = 1, 2, 3 s(e) =  0 e=0

,

51

we can give the following form for J   e 2 dx δf r t(r) s(e) − c s(r) J(dxr , xe , xf ) = λ2   f 2 −dx δer t(r) s(f ) − c s(r)

(4.53)

If we suppose that the objects that the associator is applied to can have partial derivative applied to them (such as functions, forms and vector fields) then, we can try writing (summing over a, b) ∂y ∂x ∂z ∂z φ(x, y, z) = αab (x) ⊗ + ⊗ ⊗ β (y) ⊗ ab ∂xa ∂xb ∂xa ∂xb ∂x ∂y + a⊗ ⊗ γab (z) (4.54) b ∂x ∂x Of course in general, we would have to have a covariant derivative to differentiate vector fields. However as we are just considering flat Minkowski space, we can assume that the derivative of the standard basis vectors dxi ∂ and is zero. Here we need to find what rules αab , βab and γab would need ∂xi to satisfy to make φ obey the requirements, including the given form for J and the Pentagon identity. So we have, φ(dxc , xa , xb ) = αab (dxc ) ⊗ 1 ⊗ 1

,

φ(xa , dxc , xb ) = 1 ⊗ βab (dxc ) ⊗ 1

φ(xa , xb , dxc ) = 1 ⊗ 1 ⊗ γab (dxc ) Using the first α term of this equation in the Pentagon identity we get, 0 = φ(w, x, y) ⊗ z + φ(w, x ⊗ y, z) + w ⊗ φ(x, y, z) −φ(w ⊗ x, y, z) − φ(w, x, y ⊗ z)  ∂x ∂y ∂(x ⊗ y) ∂z αab (w) ⊗ a ⊗ b ⊗ z + αab (w) ⊗ ⊗ b a  ∂x ∂x ∂x ∂x     ∂y ∂z ∂y ∂z =   +w ⊗ αab (x) ⊗ ∂xa ⊗ ∂xb − αab (w ⊗ x) ⊗ ∂xa ⊗ ∂xb     ∂x ∂(y ⊗ z) −αab (w) ⊗ a ⊗ ∂x ∂xb ∂y ∂z = αab (w) ⊗ x + w ⊗ αab (x) − αab (w ⊗ x) ⊗ a ⊗ b ∂x ∂x 52

            (4.55)

If αab (w ⊗ x) = αab (w) ⊗ x + w ⊗ αab (x), we get a solution to the pentagon identity, i.e., α acts as an element of a Lie algebra, not a Lie group. To illustrate this, we suppose that g is a group acting on a tensor product by the usual manner (corresponding to 4g = g ⊗ g in Hopf algebras) g(w ⊗ x) = gw ⊗ gx if we put g = 1 + hα, where h is small, then to first order in h, α(w ⊗ x) = α(w) ⊗ x + w ⊗ α(x) corresponding to 4(α) = α ⊗ 1 + 1 ⊗ α. Similarly for the second and third terms of (4.54), respectively they give:   βab (x ⊗ y)  ∂z ∂w    ⊗  −x ⊗ βab (y)  ⊗ b = 0 a ∂x   ∂x −βab (x) ⊗ y)   γab (y) ⊗ z  ∂w ∂x    ⊗   = 0 +y ⊗ γ (z) ab a b ∂x ∂x   −γab (y ⊗ z) Using (4.49) for A = dxr , B = xe , C = xf , as dxr is translation invariant, we ∂(dxr ) have = 0, ∂xa 1 J(dxr , xe , xf ) = αab (dxr )(δe,a δf,b − δf,a δe,b ) + βab (dxr )(δf,a δe,b − δe,a δf,b ) λ2 +γab (dxr )(δe,a δf,b − δf,a δe,b ) = (δe,a δf,b − δf,a δe,b )(αab (dxr ) − βab (dxr ) + γab (dxr )) (4.56)

53

thus, dxe δf,r t(r)s(e) − c2 s(r) − dxf δe,r t(r)s(f ) − c2 s(r) X = (δe,a δf,b − δf,a δe,b )(αab − βab + γab )(dxr ) a,b

= (αe,f − βe,f + γe,f )(dxr ) − (αf,e − βf,e + γf,e )(dxr )

(4.57)

both sides are anti-symmetric under the switch e ↔ f . We make a choice αab = γab = 0. Then (4.57) gives, (βf,e − βe,f )(dxr ) = dxe δf,r (t(r)s(e) − c2 s(r)) − dxf δe,r (t(r)s(f ) − c2 s(r)) (4.58) and it is simplest to assume that βf,e = −βe,f giving, βf,e (dxr ) =

1 dxe δf,r (t(r)s(e) − c2 s(r)) − dxf δe,r (t(r)s(f ) − c2 s(r)) (4.59) 2

Why did we make the choice αab = γab = 0? First, it leaves only one function rather than two. Second, it does not change the right and left module structure. For functions f, g, φ(f, g, x) = 0 so we still have the left action, f . (g . x) = (f . g) . x up to order O(λ2 ). This means that up to O(λ2 ) a left module remains a left module. Likewise, a right module remains a right module to O(λ2 ). But we do not have, (f . x) .g = f . (x . g) so the condition for left and right actions to commute is changed by the associator. 54

Remark 4.0.1 Since the algebra is associative, it is reasonable to set the associator to be zero on the triple tensor product of the algebra, so we will assume that α, β, γ all vanish on functions. For f, g, h in the algebra µ(µ ⊗ id) (f ⊗ g) ⊗ h = (f g)h this should be the same as µ(id ⊗ µ)Φ (f ⊗ g) ⊗ h = f (gh) + λ2 µ(id ⊗ µ)φ(f ⊗ g ⊗ h) so µ(id ⊗ µ)φ(f ⊗ g ⊗ h) = 0. As examples of (4.59), putting f = i and e = 0, we get, 1 βi0 (dxr ) = − (c2 dx0 δi,r + δ0,r dxi ) 2

(4.60)

and putting f = i and e = j for i 6= j, we get, βij (dxr ) = so β12 is given by the matrix, 

c2 (δj,r dxi − δi,r dxj ) 2

(4.61)

 0 0     2 0 0 1 0 c    2  0 −1 0 0   0 0 0 0 0

0

and β01 is given by, 

0 c2 0 0



     1 1 0 0 0   2 0 0 0 0   0 0 0 0 Note that β12 (and βij ) are infinitesimal rotations (i.e., in the lie algebra of the rotation group). Also, β01 is a Lorentz boost, (in the Lie algebra of the Lorentz group). This means that we could apply our construction to give associativity at O(λ2 ) to any objects to which we can 55

(a) apply partial derivative in Minkowski space, and (b) is a representation of the Lorentz group at each each point of Minkowski space. Proposition 4.3.2. The difference at O(λ4 ) of going clockwise around Fig.8 minus going round anticlockwise is ((u ⊗ v) ⊗ w) ⊗ x maps to   ∂ 2u ∂w ∂x ⊗ βgd βab (v) ⊗ b ⊗ d   ∂xa ∂xg ∂x ∂x   2  ∂u ∂v ∂ x 4 λ  + ⊗ g ⊗ βgd βab (w) ⊗ a b  a   ∂x2 ∂x ∂x ∂x  ∂ u ∂ 2x  − g a ⊗ βgd (v) ⊗ βab (w) ⊗ d b ∂x ∂x ∂x ∂x

(4.62)

as this is not zero, the pentagon identity is not satisfied to O(λ4 ). Proof. ((u ⊗ v) ⊗ w) ⊗ x −→ (u ⊗ (v ⊗ w)) ⊗ x + λ2 (

∂w ∂u ⊗ (βab (v) ⊗ b )) ⊗ x a ∂x ∂x

The λ4 terms appearing when applying Φ clockwise are, ∂ 2u ∂w ∂x βab (v) ∂w ∂x ∂u ⊗ βgd (βab (v) ⊗ b ) ⊗ d + λ4 a ⊗ ⊗ βgd ( b ) ⊗ d g a g ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x 2 ∂u ∂x ∂ x +λ4 a ⊗ ( g ⊗ βgd )βab (v ⊗ w) ⊗ d b , ∂x ∂x ∂x ∂x

λ4

while the term we get anticlockwise is λ4 (

∂ ∂(u ⊗ v) ∂ ∂x ⊗ βgd ) ⊗ d (βab (w) ⊗ b ) . g a ∂x ∂x ∂x ∂x

then the difference at O(λ4 ) of going to the clockwise Fig.8 minus going anticlockwise will not satisfy the Pentagon identity.

We have presented an associator which restores associativity at O(λ2 ), but has possible problems at O(λ4 ). It is possible that there might be another associator which would be exact, but it is not obvious how to find it, should it exist. We shall talk about whether it is possible to do better than this in the final section. 56

4.4

Conjugate Modules

Let the module structure for conjugate bimodule be as in section 2.6, for f ∈ algebra, v . f = f∗ v

f . v = v f∗

,

and Υ(X ⊗ Y ) = Y ⊗ X defined by, Υ(x ⊗ y) = y ⊗ x

(4.63)

This choice of bar structure is the simplest we can make for a star algebra, and in section 2.6 we showed that Υ is well defined for ⊗A and is a bimodule map. Proposition 4.4.1. Υ is a bimodule map to order λ2 , i.e., Υ (x ⊗ y) . f

= Υ(x ⊗ y) . f

Υ f . (x ⊗ y)

= f . Υ(x ⊗ y)

when the following conditions are satisfied by φ, (id ⊗ µ)φ(y ⊗ x ⊗ f ) = −Υ((µ ⊗ id)φ(f ∗ ⊗ x ⊗ y))

(4.64)

(µ ⊗ id)φ(f ⊗ y ⊗ x) = −Υ((id ⊗ µ)φ(x ⊗ y ⊗ f ∗ ))

(4.65)

Proof. Note that in the usual formula f ∗ (x⊗y) = (f ∗ x)⊗y, we have changed the order of the bracketing, so we need to introduce the associator. Υ (x ⊗ y) . f

= Υ f ∗ (x ⊗ y) = Υ (f ∗ x) ⊗ y − λ2 (µ ⊗ id)φ(f ∗ ⊗ x ⊗ y) = y ⊗ f ∗ x − λ2 Υ((µ ⊗ id)φ(f ∗ ⊗ x ⊗ y))

Υ(x ⊗ y) . f = (y ⊗ x) . f = y ⊗ (x . f ) + λ2 (id ⊗ µ)φ(y ⊗ x ⊗ f )

57

so we get, (id ⊗ µ)φ(y ⊗ x ⊗ f ) = −Υ((µ ⊗ id)φ(f ∗ ⊗ x ⊗ y)) and for the second equation, Υ (f . x ⊗ y) = Υ (x ⊗ y) f ∗ = Υ x ⊗ (y f ∗ ) + λ2 (id ⊗ µ)φ(x ⊗ y ⊗ f ∗ ) = y f ∗ ⊗ x + λ2 Υ((id ⊗ µ)φ(x ⊗ y ⊗ f ∗ )) f . Υ(x ⊗ y) = f . (x ⊗ y) = (f . y) ⊗ x − λ2 (µ ⊗ id)φ(f ⊗ y ⊗ x) so we get: (µ ⊗ id)φ(f ⊗ y ⊗ x) = −Υ((id ⊗ µ)φ(x ⊗ y ⊗ f ∗ )) . Now using the form of φ in (4.54), we use the assumption

dy dy = a. a dx dx

Proposition 4.4.2. The conditions of Proposition 4.3.1. are satisfied by φ of the form (4.54), if ∂y ∂f ∂x ∂f ∂y ∂f ∂x ∂f ⊗ βab (x) b + αab (y) ⊗ a b = − b ⊗ βab (x) a − γab (y) ⊗ b a a ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x (4.66) ∂f ∂f ∂y ∂x ∂f ∂y ∂f ∂x β (y) + ⊗ γ (x) ⊗ ⊗ = − α (x) − β (y) ⊗ ab ab ab ab ∂xa ∂xb ∂xa ∂xb ∂xb ∂xa ∂xb ∂xa (4.67) Proof. For the first relation: ∂y ∂f ∂x ∂f ⊗ βab (x) b + αab (y) ⊗ a b a ∂x ∂x ∂x ∂x ∗ ∗ ∂f ∂y ∂f ∂x (µ ⊗ id)φ(f ∗ ⊗ x ⊗ y) = βab (x) ⊗ b + a b ⊗ γab (y) ∂xa ∂x ∂x ∂x ∗ ∂y ∂f ∂f ∗ ∂x ∗ ⊗ βab (x) + γab (y) ⊗ a b Υ(µ ⊗ id)φ(f ⊗ x ⊗ y) = ∂xb ∂xa ∂x ∂x (id ⊗ µ)φ(y ⊗ x ⊗ f ) =

58

∂y ∂y ∂x ∂f ∂f ∂x ∂f ∂f ⊗ β (x) + α (y) ⊗ = − ⊗ β (x) − γ (y) ⊗ ab ab ab ab ∂xa ∂xb ∂xa ∂xb ∂xb ∂xa ∂xb ∂xa Similarly for the second relation: ∂f ∂x ∂f y βab (y) ⊗ b + a b ⊗ γab (x) a ∂x ∂x ∂x ∂x ∗ ∂f ∂y ∂x ∂f ∗ (id ⊗ µ)φ(x ⊗ y ⊗ f ∗ ) = αab (x) ⊗ a b + a ⊗ βab (y) b ∂x ∂x ∂x ∂x ∗ ∗ ∂y ∂f ∂x ∂f Υ(id ⊗ µ)φ(x ⊗ y ⊗ f ∗ ) = ⊗ αab (x) + βab (y) b ⊗ a a b ∂x ∂x ∂x ∂x (µ ⊗ id)φ(f ⊗ y ⊗ x) =

∂x ∂f ∂y ∂f ∂f ∂f ∂y ∂x βab (y) ⊗ b + a b ⊗ γab (x) = − b a ⊗ αab (x) − b βab (y) ⊗ a a ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x If we look at β, ∂y ∂y ∂f ∂f ∗ ∂y ∂f ∗ ⊗ β (x) = −Υ( . β (x) ⊗ ) = − ⊗ . βab (x) (4.68) ab ab ∂xa ∂xb ∂xa ∂xb ∂xb ∂xa We need the following compatibility condition between Φ and Υ,

(X ⊗ Y ) ⊗ Z ΦX,Y,Z ΥX⊗Y,Z X ⊗ (Y ⊗ Z)

Z ⊗X ⊗Y idZ ⊗ ΥX,Y

ΥX,Y ⊗Z

Z ⊗ (Y ⊗ X)

Y ⊗Z ⊗X −1 ΦZ,Y ,X

ΥY,Z ⊗ idX (Z ⊗ Y ) ⊗ X

Fig. 9

59

Proposition 4.4.3. φ in (4.47) and Υ in (4.63) satisfy the compatibility condition to O(λ2 ) in Fig.8 if for φ(x, y, z) = u ⊗ v ⊗ w we have, w¯ ⊗ v¯ ⊗ u¯ = −φ(¯ z , y¯, x¯)

(4.69)

Proof. Taking only the O(λ2 ) part,

(x ⊗ y) ⊗ z φ(x, y, z)

z⊗x⊗y

u ⊗ (v ⊗ w)

z ⊗ (y ⊗ x)

v⊗w⊗u −φ(z, y, x)

(w ⊗ v) ⊗ u

Fig. 10 In this diagram we have taken the original x, y, z until we apply Φ, at which point we only keep track of the O(λ2 ) corrections. So for example the full form of the top right map is (x ⊗ y) ⊗ z 7→ x ⊗ (y ⊗ z) + λ2 u ⊗ (v ⊗ w) 7→ y ⊗ z ⊗ x + v ⊗ w ⊗ λ2 u 7→ (z ⊗ y) ⊗ x + (w ⊗ v) ⊗ λ2 u , and the full form of the top left map is (x ⊗ y) ⊗ z 7→ z ⊗ (x ⊗ y) 60

7→ z ⊗ (y ⊗ x) 7→ (z ⊗ y) ⊗ x + (w ⊗ v) ⊗ λ2 u .

Next we use the formula for φ including α, β and γ. If we substitute the formula for φ in (4.47), then the condition in Proposition 4.3.2. becomes, ∂z ∂y ∂y ∂x ⊗ a ⊗ αab (x) −αab (z) ⊗ a ⊗ b b ∂x ∂x ∂x ∂x ∂z ∂x ∂z ∂x = − a ⊗ βab (y) ⊗ b + b ⊗ βab (y) ⊗ a ∂x ∂x ∂x ∂x ∂y ∂z ∂y ∂x − a ⊗ b ⊗ γab (x) +γab (z) ⊗ b ⊗ a ∂x ∂x ∂x ∂x

(4.70)

∂z ∂z = , i.e., conjugate commute with differentiation. We swap a ∂x ∂xa ab on the RHS of (4.70), the easiest way to satisfy (4.70) is to set We need

βab (y) = −βba (y)

,

γab (z) = −αba (z)

(4.71)

αab (x) = −γba (x) Note the choice of φ is not unique, we make a choice which will work and which is simpler than some other choices. Now recall that in section 4.2. we choose α = γ = 0 and βab = −βba . If we make these same choices, then (4.71) becomes the condition, βab (y) = βab (y)

(4.72)

so β commutes with conjugate. As ? (dxe ) = dxe , we have: βab (? dxe ) = βab (dxe ) = βab (dxe )

(4.73)

and as βab (dxe ) is a real number times dxe , we have: ? βab (dxe ) = βab (dxe )

(4.74)

so we see that ? βab = βab ?. We can consistently extend the definition of β on 1-forms given in section 4.2.3. to their conjugates to satisfy this, using (4.72). 61

4.5

Levi-Civita and covariant derivative

We shall look at the noncommutative analogue of the Levi-Civita connection on the deformed algebra. We suppose that E has a left covariant derivative ∇ : E −→ Ω1 A ⊗A E . The conjugate E of a bimodule E has elements e ∈ E for e ∈ E and action for a ∈ A, e . a = a∗ e + O(λ2 ),

a . e = ea∗ + O(λ2 ) ,

(4.75)

and for another bimodule F , Υ : E ⊗A F −→ F ⊗A E is defined as, Υ(e ⊗ f ) = f ⊗ e + O(λ2 )

(4.76)

Then the conjugate E has a right covariant derivative ∇ : E −→ E ⊗A Ω1 A , given by ∇(e) = f ⊗ ξ ∗ for ∇(e) = ξ ⊗ f (see [12]). This is given by the formula e −→ (id ⊗ ?−1 )Υ ∇e , and it is a right connection as, e . a −→ (id ⊗ ?−1 )Υ ∇(a∗ e) −→ (id ⊗ ?−1 )Υ (da∗ e + a∗ ∇e) −→ (id ⊗ ?−1 )(e ⊗ da∗ ) + (id ⊗ ?−1 )Υ(∇e) . a −→ e ⊗ da + (id ⊗ ?−1 )Υ(∇e) . a

(4.77)

The Levi-Civita connection preserves the inner product. For E = Ω1 A we can write this as,

62

ξ

⊗

η

(1)

∇ h, i

h, i

=

(2)

+

∇ h, i

d

Fig. 11 We write the connection ∇ in terms of Christoffel symbols, ∇dxa = Γabc dxb ⊗ dxc

(4.78)

and this gives a similar form for ∇, (id ⊗ ?−1 )Υ∇dxa = (id ⊗ ?−1 )(dxc ⊗ Γabc dxb ) = dxc ⊗ dxb (Γabc )∗

(4.79)

Apply this equation to hdxa , dxe i, where this is a number and applying d to it gives zero. When applying (1), we get, Γabc dxb ⊗ hdxc , dxe i = Γabc dxb g ce

(4.80)

hdxa , dxc i ⊗ dxb (Γebc )∗ = g ac dxb (Γebc )∗

(4.81)

applying (2), we get,

then as g is constant, Fig. 11 gives, 0 = Γabc dxb g ce + g ac dxb (Γebc )∗

(4.82)

In principle, the Γ have O(λ) corrections, however as in the classical case we can choose Γabc = 0 as the Levi-Civita connection. Of course, classically we have no alternative, but in principle we could choose to have an imaginary O(λ) correction to solve (4.82), but to preserve reality we do not do this. We now continue, using Γabc = 0. 63

4.5.1

Differentiation and Normal Orders

We need to calculate the differential of a function in a particular direction. As vector fields act on functions on the left, we order the dxi to the left of the expressions. d(x1 )2 = dx1 . x1 + x1 . dx1 = 2dx1 . x1 + [x1 , dx1 ] = 2dx1 . x1 + c2 λdx0

(4.83)

d(x1 )3 = dx1 . (x1 )2 + x1 . d(x1 )2 = dx1 . (x1 )2 + x1 (2dx1 . x1 ) + x1 c2 λdx0 = 3dx1 . (x1 )2 + 2[x1 , dx1 ]x1 + c2 λdx0 . x1 = 3dx1 . (x1 )2 + 3 c2 λdx0 . x1

(4.84)

so to order λ, d(x1 )n = n . dx1 . (x1 )n−1 +

d(x1 )n = dx1 .

n (n − 1) c2 λ . dx0 . (x1 )n−2 2

(4.85)

2 1 n ∂(x1 )n c2 0 ∂ (x ) + λ dx x1 2 ∂(x1 )2

(4.86)

∂f c2 ∂ 2f 0 + λ dx ∂x1 2 ∂(x1 )2

(4.87)

for a function f (x1 ), df (x1 ) = dx1 .

If we havef (x1 )g(x2 ), then 2 ∂f (x1 ) c2 ∂2 2 0 2 1 2 ∂g(x ) d f (x1 ) g(x2 ) = dx1 g(x ) + λ dx g(x ) + f (x ) dx ∂x1 2 (∂x1 )2 ∂x2 ∂ 2 g(x2 ) c2 + λ f (x1 ) dx0 2 ∂(x2 )2 2 c2 ∂ (f g) ∂ 2 (f g) 1 ∂(f g) 2 ∂(f g) 0 = dx + dx + λ dx + ∂x1 ∂x2 2 ∂(x1 )2 ∂(x2 )2 (4.88) 64

Similarly, we can multiply by a function of x3 . The result is that if θ(x1 , x2 , x3 ) is a function of {x1 , x2 , x3 } only (these all commute so we can forget about order), then dθ =

X

dxi

i

X ∂ 2θ ∂θ c2 0 λdx + ∂xi 2 ∂(xi )2 i

(4.89)

For φ(x0 ), we get the simpler result dφ = dx0

∂φ ∂x0

Now with normal order for ψ = φθ d(φθ) = dφ . θ + φ . dθ X ∂φ c2 X ∂ 2 θi ∂θ = dx0 ( 0 θ + λ φ ) + dxi φ i i 2 ∂x 2 ∂(x ) ∂x i i

(4.90)

The commutator of the functions of the spatial and temporal variables is, to O(λ) [θ(x1 , x2 , x3 ), φ(x0 )] = λ

∂φ X i ∂θ . x ∂x0 i ∂xi

(4.91)

If φ is a function of time and θ a function of space, then : φ θ : is an ordinary function on space time whose value is : φ θ : (x0 , x1 , x2 , x3 ) = φ(x0 ) θ(x1 , x2 , x3 ). The cost of this identification is that the product is no longer the ordinary product of functions on space time: for ψ, χ ∈ A, (4.91) gives to O(λ) : ψ χ : = : ψ : : χ : +λ

X i

Applying

xi

∂ :ψ : ∂ :χ: . ∂xi ∂x0

(4.92)

∂ to (4.90), we get ∂x0 ”

∂ψ ∂ ∂φ c2 X ∂ 2 θ i ” = eval( ⊗ dψ) = θ + λ φ ∂x0 ∂x0 ∂x0 2 ∂(xi )2 i

65

(4.93)

In terms of the normal order, :

∂Ψ ∂ : Ψ : c2 X ∂ 2 : Ψ : : = + λ ∂x0 ∂x0 2 ∂(xi )2 i

(4.94)

:

∂Ψ ∂ :Ψ: : = i ∂x ∂xi

(4.95)

: Ψ∗ : = : Ψ :∗ +λ

X

xi

∂ 2 : Ψ :∗ ∂x0 ∂xi

(4.96)

Proposition 4.5.1. There is a 1-1 correspondence between 1) ξ ∈ Ω1 A with ξ ∗ = −ξ 2) α classical anti Hermitian 1-form by: : ξ := α − λ T1 (α)

,

α=

: ξ : − : ξ :∗ 2

(4.97)

where 2T1 (α) =

2 X ∂α0 i 2 ∂αi 0 1 ∂ αa dx + c dx − x dxa ∂xi ∂xi ∂x0 ∂xi

Proof. Start with : ξ ∗ : = : ξ :∗ +λ −λ

X

xi

∂ : ξ0 :∗ i dx ∂xi

∗ ∂ 2 : ξa :∗ a 2 ∂ : ξi : dx − c λ dx0 0 i i ∂x ∂x ∂x

(4.98)

If ξ ∗ = −ξ, then: ∗ ∂ 2 : ξa :∗ a 2 ∂ : ξi : dx − c λ dx0 0 i i ∂x ∂x ∂x ∂ : ξ0 :∗ i −λ dx ∂xi

− : ξ : − : ξ :∗ = λ

X

xi

(4.99)

Now, use: : ξ :=

: ξ : − : ξ :∗ : ξ : + : ξ :∗ + 2 2

(4.100)

so we see that : ξ := α + O(λ), so in the RHS of (4.100) we may replace : ξ : by α with an error of O(λ2 ) and this gives the result. 66

Proposition 4.5.2. There is a 1-1 correspondence between 1) ω ∈ Ω2 A with ω ∗ = −ω 2) a classical antiHermitian 2-form β by: : ω := β − λ T2 (β)

(4.101)

where ∂ 2 βa ∂βi dxa ∧ dxb + c2 i dx0 ∧ dxb 0 i ∂x ∂x ∂x ∂β0 i ∂β i + i dx ∧ dxb + c2 i dxa ∧ dx0 ∂x ∂x ∂β0 a + i dx ∧ dxi ∂x

2T2 (β) = −

X

xi

Proof. Start with ∗ ∂ 2 : ωa :∗ a b 2 ∂ : ωi : dx ∧ dx − c λ dx0 ∧ dxb ∂x0 ∂xi ∂xi ∗ ∂ : ω0 :∗ i b 2 ∂ : ωi : −λ dx ∧ dx − c λ dxa ∧ dx0 ∂xi ∂xi ∂ : ω0 :∗ a −λ dx ∧ dxi (4.102) ∂xi

: ω ∗ : = : ω :∗ +λ

X

xi

If ω ∗ = −ω, then: ∗ ∂ 2 : ωa :∗ a b 2 ∂ : ωi : dx ∧ dx − c λ dx0 ∧ dxb ∂x0 ∂xi ∂xi ∗ ∂ : ω0 :∗ i b 2 ∂ : ωi : −λ dx ∧ dx − c λ dxa ∧ dx0 ∂xi ∂xi ∂ : ω 0 :∗ a −λ dx ∧ dxi (4.103) ∂xi

− : ω : − : ω :∗ = λ

X

xi

Now, use: 2β =: ω : − : ω :∗

(4.104)

so we see that : ω := β + O(λ), so in the RHS of (4.104) we may replace : ω : by β with an error of O(λ2 ) and this gives the result.

67

4.5.2

Motion Along a Constant Vector Field

Classically, motion along a constant velocity field is given by (for velocity in ∂ ∂ the x1 direction) ( + v )(f ) = 0, for a function f (t, x1 ). Classically, 1 ∂t ∂x this has solutions for any functions of x1 − v x0 . Now solve this equation in the noncommutative case, for normal order and Ψ ∈ A, we solve, :

∂Ψ ∂Ψ : +v : 1 := 0 0 ∂x ∂x

(4.105)

Using (4.94) and (4.95), this becomes, ∂ : Ψ : c2 X ∂ 2 : Ψ : ∂ :Ψ: + λ +v =0 0 i 2 ∂x 2 ∂(x ) ∂x1 i

(4.106)

1

For a plane wave solution, set : Ψ := ei(ωt+α1 x ) , then (4.106) becomes, iω −

c2 λ 2 α + ivα1 = 0 2 1

(4.107)

As the equation is linear, we can sum these plane waves with coefficients to get a more general solution. In practice, we would have a Fourier expansion of : Ψ : at t = 0, and then add the ω t term to get time evolution. For v > 0, we have the following behaviour of ω with respect to α1 , c2 λ 2 ω = −vα1 + α 2i 1

(4.108)

We plot ω along the vertical axis, and α1 along the horizontal axis. The classical case is (c).

(b)λ = −i

(a)λ = i 68

(c)λ = 0 Fig. 12 2iv . Later c2 λ in section 5.1 we shall argue for a value which is −i times a small positive The place where ω cuts the α1 axis, apart from α1 = 0 is α1 =

number, putting us in case (b).

69

Chapter 5 The Schr¨ odinger equation and propagation speeds There is much discussion in the literature about looking for variation in the velocity of light, or various cosmic rays, and about possible origins of this variation in noncommutative geometry. See [1][3][4][5] for a discussion of observations.

5.1

The Schr¨ odinger equation and effective mass

Now put Ψ ∈ A in Schrödinger equation (3.10). We use the noncommutative partial derivative in (4.94) and (4.95), and use normal order to O(λ) to get, −~2 ∂ 2 : Ψ : ∂ : Ψ : i λ c2 ~ X ∂ 2 : Ψ : + = + V :Ψ: i~ ∂x0 2 ∂(xi )2 2m ∂(xi )2 i

(5.1)

which can be rearranged to give, i~

∂ :Ψ: −~2 i λ 2 2 =( − ~ c )∇ : Ψ : +V : Ψ : ∂t 2m 2 70

(5.2)

λ is imaginary, so i λ is real.

This looks just the same as the original

Schrödinger equation, but with a shifted or effective mass m0 . We get, −~2 −~2 i λ 2 − ~c = 2m0 2m 2 1 i λ c2 1 + = m0 m ~ m 0 m = i λ c2 m 1+ ~

(5.3)

The behaviour of (5.3) gives an effective mass m0 for a given initial mass m. We will discuss this more later, now we just look at the graphs of m0 along the vertical axis vs. m along the horizontal axis.

Fig. 13 graphs of m0 vs. m ~ | = 1 and the first graph has i λ > 0 and the second i λ < 0. i λ c2 ~ Of these graphs, the i λ > 0 has a limiting value of m0 of as m −→ ∞. i λ c2 ~ The i λ < 0 graph has a singular value at m = − . Which one shall we i λ c2 choose? If we add two small masses m and n, then, We suppose |

i λ c2 m2 i λ c2 n2 2i λ c2 mn +n− − (m + n) + ~ ~ ~ 2i λ c2 mn ' (5.4) ~

m0 + n0 − (m + n)0 ' m −

This can be thought as combining two point particles to make another point particle. The equation (5.4) gives the effective mass loss by combining these particles. This corresponds to an energy loss by E = mc2 . Energy loss on 71

combining particles can be thought of as a binding energy, in this case it is 2 i λ m n c4 . One effect we get is that for large values of m, we get a limiting ~ ~ behaviour for the effective mass m0 where mCRIT = 2 . c |λ| We take the natural choice of positive binding energy, and take i λ > 0. Now ~ we have a limiting value of m0 as m −→ ∞ of . The Planck mass is i λ c2 r ~c . the supposed mass of the heaviest possible point particle, and is mP = G Putting the limiting mass equal to the Planck mass gives r ~c ~ = 2 G c |λ| r ~G |λ| = ' 5.38 × 10−44 seconds (5.5) c5 This comes from the following values: G = 6.67 × 10−11 m3 kg −1 s−2 c2 = 299792458 m/s ~ = 6.58 × 10−16 eV × seconds We also get an estimate of |λ| from (5.4). If we take masses m and n from a long distance away to a distance r apart, we get a binding energy given mnG by gravitational potential energy . If we put this equal to the binding r energy from (5.4), then: 2|λ|mnc4 mnG = ~ r G~ |λ| = 2c4 r

(5.6)

If we assume that m and n have been combined into a ”point particle”, we can assume that we have r the smallest possible distance, the Planck length, ~G , and then (5.6) becomes as a radius r = LP = c3 r G~ c3/2 1 G~ √ |λ| = = ' 2.69 × 10−44 seconds (5.7) 2c4 ~G 2 c5/2 72

The difference in these values (5.5) and (5.7) is not too surprising, the idea of Planck mass and Planck length are not precisely defined. For the sake of argument, we shall adopt a likely very error prone value λ ' −5 i × 10−44 seconds

(5.8)

For a discussion of problems with measurement at the Planck scale see [29].

5.2

Plane waves

The Fourier transform of a wave is given by Φ=

X

bα ei(ωt+α1 x1 +α2 x2 +α3 x3 ) ,

(5.9)

we use this in the standard wave equation 1 ∂ 2Φ − ∇2 Φ = 0 , c2 ∂t2

(5.10)

and observe the effect, −

ω 2 i(ωt+αx) e + (α12 + α22 + α32 )ei(ωt+αx) = 0 , c2

(5.11)

so, ω 2 = c2 (α12 + α22 + α32 ) .

(5.12)

where c is the speed of propagation of waves. In the noncommutative case, we set Φ ∈ A, and in the classical wave equation, we use the noncommutative partial derivatives (4.94) and (4.95) to O(λ), 1 ∂ c2 λ 2 ∂ c2 λ 2 ( + ∇ )( + ∇ ) : Φ : −∇2 : Φ := 0 c2 ∂x0 2 ∂x0 2

(5.13)

1 ∂2 : Φ : ∂ :Φ: − ∇2 : Φ : +λ∇2 =0 2 0 2 c ∂(x ) ∂x0

(5.14)

so we find

73

substituting : Φ := ei(ωt+αx) , we get, −

ω 2 i(ωt+αx) e + (α12 + α22 + α32 )ei(ωt+αx) − iλω(α12 + α22 + α32 )ei(ωt+αx) = 0 (5.15) c2 ω 2 = (α12 + α22 + α32 )(c2 − i λ c2 ω)

(5.16)

and this equation replaces (5.12). Now, |α2 | = α12 + α22 + α32 , so ω 2 + i λ c2 ω|α|2 − c2 |α2 | = 0 p |α|4 (i λ)2 c4 + 4 c2 |α|2 ω = − 2 p |α| (−i λ c2 |α| ± (i λ)2 c4 |α|2 + 4 c2 ) = 2 ω2 |α|2 = 2 c (1 − i λ ω) i λ c2 |α|2 ±

(5.17)

In Fig. 14, we plot |α|2 (y axis) against ω (x axis) in two cases, the left case where iλ = 1 (iλ > 0 cases are similar) and the right case where iλ = −1 (iλ < 0 cases are similar).

Fig. 14 1 . Note that |α|2 is the vector |λ| norm, not using a Hermitian product, so |α|2 < 0 is possible by definition, as

There is a critical value of ω, |ωCRIT | =

it is a sum of squares. However α imaginary gives a non-oscillatory (in fact an unbounded) result for Φ in (5.9). The wave equation (5.14) depends on : Φ : being complex values. This means that (5.14) likely does not apply to light, and we will discuss this later 74

on. But it does apply to the massless Klein - Gordon equation, where the energy and momentum operations are E = i~

∂ ∂ , Pi = −i ~ i ∂t ∂x

(5.18)

The energy of the plane wave solution is given by the eigenvalue of E applied to the plane wave, ∂Φ ∂t ∂ : Φ : c2 λ X ∂ 2 : Φ : : EΦ : = i~( + ) ∂t 2 ∂(xi )2 i2 λ c2 2 |α| : Φ :) = i~(iω : Φ : + 2 iλ = −~(ω + c2 |α|2 ) : Φ : 2 EΦ = i~

(5.19)

so the energy of the plane wave is, E = −~(ω +

iλ 2 2 c |α| ) 2

(5.20)

We may as well use the massive Klein - Gordon equation m 2 c2 1 ∂ 2Φ X ∂ 2Φ − + Φ=0 c2 ∂t2 ∂(xi )2 ~2

(5.21)

and we get, instead of (5.11), m 2 c2 −ω 2 2 2 + |α| − i λ ω|α| + =0 c2 ~2 |α|2 (1 − iλω) =

(5.22)

ω 2 m2 c2 − 2 c2 ~ (5.23) 2 4

mc ~2 1 − iλω

ω2 − c2 |α|2 =

75

(5.24)

We can put this into the energy of the plane wave to give, m2 c4 2 ω − iλ ~2 ) E = −~(ω + 2 1 − iλω iλω 2 iλm2 c4 ω− − 2 2~2 E = −~ 1 − iλω For the physical meaning of α we use P1 ei(ωt+α1 x

1 +α

2x

(5.25) 2 +α

3x

3)

= ~α1 , so ~α is

the momentum. We plot c2 |α|2 and E on the vertical axis and ω on the horizontal axis for various choices of constants. For a physical solution, we need E ≥ 0. If |α|2 < 0 we get an exponentially increasing solution.

Fig. 15: c2 |α|2 and E vs. ω, m = 0 case For m = 0, we get E ≥ 0 and |α|2 ≥ 0 for ω ≤ 0. We also get positive energy 1 above a critical value of ω = , but |α|2 < 0 here, so there are problems iλ with these solutions. Fig. 15 is a graph of c2 |α|2 (left graph) and E (right graph) vs. ω for the mass less case, with iλ = 1 and ~ = 1. r

(iλ)2 m2 c4 ~2 . We have different The zeros of E for m > 0 are at ω = iλ |λ|m c2 |λ|m c2 behaviours, depending on whether > 1 has no roots, < 1 has ~ ~ 2 |λ|m c 2 roots, and = 1 has 1 root. ~ 1±

76

1−

|λ|m c2 > 1 , we get Fig. 16, where we plot c2 |α|2 (left) and E (right) ~ on the y axis and ω on the x axis, in the case m = 2. Fig. 17 is the same,

For

except that m = 21 .

Fig. 16: c2 |α|2 and E vs. ω, m = 2 case For this case, the mass is very big-likely not realistic.

1 case 2 −mc The only cases with |α|2 ≥ 0 and E ≥ 0 are ω < . ~ Observing λ: Note that the predicted values of |λ| are very small. We want Fig. 17: c2 |α|2 and E vs. ω, m =

to observe the effect of λ 6= 0 versus λ = 0 by measuring a physical effect. The speed of a plane wave (set α2 = α3 = 0), is ei(ωt+α1 x) . For a wave moving at velocity v, we would have eik(x−vt) ,so v=−

ω α1

,

|v|2 =

77

ω2 |α|2

(5.26)

to compare small differences, we use m = 0, so if λ = 0 we get |v| = c, independently of ω. For λ 6= 0, we get, |v| =

√ ω = c 1 − iλω |α|

(5.27)

and if iλ > 0 and ω < 0, we get a wave travelling faster than c. We make the waves travel a known distance D. If 1. ω is small, time taken is

D c

D 2. ω is large, time taken is √ c 1 − iλω and the time difference measured is: 1 iλωD D D D = (1 − (1 − iλω)− 2 ) ' − √ (5.28) c c 2c c 1 − iλω D is the time taken for light to move the given distance. To measure this c

δ = time difference =

|λ||ω| × time taken to travel 2

(5.29)

The largest distance we can observe objects to in the universe have light travel times about 10 billion years. Call this travel time T .

E = −~ω for ω small compared to

1 . |λ| δ=

|λ|ET 2~

(5.30)

E is the energy of the waves that we would need to observe.

E=

2~δ |λ|T

78

(5.31)

We put in T = 1010 × 365 × 24 × 60 × 60seconds for a travel time of 1010 years. δ = 1seconds |λ| = 5 × 10−44 seconds ~ = 6.58 × 10−16 eV × seconds so, in eV (electron volts): E ' 8.34 × 1010 eV

(5.32)

The collisions at CERN occur at much higher energies than this, at about 1T eV = 1012 eV [14] There are several accounts of possible measurements which could be made using noncommutative geometry on the speed of high energy cosmic rays [1][3][4][5]. The source of most very high energy cosmic rays is likely supernova remnants [2], and there is expected to be a limit of about 5×1019 eV on energies due to interaction with the cosmic microwave background [16]. In [5] there is an analysis of wave propagation in κ-Minkowski by using Fourier transforms, which gives the speed of a massless particle v=

1 q = e−λω ' 1 − λω , → − λω 1 + |λω| λ2 k 2

(5.33)

as 1 + iλω (we have taken account of a different λ convention). Our equation is |v| √ iλω = 1 − iλω ' 1 − , c 2

(5.34)

where ω is negative to give positive energy. Since we deduce on a value iλ > 0 in (5.8), we would predict a velocity greater than the speed of light, but the difference is half of that in [5]. 79

5.3

Verifying the change in propagation speed for the massless KG equation

Fig. 18 Recalling that the phase velocity may not be the same as velocity may not be the same as the propagation speed, as pointed out in section 3.2.3, we take care to look at localised initial conditions. A standard wave equation has a finite propagation speed, illustrated by the fact that the observer at t =

1 2

can not see any waves from the stone at all. This is the ”group velocity” in the literature about the Klein-Gordon equation [15] [66]. We try to find the propagation speed of the equation by making a wave with initial shape with a zero region. We use the periodic saw tooth wave, illustrated in Fig. 19.

80

Fig. 19: Periodic Saw tooth wave, t = 0 This is given by the Fourier cosine series

a0 P π + an cos(nx), where a0 = 2 4

and  2   n odd    πn2 an = 0 4 divides n    4   n = 4m + 2 πn2 If we use the standard wave equation with c = 1, we get a time variation φ(x, t) =

a0 X + an cos(nx)e−int 2

(5.35)

Fig. 20: Plot at t = 21 , for the usual KG equation In Fig. 20, the original t = 0 graph is shown in lighter lines, and t =

1 2

graph

in darker lines. The equation which is plotted is (5.35) and was done using Mathematica command 81

Plot[{Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n /2], {n, 1, 50}]], Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n 0/2], {n, 1, 30}]]}, {x, 0, 2 Pi}, PlotPoints -> {100}, PlotStyle -> {Thickness[0.005], GrayLevel[-1]}] Now increase time to t = 1 shows that the region in the middle, where there is no signal, is smaller but still there. This is shown in Fig. 21. Note that we have chosen ω < 0 so that the energy of each component of the massless KG equation is positive. This is a principle that we also use in the deformed case.

Fig. 21 This was done using Mathematica command Plot[{Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n ], {n, 1, 100}]], Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n 0/2], {n, 1, 30}]]}, {x, 0, 2 Pi}, PlotPoints -> {200}, PlotStyle -> {Thickness[0.005], GrayLevel[-1]}] For the noncommutative case, if we put c = 1 and iλ = 1 in the wave equation, and |α| = n, we get the equation ω=

√ −n (n ∓ 4 + n2 ) . 2

82

We use the energy relation (5.20) to get E = −~(ω + n2 ) = −~

! √ n2 n 4 + n2 n~ √ ± = ∓ 4 + n2 − n . 2 2 2

To be positive energy, we must take the bottom sign, so ω=

√ −n n + 4 + n2 . 2

and we put this ω into the Mathematica code for our previous plots, to give Fig. 22 at t = 1.

Fig. 22 This was done using Mathematica command Plot[{Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n/2 (n + Sqrt[4 + n^2]) ], {n, 1, 100}]], Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n 0/2], {n, 1, 30}]]}, {x, 0, 2 Pi}, PlotPoints -> {200}, PlotStyle -> {Thickness[0.005], GrayLevel[-1]}] In the plot Fig. 22 we do not have a flat part in the middle, so it appears that we have a higher (possibly infinite) propagation speed. However this is not a proof. To make the calculation more certain, we can estimate the truncation P error in terminating the series. We have a sum an gn (x, t) where every

83

|gn (x, t)| ≤ 1. The error of truncating the series at n = 100 is error ≤ ≤ ≤

∞ X

|an ||gn (x, t)| n=101 ∞ X 4 π2

1 n2 n=101

4 × 0.00995 π2

so we have the value in the middle of Fig. 22 is definitely not zero. i.e., we do not have a propagation speed of 1. Ofcourse, we are still placing assumptions on the numerical accuracy of the computer. We also show the time variation of the solutions in the next graphs. The classical solution for 0 ≤ t ≤ 1 (time 0 at the left, time 1 at the right) is shown in Fig. 23.

Fig. 23 Standard wave equation A plot with time of the imaginary part of φ (left) and real part of φ (right) is given in Fig. 23. This was done by using the Mathematica commands Plot3D[Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n t], {n, 1, 30}]], {x, 0, 2 Pi}, {t, 0, 1}, PlotPoints -> {40, 40}] Plot3D[Im[Pi/8 + Sum [f[x, n]/2 Exp[-I n t], {n, 1, 30}]], {x, 0, 2 Pi}, {t, 0, 1}, PlotPoints -> {40, 40}]

84

The noncommutative massless KG equation is plotted in Fig. 24 with all other parameters for the graphs the same as Fig. 23.

Fig. 24 Modified wave equation This was done by using the Mathematica commands Plot3D[Re[Pi/8 + Sum [f[x, n]/2 Exp[-I n/2 (n + Sqrt[4 + n^2]) t], {n, 1, 30}]], {x, 0,2 Pi}, {t, 0, 1}, PlotPoints -> {40, 40}] Plot3D[Im[Pi/8 + Sum [f[x, n]/2 Exp[-I n/2 (n + Sqrt[4 + n^2]) t], {n, 1,30}]], {x, 0, 2 Pi}, {t, 0, 1}, PlotPoints -> {40, 40}] The value i λ = 1 was chosen, which is much larger than any physical value, but the idea is to show what the difference is on a reasonable time scale. For a nonperiodic example we use a bell shaped curve 1/(1 + x8 ) at time zero, and plot the real part of the KG solution, with time going to the right and into the paper. We need to use Fourier transformations for this. We plot the noncommutative case in Fig. 25, and the usual case in Fig. 26. We use the same numerical scheme just for comparison. We plot 0 ≤ t ≤ 5 and −10 ≤ x ≤ 10.

85

Fig. 25 Plot3D[Re[1/4 (-1)^(1/8) 1/2 (NIntegrate[(E^((-1)^(1/8) \[Alpha]) + (-1)^(1/4)E^((-1)^(3/8) \[Alpha]) - I E^(-(-1)^(5/8) \[Alpha]) - (-1)^(3/4)E^(-(-1)^(7/8) \[Alpha])) Exp[I \[Alpha]/2 ( -\[Alpha] + Sqrt[4 + \[Alpha]^2]) t] Exp[-I

\[Alpha] x],

{\[Alpha], -\[Infinity], 0}] + NIntegrate[(E^(-(-1)^(1/8) \[Alpha]) + (-1)^(1/4) E^(-(-1)^(3/8) \[Alpha]) I E^((-1)^(5/8) \[Alpha]) - (-1)^(3/4) E^((-1)^(7/8) \[Alpha])) Exp[-I \[Alpha]/ 2 ( \[Alpha] + Sqrt[4 + \[Alpha]^2]) t] Exp[-I\[Alpha] x], {\[Alpha], 0, \[Infinity]}])], {x, -10, 10}, {t, 0, 5}, PlotPoints -> {40, 40}]

86

Fig. 26 Plot3D[Re[ 1/4 (-1)^(1/8) 1/ 2 (NIntegrate[(E^((-1)^(1/8) \[Alpha]) + (-1)^(1/4) E^((-1)^(3/8) \[Alpha]) I E^(-(-1)^(5/8) \[Alpha]) - (-1)^(3/4) E^(-(-1)^(7/8) \[Alpha])) Exp[ I \[Alpha] t] Exp[-I

\[Alpha] x], {\[Alpha], -\[Infinity],

0}] + NIntegrate[(E^(-(-1)^(1/8) \[Alpha]) + (-1)^(1/4) E^(-(-1)^(3/8) \[Alpha]) I E^((-1)^(5/8) \[Alpha]) - (-1)^(3/4) E^((-1)^( 7/8) \[Alpha])) Exp[-I \[Alpha] t] Exp[-I \[Alpha], 0, \[Infinity]}])], {x, -10, 10}, {t, 0, 5}, PlotPoints -> {40, 40}]

87

\[Alpha] x], {\

Chapter 6 Electromagnetism 6.1

Curvature at O(λ)

In this chapter we shall in places quote from the joint paper [14]. There are several works on gauge theory in κ-Minkowski using the more usual 4D calculus [26][29][38][57][62]. In [52], Majid uses the 5D calculus on κ-Minkowski. We shall begin with looking at gauge theory on the trivial bundle with gauge group S 1 (Electromagnetism). For a left A module A, we write a left covariant derivative ∇(1) = ξ ⊗ 1

(6.1)

∇(a) = ∇(a . 1) = da ⊗ 1 + a . ∇(1) = da ⊗ 1 + aξ ⊗ 1

(6.2)

and then

Then the curvature R(1) is given by R(1) = dξ ⊗ 1 − ξ ∧ ∇(1) = (dξ − ξ ∧ ξ) ⊗ 1

88

(6.3)

The curvature is a left module map for all a ∈ A R(a . e) = (d ⊗ id − id ∧ ∇)(aη ⊗ f ) = d(aη) ⊗ f − aη ∧ k ⊗ g = a . dη ⊗ f − aη ∧ k ⊗ g = a . R(e)

(6.4)

where ∇E (e) = η ⊗ f, ∇E (f ) = k ⊗ g. For a 1-form ξ = ξa dxa = φa θa dxa , then to order O(λ) in the normal ordered form (with φa (t) and θa (x, y, z)) we have, ∂θa λc2 X ∂φa X ∂θa i ∂φa ∂ 2 θa 0 a θ − dx ∧dx + φ −λ dx ∧dxa dξ = φ a a a ∂x0 2 ∂(xi )2 ∂xi ∂x0 ∂xi (6.5) and ∂φb X i ∂θa ∂θb θb dxa ∧ dxb − c2 λ φi θi φb dx0 ∧ dxb x 0 i ∂x ∂x ∂xi ∂θb dxi ∧ dxb (6.6) −λ φ0 θ0 φb ∂xi

ξ ∧ ξ = λ φa

The curvature has electric and magnetic fields (3.27). Classically ∇(1) = iqAa dxa and we define Fab =

∂Ab ∂Aa − . ∂xa ∂xb

The current jb and electromagnetic fields are related by: jb =

∂F ab ∂xa

(6.7)

we raise indices using the metric g, 0 0

0

0

F a b = g aa g bb Fab

(6.8)

To make this work in noncommutative geometry, we assume that we still need a real Hermitian current, and arrange the formula to ensure this. 89

6.2

Finding the current in the noncommutative case

The following account, for the rest of the section, appeared in the archive [14] in a joint paper with my supervisor. We need a version of (6.7) that makes sense for noncommutative geometry. The guiding principle is that F is derived from the curvature R(1), which is an antiHermitian 2-form, and that j should be Hermitian, i.e. classically a real current. It is easier for us to consider everything to be a form, rather than vector fields, as there is a direct meaning of a form being Hermitian. Now we could always impose j being Hermitian, simply by giving some formula and averaging it with its star. However we would like a definition for the divergence which is natural as possible, given the Hermitian constraint. The key to this working is the choice of antisymmetry in the definition of the Fab . Classically this is obvious, but it really depends on the fact that dxa and dxb anticommute. In fact, if this did not happen it would not be at all obvious what to do. In general, this would be related to the properties of a splitting map from 2-forms to the tensor product of two 1-forms (or equivalently, the interior product of a vector field with a 2-form). This is related to other things, such as a formula for the Ricci tensor, and is not well understood. Suppose that h, i : Ω1 A ⊗A Ω1 A → A is a Hermitian form, i.e. it is an A-bimodule map and hξ, ηi∗ = hη, ξi .

(6.9)

We use the notation hdxa , dxb i = g ab . Definition 6.2.1. Define the following operations K, S : Ω1 A ⊗A Ω1 A −→

90

Ω1 A for a left covariant derivative ∇ on Ω1 A, K = (h, i ⊗ id)(∇ ⊗ id + id ⊗ ∇)

(6.10)

S = (id ⊗ h, i)(?−1 ⊗ ? ⊗ id)(∇ ⊗ id + id ⊗ ∇)

(6.11)

Note that we use the standard formula for a right covariant derivative ∇ on Ω1 A [12], ∇(η) = (id ⊗ ?−1 )Υ∇η

(6.12)

Proposition 6.2.1. For η1 ⊗ η2 ∈ Ω1 A ⊗A Ω1 A, S(η2 ⊗ η1 ) = K(η1 ⊗ η2 )∗

(6.13)

Proof. Apply K to η1 ⊗ η2 , where ∇ηi = κi ⊗ ζi , K(η1 ⊗ η2 ) = hζ1 , κ∗1 iη2 + hη1 , κ2 iζ2

(6.14)

K(η1 ⊗ η2 )∗ = η2∗ hκ∗1 , ζ1 i + ζ2∗ hκ2 , η1 i

(6.15)

Now, applying S to η2 ⊗ η1 , we get, S(η2 ⊗ η1 ) = K(η1 ⊗ η2 )∗ . Corollary 6.2.1. If η1∗ ⊗ η2 = η2∗ ⊗ η1 ∈ Ω1 A ⊗ Ω1 A, then (K − S)(η1 ⊗ η2 ) is antiHermitian. Now we can try to make a noncommutative version of (6.7). If we take a real multiple of R(1) to be η1∗ ∧ η2 , then if η1∗ ⊗ η2 obeys the condition of Corollary 6.2.1 then (K − S)(η1 ⊗ η2 ) is antiHermitian trace of the derivative. We could use this to define a Hermitian current simply by multiplying by i. Our problem is to satisfy the conditions of Corollary 6.2.1, beginning with an antiHermitian 2-form.

91

Proposition 6.2.2. If mab dxa ∧ dxb is antiHermitian, then mab dxa ∧ dxb = m?ba + (Λasp

? ∂m?bs b ∂msa + Λ ) dxa ∧ dxb sp ∂xp ∂xp

(6.16)

Proof. (mab dxa ∧ dxb )∗ = −dxb ∧ dxa m∗ab = −dxa ∧ dxb m∗ba = −dxa ∧ m∗ba dxb − dxa ∧ Λsbp

∂m∗ba s dx ∂xp

∗ ∂m∗bs b ∂msa + Λ )dxa ∧ dxb . sp p p ∂x ∂x

= −(m∗ba + Λasp

We take an approach and show that it is consistent, and gives a Hermitian result. We can not demonstrate that it is the right thing to do apart from that it gives the right classical answer. Corollary 6.2.2. If mab is antisymmetric, i.e., mab = −mba , and mab dxa ∧ dxb is an antiHermitian, then mab = −m∗ab + (Λasp

∗ ∂m∗bs b ∂msa + Λ ) sp ∂xp ∂xp

(6.17)

Corollary 6.2.3. If mab is antisymmetric and mab dxa ∧ dxb is antiHermitian, then if η1∗ ⊗ η2 = mab dxa ⊗ dxb , we have: η1∗ ⊗ η2 = η2∗ ⊗ η1 Proof. Look at η1∗ ⊗ η2 = mab dxa ⊗ dxb , η2∗ ⊗ η1 = dxb ⊗ dxa m∗ab ∗ ∂m∗sb b ∂mas = (m∗ab + Λasp + Λ )dxb ⊗ dxa sp ∂xp ∂xp from (6.17), this is η2∗ ⊗ η1 = −mab dxb ⊗ dxa = −mba dxa ⊗ dxb = mab dxa ⊗ dxb = η1∗ ⊗ η2 . 92

(6.18)

Now we have to calculate the antiHermitian form given by Corollary 6.2.1 using the result of Corollary 6.2.3. Proposition 6.2.3. If mab is antisymmetric and mab dxa ∧ dxb is antiHermitian, then (K − S)((mab dxa )∗ ⊗ dxb ) is antiHermitian, and  P ∂ 2 : mab :∗  ∂ : mab : ∂ : mab :∗ − − λ xi q q  b ∂x ∂x0 ∂xi  : (K − S)((mab dxa )∗ ⊗ dxb ) : = g aq  ∂x 2 2  dx c λ ∂ : mab : − δq0 2 ∂(xi )2 (6.19) Proof. Set η1∗ ⊗ η2 = mab dxa ⊗ dxb . Now 2 ∗ ∂m∗sb a a ∂ msb ∗ η1 = m∗ab dxa + Λasp + Λ dx , then ∇η = (dm dxq ) ⊗ dxa . 1 sp ab ∂xp ∂xp ∂xq Using hdxa , dxb i = g ab gives: hζ1 , κ∗1 iη2 = g aq (

2 ∂mab a ∂ msb − Λ )dxb sp ∂xq ∂xp ∂xq

Note that K((mab dxa )∗ ⊗ dxb ) = hζ1 , κ∗1 iη2 , as ∇dxb = 0 then by Corollary 6.2.3 and Proposition 6.2.1 we have S((mab dxa )∗ ⊗ dxb ) = (hζ1 , κ∗1 iη2 )∗ . Now, (hζ1 , κ∗1 iη2 )∗ = g aq (

2 ∗ 2 ∗ ∂mab ∗ a ∂ msb b ∂ mas ) + Λ + Λ dxb . sp ps ∂xq ∂xp ∂xq ∂xp ∂xq

 2 ∂mab ∂mab ∗ a ∂ msb − Λsp p q − ( )   q b ∂x ∂x ∂xq = g aq  ∂x  dx 2 ∗ 2 ∗ ∂ m ∂ m −Λasp p sbq − Λbps p asq ∂x ∂x ∂x ∂x 

hζ1 , κ∗1 iη2 − (hζ1 , κ∗1 iη2 )∗

Note that msb = −mbs , we use mbs = −m∗sb . Now , using (4.96), we get:  P ∂ 2 : mab :∗  ∂ : mab : ∂ : mab :∗ − − λ xi q q b aq  b ∂x ∂x ∂x0 ∂xi  a ∗ : (K − S)((mab dx ) ⊗ dx ) : = g   dx . c2 λ ∂ 2 : mab : − δq0 2 ∂(xi )2 93

We can apply Proposition 4.5.1 to the antiHermitian 1-form in Proposition 6.2.3. Corollary 6.2.4. If mab is antisymmetric and mab dxa ∧dxb is antiHermitian, then (K − S)((mab dxa )∗ ⊗ dxb ) is antiHermitian 1-form, and using the 1-1 correspondence in (4.97) it corresponds to a classical antiHermitian 1-form ∂ : mab :∗ ∂ : mab : −g aq ( − )dxb ∂xq ∂xq Proof. To zeroth order in λ, the mab are imaginary. So to order λ, the antiHermitian parts in Proposition 6.2.3 are: ∂mab ∂mab ∗ :−: : )dxb q ∂x ∂xq ∂ : mab :∗ = − ∂xq X ∂ ∂ : mab :∗ (x ) +λ i ∂xq ∂x0 ∂xi c2 λ X ∂ 2 : mab :∗ + δq,0 2 ∂xi2

AntiHermiatian parts(: (K − S)((mab dxa )∗ ⊗ dxb ) :) = −g aq (:

and on taking the antiHermitian part, the last two terms disappear, so: AntiHermiatian parts(: (K−S)((mab

dxa )∗ ⊗dxb ) :)

∂ : mab :∗ ∂ : mab : − )dxb . =g ( q q ∂x ∂x aq

Now we apply Corollary 6.2.4 to the curvature R(1) for the left covariant derivative in (6.1) from (6.5) and (6.6), R(1) = dξ − ξ ∧ ξ λ c2 ∂ 2 : ξb : 0 ∂ : ξb : a b dx ∧ dxb dx ∧ dx − : R(1) : = ∂xa 2 ∂(xi )2 ∂ 2 : ξb : −λ 0 i dxi ∧ dxb ∂x ∂x ∂ : ξa : i ∂ : ξb : a −λ x dx ∧ dxb ∂xi ∂x0 ∂ : ξb : 0 +λ c2 : ξi : dx ∧ dxb ∂xi ∂ : ξb : i dx ∧ dxb +λ : ξ0 : i ∂x 94

(6.20)

so we have, R(1) = R(1)ab dxa ∧ dxb , where ∂ : ξb : λ c2 ∂ 2 : ξb : ∂ 2 : ξb : : R(1)ab : = − δa,0 − λ 0 i δa,i ∂xa 2 ∂(xi )2 ∂x ∂x ∂ : ξa : i ∂ : ξb : ∂ : ξb : 2 −λ x + λ c : ξ : δa,0 i ∂xi ∂x0 ∂xi ∂ : ξb : δa,i +λ : ξ0 : ∂xi

(6.21)

To ensure antisymmetry, define q mab = R(1)ab − R(1)ba . Then q mab dxa ∧ dxb = 2R(1), as dxa ∧ dxb = −dxb ∧ dxa . Also R(1) is antiHermitian as ξ is antiHermitian. Remembering that λ is imaginary, as : ξa : is antiHermitian to zeroth order, the second and third terms of (6.21) are Hermitian to O(λ2 ). If we put the antiHermitian part of : ξa : equal to i q Aa , then ∂Ab ∂Aa − ) ∂xa ∂xb ∂Aa ∂Ab ∂Ab i ∂Aa +λ q ( i xi 0 − x ) ∂x ∂x ∂xi ∂x0 ∂Aa ∂Ab δb,0 ) −λ q c2 Ai ( i δa,0 − ∂x ∂xi ∂Ab ∂Aa −λ q A0 ( i δa,i − δb,i ) . (6.22) ∂x ∂xi

AntiHermitianPart(: mab :) = i (

By Corollary 6.2.4 the following formula (6.23) defines a Hermitian 1-form j: − 2 i j = (K − S)((mab dxa )∗ ⊗ dxb )

(6.23)

Further by Corollary 6.2.4, this corresponds to the following formula (6.24) for the Hermitian part of the normal order of j: − 2 i HermitianPart(: j :) = g aq Substituting (6.22) into (6.24) gives      ∂ HermitianPart(: j :) = g aq q  ∂x    

∂(: mab :∗ − : mab :) b dx ∂xq

∂Ab ∂Aa ( a − ) ∂x ∂xb ∂Aa ∂Ab ∂Ab ∂Aa −i λ q xi ( i − ) ∂x ∂x0 ∂xi ∂x0 ∂Aa ∂Ab +i λ q c2 Ai ( i δa,0 − δb,0 ) ∂x ∂xi ∂Ab ∂Aa δb,i ) +i λ q A0 ( i δa,i − ∂x ∂xi 95

(6.24)

     b  dx .    

This is simply the usual equation for the current; HermitianPart(: j :) = g aq

∂ (Fab ) dxb q ∂x

(6.25)

but for a modified field; Fab =

∂Ab ∂Aa ∂Aa ∂Ab ∂Ab ∂Aa − − i λ q xi ( i − ) a b ∂x ∂x ∂x ∂x0 ∂xi ∂x0 ∂Ab ∂Aa ∂Ab ∂Aa + i λ c2 q Ai ( i δa,0 − δb,0 ) + i λ q A0 ( i δa,i − δb,i ) . i ∂x ∂x ∂x ∂xi (6.26)

Note that [39] also gives nonlinear terms in electromagnetism. The presence of an explicit xi in the formula (6.26) for Fab may seem alarming, as after all its entries are electric and magnetic fields, and therefore possibly measurable. However this term can be considered an artefact of the normal ordering, as from (4.92) : [ψ, χ] : = λ

X i

6.3

xi

∂ :ψ : ∂ :χ: ∂ :χ: ∂ :ψ : − . ∂xi ∂x0 ∂xi ∂x0

(6.27)

Electromagnetic plane waves

The following account also appeared in the joint paper [14] with my supervisor. Set Aa = Ea sin(α0 t + α1 x1 + βa ) with β0 = 0. Then (6.26) gives the only non-zero Fab as the following, up to antisymmetry: ∂A0 ∂A1 ∂A1 − (1 + i λ q A0 ) 1 + i λ q c2 A1 , 0 ∂x ∂x ∂x1 ∂Aj ∂Aj = + i λ c2 q A1 , j 6= 0, 1 , 0 ∂x ∂x1 ∂Aj = (1 + i λ q A0 ) 1 , j 6= 0, 1 . ∂x

F01 = F0j F1j

(6.28)

For zero current we get 0 = g 00

∂ F0b ∂ F1b + g 11 . 0 ∂x ∂x1 96

(6.29)

From this we deduce that F01 is constant, so ∂A0 ∂A1 = c2 A1 , 1 ∂x ∂x1 E02 sin(2(α0 t + α1 x1 )) = c2 E12 sin(2(α0 t + α1 x1 + β1 )) . A0

(6.30)

We deduce β1 = 0 or β1 = π and E02 = c2 E12 . We also deduce that α0 E1 = ± α1 E0 . Unless E0 = E1 = 0 we must have α0 = ± c α1 , so the plane wave travels at speed c. However if E0 = E1 = 0 then (6.28) give a special case of the classical equations, so the speed is also c. We deduce that all zero current electromagnetic plane waves travel at speed c. However it would be dangerous to extrapolate to saying that all electromagnetic plane waves in vacuum travel at speed c. It would not be impossible for some mechanism to induce a current of size about |λ| in a vacuum, in fact such an effect might be likely, as the following hand waving argument shows. Fields acting on spontaneously produced charged particle pairs would accelerate them in different directions. If the field reverses (as it would in a travelling wave), they would be accelerated back together, to recombine. This would constitute a current which would likely increase as the frequency increases, as the expected time to recombination would be reduced.

97

Chapter 7 Monopole solutions of the Dirac equation The story of magnetic monopoles is, in theory, a long one, but they have never been observed. They were first mentioned by Dirac [28], and have been generalised and studied by many people, e.g. [42]. A magnetic monopole is associated to a non-trivial gauge bundle. Our approach is to solve the Dirac equation explicitly on two subsets of R3 , and then use a gauge transformation between them. We do this classically as an exercise, and then try to carry out the same construction in κ-Minkowski space. We do not quite succeed.

7.0.1

The Lorentz group

Let M be a Hermitian matrix,   t + z x + iy , M = x − iy t − z

for t, x, y, z ∈ R

detM = t2 − x2 − y 2 − z 2

98

(7.1)

(7.2)

for A ∈ GL2 (C) where A is invertible, we have A . M = AM A∗ is also Hermitian. If we set A ∈ SL2 (C), then detA = 1 and det(AM A∗ ) = detM . For a matrix A, the map from M to A . M is a linear map on the t, x, y, z coordinates which keeps t2 − x2 − y 2 − z 2 constant, i,e,. it is a Lorentz transformation. It has real coefficients as AM A∗ is also Hermitian. There are 2 choices of matrix A for every linear map. (−A) . M = (−A)M (−A)∗ = AM A∗ = A . M

(7.3)

in fact the map SL2 (C) to the Lorentz group is onto and its kernel is ±1. We want to find a formula for the matrix A which rotates the vectors (0, 0, 0, 1) and (0, 0, 0, −1) to (0, x, y, z). It is convenient to use polar coordinates. z = r cos θ, where r =

x = r sin θ cos φ,

y = r sin θ sin φ

(7.4)

p x2 + y 2 + z 2 . Proposition 7.0.1 gives matrices in SL2 (C) which

send (0, 0, 0, 1) and (0, 0, 0, −1) to (0, x, y, z) on the sphere. Proposition 7.0.1.  

cos

θ 2

− sin

sin 2θ e−iφ



− sin 2θ



cos 2θ e−iφ

 

θ iφ e 2 

.

1

0



= cos 2θ 0 −1   iφ cos θ sin θe  = sin θe−iφ − cos θ    −1 0 − cos 2θ eiφ .  θ − sin 2 0 1

in terms of 4-position vectors, using (7.1)     0 0           θ θ iφ 0 sin θ cos φ cos 2 − sin 2 e   . =     θ −iφ θ    sin 2 e cos 2 0 sin θ sin φ      1 cos θ 99

(7.5)

(7.6)

giving the points on the sphere S 2 in polar coordinates. We also get     0 0           θ θ iφ  0  sin θ cos φ − sin 2 − cos 2 e   . =     θ θ −iφ  0   sin θ sin φ  − sin 2 cos 2 e     −1 cos θ

(7.7)

We expect to run into problems because the map from SL2 (C) to the Lorentz group has a kernel. These problems can be seen in the formula in the Proposition. Looking at (7.6) the formula is 1-1 correspondence between the sphere minus (0, 0, 0, −1) if we take θ ∈ [0, π). Note that for θ = 0 and all φ the matrices fix the point (0, 0, 0, 1), but these all correspond to the identity matrix. For θ = π

 

0 e−iφ

    0 0          0  0  −eiφ . =      0  0  0     1 −1

(7.8)

so we no longer have a 1-1 correspondence. Looking at (7.7) the formula is 1-1 correspondence between the sphere minus (0, 0, 0, 1) and the matrices if we take θ ∈ (0, π]. For θ = 0 

 

0 e−iφ

  0           0  0 −eiφ . =       0  0 0     −1 1 0



(7.9)

Our plan is to construct a spinor solution to the Dirac equation which is as rotationally symmetric as possible. This is not possible to do exactly, but we can do it up to gauge equivalence.

100

Fig. 27 On the subset U, we choose the element m of SL2 given in (7.6), and use this value in the equation 

    0 0           0  0   = m . ψ   ψ m .       0 0      1 1

(7.10)

To define ψ on U. To get an almost rotationally symmetric equation. Similarly we define ψ on V using (7.7). The ψ values on the overlap are related by gauge equivalence. We first do this classically, and then try to do it in κ-Minkowski space.

7.0.2

SL2 (C) acting on spinors

We want to make a rotationally symmetric spinor field. This is not quite possible, but we make one up to gauge transformation. We want      0 0            0 0   = g . ψ   ψ g .       0 0      1 1

(7.11)

for all g ∈ SU2 . The best we can do is to make this true for a subset of g and then show that the choices are gauge equivalent. In particular, if g fixes the 101

point (0, 0, 0, 1), then g . ψ(0, 0, 0, 1) is a unit norm complex number times ψ(0, 0, 0, 1). i.e., we have ψ(0, 0, 0, 1) an eigenvector of g in the stabilizer of (0, 0, 0, 1). We use Proposition 7.0.1 to give a subset of g that we can use. Using (7.11) we get      0 0             sin θ cos φ 0 cos 2θ − sin 2θ eiφ  = ψ   .   ψ      θ  sin 2θ e−iφ  sin θ sin φ  0 cos 2      cos θ 1   0       0 cos 2θ − sin 2θ eiφ .ψ  =    0 sin 2θ e−iφ cos 2θ   1

(7.12)

which is valid for θ ∈ [0, π) and for (7.12)      0 0             − sin 2θ − cos 2θ eiφ sin θ cos φ  0   = ψ   .   ψ       cos 2θ e−iφ − sin 2θ  sin θ sin φ   0       cos θ −1   0       0 − sin 2θ − cos 2θ eiφ .ψ  =  (7.13)   θ −iφ θ   cos 2 e − sin 2 0   −1 now (7.16) and (7.17) give different values of ψ where both are valid for 0 ∈ (0, π). However, we can arrange a gauge transformation between them.

102

If

      0 0 0              0   x 0  = g0 .   =   g.        0  y  0       z −1 1

then

 −1

g0 g = 

0 −iφ

−e

iφ

e

0

(7.14)

 

    0 0         0 0 1 0    g . ψ   = (element of S ) · g . ψ    0 0     1 −1     0 0           0 0 0 eiφ 1   . ψ   = (element of S ) · ψ       −iφ 0 0 −e 0     −1 1

(7.15)

(7.16)

(7.17)

if ψ is rotationally symmetric, a rotation about the z axis only changes it up to an element of S 1 so the rotation about the z axis corresponds to   eiα 0  , so we need −iα 0 −e

 

eiα 0

    0 0          0 0 0  . ψ   = (element of S 1 ) · ψ       0 0 −e−iα     1 1

103

(7.18)

  0     0  From (7.18) we see that ψ    has to be chosen to be an eigenvalue of the 0   1 2 × 2 matrix in (7.13). Now from (7.17) and (7.18) putting φ = 0     0 0              0 0 e−iα 0 0 1  . ψ   = (element of S 1 ) · ψ     (7.19)     0 0 0 eiα −1 0     −1 1 and by (7.18) this becomes     0 0           0 0 0 1   . ψ   = (element of S 1 ) · ψ       0 0 −1 0     −1 1

(7.20)

so we may as well set   0 0           0 0 0 1   .ψ  ψ  =   0 0 −1 0     1 −1 



(7.21)

To see what (7.18) and (7.21) mean, we look at the spinor representation. It is not obvious how SL2 (C) acts on the 4-spinors. We have to go back to the Lie algebra to see this. The Lie algebra of Lorentz group acts on spinors by [63] 1 ω(ψ) = ωab [γ a , γ b ]ψ 8

(7.22)

where ωba is a 4×4 matrix and the index is lowered using the metric ωab = gac ω c b . Now we want to write the group elements used in Proposition 7.0.1 104

as exponentials of Lie algebra elements     sin |ω| cos |ω| −ω |ω| 0 −ω =  exp  sin |ω| ω 0 ω |ω| cos |ω|

(7.23)

if we put ω = 2θ e−iφ then from Proposition 7.0.1       0 − 2θ eiφ 1 0 cos θ sin θeiφ . =  exp  θ −iφ −iφ e 0 0 −1 sin θe cos θ 2

(7.24)

if we put ω = ( 2θ + π2 )e−iφ , we get       θ π iφ iφ cos θ sin θe 0 −( 2 + 2 )e −1 0  = . exp  sin θe−iφ cos θ 0 1 ( 2θ + π2 )e−iφ 0 (7.25) we can get the Lie algebra element corresponds to the matrix we exponentiated if ω = u + iv 



   0 −ω 0 −1 0 i   = u +v  ω 0 1 0 i 0 for the matrix multiplying  0 0 0   0 0 0 a ω b=  0 0 0  0 −2 0



by u in (7.26)   0 0 0     0 0 2  so, ωab =    0 0 0   0 2 0

0

(7.26)

0



  0 −2   0 0  0 0

(7.27)

and 1 1 −2 1 3 2 3 1 ωab [γ a , γ b ] = ωab γ a γ b = γ γ + γ γ = γ 3γ 1 (7.28) 8 4 4 4   0 −1  corresponds to γ 3 γ 1 , while the Lie so the Lie algebra element  1 0   0 i  corresponds to γ 3 γ 2 . The element of SL2 (C) given algebra element  i 0 105



0 −ω 0

  ω 0 −ω  acts on the 4-spinors by the matrix  by   0 ω 0  0 





cos |ω|   sin |ω| ω 0 −ω  =  |ω| exp    0 ω 0  0 





−ω sin|ω||ω| cos |ω| 0 0

0

0 0 0 0 0

   0 −1 0 1 −π   =   exp  2 1 0 −1 0



  0 0   hence  0 −ω   ω 0 

    sin |ω|  cos |ω| −ω |ω|   sin |ω| cos |ω| ω |ω| 0

0

(7.29)



(7.30)

γ 3 γ 1 ), so (7.21) gives so the group element in (7.21) correspondence to exp( −π 2       0 0 1 0 0 0              0  −1 0 0 0 0   ψ   (7.31) ψ  =      0   0 0 0 1 0       −1 0 0 −1 0 1

7.0.3

Solving Dirac equation for 4-spinors

We use the rotationally symmetric (up to gauge transformation) spinor field from section 7.0.2 to construct a solution of the Dirac equation. Suppose that the solution is independent of time, and has the formula, using the polar coordinates of (7.4)   θ θ iφ cos 2 − sin 2 e 0 0    θ −iφ  θ sin 2 e  cos 2 0 0   ψ(t, x, y, z) = g(r)  ψ 0 0 0 1  0 0 cos 2θ − sin 2θ eiφ    θ −iφ θ 0 0 sin 2 e cos 2 (7.32) 106

for region complement of (t, 0, 0, z), z ≤ 0    cos 2θ − sin 2θ eiφ 0 0 0     θ −iφ   sin 2 e  1 cos 2θ 0 0   ψ(t, x, y, z) = g(r)      0 0 cos 2θ − sin 2θ eiφ  0    θ −iφ θ 0 0 sin 2 e cos 2 1   −e−iφ sin 2θ     θ  cos 2    (7.33) = g(r)   θ −iφ −e sin 2    θ cos 2 Set 

θ 2

θ iφ e 2

cos − sin   θ −iφ sin 2 e cos 2θ B=   0 0  0 0

0 0 cos 2θ sin 2θ e−iφ

0



   0   − sin 2θ eiφ   θ cos 2

substitute in the Dirac equation (3.21)      0 0           0  µ 0   = m B g(r) ψ   i(∂µ − i q Aµ )  γ B g(r) ψ      0  0      1 1

(7.34)

(7.35)

To solve equation (7.35), we use Mathematica. The detailed calculations can be found in Appendix A ”Solution to 4-spinor Dirac Equation” where the gamma matrices in (3.22) are denoted by (g0, g1, g2, g3) and the matrix B in (7.34) is denoted by (B) and its inverse by (invB). Now, since B is a function ∂ of θ and φ, we write γ µ µ in polar coordinates. To convert between polar ∂x and cartesian coordinates it is convenient to use a matrix to implement the

107

chain rule. We have: ∂f ∂f ∂r ∂f ∂θ ∂f ∂φ = + + ∂x ∂r ∂x ∂θ ∂x ∂φ ∂x

∂f ∂f ∂x ∂f ∂y ∂f ∂z = + + (7.36) ∂r ∂x ∂r ∂y ∂r ∂z ∂r

,

and using (7.4) we get      ∂f ∂f sin θ cos φ sin θ sin φ cos θ  ∂r     ∂x   ∂f      ∂θ  =  r cos θ cos φ r cos θ sin φ −r sin θ  ∂f  ,      ∂y  ∂f ∂f −r sin θ sin φ r sin θ cos φ 0 ∂φ ∂z

(7.37)

finding the inverse of the matrix in (7.37), we get Cij in (7.38) which is called (c) in the Mathematica file,  cos θ cos φ sin θ cos φ  r  cos θ sin φ  Cij =  sin θ sin φ r  sin θ cos θ − r

 − sin φ r sin θ  cos φ   . r sin θ   0

(7.38)

The standard ∂µ (and coordinates xµ ) in the Dirac equation are with respect to the cartesian coordinates. We use indices for the polar coordinates, y 1 = r, y 2 = θ, y 3 = φ. Then we can write: ∂ ∂µ γ µ g(r)B(θ, φ) = γ i Cij j g(r)B(θ, φ) ∂y ∂B ∂B ∂g + g(r)γ i (Ci2 + Ci3 ) (7.39) = γ i Ci1 B ∂r ∂θ ∂φ In the Mathematica file, we break the calculation of (7.39) into steps (gD), (thetaD) and (phiD) to get the result of equation (7.39) which is given by (dirP1). In (7.35) we replace i q Aµ by eµ and solve the massive Dirac equation −1

g (r)B

−1

µ

(∂µ γ −

3 X

eµ )(g(r)B) = m

(7.40)

µ=0

solving (7.40) will give (Dirac) in the Mathematica file and is obtained after breaking the equation into parts (CompP1), (dirP2) and (CompP2) solving 108

each part separately then combine them to get the result of (7.40). Note: To be more exact, we want (7.40) times a constant 4-spinor on both sides, but it is convenient to keep a matrix equation here. We get a solution to the massless Dirac equation by assuming e0 = 0 and go through steps of calculations in the Mathematica file to get the solution which is denoted by (eµ ) in the file. i sin φ tan 2θ e1 = , 2r

e2 = −e1 cot φ,

e3 = 0.

(7.41)

for i = 1, 2, 3 .

(7.42)

Using the current equation (3.23) we get j0 = 2 q |g(r)|2 ,

ji = −2 q

|g(r)|2 i x r

P ∂ ja now using the current conservation law a a = 0 for a = 0, 1, 2, 3, we ∂x K get a value for g(r) = × unit norm complex valued function. By a gauge r transformation, we set the unit norm bit to be 1, and for the real constant K we may choose K = 1. In the Mathematica file the value of g(r) is called (value). The electro-magnetic field Fab =

∂ea ∂eb − b = 0 if a = 0 or b = 0. The a ∂x ∂x

non-zero ones are i cos θ iz =− = −F21 2 2qr 2qr3 i sin φ sin θ iy = = = −F31 2 2qr 2qr3 i cos φ sin θ ix = − =− = −F32 2 2qr 2qr3

F12 = − F13 F23

(7.43)

Using (3.26) we see that this is the field due to a magnetic monopole. This gives a zero current by the equation for a = 0, 1, 2, 3 ja =

X ∂Fab b

109

∂xb

(7.44)

However, this is not the value given by the spinor fields in (3.23), which we gave in (7.42) . Classically, the Dirac equation only has one charge - in quantum theory, the other charge is introduced - this may be thought as the ” absence of an electron” giving a positron, the Dirac sea. This argument does not help us to write down a positron in classical theory but we can change this by changing where φ lies and this modification will allow solutions with zero and negative charge. To summarise, we get a spinor valued field in one hemisphere given by    θ θ iφ cos 2 − sin 2 e 0 0 0     θ −iφ     1 sin 2 e cos 2θ 0 0 1    ψ(t, x, y, z) =   θ θ iφ  r  0 0 cos 2 − sin 2 e  0    0 0 sin 2θ e−iφ cos 2θ 1 (7.45) and in the other hemisphere    θ θ iφ 1 cos 2 − sin 2 e 0 0        θ −iφ  0 sin 2 e cos 2θ 0 0 1    ψ(t, x, y, z) =   θ θ iφ  r  0 0 cos 2 − sin 2 e  1    0 cos 2θ 0 0 sin 2θ e−iφ (7.46) and potentials i q Aµ = eµ given by (7.41) and fields in (7.43).

7.0.4

Solving Dirac equation for 8-spinors

The 4-spinor form of the Dirac equation always has either j0 ≥ 0 or j0 ≤ 0, it never has both signs and if the spinor is not zero, j0 6= 0. Our problem is that, for a classical field theory we need the equations for jµ from the spinor fields and the electromagnetic fields to give zero. From (7.43) and (7.44) the 110

electromagnetic field gives zero current (at least away from the singularity r = 0). We can double up the 4-spinor to give an 8-spinor by setting φ+ = φ and φ− = φ, and we get a solution for the 8-spinor equation with the same Aµ and a zero current. We take both an electron and a positron fields.   φ  1   ·     φ =  ·  ∈ C8 (7.47)     ·   φ8 We have to give a new γ matrices  γã = 

γ

a

0



0 γa

(7.48)



but we take a new different definition of   0 γ 0  φ = φ∗  0 0 −γ  then for φ = 

φ+ φ−

(7.49)

  where φ± are 4-column vectors, the current is given by 

ja = q φ∗+ φ∗−  =

q(φ∗+ γ 0 γ a φ+

γ

0



0 0



γ

a

0 −γ 0 ∗ 0 a − φ− γ γ φ− )

0 γ

a

 

φ+ φ−

  (7.50)

one obvious way to make the current zero is to put φ+ = φ− or φ+ = ζφ− where ζ ∈ S 1 and |ζ| = 1. Note in [7] there is also a use of higher than usual dimensional spinors, resulting in a monopole (this time an electric one) with zero current. However in this case, the higher dimension results from an extra 1 space and 1 time dimension in space time. 111

7.1

Trying to find a noncommutative monopole

Monopoles in noncommutative geometry have been studied largely in the context of the Hopf fibration or other Hopf-Galois extensions (e.g. [23]). We shall attempt to find a solution to the Dirac equation for our deformed space time, for the zero current case.

7.1.1

Simplifying the Dirac equation

We can generalize the result of the massless noncommutative Dirac equation up to gauge transformations by using (4.94), we get λ X ∂ 2ψ DN C ψ = (∂µ − i q Aµ )γ µ ψ + γ 0 2 ∂(xi )2 i

(7.51)

The extra bit added to the equation (7.51) in the noncommutative case is the following, which we name λ A: 

(5 + 3 cos θ) sec 2θ e−iφ tan 2θ



    3 θ 2 X  −(1 + 3 cos θ) sec 2 λ  ∂ ψ λ 0   λA = γ =  i 2 3 2 i=1 ∂(x ) 16r −(5 + 3 cos θ) sec θ e−iφ tan θ   2 2  (1 + 3 cos θ) sec 2θ

(7.52)

for c = 1 where ψ given in (7.33). We use the 3-dimensional Laplacian in polar coordinates to calculate 4 g(r)B(θ, φ) =

∂ 2 g(r) 2 ∂g(r) g(r) ∂ ∂B(θ, φ) + B(θ, φ) + (sin θ ) ∂r2 r ∂r r2 sin θ ∂θ ∂θ 1 ∂ 2 B(θ, φ) + . (7.53) sin θ ∂ 2 φ2

We suppose that the noncommutative solution is an O(λ) modification of the original solution. We set in (7.51) i q Aµ −→ e˜µ = eµ + λ fµ , 112

where eµ is the classical potential in (7.41) and fµ is a real function. We set the ψ in (7.51) to be ψnew = ψold + λξ , where ψold is the original monopole solution in (7.45) and ξ is a spinor field. Now we solve the Dirac equation to O(λ).

A − fµ γ µ ψ + (∂µ − eµ )γ µ ξ = 0

(7.54)

Equation (7.54) is called (XtraTerm) in the Mathematica file ” Finding noncommutative monopole I ” shown in Appendix B. To get (XtraTerm) we go through some steps of calculations. Similar to Appendix A, in this Mathematica file we have the same names for the gamma matrices in (3.22), B matrix in (7.34) and Cij in (7.38). (e1) and (e2) in the file are the classical potentials given by (7.41). The Laplacian in (7.53) is denoted by (Lap). The original monopole solution is denoted by (ψ) and the unknown spinor field ξ in ψnew is denoted by (ξ). Now back to (7.54), using Mathematica we can get a value for f0 setting the first component of (XtraTerm) equal zero. This value is denoted by (f0exp) and is given by 

2 −3 − 8f3 r − + 8eiφ (f1 − if2 )r2 cot( 2θ ) 1 + cos(θ) 2



      (0,0,0,1) θ θ 2 +4r − sec( )ξ3 (t, r, θ, φ) + 2 csc( ) i csc(θ)ξ3 (t, r, θ, φ) 2 2     1   (0,0,1,0) (0,1,0,0) f0 → 2  − cos(θ)ξ3 (t, r, θ, φ) − r sin(θ)ξ3 (t, r, θ, φ)   8r     +eiφ sin(θ)ξ2(0,0,1,0) (t, r, θ, φ) − r cos(θ)ξ2(0,1,0,0) (t, r, θ, φ)        (1,0,0,0) +ξ0 (t, r, θ, φ) Note that to fit with the way Mathematica works, we will frequently present equations as substitution values. 113

(7.55)

The value in (7.55) when substituting it in (7.54) gives zero in the first component as in (f0n2XTerm) in the file. We get a value for f3 by setting the second component of (f0n2XTerm) equal zero. This value is denoted by (f3exp) and is given by 

−16f1 tan( 2θ )

2

tan( 2θ )



− 16if2 r     (0,0,0,1) θ θ  +4e2iφ r2 csc( ) sec2 ( ) − (−1 + cos(θ))ξ2 (t, r, θ, φ) − 2iξ (t, r, θ, φ) 2   2 2     (0,0,1,0) θ θ   +2 cos( ) (f − if )(1 + cos(θ)) + 2 sin( )ξ (t, r, θ, φ) 1 2   2 2 2     (0,1,0,0) (1,0,0,0) θ e−iφ   ) ξ (t, r, θ, φ) + ξ (t, r, θ, φ) −2r cos( 2 0 f3 →   2 2  32r    θ iφ 2 3   +e sin(θ) − 8r csc(θ) sec( 2 )ξ3 (t, r, θ, φ) + 8 csc (θ)       (0,0,1,0) (0,0,0,1) θ 2  − 1 + cos(θ) + 4r cos( ) iξ (t, r, θ, φ)  (t, r, θ, φ) − sin(θ)ξ3 3   2     (0,1,0,0) (1,0,0,0) +r(−1 + cos(θ)) ξ3 (t, r, θ, φ) + ξ1 (t, r, θ, φ) (7.56) This value in (7.56) will give a zero in the second component of (f0n2XTerm) and this is shown in (f3n2f0XTerm). Now equation (7.55) and (7.56) are solutions to the first two components of equation (7.54). The other two component of (XtraTerm) are denoted by (zer0) and (zer1) and are given receptively by

114





(−1 + cos(θ))ξ1 (t, r, θ, φ) − (−1 + cos(θ))ξ3 (t, r, θ, φ)     (0,0,0,1) (0,0,0,1)   +2iξ1 (t, r, θ, φ) − 2iξ3 (t, r, θ, φ)     (0,0,1,0) (0,0,1,0)   + sin(2θ) − ξ1 (t, r, θ, φ) + ξ3 (t, r, θ, φ)       (0,1,0,0) (0,1,0,0) e−iφ csc( 2θ ) sec( 2θ )  (t, r, θ, φ) + ξ3 (t, r, θ, φ)  +2 sin(θ) r sin(θ) − ξ1 =0   4r   (0,0,1,0) (0,0,1,0) iφ   sin(θ) ξ0 (t, r, θ, φ) − ξ2 (t, r, θ, φ) +e         +r cos(θ) − ξ (0,1,0,0) (t, r, θ, φ) + ξ (0,1,0,0) (t, r, θ, φ) 0 2       (1,0,0,0) (1,0,0,0) +ξ0 (t, r, θ, φ) − ξ2 (t, r, θ, φ) (7.57) 

 (0,0,0,1) (0,0,0,1) + sin(φ)) ξ0 (t, r, θ, φ) − ξ2 (t, r, θ, φ)    (0,0,1,0) (0,0,1,0)  (t, r, θ, φ) (t, r, θ, φ) + ξ3 +2 sin(θ) − ξ1    θ iφ  +e tan( 2 ) ξ0 (t, r, θ, φ) − ξ2 (t, r, θ, φ)    (0,0,0,1) (0,0,0,1)  −iξ0 (t, r, θ, φ) + iξ2 (t, r, θ, φ)  =0  (0,0,1,0) (0,0,1,0)  +2 cos(θ) − ξ0 (t, r, θ, φ) + ξ2 (t, r, θ, φ)    (0,1,0,0) (0,1,0,0)  +2r sin(θ) − ξ0 (t, r, θ, φ) + ξ2 (t, r, θ, φ)    (0,1,0,0) (0,1,0,0)  +2r cos(θ) ξ1 (t, r, θ, φ) − ξ3 (t, r, θ, φ)    (1,0,0,0) (1,0,0,0) +ξ1 (t, r, θ, φ) − ξ3 (t, r, θ, φ)

θ cot( 2 )(−i cos(φ)

           1   2r             

Looking at the values of (zer0) and (zer1) in (7.57) and (7.58), we can define: xx = ξ2 − ξ0 , yy = ξ3 − ξ1 ,

(7.59)

then (7.57) and (7.58) can be simplified into (zero0) and (zero1) respectively,

115

(7.58)

given by 

 csc(θ) tan( 2θ )yy(t, r, θ, φ) − 2i csc2 (θ)yy(0,0,0,1) (t, r, θ, φ)     (0,0,1,0) (0,1,0,0)  +2 cot(θ)yy (t, r, θ, φ) + 2ryy (t, r, θ, φ)  −iφ   e sin(θ)    −2eiφ xx(0,0,1,0) (t, r, θ, φ) − r cot(θ)xx(0,1,0,0) (t, r, θ, φ)  = 0 2r       +r csc(θ)xx(1,0,0,0) (t, r, θ, φ)



iφ

tan( 2θ )xx(t, r, θ, φ)+

(7.60)



−e     iφ (0,0,0,1) iφ (0,0,1,0) 2e csc(θ)xx (t, r, θ, φ) + 2e cos(θ)xx (t, r, θ, φ) 1  =0   2r  +2 sin(θ)yy(0,0,1,0) (t, r, θ, φ) + 2eiφ r sin(θ)xx(0,1,0,0) (t, r, θ, φ)   (0,1,0,0) (1,0,0,0) −2r cos(θ)yy (t, r, θ, φ) − 2ryy (t, r, θ, φ)

(7.61)

We can write, eiφ (ξ2 − ξ0 ) = eiφ xx(t, r, θ, φ) −→ zz(t, r, θ, φ) ,

(7.62)

in both equations then (7.60) and (7.61) will be simplified to the equations (zero00) and (zero11) given below,   θ (1,0,0,0) (t, r, θ, φ)   tan( 2 )yy(t, r, θ, φ) + 2 − r cos(θ)zz     −iφ  − sin(θ)zz(0,0,1,0) (t, r, θ, φ) + r cos(θ)zz(0,1,0,0) (t, r, θ, φ)  e   = 0 (7.63)  2r  +r sin(θ)yy(0,1,0,0) (t, r, θ, φ) + cos(θ)yy(0,0,1,0) (t, r, θ, φ)     (0,0,0,1) −i csc(θ)yy (t, r, θ, φ)

  (cot(θ) + csc(θ))zz(t, r, θ, φ) + 2i csc(θ)zz(0,0,0,1) (t, r, θ, φ)     (0,0,1,0) (0,1,0,0) +2 cos(θ)zz (t, r, θ, φ) − 2r cos(θ)yy (t, r, θ, φ) 1   =0  (0,0,1,0) (0,1,0,0) 2r   +2 sin(θ)yy (t, r, θ, φ) + 2r sin(θ)zz (t, r, θ, φ)    (1,0,0,0) −2ryy (t, r, θ, φ)

116

(7.64)

The φ derivative of yy(t, r, θ, φ) and zz(t, r, θ, φ) using the above two equations are given by (0,0,0,1)

yy

(t, r, θ, φ) →

i sin(θ) − 2 r sin(θ)yy(0,1,0,0) (t, r, θ, φ) 2 + cos(θ)yy(0,0,1,0) (t, r, θ, φ) − rzz(1,0,0,0) (t, r, θ, φ) − sin(θ)zz(0,0,1,0) (t, r, θ, φ) + r cos(θ)zz(0,1,0,0) (t, r, θ, φ) θ (7.65) − tan( )yy(t, r, θ, φ) . 2

(0,0,0,1)

zz

(t, r, θ, φ) →

i 2 sin(θ) − r(yy(1,0,0,0) (t, r, θ, φ) 2 + cos(θ)yy(0,1,0,0) (t, r, θ, φ) − sin(θ)zz(0,1,0,0) (t, r, θ, φ)) + sin(θ)yy(0,0,1,0) (t, r, θ, φ)+ cos(θ)zz(0,0,1,0) (t, r, θ, φ) +(cos(θ) + 1)zz(t, r, θ, φ) . (7.66)

So, the following equations (7.55), (7.56), (7.65) and (7.66) solves (7.54). We see from the Mathematica file that the entries of (XtraTerm) depends on f0 ± f3 and f1 ± if2 , with only one choice of sign in any entry. So we express, for a real valued functions a, b and complex valued functions v, w: a+b a−b , f3 = , a, b ∈ R 2 2 v+w v−w f1 = , f2 = , v∗ = w . 2 2i f0 =

(7.67)

This brings us to (fmus) where we use these expressions in (7.67) in (XtraTerm) equation and we solve to get values for a and b as in (Solve4ab) in Mathematica and we get (aexp) and (bexp) given by, sec2 ( 2θ ) 1 3 θ θ iφ + e v cos( ) − − sec( )ξ3 (t, r, θ, φ) 8r2 2 8r2 2 2 θ θ (0,0,1,0) (0,0,0,1) +i csc( ) csc(θ)ξ3 (t, r, θ, φ) + 2eiφ cos( )ξ2 (t, r, θ, φ) 2 2 θ (0,0,1,0) θ (0,1,0,0) − cos(θ) csc( )ξ3 (t, r, θ, φ) − eiφ r cos(θ) csc( )ξ2 (t, r, θ, φ) 2 2 θ (0,1,0,0) θ (1,0,0,0) (t, r, θ, φ) − eiφ r csc( )ξ0 (t, r, θ, φ) , (7.68) −2r cos( )ξ3 2 2

a → −

117

sec2 ( 2θ ) 3 cos(θ) sec2 ( 2θ ) θ iφ 2 3 θ −iφ ) − 2e csc (θ) sin ( )ξ2 (t, r, θ, φ) − + e w tan( 16r2 16r2 2 2 θ (0,0,1,0) θ (0,0,0,1) iφ iφ (t, r, θ, φ) + e cos(θ) sec( )ξ2 (t, r, θ, φ) +ie csc(θ) sec( )ξ2 2 2 θ (0,0,1,0) θ (0,1,0,0) +2 sin( )ξ3 (t, r, θ, φ) + 2eiφ r sin( )ξ2 (t, r, θ, φ) 2 2 θ (0,1,0,0) θ (1,0,0,0) −r cos(θ) sec( )ξ3 (t, r, θ, φ) + r sec( )ξ1 (t, r, θ, φ) . (7.69) 2 2

b → −

Applying xx, yy then zz as defined previously also (7.65) and (7.66) to the values we got for a and b, gives the following new a and b which is given by (sub3a) and (sub3b) in the Mathematica file after some steps. sec2 ( 2θ ) 1 3 θ θ iφ a → − 2 + e v cos( ) − − sec( )ξ1 (t, r, θ, φ) 2 8r 2 8r 2 2 θ (0,0,1,0) θ (0,0,0,1) iφ (t, r, θ, φ) + 2e cos( )ξ0 (t, r, θ, φ) +i csc( ) csc(θ)ξ1 2 2 θ θ (0,0,1,0) θ θ (0,1,0,0) − cos( ) cot( )ξ1 (t, r, θ, φ) − eiφ r cos( ) cot( )ξ0 (t, r, θ, φ) 2 2 2 2 θ (0,1,0,0) θ (1,0,0,0) −2r cos( )ξ1 (t, r, θ, φ) − eiφ r csc( )ξ0 (t, r, θ, φ) 2 2 θ (0,0,1,0) θ (0,1,0,0) + sin( )ξ1 (t, r, θ, φ) + eiφ r sin( )ξ0 (t, r, θ, φ) 2 2 θ −r csc( )zz(1,0,0,0) (t, r, θ, φ) , (7.70) 2 

2

w cos( 2θ )

2

w cos( 3θ ) 2

iφ

sin( 2θ )



−8r + 8r −e     3θ iφ 2iφ 2   +3e sin( 2 ) − 8e r (−1 + cos(θ))ξ0 (t, r, θ, φ)      −16ie2iφ r2 ξ2(0,0,0,1) (t, r, θ, φ) − 8e2iφ r2 sin(2θ)ξ0(0,0,1,0) (t, r, θ, φ)   θ 2 θ  csc( 2 ) sec ( 2 )   (0,0,1,0) (0,0,1,0) iφ 2 iφ 2 b→−   (t, r, θ, φ) (t, r, θ, φ) + 8e r cos(2θ)ξ1 −8e r ξ1 2   32r    −8e2iφ r3 ξ0(0,1,0,0) (t, r, θ, φ) + 8e2iφ r3 cos(2θ)ξ0(0,1,0,0) (t, r, θ, φ)      (0,1,0,0) iφ 3 iφ 3 (1,0,0,0) +8e r sin(2θ)ξ1 (t, r, θ, φ) − 16e r sin(θ)yy (t, r, θ, φ)   (1,0,0,0) iφ 3 −16e r sin(θ)ξ1 (t, r, θ, φ) (7.71) Thus the content of the Dirac equation is summarized in the values of a and b above in (7.70) and (7.71), and in (7.63) and (7.64). 118

Now we have equations (7.63), (7.64), (7.70) and (7.71) in variables zz, yy, ξ0 , ξ1 , v, w, and these are connected to the original variables by (7.67), (7.59) and (7.62). Then by (7.67) using the values of a, b in (7.70) and (7.71), f0 will be given by, which is (f0BYab) in Mathematica,   −4e2iφ r2 sec( 2θ ) tan( 2θ )ξ0 (t, r, θ, φ) − csc2 (θ)    iφ 2   8e r sin( 2θ ) sin(θ)ξ1 (t, r, θ, φ) − 16ie2iφ r2 sin( 2θ )ξ0(0,0,0,1) (t, r, θ, φ)      (0,0,0,1) θ θ iφ 2   +2 cos( 2 ) − 8ie r ξ0 (t, r, θ, φ) + 2 sin( 2 )     2iφ 2 2 2iφ 2 2 iφ −iφ   − 4e r v − 4r w − 4e r v cos(θ) + 4r w cos(θ) + 3e sin(θ) e   f0 →  (0,0,1,0) (0,0,1,0) 3θ 3θ 2iφ 2 iφ 2 16r2   −8e r sin( 2 )ξ0 (t, r, θ, φ) + 8e r cos( 2 )ξ1 (t, r, θ, φ)      (0,1,0,0) (0,1,0,0) 3θ 3θ 2iφ 3 iφ 3  +8e r cos( 2 )ξ0 (t, r, θ, φ) + 8e r sin( 2 )ξ1 (t, r, θ, φ)       −8eiφ r3 sin( 2θ )yy(1,0,0,0) (t, r, θ, φ) + 8eiφ r3 cos( 2θ )zz(1,0,0,0) (t, r, θ, φ)    θ (1,0,0,0) θ (1,0,0,0) 2iφ 3 iφ 3 +8e r cos( 2 )ξ0 (t, r, θ, φ) − 8e r sin( 2 )ξ1 (t, r, θ, φ) (7.72)

119

and f3 will be, which is (f3BYab) in Mathematica,   iφ −iφ iφ −iφ 2e v − 2e w + 2e v cos(θ) + 2e w cos(θ)     cot(θ) csc(θ) 3 θ iφ   + 2 − + 4e csc(θ) sin ( )ξ (t, r, θ, φ) 0   2 r r2     θ (0,0,0,1) θ iφ )ξ (t, r, θ, φ) − 2ie sec( )ξ (t, r, θ, φ) −2 sin(   2 1 2 0   (0,0,0,1) (0,0,1,0)   +2i csc( 2θ )ξ1 (t, r, θ, φ) + eiφ csc( 2θ )ξ0 (t, r, θ, φ)      −2eiφ cos( 3θ ) csc(θ)ξ (0,0,1,0) (t, r, θ, φ) − sec( θ )ξ (0,0,1,0) (t, r, θ, φ)    0 2 2 1     (0,0,1,0) (0,1,0,0) 3θ θ iφ 1 (t, r, θ, φ) − e r sec( 2 )ξ0 (t, r, θ, φ)   −2 sec(θ) sin( 2 )ξ1 f3 → csc(θ)     4 (0,1,0,0) (0,1,0,0) θ )ξ (t, r, θ, φ) − r csc( )ξ (t, r, θ, φ)  −2eiφ r csc(θ) sin( 3θ  0 2 2 1     (0,1,0,0) 3θ θ (1,0,0,0) (t, r, θ, φ) − r csc( 2 )yy (t, r, θ, φ)   +2r cos( 2 ) csc(θ)ξ1     θ (1,0,0,0) (1,0,0,0) ) csc(θ)yy (t, r, θ, φ) − r sec( )zz (t, r, θ, φ)  +2r cos( 3θ  2 2     (1,0,0,0) θ (1,0,0,0) iφ )zz (t, r, θ, φ) − e r sec( )ξ (t, r, θ, φ) −2r csc(θ) sin(( 3θ  2 2 0     (1,0,0,0) (1,0,0,0) θ 3θ iφ (t, r, θ, φ) − r csc( 2 )ξ1 (t, r, θ, φ)   −2e r csc(θ) sin( 2 )ξ0   (1,0,0,0) +2r cos( 3θ ) csc(θ)ξ (t, r, θ, φ) 1 2 (7.73) Note that we have so far not made any assumption that can not be reversible. The content of the Dirac equation is summarized by the f0 and f3 given in (7.72) and (7.73) by using the values of a, b in (7.70) and (7.71), and also by equations (7.63) and (7.64).

7.1.2

Solving the zero current equations.

From the current equation (6.26) we get, where Fab is the commutative case, iq

NC Fab

∂fb ∂fa λc2 ∂eb ∂ea = i q Fab + λ( a − b ) + ei ( i δa,0 − i δb,0 ). ∂x ∂x q ∂x ∂x

(7.74)

The noncommutative currents for b 6= 0 j0N C =

λ X aa ∂ ∂f0 ∂fa λ X aa ∂ i ∂ea ( − g (e ) g ) − i q a6=0 ∂xa ∂xa ∂t i q 2 a=1,2 ∂xa ∂xi 120

(7.75)

jbN C =

λ X aa ∂ ∂fb ∂fa g ( a − b ). a i q a6=b ∂x ∂x ∂x

(7.76)

as we know ea from (7.41), for (7.75) and (7.76) we have for b 6= 0, putting jN C = 0 gives θ (−1 + 2 cos θ + cos 2θ) sec4 2 ∂f ∂ f ∂ a 0 2. g aa ( − ( a )) = − a 2 4 ∂(x ) ∂t ∂x 16 q r a6=0

X

X ∂ ∂fb ∂fa ∂ ∂fb ∂f0 ( − b) − ( a − b) = 0 a ∂t ∂t ∂x ∂x ∂x ∂x a6=0

(7.77)

(7.78)

Now we have simplified these equations, f0 =

k(θ) + f˜0 r2

(7.79)

where equations (7.77) and (7.78) becomes for b 6= 0, θ (−1 + 2 cos θ + cos 2θ) sec4 2˜ f ∂ ∂f ∂ 0 a 2 − ( a )) = − g aa ( a 2 4 ∂(x ) ∂t ∂x 16 q r a6=0 2 X ∂ k(θ) aa − g , (7.80) a )2 2 ∂(x r a6=0

X

X ∂ ∂fb ∂fa ∂ ∂fb ∂ f˜0 ( − b) − ( a − b) = 0 . a ∂t ∂t ∂x ∂x ∂x ∂x a6=0

(7.81)

we choose k(θ) such that the right hand side of equation (7.80) is zero. The function k(θ) is solved in the equation (kForf0), where we have used (7.53) to calculate the Laplacian in polars. Mathematica gives k(θ) =

1 · √ 2 4 2 (cos2 θ + cos θ − 1) 2q sin θ − sin θ cos θ(cos θ + 1)  p p √ 4 2 4 2 2 2 − sin θ(−2c1 q − sin θ cos θ cos θ(cos θ + 1) (cos θ + cos θ − 1)   √     tanh−1 (cos θ) − 21 (2 cos θ + cos 2θ − 1) cos4 ( 2θ )( cos θ(3 cos 2θ      + cos 3θ + cos θ 2 sin2 θ(log(− cos θ − 1) − log(cos θ − 1)) − 3 − 1)    p √ √ 4 4 2 2 +8q − sin θ cos θ −c1 − ic2 cos θ √

p

(7.82) 121

which with some simplification giving the equation (checking that it still solves the differential equation). k(θ) =

1 · 4q(cos θ + 1)2 (cos2 θ + cos θ − 1) 



csc2 θ(−(2 cos θ + cos 2θ − 1) cos4 2θ (−8ic1 q sin2 θ

    2 + cos θ(2 sin θ(4c2 q + log(− cos θ − 1) − log(cos θ − 1)) − 3)     2 2 2  +3 cos 2θ + cos 3θ − 1) − 4ic1 q sin θ cos θ(cos θ + 1) (cos θ    −1 + cos θ − 1) tanh (cos θ)) (7.83) The problem with this is that it can be singular at θ = 0. To avoid this i singularity we get c1 = giving a value for k(θ) solving (7.80) given in the 2q Mathematica file as (kValue), of cos(θ) −8 c2 q + sec2 2θ + 4 k(θ) = (7.84) 8q Of course (7.84) has a singularity at θ = 180o , but we expect that we will only be able to solve these equations separately on charts θ 6= 180◦ and θ 6= 0◦ , as in the classical case. Note that as f0 is real, we need c2 a real constant. So, using the definition of f0 in (7.79) and by (7.72) and the value of k(θ)

122

in (7.84), f˜0 which is (f0TILDA) in Mathematica file will be,   −3 4 cos(θ) 8c2 cos(θ) iφ −iφ + 4e v cot(θ) − 4e w cot(θ)   r2 − qr2 + r2     cos(θ) sec( 2θ ) iφ −iφ   +4e v csc(θ) + 4e w csc(θ) −   2 qr       −2eiφ sec( 2θ ) tan( 2θ )ξ0 (t, r, θ, φ) − 2 sec( 2θ )ξ1 (t, r, θ, φ)     (0,0,0,1) θ iφ 2   +8ie csc (θ) sin( 2 )ξ0 (t, r, θ, φ)     (0,0,0,1) θ   ) csc(θ)ξ (t, r, θ, φ) +4i csc( 1 2 1  ˜ f0 →   (7.85) (0,0,1,0) θ iφ  8 +8e csc(θ) sin(3 2 )ξ0 (t, r, θ, φ)     (0,0,1,0) θ   −4 csc(θ) csc(3 2 ) sin(3θ)ξ1 (t, r, θ, φ)     (0,1,0,0) θ iφ   −4e r csc(θ) csc(3 2 ) sin(3θ)ξ0 (t, r, θ, φ)     (0,1,0,0) θ   (t, r, θ, φ) −8r csc(θ) sin(3 2 )ξ1      +4r sec( 2θ )yy(1,0,0,0) (t, r, θ, φ) − 4r csc( 2θ )zz(1,0,0,0) (t, r, θ, φ)    (1,0,0,0) (1,0,0,0) θ θ iφ −4e r csc( 2 )ξ0 (t, r, θ, φ) + 4r sec( 2 )ξ1 (t, r, θ, φ) Now (7.85) replaces (7.72), and the zero current equations become X a6=0

g aa (

∂ 2 f˜0 ∂ ∂fa − ( a )) = 0 , a 2 ∂(x ) ∂t ∂x

X ∂ ∂fb ∂fa ∂ ∂fb ∂ f˜0 ( − b) − ( a − b) = 0 . a ∂t ∂t ∂x ∂x ∂x ∂x a6=0

7.1.3

(7.86)

(7.87)

Solving the Dirac equation with zero current

In this section, we go through Appendix C for the Mathematica file ” Finding noncommutative monopole II ”. In (7.72) and (7.73), the functions zz and yy enter only through their time derivatives. If we make the assumption that they have no time dependence, the only place where zz and yy then appear is in (7.63) and (7.64). To solve these we may as well assume zz = 0 and yy = 0, with no effect on anything else. Looking at (7.85) we can substitute the following, to remove the explicit φ dependence (we still have a dependence in the function parameters, but 123

we remove the e±iφ ). The n is the same for vv and ww as in fact w is the conjugate of v. v → rn e−iφ vv(t, r, θ, φ)

,

w → rn eiφ ww(t, r, θ, φ)

ξ1 (t, r, θ, φ) → rn hhh(t, r, θ, φ) ξ0 (t, r, θ, φ) → e−iφ η0 (t, r, θ, φ)

(7.88)

using (7.88), equation (7.85) becomes (f0T) in Mathematica and we get it after some steps of calculations, which gives f˜0 equal to   (0,0,0,1) θ 2 n 2qr 2ir csc 2 csc(θ)hhh (t, r, θ, φ)       (0,0,1,0) 3θ n   (t, r, θ, φ) −2r csc 2 csc(θ) sin(3θ)hhh     (0,1,0,0)   −4r1+n sin 3θ csc(θ)hhh (t, r, θ, φ) 2     θ (1,0,0,0)   (t, r, θ, φ) −2r csc 2 η0     3θ (0,1,0,0)   −2r csc 2 csc(θ) sin(3θ)η0 (t, r, θ, φ)   1   3θ (0,0,1,0) f˜0 →   (7.89) +4 sin 2 csc(θ)η0 (t, r, θ, φ)  8qr2  (0,0,0,1)   θ 2   +4i csc (θ) sin 2 η0 (t, r, θ, φ)       +(4 csc2 (θ) sin 2θ − sec 2θ tan 2θ )η0(t, r, θ, φ)     +2rn (cot (θ) + csc (θ))vv(t, r, θ, φ) + 2rn tan θ ww(t, r, θ, φ)    2     θ 2+n   −2qr (1 + 2n + 4n cos(θ)) sec 2 hhh(t, r, θ, φ)   + (8c2 q − 4) cos(θ) + sec2 2θ − 3q − 2 1 dependence and no t dependence, so hhh rn is a function hhh(θ, φ) only and also that η0 has no time dependence. Under Now we assume that ξ1 has a

124

this assumption, we can solve (7.89) for hhh(0,1) we get:   −4iqr2+n sin( 2θ ) (1 + 2n + 4n cos(θ))hhh(θ, φ) sin( 2θ )     −(1 + cos(θ))vv(t, r, θ, φ) + (−1 + cos(θ))ww(t, r, θ, φ)     i sin(θ) (1,0) 3θ   +2 cos( )hhh (θ, φ) + 2   2(1 + cos(θ))     −2−n θ 3θ r   (2 − 3q) sin( 2 ) + (−4 + q(−3 + 4c2 )) sin( 2 ) (0,1) hhh (θ, φ) →  . (7.90)   4q    ) +2(−1 + 2qc2 ) sin( 5θ 2     2 (0,0,1)   +4qr ((1 + cos(θ))η0 (r, θ, φ) + 2iη0 (r, θ, φ)       2(sin(θ) + sin(2θ))η0 (0,1,0) (r, θ, φ)   (1,0,0) −2r(cos(θ) + cos(2θ))η0 (r, θ, φ)) Now applying the same substitutions in (7.88) and then using (7.90), equation (7.73) becomes (f3) in Mathematica which we get after some steps of calculations and is given by,   2 + q − 4qc2 + (6 + q(3 − 8c2 )) cos(θ) + (2 − 4qc2 ) cos(2θ)     θ θ 2 n  +4qr 4r cos( 2 ) n cos(θ)hhh(θ, φ) − sin( 2 )ww(t, r, θ, φ)    1   (1,0) θ f3 →  − sin(θ)hhh (θ, φ) − csc( 2 ) (1 + cos(θ))η0 (r, θ, φ)  (7.91)  8qr2 (1 + cos(θ))    (0,1,0) (0,0,1)   (r, θ, φ) (r, θ, φ) + 2 sin(θ) cos(θ)η0 +2iη0   (1,0,0) +r sin(θ)η0 (r, θ, φ) Then if (7.90) and (7.91) are satisfied, the Dirac equation is solved. At this stage we have the following unknowns: hhh(θ, φ), c2 , η0 (r, θ, φ), ww(t, r, θ, φ) and vv(t, r, θ, φ) where vv = ww∗ and only the current equations. Returning to (7.80) and (7.81) with relations in (7.67) and the assumptions of v, w given in (7.88), we write ∇·f =

∂f1 ∂f2 ∂f3 + + , ∂x ∂y ∂z

(7.92)

∂ (∇ · f ) , ∂z

(7.93)

and get ∇ 2 f3 =

125

∇2 v = (

∂ ∂ + i )(∇ · f ) , ∂x ∂y

(7.94)

∂ ∂ − i )(∇ · f ) . ∂x ∂y

(7.95)

∇2 w = ( which gives, ∂2 ∂2 1 ∂ ( 2 + 2 )f3 = ∂x ∂y 2 ∂z

∂ ∂ ∂ ∂ ( − i )v + ( + i )w ∂x ∂y ∂x ∂y

,

(7.96)

also, (

∂ ∂f3 1 ∂2 ∂2 ∂ 2v ∂ 2w ∂ + i )( ) = ( 2 + 2 )(v − w) + 2 − i , ∂x ∂y ∂z 2 ∂x ∂y ∂z ∂x∂y

(7.97)

(

∂ ∂ ∂f3 1 ∂2 ∂2 ∂ 2v ∂ 2w − i )( ) = ( 2 + 2 )(v − w) + 2 + i . ∂x ∂y ∂z 2 ∂x ∂y ∂z ∂x∂y

(7.98)

In the Appendix we list these equations as (Eqn796=0) for (7.96), while (Eqn797=0) for (7.97) and (Eqn798=0) for (7.98). These equations are very complicated, and there is currently no way that we can see to solve them apart from the trivial solution f3 = v = w = 0. Then we have to solve (7.91) equal to zero (these are related by a substitution, which is then automatically satisfied). However, there seems no way to separate hhh and η0 in these equations, and they contain terms which are definitely non-zero. We see no way to proceed with this calculation, and for now leave the existence of a deformed monopole as an open question. (Of course, if there were no solutions to (7.91)=0, the fault might lie in an earlier assumption.)

126

Chapter 8 Noncommutative geometry on a Cubic Lattice The differential geometry of finite groups and lattices has been around for some time [17] [54]. More recently proposals have been made to use the differential geometry on lattices, which must be noncommutative, in numerical analysis. Usual numerical analysis is only approximate, and having only approximately conserved energy and other quantities leads to trouble. There are numerical schemes to exactly conserve quantities [6], but a more exotic solution was proposed in [36][37][46]. This uses the fact that the noncommutative finite difference calculus is exact, and so has exactly conserved quantities. However these papers left a problem, one that was already well known for calculi on graphs. The problem is that the differential calculi on lattices are not star algebras. To be more precise, to have a star calculus on a discrete group, we need to have for every step generating the calculus g → g 0 we need the inverse step g 0 → g. For a lattice hZ3 we would need to have steps x 7→ x + h, y 7→ y + h, z 7→ z + h and also x 7→ x − h, y 7→ y − h and z 7→ z − h. We would get a 6 dimensional calculus on hZ3 , 127

which would not be consistent with usual numerical analysis. In this chapter we use bar categories, already used in chapter 4, to construct a star calculus of the correct dimension for the lattice.

8.1

Numerical Partial Derivatives and Noncommutative Geometry

8.1.1

Finite difference calculus on R

A commonly used approximation to the derivative of a function at x is: 0

f (x) '

f (x + h) − f (x) h

(8.1)

if we put a fixed value of h, rather than let h −→ 0, we get an approximation 0

for f (x) with a step length h. In one dimension, if we have a lattice hZ2 ⊂ R, we have partial finite difference derivatives. f h(n + 1) − f (hn) (Dx f )(hn) = h

(8.2)

The problem is that this operator Dx does not satisfy the Leibniz rule. If we write a differential form df = dx.Dx (f ), then we have: d(f g) = dx . Dx (f g) f (hn + h)g(hn + h) − f (hn)g(hn) Dx (f g)(hn) = h f (hn + h) g(hn + h) − g(hn) = h f (hn + h) − f (hn) g(hn) + h = f (hn + h)Dx (g)(hn) + Dx (f )(hn)g(hn)

(8.3)

To simplify this we need a translation, (Th f )(hn) = f (hn + h) 128

(8.4)

then, Dx (f g) = Th (f ) . Dx (g) + Dx (f ) . g

(8.5)

We would like the following Leibniz rule to hold: d(f g) = df . g + f . dg

(8.6)

now using (8.5), dx Th (f ) . Dx (g) + Dx (f ) . g

= dx Dx (f ) g + f dx Dx (g)

(8.7)

we deduce that: dx . Th f = f . dx The basis element

(8.8)

∂ of vector fields can be defined by: ∂x evaluation(

∂ , dx) = 1 ∂x

(8.9)

and for consistency, we require, ∂ ∂ . T−h (f ) = f . ∂x ∂x

(8.10)

This is not partial derivative applied to f , it is the commutation relation relating right and left multiplication on vector fields! We generalize this to n dimensions as follows: Let el be the basis vector of Zn given by (0, ..., 0, 1, 0, ..., 0) where 1 is in position l. Suppose that f : hZn → C, where we keep the h to emphasise that we view this as a discrete approximation to Rn , which arises in the limit h → 0. Define: Dl (f )(x) =

f (x + h el ) − f (x) . h

(8.11)

Dl (f )(x + h em ) − Dl (f )(x) h f (x + h el + h em ) − f (x + h em ) − f (x + h el ) + f (x) = h2

Dm Dl (f )(x) =

129

= Dl Dm (f )(x) ,

(8.12)

which is our version of commutation of partial derivatives. We can collect the previous discussion into the following well known result: Proposition 8.1.1. There is a differential calculus on C(hZn , C) given by: df (h a1 , ..., h an ) =

X

dxi .

i

f (h a1 , ..., h ai + h, ..., h an ) − f (h a1 , ..., h an ) , h (8.13)

for ai ∈ Z. Or in terms of translations, df = =

X i X

dxi . h−1 Th ei (f ) − f

f − T−h ei (f ) dxi . h−1 ,

(8.14)

i

This is extended to high forms by d(f . dxi1 ∧ ... ∧ dxik ) = df ∧ dxi1 ∧ ... ∧ dxik . We have relations: dxi ∧ dxj = −dxj ∧ dxi , dxi . f = T−h ei (f ) dxi .

(8.15) (8.16)

Proof. For a function f (x) we have: d f = dxl . Dl (f ) ,

(8.17)

d2 f = −dxl ∧ d Dl (f ) = −dxl ∧ dxm Dm Dl (f ) .

(8.18)

but by (8.15) this is zero, since dxl ∧ dxm = −dxm ∧ dxl . The same applies for f times a product of dxi . i.e., d(f . dxi1 ∧ ... ∧ dxik ) = df ∧ dxi1 ∧ ... ∧ dxik , d2 (f . dxi1 ∧ ... ∧ dxik ) = d2 f ∧ dxi1 ∧ ... ∧ dxik = 0 .

130

we also need to check the product ξ = f . dxi1 ∧ ... ∧ dxik

,

η = g . dxj1 ∧ ... ∧ dxjs

(8.19)

d(ξ ∧ η) = d(f . dxi1 ∧ ... ∧ dxik ∧ g . dxj1 ∧ ... ∧ dxjs ) = d(f T−h(ei1 +...+eik ) (g)dxi1 ∧ ... ∧ dxik ∧ dxj1 ∧ ... ∧ dxjs ) = dxr Dr (f T−h(ei1 +...+eik ) (g)) ∧ dxi1 ∧ ... ∧ dxik ∧ dxj1 ∧ ... ∧ dxjs , (8.20) and also dξ ∧ η + (−1)ks ξ ∧ dη = dxr Dr (f )dxi1 ∧ ... ∧ dxik ∧ g dxj1 ∧ ... ∧ dxjs + (−1)k f dxi1 ∧ ... ∧ dxik ∧ dxr Dr (g) dxj1 ∧ ... ∧ dxjs = dxr Dr (f )T−h (ei1 +...+eik ) (g)dxi1 ∧ ... ∧ dxik ∧ dxj1 ∧ ... ∧ dxjs + dxr Th er (f ) T−h (ei1 +...+eik ) Dr (g) dxi1 ∧ ... ∧ dxik ∧ dxj1 ∧ ... ∧ dxjs .

(8.21)

so we need to show, for all v Dr (f T−h v (g)) = Dr (f )T−h v (g) + Th er (f )T−h v (Dr (g)) .

(8.22)

Now first we show that T−h v Dr (g) = Dr (T−h v g), Dr (T−h v g) =

Th er −h v (g) − T−h v (g) Th er (g) − g = T−h v = T−h v Dr (g) h h (8.23)

so now we just have to show Dr (f g) = Dr (f )g + Th er (f )Dr (g). To do this: Th er (f )g − f g Th er (f )Th er (g) − Th er (f ) g + h h Th er (f g) − f g = = Dr (f g) . (8.24) h

Dr (f )g + Th er (f )Dr (g) =

This differential calculus is mentioned in the context of numerical analysis with exactly conserved quantise in [36][37][46]. However the calculus itself is much older, going back to calculus on graphs and finite groups [17]. 131

8.1.2

The problem with star operation for a lattice

Defining A = C(hZn , C), for f ∈ A, we have (f ∗ )(x) = f (x)∗ . This star on functions is translation invariant, as: ∗ (Tk f )∗ (x) = Tk f (x) = f (x + kh)∗ = Tk (f ∗ )(x) .

(8.25)

If n = 1, then the star operation applied to dx ∈ Ω1 A has to be dx . g for some g ∈ A, but we expect this should be translation invariant as dx is, so g is a constant. Hence: (dx)∗ = dx

(8.26)

The star operation reverses order, so: (f . dx)∗ = dx . f ∗ , (dx . g)∗ = g ∗ . dx

(8.27)

Applying the star operation on (8.8), we get: (Th f )∗ dx = dx . f ∗

(8.28)

dx . f ∗ = T−h (f ∗ ) . dx

(8.29)

(Th f )∗ = T−h (f ∗ )

(8.30)

But we have,

so,

which is, in general, not true. This means that we have a problem in defining star for 1-forms. Now if we look at the differential calculi on graphs, we can get a star structure, if every single edge has two directions in the standard way − → ← − − → ← − [50] . So, we get dx and dx in the 1-dimensional case. We define (dx)∗ = dx, but this is not like the classical case, as in C(hZ, C) we have a 2-dimensional − → ← − differential calculus dx . f + dx . g Our aim was to have something should be like functions on real line, but this is not! For example, in lattice calculations on fluid flow it would be rather difficult to explain to the engineers why they − → ← − had to have dx and dx in their fluid models. 132

8.2

A Bar Category

We define a bar category in section 2.6.

8.2.1

A star calculus in a bar category

The rules for a usual ? algebra are: 1) d(ξ ∗ ) = (dξ)∗

for ξ ∈ Ωk ,

,

2) (ξ ∧ η)∗ = (−1)ks η ∗ ∧ ξ ∗

,

for η ∈ Ωs .

When we convert these to a bar category where ?(ξ) = ξ ∗ , we get: 1’) d ? = ? d : Ωk → Ωk , 2’) k

s

Ω ⊗Ω

?⊗?

Ωk

⊗

Ωs

Υ−1

∧(−1)ks Ωk+s

Ωs ⊗ Ωk ∧

?

Ωk+s

Fig. 28 We need to show that we can find a bar category structure so that these are satisfied for our C(hZn , C) example. First we need describe the category.

8.2.2

A category of graded modules

One way to fix the star problem is to start with monoidal category which is a category of representations of the algebra A = C(hZn , C) where every object is an A-bimodule. Now for dxi ∈ Ω1 A where 1 ≤ i ≤ n, the right action and left action are related by: dxi . Thei (f ) = f . dxi

133

(8.31)

For vector fields, the relations between the right action and the left one are: ∂ ∂ . f = Thej (f ) . j j ∂x ∂x ∂ ∂ . T−hei (f ) = f . i ∂xi ∂x

(8.32) (8.33)

and if we look at Ω1 A ⊗ Ω1 A , the relation is: f . (dxi ⊗ dxj ) = (dxi ⊗ dxj ) . Th(ei +ej ) (f )

(8.34)

We will generalise commutation relation using a graded vector space: A graded vector space V is a direct sum of subspaces V = ⊕i∈S Vi , where S is the set of grades. An element of one of the subspaces (whose grade exist) is called homogeneous, and every element can be written as a sum of homogeneous elements[9]. In our case, we have a direct sum for every object V in the monoidal category V = ⊕r∈Zn Vr . The grade says how the left and right actions are related, for |v| ∈ Zn , we have: v . f = Th|v| (f ) . v

(8.35)

∂ | = ej ∂xj

(8.36)

|dxi | = −ei

(8.37)

|dxi ⊗ dxj | = −ei − ej

(8.38)

For vector fields, | for 1-form,

and for |dxi ⊗ dxj |,

To any vector field or form, or tensor product of these, we can assign a grade ∈ Zn . If we look at the grade of dxi . a ∈ Ω1 A for all a ∈ A, (dxi . a)f = dxi . f . a = T−hei (f ) . dxi . a |dxi . a| = −ei

(8.39) 134

so multiplying by a function does not change the grade. In general, for homogeneous v, w, |v ⊗ w| = |v| + |w|

8.2.3

(8.40)

The monoidal category C

The objects of C are Zn graded. Let A = C(hZn , C) be the usual algebra, and take V an A-bimodule, with an action of the group (hZn , +) defined, satisfying the following conditions: Given an object V , the grade of v ∈ V is written |v| ∈ Zn . Tk (v) ∈ V denotes the action ”Translation” by k ∈ hZn . a.v ∈ V and v.a ∈ V denote the left and right A actions, for a ∈ A. 1) |a.v| = |v| = |Tk (v)| = |v.a|, 2) Tk (a.v) = Tk (a).Tk (v) , Tk (v.a) = Tk (v).Tk (a), 3) v.a = Th|v| (a).v. A definition of morphism which we can use is θ : V → V , which is a morphism when, 1) θ : V → V is an A-bimodule map, 2) |θ(v)| = |v|, 3) θ(Tk (v)) = Tk (θ(v)). Example 8.2.1. The object A in C is giving the usual bimodule structure by product on A, |a| = 0 and the usual translation of functions (Tv (a))(x) = f (x + v). Ω1 A is an object in C with |dxi .a| = −ei and Tv (dxi .a) = dxi .Tv (a). ∂ ∂ ∂ VecA is an object in C with |a i | = ei and Tv (a i ) = Tv (a) i ∂x ∂x ∂x Given objects E, F in C, we define a new object E ⊗A F with grade and actions as following: 1) a.(e ⊗ f ) = a.e ⊗ f ,

2) (e ⊗ f ).a = e ⊗ f.a

135

3) |e ⊗ f | = |e| + |f |,

4) Tv (e ⊗ f ) = Tv (e) ⊗ Tv (f )

Note that ⊗ has A as the identity object 1C . For any object E in C, lE : E → E ⊗A A and rE : E → A ⊗A E, we have lE (e) = e ⊗ 1 and rE (e) = 1 ⊗ e. We need to check the consistency of this with rule (v.a = Th|v| (a).v), (e ⊗ f ) . a = e ⊗ f a = e ⊗ Th|f | (a) f = e . Th|f | (a) ⊗ f = Th(|e|+|f |) (a) . e ⊗ f . (8.41)

8.3

A Conjugate on C

We need to describe the object X, for X ∈ C. We take X to be the conjugate vector space to X, with elements x ∈ X for x ∈ X. that means x+y =x+y

,

λ . x = λ∗ x for λ ∈ C .

(8.42)

The grade is given by |x| = |x|, and Tv (x) = Tv x. The module structure we shall find later. Rule (b) of Definition 2.6.1 can be written as the commutating diagram:

136

¯ Φ

(X ⊗ Y ) ⊗ Z

X ⊗ (Y ⊗ Z) ΥX,Y ⊗Z

ΥX⊗Y,Z Z¯ ⊗ X ⊗ Y

¯ Y ⊗Z ⊗X idZ¯ ⊗ ΥX,Y ΥY,Z ⊗ idX¯

¯ Z¯ ⊗ (Y¯ ⊗ X)

ΦZ, ¯ Y¯ ,X ¯

¯ (Z¯ ⊗ Y¯ ) ⊗ X

Fig. 29 Note in our case, Φ is just the identity. But taking Υ trivial leads us to the standard way which does not give the star operation. So, we work on modifying this condition by translation of functions. We guess that Υ(x ⊗ y) = T.. (y) ⊗ T... (x). These dots indicate h multiplied by the grade of some elements of Zn . To find these elements, we start by assuming that for |x|, |y| ∈ Zn and p, q, r, s ∈ Z, we have: Υ(x ⊗ y) = Th(p|y|+q|x|) (y) ⊗ Th(r|y|+s|x|) (x) .

(8.43)

If we start by the right side of the commutative diagram, we have: (x ⊗ y) ⊗ z

Φ

−→ ΥX,(Y ⊗Z)

−→

x ⊗ (y ⊗ z) Th(p|y|+p|z|+q|x|) (y ⊗ z) ⊗ Th(r|y|+r|z|+s|x|) (x)

ΥY,Z⊗id

−→ X Th((p+q)|y|+2p|z|+q|x|) (z) ⊗ Th((p+s)|y|+(p+r)|z|+q|x|) (y) ⊗ Th(r|y|+r|z|+s|x|) (x)

137

(8.44)

regarding that Te (y ⊗ z) = Te (y) ⊗ Te (z) and |Te (x)| = |x|. For the left side, we get: (x ⊗ y) ⊗ z

Υ(X⊗Y ),Z

−→

Υid

⊗X,Y

Z −→

Th(p|z|+q|x|+q|y| (z) ⊗ Th(r|z|+s|x|+s|y|) (x ⊗ y) Th(p|z|+q|x|+q|y|) (z) ⊗ Th(r|z|+(s+q)|x|+(s+p)|y|) (y) ⊗ Th(r|z|+2s|x|+(s+r)|y|) (x)

(8.45)

but to make both sides equal, we require p = s = 0. Hence, Υ(x ⊗ y) = Thq|x| (y) ⊗ Thr|y| (x)

(8.46)

Υ−1 (y ⊗ x) = T−hr|y| (x) ⊗ T−hq|x| (y)

(8.47)

The left and the right modules structure for the conjugates are given by: l−1

Υ−1

?⊗id

E A ⊗A E −→ A ⊗A E −→ E ⊗A A −→ E

Υ−1

id⊗?

r−1

E E ⊗ A −→ E ⊗ A −→ A ⊗ E −→ E

(8.48) (8.49)

then the left and right actions on X respectively : a . x = x . T−hq|x| (a∗ ) ,

x . a = T−hr|x| (a∗ ) . x

(8.50)

The ? operation on A is just ?(a) = a∗ e ⊗ a 7−→ e ⊗ a∗ 7−→ T−hr|e| (a∗ ) ⊗ e 7−→ T−hr|e| (a∗ ) . e

(8.51)

Now we check that Υ is a right module map: Υ(x ⊗ y.a) = Υ(T−hr(|x|+|y|) (a∗ ).(x ⊗ y)) = Thq|x| (y) ⊗ T−hr|x| (a∗ ).Thr|y| (x) = Thq|x| (y) ⊗ Thr|y| (x).(a) We define ?(dxi ) = dxi ,

(8.52)

there is not much choice here, if we choose to

preserve translation invariance. Then, as ? is a bimodule map: ? (da) = ? (dxi . Di (a)) = ? (dxi ) . Di (a) = dxi . Di (a) = Threi (Di (a∗ )) . dxi = dxi . Th(r+1)ei (Di (a∗ )) 138

(8.53)

This is not in general da∗ , it is a translation. ? (da) = T−h(r+1)|da| (da∗ )

(8.54)

However we wish to have ? commute with d to have a ? - calculus, so we set r = −1 . We also want the following for a ? - calculus. Classically we want (ξ ∧ η)∗ = (−1)ks η ∗ ∧ ξ ∗ , for ξ ∈ Ωk and η ∈ Ωs . In a bar category, Ωk ⊗ Ωs

?⊗?

Ωk ⊗ Ωs

Υ−1

∧(−1)ks

Ωs ⊗ Ωk ∧

?

Ωk+s

Ωk+s

Fig. 30 For particular cases of k = 0, s = 1 and the other way round, we have: a ⊗ dxi

?⊗?

a∗ ⊗ dxi

Υ−1

∧(−1)ks

dxi ⊗ Thqei (a∗ ) ∧

?

a dxi

dxi Thqei (a∗ )

Fig. 31 i

dx ⊗ a

?⊗?

dxi

⊗

a∗

∧(−1)ks dxi a

Υ−1

Threi (a∗ ) ⊗ dxi ∧

?

Threi (a∗ ) dxi

Fig. 32 From Fig. 30 we have ?(a dxi ) = a (?dxi ) = a dxi = dxi Thqei (a∗ ) , 139

(8.55)

and from Fig. 31 we have ?(a dxi ) = ?(dxi Thei (a)) = ?(dxi ) Thei (a) = dxi Thei (a) = Threi (Thei (a∗ )) dxi , (8.56) now from (8.55) and (8.56), dxi Thqei (a∗ ) = Th(r+1)ei (a∗ ) dxi , dxi Thqei (a∗ ) = dxi Thei (Th(r+1)ei (a∗ )) = Th(r+2)ei (a∗ ) dxi .

(8.57) (8.58)

For all a, q = r + 2 but r = −1, then q = 1 and the conjugation works! Proposition 8.3.1. The category C is a bar category, with the operations bb(x) = x a . x = x . T−h|x| (a∗ )

,

x . a = Th|x| (a∗ ) . x

Υ(x ⊗ y) = Th|x| (y) ⊗ T−h|y| (x)

(8.59)

Proof. The form of Υ has been chosen to satisfy Fig. 28. Now, Equation (8.52) shows that Υ is a right module map. We now show that Υ is left module map and bb is a bimodule map. Υ(a.x ⊗ y) = Υ((x ⊗ y).T−h(|x|+|y|) (a∗ )) = Th|x| (y).Th|y| (a∗ ) ⊗ T−h|y| (x) = (a).Th|x| (y) ⊗ T−h|y| (x)

(8.60)

bb(a.x) = a.x = x.T−h|x| (a∗ ) = a∗ .x = a.bb(x) bb(x.a) = x.a = Th|x| (a∗ ).x = x.a∗ = bb(x).a (a) of Definition 2.6.1. is just (8.48) and (8.49), and this gives the displayed formula for the bimodule action on X.

140

Theorem 8.3.1. The algebra C(hZn , C) has a star differential calculus in the bar category C. This is given for f ∈ C(hZn , C) by, 1)

(f dxi1 ∧ ... ∧ dxik )∗ = f ∗ dxi1 ∧ ... ∧ dxik ,

2)

|f dxi1 ∧ ... ∧ dxik | = −ei1 − ... − eik ,

3)

Tv (f dxi1 ∧ ... ∧ dxik ) = Tv (f ) dxi1 ∧ ... ∧ dxik .

Proof. For ξ = f . dxi1 ∧ ... ∧ dxik ∈ Ωk , we have dξ = d(f . dxi1 ∧ ... ∧ dxik )

,

ξ ∗ = f ∗ . dxi1 ∧ ... ∧ dxik

d ? ξ = d(f ∗ ) . dxi1 ∧ ... ∧ dxik

(8.61)

So we need to show that: ?(d f . dxi1 ∧ ... ∧ dxik ) = d(f ∗ ) . dxi1 ∧ ... ∧ dxik 1 ? (T−h ei (f ) − f )dxi ∧ dxi1 ∧ ... ∧ dxik ?(d f . dxi1 ∧ ... ∧ dxik ) = h 1 = (T−h ei (f ∗ ) − f ∗ )dxi1 ∧ ... ∧ dxik . h (8.62) For η = g . dxj1 ∧ ... ∧ dxjs ∈ Ωs and if we apply Fig. 28 to ξ ⊗ η. ∧Υ−1 (? ⊗ ?)(ξ ⊗ η) = ∧Υ−1 (f ∗ . dxi1 ∧ ... ∧ dxik ⊗ g ∗ . dxj1 ∧ ... ∧ dxjs ) = ∧(T−h(ei1 +...+eik ) g ∗ dxj1 ∧ ... ∧ dxjs ⊗ Th(ej1 +...+ejs ) f ∗ dxi1 ∧ ... ∧ dxik ) = T−h(ei1 +...+eik ) g ∗ dxj1 ∧ ... ∧ dxjs ∧ Th(ej1 +...+ejs ) f ∗ dxi1 ∧ ... ∧ dxik = T−h(ei1 +...+eik ) g ∗ f ∗ dxj1 ∧ ... ∧ dxjs ∧ dxi1 ∧ ... ∧ dxik ξ ∧ η = f . dxi1 ∧ ... ∧ dxik ∧ g . dxj1 ∧ ... ∧ dxjs = f T−h(ei1 +...+eik ) (g)dxi1 ∧ ... ∧ dxik ∧ dxj1 ∧ ... ∧ dxjs = (−1)sk f T−h(ei1 +...+eik ) (g)dxj1 ∧ ... ∧ dxjs ∧ dxi1 ∧ ... ∧ dxik (8.64) Note that d is not a morphism in the category, but this is also occurs in the ordinary algebra bimodule category, where d is not a bimodule map, so it is also not a morphism in that case. 141

(8.63)

To go from here to looking at Riemannian structure, we would need an inner product h i : Ω1 ⊗A Ω1 → A. However, we have |dxi | = −ei , so |dxi ⊗ dxj | = −ei − ej which is never zero! So the Riemannian structure can not be this simple. It is not obvious how to proceed here.

142

Chapter 9 Conclusion and possible further work We have already summarised the contents in the introduction, but we use this section to give possible further directions for research work. In general, the interaction between between gravity, quantum theory and noncommutative geometry remain only partly understood. In chapter 4 we introduce an O(λ2 ) associator, only to find that the pentagon equation fails to be satisfied at O(λ4 ). Can we continue to introduce higher order corrections? The idea of successive patching of nonassoiciativity structure to make associative structure may be linked to the theory of A∞ algebras (see [44][45]). This might allow us to add to the associator to give higher order corrections to make on A∞ algebra. How does the differential calculus in chapter 4 relate to a deformed Poincaré algebra and its differential calculus - this is so far not known. It would be expected that the resulting deformation of the Poincaré algebra would also have a nontrivial associator. Can we introduce a Hermitian metric for the lattice in chapter 8? How 143

to make such a metric central for the algebra is not obvious. But what sort of construction would be useful for numerical calculations? In chapter 5 we calculate a variation in propagation speed. However does this variation depend on assumptions, such as a choice of normal order? Is there a way to make a physical choice of equation which is natural and does not depend on such assumptions? The presence of a term with xj multiplying it in the fields in (6.26) indicates how normal orders may have strange effects on results. In chapter 7, it would be nice to find if there really is a monopole solution to the noncommutative Dirac equation (including the current equation) for our central metric calculus. It may well be that the failure to solve the differential equation shows up a real theoretical problem, in that we may not have the correct equations. In particular, should the γ µ matrices be deformed? A more theoretical analysis might give some answers. In particular, it might be useful to ask whether our non-standard calculus comes from a calculus on the κ-Poincaré Hopf algebra (again to first order, with extension to second order via an associator). If this could be done, a Hopf-algebraic construction of the monopole might be possible. Note. There tend to be two approaches to making classical monopoles, solving the Dirac equation and group theoretic methods. These are, of course, related. However, the distinction may be useful in the noncommutative case where we are uncertain exactly which Dirac equation to solve!

144

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152

Appendix A Solution to 4-spinor Dirac equation

153

Here, we solve the Massive Dirac equation for 4 -− spinors using the rotationally symmetric spinor fields (up to gauge transformation) and we get a solution to the Mass -− less Dirac equation. we solve : ⅈ γμ ∂μ -− ⅈ q Aμ  ϕ = m ϕ where ϕ = g (r) B ψ.

Define: (*⋆ Gamma matrices γμ , EQuation (3.22)*⋆) g0 := {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -−1, 0}, {0, 0, 0, -−1}} g1 := {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, -−1, 0, 0}, {-−1, 0, 0, 0}} g2 := {{0, 0, 0, -−I}, {0, 0, I, 0}, {0, I, 0, 0}, {-−I, 0, 0, 0}} g3 := {{0, 0, 1, 0}, {0, 0, 0, -−1}, {-−1, 0, 0, 0}, {0, 1, 0, 0}}

(*⋆ Matrix B and its inverse are needed to solve, EQuation (7.34) *⋆) B := {{Cos[θ /∕ 2], (-−Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2], 0, 0}, {(Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2], Cos[θ /∕ 2], 0, 0}, {0, 0, Cos[θ /∕ 2], (-−Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2]}, {0, 0, (Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2], Cos[θ /∕ 2]}} θ θ invB := Cos , Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 0, 0, 2 2 θ θ -−Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), Cos , 0, 0, 2 2 θ θ 0, 0, Cos , Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 2 2 θ θ 0, 0, -−Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), Cos  2 2

(*⋆ Ploar coordinates matrix cij , EQuation (7.38) *⋆) Cos[θ] Cos[ϕ] Csc[θ] Sin[ϕ] c := Cos[ϕ] Sin[θ], , -− , r r Cos[θ] Sin[ϕ] Cos[ϕ] Csc[θ] Sin[θ] Sin[θ] Sin[ϕ], , , Cos[θ], -− , 0 r r r

dirP1 = Solving the first bit of the Dirac : ∂ μ γμ (g (r) B) = γi ci1 B D[g (r), r] + g (r) γi (ci2 D[B, θ] + ci3 D[B, ϕ]) . Then, multiply by invB and invg (r) coming form R.H.S to get CompP1

Printed by Wolfram Mathematica Student Edition

2

Solution to 4-spinor Dirac equation.nb

(*⋆ g(r) derivative. The value of g(r) is unknown! *⋆) gD = Simplify[(c[[1, 1]] g1 + c[[2, 1]] g2 + c[[3, 1]] g3).B D[g[r], r]] θ θ 0, 0, Cos  g′ [r], Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]) g′ [r], 2 2 θ θ 0, 0, Sin  (Cos[ϕ] + ⅈ Sin[ϕ]) g′ [r], -− Cos  g′ [r], 2 2 θ θ -− Cos  g′ [r], -− Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]) g′ [r], 0, 0, 2 2 θ θ -− Sin  (Cos[ϕ] + ⅈ Sin[ϕ]) g′ [r], Cos  g′ [r], 0, 0 2 2

(*⋆ θ derivative. *⋆) thetaD = Simplify[(c[[1, 2]] g1 + c[[2, 2]] g2 + c[[3, 2]] g3).D[B, θ]] θ 2

Cos  0, 0,

2r

θ 2

Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]) ,

, 2r

θ 2

Sin  (Cos[ϕ] + ⅈ Sin[ϕ]) 0, 0,

, -−

2r θ 2

Cos  -−

2r

,

2r

θ 2

Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]) , -−

, 0, 0,

2r

θ 2

Sin  (Cos[ϕ] + ⅈ Sin[ϕ]) -−

θ 2

Cos 

θ 2

Cos  ,

2r

, 0, 0

2r

(*⋆ ϕ derivative. *⋆) phiD = Simplify[(c[[1, 3]] g1 + c[[2, 3]] g2 + c[[3, 3]] g3).D[B, ϕ]] θ 2

Csc[θ] Sin  0, 0,

r θ 2

, 0, 0, 0, 0, -−

Sec  -−

2r

θ 2

Sec  θ 2

,

2r

Csc[θ] Sin  , 0, 0, 0, 0,

r

, 0, 0



3

(*⋆ EQUATION (7.39) *⋆) dirP1 = Simplify[gD + g[r] (thetaD + phiD)] θ 2

2

θ 2

Sec  (3 + Cos[θ]) g[r] + 4 r Cos  g′ [r] 0, 0,

,

4r θ 2

Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]) (g[r] + 2 r g′ [r]) , 2r θ 2

Sin  (Cos[ϕ] + ⅈ Sin[ϕ]) (g[r] + 2 r g′ [r]) 0, 0,

,

2r θ 2

θ 2

2

Cos  -− g[r] 1 + Sec   -− 2 r g′ [r]

θ 2

θ 2

2

Cos  -− g[r] 1 + Sec   -− 2 r g′ [r] , 

2r

2r

θ 2

Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]) (g[r] + 2 r g′ [r]) -−

, 0, 0,

2r θ 2

Sin  (Cos[ϕ] + ⅈ Sin[ϕ]) (g[r] + 2 r g′ [r]) -−

,

2r θ 2

θ 2

2

Sec  (3 + Cos[θ]) g[r] + 4 r Cos  g′ [r] , 0, 0

4r

CompP1 = Simplify[( invB.dirP1) /∕ (g[r])] 0, 0,

1 r

0, 0, -−

-−



1 r

-−

+

g′ [r] g[r]

, -−

(Cos[ϕ] -− ⅈ Sin[ϕ]) Tan θ2  2r

(Cos[ϕ] + ⅈ Sin[ϕ]) Tan θ2 

g′ [r] g[r]

2r ,

, -−

1 r

(Cos[ϕ] -− ⅈ Sin[ϕ]) Tan θ2  2r

(Cos[ϕ] + ⅈ Sin[ϕ]) Tan θ2  2r

,

1 r

+

g′ [r] g[r]

-−

,

g′ [r] g[r]

,

, 0, 0,

, 0, 0

dirP2 = Solving the second bit of the Dirac for : ⅈ q Aμ → ∑ eμ γμ and multiply it by invB coming form the R.H.S to get CompP2


,

4


dirP2 = Simplify[(e0 g0 + e1 g1 + e2 g2 + e3 g3).B ] θ θ e0 Cos , -−e0 Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 2 2 θ θ e3 Cos  + (e1 -− ⅈ e2) Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), 2 2 θ θ (e1 -− ⅈ e2) Cos  -− e3 Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 2 2 θ θ e0 Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), e0 Cos , 2 2 θ θ (e1 + ⅈ e2) Cos  -− e3 Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), 2 2 θ θ -−e3 Cos  + (e1 + ⅈ e2) Sin  (-−Cos[ϕ] + ⅈ Sin[ϕ]), 2 2 θ θ -−e3 Cos  -− (e1 -− ⅈ e2) Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), 2 2 θ θ -−(e1 -− ⅈ e2) Cos  + e3 Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 2 2 θ θ -−e0 Cos , e0 Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 2 2 θ θ -−(e1 + ⅈ e2) Cos  + e3 Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), 2 2 θ θ e3 Cos  + (e1 + ⅈ e2) Sin  (Cos[ϕ] -− ⅈ Sin[ϕ]), 2 2 θ θ -−e0 Sin  (Cos[ϕ] + ⅈ Sin[ϕ]), -−e0 Cos  2 2

CompP2 = Simplify[invB.dirP2] {{e0, 0, e3 Cos[θ] + Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]), (Cos[ϕ] -− ⅈ Sin[ϕ]) ((-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− e3 Sin[θ] + (ⅈ e1 + e2 Cos[θ]) Sin[ϕ])}, {0, e0, (Cos[ϕ] + ⅈ Sin[ϕ]) ((ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− e3 Sin[θ] + (-−ⅈ e1 + e2 Cos[θ]) Sin[ϕ]), -−e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ])}, {-−e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]), -−(Cos[ϕ] -− ⅈ Sin[ϕ]) ((-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− e3 Sin[θ] + (ⅈ e1 + e2 Cos[θ]) Sin[ϕ]), -−e0, 0}, {-−(Cos[ϕ] + ⅈ Sin[ϕ]) ((ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− e3 Sin[θ] + (-−ⅈ e1 + e2 Cos[θ]) Sin[ϕ]), e3 Cos[θ] + Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]), 0, -−e0}}

Subtracting the two parts to get the Dirac equation in EQUATION (7.40)



Dirac = Simplify[CompP1 -− CompP2] -−e0, 0, -−

1 2r

1 r

-− e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]) +

g′ [r] g[r]

,

(Cos[ϕ] -− ⅈ Sin[ϕ]) 2 r (-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -−

θ 2 e3 r Sin[θ] + 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 2 1 0, -−e0, -− (Cos[ϕ] + ⅈ Sin[ϕ]) 2 r (ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2r θ 2 e3 r Sin[θ] -− 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 2 1 g′ [r] -− + e3 Cos[θ] + e1 Cos[ϕ] Sin[θ] + e2 Sin[θ] Sin[ϕ] -− , r g[r] -−

1 r

+ e3 Cos[θ] + e1 Cos[ϕ] Sin[θ] + e2 Sin[θ] Sin[ϕ] -−

g′ [r] g[r]

,

1 2r

(Cos[ϕ] -− ⅈ Sin[ϕ]) 2 r (-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2 e3 r Sin[θ] + θ 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , e0, 0, 2 

1 2r

(Cos[ϕ] + ⅈ Sin[ϕ]) 2 r (ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2 e3 r Sin[θ] -−

θ 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 2 1 g′ [r] -− e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]) + , 0, e0 r g[r]

(*⋆ Solving for eμ 's *⋆) Let = {e0 → 0};


5

6


COMP = Simplify[Dirac /∕/∕. Let] 0, 0, 1

-−

2r

1 r

-− e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]) +

g′ [r] g[r]

,

(Cos[ϕ] -− ⅈ Sin[ϕ]) 2 r (-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -−

θ 2 e3 r Sin[θ] + 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 2 1 0, 0, -− (Cos[ϕ] + ⅈ Sin[ϕ]) 2 r (ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2r θ 2 e3 r Sin[θ] -− 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 2 1 g′ [r] -− + e3 Cos[θ] + e1 Cos[ϕ] Sin[θ] + e2 Sin[θ] Sin[ϕ] -− , r g[r] 1

-−

r

+ e3 Cos[θ] + e1 Cos[ϕ] Sin[θ] + e2 Sin[θ] Sin[ϕ] -−

g′ [r] g[r]

,

1 2r

(Cos[ϕ] -− ⅈ Sin[ϕ]) 2 r (-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2 e3 r Sin[θ] + θ 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 0, 0, 2 

1 2r

(Cos[ϕ] + ⅈ Sin[ϕ]) 2 r (ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2 e3 r Sin[θ] -−

θ 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  , 2 1 g′ [r] -− e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]) + , 0, 0 r g[r]

whatise3 = Simplify[Solve[-−e3 Cos[θ] -− Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]) ⩵ 0, e3]] {{e3 → -−(e1 Cos[ϕ] + e2 Sin[ϕ]) Tan[θ]}}

Simp = Simplify-−

1 2r

(Cos[ϕ] -− ⅈ Sin[ϕ]) 2 r (-−ⅈ e2 + e1 Cos[θ]) Cos[ϕ] -− 2 e3 r Sin[θ] +

θ 2 ⅈ e1 r Sin[ϕ] + 2 e2 r Cos[θ] Sin[ϕ] + Tan  /∕/∕. 2 e3 → -−(e1 Cos[ϕ] + e2 Sin[ϕ]) Tan[θ] -−

1 2r

(Cos[ϕ] -− ⅈ Sin[ϕ])

θ 2 r (e1 -− ⅈ e2 Cos[θ]) Cos[ϕ] Sec[θ] + 2 r (ⅈ e1 + e2 Sec[θ]) Sin[ϕ] + Tan  2



7

compute2 = Simplify[Solve[-−Sin[θ] (e1 Cos[ϕ] + e2 Sin[ϕ]) ⩵ 0, e2]] {{e2 → -−e1 Cot[ϕ]}}

Simp1 = Simplify[Simp /∕/∕. e2 → -−e1 Cot[ϕ]] -−

1

θ θ 2 e1 r + 2 ⅈ e1 r Cot[ϕ] + Cos[ϕ] Tan  -− ⅈ Sin[ϕ] Tan  2r 2 2

compute1 = SimplifySolve -−

1

θ ⅈ (Cos[ϕ] -− ⅈ Sin[ϕ]) 2 e1 r Cos[ϕ] Cot[ϕ] + 2 e1 r Sin[ϕ] -− ⅈ Tan  ⩵ 0, 2r 2

e1 e1 →

ⅈ Sin[ϕ] Tan θ2  2r



compute3 = Simplifywhatise3 /∕/∕. e2 → -−e1 Cot[ϕ], e1 →

ⅈ Sin[ϕ] Tan θ2 

{{e3 → 0}} (*⋆ Solution, EQUATION (7.40) *⋆)

eμ = e0 → 0, e1 →

e0 → 0, e1 →



, e2 → -−e1 Cot[ϕ], e3 → 0

, e2 → -−e1 Cot[ϕ], e3 → 0

value = {g[r] -−> 1 /∕ r , g′ [r] → -−1 /∕ r^2} g[r] →

1 r

, g′ [r] → -−

1 r2




2r



8


subDirac = SimplifySimplifyDirac /∕/∕. e0 → 0, e1 →

ⅈ Sin[ϕ] Tan θ2 

{0, 0, 0, 0}, 0, 0, -−

{0, 0, 0, 0}, 

2r

, e2 → -−e1 Cot[ϕ], e3 → 0 /∕/∕. value

(Cos[ϕ] + ⅈ Sin[ϕ]) Tan θ2  r

(Cos[ϕ] + ⅈ Sin[ϕ]) Tan θ2  r

, 0,

, 0, 0, 0

(*⋆ ψ could be also {{1},{0},{1},{0}} *⋆) DIRAC = subDirac.{{0}, {1}, {0}, {1}} {{0}, {0}, {0}, {0}}


Appendix B Finding a noncommutative monopole I

162

We solve the general massless noncommutative Dirac equation : D_NC ψ = γμ ∂μ -− ⅈ q Aμ  ψ +

λ 2

γ0 

∂2 ψ ∂2 xi

.

Given: (*⋆ g0 g1 g2 g3

Gamma matrices γμ , EQuation (3.22)*⋆) := {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -− 1, 0}, {0, 0, 0, -− 1}} := {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, -− 1, 0, 0}, {-− 1, 0, 0, 0}} := {{0, 0, 0, -− I}, {0, 0, I, 0}, {0, I, 0, 0}, {-− I, 0, 0, 0}} := {{0, 0, 1, 0}, {0, 0, 0, -− 1}, {-− 1, 0, 0, 0}, {0, 1, 0, 0}}

(*⋆ Matrix B, EQuation (7.34) *⋆) B := {{Cos[θ /∕ 2], (-−Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2], 0, 0}, {(Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2], Cos[θ /∕ 2], 0, 0}, {0, 0, Cos[θ /∕ 2], (-−Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2]}, {0, 0, (Cos[ϕ] + I Sin[ϕ]) Sin[θ /∕ 2], Cos[θ /∕ 2]}}

(*⋆ Ploar coordinates matrix cij , EQuation (7.38) *⋆) Cos[θ] Cos[ϕ] Csc[θ] Sin[ϕ] c := Cos[ϕ] Sin[θ], , -− , r r Cos[θ] Sin[ϕ] Cos[ϕ] Csc[θ] Sin[θ] Sin[θ] Sin[ϕ], , , Cos[θ], -− , 0 r r r (*⋆ Solution, where e0=e3=0, EQUATION (7.41) *⋆) θ 2

ⅈ Sin[ϕ] Tan  e1 =

;

2r θ 2

ⅈ Sin[ϕ] Tan  e2 = -−

2r

Cot[ϕ];

(*⋆ Laplacian, EQUATION (7.53) *⋆) θ 2

θ 2

ⅇ-−ⅈ ϕ (5 + 3 Cos[θ]) Sec  Tan  Lap = 

θ 2

(1 + 3 Cos[θ]) Sec  , -−

16 r3 θ 2

θ 2

ⅇ-−ⅈ ϕ (5 + 3 Cos[θ]) Sec  Tan  -−

,

16 r3 θ 2

(1 + 3 Cos[θ]) Sec  , 

16 r3

;

16 r3

(*⋆ Original monopole solution with g(r)=1/∕r, EQUATION (7.33) *⋆) ψ = FullSimplify[(1 /∕ r) B.{{0}, {1}, {0}, {1}}] θ 2

ⅇ-−ⅈ ϕ Sin  -− r

θ 2

,  r

θ 2

ⅇ-−ⅈ ϕ Sin 

Cos  , -−

r

θ 2

Cos  , 



r

(*⋆ unknown spinor field defined in ψ_new *⋆) ξ = {{ξ0[t, r, θ, ϕ]}, {ξ1[t, r, θ, ϕ]}, {ξ2[t, r, θ, ϕ]}, {ξ3[t, r, θ, ϕ]}} ;

Now the Extra terms in the General NC Dirac: Lap + partial xi gamma - emus xi gamma - fmus gamma psi (NO Lambdas included) Printed by Wolfram Mathematica Student Edition

2

Finding noncommutative monopole I.nb

Now the Extra terms in the General NC Dirac: Lap + partial xi gamma - emus xi gamma - fmus gamma psi (NO Lambdas included) (*⋆ coordinates changed to cartizian *⋆) one = FullSimplify[ g0.D[ ξ, t] + g1.(c[[1, 1]] D[ ξ, r] + c[[1, 2]] D[ξ, θ] + c[[1, 3]] D[ξ, ϕ]) + g2.(c[[2, 1]] D[ ξ, r] + c[[2, 2]] D[ξ, θ] + c[[2, 3]] D[ξ, ϕ]) + g3.(c[[3, 1]] D[ ξ, r] + c[[3, 2]] D[ξ, θ] + c[[3, 3]] D[ξ, ϕ])] 

1 r

Csc[θ] (-− ⅈ Cos[ϕ] -− Sin[ϕ]) ξ3(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇ-−ⅈ ϕ Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ],



1 r

Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ⅈ Csc[θ]2 ξ2(0,0,0,1) [t, r, θ, ϕ] + Cot[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + r ξ2(0,1,0,0) [t, r, θ, ϕ] -− r Cot[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ],



1 r

Csc[θ] (ⅈ Cos[ϕ] + Sin[ϕ]) ξ1(0,0,0,1) [t, r, θ, ϕ] + Sin[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ⅇ-−ⅈ ϕ Cos[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r Cos[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(1,0,0,0) [t, r, θ, ϕ],



1 r

Csc[θ] (-− ⅈ Cos[ϕ] + Sin[ϕ]) ξ0(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r ξ3(1,0,0,0) [t, r, θ, ϕ]

two = FullSimplify[e1 g1. ξ + e2 g2. ξ] θ 2

ⅇ-−ⅈ ϕ Tan  ξ3[t, r, θ, ϕ] -−

θ 2

ⅇⅈ ϕ Tan  ξ2[t, r, θ, ϕ] , 

2r θ 2

ⅇ-−ⅈ ϕ Tan  ξ1[t, r, θ, ϕ] 

θ 2

ⅇⅈ ϕ Tan  ξ0[t, r, θ, ϕ] , -−

2r

, 2r  2r

three = FullSimplify[(f0 g0.ψ + f1 g1.ψ + f2 g2.ψ + f3 g3.ψ)]    

1 r 1 r 1 r 1 r

θ θ (f1 -− ⅈ f2) Cos  -− ⅇ-−ⅈ ϕ (f0 + f3) Sin  , 2 2 θ θ (f0 -− f3) Cos  -− ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  , 2 2 θ θ -− (f1 -− ⅈ f2) Cos  + ⅇ-−ⅈ ϕ (f0 + f3) Sin  , 2 2 θ θ (-− f0 + f3) Cos  + ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin   2 2



Simplify[one -− two -− three] 

1

θ θ θ -− 2 (f1 -− ⅈ f2) Cos  -− ⅇ-−ⅈ ϕ (f0 + f3) Sin  + ⅇ-−ⅈ ϕ Tan  ξ3[t, r, θ, ϕ] + 2r 2 2 2 2 Csc[θ] (-− ⅈ Cos[ϕ] -− Sin[ϕ]) ξ3(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇ-−ⅈ ϕ Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] ,

-−

1

θ θ θ 2 (f0 -− f3) Cos  -− ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  + ⅇⅈ ϕ Tan  ξ2[t, r, θ, ϕ] -− 2r 2 2 2 2 Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ⅈ Csc[θ]2 ξ2(0,0,0,1) [t, r, θ, ϕ] + Cot[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + r ξ2(0,1,0,0) [t, r, θ, ϕ] -− r Cot[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] ,

-−

1

θ θ θ 2 -− (f1 -− ⅈ f2) Cos  + ⅇ-−ⅈ ϕ (f0 + f3) Sin  + ⅇ-−ⅈ ϕ Tan  ξ1[t, r, θ, ϕ] -− 2r 2 2 2 (0,0,0,1) 2 Csc[θ] (ⅈ Cos[ϕ] + Sin[ϕ]) ξ1 [t, r, θ, ϕ] + Sin[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ⅇ-−ⅈ ϕ Cos[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r Cos[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(1,0,0,0) [t, r, θ, ϕ] ,



1

θ θ θ -− 2 (-− f0 + f3) Cos  + ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  + ⅇⅈ ϕ Tan  ξ0[t, r, θ, ϕ] + 2r 2 2 2 2 Csc[θ] (-− ⅈ Cos[ϕ] + Sin[ϕ]) ξ0(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r ξ3(1,0,0,0) [t, r, θ, ϕ] 


3

4


Simplify[Lap + one -− two -− three] 

1

θ θ -− 16 r2 (f1 -− ⅈ f2) Cos  -− ⅇ-−ⅈ ϕ (f0 + f3) Sin  + 2 2

16 r3

θ θ θ ⅇ-−ⅈ ϕ (5 + 3 Cos[θ]) Sec  Tan  + 8 ⅇ-−ⅈ ϕ r2 Tan  ξ3[t, r, θ, ϕ] + 2 2 2 16 r2 Csc[θ] (-− ⅈ Cos[ϕ] -− Sin[ϕ]) ξ3(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇ-−ⅈ ϕ Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] , θ 2

-−

θ 2

-−

16 r3

θ 2

(f0 -− f3) Cos  -− ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin 

(1 + 3 Cos[θ]) Sec 

-−

r

θ 2

ⅇⅈ ϕ Tan  ξ2[t, r, θ, ϕ] + 2r 1 r

Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ⅈ Csc[θ]2 ξ2(0,0,0,1) [t, r, θ, ϕ] + Cot[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + r ξ2(0,1,0,0) [t, r, θ, ϕ] -− r Cot[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ], θ 2

θ 2

-− (f1 -− ⅈ f2) Cos  + ⅇ-−ⅈ ϕ (f0 + f3) Sin  -−

-−

r θ 2

θ 2

θ 2

ⅇ-−ⅈ ϕ (5 + 3 Cos[θ]) Sec  Tan  16 r3

-−

ⅇ-−ⅈ ϕ Tan  ξ1[t, r, θ, ϕ] + 2r 1 r

Csc[θ] (ⅈ Cos[ϕ] + Sin[ϕ]) ξ1(0,0,0,1) [t, r, θ, ϕ] + Sin[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ⅇ-−ⅈ ϕ Cos[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r Cos[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(1,0,0,0) [t, r, θ, ϕ],



1 16 r3

θ θ θ (1 + 3 Cos[θ]) Sec  -− 16 r2 (-− f0 + f3) Cos  + ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  + 2 2 2

θ 8 ⅇⅈ ϕ r2 Tan  ξ0[t, r, θ, ϕ] + 2 16 r2 Csc[θ] (-− ⅈ Cos[ϕ] + Sin[ϕ]) ξ0(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r ξ3(1,0,0,0) [t, r, θ, ϕ] 



5

(*⋆ EQUATION (7.54) *⋆) XtraTerm = 1 θ θ FullSimplify -− 16 r2 (f1 -− ⅈ f2) Cos  -− ⅇ-−ⅈ ϕ (f0 + f3) Sin  + ⅇ-−ⅈ ϕ 3 2 2 16 r θ θ θ (5 + 3 Cos[θ]) Sec  Tan  + 8 ⅇ-−ⅈ ϕ r2 Tan  ξ3[t, r, θ, ϕ] + 16 r2 2 2 2 (0,0,0,1) Csc[θ] (-− ⅈ Cos[ϕ] -− Sin[ϕ]) ξ3 [t, r, θ, ϕ] -− Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇ-−ⅈ ϕ Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] , θ 2

(1 + 3 Cos[θ]) Sec  -− 16 r3 θ ⅇⅈ ϕ Tan  2

-−

1 r

θ θ (f0 -− f3) Cos  -− ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  -− 2 2

ξ2[t, r, θ, ϕ] + 2r

1 r


-−

1

θ θ -− (f1 -− ⅈ f2) Cos  + ⅇ-−ⅈ ϕ (f0 + f3) Sin  -− 2 2

r 1

ⅇ

-−ⅈ ϕ

θ

θ

θ 2

ⅇ-−ⅈ ϕ Tan  ξ1[t, r, θ, ϕ]

(5 + 3 Cos[θ]) Sec  Tan  -− 2 2

+

1

2r r 16 r3 (0,0,0,1) (0,0,1,0) Csc[θ] (ⅈ Cos[ϕ] + Sin[ϕ]) ξ1 [t, r, θ, ϕ] + Sin[θ] ξ0 [t, r, θ, ϕ] -− ⅇ-−ⅈ ϕ Cos[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r Cos[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(1,0,0,0) [t, r, θ, ϕ], 

1

θ θ θ (1 + 3 Cos[θ]) Sec  -− 16 r2 (-− f0 + f3) Cos  + ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  + 2 2 2 16 r3 θ 8 ⅇⅈ ϕ r2 Tan  ξ0[t, r, θ, ϕ] + 16 r2 2 Csc[θ] (-− ⅈ Cos[ϕ] + Sin[ϕ]) ξ0(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− r ξ3(1,0,0,0) [t, r, θ, ϕ] 


6




1 8 r3 θ θ θ θ ⅇ-−ⅈ ϕ -− 8 ⅇⅈ ϕ (f1 -− ⅈ f2) r2 Cos  + 3 + 8 (f0 + f3) r2  Sin  + Sec  Tan  + 4 r2 2 2 2 2 θ Tan  ξ3[t, r, θ, ϕ] -− 2 ⅈ Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 Cos[θ] ξ3(0,0,1,0) [ 2 t, r, θ, ϕ] + 2 r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ -− Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] θ 2

(1 + 3 Cos[θ]) Sec  -− 16 r θ ⅇⅈ ϕ Tan  2

3

-−

1 r

,

θ θ (f0 -− f3) Cos  -− ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  -− 2 2

ξ2[t, r, θ, ϕ] + 2r

1 r




1

θ θ θ 2 ⅇ-−ⅈ ϕ Sin[θ] 8 ⅇⅈ ϕ (f1 -− ⅈ f2) r2 Csc  -− Sec  3 + 8 (f0 + f3) r2 + Sec  -− 2 2 2 16 r3 8 r2 ξ1[t, r, θ, ϕ] 1 + Cos[θ]

+ 16 r2 ⅈ Csc[θ]2 ξ1(0,0,0,1) [t, r, θ, ϕ] -− Cot[θ]

ξ1(0,0,1,0) [t, r, θ, ϕ] -− r ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] -− r Csc[θ] Cos[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(1,0,0,0) [t, r, θ, ϕ] ,



1 16 r3

θ θ θ 2 3 + 8 (f0 -− f3) r2  Cos  -− 2 Sec  -− 16 ⅇ-−ⅈ ϕ (f1 + ⅈ f2) r2 Sin  + 2 2 2

8 r2 2 Csc[θ] (-− ⅈ Cos[ϕ] + Sin[ϕ]) ξ0(0,0,0,1) [t, r, θ, ϕ] + θ ⅇⅈ ϕ Tan  ξ0[t, r, θ, ϕ] -− 2 Cos[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] + 2 r Sin[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 Sin[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + r -− Cos[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ξ3(1,0,0,0) [t, r, θ, ϕ]





Simplify[Solve[{XtraTerm[[1, Simplify[Solve[{XtraTerm[[1, Simplify[Solve[{XtraTerm[[1, Simplify[Solve[{XtraTerm[[1, 3

f3 → -− f0 -−

8 r2

+ ⅇ

ⅈ ϕ

1]] 1]] 1]] 1]]

⩵ ⩵ ⩵ ⩵

θ

f1 Cot  -− ⅈ ⅇ 2

0}, 0}, 0}, 0},

ⅈ ϕ

7

f3]] f2]] f1]] f0]] θ 2

Sec 

θ

f2 Cot  -− 2

2

-−

1

θ Sec  ξ3[t, r, θ, ϕ] + 2 2

8 r2 θ θ ⅈ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ Cos  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ θ Cos[θ] Csc  ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[θ] Csc  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 2 r Cos  ξ3(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] 2 2

f2 → 1

θ θ θ ⅇ-−ⅈ ϕ 4 ⅈ Sec  Tan  ξ3[t, r, θ, ϕ] + 8 Csc[θ] Sec  ξ3(0,0,0,1) [t, r, θ, ϕ] -− 8 2 2 2

ⅈ 8 ⅇ

ⅈ ϕ

θ 2

8 Csc[θ]3 Sin  f1 -−

4

r2

θ

θ

-− 8 f0 Tan  -− 8 f3 Tan  -− 2 2

θ 2

3 Tan  r2

+

θ θ 16 ⅇⅈ ϕ Sin  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 8 Cos[θ] Sec  ξ3(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 8 ⅇⅈ ϕ r Cos[θ] Sec  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 θ θ 16 r Sin  ξ3(0,1,0,0) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r Sec  ξ0(1,0,0,0) [t, r, θ, ϕ] 2 2

f1 →



1

θ ⅇ-−ⅈ ϕ 8 ⅈ ⅇⅈ ϕ f2 -− Csc[θ] Sec  ξ3(0,0,0,1) [t, r, θ, ϕ] + 8 2 1 θ θ θ θ θ θ Sec  3 Sin  + 8 f0 r2 Sin  + 8 f3 r2 Sin  + Sec  Tan  + 2 2 2 2 2 2 r2 θ 4 r2 Tan  ξ3[t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r2 Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + 2 8 r2 Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + 8 r3 Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 ξ0(1,0,0,0) [t, r, θ, ϕ]

f0 → -− f3 -−

3 8r

2

+ ⅇ

ⅈ ϕ

θ

f1 Cot  -− ⅈ ⅇ 2

ⅈ ϕ

θ

θ 2

Sec 

f2 Cot  -− 2

2

2

-−



1

θ Sec  ξ3[t, r, θ, ϕ] + 2 2

8r θ θ ⅈ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ Cos  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ θ Cos[θ] Csc  ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[θ] Csc  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 2 r Cos  ξ3(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] 2 2


8


(*⋆ EQUATION (7.55) *⋆) f0exp = FullSimplify f0 → -− f3 -−

3 8r

f0 →

+ ⅇ

ⅈ ϕ

2

θ

f1 Cot  -− ⅈ ⅇ 2

ⅈ ϕ

θ

θ 2

Sec 

f2 Cot  -− 2

2

2

-−

1

θ Sec  ξ3[t, r, θ, ϕ] + 2 2

8r θ θ ⅈ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ Cos  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ θ Cos[θ] Csc  ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[θ] Csc  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 2 r Cos  ξ3(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] 2 2 1 8 r2

-− 3 -− 8 f3 r2 -−

2

θ θ + 8 ⅇⅈ ϕ (f1 -− ⅈ f2) r2 Cot  + 4 r2 -− Sec  ξ3[t, r, θ, ϕ] + 1 + Cos[θ] 2 2

θ 2 Csc  ⅈ Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] -− Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− 2 r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] -− r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ]



9

f0n2XTerm = Simplify[XtraTerm /∕/∕. f0exp] {0}, θ 2

-−

-−

3

16 r θ 1 Cos  -− f3 + 2 8 r2

θ 2

ⅇⅈ ϕ Tan  ξ2[t, r, θ, ϕ]

(1 + 3 Cos[θ]) Sec 

-− 2r

-− 3 -− 8 f3 r2 -−

1 r

θ -− ⅇ-−ⅈ ϕ (f1 + ⅈ f2) Sin  + 2

2

θ + 8 ⅇⅈ ϕ (f1 -− ⅈ f2) r2 Cot  + 1 + Cos[θ] 2

θ θ 4 r2 -− Sec  ξ3[t, r, θ, ϕ] + 2 Csc  ⅈ Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] -− 2 2 (0,0,1,0) Cos[θ] ξ3 [t, r, θ, ϕ] -− r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] -− r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] 1 r

+




1

θ θ ⅇ-−ⅈ ϕ Csc  Sec  (-− 1 + Cos[θ]) ξ1[t, r, θ, ϕ] -− 4r 2 2 (-− 1 + Cos[θ]) ξ3[t, r, θ, ϕ] + 2 ⅈ ξ1(0,0,0,1) [t, r, θ, ϕ] -− 2 ⅈ ξ3(0,0,0,1) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− Sin[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Cos[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + Sin[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Sin[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− r ξ1(0,1,0,0) [t, r, θ, ϕ] + r Cos[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Sin[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + r ξ3(0,1,0,0) [t, r, θ, ϕ] -− r Cos[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ r Sin[θ] ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Sin[θ] ξ2(1,0,0,0) [t, r, θ, ϕ],



1 16 r

3

θ θ θ -− 2 Sec  -− 16 ⅇ-−ⅈ ϕ (f1 + ⅈ f2) r2 Sin  + 2 Cos  2 2 2 3 + 8 r2 -− f3 +

1

-− 3 -− 8 f3 r2 -−

2

θ + 8 ⅇⅈ ϕ (f1 -− ⅈ f2) r2 Cot  + 4 r2 1 + Cos[θ] 2

8 r2 θ θ -− Sec  ξ3[t, r, θ, ϕ] + 2 Csc  ⅈ Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] -− 2 2 (0,0,1,0) Cos[θ] ξ3 [t, r, θ, ϕ] -− r Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] -− r Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ]





+

10


Simplify[Solve[{f0n2XTerm[[2, 1]] ⩵ 0}, f3]] f3 → 1 2

θ θ θ -− ⅈ ⅇ-−ⅈ ϕ ⅇ2 ⅈ ϕ f2 Cot  + f2 Tan  + ⅇ2 ⅈ ϕ Csc[θ] Sec  ξ2(0,0,0,1) [t, r, θ, ϕ] -− 2 2 2 θ ⅇⅈ ϕ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 1 θ θ θ θ Sec  -− 6 Cos  -− 2 Sec  + (1 + 3 Cos[θ]) Sec  -− 16 ⅇ-−ⅈ ϕ f1 2 2 2 2 16 r2 θ θ 3 θ r2 Sin  + 4 ⅇⅈ ϕ f1 r2 Csc  Sin[θ]2 + 8 ⅇⅈ ϕ r2 Tan  ξ2[t, r, θ, ϕ] -− 2 2 2 θ 8 r2 ξ3[t, r, θ, ϕ] -− 2 Cot  ⅇⅈ ϕ Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− r ⅇⅈ ϕ Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(1,0,0,0) [t, r, θ, ϕ] -− 16 r2 ⅇⅈ ϕ Cos[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + r ⅇⅈ ϕ Sin[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] -− Cos[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ξ1(1,0,0,0) [t, r, θ, ϕ]





(*⋆ EQUATION (7.56) *⋆) f3exp = 1 θ θ θ FullSimplifyf3 → -− ⅈ ⅇ-−ⅈ ϕ ⅇ2 ⅈ ϕ f2 Cot  + f2 Tan  + ⅇ2 ⅈ ϕ Csc[θ] Sec  2 2 2 2 θ ξ2(0,0,0,1) [2 t, r, θ, ϕ] -− ⅇⅈ ϕ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 1 θ θ θ θ Sec  -− 6 Cos  -− 2 Sec  + (1 + 3 Cos[θ]) Sec  -− 16 ⅇ-−ⅈ ϕ f1 2 2 2 2 2 16 r 3 θ θ θ r2 Sin  + 4 ⅇⅈ ϕ f1 r2 Csc  Sin[θ]2 + 8 ⅇⅈ ϕ r2 Tan  ξ2[t, r, θ, ϕ] -− 2 2 2 θ 8 r2 ξ3[t, r, θ, ϕ] -− 2 Cot  ⅇⅈ ϕ Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] -− Cos[θ] 2 ξ3(0,0,1,0) [t, r, θ, ϕ] -− r ⅇⅈ ϕ Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(1,0,0,0) [t, r, θ, ϕ] -− 16 r2 ⅇⅈ ϕ Cos[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + r ⅇⅈ ϕ Sin[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] -− Cos[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ξ1(1,0,0,0) [t, r, θ, ϕ] f3 →



1

θ θ θ ⅇ-−ⅈ ϕ -− 16 f1 r2 Tan  -− 16 ⅈ f2 r2 Tan  + 4 ⅇ2 ⅈ ϕ r2 Csc  2 2 2 32 r θ 2 Sec  -− (-− 1 + Cos[θ]) ξ2[t, r, θ, ϕ] -− 2 ⅈ ξ2(0,0,0,1) [t, r, θ, ϕ] + 2 θ θ 2 Cos  (f1 -− ⅈ f2) (1 + Cos[θ]) + 2 Sin  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 2 r Cos  ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] + 2 θ ⅇⅈ ϕ Sin[θ] -− 8 r2 Csc[θ] Sec  ξ3[t, r, θ, ϕ] + 8 Csc[θ]3 -− 1 + Cos[θ] + 2 θ 4 r2 Cos  ⅈ ξ3(0,0,0,1) [t, r, θ, ϕ] -− Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + 2 2

r (-− 1 + Cos[θ]) ξ3(0,1,0,0) [t, r, θ, ϕ] + ξ1(1,0,0,0) [t, r, θ, ϕ]


11

12


f3n2f0XTerm = Simplify[f0n2XTerm /∕/∕. f3exp] {0}, {0}, 

1

θ θ ⅇ-−ⅈ ϕ Csc  Sec  (-− 1 + Cos[θ]) ξ1[t, r, θ, ϕ] -− (-− 1 + Cos[θ]) ξ3[t, r, θ, ϕ] + 4r 2 2 (0,0,0,1) 2 ⅈ ξ1 [t, r, θ, ϕ] -− 2 ⅈ ξ3(0,0,0,1) [t, r, θ, ϕ] + ⅈ ϕ (0,0,1,0) ⅇ ξ0 [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− Sin[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Cos[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + Sin[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Sin[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− r ξ1(0,1,0,0) [t, r, θ, ϕ] + r Cos[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Sin[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + r ξ3(0,1,0,0) [t, r, θ, ϕ] -− r Cos[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ r Sin[θ] ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Sin[θ] ξ2(1,0,0,0) [t, r, θ, ϕ],



1

θ θ Csc  Sec  -− ⅇⅈ ϕ (-− 1 + Cos[θ]) ξ0[t, r, θ, ϕ] + ⅇⅈ ϕ (-− 1 + Cos[θ]) 4r 2 2 ξ2[t, r, θ, ϕ] -− 2 ⅈ ⅇⅈ ϕ ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ ⅇⅈ ϕ ξ2(0,0,0,1) [t, r, θ, ϕ] -− ⅇⅈ ϕ Sin[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ξ1(0,0,1,0) [t, r, θ, ϕ] + Cos[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ξ3(0,0,1,0) [t, r, θ, ϕ] -− Cos[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r ξ0(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Cos[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Sin[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r ξ2(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] -− r Sin[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 r Sin[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] -− 2 r Sin[θ] ξ3(1,0,0,0) [t, r, θ, ϕ]

(*⋆ EQUATION (7.57) *⋆) zer0 = FullSimplify 1

θ θ ⅇ-−ⅈ ϕ Csc  Sec  (-− 1 + Cos[θ]) ξ1[t, r, θ, ϕ] -− (-− 1 + Cos[θ]) ξ3[t, r, θ, ϕ] + 4r 2 2 2 ⅈ ξ1(0,0,0,1) [t, r, θ, ϕ] -− 2 ⅈ ξ3(0,0,0,1) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− Sin[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Cos[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + Sin[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Sin[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− r ξ1(0,1,0,0) [t, r, θ, ϕ] + r Cos[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Sin[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + r ξ3(0,1,0,0) [t, r, θ, ϕ] -− r Cos[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ r Sin[θ] ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Sin[θ] ξ2(1,0,0,0) [t, r, θ, ϕ] 1

θ θ ⅇ-−ⅈ ϕ Csc  Sec  (-− 1 + Cos[θ]) ξ1[t, r, θ, ϕ] -− 4r 2 2 (-− 1 + Cos[θ]) ξ3[t, r, θ, ϕ] + 2 ⅈ ξ1(0,0,0,1) [t, r, θ, ϕ] -− 2 ⅈ ξ3(0,0,0,1) [t, r, θ, ϕ] + Sin[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + 2 Sin[θ] -− Cos[θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] -− ξ1(0,1,0,0) [t, r, θ, ϕ] + ξ3(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ξ2(0,0,1,0) [t, r, θ, ϕ] + r Cos[θ] -− ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(0,1,0,0) [t, r, θ, ϕ] + ξ0(1,0,0,0) [t, r, θ, ϕ] -− ξ2(1,0,0,0) [t, r, θ, ϕ]



13

(*⋆ EQUATION (7.58) *⋆) 1 θ θ zer1 = FullSimplify Csc  Sec  4r 2 2 -− ⅇⅈ ϕ (-− 1 + Cos[θ]) ξ0[t, r, θ, ϕ] + ⅇⅈ ϕ (-− 1 + Cos[θ]) ξ2[t, r, θ, ϕ] -− 2 ⅈ ⅇⅈ ϕ ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ ⅇⅈ ϕ ξ2(0,0,0,1) [t, r, θ, ϕ] -− ⅇⅈ ϕ Sin[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ξ1(0,0,1,0) [t, r, θ, ϕ] + Cos[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ξ3(0,0,1,0) [t, r, θ, ϕ] -− Cos[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r ξ0(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Cos[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Sin[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r ξ2(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] -− r Sin[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 r Sin[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] -− 2 r Sin[θ] ξ3(1,0,0,0) [t, r, θ, ϕ] 1 2r θ Cot  (-− ⅈ Cos[ϕ] + Sin[ϕ]) ξ0(0,0,0,1) [t, r, θ, ϕ] -− ξ2(0,0,0,1) [t, r, θ, ϕ] + 2 Sin[θ] 2 θ -− ξ1(0,0,1,0) [t, r, θ, ϕ] + ξ3(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Tan  ξ0[t, r, θ, ϕ] -− 2 ξ2[t, r, θ, ϕ] -− ⅈ ξ0(0,0,0,1) [t, r, θ, ϕ] -− ξ2(0,0,0,1) [t, r, θ, ϕ] + 2 Cos[θ] -− ξ0(0,0,1,0) [t, r, θ, ϕ] + ξ2(0,0,1,0) [t, r, θ, ϕ] + 2 r Sin[θ] -− ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(0,1,0,0) [t, r, θ, ϕ] + 2 r Cos[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− ξ3(0,1,0,0) [t, r, θ, ϕ] + ξ1(1,0,0,0) [t, r, θ, ϕ] -− ξ3(1,0,0,0) [t, r, θ, ϕ] (*⋆ EQUATION (7.60) *⋆) zero0 = FullSimplifyzer0 /∕/∕. ξ2[t, r, θ, ϕ] -−> ξ0[t, r, θ, ϕ] + xx[t, r, θ, ϕ], ξ3[t, r, θ, ϕ] -−> ξ1[t, r, θ, ϕ] + yy[t, r, θ, ϕ], ξ3(0,0,1,0) [t, r, θ, ϕ] -−> ξ1(0,0,1,0) [t, r, θ, ϕ] + yy(0,0,1,0) [t, r, θ, ϕ], ξ2(0,0,0,1) [t, r, θ, ϕ] -−> ξ0(0,0,0,1) [t, r, θ, ϕ] + xx(0,0,0,1) [t, r, θ, ϕ], ξ2(0,0,1,0) [t, r, θ, ϕ] -−> ξ0(0,0,1,0) [t, r, θ, ϕ] + xx(0,0,1,0) [t, r, θ, ϕ], ξ2(0,1,0,0) [t, r, θ, ϕ] -−> ξ0(0,1,0,0) [t, r, θ, ϕ] + xx(0,1,0,0) [t, r, θ, ϕ], ξ3(0,1,0,0) [t, r, θ, ϕ] -−> ξ1(0,1,0,0) [t, r, θ, ϕ] + yy(0,1,0,0) [t, r, θ, ϕ], ξ3(1,0,0,0) [t, r, θ, ϕ] -−> ξ1(1,0,0,0) [t, r, θ, ϕ] + yy(1,0,0,0) [t, r, θ, ϕ], ξ3(0,0,0,1) [t, r, θ, ϕ] -−> ξ1(0,0,0,1) [t, r, θ, ϕ] + yy(0,0,0,1) [t, r, θ, ϕ], ξ2(1,0,0,0) [t, r, θ, ϕ] -−> ξ0(1,0,0,0) [t, r, θ, ϕ] + xx(1,0,0,0) [t, r, θ, ϕ](*⋆polar*⋆) 1 2r θ ⅇ-−ⅈ ϕ Sin[θ] Csc[θ] Tan  yy[t, r, θ, ϕ] -− 2 ⅈ Csc[θ]2 yy(0,0,0,1) [t, r, θ, ϕ] + 2 Cot[θ] 2 yy(0,0,1,0) [t, r, θ, ϕ] + 2 r yy(0,1,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ xx(0,0,1,0) [t, r, θ, ϕ] -− r Cot[θ] xx(0,1,0,0) [t, r, θ, ϕ] + r Csc[θ] xx(1,0,0,0) [t, r, θ, ϕ]


14


(*⋆ EQUATION (7.61) *⋆) zero1 = Simplifyzer1 /∕/∕. ξ2[t, r, θ, ϕ] -−> ξ0[t, r, θ, ϕ] + xx[t, r, θ, ϕ], ξ3[t, r, θ, ϕ] -−> ξ1[t, r, θ, ϕ] + yy[t, r, θ, ϕ], ξ3(0,0,1,0) [t, r, θ, ϕ] -−> ξ1(0,0,1,0) [t, r, θ, ϕ] + yy(0,0,1,0) [t, r, θ, ϕ], ξ2(0,0,0,1) [t, r, θ, ϕ] -−> ξ0(0,0,0,1) [t, r, θ, ϕ] + xx(0,0,0,1) [t, r, θ, ϕ], ξ2(0,0,1,0) [t, r, θ, ϕ] -−> ξ0(0,0,1,0) [t, r, θ, ϕ] + xx(0,0,1,0) [t, r, θ, ϕ], ξ2(0,1,0,0) [t, r, θ, ϕ] -−> ξ0(0,1,0,0) [t, r, θ, ϕ] + xx(0,1,0,0) [t, r, θ, ϕ], ξ3(0,1,0,0) [t, r, θ, ϕ] -−> ξ1(0,1,0,0) [t, r, θ, ϕ] + yy(0,1,0,0) [t, r, θ, ϕ], ξ3(1,0,0,0) [t, r, θ, ϕ] -−> ξ1(1,0,0,0) [t, r, θ, ϕ] + yy(1,0,0,0) [t, r, θ, ϕ], ξ3(0,0,0,1) [t, r, θ, ϕ] -−> ξ1(0,0,0,1) [t, r, θ, ϕ] + yy(0,0,0,1) [t, r, θ, ϕ], ξ2(1,0,0,0) [t, r, θ, ϕ] -−> ξ0(1,0,0,0) [t, r, θ, ϕ] + xx(1,0,0,0) [t, r, θ, ϕ](*⋆polar*⋆) 1

θ -− ⅇⅈ ϕ Tan  xx[t, r, θ, ϕ] + 2r 2 2 ⅈ (ⅈ + Cot[ϕ]) Csc[θ] Sin[ϕ] xx(0,0,0,1) [t, r, θ, ϕ] + ⅇⅈ ϕ Cos[θ] xx(0,0,1,0) [t, r, θ, ϕ] + Sin[θ] yy(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Sin[θ] xx(0,1,0,0) [t, r, θ, ϕ] -− r Cos[θ] yy(0,1,0,0) [t, r, θ, ϕ] -− r yy(1,0,0,0) [t, r, θ, ϕ] (*⋆ EQUATION (7.63) *⋆) zero00 = FullSimplifyzero0 /∕/∕.  xx[t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,0,1) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,0,1) [t, r, θ, ϕ] -− ⅈ ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,1,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,1,0) [t, r, θ, ϕ], xx(0,1,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,1,0,0) [t, r, θ, ϕ], xx(1,0,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(1,0,0,0) [t, r, θ, ϕ] 1

θ ⅇ-−ⅈ ϕ Tan  yy[t, r, θ, ϕ] + 2 -− ⅈ Csc[θ] yy(0,0,0,1) [t, r, θ, ϕ] + 2r 2 Sin[θ] -− zz(0,0,1,0) [t, r, θ, ϕ] + r yy(0,1,0,0) [t, r, θ, ϕ] + Cos[θ] yy(0,0,1,0) [t, r, θ, ϕ] + r zz(0,1,0,0) [t, r, θ, ϕ] -− r zz(1,0,0,0) [t, r, θ, ϕ] (*⋆ EQUATION (7.64) *⋆) zero11 = FullSimplifyzero1 /∕/∕. xx[t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,0,1) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,0,1) [t, r, θ, ϕ] -− ⅈ ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,1,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,1,0) [t, r, θ, ϕ], xx(0,1,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,1,0,0) [t, r, θ, ϕ], xx(1,0,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(1,0,0,0) [t, r, θ, ϕ] 1 2r

(Cot[θ] + Csc[θ]) zz[t, r, θ, ϕ] + 2 ⅈ Csc[θ] zz(0,0,0,1) [t, r, θ, ϕ] + 2 Cos[θ] zz(0,0,1,0) [t, r, θ, ϕ] -− r yy(0,1,0,0) [t, r, θ, ϕ] + 2 Sin[θ] yy(0,0,1,0) [t, r, θ, ϕ] + r zz(0,1,0,0) [t, r, θ, ϕ] -− 2 r yy(1,0,0,0) [t, r, θ, ϕ]



(*⋆EQUATION (7.65)*⋆) Simplify Solve

1

θ ⅇ-−ⅈ ϕ Tan  yy[t, r, θ, ϕ] + 2 -− ⅈ Csc[θ] yy(0,0,0,1) [t, r, θ, ϕ] + Sin[θ] 2r 2 -− zz(0,0,1,0) [t, r, θ, ϕ] + r yy(0,1,0,0) [t, r, θ, ϕ] + Cos[θ] yy(0,0,1,0) [t, r, θ, ϕ] + r zz(0,1,0,0) [t, r, θ, ϕ] -− r zz(1,0,0,0) [t, r, θ, ϕ] ⩵ 0, yy(0,0,0,1) [t, r, θ, ϕ]

yy(0,0,0,1) [t, r, θ, ϕ] → 1

θ ⅈ Sin[θ] -− Tan  yy[t, r, θ, ϕ] -− 2 Cos[θ] yy(0,0,1,0) [t, r, θ, ϕ] -− 2 2 Sin[θ] zz(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] yy(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] zz(0,1,0,0) [t, r, θ, ϕ] -− r zz(1,0,0,0) [t, r, θ, ϕ]  (*⋆EQUATION (7.66)*⋆) Simplify 1

(Cot[θ] + Csc[θ]) zz[t, r, θ, ϕ] + 2 ⅈ Csc[θ] zz(0,0,0,1) [t, r, θ, ϕ] + 2r 2 Cos[θ] zz(0,0,1,0) [t, r, θ, ϕ] -− r yy(0,1,0,0) [t, r, θ, ϕ] +

Solve

2 Sin[θ] yy(0,0,1,0) [t, r, θ, ϕ] + r zz(0,1,0,0) [t, r, θ, ϕ] -− 2 r yy(1,0,0,0) [t, r, θ, ϕ] ⩵ 0, zz(0,0,0,1) [t, r, θ, ϕ] zz(0,0,0,1) [t, r, θ, ϕ] → 1 2

ⅈ (1 + Cos[θ]) zz[t, r, θ, ϕ] + 2 Sin[θ] Sin[θ] yy(0,0,1,0) [t, r, θ, ϕ] + Cos[θ] zz(0,0,1,0) [t, r, θ, ϕ] -− r Cos[θ] yy(0,1,0,0) [t, r, θ, ϕ] -− Sin[θ] zz(0,1,0,0) [t, r, θ, ϕ] + yy(1,0,0,0) [t, r, θ, ϕ]

(*⋆ end *⋆)


15

16


(*⋆ Looking at f_mu entries of (XtraTerm), we can choose as in EQUATION (7.67)*⋆)

fmus = Simplify[ XtraTerm /∕/∕. {f0 → (a + b) /∕ 2, f3 → (a -− b) /∕ 2, f1 → (w + v) /∕ 2, f2 → (w -− v) /∕ (2 I)}] 

1

θ θ ⅇ-−ⅈ ϕ -− 8 ⅇⅈ ϕ r2 v Cos  + 3 Sin  + 2 2 8 r3 θ θ θ θ 8 a r2 Sin  + Sec  Tan  + 4 r2 Tan  ξ3[t, r, θ, ϕ] -− 2 2 2 2 8 ⅈ r2 Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r2 Sin[θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + 8 r2 Cos[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 Cos[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + 8 r3 Sin[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 ξ0(1,0,0,0) [t, r, θ, ϕ] , θ 2

b Cos  -−

-−

r

θ 2

Sec  16 r3

θ 2

-−

θ 2

16 r3

ⅇⅈ ϕ Tan  ξ2[t, r, θ, ϕ] +

θ 2

ⅇ-−ⅈ ϕ w Sin 

3 Cos[θ] Sec  +

-−

r

ⅈ ⅇⅈ ϕ Csc[θ] ξ2(0,0,0,1) [t, r, θ, ϕ]

2r

+

r

ⅇⅈ ϕ Cos[θ] ξ2(0,0,1,0) [t, r, θ, ϕ]

+

r Sin[θ] ξ3(0,0,1,0) [t, r, θ, ϕ] r

+ ⅇⅈ ϕ Sin[θ] ξ2(0,1,0,0) [t, r, θ, ϕ] -−

Cos[θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + ξ1(1,0,0,0) [t, r, θ, ϕ], 

1

θ θ θ 2 ⅇ-−ⅈ ϕ Sin[θ] 8 ⅇⅈ ϕ r2 v Csc  -− Sec  3 + 8 a r2 + Sec  -− 2 2 2 16 r3 8 r2 ξ1[t, r, θ, ϕ] 1 + Cos[θ]

+ 16 r2 ⅈ Csc[θ]2 ξ1(0,0,0,1) [t, r, θ, ϕ] -− Cot[θ]

ξ1(0,0,1,0) [t, r, θ, ϕ] -− r ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] -− r Csc[θ] Cos[θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + ξ2(1,0,0,0) [t, r, θ, ϕ] ,



1 16 r3

θ θ θ 2 3 + 8 b r2  Cos  -− 2 Sec  -− 16 ⅇ-−ⅈ ϕ r2 w Sin  + 2 2 2






(*⋆ note that a,b are in polar coordinates! *⋆)

Solve4ab = Simplify[Solve[{fmus[[1, 1]] ⩵ 0, fmus[[2, 1]] ⩵ 0}, {a, b}]] a → -−

3 8 r2

+ ⅇ

ⅈ ϕ

θ

θ 2

Sec 

v Cot  -− 2

2

-−

8 r2

1

θ Sec  ξ3[t, r, θ, ϕ] + 2 2

θ θ ⅈ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ Cos  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ θ Cos[θ] Csc  ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[θ] Csc  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 2 r Cos  ξ3(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ], 2 2 θ 2

Sec  b → -−

16 r2

2

θ 2

3 Cos[θ] Sec  -−

16 r2

2

θ + ⅇ-−ⅈ ϕ w Tan  -− 2

θ 3 θ 2 ⅇⅈ ϕ Csc[θ]2 Sin  ξ2[t, r, θ, ϕ] + ⅈ ⅇⅈ ϕ Csc[θ] Sec  ξ2(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ ⅇⅈ ϕ Cos[θ] Sec  ξ2(0,0,1,0) [t, r, θ, ϕ] + 2 θ θ 2 Sin  ξ3(0,0,1,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ r Sin  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ r Cos[θ] Sec  ξ3(0,1,0,0) [t, r, θ, ϕ] + r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2


17

18


Simplifyfmus /∕/∕. a → -−

3

+ ⅇ

ⅈ ϕ

8 r2

θ

v Cot  -− 2

θ 2

Sec 

2

8 r2

-−

1

θ Sec  ξ3[t, r, θ, ϕ] + 2 2

θ θ ⅈ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ Cos  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ θ Cos[θ] Csc  ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[θ] Csc  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 2 r Cos  ξ3(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ], 2 2 θ 2

Sec  b → -−

16 r

2

θ 2

3 Cos[θ] Sec  -−

2

16 r

2

2

θ + ⅇ-−ⅈ ϕ w Tan  -− 2

θ

3 θ 2 ⅇⅈ ϕ Csc[θ]2 Sin  ξ2[t, r, θ, ϕ] + ⅈ ⅇⅈ ϕ Csc[θ] Sec  ξ2(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ ⅇⅈ ϕ Cos[θ] Sec  ξ2(0,0,1,0) [t, r, θ, ϕ] + 2 θ θ 2 Sin  ξ3(0,0,1,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ r Sin  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ r Cos[θ] Sec  ξ3(0,1,0,0) [t, r, θ, ϕ] + r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2 1 {0}, {0},  2 r (1 + Cos[θ]) θ ⅇ-−ⅈ ϕ Cot  (-− 1 + Cos[θ]) ξ1[t, r, θ, ϕ] -− (-− 1 + Cos[θ]) ξ3[t, r, θ, ϕ] + 2 2 ⅈ ξ1(0,0,0,1) [t, r, θ, ϕ] -− 2 ⅈ ξ3(0,0,0,1) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ Cos[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− Sin[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ ξ2(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Cos[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + Sin[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Sin[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− r ξ1(0,1,0,0) [t, r, θ, ϕ] + r Cos[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Sin[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] + r ξ3(0,1,0,0) [t, r, θ, ϕ] -− r Cos[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] +

2 ⅇⅈ ϕ r Sin[θ] ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Sin[θ] ξ2(1,0,0,0) [t, r, θ, ϕ], 

1

θ θ Csc  Sec  -− ⅇⅈ ϕ (-− 1 + Cos[θ]) ξ0[t, r, θ, ϕ] + ⅇⅈ ϕ (-− 1 + Cos[θ]) 4r 2 2 ξ2[t, r, θ, ϕ] -− 2 ⅈ ⅇⅈ ϕ ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ ⅇⅈ ϕ ξ2(0,0,0,1) [t, r, θ, ϕ] -− ⅇⅈ ϕ Sin[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− ξ1(0,0,1,0) [t, r, θ, ϕ] + Cos[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ Sin[2 θ] ξ2(0,0,1,0) [t, r, θ, ϕ] + ξ3(0,0,1,0) [t, r, θ, ϕ] -− Cos[2 θ] ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r ξ0(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r Cos[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + r Sin[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ r ξ2(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[2 θ] ξ2(0,1,0,0) [t, r, θ, ϕ] -− r Sin[2 θ] ξ3(0,1,0,0) [t, r, θ, ϕ] + 2 r Sin[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] -− 2 r Sin[θ] ξ3(1,0,0,0) [t, r, θ, ϕ]



3

aexp = a → -−

+ ⅇ

ⅈ ϕ

θ 2

Sec 

θ

v Cot  -− 2

2

-−

1

θ Sec  ξ3[t, r, θ, ϕ] + 2 2

8 r2 8 r2 θ θ ⅈ Csc  Csc[θ] ξ3(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ Cos  ξ2(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ θ Cos[θ] Csc  ξ3(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos[θ] Csc  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 2 r Cos  ξ3(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ]; 2 2 θ 2

Sec  bexp = b → -−

16 r

2

θ 2

3 Cos[θ] Sec  -−

2

16 r θ

2

2

θ + ⅇ-−ⅈ ϕ w Tan  -− 2

θ 2 ⅇⅈ ϕ Csc[θ]2 Sin  ξ2[t, r, θ, ϕ] + ⅈ ⅇⅈ ϕ Csc[θ] Sec  ξ2(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ ⅇⅈ ϕ Cos[θ] Sec  ξ2(0,0,1,0) [t, r, θ, ϕ] + 2 θ θ 2 Sin  ξ3(0,0,1,0) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ r Sin  ξ2(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ r Cos[θ] Sec  ξ3(0,1,0,0) [t, r, θ, ϕ] + r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ]; 2 2 3

sub1a = Simplifyaexp /∕/∕. ξ2[t, r, θ, ϕ] -−> ξ0[t, r, θ, ϕ] + xx[t, r, θ, ϕ], ξ3[t, r, θ, ϕ] -−> ξ1[t, r, θ, ϕ] + yy[t, r, θ, ϕ], ξ3(0,0,1,0) [t, r, θ, ϕ] -−> ξ1(0,0,1,0) [t, r, θ, ϕ] + yy(0,0,1,0) [t, r, θ, ϕ], ξ2(0,0,0,1) [t, r, θ, ϕ] -−> ξ0(0,0,0,1) [t, r, θ, ϕ] + xx(0,0,0,1) [t, r, θ, ϕ], ξ2(0,0,1,0) [t, r, θ, ϕ] -−> ξ0(0,0,1,0) [t, r, θ, ϕ] + xx(0,0,1,0) [t, r, θ, ϕ], ξ2(0,1,0,0) [t, r, θ, ϕ] -−> ξ0(0,1,0,0) [t, r, θ, ϕ] + xx(0,1,0,0) [t, r, θ, ϕ], ξ3(0,1,0,0) [t, r, θ, ϕ] -−> ξ1(0,1,0,0) [t, r, θ, ϕ] + yy(0,1,0,0) [t, r, θ, ϕ], ξ3(1,0,0,0) [t, r, θ, ϕ] -−> ξ1(1,0,0,0) [t, r, θ, ϕ] + yy(1,0,0,0) [t, r, θ, ϕ], ξ3(0,0,0,1) [t, r, θ, ϕ] -−> ξ1(0,0,0,1) [t, r, θ, ϕ] + yy(0,0,0,1) [t, r, θ, ϕ], ξ2(1,0,0,0) [ t, r, θ, ϕ] -−> ξ0(1,0,0,0) [t, r, θ, ϕ] + xx(1,0,0,0) [t, r, θ, ϕ](*⋆ polar *⋆) a → -−

3 8r

2

+ ⅇ

ⅈ ϕ

θ

v Cot  -− 2

θ 2

Sec  8r

2

2

-−

1

θ Sec  (yy[t, r, θ, ϕ] + ξ1[t, r, θ, ϕ]) + 2 2

θ

ⅈ Csc  Csc[θ] yy(0,0,0,1) [t, r, θ, ϕ] + ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 θ 2 ⅇⅈ ϕ Cos  xx(0,0,1,0) [t, r, θ, ϕ] + ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 θ Cos[θ] Csc  yy(0,0,1,0) [t, r, θ, ϕ] + ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 θ ⅇⅈ ϕ r Cos[θ] Csc  xx(0,1,0,0) [t, r, θ, ϕ] + ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 θ 2 r Cos  yy(0,1,0,0) [t, r, θ, ϕ] + ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 θ ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] 2


19

20


sub2a = Simplifysub1a /∕/∕. xx[t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,0,1) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,0,1) [t, r, θ, ϕ] -− ⅈ ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,1,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,1,0) [t, r, θ, ϕ], xx(0,1,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,1,0,0) [t, r, θ, ϕ], xx(1,0,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(1,0,0,0) [t, r, θ, ϕ] 3

a → -−

8r

2

+ ⅇ

ⅈ ϕ

θ

v Cot  -− 2

θ 2

Sec  8r

2

-−

2

1

θ Sec  (yy[t, r, θ, ϕ] + ξ1[t, r, θ, ϕ]) + 2 2

θ ⅈ Csc  Csc[θ] yy(0,0,0,1) [t, r, θ, ϕ] + ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 θ 2 Cos  zz(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 θ Cos[θ] Csc  yy(0,0,1,0) [t, r, θ, ϕ] + ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 θ r Cos[θ] Csc  zz(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 θ 2 r Cos  yy(0,1,0,0) [t, r, θ, ϕ] + ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 θ ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] 2

(*⋆EQUATION (7.70)*⋆) sub3a = Simplifysub2a /∕/∕. yy(0,0,0,1) [t, r, θ, ϕ] → 1

θ ⅈ Sin[θ] -− Tan  yy[t, r, θ, ϕ] -− 2 Cos[θ] yy(0,0,1,0) [t, r, θ, ϕ] -− 2 2 Sin[θ] zz(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] yy(0,1,0,0) [t, r, θ, ϕ] + r Cos[θ] zz(0,1,0,0) [t, r, θ, ϕ] -− r zz(1,0,0,0) [t, r, θ, ϕ] 

a → -−

3 8 r2

+ ⅇ

ⅈ ϕ

θ

v Cot  -− 2

θ 2

Sec  8 r2

2

-−

1

θ θ Sec  ξ1[t, r, θ, ϕ] + ⅈ Csc  Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 2 2 θ θ θ 2 ⅇⅈ ϕ Cos  ξ0(0,0,1,0) [t, r, θ, ϕ] -− Cos  Cot  ξ1(0,0,1,0) [t, r, θ, ϕ] + 2 2 2 θ θ θ Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Cos  Cot  ξ0(0,1,0,0) [t, r, θ, ϕ] + 2 2 2 θ θ ⅇⅈ ϕ r Sin  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Cos  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ r Csc  zz(1,0,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] 2 2



sub1b = Simplifybexp /∕/∕. ξ2[t, r, θ, ϕ] -−> ξ0[t, r, θ, ϕ] + xx[t, r, θ, ϕ], ξ3[t, r, θ, ϕ] -−> ξ1[t, r, θ, ϕ] + yy[t, r, θ, ϕ], ξ3(0,0,1,0) [t, r, θ, ϕ] -−> ξ1(0,0,1,0) [t, r, θ, ϕ] + yy(0,0,1,0) [t, r, θ, ϕ], ξ2(0,0,0,1) [t, r, θ, ϕ] -−> ξ0(0,0,0,1) [t, r, θ, ϕ] + xx(0,0,0,1) [t, r, θ, ϕ], ξ2(0,0,1,0) [t, r, θ, ϕ] -−> ξ0(0,0,1,0) [t, r, θ, ϕ] + xx(0,0,1,0) [t, r, θ, ϕ], ξ2(0,1,0,0) [t, r, θ, ϕ] -−> ξ0(0,1,0,0) [t, r, θ, ϕ] + xx(0,1,0,0) [t, r, θ, ϕ], ξ3(0,1,0,0) [t, r, θ, ϕ] -−> ξ1(0,1,0,0) [t, r, θ, ϕ] + yy(0,1,0,0) [t, r, θ, ϕ], ξ3(1,0,0,0) [t, r, θ, ϕ] -−> ξ1(1,0,0,0) [t, r, θ, ϕ] + yy(1,0,0,0) [t, r, θ, ϕ], ξ3(0,0,0,1) [t, r, θ, ϕ] -−> ξ1(0,0,0,1) [t, r, θ, ϕ] + yy(0,0,0,1) [t, r, θ, ϕ], ξ2(1,0,0,0) [t, r, θ, ϕ] -−> ξ0(1,0,0,0) [t, r, θ, ϕ] + xx(1,0,0,0) [t, r, θ, ϕ] θ 2

Sec  b → -−

16 r2

2

θ 2

3 Cos[θ] Sec  -−

16 r2

2

θ + ⅇ-−ⅈ ϕ w Tan  -− 2

θ 3 2 ⅇⅈ ϕ Csc[θ]2 Sin  (xx[t, r, θ, ϕ] + ξ0[t, r, θ, ϕ]) + 2 θ ⅈ ⅇⅈ ϕ Csc[θ] Sec  xx(0,0,0,1) [t, r, θ, ϕ] + ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 θ ⅇⅈ ϕ Cos[θ] Sec  xx(0,0,1,0) [t, r, θ, ϕ] + ξ0(0,0,1,0) [t, r, θ, ϕ] + 2 θ 2 Sin  yy(0,0,1,0) [t, r, θ, ϕ] + ξ1(0,0,1,0) [t, r, θ, ϕ] + 2 θ 2 ⅇⅈ ϕ r Sin  xx(0,1,0,0) [t, r, θ, ϕ] + ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 θ r Cos[θ] Sec  yy(0,1,0,0) [t, r, θ, ϕ] + ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 θ r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2


21

22


sub2b = Simplifysub1b /∕/∕. xx[t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,0,1) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,0,1) [t, r, θ, ϕ] -− ⅈ ⅇ-−ⅈ ϕ zz[t, r, θ, ϕ], xx(0,0,1,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,0,1,0) [t, r, θ, ϕ], xx(0,1,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(0,1,0,0) [t, r, θ, ϕ], xx(1,0,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ zz(1,0,0,0) [t, r, θ, ϕ] θ 2

Sec  b → -−

16 r

2

2

θ 2

3 Cos[θ] Sec  -− 16 r

2

2

θ + ⅇ-−ⅈ ϕ w Tan  -− 2

θ θ 2 Csc[θ]2 Sin  zz[t, r, θ, ϕ] + ⅇⅈ ϕ ξ0[t, r, θ, ϕ] + Csc[θ] Sec  2 2 zz[t, r, θ, ϕ] + ⅈ zz(0,0,0,1) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,0,1) [t, r, θ, ϕ] + 3

θ Cos[θ] Sec  zz(0,0,1,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,0,1,0) [t, r, θ, ϕ] + 2 θ 2 Sin  yy(0,0,1,0) [t, r, θ, ϕ] + ξ1(0,0,1,0) [t, r, θ, ϕ] + 2 θ 2 r Sin  zz(0,1,0,0) [t, r, θ, ϕ] + ⅇⅈ ϕ ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 θ r Cos[θ] Sec  yy(0,1,0,0) [t, r, θ, ϕ] + ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 θ r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2

(*⋆EQUATION (7.71)*⋆) sub3b = Simplifysub2b /∕/∕. zz(0,0,0,1) [t, r, θ, ϕ] → 1 2

ⅈ (1 + Cos[θ]) zz[t, r, θ, ϕ] + 2 Sin[θ] Sin[θ] yy(0,0,1,0) [t, r, θ, ϕ] + Cos[θ] zz(0,0,1,0) [t, r, θ, ϕ] -− r Cos[θ] yy(0,1,0,0) [t, r, θ, ϕ] -− Sin[θ] zz(0,1,0,0) [t, r, θ, ϕ] + yy(1,0,0,0) [t, r, θ, ϕ]

b → -−

1 32 r2

θ θ 2 θ 3θ θ 3θ ⅇ-−ⅈ ϕ Csc  Sec  -− 8 r2 w Cos  + 8 r2 w Cos  -− ⅇⅈ ϕ Sin  + 3 ⅇⅈ ϕ Sin  -− 2 2 2 2 2 2 8 ⅇ2 ⅈ ϕ r2 (-− 1 + Cos[θ]) ξ0[t, r, θ, ϕ] -− 16 ⅈ ⅇ2 ⅈ ϕ r2 ξ0(0,0,0,1) [t, r, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ r2 Sin[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r2 ξ1(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r2 Cos[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ r3 ξ0(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇ2 ⅈ ϕ r3 Cos[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 Sin[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− 16 ⅇⅈ ϕ r3 Sin[θ] yy(1,0,0,0) [t, r, θ, ϕ] -− 16 ⅇⅈ ϕ r3 Sin[θ] ξ1(1,0,0,0) [t, r, θ, ϕ]



23

(*⋆EQUATION (7.72)*⋆) f0BYab = Simplify(a + b) /∕ 2 /∕/∕. a → -−

3 8r

+ ⅇ 2

ⅈ ϕ

θ

v Cot  -− 2

θ

θ 2

Sec  2

2

-−

1

θ Sec  ξ1[t, r, θ, ϕ] + 2 2

8r θ Cos  ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2

ⅈ Csc  Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ 2 θ θ θ Cos  Cot  ξ1(0,0,1,0) [t, r, θ, ϕ] + Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 2 2 θ θ ⅇⅈ ϕ r Cos  Cot  ξ0(0,1,0,0) [t, r, θ, ϕ] + 2 2 θ θ ⅇⅈ ϕ r Sin  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Cos  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ r Csc  zz(1,0,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ], 2 2 2 1 θ θ θ 3θ b → -− ⅇ-−ⅈ ϕ Csc  Sec  -− 8 r2 w Cos  + 8 r2 w Cos  -− 2 2 2 2 32 r2 θ 3θ ⅇⅈ ϕ Sin  + 3 ⅇⅈ ϕ Sin  -− 8 ⅇ2 ⅈ ϕ r2 (-− 1 + Cos[θ]) ξ0[t, r, θ, ϕ] -− 2 2 16 ⅈ ⅇ2 ⅈ ϕ r2 ξ0(0,0,0,1) [t, r, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ r2 Sin[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r2 ξ1(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r2 Cos[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ r3 ξ0(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇ2 ⅈ ϕ r3 Cos[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 Sin[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− 16 ⅇⅈ ϕ r3 Sin[θ] yy(1,0,0,0) [t, r, θ, ϕ] -− 16 ⅇⅈ ϕ r3 Sin[θ] ξ1(1,0,0,0) [t, r, θ, ϕ]  1

θ θ ⅇ-−ⅈ ϕ -− 4 ⅇ2 ⅈ ϕ r2 Sec  Tan  ξ0[t, r, θ, ϕ] -− 2 2 16 r θ θ Csc[θ]2 8 ⅇⅈ ϕ r2 Sin  Sin[θ] ξ1[t, r, θ, ϕ] -− 16 ⅈ ⅇ2 ⅈ ϕ r2 Sin  2 2 θ ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 Cos  -− 8 ⅈ ⅇⅈ ϕ r2 ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 θ 2 Sin  -− 4 ⅇ2 ⅈ ϕ r2 v -− 4 r2 w -− 4 ⅇ2 ⅈ ϕ r2 v Cos[θ] + 4 r2 w Cos[θ] + 2 3θ 3θ 3 ⅇⅈ ϕ Sin[θ] -− 8 ⅇ2 ⅈ ϕ r2 Sin  ξ0(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r2 Cos  2 2 3θ ξ1(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇ2 ⅈ ϕ r3 Cos  ξ0(0,1,0,0) [t, r, θ, ϕ] + 2 3θ θ 8 ⅇⅈ ϕ r3 Sin  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r3 Sin  yy(1,0,0,0) [ 2 2 θ θ t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 Cos  zz(1,0,0,0) [t, r, θ, ϕ] + 8 ⅇ2 ⅈ ϕ r3 Cos  2 2 θ ξ0(1,0,0,0) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r3 Sin  ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2


24


(*⋆EQUATION (7.73)*⋆) f3BYab = Simplify(a -− b) /∕ 2 /∕/∕. a → -−

3 8r

+ ⅇ 2

ⅈ ϕ

θ

v Cot  -− 2

θ

θ 2

Sec  2

2

-−

1

θ Sec  ξ1[t, r, θ, ϕ] + 2 2

8r θ Cos  ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2

ⅈ Csc  Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 ⅇⅈ ϕ 2 θ θ θ Cos  Cot  ξ1(0,0,1,0) [t, r, θ, ϕ] + Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 2 2 θ θ ⅇⅈ ϕ r Cos  Cot  ξ0(0,1,0,0) [t, r, θ, ϕ] + 2 2 θ θ ⅇⅈ ϕ r Sin  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Cos  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ θ r Csc  zz(1,0,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ], 2 2 2 1 θ θ θ 3θ b → -− ⅇ-−ⅈ ϕ Csc  Sec  -− 8 r2 w Cos  + 8 r2 w Cos  -− 2 2 2 2 32 r2 θ 3θ ⅇⅈ ϕ Sin  + 3 ⅇⅈ ϕ Sin  -− 8 ⅇ2 ⅈ ϕ r2 (-− 1 + Cos[θ]) ξ0[t, r, θ, ϕ] -− 2 2 16 ⅈ ⅇ2 ⅈ ϕ r2 ξ0(0,0,0,1) [t, r, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ r2 Sin[2 θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅇⅈ ϕ r2 ξ1(0,0,1,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r2 Cos[2 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ r3 ξ0(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇ2 ⅈ ϕ r3 Cos[2 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ r3 Sin[2 θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− 16 ⅇⅈ ϕ r3 Sin[θ] yy(1,0,0,0) [t, r, θ, ϕ] -− 16 ⅇⅈ ϕ r3 Sin[θ] ξ1(1,0,0,0) [t, r, θ, ϕ]  1 4

Csc[θ] 2 ⅇⅈ ϕ v -− 2 ⅇ-−ⅈ ϕ w + 2 ⅇⅈ ϕ v Cos[θ] + 2 ⅇ-−ⅈ ϕ w Cos[θ] + Csc[θ]

Cot[θ] r2

-−

θ 3 θ + 4 ⅇⅈ ϕ Csc[θ] Sin  ξ0[t, r, θ, ϕ] -− 2 Sin  ξ1[t, r, θ, ϕ] -− 2 2 r2 θ θ 2 ⅈ ⅇⅈ ϕ Sec  ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ Csc  ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ 3 θ ⅇⅈ ϕ Csc  ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ Cos  Csc[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ Sec  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 Csc[θ] Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ ⅇⅈ ϕ r Sec  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Csc[θ] Sin  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Csc  ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Csc  yy(1,0,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] yy(1,0,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Sec  zz(1,0,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Sin  zz(1,0,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ ⅇⅈ ϕ r Sec  ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Csc[θ] Sin  ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 2 θ 3 θ r Csc  ξ1(1,0,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2



(*⋆ end *⋆)

Solving the zero current equations part..

The following part is to compute one of the parts of j_0^NC which is known and needs change of coordinates. p1 = FullSimplify[e1 (c[[1, 1]] D[e1, r] + c[[1, 2]] D[e1, θ] + c[[1, 3]] D[e1, ϕ]) + e2 (c[[2, 1]] D[e1, r] + c[[2, 2]] D[e1, θ] + c[[2, 3]] D[e1, ϕ])] θ 2

Cos[ϕ] Csc[θ] Tan 

2

4 r3 p2 = FullSimplify[e1 (c[[1, 1]] D[e2, r] + c[[1, 2]] D[e2, θ] + c[[1, 3]] D[e2, ϕ]) + e2 (c[[2, 1]] D[e2, r] + c[[2, 2]] D[e2, θ] + c[[2, 3]] D[e2, ϕ])] θ 2

4

Csc[θ]3 Sin  Sin[ϕ] r3 partOFj0nc = Simplify[λ /∕ q (c[[1, 1]] D[p1, r] + c[[1, 2]] D[p1, θ] + c[[1, 3]] D[p1, ϕ] + c[[2, 1]] D[p2, r] + c[[2, 2]] D[p2, θ] + c[[2, 3]] D[p2, ϕ])](*⋆ this is the value of f0!!!*⋆) θ 2

λ (-− 1 + 2 Cos[θ] + Cos[2 θ]) Sec 

4

16 q r4

kForf0 = FullSimplifyDSolve2 k[θ] + Cot[θ] D[k[θ], θ] + D[D[k[θ], θ], θ] == θ 2

(-− 1 + 2 Cos[θ] + Cos[2 θ]) Sec  16 q

4

, k[θ], θ

$Aborted


25

26


(*⋆ made subs. for the roots *⋆) Root1 = Simplify

-− Sin[θ]2

-− 2 q ArcTanh[Cos[θ]] C[1] Cos[θ] Cos[θ]2 

(1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2 

-− Sin[θ]4 -−

(-− 1 + 2 Cos[θ] + Cos[2 θ]) 8 q Cos[θ]2  -− Sin[θ]4 +

1/∕4

1/∕4

1

θ 4 Cos  2 2

-− C[1] -− ⅈ C[2]

Cos[θ]2

Cos[θ] -− 1 + 3 Cos[2 θ] + Cos[3 θ] +

Cos[θ] -− 3 + 2 (Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2  2q

Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2 

-− Sin[θ]2 → ⅈ Sin[θ],



Sin[θ]2

-− Sin[θ]4

 /∕/∕.

-− Sin[θ]4 → ⅈ Sin[θ] ^ 2,

Sin[θ]2 → Sin[θ] ⅈ Sin[θ] -− 2 ⅈ q ArcTanh[Cos[θ]] C[1] Cos[θ] Cos[θ]2 

1/∕4

(1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  Sin[θ]2 -−

1

θ 4 1/∕4 Cos  (-− 1 + 2 Cos[θ] + Cos[2 θ]) 8 q Cos[θ]2  2 2 -− ⅈ C[1] + C[2]

Cos[θ]2

Sin[θ]2 +

Cos[θ] -− 1 + 3 Cos[2 θ] + Cos[3 θ] +

Cos[θ] -− 3 + 2 (Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2  2q

Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2 



Sin[θ]2

-− Sin[θ]4 R1MMod = Simplify ⅈ

-− 2 ⅈ q ArcTanh[Cos[θ]] C[1] Cos[θ]

Cos[θ]2  Sin[θ]2 -−

Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] +

1

θ 4 Cos  (-− 1 + 2 Cos[θ] + Cos[2 θ]) 8 q 2 2

(-− ⅈ C[1] + C[2] Cos[θ]) Sin[θ]2 +

Cos[θ]

Cos[θ] -− 1 + 3 Cos[2 θ] + Cos[3 θ] +

Cos[θ] -− 3 + 2 (Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2  2 q



Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  ⅈ Sin[θ] ^ 2

(*⋆ subs. made manually *⋆) Csc[θ]2 -− 4 ⅈ q ArcTanh[Cos[θ]] C[1] Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  Sin[θ]2 -− θ 4 Cos  (-− 1 + 2 Cos[θ] + Cos[2 θ]) -− 1 + 3 Cos[2 θ] + Cos[3 θ] -− 8 ⅈ q C[1] Sin[θ]2 + 2 Cos[θ] -− 3 + 2 (4 q C[2] + Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2  4 q (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2 





27

R1N2k = Simplify 2

Csc[θ]2 -− 4 ⅈ q ArcTanh[Cos[θ]] C[1] Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  θ 4 Sin[θ]2 -− Cos  (-− 1 + 2 Cos[θ] + Cos[2 θ]) 2 -− 1 + 3 Cos[2 θ] + Cos[3 θ] -− 8 ⅈ q C[1] Sin[θ]2 + Cos[θ] -− 3 + 2 (4 q C[2] + Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2 



4 q (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  + 1 /∕ Sin[θ] DSin[θ] D

Csc[θ]2 -− 4 ⅈ q ArcTanh[Cos[θ]] C[1] Cos[θ] (1 + Cos[θ])2 θ 4 -− 1 + Cos[θ] + Cos[θ]2  Sin[θ]2 -− Cos  (-− 1 + 2 Cos[θ] + Cos[2 θ]) 2 -− 1 + 3 Cos[2 θ] + Cos[3 θ] -− 8 ⅈ q C[1] Sin[θ]2 + Cos[θ] -− 3 + 2 (4 q C[2] + Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2 



4 q (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  , θ, θ -− θ 2

(-− 1 + 2 Cos[θ] + Cos[2 θ]) Sec 

4



16 q 0 kk[θ_] := Csc[θ]2 -− 4 ⅈ q ArcTanh[Cos[θ]] C[1] Cos[θ] (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  θ 4 Sin[θ]2 -− Cos  (-− 1 + 2 Cos[θ] + Cos[2 θ]) 2 -− 1 + 3 Cos[2 θ] + Cos[3 θ] -− 8 ⅈ q C[1] Sin[θ]2 + Cos[θ] -− 3 + 2 (4 q C[2] + Log[-− 1 -− Cos[θ]] -− Log[-− 1 + Cos[θ]]) Sin[θ]2  4 q (1 + Cos[θ])2 -− 1 + Cos[θ] + Cos[θ]2  Modkk = Simplify[kk[θ] /∕/∕. Log[-− 1 -− Cos[θ]] → 2 ArcTanh[Cos[θ]] + Log[-− 1 + Cos[θ]]] 1 8q

3 + 4 ⅈ q C[1] -− 2 q C[2] + (5 + 4 ⅈ q C[1] -− 4 q C[2]) Cos[θ] -− 4 ⅈ ArcTanh[Cos[θ]] θ 2 θ 2 (-− ⅈ + 2 q C[1]) Cos  Cos[θ] + Cos[2 θ] -− 2 q C[2] Cos[2 θ] Sec  2 2

C1 = Solve[-− ⅈ + 2 q C[1] ⩵ 0, C[1]] ⅈ C[1] →

 2q




28


kValue = SimplifyModkk /∕/∕. C[1] →

ⅈ  2q θ 2

Cos[θ] (-− 3 + 4 q C[2] + (-− 2 + 4 q C[2]) Cos[θ]) Sec  -−

2

8q Simplify2 -− Cot[θ] D-− DD-−

1

θ 2 Cos[θ] (-− 3 + 4 q C[2] + (-− 2 + 4 q C[2]) Cos[θ]) Sec  + 8 q r^4 2 1

θ 2 Cos[θ] (-− 3 + 4 q C[2] + (-− 2 + 4 q C[2]) Cos[θ]) Sec  , θ + 8 q r^4 2

1

θ 2 Cos[θ] (-− 3 + 4 q C[2] + (-− 2 + 4 q C[2]) Cos[θ]) Sec  , θ, θ 8 q r^4 2 θ 2

(-− 1 + 2 Cos[θ] + Cos[2 θ]) Sec 

4

16 q r4

(*⋆EQUATION (7.85)*⋆) θ 2

2

Cos[θ] 4 -− 8 q C[2] + Sec   f0TILDA = Simplifyf0BYab -− 1 8

-−

3 r2

4 ⅇ

-−

-−ⅈ ϕ

4 Cos[θ] q r2

+

8 C[2] Cos[θ] r2 θ 2

Cos[θ] Sec  w Csc[θ] -− qr



8 q r^2

2

+ 4 ⅇⅈ ϕ v Cot[θ] -− 4 ⅇ-−ⅈ ϕ w Cot[θ] + 4 ⅇⅈ ϕ v Csc[θ] + 2

θ θ -− 2 ⅇⅈ ϕ Sec  Tan  ξ0[t, r, θ, ϕ] -− 2 2

θ θ 2 Sec  ξ1[t, r, θ, ϕ] + 8 ⅈ ⅇⅈ ϕ Csc[θ]2 Sin  ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ 3θ 4 ⅈ Csc  Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ Csc[θ] Sin  2 2 3θ ξ0(0,0,1,0) [t, r, θ, ϕ] -− 4 Csc[θ] Csc  Sin[3 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 4 ⅇⅈ ϕ r Csc[θ] Csc  Sin[3 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 3θ 8 r Csc[θ] Sin  ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 θ θ 4 r Sec  yy(1,0,0,0) [t, r, θ, ϕ] -− 4 r Csc  zz(1,0,0,0) [t, r, θ, ϕ] -− 2 2 θ θ 4 ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] + 4 r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2 (*⋆ end . so far no assumption that can not be reversable is made . *⋆)


Appendix C Finding a noncommutative monopole II

191

Section (7.1.3) (*⋆EQUATION (7.85)*⋆) f0TILDA = 1

-−

3

-−

4 Cos[θ]

r2

8

+

8 C[2] Cos[θ]

q r2

r2 θ 2

+ 4 ⅇⅈ ϕ v Cot[θ] -− 4 ⅇ-−ⅈ ϕ w Cot[θ] + 4 ⅇⅈ ϕ v Csc[θ] +

Cos[θ] Sec  4 ⅇ

-−ⅈ ϕ

w Csc[θ] -− qr

2

2

θ θ -− 2 ⅇⅈ ϕ Sec  Tan  ξ0[t, r, θ, ϕ] -− 2 2

θ θ θ 2 Sec  ξ1[t, r, θ, ϕ] + 8 ⅈ ⅇⅈ ϕ Csc[θ]2 Sin  ξ0(0,0,0,1) [t, r, θ, ϕ] + 4 ⅈ Csc  2 2 2 3 θ Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ Csc[θ] Sin  ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 3θ 4 Csc[θ] Csc  Sin[3 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 4 ⅇⅈ ϕ r Csc[θ] Csc  2 2 3θ Sin[3 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− 8 r Csc[θ] Sin  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 θ θ 4 ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] + 4 r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] ; 2 2 f0TILD = Simplifyf0TILDA /∕/∕. v -−> ⅇ-−ⅈ ϕ vv[r, t, θ, ϕ] r ^ n, w -−> ⅇⅈ ϕ ww[r, t, θ, ϕ] r ^ n (*⋆ but f0TILDA does not have any zz, yy the time derivative ones we take to zero! *⋆) 1 8

-−

3 r2

-−

4 Cos[θ] q r2 θ 2

Cos[θ] Sec  q r2

+

8 C[2] Cos[θ] r2

-−

2

+ 4 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] +

θ θ θ 4 rn Tan  ww[r, t, θ, ϕ] -− 2 ⅇⅈ ϕ Sec  Tan  ξ0[t, r, θ, ϕ] -− 2 2 2 θ θ 2 Sec  ξ1[t, r, θ, ϕ] + 8 ⅈ ⅇⅈ ϕ Csc[θ]2 Sin  ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ 3θ 4 ⅈ Csc  Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 8 ⅇⅈ ϕ Csc[θ] Sin  2 2 3θ ξ0(0,0,1,0) [t, r, θ, ϕ] -− 4 Csc[θ] Csc  Sin[3 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 3 θ 4 ⅇⅈ ϕ r Csc[θ] Csc  Sin[3 θ] ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 3θ 8 r Csc[θ] Sin  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 θ θ 4 ⅇⅈ ϕ r Csc  ξ0(1,0,0,0) [t, r, θ, ϕ] + 4 r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2


2

Finding noncommutative monopole II.nb

f0TIL = Simplifyf0TILD /∕/∕. ξ0[t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0[t, r, θ, ϕ], ξ0(1,0,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0(1,0,0,0) [t, r, θ, ϕ], ξ0(0,1,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0(0,1,0,0) [t, r, θ, ϕ], ξ0(0,0,1,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0(0,0,1,0) [t, r, θ, ϕ], ξ0(0,0,0,1) [t, r, θ, ϕ] -−> ⅇ-−ⅈ ϕ η0(0,0,0,1) [t, r, θ, ϕ] -− ⅈ ⅇ-−ⅈ ϕ η0[t, r, θ, ϕ] 1 8

-−

3 r2

-−

4 Cos[θ] q r2

+

8 C[2] Cos[θ] r2

θ 2

Cos[θ] Sec  -−

q r2

2

+

θ 4 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] + 4 rn Tan  ww[r, t, θ, ϕ] + 2 θ θ θ 8 Csc[θ]2 Sin  η0[t, r, θ, ϕ] -− 2 Sec  Tan  η0[t, r, θ, ϕ] -− 2 2 2 θ θ 2 Sec  ξ1[t, r, θ, ϕ] + 8 ⅈ Csc[θ]2 Sin  η0(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ 3θ 4 ⅈ Csc  Csc[θ] ξ1(0,0,0,1) [t, r, θ, ϕ] + 8 Csc[θ] Sin  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 2 3θ 4 Csc[θ] Csc  Sin[3 θ] ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 4 r Csc[θ] Csc  Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 3θ 8 r Csc[θ] Sin  ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 θ θ 4 r Csc  η0(1,0,0,0) [t, r, θ, ϕ] + 4 r Sec  ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2



f0TI = Simplifyf0TIL /∕/∕. ξ1[t, r, θ, ϕ] → r ^ n hhh[t, r, θ, ϕ], ξ1(1,0,0,0) [t, r, θ, ϕ] → r ^ n hhh(1,0,0,0) [t, r, θ, ϕ] 0, ξ1(0,1,0,0) [t, r, θ, ϕ] → n r ^ (n -− 1) hhh[t, r, θ, ϕ] + r ^ n hhh(0,1,0,0) [t, r, θ, ϕ], ξ1(0,0,1,0) [t, r, θ, ϕ] → r ^ n hhh(0,0,1,0) [t, r, θ, ϕ], ξ1(0,0,0,1) [t, r, θ, ϕ] -−> r ^ n hhh(0,0,0,1) [t, r, θ, ϕ] 1 8

-−

3 r2

-−

4 Cos[θ] q r2 θ 2

Cos[θ] Sec  q r2

+

8 C[2] Cos[θ] r2

θ -− 2 rn hhh[t, r, θ, ϕ] Sec  -− 2

2

+ 4 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] +

θ θ 4 rn Tan  ww[r, t, θ, ϕ] + 8 Csc[θ]2 Sin  η0[t, r, θ, ϕ] -− 2 2 θ θ θ 2 Sec  Tan  η0[t, r, θ, ϕ] + 4 ⅈ rn Csc  Csc[θ] hhh(0,0,0,1) [t, r, θ, ϕ] + 2 2 2 θ 3θ 8 ⅈ Csc[θ]2 Sin  η0(0,0,0,1) [t, r, θ, ϕ] -− 4 rn Csc[θ] Csc  Sin[3 θ] 2 2 3θ hhh(0,0,1,0) [t, r, θ, ϕ] + 8 Csc[θ] Sin  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 8 rn Csc[θ] Sin  n hhh[t, r, θ, ϕ] + r hhh(0,1,0,0) [t, r, θ, ϕ] -− 2 3θ 4 r Csc[θ] Csc  Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 θ 4 r Csc  η0(1,0,0,0) [t, r, θ, ϕ] 2


3

4


FullSimplify 1

-−

3

-−

r2

8

4 Cos[θ] q r2

+

8 C[2] Cos[θ] r2

n

θ

-− 2 r hhh[t, r, θ, ϕ] Sec  -− 2

θ 2

Cos[θ] Sec 

2

+

q r2

θ 4 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] + 4 rn Tan  ww[r, t, θ, ϕ] + 2 θ θ θ 8 Csc[θ]2 Sin  η0[t, r, θ, ϕ] -− 2 Sec  Tan  η0[t, r, θ, ϕ] + 2 2 2 θ 4 ⅈ rn Csc  Csc[θ] hhh(0,0,0,1) [t, r, θ, ϕ] + 2 θ 3θ 8 ⅈ Csc[θ]2 Sin  η0(0,0,0,1) [t, r, θ, ϕ] -− 4 rn Csc[θ] Csc  Sin[3 θ] 2 2 3θ hhh(0,0,1,0) [t, r, θ, ϕ] + 8 Csc[θ] Sin  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 8 rn Csc[θ] Sin  n hhh[t, r, θ, ϕ] + r hhh(0,1,0,0) [t, r, θ, ϕ] -− 4 r Csc[θ] 2 Csc

3θ

θ  Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 4 r Csc  η0(1,0,0,0) [t, r, θ, ϕ]  2 2

1 8 q r2 θ -− 2 -− 3 q + (-− 4 + 8 q C[2]) Cos[θ] -− 2 q r2+n (1 + 2 n + 4 n Cos[θ]) hhh[t, r, θ, ϕ] Sec  + 2 θ 2 Sec  + 2 q r2 2 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] + 2 θ θ 2 rn Tan  ww[r, t, θ, ϕ] + 4 Csc[θ]2 Sin  η0[t, r, θ, ϕ] -− 2 2 θ θ θ Sec  Tan  η0[t, r, θ, ϕ] + 2 ⅈ rn Csc  Csc[θ] hhh(0,0,0,1) [t, r, θ, ϕ] + 2 2 2 θ 3θ 2 (0,0,0,1) 4 ⅈ Csc[θ] Sin  η0 [t, r, θ, ϕ] -− 2 rn Csc[θ] Csc  Sin[3 θ] 2 2 3θ hhh(0,0,1,0) [t, r, θ, ϕ] + 4 Csc[θ] Sin  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 3θ 4 r1+n Csc[θ] Sin  hhh(0,1,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Csc  2 2 θ Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Csc  η0(1,0,0,0) [t, r, θ, ϕ] 2



(*⋆EQUATION (7.89)*⋆) f0T = 1 -− 2 -− 3 q + (-− 4 + 8 q C[2]) Cos[θ] -− 2 q r2+n (1 + 2 n + 4 n Cos[θ]) hhh[t, r, θ, ϕ] 8 q r2 θ θ 2 Sec  + Sec  + 2 q r2 2 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] + 2 2 θ θ 2 rn Tan  ww[r, t, θ, ϕ] + 4 Csc[θ]2 Sin  η0[t, r, θ, ϕ] -− 2 2 θ θ θ Sec  Tan  η0[t, r, θ, ϕ] + 2 ⅈ rn Csc  Csc[θ] hhh(0,0,0,1) [t, r, θ, ϕ] + 2 2 2 θ 3θ 4 ⅈ Csc[θ]2 Sin  η0(0,0,0,1) [t, r, θ, ϕ] -− 2 rn Csc[θ] Csc  Sin[3 θ] 2 2 3θ hhh(0,0,1,0) [t, r, θ, ϕ] + 4 Csc[θ] Sin  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 3θ 3θ 4 r1+n Csc[θ] Sin  hhh(0,1,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Csc  2 2 θ Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Csc  η0(1,0,0,0) [t, r, θ, ϕ] ; 2


5

6


Simplify Solve

1 8 q r2

-− 2 -− 3 q + (-− 4 + 8 q C[2]) Cos[θ] -− 2 q r2+n (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ]

θ θ 2 Sec  + Sec  + 2 q r2 2 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] + 2 2 θ θ 2 rn Tan  ww[r, t, θ, ϕ] + 4 Csc[θ]2 Sin  η0[r, θ, ϕ] -− 2 2 θ θ θ Sec  Tan  η0[r, θ, ϕ] + 2 ⅈ rn Csc  Csc[θ] hhh(0,1) [θ, ϕ] + 2 2 2 θ 3θ 4 ⅈ Csc[θ]2 Sin  η0(0,0,1) [r, θ, ϕ] -− 2 rn Csc[θ] Csc  2 2 3θ Sin[3 θ] hhh(1,0) [θ, ϕ] + 4 Csc[θ] Sin  η0(0,1,0) [r, θ, ϕ] -− 2 3θ 2 r Csc[θ] Csc  Sin[3 θ] η0(1,0,0) [r, θ, ϕ] ⩵ 0, hhh(0,1) [θ, ϕ] 2 hhh(0,1) [θ, ϕ] → -−

1

θ θ θ ⅈ r-−2-−n Sin[θ] 2 Sin  + 3 q Sin  + 4 Cos[θ] Sin  -− 8 q C[2] Cos[θ] 4q 2 2 2 θ θ θ θ Sin  + 2 q r2+n (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ] Tan  -− Sec  Tan  -− 2 2 2 2 3 θ θ 4 q r2+n Cos  vv[r, t, θ, ϕ] -− 8 q r2+n Csc[θ] Sin  ww[r, t, θ, ϕ] -− 2 2 2 2 θ θ 2 q r2 Sec  η0[r, θ, ϕ] + 2 q r2 Tan  η0[r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] + 2 2 θ 2 8 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] -− 2 ⅈ q r2 Sec  η0(0,0,1) [r, θ, ϕ] -− 2 θ θ 4 q r2 Tan  η0(0,1,0) [r, θ, ϕ] -− 8 q r2 Cos[θ] Tan  η0(0,1,0) [r, θ, ϕ] -− 2 2 4 q r3 η0(1,0,0) [r, θ, ϕ] + 8 q r3 Cos[θ] η0(1,0,0) [r, θ, ϕ] 



7

(*⋆ EQUATION (7.90)*⋆) hphi = FullSimplify 1

θ θ θ ⅈ r-−2-−n Sin[θ] 2 Sin  + 3 q Sin  + 4 Cos[θ] Sin  -− 4q 2 2 2 θ θ θ 8 q C[2] Cos[θ] Sin  + 2 q r2+n (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ] Tan  -− Sec  2 2 2 3 θ θ θ Tan  -− 4 q r2+n Cos  vv[r, t, θ, ϕ] -− 8 q r2+n Csc[θ] Sin  ww[r, t, θ, ϕ] -− 2 2 2 2 2 θ θ 2 q r2 Sec  η0[r, θ, ϕ] + 2 q r2 Tan  η0[r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] + 2 2 θ 2 8 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] -− 2 ⅈ q r2 Sec  η0(0,0,1) [r, θ, ϕ] -− 2 θ θ 4 q r2 Tan  η0(0,1,0) [r, θ, ϕ] -− 8 q r2 Cos[θ] Tan  η0(0,1,0) [r, θ, ϕ] -− 2 2

hhh(0,1) [θ, ϕ] → -−

4 q r3 η0(1,0,0) [r, θ, ϕ] + 8 q r3 Cos[θ] η0(1,0,0) [r, θ, ϕ]  hhh(0,1) [θ, ϕ] → 1 -−2-−n θ θ r -− 4 ⅈ q r2+n Sin  (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ] Sin  -− (1 + Cos[θ]) 4q 2 2 3θ vv[r, t, θ, ϕ] + (-− 1 + Cos[θ]) ww[r, t, θ, ϕ] + 2 Cos  hhh(1,0) [θ, ϕ] + 2 1 θ 3θ ⅈ Sin[θ] (2 -− 3 q) Sin  + (-− 4 + q (-− 3 + 4 C[2])) Sin + 2 (1 + Cos[θ]) 2 2 5θ 2 (-− 1 + 2 q C[2]) Sin + 2 4 q r2 (1 + Cos[θ]) η0[r, θ, ϕ] + 2 ⅈ η0(0,0,1) [r, θ, ϕ] + 2 (Sin[θ] + Sin[2 θ]) η0(0,1,0) [r, θ, ϕ] -− 2 r (Cos[θ] + Cos[2 θ]) η0(1,0,0) [r, θ, ϕ]


8


Simplify 1 8 q r2

θ -− 2 -− 3 q + (-− 4 + 8 q C[2]) Cos[θ] -− 2 q r2+n (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ] Sec  + 2

θ 2 θ Sec  + 2 q r2 2 rn (Cot[θ] + Csc[θ]) vv[r, t, θ, ϕ] + 2 rn Tan  2 2 θ θ θ ww[r, t, θ, ϕ] + 4 Csc[θ]2 Sin  η0[r, θ, ϕ] -− Sec  Tan  η0[r, θ, ϕ] + 2 2 2 θ θ 2 ⅈ rn Csc  Csc[θ] hhh(0,1) [θ, ϕ] + 4 ⅈ Csc[θ]2 Sin  η0(0,0,1) [r, θ, ϕ] -− 2 2 3 θ 3θ 2 rn Csc[θ] Csc  Sin[3 θ] hhh(1,0) [θ, ϕ] + 4 Csc[θ] Sin  2 2 3θ η0(0,1,0) [r, θ, ϕ] -− 2 r Csc[θ] Csc  Sin[3 θ] η0(1,0,0) [r, θ, ϕ] /∕/∕. 2 1 -−2-−n θ θ hhh(0,1) [θ, ϕ] → r -− 4 ⅈ q r2+n Sin  (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ] Sin  -− 4q 2 2 (1 + Cos[θ]) vv[r, t, θ, ϕ] + (-− 1 + Cos[θ]) ww[r, t, θ, ϕ] + 3θ 1 2 Cos  hhh(1,0) [θ, ϕ] + ⅈ Sin[θ] 2 2 (1 + Cos[θ]) θ 3θ 5θ (2 -− 3 q) Sin  + (-− 4 + q (-− 3 + 4 C[2])) Sin  + 2 (-− 1 + 2 q C[2]) Sin + 2 2 2 4 q r2 (1 + Cos[θ]) η0[r, θ, ϕ] + 2 ⅈ η0(0,0,1) [r, θ, ϕ] + 2 (Sin[θ] + Sin[2 θ]) η0(0,1,0) [r, θ, ϕ] -− 2 r (Cos[θ] + Cos[2 θ]) η0(1,0,0) [r, θ, ϕ] 0





(*⋆EQUATION (7.73)*⋆) 1 f3BYab = Csc[θ] 2 ⅇⅈ ϕ v -− 2 ⅇ-−ⅈ ϕ w + 2 ⅇⅈ ϕ v Cos[θ] + 4 Cot[θ] Csc[θ] θ 3 2 ⅇ-−ⅈ ϕ w Cos[θ] + -− + 4 ⅇⅈ ϕ Csc[θ] Sin  ξ0[t, r, θ, ϕ] -− 2 r2 r2 θ θ 2 Sin  ξ1[t, r, θ, ϕ] -− 2 ⅈ ⅇⅈ ϕ Sec  ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ θ 2 ⅈ Csc  ξ1(0,0,0,1) [t, r, θ, ϕ] + ⅇⅈ ϕ Csc  ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 2 3 θ θ 2 ⅇⅈ ϕ Cos  Csc[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− Sec  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 2 3θ θ 2 Csc[θ] Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Sec  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 2 3 θ θ 2 ⅇⅈ ϕ r Csc[θ] Sin  ξ0(0,1,0,0) [t, r, θ, ϕ] -− r Csc  ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 2 3θ θ 2 r Cos  Csc[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− ⅇⅈ ϕ r Sec  ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 2 3 θ 2 ⅇⅈ ϕ r Csc[θ] Sin  ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 θ 3θ r Csc  ξ1(1,0,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] ; 2 2 f3n = Simplifyf3BYab /∕/∕. v -−> ⅇ-−ⅈ ϕ vv[r, t, θ, ϕ] r ^ n, w -−> ⅇⅈ ϕ ww[r, t, θ, ϕ] r ^ n 1 4

Csc[θ] Cot[θ] r

2

-−

Csc[θ] r2

+ 2 rn (1 + Cos[θ]) vv[r, t, θ, ϕ] + 2 rn (-− 1 + Cos[θ]) ww[r, t, θ, ϕ] +

θ 3 θ 4 ⅇⅈ ϕ Csc[θ] Sin  ξ0[t, r, θ, ϕ] -− 2 Sin  ξ1[t, r, θ, ϕ] -− 2 2 θ θ 2 ⅈ ⅇⅈ ϕ Sec  ξ0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ Csc  ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ 3 θ ⅇⅈ ϕ Csc  ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ Cos  Csc[θ] ξ0(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ Sec  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 Csc[θ] Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ ⅇⅈ ϕ r Sec  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Csc[θ] Sin  ξ0(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Csc  ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ ⅇⅈ ϕ r Sec  ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 ⅇⅈ ϕ r Csc[θ] Sin  ξ0(1,0,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Csc  ξ1(1,0,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2


9

10


f3ne = Simplifyf3n /∕/∕. ξ0[t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0[t, r, θ, ϕ], ξ0(1,0,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0(1,0,0,0) [t, r, θ, ϕ], ξ0(0,1,0,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0(0,1,0,0) [t, r, θ, ϕ], ξ0(0,0,1,0) [t, r, θ, ϕ] → ⅇ-−ⅈ ϕ η0(0,0,1,0) [t, r, θ, ϕ], ξ0(0,0,0,1) [t, r, θ, ϕ] -−> ⅇ-−ⅈ ϕ η0(0,0,0,1) [t, r, θ, ϕ] -− ⅈ ⅇ-−ⅈ ϕ η0[t, r, θ, ϕ] 1 4

Csc[θ] Cot[θ] r

2

-−

Csc[θ] r2

+ 2 rn (1 + Cos[θ]) vv[r, t, θ, ϕ] + 2 rn (-− 1 + Cos[θ]) ww[r, t, θ, ϕ] -−

θ θ 3 θ 2 Sec  η0[t, r, θ, ϕ] + 4 Csc[θ] Sin  η0[t, r, θ, ϕ] -− 2 Sin  ξ1[t, r, θ, ϕ] -− 2 2 2 θ θ 2 ⅈ Sec  η0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ Csc  ξ1(0,0,0,1) [t, r, θ, ϕ] + 2 2 θ 3 θ Csc  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 Cos  Csc[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ Sec  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 Csc[θ] Sin  ξ1(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Sec  η0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Sin  η0(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Csc  ξ1(0,1,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Sec  η0(1,0,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Sin  η0(1,0,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Csc  ξ1(1,0,0,0) [t, r, θ, ϕ] + 2 r Cos  Csc[θ] ξ1(1,0,0,0) [t, r, θ, ϕ] 2 2 f3new = Simplifyf3ne /∕/∕. ξ1[t, r, θ, ϕ] → r ^ n hhh[t, r, θ, ϕ], ξ1(1,0,0,0) [t, r, θ, ϕ] → r ^ n hhh(1,0,0,0) [t, r, θ, ϕ] 0, ξ1(0,1,0,0) [t, r, θ, ϕ] → n r ^ (n -− 1) hhh[t, r, θ, ϕ] + r ^ n hhh(0,1,0,0) [t, r, θ, ϕ], ξ1(0,0,1,0) [t, r, θ, ϕ] → r ^ n hhh(0,0,1,0) [t, r, θ, ϕ], ξ1(0,0,0,1) [t, r, θ, ϕ] -−> r ^ n hhh(0,0,0,1) [t, r, θ, ϕ] 1 4

Cot[θ]

Csc[θ]

θ -− 2 (1 + 2 n) rn hhh[t, r, θ, ϕ] Sin  + 2 r2 r2 2 rn (1 + Cos[θ]) vv[r, t, θ, ϕ] -− 2 rn ww[r, t, θ, ϕ] + 2 rn Cos[θ] ww[r, t, θ, ϕ] -− θ θ 3 2 Sec  η0[t, r, θ, ϕ] + 4 Csc[θ] Sin  η0[t, r, θ, ϕ] + 2 2 θ θ 2 ⅈ rn Csc  hhh(0,0,0,1) [t, r, θ, ϕ] -− 2 ⅈ Sec  η0(0,0,0,1) [t, r, θ, ϕ] -− 2 2 θ 3θ rn Sec  hhh(0,0,1,0) [t, r, θ, ϕ] -− 2 rn Csc[θ] Sin  hhh(0,0,1,0) [t, r, θ, ϕ] + 2 2 θ 3θ Csc  η0(0,0,1,0) [t, r, θ, ϕ] -− 2 Cos  Csc[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r1+n Csc  hhh(0,1,0,0) [t, r, θ, ϕ] + 2 r1+n Cos  Csc[θ] hhh(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Sec  η0(0,1,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Sin  η0(0,1,0,0) [t, r, θ, ϕ] -− 2 2 θ 3θ r Sec  η0(1,0,0,0) [t, r, θ, ϕ] -− 2 r Csc[θ] Sin  η0(1,0,0,0) [t, r, θ, ϕ] 2 2

Csc[θ]

-−



Simplify

1

4 Cot[θ]

11

Csc[θ] Csc[θ]

θ -− 2 (1 + 2 n) rn hhh[θ, ϕ] Sin  + 2 rn (1 + Cos[θ]) vv[r, t, θ, ϕ] -− 2 r r θ 2 rn ww[r, t, θ, ϕ] + 2 rn Cos[θ] ww[r, t, θ, ϕ] -− 2 Sec  η0[r, θ, ϕ] + 2 θ 3 θ θ 4 Csc[θ] Sin  η0[r, θ, ϕ] + 2 ⅈ rn Csc  hhh(0,1) [θ, ϕ] -− 2 ⅈ Sec  2 2 2 θ 3 θ η0(0,0,1) [r, θ, ϕ] -− rn Sec  hhh(1,0) [θ, ϕ] -− 2 rn Csc[θ] Sin  hhh(1,0) [θ, ϕ] + 2 2 θ 3θ Csc  η0(0,1,0) [r, θ, ϕ] -− 2 Cos  Csc[θ] η0(0,1,0) [r, θ, ϕ] -− 2 2 θ 3θ r Sec  η0(1,0,0) [r, θ, ϕ] -− 2 r Csc[θ] Sin  η0(1,0,0) [r, θ, ϕ] /∕/∕. 2 2 1 -−2-−n θ θ hhh(0,1) [θ, ϕ] → r -− 4 ⅈ q r2+n Sin  (1 + 2 n + 4 n Cos[θ]) hhh[θ, ϕ] Sin  -− 4q 2 2 (1 + Cos[θ]) vv[r, t, θ, ϕ] + (-− 1 + Cos[θ]) ww[r, t, θ, ϕ] + 3θ 1 2 Cos  hhh(1,0) [θ, ϕ] + ⅈ Sin[θ] 2 2 (1 + Cos[θ]) θ 3θ 5θ (2 -− 3 q) Sin  + (-− 4 + q (-− 3 + 4 C[2])) Sin  + 2 (-− 1 + 2 q C[2]) Sin + 2 2 2 4 q r2 (1 + Cos[θ]) η0[r, θ, ϕ] + 2 ⅈ η0(0,0,1) [r, θ, ϕ] + 2 (Sin[θ] + Sin[2 θ]) 2

-−

2

η0(0,1,0) [r, θ, ϕ] -− 2 r (Cos[θ] + Cos[2 θ]) η0(1,0,0) [r, θ, ϕ] -−



1 16 q r2 (1 + Cos[θ]) θ θ θ 3θ 3θ 3θ Csc  2 Sin  + q Sin  -− 4 Sin  -− 3 q Sin  + 4 q C[2] Sin  -− 2 2 2 2 2 2 5θ 5θ 8 n q r2+n hhh[θ, ϕ] Sin[2 θ] -− 2 Sin  + 4 q C[2] Sin + 2 2 θ 16 q r2+n Sin  Sin[θ] ww[r, t, θ, ϕ] + 8 q r2 η0[r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[r, θ, ϕ] + 8 q r2+n hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 16 ⅈ q r2 η0(0,0,1) [r, θ, ϕ] + 8 q r2 Sin[2 θ] η0(0,1,0) [r, θ, ϕ] + 8 q r3 η0(1,0,0) [r, θ, ϕ] -− 8 q r3 Cos[2 θ] η0(1,0,0) [r, θ, ϕ]


12


FullSimplify -−

1

θ θ θ 3θ 3θ Csc  2 Sin  + q Sin  -− 4 Sin  -− 3 q Sin + 2 2 2 2 2 16 q r2 (1 + Cos[θ]) 4 q C[2] Sin

3θ 2 5θ

 -− 8 n q r2+n hhh[θ, ϕ] Sin[2 θ] -− 2 Sin

5θ

+

2

θ  + 16 q r2+n Sin  Sin[θ] ww[r, t, θ, ϕ] + 8 q r2 η0[r, θ, ϕ] + 2 2 8 q r2 Cos[θ] η0[r, θ, ϕ] + 8 q r2+n hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 16 ⅈ q r2 η0(0,0,1) [r, θ, ϕ] + 8 q r2 Sin[2 θ] η0(0,1,0) [r, θ, ϕ] + 4 q C[2] Sin

8 q r3 η0(1,0,0) [r, θ, ϕ] -− 8 q r3 Cos[2 θ] η0(1,0,0) [r, θ, ϕ]  1 8 q r2 (1 + Cos[θ]) 2 + q -− 4 q C[2] + (6 + q (3 -− 8 C[2])) Cos[θ] + (2 -− 4 q C[2]) Cos[2 θ] + 4 q r2 θ θ 4 rn Cos  n Cos[θ] hhh[θ, ϕ] -− Sin  ww[r, t, θ, ϕ] -− Sin[θ] hhh(1,0) [θ, ϕ] -− 2 2 θ Csc  (1 + Cos[θ]) η0[r, θ, ϕ] + 2 ⅈ η0(0,0,1) [r, θ, ϕ] + 2 2 Sin[θ] Cos[θ] η0(0,1,0) [r, θ, ϕ] + r Sin[θ] η0(1,0,0) [r, θ, ϕ] (*⋆ EQUATION (7.91) *⋆) f3 = 1 2 + q -− 4 q C[2] + (6 + q (3 -− 8 C[2])) Cos[θ] + (2 -− 4 q C[2]) Cos[2 θ] + 2 8 q r (1 + Cos[θ]) θ θ 4 q r2 4 rn Cos  n Cos[θ] hhh[θ, ϕ] -− Sin  ww[r, t, θ, ϕ] -− Sin[θ] 2 2 θ hhh(1,0) [θ, ϕ] -− Csc  (1 + Cos[θ]) η0[r, θ, ϕ] + 2 ⅈ η0(0,0,1) [r, θ, ϕ] + 2 2 Sin[θ] Cos[θ] η0(0,1,0) [r, θ, ϕ] + r Sin[θ] η0(1,0,0) [r, θ, ϕ]

(*⋆ We try to solve equations (7.96),(7.97)and (7.98)

;

*⋆)

f1v = 1 /∕ (2 r ^ 2) ⅇ-−ⅈ ϕ vv[r, t, θ, ϕ] + ⅇⅈ ϕ ww[r, t, θ, ϕ]; f2v = 1 /∕ (2 I r ^ 2) ⅇ-−ⅈ ϕ vv[r, t, θ, ϕ] -− ⅇⅈ ϕ ww[r, t, θ, ϕ];

(*⋆ Ploar coordinates matrix cij , EQuation (7.38) *⋆) Cos[θ] Cos[ϕ] Csc[θ] Sin[ϕ] c := Cos[ϕ] Sin[θ], , -− , r r Cos[θ] Sin[ϕ] Cos[ϕ] Csc[θ] Sin[θ] Sin[θ] Sin[ϕ], , , Cos[θ], -− , 0 r r r (*⋆try to calc EQUATION (7.96)*⋆)



wx = FullSimplifyc[[1, 1]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], r + c[[1, 2]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], θ + c[[1, 3]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], ϕ 1 r

ⅇⅈ ϕ Csc[θ] -− ⅈ Sin[ϕ] ww[r, t, θ, ϕ] -− Sin[ϕ] ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] Cos[θ] ww(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ]

wiy = FullSimplifyI c[[2, 1]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], r + c[[2, 2]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], θ + c[[2, 3]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], ϕ 1 r

ⅈ ⅇⅈ ϕ Csc[θ] Cos[ϕ] ⅈ ww[r, t, θ, ϕ] + ww(0,0,0,1) [r, t, θ, ϕ] + Sin[θ] Sin[ϕ] Cos[θ] ww(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ]

wTerm = FullSimplify[wx + wiy] 1 r

ⅇ2 ⅈ ϕ Csc[θ] -− ww[r, t, θ, ϕ] + ⅈ ww(0,0,0,1) [r, t, θ, ϕ] + Sin[θ] Cos[θ] ww(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ]

vx = FullSimplifyc[[1, 1]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], r + c[[1, 2]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], θ + c[[1, 3]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], ϕ 1 r

ⅇ-−ⅈ ϕ ⅈ Csc[θ] Sin[ϕ] vv[r, t, θ, ϕ] + ⅈ vv(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Cos[θ] vv(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] vv(1,0,0,0) [r, t, θ, ϕ]

viy = FullSimplifyI c[[2, 1]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], r + c[[2, 2]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], θ + c[[2, 3]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], ϕ 1 r

ⅇ-−ⅈ ϕ Cos[ϕ] Csc[θ] vv[r, t, θ, ϕ] + ⅈ vv(0,0,0,1) [r, t, θ, ϕ] + ⅈ Sin[ϕ] Cos[θ] vv(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] vv(1,0,0,0) [r, t, θ, ϕ]

vTerm = FullSimplify[vx -− viy] 1 r

ⅇ-−2 ⅈ ϕ -− Csc[θ] vv[r, t, θ, ϕ] + ⅈ vv(0,0,0,1) [r, t, θ, ϕ] + Cos[θ] vv(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] vv(1,0,0,0) [r, t, θ, ϕ]

brac = FullSimplify[vTerm + wTerm] 1 r

ⅇ-−2 ⅈ ϕ -− Csc[θ] vv[r, t, θ, ϕ] + ⅈ vv(0,0,0,1) [r, t, θ, ϕ] + Cos[θ] vv(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] vv(1,0,0,0) [r, t, θ, ϕ] + ⅇ4 ⅈ ϕ Csc[θ] -− ww[r, t, θ, ϕ] + ⅈ ww(0,0,0,1) [r, t, θ, ϕ] + Sin[θ] Cos[θ] ww(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ]


13

14


bracz = Simplify[c[[3, 1]] D[brac, r] + c[[3, 2]] D[brac, θ] + c[[3, 3]] D[brac, ϕ]] 1

ⅇ-−2 ⅈ ϕ 2 Sin[θ]2 vv(0,0,1,0) [r, t, θ, ϕ] +

r2

2 ⅇ4 ⅈ ϕ Sin[θ]2 ww(0,0,1,0) [r, t, θ, ϕ] + ⅈ vv(0,0,1,1) [r, t, θ, ϕ] -− ⅈ ⅇ4 ⅈ ϕ ww(0,0,1,1) [r, t, θ, ϕ] -− Cos[θ] Sin[θ] vv(0,0,2,0) [r, t, θ, ϕ] -− ⅇ4 ⅈ ϕ Cos[θ] Sin[θ] ww(0,0,2,0) [r, t, θ, ϕ] -− r Cot[θ] vv(1,0,0,0) [r, t, θ, ϕ] -− r Cos[θ] Sin[θ] vv(1,0,0,0) [r, t, θ, ϕ] -− ⅇ4 ⅈ ϕ r Cot[θ] ww(1,0,0,0) [r, t, θ, ϕ] -− ⅇ4 ⅈ ϕ r Cos[θ] Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ] -− ⅈ r Cot[θ] vv(1,0,0,1) [r, t, θ, ϕ] + ⅈ ⅇ4 ⅈ ϕ r Cot[θ] ww(1,0,0,1) [r, t, θ, ϕ] + r Cos[θ]2 vv(1,0,1,0) [r, t, θ, ϕ] -− r Sin[θ]2 vv(1,0,1,0) [r, t, θ, ϕ] + ⅇ4 ⅈ ϕ r Cos[θ]2 ww(1,0,1,0) [r, t, θ, ϕ] -− ⅇ4 ⅈ ϕ r Sin[θ]2 ww(1,0,1,0) [r, t, θ, ϕ] + r2 Cos[θ] Sin[θ] vv(2,0,0,0) [r, t, θ, ϕ] + ⅇ4 ⅈ ϕ r2 Cos[θ] Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ] RHS796 = FullSimplify[1 /∕ 2 (bracz)] 1 4 r2

ⅇ-−2 ⅈ ϕ 4 Sin[θ]2 vv(0,0,1,0) [r, t, θ, ϕ] + 2 ⅈ vv(0,0,1,1) [r, t, θ, ϕ] -− 2 Cos[θ] Sin[θ] vv(0,0,2,0) [r, t, θ, ϕ] -− 2 r Cot[θ] vv(1,0,0,0) [r, t, θ, ϕ] + r -− 2 ⅈ Cot[θ] vv(1,0,0,1) [r, t, θ, ϕ] + 2 Cos[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + Sin[2 θ] -− vv(1,0,0,0) [r, t, θ, ϕ] + r vv(2,0,0,0) [r, t, θ, ϕ] + ⅇ4 ⅈ ϕ 4 Sin[θ]2 ww(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅈ ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[2 θ] ww(0,0,2,0) [r, t, θ, ϕ] + r -− (2 Cot[θ] + Sin[2 θ]) ww(1,0,0,0) [r, t, θ, ϕ] + 2 ⅈ Cot[θ] ww(1,0,0,1) [r, t, θ, ϕ] + 2 Cos[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[2 θ] ww(2,0,0,0) [r, t, θ, ϕ]

(*⋆End RHS of (7.96)*⋆) (*⋆7.96 LHS begins*⋆) (*⋆ EQUATION (7.91) *⋆) 1 f3 = 8 q r2 (1 + Cos[θ]) 2 + q -− 4 q C[2] + (6 + q (3 -− 8 C[2])) Cos[θ] + (2 -− 4 q C[2]) Cos[2 θ] + 4 q r2 4 rn θ θ Cos  n Cos[θ] hhh[θ, ϕ] -− Sin  ww[r, t, θ, ϕ] -− Sin[θ] hhh(1,0) [θ, ϕ] -− 2 2 θ Csc  (1 + Cos[θ]) η0[t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,1) [t, r, θ, ϕ] + 2 Sin[θ] 2 Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ]

;

f3x = Simplify[c[[1, 1]] D[f3, r] + c[[1, 2]] D[f3, θ] + c[[1, 3]] D[f3, ϕ]] 1 3

8 r (1 + Cos[θ])2 θ -− 4 r2 (1 + Cos[θ]) Csc[θ] Sin[ϕ] 4 rn Cos  n Cos[θ] hhh(0,1) [θ, ϕ] -− 2 θ Sin[θ] hhh(1,1) [θ, ϕ] -− Sin  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 θ Csc  (1 + Cos[θ]) η0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,2) [t, r, θ, ϕ] + 2  + Printed by Wolfram Mathematica Student Edition


15

2 Sin[θ] Cos[θ] η0(0,0,1,1) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 1 q

Cos[θ] Cos[ϕ] Sin[θ] 2 + q -− 4 q C[2] + (6 + q (3 -− 8 C[2])) Cos[θ] + θ (2 -− 4 q C[2]) Cos[2 θ] + 4 q r2 4 rn Cos  2 θ n Cos[θ] hhh[θ, ϕ] -− Sin  ww[r, t, θ, ϕ] -− Sin[θ] hhh(1,0) [θ, ϕ] -− 2 θ Csc  (1 + Cos[θ]) η0[t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,1) [t, r, θ, ϕ] + 2 Sin[θ] 2 Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ]

+

(1 + Cos[θ]) (-− 6 + q (-− 3 + 8 C[2])) Sin[θ] + 4 (-− 1 + 2 q C[2]) Sin[2 θ] + θ θ 4 q r2 2 rn Sin  -− n Cos[θ] hhh[θ, ϕ] + Sin  ww[r, t, θ, ϕ] + Sin[θ] 2 2 θ θ hhh(1,0) [θ, ϕ] -− 2 rn Cos  2 n hhh[θ, ϕ] Sin[θ] + Cos  ww[r, t, 2 2 θ, ϕ] + 2 Cos[θ] hhh(1,0) [θ, ϕ] -− n Cos[θ] hhh(1,0) [θ, ϕ] + Sin[θ] θ hhh(2,0) [θ, ϕ] + Sin  ww(0,0,1,0) [r, t, θ, ϕ] 2 θ 2

+

1

θ Cot  Csc 2 2

 (1 + Cos[θ]) η0[t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,1) [t, r, θ, ϕ] + 2 Sin[θ] Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ] -−

θ Csc  -− Sin[θ] η0[t, r, θ, ϕ] + (1 + Cos[θ] + 2 Cos[2 θ]) 2 η0(0,0,1,0) [t, r, θ, ϕ] + 2 ⅈ η0(0,0,1,1) [t, r, θ, ϕ] + Sin[2 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 2 r Sin[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] + r η0(0,1,1,0) [t, r, θ, ϕ] -− r Cos[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] 1 q

(1 + Cos[θ]) Csc[ϕ] Sin[θ] Sin[2 ϕ] 2 + q -− 4 q C[2] + 6 Cos[θ] + 3 q Cos[θ] -− 8 q C[2] Cos[θ] + 2 Cos[2 θ] -− 4 q C[2] Cos[2 θ] -− θ 4 n2 q r2+n Cos[θ] Csc  hhh[θ, ϕ] Sin[θ] + 2 4 n q r2+n Sin[θ] ww[r, t, θ, ϕ] + θ 4 n q r2+n Csc  Sin[θ]2 hhh(1,0) [θ, ϕ] + 2 θ 2 q r3 Csc  η0(0,1,0,0) [t, r, θ, ϕ] + 2 θ 2 q r3 Cos[θ] Csc  η0(0,1,0,0) [t, r, θ, ϕ] + 2 θ 4 q r3 Csc  Sin[θ]2 η0(0,1,0,0) [t, r, θ, ϕ] + 2 θ 4 ⅈ q r3 Csc  η0(0,1,0,1) [t, r, θ, ϕ] + 2 + Printed by Wolfram Mathematica Student Edition

-−

16


θ 4 q r3 Cos[θ] Csc  Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 2 θ 4 q r4 Csc  Sin[θ]2 η0(0,2,0,0) [t, r, θ, ϕ] + 2 4 q r3+n Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ] f3xx = c[[1, 1]] D[f3x, r] + c[[1, 2]] D[f3x, θ] + c[[1, 3]] D[f3x, ϕ]

-−

Csc[θ] Sin[ϕ]  ⋯ 1 ⋯  8 r4 (1+Cos[θ])2

Cos[ϕ] Sin[θ] -− large output

Cos[θ] Cos[ϕ]

+

4 r3

⋯ 1 ⋯

3



+

⋯ 1 ⋯ ⋯ 1 ⋯

r

3  ⋯ 1 ⋯  8 r4 (1+Cos[θ])2

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Sin[θ]  ⋯ 1 ⋯

+

show more

⋯ 1 ⋯

8 r3 (1+Cos[θ])2

show all

+

 set size limit...

f3y = Simplify[c[[2, 1]] D[f3, r] + c[[2, 2]] D[f3, θ] + c[[2, 3]] D[f3, ϕ]] 1 8 r3 (1 + Cos[θ])2

θ 4 r2 (1 + Cos[θ]) Cos[ϕ] Csc[θ] 4 rn Cos  2

θ n Cos[θ] hhh(0,1) [θ, ϕ] -− Sin[θ] hhh(1,1) [θ, ϕ] -− Sin  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 θ Csc  (1 + Cos[θ]) η0(0,0,0,1) [t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,2) [t, r, θ, ϕ] + 2 2 Sin[θ] Cos[θ] η0(0,0,1,1) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 1 q

Cos[θ] Sin[ϕ] Sin[θ] 2 + q -− 4 q C[2] + (6 + q (3 -− 8 C[2])) Cos[θ] + θ (2 -− 4 q C[2]) Cos[2 θ] + 4 q r2 4 rn Cos  2 θ n Cos[θ] hhh[θ, ϕ] -− Sin  ww[r, t, θ, ϕ] -− Sin[θ] hhh(1,0) [θ, ϕ] -− 2 θ Csc  (1 + Cos[θ]) η0[t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,1) [t, r, θ, ϕ] + 2 Sin[θ] 2 Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ]

+

(1 + Cos[θ]) (-− 6 + q (-− 3 + 8 C[2])) Sin[θ] + 4 (-− 1 + 2 q C[2]) Sin[2 θ] + θ θ 4 q r2 2 rn Sin  -− n Cos[θ] hhh[θ, ϕ] + Sin  ww[r, t, θ, ϕ] + Sin[θ] 2 2 θ θ hhh(1,0) [θ, ϕ] -− 2 rn Cos  2 n hhh[θ, ϕ] Sin[θ] + Cos  ww[r, t, 2 2 θ, ϕ] + 2 Cos[θ] hhh(1,0) [θ, ϕ] -− n Cos[θ] hhh(1,0) [θ, ϕ] + Sin[θ] θ hhh(2,0) [θ, ϕ] + Sin  ww(0,0,1,0) [r, t, θ, ϕ] 2 θ 2

+

1

θ Cot  Csc 2 2

 (1 + Cos[θ]) η0[t, r, θ, ϕ] + 2 ⅈ η0(0,0,0,1) [t, r, θ, ϕ] + 2 Sin[θ] Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] + r Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ] -−



θ Csc  -− Sin[θ] η0[t, r, θ, ϕ] + (1 + Cos[θ] + 2 Cos[2 θ]) 2 η0(0,0,1,0) [t, r, θ, ϕ] + 2 ⅈ η0(0,0,1,1) [t, r, θ, ϕ] + Sin[2 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 2 r Sin[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] + r η0(0,1,1,0) [t, r, θ, ϕ] -− r Cos[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] 1 q

2 (1 + Cos[θ]) Sin[θ] Sin[ϕ] 2 + q -− 4 q C[2] + 6 Cos[θ] + 3 q Cos[θ] -− 8 q C[2] Cos[θ] + 2 Cos[2 θ] -− 4 q C[2] Cos[2 θ] -− θ 4 n2 q r2+n Cos[θ] Csc  hhh[θ, ϕ] Sin[θ] + 2 4 n q r2+n Sin[θ] ww[r, t, θ, ϕ] + θ 4 n q r2+n Csc  Sin[θ]2 hhh(1,0) [θ, ϕ] + 2 θ 2 q r3 Csc  η0(0,1,0,0) [t, r, θ, ϕ] + 2 θ 2 q r3 Cos[θ] Csc  η0(0,1,0,0) [t, r, θ, ϕ] + 2 θ 4 q r3 Csc  Sin[θ]2 η0(0,1,0,0) [t, r, θ, ϕ] + 2 θ 4 ⅈ q r3 Csc  η0(0,1,0,1) [t, r, θ, ϕ] + 2 θ 4 q r3 Cos[θ] Csc  Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 2 θ 4 q r4 Csc  Sin[θ]2 η0(0,2,0,0) [t, r, θ, ϕ] + 2 4 q r3+n Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ]

f3yy = c[[2, 1]] D[f3y, r] + c[[2, 2]] D[f3y, θ] + c[[2, 3]] D[f3y, ϕ]

Cos[ϕ] Csc[θ]  ⋯ 1 ⋯  4

8 r (1+Cos[θ])

2

Cos[θ] Sin[ϕ]

+

⋯ 1 ⋯ ⋯ 1 ⋯

+ ⋯ 1 ⋯

r

+ Sin[θ] Sin[ϕ]

-− 3 4 r2 (1 + Cos[θ]) Cos[ϕ] Csc[θ]  ⋯ 1 ⋯ ⋯ 1 ⋯

q

large output

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-−

⋯ 1 ⋯

q

  8 r4 (1 + Cos[θ])2  +

show more

show all

θ 2

-− Csc   ⋯ 1 ⋯  + ⋯ 1 ⋯

8 r3  ⋯ 1 ⋯ 

2

set size limit...

f3left = Simplify[f3xx + f3yy] -−

1 128 q r4 (1 + Cos[θ])3 θ 2 θ θ θ θ 3θ Cot  Csc  96 Sin  + 80 q Sin  + 16 n q r2+n Cos  + Cos  2 2 2 2 2 2 23 -− 20 n -− 4 n2  Cos[θ] + 2 9 -− 16 n + 4 n2  Cos[2 θ] + 3

 Printed by Wolfram Mathematica Student Edition

-−

-−

17

18


θ 3 (-− 1 + 2 n) (-− 2 -− 4 n + (-− 5 + 2 n) Cos[3 θ]) hhh[θ, ϕ] Sin  -− 2 3θ 3θ 3θ 5θ 5θ 87 Sin  -− 76 q Sin  + 54 q C[2] Sin  -− 31 Sin  + 8 q Sin + 2 2 2 2 2 5θ 7θ 7θ 7θ 54 q C[2] Sin  + 66 Sin  + 42 q Sin  -− 72 q C[2] Sin + 2 2 2 2 9θ 9θ 9θ 11 θ 34 Sin  -− 6 q Sin  -− 72 q C[2] Sin  -− 23 Sin  -− 2 2 2 2 11 θ 11 θ 13 θ 13 θ 12 q Sin  + 30 q C[2] Sin  -− 15 Sin  + 30 q C[2] Sin + 2 2 2 2 θ 32 q r2+n Cos  4 + 2 n + 3 n2 + (-− 2 + 4 n) Cos[θ] -− 4 -− 1 + n2  Cos[2 θ] + 2 θ 2 2 Cos[3 θ] -− 4 n Cos[3 θ] -− 2 n Cos[4 θ] + n2 Cos[4 θ] Sin  ww[r, t, θ, ϕ] + 2 38 q r2 η0[t, r, θ, ϕ] + 62 q r2 Cos[θ] η0[t, r, θ, ϕ] + 32 q r2 Cos[2 θ] η0[t, r, θ, ϕ] + 5 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[4 θ] η0[t, r, θ, ϕ] -− 3 q r2 Cos[5 θ] η0[t, r, θ, ϕ] -− 64 n q r2+n Sin[2 θ] hhh(0,2) [θ, ϕ] -− 2 q r2+n hhh(1,0) [θ, ϕ] + 48 n q r2+n hhh(1,0) [θ, ϕ] + 24 n2 q r2+n hhh(1,0) [θ, ϕ] + 5 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 24 n q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 52 n2 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 8 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 16 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 48 n q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] + 40 n2 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 16 n q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 5 q r2+n Cos[6 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Cos[6 θ] hhh(1,0) [θ, ϕ] -− 12 n2 q r2+n Cos[6 θ] hhh(1,0) [θ, ϕ] + 64 q r2+n hhh(1,2) [θ, ϕ] -− 64 q r2+n Cos[2 θ] hhh(1,2) [θ, ϕ] + 16 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 28 n q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 8 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] + 16 q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 32 n q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 8 q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[6 θ] hhh(2,0) [θ, ϕ] + 12 n q r2+n Sin[6 θ] hhh(2,0) [θ, ϕ] + 8 q r2+n hhh(3,0) [θ, ϕ] -− 4 q r2+n Cos[2 θ] hhh(3,0) [θ, ϕ] -− 8 q r2+n Cos[4 θ] hhh(3,0) [θ, ϕ] + 4 q r2+n Cos[6 θ] hhh(3,0) [θ, ϕ] + 172 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] -− 160 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] + 160 ⅈ q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 32 ⅈ q r2 Cos[3 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− θ 12 ⅈ q r2 Cos[4 θ] η0(0,0,0,1) [t, r, θ, ϕ] + 64 q r2+n Cos  ww(0,0,0,2) [r, t, θ, ϕ] -− 2 3 θ 64 q r2+n Cos  ww(0,0,0,2) [r, t, θ, ϕ] + 64 q r2 η0(0,0,0,2) [t, r, θ, ϕ] + 2 64 q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] + 128 ⅈ q r2 η0(0,0,0,3) [t, r, θ, ϕ] + θ θ 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 3 θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 5θ 5θ 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 24 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 7θ 24 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 -−


-−


24 n q r2+n Sin 8 n q r2+n Sin

7θ

2 9θ

 ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin

9θ

2 11 θ

19

 ww(0,0,1,0) [r, t, θ, ϕ] -−

 ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 11 θ 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 56 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] + q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 44 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 8 q r2 Sin[4 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 12 q r2 Sin[5 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 5 q r2 Sin[6 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 32 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 96 ⅈ q r2 Sin[2 θ] η0(0,0,1,1) [t, r, θ, ϕ] + 32 ⅈ q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[4 θ] η0(0,0,1,1) [t, r, θ, ϕ] + θ 64 q r2 Sin[2 θ] η0(0,0,1,2) [t, r, θ, ϕ] + 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 3 θ 5θ 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 2 7θ 9θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 11 θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 16 q r2 η0(0,0,2,0) [t, r, θ, ϕ] + 2 24 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 12 q r2 Cos[2 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 16 q r2 Cos[4 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[6 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 16 ⅈ q r2 η0(0,0,2,1) [t, r, θ, ϕ] -− 16 ⅈ q r2 Cos[4 θ] η0(0,0,2,1) [t, r, θ, ϕ] + 12 q r2 Sin[2 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 4 q r2 Sin[6 θ] η0(0,0,3,0) [t, r, θ, ϕ] + 86 q r3 η0(0,1,0,0) [t, r, θ, ϕ] + 16 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 99 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 16 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 10 q r3 Cos[4 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 3 q r3 Cos[6 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 32 ⅈ q r3 η0(0,1,0,1) [t, r, θ, ϕ] + 32 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 64 ⅈ q r3 Cos[2 θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 32 ⅈ q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 32 ⅈ q r3 Cos[4 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 64 q r3 η0(0,1,0,2) [t, r, θ, ϕ] -− 64 q r3 Cos[2 θ] η0(0,1,0,2) [t, r, θ, ϕ] + 48 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 68 q r3 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 24 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 16 q r3 Sin[4 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 24 q r3 Sin[5 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 12 q r3 Sin[6 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 64 ⅈ q r3 Sin[2 θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 32 ⅈ q r3 Sin[4 θ] η0(0,1,1,1) [t, r, θ, ϕ] + 24 q r3 η0(0,1,2,0) [t, r, θ, ϕ] -− 12 q r3 Cos[2 θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 24 q r3 Cos[4 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 12 q r3 Cos[6 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 168 q r4 η0(0,2,0,0) [t, r, θ, ϕ] + 24 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 224 q r4 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 36 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 56 q r4 Cos[4 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Cos[5 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 48 ⅈ q r4 η0(0,2,0,1) [t, r, θ, ϕ] -− 64 ⅈ q r4 Cos[2 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 16 ⅈ q r4 Cos[4 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 60 q r4 Sin[2 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 48 q r4 Sin[4 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 12 q r4 Sin[6 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 40 q r5 η0(0,3,0,0) [t, r, θ, ϕ] -− 60 q r5 Cos[2 θ] η0(0,3,0,0) [t, r, θ, ϕ] + 24 q r5 Cos[4 θ] η0(0,3,0,0) [t, r, θ, ϕ] -− θ 4 q r5 Cos[6 θ] η0(0,3,0,0) [t, r, θ, ϕ] + 56 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 θ 3θ 80 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 24 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 -−


20


80 n q r3+n Cos

3θ 2

 ww(1,0,0,0) [r, t, θ, ϕ] -− 44 q r3+n Cos

ww(1,0,0,0) [r, t, θ, ϕ] -− 40 n q r3+n Cos 4 q r3+n Cos

7θ

12 q r3+n Cos

5θ 2

5θ



2

 ww(1,0,0,0) [r, t, θ, ϕ] -−

 ww(1,0,0,0) [r, t, θ, ϕ] + 40 n q r3+n Cos

2 9θ

7θ 2

 ww(1,0,0,0) [r, t, θ, ϕ] +

 ww(1,0,0,0) [r, t, θ, ϕ] + 2 9θ 11 θ 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 11 θ 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 θ 3θ 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 7θ 24 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 24 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 2 9θ 11 θ 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 θ 3θ 40 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 40 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 2 5θ 7θ 20 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 20 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 2 2 9θ 11 θ 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] 2 2 (*⋆End LHS of (7.96)*⋆) Eqn796 = Simplify[f3left -− RHS796] -−

1 128 q r4 (1 + Cos[θ])3 θ 2 θ θ θ θ 3θ Cot  Csc  96 Sin  + 80 q Sin  + 16 n q r2+n Cos  + Cos  2 2 2 2 2 2 23 -− 20 n -− 4 n2  Cos[θ] + 2 9 -− 16 n + 4 n2  Cos[2 θ] + θ 3 3θ (-− 1 + 2 n) (-− 2 -− 4 n + (-− 5 + 2 n) Cos[3 θ]) hhh[θ, ϕ] Sin  -− 87 Sin  -− 2 2 3θ 3θ 5θ 5θ 76 q Sin  + 54 q C[2] Sin  -− 31 Sin  + 8 q Sin + 2 2 2 2 5θ 7θ 7θ 7θ 54 q C[2] Sin  + 66 Sin  + 42 q Sin  -− 72 q C[2] Sin + 2 2 2 2 9θ 9θ 9θ 11 θ 34 Sin  -− 6 q Sin  -− 72 q C[2] Sin  -− 23 Sin  -− 2 2 2 2 11 θ 11 θ 13 θ 13 θ 12 q Sin  + 30 q C[2] Sin  -− 15 Sin  + 30 q C[2] Sin + 2 2 2 2 θ 32 q r2+n Cos  4 + 2 n + 3 n2 + (-− 2 + 4 n) Cos[θ] -− 4 -− 1 + n2  Cos[2 θ] + 2 θ 2 2 Cos[3 θ] -− 4 n Cos[3 θ] -− 2 n Cos[4 θ] + n2 Cos[4 θ] Sin  ww[r, t, θ, ϕ] + 2 38 q r2 η0[t, r, θ, ϕ] + 62 q r2 Cos[θ] η0[t, r, θ, ϕ] + 32 q r2 Cos[2 θ] η0[t, r, θ, ϕ] + 5 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[4 θ] η0[t, r, θ, ϕ] -− -− -− + + Printed by Wolfram Mathematica Student Edition


21

η0[t, r, θ, ϕ] + 5 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[4 θ] η0[t, r, θ, ϕ] -− 3 q r2 Cos[5 θ] η0[t, r, θ, ϕ] -− 64 n q r2+n Sin[2 θ] hhh(0,2) [θ, ϕ] -− 2 q r2+n hhh(1,0) [θ, ϕ] + 48 n q r2+n hhh(1,0) [θ, ϕ] + 24 n2 q r2+n hhh(1,0) [θ, ϕ] + 5 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 24 n q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 52 n2 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 8 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 16 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 48 n q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] + 40 n2 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 16 n q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 5 q r2+n Cos[6 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Cos[6 θ] hhh(1,0) [θ, ϕ] -− 12 n2 q r2+n Cos[6 θ] hhh(1,0) [θ, ϕ] + 64 q r2+n hhh(1,2) [θ, ϕ] -− 64 q r2+n Cos[2 θ] hhh(1,2) [θ, ϕ] + 16 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 28 n q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 8 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] + 16 q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 32 n q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 8 q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[6 θ] hhh(2,0) [θ, ϕ] + 12 n q r2+n Sin[6 θ] hhh(2,0) [θ, ϕ] + 8 q r2+n hhh(3,0) [θ, ϕ] -− 4 q r2+n Cos[2 θ] hhh(3,0) [θ, ϕ] -− 8 q r2+n Cos[4 θ] hhh(3,0) [θ, ϕ] + 4 q r2+n Cos[6 θ] hhh(3,0) [θ, ϕ] + 172 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] -− 160 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] + 160 ⅈ q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 32 ⅈ q r2 Cos[3 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 12 ⅈ q r2 Cos[4 θ] η0(0,0,0,1) [t, r, θ, ϕ] + θ 3θ 64 q r2+n Cos  ww(0,0,0,2) [r, t, θ, ϕ] -− 64 q r2+n Cos  ww(0,0,0,2) [r, t, θ, ϕ] + 2 2 64 q r2 η0(0,0,0,2) [t, r, θ, ϕ] + 64 q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] + θ 128 ⅈ q r2 η0(0,0,0,3) [t, r, θ, ϕ] + 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 θ 3θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 16 n q r2+n Sin  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 5θ 7θ 24 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 24 q r2+n Sin  2 2 7θ ww(0,0,1,0) [r, t, θ, ϕ] -− 24 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 9θ 9θ 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 2 11 θ 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 11 θ 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 56 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] + 2 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 44 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 8 q r2 Sin[4 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 12 q r2 Sin[5 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 5 q r2 Sin[6 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 32 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 96 ⅈ q r2 Sin[2 θ] η0(0,0,1,1) [t, r, θ, ϕ] + 32 ⅈ q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[4 θ] η0(0,0,1,1) [t, r, θ, ϕ] + 64 q r2 Sin[2 θ] η0(0,0,1,2) [t, r, θ, ϕ] + θ 3θ 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 5θ 7θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 2 9θ 11 θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 2 + + -− + Printed by Wolfram Mathematica Student Edition

22


16 q r2 η0(0,0,2,0) [t, r, θ, ϕ] + 24 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 12 q r2 Cos[2 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 16 q r2 Cos[4 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[6 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 16 ⅈ q r2 η0(0,0,2,1) [t, r, θ, ϕ] -− 16 ⅈ q r2 Cos[4 θ] η0(0,0,2,1) [t, r, θ, ϕ] + 12 q r2 Sin[2 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 4 q r2 Sin[6 θ] η0(0,0,3,0) [t, r, θ, ϕ] + 86 q r3 η0(0,1,0,0) [t, r, θ, ϕ] + 16 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 99 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 16 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 10 q r3 Cos[4 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 3 q r3 Cos[6 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 32 ⅈ q r3 η0(0,1,0,1) [t, r, θ, ϕ] + 32 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 64 ⅈ q r3 Cos[2 θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 32 ⅈ q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 32 ⅈ q r3 Cos[4 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 64 q r3 η0(0,1,0,2) [t, r, θ, ϕ] -− 64 q r3 Cos[2 θ] η0(0,1,0,2) [t, r, θ, ϕ] + 48 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 68 q r3 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 24 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 16 q r3 Sin[4 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 24 q r3 Sin[5 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 12 q r3 Sin[6 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 64 ⅈ q r3 Sin[2 θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 32 ⅈ q r3 Sin[4 θ] η0(0,1,1,1) [t, r, θ, ϕ] + 24 q r3 η0(0,1,2,0) [t, r, θ, ϕ] -− 12 q r3 Cos[2 θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 24 q r3 Cos[4 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 12 q r3 Cos[6 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 168 q r4 η0(0,2,0,0) [t, r, θ, ϕ] + 24 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 224 q r4 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 36 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 56 q r4 Cos[4 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Cos[5 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 48 ⅈ q r4 η0(0,2,0,1) [t, r, θ, ϕ] -− 64 ⅈ q r4 Cos[2 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 16 ⅈ q r4 Cos[4 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 60 q r4 Sin[2 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 48 q r4 Sin[4 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 12 q r4 Sin[6 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 40 q r5 η0(0,3,0,0) [t, r, θ, ϕ] -− 60 q r5 Cos[2 θ] η0(0,3,0,0) [t, r, θ, ϕ] + 24 q r5 Cos[4 θ] η0(0,3,0,0) [t, r, θ, ϕ] -− 4 q r5 Cos[6 θ] η0(0,3,0,0) [t, r, θ, ϕ] + θ θ 56 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 80 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 3 θ 24 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 3θ 5θ 80 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 44 q r3+n Cos  2 2 5θ ww(1,0,0,0) [r, t, θ, ϕ] -− 40 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 7θ 7θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 40 n q r3+n Cos  2 2 9θ ww(1,0,0,0) [r, t, θ, ϕ] + 12 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 9θ 11 θ 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 4 q r3+n Cos  2 2 11 θ ww(1,0,0,0) [r, t, θ, ϕ] -− 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 θ 3θ 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 24 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 7θ 24 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 +


+


8 q r3+n Sin

9θ 2 θ

 ww(1,0,1,0) [r, t, θ, ϕ] + 8 q r3+n Sin

11 θ 2 3θ

23

 ww(1,0,1,0) [r, t, θ, ϕ] +

40 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 40 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 2 5θ 20 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 2 7 θ 9θ 20 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 2 11 θ 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 1 4 r2

ⅇ-−2 ⅈ ϕ 4 Sin[θ]2 vv(0,0,1,0) [r, t, θ, ϕ] + 2 ⅈ vv(0,0,1,1) [r, t, θ, ϕ] -− 2 Cos[θ] Sin[θ] vv(0,0,2,0) [r, t, θ, ϕ] -− 2 r Cot[θ] vv(1,0,0,0) [r, t, θ, ϕ] + r -− 2 ⅈ Cot[θ] vv(1,0,0,1) [r, t, θ, ϕ] + 2 Cos[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + Sin[2 θ] -− vv(1,0,0,0) [r, t, θ, ϕ] + r vv(2,0,0,0) [r, t, θ, ϕ] + ⅇ4 ⅈ ϕ 4 Sin[θ]2 ww(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅈ ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[2 θ] ww(0,0,2,0) [r, t, θ, ϕ] + r -− (2 Cot[θ] + Sin[2 θ]) ww(1,0,0,0) [r, t, θ, ϕ] + 2 ⅈ Cot[θ] ww(1,0,0,1) [r, t, θ, ϕ] + 2 Cos[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[2 θ] ww(2,0,0,0) [r, t, θ, ϕ]

(*⋆eqn 7.97*⋆) vxx = Simplify[c[[1, 1]] D[vx, r] + c[[1, 2]] D[vx, θ] + c[[1, 3]] D[vx, ϕ]] 1 r2

ⅇ-−ⅈ ϕ 1 2

Csc[θ]2 (-− 1 + Cos[2 ϕ] -− 2 ⅈ Sin[2 ϕ]) vv[r, t, θ, ϕ] + 2 Csc[θ]2 (Cos[ϕ] -− ⅈ Sin[ϕ])

Sin[ϕ] vv(0,0,0,1) [r, t, θ, ϕ] + Csc[θ]2 Sin[ϕ]2 vv(0,0,0,2) [r, t, θ, ϕ] -− 2 Cos[θ] Cos[ϕ]2 Sin[θ] vv(0,0,1,0) [r, t, θ, ϕ] + 2 ⅈ Cos[ϕ] Cot[θ] Sin[ϕ] vv(0,0,1,0) [r, t, θ, ϕ] + Cot[θ] Sin[ϕ]2 vv(0,0,1,0) [r, t, θ, ϕ] -− 2 Cos[ϕ] Cot[θ] Sin[ϕ] vv(0,0,1,1) [r, t, θ, ϕ] + Cos[θ]2 Cos[ϕ]2 vv(0,0,2,0) [r, t, θ, ϕ] + r Cos[θ]2 Cos[ϕ]2 vv(1,0,0,0) [r, t, θ, ϕ] + 2 ⅈ r Cos[ϕ] Sin[ϕ] vv(1,0,0,0) [r, t, θ, ϕ] + r Sin[ϕ]2 vv(1,0,0,0) [r, t, θ, ϕ] -− 2 r Cos[ϕ] Sin[ϕ] vv(1,0,0,1) [r, t, θ, ϕ] + r Cos[ϕ]2 Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + r2 Cos[ϕ]2 Sin[θ]2 vv(2,0,0,0) [r, t, θ, ϕ] vy = FullSimplifyc[[2, 1]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], r + c[[2, 2]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], θ + c[[2, 3]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], ϕ 1 r

ⅇ-−ⅈ ϕ Cos[ϕ] Csc[θ] -− ⅈ vv[r, t, θ, ϕ] + vv(0,0,0,1) [r, t, θ, ϕ] + Cos[θ] Sin[ϕ] vv(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] Sin[ϕ] vv(1,0,0,0) [r, t, θ, ϕ]


24


vyy = Simplify[c[[2, 1]] D[vy, r] + c[[2, 2]] D[vy, θ] + c[[2, 3]] D[vy, ϕ]] 1

ⅇ-−ⅈ ϕ -−

r2

1 2

Csc[θ]2 (1 + Cos[2 ϕ] -− 2 ⅈ Sin[2 ϕ]) vv[r, t, θ, ϕ] +

Cos[ϕ] Csc[θ]2 (-− 2 ⅈ Cos[ϕ] -− 2 Sin[ϕ]) vv(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ]2 Csc[θ]2 vv(0,0,0,2) [r, t, θ, ϕ] + Cos[ϕ]2 Cot[θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅈ Cos[ϕ] Cot[θ] Sin[ϕ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 Cos[θ] Sin[θ] Sin[ϕ]2 vv(0,0,1,0) [r, t, θ, ϕ] + Cot[θ] Sin[2 ϕ] vv(0,0,1,1) [r, t, θ, ϕ] + Cos[θ]2 Sin[ϕ]2 vv(0,0,2,0) [r, t, θ, ϕ] + r Cos[ϕ]2 vv(1,0,0,0) [r, t, θ, ϕ] -− 2 ⅈ r Cos[ϕ] Sin[ϕ] vv(1,0,0,0) [r, t, θ, ϕ] + r Cos[θ]2 Sin[ϕ]2 vv(1,0,0,0) [r, t, θ, ϕ] + r Sin[2 ϕ] vv(1,0,0,1) [r, t, θ, ϕ] + r Sin[2 θ] Sin[ϕ]2 vv(1,0,1,0) [r, t, θ, ϕ] + r2 Sin[θ]2 Sin[ϕ]2 vv(2,0,0,0) [r, t, θ, ϕ] wxx = Simplify[c[[1, 1]] D[wx, r] + c[[1, 2]] D[wx, θ] + c[[1, 3]] D[wx, ϕ]] 1 2 r2

ⅇⅈ ϕ Csc[θ]2 (-− 1 + Cos[2 ϕ] + 2 ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] + 2 2 ⅈ Sin[ϕ]2 + Sin[2 ϕ] ww(0,0,0,1) [r, t, θ, ϕ] + Sin[ϕ]2 ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] -− 2 Cos[ϕ]2 Sin[θ]2 -− 2 ⅈ Cos[ϕ] Sin[ϕ] + Sin[ϕ]2  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 Cos[θ] Cos[ϕ] Sin[ϕ] ww(0,0,1,1) [r, t, θ, ϕ] + Sin[θ] Cos[θ]2 Cos[ϕ]2 ww(0,0,2,0) [r, t, θ, ϕ] + r Cos[θ]2 Cos[ϕ]2 + Sin[ϕ] (-− 2 ⅈ Cos[ϕ] + Sin[ϕ]) ww(1,0,0,0) [r, t, θ, ϕ] + Cos[ϕ] -− 2 Sin[ ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] 2 Cos[θ] ww(1,0,1,0) [ r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ]

wy = FullSimplifyc[[2, 1]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], r + c[[2, 2]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], θ + c[[2, 3]] Dⅇⅈ ϕ ww[r, t, θ, ϕ], ϕ 1 r

ⅇⅈ ϕ Csc[θ] Cos[ϕ] ⅈ ww[r, t, θ, ϕ] + ww(0,0,0,1) [r, t, θ, ϕ] + Sin[θ] Sin[ϕ] Cos[θ] ww(0,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ]

wyy = Simplify[c[[2, 1]] D[wy, r] + c[[2, 2]] D[wy, θ] + c[[2, 3]] D[wy, ϕ]] 1 2

ⅇⅈ ϕ Csc[θ]2

2r -− (1 + Cos[2 ϕ] + 2 ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] + 2 2 ⅈ Cos[ϕ] (Cos[ϕ] + ⅈ Sin[ϕ]) ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ]2 ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] Cos[ϕ]2 + 2 ⅈ Cos[ϕ] Sin[ϕ] -− 2 Sin[θ]2 Sin[ϕ]2  ww(0,0,1,0) [ r, t, θ, ϕ] + Cos[θ] Sin[2 ϕ] ww(0,0,1,1) [r, t, θ, ϕ] + Sin[θ] Cos[θ]2 Sin[ϕ]2 ww(0,0,2,0) [r, t, θ, ϕ] + r Cos[ϕ]2 + 2 ⅈ Cos[ϕ] Sin[ϕ] + Cos[θ]2 Sin[ϕ]2  ww(1,0,0,0) [r, t, θ, ϕ] + Sin[2 ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Sin[θ] Sin[ϕ]2 2 Cos[θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ]



trm1 = Simplify[1 /∕ 2 (vxx + vyy -− wxx -− wyy)] -−

1 16 r2

ⅇ-−ⅈ ϕ Csc[θ]2

8 vv[r, t, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ ww[r, t, θ, ϕ] + 16 ⅈ vv(0,0,0,1) [r, t, θ, ϕ] + 16 ⅈ ⅇ2 ⅈ ϕ ww(0,0,0,1) [r, t, θ, ϕ] -− 8 vv(0,0,0,2) [r, t, θ, ϕ] + 8 ⅇ2 ⅈ ϕ ww(0,0,0,2) [r, t, θ, ϕ] -− 2 Sin[4 θ] vv(0,0,1,0) [r, t, θ, ϕ] + 2 ⅇ2 ⅈ ϕ Sin[4 θ] ww(0,0,1,0) [r, t, θ, ϕ] -− vv(0,0,2,0) [r, t, θ, ϕ] + Cos[4 θ] vv(0,0,2,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ ww(0,0,2,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ Cos[4 θ] ww(0,0,2,0) [r, t, θ, ϕ] -− 5 r vv(1,0,0,0) [r, t, θ, ϕ] + 4 r Cos[2 θ] vv(1,0,0,0) [r, t, θ, ϕ] + r Cos[4 θ] vv(1,0,0,0) [r, t, θ, ϕ] + 5 ⅇ2 ⅈ ϕ r ww(1,0,0,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r Cos[2 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ r Cos[4 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 4 r Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 2 r Sin[4 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r Sin[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] -− 2 ⅇ2 ⅈ ϕ r Sin[4 θ] ww(1,0,1,0) [r, t, θ, ϕ] -− 3 r2 vv(2,0,0,0) [r, t, θ, ϕ] + 4 r2 Cos[2 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− r2 Cos[4 θ] vv(2,0,0,0) [r, t, θ, ϕ] + 3 ⅇ2 ⅈ ϕ r2 ww(2,0,0,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r2 Cos[2 θ] ww(2,0,0,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ r2 Cos[4 θ] ww(2,0,0,0) [r, t, θ, ϕ] vz = FullSimplifyc[[3, 1]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], r + c[[3, 2]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], θ + c[[3, 3]] Dⅇ-−ⅈ ϕ vv[r, t, θ, ϕ], ϕ 1 r

ⅇ-−ⅈ ϕ -− Sin[θ] vv(0,0,1,0) [r, t, θ, ϕ] + r Cos[θ] vv(1,0,0,0) [r, t, θ, ϕ]

vzz = FullSimplify[c[[3, 1]] D[vz, r] + c[[3, 2]] D[vz, θ] + c[[3, 3]] D[vz, ϕ]] 1 r2

ⅇ-−ⅈ ϕ Sin[2 θ] vv(0,0,1,0) [r, t, θ, ϕ] + Sin[θ] Sin[θ] vv(0,0,2,0) [r, t, θ, ϕ] + r vv(1,0,0,0) [r, t, θ, ϕ] -− 2 r Cos[θ] vv(1,0,1,0) [r, t, θ, ϕ] + r2 Cos[θ]2 vv(2,0,0,0) [r, t, θ, ϕ]

trm2 = Simplify[trm1 + vzz] -−

1 16 r2

ⅇ-−ⅈ ϕ Csc[θ]2

8 vv[r, t, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ ww[r, t, θ, ϕ] + 16 ⅈ vv(0,0,0,1) [r, t, θ, ϕ] + 16 ⅈ ⅇ2 ⅈ ϕ ww(0,0,0,1) [r, t, θ, ϕ] -− 8 vv(0,0,0,2) [r, t, θ, ϕ] + 8 ⅇ2 ⅈ ϕ ww(0,0,0,2) [r, t, θ, ϕ] -− 8 Sin[2 θ] vv(0,0,1,0) [r, t, θ, ϕ] + 2 Sin[4 θ] vv(0,0,1,0) [r, t, θ, ϕ] + 2 ⅇ2 ⅈ ϕ Sin[4 θ] ww(0,0,1,0) [r, t, θ, ϕ] -− 7 vv(0,0,2,0) [r, t, θ, ϕ] + 8 Cos[2 θ] vv(0,0,2,0) [r, t, θ, ϕ] -− Cos[4 θ] vv(0,0,2,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ ww(0,0,2,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ Cos[4 θ] ww(0,0,2,0) [r, t, θ, ϕ] -− 11 r vv(1,0,0,0) [r, t, θ, ϕ] + 12 r Cos[2 θ] vv(1,0,0,0) [r, t, θ, ϕ] -− r Cos[4 θ] vv(1,0,0,0) [r, t, θ, ϕ] + 5 ⅇ2 ⅈ ϕ r ww(1,0,0,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r Cos[2 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ r Cos[4 θ] ww(1,0,0,0) [r, t, θ, ϕ] + 4 r Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] -− 2 r Sin[4 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r Sin[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] -− 2 ⅇ2 ⅈ ϕ r Sin[4 θ] ww(1,0,1,0) [r, t, θ, ϕ] -− 5 r2 vv(2,0,0,0) [r, t, θ, ϕ] + 4 r2 Cos[2 θ] vv(2,0,0,0) [r, t, θ, ϕ] + r2 Cos[4 θ] vv(2,0,0,0) [r, t, θ, ϕ] + 3 ⅇ2 ⅈ ϕ r2 ww(2,0,0,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r2 Cos[2 θ] ww(2,0,0,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ r2 Cos[4 θ] ww(2,0,0,0) [r, t, θ, ϕ]


25

26


wiyx = Simplify[c[[1, 1]] D[wiy, r] + c[[1, 2]] D[wiy, θ] + c[[1, 3]] D[wiy, ϕ]] 1 r2

ⅇⅈ ϕ Csc[θ]2

1 2

(2 Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] -−

ⅈ (Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[ϕ] ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] -− ⅈ Cos[ϕ]2 + Cos[ϕ] Sin[ϕ] + ⅈ Sin[ϕ]2 + Sin[θ]2 Sin[2 ϕ] ww(0,0,1,0) [r, t, θ, ϕ] -− Cos[θ] Cos[2 ϕ] ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[θ] Cos[θ]2 Cos[ϕ] Sin[ϕ] ww(0,0,2,0) [r, t, θ, ϕ] + r ⅈ Cos[ϕ]2 + ⅈ Cos[ϕ] Sin[θ]2 Sin[ϕ] -− Sin[ϕ]2  ww(1,0,0,0) [r, t, θ, ϕ] + Cos[2 ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] Sin[ϕ] 2 Cos[θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ] RHS797 = Simplify[trm2 -− wiyx] 1 16 r2 ⅇ-−ⅈ ϕ Csc[θ]2 -− 8 vv[r, t, θ, ϕ] + 8 ⅇ2 ⅈ ϕ ww[r, t, θ, ϕ] -− 16 ⅈ vv(0,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ ⅇ2 ⅈ ϕ ww(0,0,0,1) [r, t, θ, ϕ] + 8 vv(0,0,0,2) [r, t, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ ww(0,0,0,2) [r, t, θ, ϕ] + 8 Sin[2 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 Sin[4 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅇ2 ⅈ ϕ Sin[4 θ] ww(0,0,1,0) [r, t, θ, ϕ] + 7 vv(0,0,2,0) [r, t, θ, ϕ] -− 8 Cos[2 θ] vv(0,0,2,0) [r, t, θ, ϕ] + Cos[4 θ] vv(0,0,2,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ ww(0,0,2,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ Cos[4 θ] ww(0,0,2,0) [r, t, θ, ϕ] + 11 r vv(1,0,0,0) [r, t, θ, ϕ] -− 12 r Cos[2 θ] vv(1,0,0,0) [r, t, θ, ϕ] + r Cos[4 θ] vv(1,0,0,0) [r, t, θ, ϕ] -− 5 ⅇ2 ⅈ ϕ r ww(1,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r Cos[2 θ] ww(1,0,0,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ r Cos[4 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 4 r Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 2 r Sin[4 θ] vv(1,0,1,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r Sin[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 2 ⅇ2 ⅈ ϕ r Sin[4 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 5 r2 vv(2,0,0,0) [r, t, θ, ϕ] -− 4 r2 Cos[2 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− r2 Cos[4 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− 3 ⅇ2 ⅈ ϕ r2 ww(2,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r2 Cos[2 θ] ww(2,0,0,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ r2 Cos[4 θ] ww(2,0,0,0) [r, t, θ, ϕ] -− 1 16 ⅇ2 ⅈ ϕ (2 Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] -− 2 ⅈ (Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[ϕ] ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] -− ⅈ Cos[ϕ]2 + Cos[ϕ] Sin[ϕ] + ⅈ Sin[ϕ]2 + Sin[θ]2 Sin[2 ϕ] ww(0,0,1,0) [r, t, θ, ϕ] -− Cos[θ] Cos[2 ϕ] ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[θ] Cos[θ]2 Cos[ϕ] Sin[ϕ] ww(0,0,2,0) [r, t, θ, ϕ] + r ⅈ Cos[ϕ]2 + ⅈ Cos[ϕ] Sin[θ]2 Sin[ϕ] -− Sin[ϕ]2  ww(1,0,0,0) [r, t, θ, ϕ] + Cos[2 ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] Sin[ϕ] 2 Cos[θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ]



27

f3z = Simplify[c[[3, 1]] D[f3, r] + c[[3, 2]] D[f3, θ] + c[[3, 3]] D[f3, ϕ]] -−

1 8 q r3 (1 + Cos[θ])2 θ θ θ θ θ Cos  Cot  -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 4 n q r2+n 2 2 2 2 2 3θ 3θ (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− Sin  + q Sin + 2 2 3θ 5θ 5θ 5θ 7θ 2 q C[2] Sin  + 5 Sin  + 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 2 2 2 2 2 7θ θ 6 q C[2] Sin  + 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 2 2 2 2 6 q r η0[t, r, θ, ϕ] + 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + θ 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 3 θ 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 3θ 5θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] 2 2

f3zx = c[[1, 1]] D[f3z, r] + c[[1, 2]] D[f3z, θ] + c[[1, 3]] D[f3z, ϕ] 1

θ θ Cos  Cot  Csc[θ] Sin[ϕ] 2 2 8 q r4 (1 + Cos[θ])2 -− 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) Sin[θ] hhh(0,1) [θ, ϕ] -− 4 q r2+n hhh(1,1) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,1) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,1) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,1) [θ, ϕ] + θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww(0,0,0,1) [r, t, θ, ϕ] + 2 6 q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 8 q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,2) [t, r, θ, ϕ] + θ 24 ⅈ q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 2 3 θ 4 q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 2 + Printed by Wolfram Mathematica Student Edition

-−

28


4 q r2+n Sin

5θ

 ww(0,0,1,1) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,2) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,1) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,1) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,2) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,1) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,1) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,1) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] -− 2 5θ 1 4 q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] + Cos[θ] Cos[ϕ] 2 r 1 θ θ θ θ Cos  -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 3 2 2 2 2 2 16 q r (1 + Cos[θ]) 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− Sin q Sin

3θ 2 7θ

 + 2 q C[2] Sin

3θ 2 7θ

 + 5 Sin

5θ 2

 + 3 q Sin

5θ 2

 -− 6 q C[2] Sin

3θ 2 5θ

+ +

2

θ  + 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, 2 2 2 t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 8 q r2 Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q θ 3θ r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 1 θ θ θ θ θ Cot  Csc  -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 2 2 2 2 2 16 q r3 (1 + Cos[θ])2 3 Sin

 -− 6 q C[2] Sin

4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− 3θ 3θ 3θ 5θ Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 +


+


29

θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q θ 3θ r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 1 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 4 q r3 (1 + Cos[θ])3 θ θ θ θ θ Cos  Cot  Sin[θ] -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 2 2 2 2 2 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− 3θ 3θ 3θ 5θ Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q θ 3θ r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 -− Printed by Wolfram Mathematica Student Edition

30


4 q r3+n Cos

5θ 2

 ww(1,0,0,0) [r, t, θ, ϕ] -−

1

θ θ 3 θ θ θ Cos  Cot  -− Cos  -− q Cos  + q C[2] Cos  -− 2 2 2 2 2 2 8 q r3 (1 + Cos[θ])2 3

Cos

2 15

3θ

+

2

q Cos

3 2

5θ

q Cos

3θ 2

 + 3 q C[2] Cos

 -− 15 q C[2] Cos

5θ

+

21

3θ

+

2

Cos

25 2

7θ

Cos

5θ

+

2

 -− 21 q C[2] Cos

7θ

 -− 2 2 2 2 2 2 4 n q r2+n Cos[θ] (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] -− 4 n q r2+n hhh[θ, ϕ] Sin[θ] (2 Sin[θ] -− 2 (-− 1 + 2 n) Sin[2 θ]) + θ 16 q r2+n Cos[θ] (-− 1 + n Cos[θ]) Sin  ww[r, t, θ, ϕ] + 2 θ 8 q r2+n Cos  (-− 1 + n Cos[θ]) Sin[θ] ww[r, t, θ, ϕ] -− 2 θ 16 n q r2+n Sin  Sin[θ]2 ww[r, t, θ, ϕ] -− 2 8 q r2 Sin[θ] η0[t, r, θ, ϕ] -− 4 q r2 Sin[2 θ] η0[t, r, θ, ϕ] + 2 q r2+n Sin[θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Sin[θ] hhh(1,0) [θ, ϕ] -− 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) Sin[θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Sin[2 θ] hhh(1,0) [θ, ϕ] -− 6 q r2+n Sin[3 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Sin[3 θ] hhh(1,0) [θ, ϕ] -− 4 q r2+n hhh(2,0) [θ, ϕ] -− 14 q r2+n Cos[θ] hhh(2,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(2,0) [θ, ϕ] + 14 q r2+n Cos[3 θ] hhh(2,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(3,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(3,0) [θ, ϕ] -− 24 ⅈ q r2 Sin[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− θ 3θ 4 q r2+n Cos  ww(0,0,1,0) [r, t, θ, ϕ] -− 6 q r2+n Cos  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] + 10 q r2+n Cos  ww(0,0,1,0) [r, t, θ, ϕ] + 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 (0,0,1,0) 6 q r η0 [t, r, θ, ϕ] + 14 q r2 Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 14 q r Cos[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅈ q r2 η0(0,0,1,1) [t, r, θ, ϕ] + 8 ⅈ q r2 Cos[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 8 q r2+n θ 3θ Sin  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,2,0) [r, t, θ, ϕ] + 10 q r2 Sin[θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 14 q r2 Sin[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,2,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,3,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 10 q r3 Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 16 q r3 Sin[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 6 q r3 Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 16 ⅈ q r3 Sin[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 2 q r3 Cos[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 22 q r3 Cos[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,2,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 4 q r4 Sin[θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Sin[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 6 q r3+n Sin  ww(1,0,0,0) [r, t, θ, ϕ] + 2 + Printed by Wolfram Mathematica Student Edition


10 q r3+n Sin

5θ 2

 ww(1,0,0,0) [r, t, θ, ϕ] + 4 q r3+n Cos

ww(1,0,1,0) [r, t, θ, ϕ] -− 4 q r3+n Cos Cos[ϕ] Sin[θ]

5θ 2

3θ

31



2

 ww(1,0,1,0) [r, t, θ, ϕ]

+

1

θ θ 3 Cos  Cot  2 2 8 q r (1 + Cos[θ]) 4

2

θ θ θ -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 2 2 2 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− 3θ 3θ 3θ 5θ Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 8 q r2 Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− θ 3θ 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5 θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5 θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 1 θ θ Cos  Cot  3 2 2 2 8 q r (1 + Cos[θ]) -− 4 n (2 + n) q r1+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] + θ 16 (2 + n) q r1+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 2 12 q r η0[t, r, θ, ϕ] + 16 q r Cos[θ] η0[t, r, θ, ϕ] + 4 q r Cos[2 θ] η0[t, r, θ, ϕ] -− 4 (2 + n) q r1+n hhh(1,0) [θ, ϕ] -− 2 (2 + n) q r1+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n (2 + n) q r1+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 (2 + n) q r1+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 (2 + n) q r1+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n (2 + n) q r1+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 (2 + n) q r1+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 (2 + n) q r1+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 16 ⅈ q r η0(0,0,0,1) [t, r, θ, ϕ] + 48 ⅈ q r Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− -− Printed by Wolfram Mathematica Student Edition

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θ 3θ 8 (2 + n) q r1+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 (2 + n) q r1+n Sin  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] + 4 (2 + n) q r1+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 12 q r Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 q r Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 4 q r Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 32 ⅈ q r Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 8 q r Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 8 q r Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 6 q r2 η0(0,1,0,0) [t, r, θ, ϕ] + 38 q r2 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 26 q r2 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 8 ⅈ q r2 η0(0,1,0,1) [t, r, θ, ϕ] + 72 ⅈ q r2 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 18 q r2 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 8 q r2 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 22 q r2 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,1,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 26 q r3 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 18 q r3 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,2,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,2,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,3,0,0) [t, r, θ, ϕ] -− 4 q r4 Cos[3 θ] η0(0,3,0,0) [t, r, θ, ϕ] + 3θ 4 (3 + n) q r2+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5 θ 4 (3 + n) q r2+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 2 θ 3θ 8 q r2+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 3θ 4 q r2+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 4 q r3+n Cos  2 2 5θ ww(2,0,0,0) [r, t, θ, ϕ] -− 4 q r3+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] 2 f3zy = c[[2, 1]] D[f3z, r] + c[[2, 2]] D[f3z, θ] + c[[2, 3]] D[f3z, ϕ] -−

1

θ θ Cos  Cos[ϕ] Cot  Csc[θ] 2 2 8 q r4 (1 + Cos[θ])2 -− 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) Sin[θ] hhh(0,1) [θ, ϕ] -− 4 q r2+n hhh(1,1) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,1) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,1) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,1) [θ, ϕ] + θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww(0,0,0,1) [r, t, θ, ϕ] + 2 6 q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 8 q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,2) [t, r, θ, ϕ] + θ 24 ⅈ q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 2 3 θ 5θ 4 q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 4 q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 2 2 6 q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 8 q r2 Sin[2 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,2) [t, r, θ, ϕ] -− + + + -− Printed by Wolfram Mathematica Student Edition


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2 q r2 Sin[3 θ] η0 [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0 [t, r, θ, ϕ] -− 2 (0,0,2,1) 2 (0,0,2,1) 4 q r Cos[θ] η0 [t, r, θ, ϕ] + 4 q r Cos[3 θ] η0 [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,2) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,1) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,1) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,1) [t, r, θ, ϕ] -− 4 q r4 Cos[3 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 3θ 5θ 4 q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] -− 4 q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] + 2 2 1 1 θ θ θ Cos[θ] Sin[ϕ] Cos  -− 3 Sin  -− 2 q Sin  + r 2 2 2 16 q r3 (1 + Cos[θ])2 θ 2 q C[2] Sin  -− 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) 2 3θ 3θ 3θ 5θ hhh[θ, ϕ] Sin[θ] -− Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q θ 3θ r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 1 θ θ θ θ θ Cot  Csc  -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 3 2 2 2 2 2 2 16 q r (1 + Cos[θ]) 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− 3θ 3θ 3θ 5θ Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + + + -− -− Printed by Wolfram Mathematica Student Edition

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4qr hhh [θ, ϕ] -− 2 q r Cos[θ] hhh [θ, ϕ] + 2+n (1,0) 2+n 8nqr Cos[θ] hhh [θ, ϕ] + 4 q r Cos[2 θ] hhh(1,0) [θ, ϕ] + 2+n (1,0) 2+n 2qr Cos[3 θ] hhh [θ, ϕ] -− 8 n q r Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q θ 3θ r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 1 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 4 q r3 (1 + Cos[θ])3 θ θ θ θ θ Cos  Cot  Sin[θ] -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 2 2 2 2 2 2+n 4nqr (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− 3θ 3θ 3θ 5θ Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 8 q θ 3θ r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 1 θ θ 3 θ θ θ Cos  Cot  -− Cos  -− q Cos  + q C[2] Cos  -− 2 2 2 2 2 2 8 q r3 (1 + Cos[θ])2 +

+ Printed by Wolfram Mathematica Student Edition

+

+


3

Cos

2 15

3θ

+

2

q Cos

3 2

5θ

q Cos

3θ 2

 + 3 q C[2] Cos

 -− 15 q C[2] Cos

5θ

+

21

3θ

+

2

Cos

25 2

7θ

Cos

5θ

35

+

2

 -− 21 q C[2] Cos

7θ

 -− 2 2 2 2 2 2 4 n q r2+n Cos[θ] (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] -− 4 n q r2+n hhh[θ, ϕ] Sin[θ] (2 Sin[θ] -− 2 (-− 1 + 2 n) Sin[2 θ]) + θ 16 q r2+n Cos[θ] (-− 1 + n Cos[θ]) Sin  ww[r, t, θ, ϕ] + 2 θ 8 q r2+n Cos  (-− 1 + n Cos[θ]) Sin[θ] ww[r, t, θ, ϕ] -− 2 θ 16 n q r2+n Sin  Sin[θ]2 ww[r, t, θ, ϕ] -− 2 8 q r2 Sin[θ] η0[t, r, θ, ϕ] -− 4 q r2 Sin[2 θ] η0[t, r, θ, ϕ] + 2 q r2+n Sin[θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Sin[θ] hhh(1,0) [θ, ϕ] -− 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) Sin[θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Sin[2 θ] hhh(1,0) [θ, ϕ] -− 6 q r2+n Sin[3 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Sin[3 θ] hhh(1,0) [θ, ϕ] -− 4 q r2+n hhh(2,0) [θ, ϕ] -− 14 q r2+n Cos[θ] hhh(2,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(2,0) [θ, ϕ] + 14 q r2+n Cos[3 θ] hhh(2,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(3,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(3,0) [θ, ϕ] -− 24 ⅈ q r2 Sin[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− θ 3θ 4 q r2+n Cos  ww(0,0,1,0) [r, t, θ, ϕ] -− 6 q r2+n Cos  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] + 10 q r2+n Cos  ww(0,0,1,0) [r, t, θ, ϕ] + 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 (0,0,1,0) 6 q r η0 [t, r, θ, ϕ] + 14 q r2 Cos[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 14 q r Cos[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 8 ⅈ q r2 η0(0,0,1,1) [t, r, θ, ϕ] + 8 ⅈ q r2 Cos[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 8 q r2+n θ 3θ Sin  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,2,0) [r, t, θ, ϕ] + 10 q r2 Sin[θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 14 q r2 Sin[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,2,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,3,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 10 q r3 Sin[θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 16 q r3 Sin[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 6 q r3 Sin[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 16 ⅈ q r3 Sin[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 2 q r3 Cos[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 22 q r3 Cos[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,2,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 4 q r4 Sin[θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Sin[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 6 q r3+n Sin  ww(1,0,0,0) [r, t, θ, ϕ] + 2 5θ 3θ 10 q r3+n Sin  ww(1,0,0,0) [r, t, θ, ϕ] + 4 q r3+n Cos  2 2 5θ ww(1,0,1,0) [r, t, θ, ϕ] -− 4 q r3+n Cos  ww(1,0,1,0) [r, t, θ, ϕ] + 2


36


Sin[θ] Sin[ϕ]

1

θ θ 3 Cos  Cot  2 2 8 q r4 (1 + Cos[θ])2

θ θ θ -− 3 Sin  -− 2 q Sin  + 2 q C[2] Sin  -− 2 2 2 4 n q r2+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] -− 3θ 3θ 3θ 5θ Sin  + q Sin  + 2 q C[2] Sin  + 5 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 3 q Sin  -− 6 q C[2] Sin  + 3 Sin  -− 6 q C[2] Sin + 2 2 2 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 6 q r2 η0[t, r, θ, ϕ] + 2 2 8 q r Cos[θ] η0[t, r, θ, ϕ] + 2 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 4 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 24 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− θ 3θ 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 4 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 6 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 8 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 10 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 2 q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 3θ 4 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 1 θ θ Cos  Cot  3 2 2 2 8 q r (1 + Cos[θ]) -− 4 n (2 + n) q r1+n (3 + 2 n -− 2 Cos[θ] + (-− 1 + 2 n) Cos[2 θ]) hhh[θ, ϕ] Sin[θ] + θ 16 (2 + n) q r1+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww[r, t, θ, ϕ] + 2 12 q r η0[t, r, θ, ϕ] + 16 q r Cos[θ] η0[t, r, θ, ϕ] + 4 q r Cos[2 θ] η0[t, r, θ, ϕ] -− 4 (2 + n) q r1+n hhh(1,0) [θ, ϕ] -− 2 (2 + n) q r1+n Cos[θ] hhh(1,0) [θ, ϕ] + 8 n (2 + n) q r1+n Cos[θ] hhh(1,0) [θ, ϕ] + 4 (2 + n) q r1+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 2 (2 + n) q r1+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 n (2 + n) q r1+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 12 (2 + n) q r1+n Sin[θ] hhh(2,0) [θ, ϕ] + 4 (2 + n) q r1+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 16 ⅈ q r η0(0,0,0,1) [t, r, θ, ϕ] + 48 ⅈ q r Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− θ 3θ 8 (2 + n) q r1+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 4 (2 + n) q r1+n Sin  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] + 4 (2 + n) q r1+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 12 q r Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 16 q r Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− -− -− + + Printed by Wolfram Mathematica Student Edition


37

12 q r Sin[θ] η0 [t, r, θ, ϕ] -− 16 q r Sin[2 θ] η0 [t, r, θ, ϕ] -− (0,0,1,0) (0,0,1,1) 4 q r Sin[3 θ] η0 [t, r, θ, ϕ] -− 32 ⅈ q r Sin[θ] η0 [t, r, θ, ϕ] -− 8 q r Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 8 q r Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 6 q r2 η0(0,1,0,0) [t, r, θ, ϕ] + 38 q r2 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 26 q r2 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− 8 ⅈ q r2 η0(0,1,0,1) [t, r, θ, ϕ] + 72 ⅈ q r2 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 18 q r2 Sin[θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 8 q r2 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 22 q r2 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 16 ⅈ q r2 Sin[θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 4 q r2 Cos[θ] η0(0,1,2,0) [t, r, θ, ϕ] + 4 q r2 Cos[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 26 q r3 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] + 8 q r3 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 18 q r3 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 16 ⅈ q r3 Cos[θ] η0(0,2,0,1) [t, r, θ, ϕ] -− 8 q r3 Sin[θ] η0(0,2,1,0) [t, r, θ, ϕ] + 8 q r3 Sin[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 4 q r4 Cos[θ] η0(0,3,0,0) [t, r, θ, ϕ] -− 4 q r4 Cos[3 θ] η0(0,3,0,0) [t, r, θ, ϕ] + 3θ 4 (3 + n) q r2+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 4 (3 + n) q r2+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 θ 16 q r2+n (-− 1 + n Cos[θ]) Sin  Sin[θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 2 θ 3θ 8 q r2+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 4 q r2+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 5θ 3θ 4 q r2+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 4 q r3+n Cos  2 2 5θ ww(2,0,0,0) [r, t, θ, ϕ] -− 4 q r3+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] 2 LHS797 = Simplify[f3zx + I f3zy] 1 256 q r4 θ 2 θ 3 θ θ θ Csc  Sec  (Cos[ϕ] + ⅈ Sin[ϕ]) -− 24 Sin  -− 18 q Sin  + 12 q C[2] Sin  -− 2 2 2 2 2 3θ 3θ 3θ 5θ 5θ 8 Sin  + 6 q Sin  + 12 q C[2] Sin  + 49 Sin  + 30 q Sin  -− 2 2 2 2 2 5θ 7θ 7θ 7θ 54 q C[2] Sin  + 25 Sin  -− 6 q Sin  -− 54 q C[2] Sin  -− 2 2 2 2 9θ 9θ 9θ 23 Sin  -− 12 q Sin  + 30 q C[2] Sin  + n q r2+n hhh[θ, ϕ] 2 2 2 2 -− 23 + 8 n + 4 n2  Sin[θ] + 32 Sin[2 θ] + 13 Sin[3 θ] -− 36 n Sin[3 θ] + 4 n2 Sin[3 θ] -− 8 Sin[4 θ] + 8 n Sin[4 θ] -− 5 Sin[5 θ] + 12 n Sin[5 θ] -− 4 n2 Sin[5 θ] -− 11 θ

11 θ

θ  + 32 q r2+n Cos  2 2 2 2 -− 2 + 4 + 2 n -− n  Cos[θ] + (2 -− 4 n) Cos[2 θ] -− 2 n Cos[3 θ] + n2 Cos[3 θ]

15 Sin

 + 30 q C[2] Sin

θ 2 Sin  ww[r, t, θ, ϕ] + 27 q r2 η0[t, r, θ, ϕ] + 38 q r2 Cos[θ] η0[t, r, θ, ϕ] + 2 2 8 q r Cos[2 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 3 q r2 Cos[4 θ] η0[t, r, θ, ϕ] + 56 ⅈ n q r2+n Sin[θ] hhh(0,1) [θ, ϕ] + 16 ⅈ n2 q r2+n Sin[θ] hhh(0,1) [θ, ϕ] -− 16 ⅈ n q r2+n Sin[2 θ] hhh(0,1) [θ, ϕ] -− 8 ⅈ n q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] + 16 ⅈ n2 q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] -− 8 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + -− + -− + Printed by Wolfram Mathematica Student Edition

38


8qr hhh [θ, ϕ] -− 2 q r Cos[θ] hhh [θ, ϕ] + 2+n (1,0) 2 2+n 40 n q r Cos[θ] hhh [θ, ϕ] -− 8 n q r Cos[θ] hhh(1,0) [θ, ϕ] + 2+n (1,0) 2+n 16 q r Cos[2 θ] hhh [θ, ϕ] -− 16 n q r Cos[2 θ] hhh(1,0) [θ, ϕ] + 7 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 64 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] + 20 n2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] + 16 n q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 5 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 12 n2 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 16 ⅈ q r2+n hhh(1,1) [θ, ϕ] + 8 ⅈ q r2+n Cos[θ] hhh(1,1) [θ, ϕ] -− 32 ⅈ n q r2+n Cos[θ] hhh(1,1) [θ, ϕ] -− 16 ⅈ q r2+n Cos[2 θ] hhh(1,1) [θ, ϕ] -− 8 ⅈ q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] + 32 ⅈ n q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 24 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 24 n q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 16 q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 28 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 28 n q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] + 12 n q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] + 48 ⅈ q r2+n Sin[θ] hhh(2,1) [θ, ϕ] -− 16 ⅈ q r2+n Sin[3 θ] hhh(2,1) [θ, ϕ] + 8 q r2+n Cos[θ] hhh(3,0) [θ, ϕ] -− 12 q r2+n Cos[3 θ] hhh(3,0) [θ, ϕ] + θ 4 q r2+n Cos[5 θ] hhh(3,0) [θ, ϕ] + 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 3 θ 3θ 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ n q r2+n Cos  2 2 5θ ww(0,0,0,1) [r, t, θ, ϕ] + 16 ⅈ n q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 88 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 140 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 40 ⅈ q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 12 ⅈ q r2 Cos[3 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 32 q r2 η0(0,0,0,2) [t, r, θ, ϕ] + 96 q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] -− θ θ 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 3 θ 5θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 2 5θ 7θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 2 7θ 9θ 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 9θ 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 14 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 56 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 13 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 12 q r2 Sin[4 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 5 q r2 Sin[5 θ] η0(0,0,1,0) [t, r, θ, ϕ] + θ 3θ 32 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 16 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 2 2 5θ 16 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 104 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] + 2 64 ⅈ q r2 Sin[2 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 8 ⅈ q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 3θ 64 q r2 Sin[θ] η0(0,0,1,2) [t, r, θ, ϕ] + 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 5θ 7θ 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 9θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 12 q r2 η0(0,0,2,0) [t, r, θ, ϕ] -− 2 16 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 28 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[4 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0(0,0,2,0) [t, r, θ, ϕ] + -− + + -− Printed by Wolfram Mathematica Student Edition


39

12 q r2 Cos[4 θ] η0 [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0 [t, r, θ, ϕ] + 2 (0,0,2,1) 2 (0,0,2,1) 32 ⅈ q r Cos[θ] η0 [t, r, θ, ϕ] -− 32 ⅈ q r Cos[3 θ] η0 [t, r, θ, ϕ] + 8 q r2 Sin[θ] η0(0,0,3,0) [t, r, θ, ϕ] + 4 q r2 Sin[3 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 4 q r2 Sin[5 θ] η0(0,0,3,0) [t, r, θ, ϕ] + 8 q r3 η0(0,1,0,0) [t, r, θ, ϕ] + 30 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 24 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 3 q r3 Cos[5 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 24 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 64 ⅈ q r3 Cos[2 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 40 ⅈ q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 64 q r3 Cos[θ] η0(0,1,0,2) [t, r, θ, ϕ] + 16 q r3 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 20 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 24 q r3 Sin[4 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 12 q r3 Sin[5 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 64 ⅈ q r3 Sin[θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 64 ⅈ q r3 Sin[3 θ] η0(0,1,1,1) [t, r, θ, ϕ] + 8 q r3 Cos[θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 20 q r3 Cos[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 12 q r3 Cos[5 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 4 q r4 η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 16 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Cos[4 θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 32 ⅈ q r4 Cos[θ] η0(0,2,0,1) [t, r, θ, ϕ] + 32 ⅈ q r4 Cos[3 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 24 q r4 Sin[θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 28 q r4 Sin[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 12 q r4 Sin[5 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 8 q r5 Cos[θ] η0(0,3,0,0) [t, r, θ, ϕ] + 12 q r5 Cos[3 θ] η0(0,3,0,0) [t, r, θ, ϕ] -− 4 q r5 Cos[5 θ] η0(0,3,0,0) [t, r, θ, ϕ] + θ 3θ 16 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 8 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 3θ 16 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 5θ 24 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 16 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 2 7θ 7θ 12 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 2 9θ 9θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 3θ 16 ⅈ q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] + 2 5θ θ 16 ⅈ q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] + 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 3 θ 5 θ 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 2 7 θ 9 θ 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 2 3θ 5θ 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 2 2 7θ 9θ 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] 2 2 Eqn797 = Simplify[LHS797 -− RHS797] 1 256 r4 1 θ 2 θ 3 θ θ θ Csc  Sec  (Cos[ϕ] + ⅈ Sin[ϕ]) -− 24 Sin  -− 18 q Sin  + 12 q C[2] Sin  -− q 2 2 2 2 2 3θ 3θ 3θ 5θ 8 Sin  + 6 q Sin  + 12 q C[2] Sin  + 49 Sin + 2 2 2 2 -−

+


-−

-−

40


30 q Sin

5θ

 -− 54 q C[2] Sin

5θ

 + 25 Sin

7θ

 -− 6 q Sin

7θ

 -− 2 2 2 7θ 9θ 9θ 9θ 54 q C[2] Sin  -− 23 Sin  -− 12 q Sin  + 30 q C[2] Sin + 2 2 2 2 n q r2+n hhh[θ, ϕ] 2 -− 23 + 8 n + 4 n2  Sin[θ] + 32 Sin[2 θ] + 13 Sin[3 θ] -− 2

36 n Sin[3 θ] + 4 n2 Sin[3 θ] -− 8 Sin[4 θ] + 8 n Sin[4 θ] -− 5 Sin[5 θ] + 11 θ 11 θ 12 n Sin[5 θ] -− 4 n2 Sin[5 θ] -− 15 Sin  + 30 q C[2] Sin  + 32 q r2+n 2 2 θ Cos  -− 2 + 4 + 2 n -− n2  Cos[θ] + (2 -− 4 n) Cos[2 θ] -− 2 n Cos[3 θ] + n2 Cos[3 θ] 2 θ 2 Sin  ww[r, t, θ, ϕ] + 27 q r2 η0[t, r, θ, ϕ] + 38 q r2 Cos[θ] η0[t, r, θ, ϕ] + 2 8 q r2 Cos[2 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 3 q r2 Cos[4 θ] η0[t, r, θ, ϕ] + 56 ⅈ n q r2+n Sin[θ] hhh(0,1) [θ, ϕ] + 16 ⅈ n2 q r2+n Sin[θ] hhh(0,1) [θ, ϕ] -− 16 ⅈ n q r2+n Sin[2 θ] hhh(0,1) [θ, ϕ] -− 8 ⅈ n q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] + 16 ⅈ n2 q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] -− 8 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 40 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] -− 8 n2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 16 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 16 n q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 7 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 64 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] + 20 n2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] + 16 n q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 5 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 12 n2 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 16 ⅈ q r2+n hhh(1,1) [θ, ϕ] + 8 ⅈ q r2+n Cos[θ] hhh(1,1) [θ, ϕ] -− 32 ⅈ n q r2+n Cos[θ] hhh(1,1) [θ, ϕ] -− 16 ⅈ q r2+n Cos[2 θ] hhh(1,1) [θ, ϕ] -− 8 ⅈ q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] + 32 ⅈ n q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 24 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 24 n q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 16 q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 28 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 28 n q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] + 12 n q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] + 48 ⅈ q r2+n Sin[θ] hhh(2,1) [θ, ϕ] -− 16 ⅈ q r2+n Sin[3 θ] hhh(2,1) [θ, ϕ] + 8 q r2+n Cos[θ] hhh(3,0) [θ, ϕ] -− 12 q r2+n Cos[3 θ] hhh(3,0) [θ, ϕ] + θ 4 q r2+n Cos[5 θ] hhh(3,0) [θ, ϕ] + 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 3 θ 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 3θ 16 ⅈ n q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] + 2 5θ 16 ⅈ n q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 88 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 2 140 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 40 ⅈ q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 12 ⅈ q r2 Cos[3 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 32 q r2 η0(0,0,0,2) [t, r, θ, ϕ] + θ 96 q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] -− 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 θ 3θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 16 n q r2+n Sin  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] + 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 -−



16 n q r2+n Sin

5θ 2

 ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin

ww(0,0,1,0) [r, t, θ, ϕ] -− 8 n q r2+n Sin Sin

9θ

7θ 2

7θ

41



2

 ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n

 ww(0,0,1,0) [r, t, θ, ϕ] + 8 n q r2+n Sin

9θ

 ww(0,0,1,0) [r, t, θ, ϕ] +

2 2 14 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 56 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 13 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 12 q r2 Sin[4 θ] η0(0,0,1,0) [t, r, θ, ϕ] + θ 5 q r2 Sin[5 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 32 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 2 3 θ 5θ 16 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 16 ⅈ q r2+n Sin  2 2 ww(0,0,1,1) [r, t, θ, ϕ] -− 104 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] + 64 ⅈ q r2 Sin[2 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 8 ⅈ q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 3θ 64 q r2 Sin[θ] η0(0,0,1,2) [t, r, θ, ϕ] + 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 5θ 7θ 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 9θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 12 q r2 η0(0,0,2,0) [t, r, θ, ϕ] -− 2 16 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 28 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[4 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 32 ⅈ q r2 Cos[θ] η0(0,0,2,1) [t, r, θ, ϕ] -− 32 ⅈ q r2 Cos[3 θ] η0(0,0,2,1) [t, r, θ, ϕ] + 8 q r2 Sin[θ] η0(0,0,3,0) [t, r, θ, ϕ] + 4 q r2 Sin[3 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 4 q r2 Sin[5 θ] η0(0,0,3,0) [t, r, θ, ϕ] + 8 q r3 η0(0,1,0,0) [t, r, θ, ϕ] + 30 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 24 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 3 q r3 Cos[5 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 24 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 64 ⅈ q r3 Cos[2 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 40 ⅈ q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] + 64 q r3 Cos[θ] η0(0,1,0,2) [t, r, θ, ϕ] + 16 q r3 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 20 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 24 q r3 Sin[4 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 12 q r3 Sin[5 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 64 ⅈ q r3 Sin[θ] η0(0,1,1,1) [t, r, θ, ϕ] -− 64 ⅈ q r3 Sin[3 θ] η0(0,1,1,1) [t, r, θ, ϕ] + 8 q r3 Cos[θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 20 q r3 Cos[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 12 q r3 Cos[5 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 4 q r4 η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 16 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Cos[4 θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 32 ⅈ q r4 Cos[θ] η0(0,2,0,1) [t, r, θ, ϕ] + 32 ⅈ q r4 Cos[3 θ] η0(0,2,0,1) [t, r, θ, ϕ] + 24 q r4 Sin[θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 28 q r4 Sin[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 12 q r4 Sin[5 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 8 q r5 Cos[θ] η0(0,3,0,0) [t, r, θ, ϕ] + 12 q r5 Cos[3 θ] η0(0,3,0,0) [t, r, θ, ϕ] -− 4 q r5 Cos[5 θ] η0(0,3,0,0) [t, r, θ, ϕ] + θ 3θ 16 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 8 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 3 θ 16 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 5θ 24 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 16 n q r3+n Cos  2 2 7θ ww(1,0,0,0) [r, t, θ, ϕ] + 12 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 +


42


8 n q r3+n Cos

7θ 2

 ww(1,0,0,0) [r, t, θ, ϕ] + 4 q r3+n Cos

ww(1,0,0,0) [r, t, θ, ϕ] -− 8 n q r3+n Cos 16 ⅈ q r3+n Cos

3θ 2

9θ 2

9θ



2

 ww(1,0,0,0) [r, t, θ, ϕ] -−

 ww(1,0,0,1) [r, t, θ, ϕ] + 16 ⅈ q r3+n Cos

5θ



2

θ ww(1,0,0,1) [r, t, θ, ϕ] + 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 3 θ 5θ 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 16 q r3+n Sin  2 2 7θ ww(1,0,1,0) [r, t, θ, ϕ] -− 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 9θ 3θ 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 2 2 5θ 7θ 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 2 9θ 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 16 ⅇ-−ⅈ ϕ r2 Csc[θ]2 -− 8 vv[r, t, θ, ϕ] + 8 ⅇ2 ⅈ ϕ ww[r, t, θ, ϕ] -− 16 ⅈ vv(0,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ ⅇ2 ⅈ ϕ ww(0,0,0,1) [r, t, θ, ϕ] + 8 vv(0,0,0,2) [r, t, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ ww(0,0,0,2) [r, t, θ, ϕ] + 8 Sin[2 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 Sin[4 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅇ2 ⅈ ϕ Sin[4 θ] ww(0,0,1,0) [r, t, θ, ϕ] + 7 vv(0,0,2,0) [r, t, θ, ϕ] -− 8 Cos[2 θ] vv(0,0,2,0) [r, t, θ, ϕ] + Cos[4 θ] vv(0,0,2,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ ww(0,0,2,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ Cos[4 θ] ww(0,0,2,0) [r, t, θ, ϕ] + 11 r vv(1,0,0,0) [r, t, θ, ϕ] -− 12 r Cos[2 θ] vv(1,0,0,0) [r, t, θ, ϕ] + r Cos[4 θ] vv(1,0,0,0) [r, t, θ, ϕ] -− 5 ⅇ2 ⅈ ϕ r ww(1,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r Cos[2 θ] ww(1,0,0,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ r Cos[4 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 4 r Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 2 r Sin[4 θ] vv(1,0,1,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r Sin[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 2 ⅇ2 ⅈ ϕ r Sin[4 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 5 r2 vv(2,0,0,0) [r, t, θ, ϕ] -− 4 r2 Cos[2 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− r2 Cos[4 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− 3 ⅇ2 ⅈ ϕ r2 ww(2,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r2 Cos[2 θ] ww(2,0,0,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ r2 Cos[4 θ] ww(2,0,0,0) [r, t, θ, ϕ] -− 1 16 ⅇ2 ⅈ ϕ (2 Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] -− 2 ⅈ (Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[ϕ] ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] -− ⅈ Cos[ϕ]2 + Cos[ϕ] Sin[ϕ] + ⅈ Sin[ϕ]2 + Sin[θ]2 Sin[2 ϕ] ww(0,0,1,0) [r, t, θ, ϕ] -− Cos[θ] Cos[2 ϕ] ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[θ] Cos[θ]2 Cos[ϕ] Sin[ϕ] ww(0,0,2,0) [ r, t, θ, ϕ] + r ⅈ Cos[ϕ]2 + ⅈ Cos[ϕ] Sin[θ]2 Sin[ϕ] -− Sin[ϕ]2  ww(1,0,0,0) [r, t, θ, ϕ] + Cos[2 ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] Sin[ϕ] 2 Cos[θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ] (*⋆End Eqn (7.97)*⋆)



43

(*⋆Eqn 7.98 begins *⋆) RHS798 = Simplify[trm2 + wiyx] 1 16 r2 ⅇ-−ⅈ ϕ Csc[θ]2 -− 8 vv[r, t, θ, ϕ] + 8 ⅇ2 ⅈ ϕ ww[r, t, θ, ϕ] -− 16 ⅈ vv(0,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ ⅇ2 ⅈ ϕ ww(0,0,0,1) [r, t, θ, ϕ] + 8 vv(0,0,0,2) [r, t, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ ww(0,0,0,2) [r, t, θ, ϕ] + 8 Sin[2 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 Sin[4 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅇ2 ⅈ ϕ Sin[4 θ] ww(0,0,1,0) [r, t, θ, ϕ] + 7 vv(0,0,2,0) [r, t, θ, ϕ] -− 8 Cos[2 θ] vv(0,0,2,0) [r, t, θ, ϕ] + Cos[4 θ] vv(0,0,2,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ ww(0,0,2,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ Cos[4 θ] ww(0,0,2,0) [r, t, θ, ϕ] + 11 r vv(1,0,0,0) [r, t, θ, ϕ] -− 12 r Cos[2 θ] vv(1,0,0,0) [r, t, θ, ϕ] + r Cos[4 θ] vv(1,0,0,0) [r, t, θ, ϕ] -− 5 ⅇ2 ⅈ ϕ r ww(1,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r Cos[2 θ] ww(1,0,0,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ r Cos[4 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 4 r Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 2 r Sin[4 θ] vv(1,0,1,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r Sin[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 2 ⅇ2 ⅈ ϕ r Sin[4 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 5 r2 vv(2,0,0,0) [r, t, θ, ϕ] -− 4 r2 Cos[2 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− r2 Cos[4 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− 3 ⅇ2 ⅈ ϕ r2 ww(2,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r2 Cos[2 θ] ww(2,0,0,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ r2 Cos[4 θ] ww(2,0,0,0) [r, t, θ, ϕ] + 1 16 ⅇ2 ⅈ ϕ (2 Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] -− 2 ⅈ (Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[ϕ] ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] -− ⅈ Cos[ϕ]2 + Cos[ϕ] Sin[ϕ] + ⅈ Sin[ϕ]2 + Sin[θ]2 Sin[2 ϕ] ww(0,0,1,0) [r, t, θ, ϕ] -− Cos[θ] Cos[2 ϕ] ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[θ] Cos[θ]2 Cos[ϕ] Sin[ϕ] ww(0,0,2,0) [r, t, θ, ϕ] + r ⅈ Cos[ϕ]2 + ⅈ Cos[ϕ] Sin[θ]2 Sin[ϕ] -− Sin[ϕ]2  ww(1,0,0,0) [r, t, θ, ϕ] + Cos[2 ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] Sin[ϕ] 2 Cos[θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ] LHS798 = Simplify[f3zx -− I f3zy] 1 256 q r4 θ 2 θ 3 θ θ θ Csc  Sec  (Cos[ϕ] -− ⅈ Sin[ϕ]) -− 24 Sin  -− 18 q Sin  + 12 q C[2] Sin  -− 2 2 2 2 2 3θ 3θ 3θ 5θ 5θ 8 Sin  + 6 q Sin  + 12 q C[2] Sin  + 49 Sin  + 30 q Sin  -− 2 2 2 2 2 5θ 7θ 7θ 7θ 54 q C[2] Sin  + 25 Sin  -− 6 q Sin  -− 54 q C[2] Sin  -− 2 2 2 2 9θ 9θ 9θ 23 Sin  -− 12 q Sin  + 30 q C[2] Sin  + n q r2+n hhh[θ, ϕ] 2 2 2 2 -− 23 + 8 n + 4 n2  Sin[θ] + 32 Sin[2 θ] + 13 Sin[3 θ] -− 36 n Sin[3 θ] + 4 n2 Sin[3 θ] -− 8 Sin[4 θ] + 8 n Sin[4 θ] -− 5 Sin[5 θ] + 12 n Sin[5 θ] -− 4 n2 Sin[5 θ] -− +

+ Printed by Wolfram Mathematica Student Edition

44


11 θ

11 θ

θ  + 32 q r2+n Cos  2 2 2 2 -− 2 + 4 + 2 n -− n  Cos[θ] + (2 -− 4 n) Cos[2 θ] -− 2 n Cos[3 θ] + n2 Cos[3 θ]

15 Sin

 + 30 q C[2] Sin

θ 2 Sin  ww[r, t, θ, ϕ] + 27 q r2 η0[t, r, θ, ϕ] + 38 q r2 Cos[θ] η0[t, r, θ, ϕ] + 2 2 8 q r Cos[2 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 3 q r2 Cos[4 θ] η0[t, r, θ, ϕ] -− 56 ⅈ n q r2+n Sin[θ] hhh(0,1) [θ, ϕ] -− 16 ⅈ n2 q r2+n Sin[θ] hhh(0,1) [θ, ϕ] + 16 ⅈ n q r2+n Sin[2 θ] hhh(0,1) [θ, ϕ] + 8 ⅈ n q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] -− 16 ⅈ n2 q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] -− 8 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 40 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] -− 8 n2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 16 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 16 n q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 7 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 64 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] + 20 n2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] + 16 n q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 5 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 12 n2 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 16 ⅈ q r2+n hhh(1,1) [θ, ϕ] -− 8 ⅈ q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 32 ⅈ n q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 16 ⅈ q r2+n Cos[2 θ] hhh(1,1) [θ, ϕ] + 8 ⅈ q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 32 ⅈ n q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 24 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 24 n q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 16 q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 28 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 28 n q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] + 12 n q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] -− 48 ⅈ q r2+n Sin[θ] hhh(2,1) [θ, ϕ] + 16 ⅈ q r2+n Sin[3 θ] hhh(2,1) [θ, ϕ] + 8 q r2+n Cos[θ] hhh(3,0) [θ, ϕ] -− 12 q r2+n Cos[3 θ] hhh(3,0) [θ, ϕ] + θ 4 q r2+n Cos[5 θ] hhh(3,0) [θ, ϕ] -− 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] + 2 3 θ 3θ 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] + 16 ⅈ n q r2+n Cos  2 2 5θ ww(0,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ n q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 40 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 204 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 24 ⅈ q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 12 ⅈ q r2 Cos[3 θ] η0(0,0,0,1) [t, r, θ, ϕ] + 32 q r2 η0(0,0,0,2) [t, r, θ, ϕ] -− 96 q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] -− θ θ 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 3 θ 5θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 2 5θ 7θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 2 7θ 9θ 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 9θ 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 14 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 2 56 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 13 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 12 q r2 Sin[4 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 5 q r2 Sin[5 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− θ 3θ 32 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 16 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 2 2 -−



16 ⅈ q r2+n Sin

5θ

45

 ww(0,0,1,1) [r, t, θ, ϕ] -− 2 56 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 24 ⅈ q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] + 3θ 64 q r2 Sin[θ] η0(0,0,1,2) [t, r, θ, ϕ] + 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 5θ 7θ 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 9θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 12 q r2 η0(0,0,2,0) [t, r, θ, ϕ] -− 2 16 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 28 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[4 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 8 q r2 Sin[θ] η0(0,0,3,0) [t, r, θ, ϕ] + 4 q r2 Sin[3 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 4 q r2 Sin[5 θ] η0(0,0,3,0) [t, r, θ, ϕ] + 8 q r3 η0(0,1,0,0) [t, r, θ, ϕ] + 30 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 24 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 3 q r3 Cos[5 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 104 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 24 ⅈ q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 64 q r3 Cos[θ] η0(0,1,0,2) [t, r, θ, ϕ] + 16 q r3 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 20 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 24 q r3 Sin[4 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 12 q r3 Sin[5 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Cos[θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 20 q r3 Cos[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 12 q r3 Cos[5 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 4 q r4 η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 16 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Cos[4 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 24 q r4 Sin[θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 28 q r4 Sin[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 12 q r4 Sin[5 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 8 q r5 Cos[θ] η0(0,3,0,0) [t, r, θ, ϕ] + 12 q r5 Cos[3 θ] η0(0,3,0,0) [t, r, θ, ϕ] -− θ 4 q r5 Cos[5 θ] η0(0,3,0,0) [t, r, θ, ϕ] + 16 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 3 θ 3θ 8 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 16 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 5θ 5θ 24 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 16 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 2 7θ 7θ 12 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 2 9θ 9θ 4 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 8 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 2 3θ 16 ⅈ q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] -− 2 5θ θ 16 ⅈ q r3+n Cos  ww(1,0,0,1) [r, t, θ, ϕ] + 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 2 3 θ 5 θ 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 2 7 θ 9 θ 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 2 2 3θ 5θ 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 2 2 7θ 9θ 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] 2 2 Eqn798 = Simplify[LHS798 -− RHS798]


46


1 256 r4 1 θ 2 θ 3 θ θ θ Csc  Sec  (Cos[ϕ] -− ⅈ Sin[ϕ]) -− 24 Sin  -− 18 q Sin  + 12 q C[2] Sin  -− q 2 2 2 2 2 3θ 3θ 3θ 5θ 8 Sin  + 6 q Sin  + 12 q C[2] Sin  + 49 Sin + 2 2 2 2 5θ 5θ 7θ 7θ 30 q Sin  -− 54 q C[2] Sin  + 25 Sin  -− 6 q Sin  -− 2 2 2 2 7θ 9θ 9θ 9θ 54 q C[2] Sin  -− 23 Sin  -− 12 q Sin  + 30 q C[2] Sin + 2 2 2 2 n q r2+n hhh[θ, ϕ] 2 -− 23 + 8 n + 4 n2  Sin[θ] + 32 Sin[2 θ] + 13 Sin[3 θ] -− 36 n Sin[3 θ] + 4 n2 Sin[3 θ] -− 8 Sin[4 θ] + 8 n Sin[4 θ] -− 5 Sin[5 θ] + 11 θ 11 θ 12 n Sin[5 θ] -− 4 n2 Sin[5 θ] -− 15 Sin  + 30 q C[2] Sin  + 32 q r2+n 2 2 θ Cos  -− 2 + 4 + 2 n -− n2  Cos[θ] + (2 -− 4 n) Cos[2 θ] -− 2 n Cos[3 θ] + n2 Cos[3 θ] 2 θ 2 Sin  ww[r, t, θ, ϕ] + 27 q r2 η0[t, r, θ, ϕ] + 38 q r2 Cos[θ] η0[t, r, θ, ϕ] + 2 2 8 q r Cos[2 θ] η0[t, r, θ, ϕ] -− 6 q r2 Cos[3 θ] η0[t, r, θ, ϕ] -− 3 q r2 Cos[4 θ] η0[t, r, θ, ϕ] -− 56 ⅈ n q r2+n Sin[θ] hhh(0,1) [θ, ϕ] -− 16 ⅈ n2 q r2+n Sin[θ] hhh(0,1) [θ, ϕ] + 16 ⅈ n q r2+n Sin[2 θ] hhh(0,1) [θ, ϕ] + 8 ⅈ n q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] -− 16 ⅈ n2 q r2+n Sin[3 θ] hhh(0,1) [θ, ϕ] -− 8 q r2+n hhh(1,0) [θ, ϕ] -− 2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 40 n q r2+n Cos[θ] hhh(1,0) [θ, ϕ] -− 8 n2 q r2+n Cos[θ] hhh(1,0) [θ, ϕ] + 16 q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] -− 16 n q r2+n Cos[2 θ] hhh(1,0) [θ, ϕ] + 7 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 64 n q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] + 20 n2 q r2+n Cos[3 θ] hhh(1,0) [θ, ϕ] -− 8 q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] + 16 n q r2+n Cos[4 θ] hhh(1,0) [θ, ϕ] -− 5 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] + 24 n q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 12 n2 q r2+n Cos[5 θ] hhh(1,0) [θ, ϕ] -− 16 ⅈ q r2+n hhh(1,1) [θ, ϕ] -− 8 ⅈ q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 32 ⅈ n q r2+n Cos[θ] hhh(1,1) [θ, ϕ] + 16 ⅈ q r2+n Cos[2 θ] hhh(1,1) [θ, ϕ] + 8 ⅈ q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 32 ⅈ n q r2+n Cos[3 θ] hhh(1,1) [θ, ϕ] -− 24 q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 24 n q r2+n Sin[θ] hhh(2,0) [θ, ϕ] + 16 q r2+n Sin[2 θ] hhh(2,0) [θ, ϕ] + 28 q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 28 n q r2+n Sin[3 θ] hhh(2,0) [θ, ϕ] -− 8 q r2+n Sin[4 θ] hhh(2,0) [θ, ϕ] -− 12 q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] + 12 n q r2+n Sin[5 θ] hhh(2,0) [θ, ϕ] -− 48 ⅈ q r2+n Sin[θ] hhh(2,1) [θ, ϕ] + 16 ⅈ q r2+n Sin[3 θ] hhh(2,1) [θ, ϕ] + 8 q r2+n Cos[θ] hhh(3,0) [θ, ϕ] -− 12 q r2+n Cos[3 θ] hhh(3,0) [θ, ϕ] + θ 4 q r2+n Cos[5 θ] hhh(3,0) [θ, ϕ] -− 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] + 2 3 θ 32 ⅈ q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] + 2 3θ 16 ⅈ n q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 2 5θ 16 ⅈ n q r2+n Cos  ww(0,0,0,1) [r, t, θ, ϕ] -− 40 ⅈ q r2 η0(0,0,0,1) [t, r, θ, ϕ] + 2 204 ⅈ q r2 Cos[θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 24 ⅈ q r2 Cos[2 θ] η0(0,0,0,1) [t, r, θ, ϕ] -− 12 ⅈ q r2 Cos[3 θ] η0(0,0,0,1) [t, r, θ, ϕ] + 32 q r2 η0(0,0,0,2) [t, r, θ, ϕ] -− -−


+


47

θ 96 q r2 Cos[θ] η0(0,0,0,2) [t, r, θ, ϕ] -− 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 θ 3θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 16 n q r2+n Sin  2 2 5θ ww(0,0,1,0) [r, t, θ, ϕ] + 32 q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 2 5θ 7θ 16 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n Sin  2 2 7θ ww(0,0,1,0) [r, t, θ, ϕ] -− 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] -− 8 q r2+n 2 9θ 9θ Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 8 n q r2+n Sin  ww(0,0,1,0) [r, t, θ, ϕ] + 2 2 14 q r2 Sin[θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 56 q r2 Sin[2 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 13 q r2 Sin[3 θ] η0(0,0,1,0) [t, r, θ, ϕ] + 12 q r2 Sin[4 θ] η0(0,0,1,0) [t, r, θ, ϕ] + θ 5 q r2 Sin[5 θ] η0(0,0,1,0) [t, r, θ, ϕ] -− 32 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] -− 2 3 θ 5θ 16 ⅈ q r2+n Sin  ww(0,0,1,1) [r, t, θ, ϕ] + 16 ⅈ q r2+n Sin  2 2 ww(0,0,1,1) [r, t, θ, ϕ] -− 56 ⅈ q r2 Sin[θ] η0(0,0,1,1) [t, r, θ, ϕ] -− 24 ⅈ q r2 Sin[3 θ] η0(0,0,1,1) [t, r, θ, ϕ] + 64 q r2 Sin[θ] η0(0,0,1,2) [t, r, θ, ϕ] + 3θ 5θ 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 8 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] -− 2 2 7θ 9θ 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 4 q r2+n Cos  ww(0,0,2,0) [r, t, θ, ϕ] + 2 2 12 q r2 η0(0,0,2,0) [t, r, θ, ϕ] -− 16 q r2 Cos[θ] η0(0,0,2,0) [t, r, θ, ϕ] + 28 q r2 Cos[3 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[4 θ] η0(0,0,2,0) [t, r, θ, ϕ] -− 12 q r2 Cos[5 θ] η0(0,0,2,0) [t, r, θ, ϕ] + 8 q r2 Sin[θ] η0(0,0,3,0) [t, r, θ, ϕ] + 4 q r2 Sin[3 θ] η0(0,0,3,0) [t, r, θ, ϕ] -− 4 q r2 Sin[5 θ] η0(0,0,3,0) [t, r, θ, ϕ] + 8 q r3 η0(0,1,0,0) [t, r, θ, ϕ] + 30 q r3 Cos[θ] η0(0,1,0,0) [t, r, θ, ϕ] + 24 q r3 Cos[2 θ] η0(0,1,0,0) [t, r, θ, ϕ] -− q r3 Cos[3 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 3 q r3 Cos[5 θ] η0(0,1,0,0) [t, r, θ, ϕ] + 104 ⅈ q r3 Cos[θ] η0(0,1,0,1) [t, r, θ, ϕ] + 24 ⅈ q r3 Cos[3 θ] η0(0,1,0,1) [t, r, θ, ϕ] -− 64 q r3 Cos[θ] η0(0,1,0,2) [t, r, θ, ϕ] + 16 q r3 Sin[2 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 20 q r3 Sin[3 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 24 q r3 Sin[4 θ] η0(0,1,1,0) [t, r, θ, ϕ] -− 12 q r3 Sin[5 θ] η0(0,1,1,0) [t, r, θ, ϕ] + 8 q r3 Cos[θ] η0(0,1,2,0) [t, r, θ, ϕ] -− 20 q r3 Cos[3 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 12 q r3 Cos[5 θ] η0(0,1,2,0) [t, r, θ, ϕ] + 4 q r4 η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[θ] η0(0,2,0,0) [t, r, θ, ϕ] -− 16 q r4 Cos[2 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 16 q r4 Cos[3 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 12 q r4 Cos[4 θ] η0(0,2,0,0) [t, r, θ, ϕ] + 24 q r4 Sin[θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 28 q r4 Sin[3 θ] η0(0,2,1,0) [t, r, θ, ϕ] + 12 q r4 Sin[5 θ] η0(0,2,1,0) [t, r, θ, ϕ] -− 8 q r5 Cos[θ] η0(0,3,0,0) [t, r, θ, ϕ] + 12 q r5 Cos[3 θ] η0(0,3,0,0) [t, r, θ, ϕ] -− 4 q r5 Cos[5 θ] η0(0,3,0,0) [t, r, θ, ϕ] + θ 3θ 16 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 8 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 2 3θ 16 n q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] -− 2 5θ 5θ 24 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 16 n q r3+n Cos  2 2 7θ ww(1,0,0,0) [r, t, θ, ϕ] + 12 q r3+n Cos  ww(1,0,0,0) [r, t, θ, ϕ] + 2 + Printed by Wolfram Mathematica Student Edition

48


8 n q r3+n Cos

7θ 2

 ww(1,0,0,0) [r, t, θ, ϕ] + 4 q r3+n Cos

ww(1,0,0,0) [r, t, θ, ϕ] -− 8 n q r3+n Cos 16 ⅈ q r3+n Cos

3θ 2

9θ 2

9θ



2

 ww(1,0,0,0) [r, t, θ, ϕ] +

 ww(1,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ q r3+n Cos

5θ



2

θ ww(1,0,0,1) [r, t, θ, ϕ] + 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 3 θ 5θ 16 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 16 q r3+n Sin  2 2 7θ ww(1,0,1,0) [r, t, θ, ϕ] -− 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] + 2 9θ 3θ 8 q r3+n Sin  ww(1,0,1,0) [r, t, θ, ϕ] -− 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 2 2 5θ 7θ 8 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] + 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 2 9θ 4 q r4+n Cos  ww(2,0,0,0) [r, t, θ, ϕ] -− 2 16 ⅇ-−ⅈ ϕ r2 Csc[θ]2 -− 8 vv[r, t, θ, ϕ] + 8 ⅇ2 ⅈ ϕ ww[r, t, θ, ϕ] -− 16 ⅈ vv(0,0,0,1) [r, t, θ, ϕ] -− 16 ⅈ ⅇ2 ⅈ ϕ ww(0,0,0,1) [r, t, θ, ϕ] + 8 vv(0,0,0,2) [r, t, θ, ϕ] -− 8 ⅇ2 ⅈ ϕ ww(0,0,0,2) [r, t, θ, ϕ] + 8 Sin[2 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 Sin[4 θ] vv(0,0,1,0) [r, t, θ, ϕ] -− 2 ⅇ2 ⅈ ϕ Sin[4 θ] ww(0,0,1,0) [r, t, θ, ϕ] + 7 vv(0,0,2,0) [r, t, θ, ϕ] -− 8 Cos[2 θ] vv(0,0,2,0) [r, t, θ, ϕ] + Cos[4 θ] vv(0,0,2,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ ww(0,0,2,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ Cos[4 θ] ww(0,0,2,0) [r, t, θ, ϕ] + 11 r vv(1,0,0,0) [r, t, θ, ϕ] -− 12 r Cos[2 θ] vv(1,0,0,0) [r, t, θ, ϕ] + r Cos[4 θ] vv(1,0,0,0) [r, t, θ, ϕ] -− 5 ⅇ2 ⅈ ϕ r ww(1,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r Cos[2 θ] ww(1,0,0,0) [r, t, θ, ϕ] + ⅇ2 ⅈ ϕ r Cos[4 θ] ww(1,0,0,0) [r, t, θ, ϕ] -− 4 r Sin[2 θ] vv(1,0,1,0) [r, t, θ, ϕ] + 2 r Sin[4 θ] vv(1,0,1,0) [r, t, θ, ϕ] -− 4 ⅇ2 ⅈ ϕ r Sin[2 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 2 ⅇ2 ⅈ ϕ r Sin[4 θ] ww(1,0,1,0) [r, t, θ, ϕ] + 5 r2 vv(2,0,0,0) [r, t, θ, ϕ] -− 4 r2 Cos[2 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− r2 Cos[4 θ] vv(2,0,0,0) [r, t, θ, ϕ] -− 3 ⅇ2 ⅈ ϕ r2 ww(2,0,0,0) [r, t, θ, ϕ] + 4 ⅇ2 ⅈ ϕ r2 Cos[2 θ] ww(2,0,0,0) [r, t, θ, ϕ] -− ⅇ2 ⅈ ϕ r2 Cos[4 θ] ww(2,0,0,0) [r, t, θ, ϕ] + 1 16 ⅇ2 ⅈ ϕ (2 Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww[r, t, θ, ϕ] -− 2 ⅈ (Cos[2 ϕ] + ⅈ Sin[2 ϕ]) ww(0,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[ϕ] ww(0,0,0,2) [r, t, θ, ϕ] + Sin[θ] Cos[θ] -− ⅈ Cos[ϕ]2 + Cos[ϕ] Sin[ϕ] + ⅈ Sin[ϕ]2 + Sin[θ]2 Sin[2 ϕ] ww(0,0,1,0) [r, t, θ, ϕ] -− Cos[θ] Cos[2 ϕ] ww(0,0,1,1) [r, t, θ, ϕ] -− Sin[θ] Cos[θ]2 Cos[ϕ] Sin[ϕ] ww(0,0,2,0) [ r, t, θ, ϕ] + r ⅈ Cos[ϕ]2 + ⅈ Cos[ϕ] Sin[θ]2 Sin[ϕ] -− Sin[ϕ]2  ww(1,0,0,0) [r, t, θ, ϕ] + Cos[2 ϕ] ww(1,0,0,1) [r, t, θ, ϕ] + Cos[ϕ] Sin[θ] Sin[ϕ] 2 Cos[θ] ww(1,0,1,0) [r, t, θ, ϕ] + r Sin[θ] ww(2,0,0,0) [r, t, θ, ϕ] (*⋆End Eqn (7.98)*⋆)




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