tensor analysis and curvilinear coordinates, appendices

1 Tensor Analysis and Curvilinear Coordinates: Appendices Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah 84103 last update: March 31, 2012 These Appendices belong with the document "Tensor Analysis and Curvilinear Coordinates". All Maple code referred to is available to any interested party for the asking. Appendix A: Reciprocal Base Vectors the Hard Way ........................................................................ 4 (a) Definition of E............................................................................................................ ..................... 4 (b) Simpler notation ........................................................................................................... ................... 5 (c) Generalized Cross Product of N-1 vectors of dimension N ............................................................ 5 (d) Missing Man Formation ...................................................................................................... ............ 7 (e) Apply this Notation to E.................................................................................................................. 7 (f) Compute E m • en............................................................................................................................... 8 (g) Compute E n • Em............................................................................................................................. 9 (h) Summary of relationship between the tangent and reciprocal base vectors .................................... 9 (i) Another Cross Product Notati on and another expression for E ..................................................... 10 Appendix B: The Geometry of Pa rallelepipeds in N dimensions..................................................... 11 (a) Preliminary: Equation of a plane in N dimensions....................................................................... 11 (b) N-pipeds and their F aces in Various Dimensions ......................................................................... 12 The 1-piped ..................................................................................................................................... 12 The 2-piped ..................................................................................................................................... 12 The 3-piped ..................................................................................................................................... 14 The N-piped .................................................................................................................... ................ 15 (c) The question of inward versus outward facing normal vectors..................................................... 16 (d) The Face Area and Volume of N-pipeds in Various Dimensions ................................................ 17 The 2-piped ..................................................................................................................................... 17 The 3-piped ..................................................................................................................................... 18 The 4-piped ..................................................................................................................................... 20 The N-piped .................................................................................................................... ................ 22 (e) Summary of Main Resu lts of this Appendix ................................................................................. 24 Appendix C: Elliptical Polar Coordinates ( N=2, non-orthogonal)................................................ 26 (a) Elliptical polar coordinates............................................................................................... ............. 26 (b) Forward coordinate lines................................................................................................... ............ 27 (c) Inverse coordinate lines................................................................................................... .............. 27 (d) Drawing a contravariant vector V in x-space: the meaning of V' n .............................................. 28 (e) Drawing a contravariant vector V' in x'-space: two "Views" ....................................................... 29 (f) Drawing the specific contravariant vector dx in x-space and x'-space .......................................... 32 (g) Study of how dx transforms in th e mapping between x-space and x'-space ................................. 32 (h) Derivation of the Jaco bian Integration Rule................................................................................ ..34 2 Appendix D. Tensor Densities and the ε tensor .................................................................................. 37 (a) Definition of a tensor density ............................................................................................. ........... 37 (b) A few facts about tensor densities................................................................................................. 38 (c) Theorem about Totally Antisymmetric Tensors: there is really only one: εabc......................... 40 (d) The contravariant ε tensor ............................................................................................................. 41 (e) Some facts about the ε tensor ........................................................................................................ 42 (f) The covariant ε tensor : repeat section (d) as if its weight were not known ................................. 44 (g) Generalized cross products................................................................................................. ........... 45 (h) The tensorial nature of curl B............................................................................................. ........... 45 (i) Tensor E as a weight 0 version of ε : three conventions................................................................ 46 (j) Representation of ε, εε and contracted εε as determinants............................................................ 49 (k) Covariant forms of th e previous section results ............................................................................ 55 Appendix E: Tensor Expansions: direct product, polyadic and operator notation...................... 57 (a) Direct Product Notation.................................................................................................... ............. 57 (b) Tensor Expans ions and Bases ....................................................................................................... 58 (c) Polyadic Notation .......................................................................................................................... 60 (d) Dyadic Products ............................................................................................................ ................ 61 (e) Transpose notation for dyadics............................................................................................. ......... 62 (f) Large and small dots used with dyadics..................................................................................... ....63 (g) Operators and Matrices for Rank-2 tensors.................................................................................. .63 Appendix F: The Affine Connection Γc ab and Covariant Derivatives ............................................ 68 (a) Definition and Interpretation of Γ : Γc ab = ec • (∂aeb) = Rc i(∂aRbi) ........................................ 68 (b) Identities of the form ( ∂aRd n) = – Re n Rd m (∂aRem) ...................................................................... 69 (c) Identities of the form ( ∂cgab) = – [gan Γ b cn + gbn Γa cn] ............................................................ 70 (d) Identity: Γd ab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab].................................................................... 71 (e) Picture D1 Context ......................................................................................................... ............... 73 (f) Relations between Γ and Γ ' ........................................................................................................... 74 (g) Statement and Proof of the Covariant Derivative Theorem .......................................................... 75 (h) Rule for raising any index on a covarian t derivative of a covariant tensor density....................... 80 (i) Examples of covari ant derivative expressions ............................................................................... 81 (j) The Leibnitz rule for the covariant derivative of the product of two tensor densities ................... 84 Appendix G: Expansion of ( ∇v) in curvilinear coordinates............................................................. 88 (a) Continuum Mechanics motivation............................................................................................. ....88 (b) Expansion of ∇ v on ei⊗ej by Method 1: Use fact that v b;a is a tensor....................................... 88 (c) Expansion of ∇v on ei⊗ej by Method 2: Use brute force. .......................................................... 90 (d) Expansion on e i⊗ej and e ^i⊗e^j..................................................................................................... 92 (e) Orthogonal coordinate systems .............................................................................................. ....... 93 (f) Maple evaluation of ( ∇v) in several coordinate systems ............................................................... 93 Appendix H: Expansion of div(T) in curvilinear coordinates.......................................................... 97 (a) Continuum Mechanics motivation............................................................................................. ....97 (b) Expansion of divT on e n by Method 1: Use fact that Tab ;α is a tensor......................................... 97 (c) Expansion of divT on e n by Method 2: Use brute force ............................................................... 98 (d) Adjustment for T expanded on ( e^i⊗e^j) and divT expanded on e^a............................................ 100 (e) Maple: divT in cylindri cal and spherical coordinates................................................................. 101 3 Appendix I : The Vector Laplacian in Spherical and Cylindrical Coordinates........................... 103 (a) The first met hod : a review............................................................................................... .......... 103 (b) The first method in spherical coordinates: Maple speaks .......................................................... 105 (c) The first method in spherical coordina tes: putting results in traditional form............................ 108 (d) The second method : Part I.......................................................................................................... 110 (e) The second method : Part II................................................................................................ ......... 111 (f) The second method in spherical coordinates: Maple speaks again............................................. 112 (g) Results for Cylindrical Coordinates from both methods............................................................. 115 4 Appendix A: Reciprocal Base Vectors the Hard Way Note : This Appendix is written in the development not ation, not the Standard Notation, though a few equations are translated to the latter form. The rules for translation to Standard Notation are E n → en Rij → Ri j Sij → Si j g¯'nm → g'nm g'nm → g'nm . Introduction In Section 6 of the main text the reciprocal base vectors are defined as E n ≡ g'ni ei , and the results given in that section, (en)k = Skn en • em = g¯'nm | en| = g¯'nn = h'n S = [ e1, e2, e3 .... eN ] ( En)i ≡ gia Rna En • Em = g'nm |En| = g'nn R = [E ¯1, E¯2, E¯3 .... E¯N ]T = g' na Sia en • Em = δn,m En ≡ g'ni ei en = g¯'ni Ei , are all applicable in the Picture A context with arbitrary metric tensors g' and g, This Appendix begins with a differe nt definition of something called Ek. Although the definition is meaningful in the general Picture A context, the object so defined only agrees with the Ek of Section 6 if x-space is Cartesian (g = 1). The reason can be traced to the fact that the dot product rule en • Em = δn,m is only valid for the Appendix A definition of Em when g = 1 because only then is a cross product orthogonal to all its component vectors. One application of the reciprocal base vectors is in the study of curvilinear coordinates where one always takes g = 1, and g' is then the curvilinear coordinates metric tensor of interest. Therefore, the reader should think of this Appendix in the context of Picture B (a) Definition of E n The reciprocal base vectors are defined in the following very strange looking and clumsy manner, 5 (Ek)α ≡ det(R) (-1)k-1 εαi1i2i3...ik...iN (e1)i1 (e2)i2 ...... ( ek)ik.......... ( eN)iN where N is the number of dimensi ons of the Cartesian x-space RN in which the vectors en and En exist. Notice that the ε subscript i k is "crossed out" and the same for factor ( ek)ik . Crossed out means they are simply missing, they are omitted. Thus, in the above expression there are N-1 implied summation indices (α is fixed) and there are N-1 factors of the form ( en)in . The object ε has N subscripts and is the "totally antisymmetric tensor" in N dimensions: ε123...N ≡ +1, and each time any two indices on ε are swapped, ε negates. For example, ε 1234 = 1 but ε1432= -1. If two indices are the same, then ε = 0. (b) Simpler notation To avoid dealing with subscripts on subscripts, one can rewrite the above definition in a less precise but simpler notation ( Ek)α ≡ det(R) (-1)k-1εαabc...x (e1)a(e2)b ...... ( eN)x // κ(k) and ( ek)κ are missing In this notation, subscript x stands for the N th letter of the alphabet (imagine N ≤ 26). If κ is the kth letter of the alphabet, then κ is missing from the indices on ε, and the factor ( ek)κ is missing from the product of factors. For example, if k = 2, then summation index κ = b is missing from the ε. Now take the ε subscript α and slide it right to the "hole" where κ is missing, picking up a minus sign for each step of this slide. Moving k-1 positions results in (-1)k-1. Thus the above becomes, ( E k)α ≡ det(R) εabc..α..x (e1)a(e2)b ...... ( eN)x // (e k)κ is missing, α in κ position (the kth) Example : For N = 3 the above becomes, ( E1)α ≡ det(R) εαbc(e2)b(e3)c => E1 = det(R) e 2 x e3 a is missing ( E2)α ≡ det(R) εaαc(e1)a(e3)c => E2 = det(R) e 3 x e1 b is missing ( E3)α ≡ det(R) εabα(e1)a(e2)b => E3 = det(R) e 1 x e2 c is missing and the results are cyclic. Here is a detail from the middle line εaαc(e1)a(e3)c = – εαac(e1)a(e3)c = + εαca(e1)a(e3)c = εαca(e3)c (e1)a = [ e3 x e1]α (c) Generalized Cross Product of N-1 vectors of dimension N One can defi ne a generalized "cross product" of N-1 v ectors, each of dimension N, in this fashion: Qa ≡ εabc...x BbCcDd.....Xx where x and X represent the N th letter of the alphabet. The ε object is again the totally antisymmetric tensor with N indices. In vector notation one writes this symbolically as 6 Q = B x C x D x ... x X / N-1 factors, N-2 crosses This vector notation is defined by the previous line. The vector Q is orthogonal to all the vectors from which it is constructed! For example (here is the point where Q • C ≡ gabQaCb needs to be QaCa, so g = 1 is required in x-space) Q • C = CaQa = Ca εabc...x BbCcDd.....Xx = BbDd...Xx { εabc...x CaCc } But {..} is the contraction of something symmetric under a ↔c (CaCc) with something antisymmetric under a↔c (εαabc...x ) and therefore {..} = 0. In general S acAac = Sca Aca // relabel both dummy summation indices = S ac (-Aac) // S is Symmetric, A is antisymmetric = - S ac Aac // = the negative of the starting expression = 0 Similarly, Q•A = 0, Q•B = 0 and so on. Swapping the position of any two vector s in the generalized cross product causes Q to change sign. For example, swapping B and C, Qa ≡ εabc...x CbBcDd.....Xx = εacb...x CcBbDd.....Xx // b ↔ c = - εabc...x BbCcDd.....Xx = -Q a // swap indices on ε Thus, the notions of orthogonality and interchange are consistent with the regular Q = B x C cross product for N=3. When N=2, one must be a little careful with this notation. The component equation is Q a ≡ εab Bb => Q 1 = B2 and Q2 = -B1 One might be tempted to express the vector equation as Q = B since there are no "no crosses". This vector equation is wrong , while the component equation is correct. One can rescue the vector notation by a simple trick. When N=2 the vector B can be represented of course as B = B11^ + B2 2^. Imagine this 2D space to be embedded in the usual 3D space with a third axis 3^. Then consider this 3D cross product: Q = B x 3^ => Q a ≡ εabc Bb(3^)c = εabc Bbδ3,c = εab3 Bb = εabBb Thus, this trick reproduces the correct component equation, and it makes more obvious the fact that Q is orthogonal to B. Summary : The generalized cross product Q of N-1 vectors each of dimension N can be expressed in both component and vector notation: Q a ≡ εabc...x BbCcDd.....Xx Q = B x C x D x ... x X / N-1 factors, N-2 crosses 7 Q is orthogonal to all the vectors from which it is composed. Swapping any two vectors negates Q. When N=2, one can rescue the otherwise failing v ector notation by thi nking of it as saying Q = B x 3^. Comment: Notice that Q = B x C x D is defined for 4-vectors only. This is a completely different animal from the object Q = B x ( C x D ) which is defined for 3-vectors onl y. This latter object contains two ε factors, while the former only one. (d) Missing Man Formation We now make a small variation in the no tation. Start with the above equation, Q a ≡ εabc...x BbCcDd.....Xx , then change a to α, back up all the Latin letters by one (but leave the last as "unknown" x), and assume that some subscript κ and factor K κ are "missing". The result is, Qα ≡ εαac...x AaBbCc.....Xx // κ and Kκ are missing There are still N-1 factors, and one can still write this in vector notation Q = A x B x C x ... x X // K is missing and of course it is still true that Q •C = 0, etc. For N=2 the vector notation is rescued as in (c) above. (e) Apply this Notation to E Compare the above Q α to the section (a) definition of ( Ek)α , ( Ek)α ≡ det(R) (-1)k-1{ εαabc...x (e1)a(e2)b ...... ( eN)x } // κ (k) and ( ek)κ are missing; N ≥ 2 Therefore, the definition of E κ for N > 2 can be written in this vector notation, E k ≡ det(R) (-1)k-1 e1 x e2 x ......x e N // ek missing; N > 2 The reciprocal base vector E k is thus orthogonal to all the tangent base vectors from which it is constructed (remember ek is missing)! For example, for N=3 the three E vectors are given by E 1 = det(R) (-1)1-1 e2 x e3 = det(R) e2 x e3 E2 = det(R) (-1)2-1 e1 x e3 = det(R) e3 x e1 E3 = det(R) (-1)3-1 e1 x e2 = det(R) e1 x e2 which agrees with the results quoted above. For N =2 ( E's label corresponds to the missing e's label ), 8 E1 = det(R) (-1)1-1 e2 x 3^ = det(R) e2 x 3^ or ( E1)k = det(R) εka(e2)a E2 = det(R) (-1)2-1 e1 x 3^ = - det(R) e1 x 3^ or ( E2)k = -det(R) ε ka(e1)a One can combine these two lines in to one as follows ( eg, k = 1, then 3-1 = 2, etc) Ek = det(R) (-1)k-1 e3-k x 3^ = det(R) e3-k x 3^ or ( E1)k = det(R) (-1)k-1εka(e3-k)a The vector "trick" notation shows that E1•e2 = 0 and E2•e1 = 0, E1•e2 = det(R) e2 x 3^ • e2 = 0 E2•e1 = -det(R) e 1 x 3^ • e1 = 0 and also E 1•e1 = det(R) εka(e2)a (e1)k = det(R)det[ e1, e2] = det(R)det(S) = 1 E2•e2 = -det(R) ε ka(e1)a (e2)k = -det(R)det[ e2, e1] = det(R)det(S) = 1 It is shown next that these N=2 results are special cases of a general fact: E m • en = δm,n . Section 5 (j) showed that e m • en = g¯'mn . The other two dot products are now considered. (f) Compute E m • en One can now compute, for general N, E k • ek = ( Ek)α(ek)α = { det(R) (-1)k-1εαabc...x (e1)a(e2)b ...... ( eN)x } ( ek)α k is missing Slide α to the right in the ε subscript field and put it into the hole of the missing subscript κ , picking up (-1) k-1. At the same time, move the ( eκ)α to the left and position it in its proper place in the product of factors, Ek • ek = ( Ek)α(ek)α = { det(R) εabc..α..x (e1)a(e2)b ... (eκ)α ... (eN)x } = det(R) det [ e1, e2, e3 .... eN ] = det(R) det(S) = 1 // since RS = 1 We already know that E k is orthogonal to all the en which form the generalized cross product, therefore Em • en = δm,n which is the "duality relation" discussed more generally in Section 6 (b) 9 (g) Compute E n • Em Since the vectors { en } are linearly independent and thus form a basis in RN, Em can be expanded onto the en , E m = Σn An(m) en δm,k = Em • ek = Σn An(m) en • ek = Σn An(m) g¯'nk Multiplying both sides by g' ki and summing on k gives LHS = Σk g'ki δm,k = g'mi RHS = Σn An(m) (Σk g¯'nk g'ki) = Σn An(m) (g¯'g')ni = Σn An(m)δn,i = Ai(m) Therefore A i(m) = g'mi so, E m = Σn An(m) en = Σn g'mn en which is to say E k is this linear combination of the ei (this is the definition used in Section 6 (a)) E k = Σi g'ki ei = g'ki ei // implied sum on i // Std Notation: ek = Σi g'ki ei which may be compared with the previous result E k ≡ det(R) (-1)k-1 e1 x e2 x ......x e N // ek missing; It seems rather impressive that these two dissimilar ways of writing E are equal. Finally, En • Em = En • (g'mi ei) = g'mi (En • ei) = g'mi δn,i = g'mn = g'nm // recall g' symmetric (h) Summary of relationship between the tangent and reciproca l base vectors en • em = g¯'nm En • Em = g'nm en • Em = δn,m En = Σi g'ni ei en = Σi g¯'ni Ei g¯' = g'-1 Although these results have just been derived in the Pi cture B context, they are also valid in the more general Picture A context, as shown in Section 6 in which the equation En = Σi g'ni ei is used as the definition of En . As a reminder, the cross product expression for En is only valid in Picture B. In Standard Notation, the summa ry above can be restated as e n • em = g'nm en • em = g'nm en • em = δnm en = Σi g'ni ei en = Σi g'ni ei g'ab = (g'ab) -1 10 (i) Another Cross Product Notation and another expression for E Go back to th e general cross product of N-1 vectors each of dimension N, Q = B x C x D x ... x X // N-1 factors, N-2 crosses Replace B,C,D ... by vectors A(n), Q = A (1) x A(2) x A(3) x ... x A(N-1) // N-1 factors, N-2 crosses It is convenient to write this as Q = Π x i=1N-1 A(i) = Πx i A(i) where in the second form it is understood that i takes on all values i = 1 to N-1. The superscript x means that this is not a regular product, it is our generalized cross product. This Πx symbol also implies correct handling of the special case N=2 such that Q = Πx i=11 A(i) = A(1) x 3^ // ≠ A(1) as discussed in section (c) above. This same Πx notation can be applied to the "missing man formation" of (d) above. Suppose Q = A (1) x A(2) x A(3) x ... x A(N) // A(n) is missing One can write this as Q = Π x i=1..N,i ≠n A(i) ≡ Πx i≠n A(i) And of course this idea can be applied to the expression for Ek Ek = det(R) (-1)k-1 e1 x e2 x ......x e N // ek missing; E k = det(R) (-1)k-1 Πx i≠k ei Once again, for N=2 the Π x symbol implies that ( ek = "missing", e 3-k = the one not missing) Πx i≠k ei = Πx i=1..2,i ≠k ei = e3-k x 3^ E k = det(R) (-1)k-1 e3-k x 3^ which is the "trick" notation of section (c) above for the N=2 case. 11 Appendix B: The Geometry of Parallelepipeds in N dimensions Introduction This Appendi x presents a simple method for construc ting an N dimensional parallelepiped, which name we shorten to "N-piped". It is found that an N-piped has 2N vertices and N pairs of faces for a total of 2N faces, and the locus of points that make up each of these faces is stated. Each face of an N-piped is in fact an (N-1)-piped which has 2N-1 vertices and is planar in N dimensions (meaning it lies on an N-1 dimensional flat surface). The two faces which make up each face pair lie on parallel planes in RN. For example, for N=3 each face is a 2-piped having 23-1 = 4 vertices, and there are N=3 face pairs for a total of 6 faces, and each pair of faces is planar in 3 dimensions. For N=4 there are 4 pairs of faces for a total of 8 faces. Each face is a 3-piped having 24-1 = 8 vertices. For example, one would say that each face of a 4-cube is a 3-cube. It is not intuitively obvious that two faces each of which is a regular cube can in fact lie on surfaces which are planar and parallel in 4 dimensions, but we show how this works below. It is then shown that, if the N-piped is spanned by the N tangent base vectors en of Section 3, the normal vectors for the pairs of parallel faces are just the reciprocal base vectors En of Section 6. Section (d) focuses on the area and volume of N-pipeds in various dimensions, and simple expressions for the volume and vector areas of the faces of an N-piped are obtained. Rather than just state the results in N dimensi ons, we attempt an inductive approach to provide motivation for the N dimensional results. In this appro ach, cases N = 2,3.. are treated with nearly identical boilerplate templates to build up the inductive case. All major results of this Appendix are concisely st ated in Summary section (e). Since this is a very long section (~15 p), a reader not interested in details would do well to simply read that summary and skip the rest of this Appendix. (a) Preliminary: Equation of a plane in N dimensions Consider an a rbitrary plane drawn in N space which do es not pass through the origin. There is some point on that plane which lies closer to the orig in than all other points on the plane. Let p be a vector from the origin to that closest point, and let r represent a point lying on the plane, Since p is normal to the plane, and since r-p is a vector lying in the plane, it follows that p•(r-p) = 0 => r•p = p2 => r•p^ = p Therefore, one way to write the equation of a plane in N dimensions is 12 r•p^ = p r = (x1, x2, .....xN) where p^ is the unit vector normal to the plane which points "away from the origin", and where p > 0 is the distance of closest approach of the plane to the origin. In the limit p →0, the plane passes through the origin and the equation is then r•p^ = 0 where p^ is either normal to the plane. (b) N-pipeds and their Faces in Various Dimensions The 1-piped Start with N = 1 where the piped is some arbitrary line segment e1 in direction e^1 having length e 1, with one end affixed to the origin of the real axis. N=1 rvolume1 = α1 e1 0 ≤ α1 ≤ 1 This piped has two vertices located at v 1 = 0 and v2 = e1. These two vertices are also the "faces" of this 1-piped, so there are two faces (one pair of faces) . These faces are 0 dimensi onal and therefore don't point in any direction (they are the endpoints of the segment). The 1-piped is a piece of a plane in 1 dimension (a line). One can think of the vertex at the origin as the "generator 0- piped" and the other vertex as the partner face of the generator, in the sense of the generator idea described below. The volume of this 1-piped is e 1. The 2-piped Now add another dimension, goin g to N=2. Introduce a unit vector e2 in some arbitrary direction in R2 other than e 1 so that e1 and e2 are linearly independent. Take the 1-piped described above (line segment) and translate it by e2 to create a new copy of the line segment. The original 1-piped we call the generator piped, and the copy is the partner of the generator piped which, it will be shown, lies on a plane (a 1-plane = line) which is parallel to the plane of the generator piped, but its plane does not pass through the origin. In N=2 dimensions, the generator 1-piped and its partner are now "faces" of a 2-dimensional object, a parallelogram = a 2-piped. Draw line segments from a ll the vertices of the generator piped to matching vertices of its partner piped (add 2 line segments) to make 2 additional "side" faces. One of these faces necessarily touches the origin, and the other face does not. Faces always occur in parallel pairs one of which touches the origin, and one of which does not, the latter we will call the "partner" face. For our 2- piped, each face is a 1-piped. There are now four faces, each is a line segment. 13 The loci of the 2-piped's volume and of its four 1-piped faces are given by rvolume2 = α1e1 + α2 e2 0 ≤ α1,α2 ≤ 1 r face2 = α1e1 0 ≤ α1 ≤ 1 // the generator face rface2p = α1e1 + e2 0 ≤ α1 ≤ 1 // partner of the generator face rface1 = α2e2 0 ≤ α2 ≤ 1 // side face touching the origin rface1p = α2e2 + e1 0 ≤ α2 ≤ 1 // partner of the above side face The origin-touching faces are numbered using the index of the e n vector that does not appear in the locus for the face. This seems strange but for N > 2 it will be clear why this is done. It is possible to construct vectors E1 and E2 as linear combinations of e1 and e2 such that the following is true (see Section 6 (b)) E i• ej = δi,j / / Ek = Σi=12 g'ki ei , see Section 6 (a) If one interprets the e n vectors as tangent base vectors for some transformation F, then the two vectors En are the corresponding reciprocal base vectors which are discussed in Section 6 and Appendix A. Consider now these dot products: E 2 • rface2 = E2 • α1e1 = 0 => E^2 • rface2 = 0 E2 • rface2p = E2 • [ α1e1+ e2] = 1 => E^2 • rface2p = 1/E2 The first line says (section (a) above) that face 2 lies on a plane which passes through the origin and which has normal vector E^2. The second line says that face 2p has the same normal and its plane is therefore parallel to face 1 but misses the origin by distance 1/|E 1|. Similarly, E 1 • rface1 = E1 • α2e2 = 0 => E^1 • rface1 = 0 E1 • rface1p = E1 • [ α2e2+ e1] = 1 => E^1 • rface1p = 1/E1 These two faces are also parallel, both having normal E^1. The first touches the origin while the partner's plane misses the origin by distance 1/|E 1| . 14 The conclusions that En is normal to face n and that the pair of faces n and np are parallel do not depend on the specific upper endpoints of the ranges of α1 and α2 which happen to be given as 1 above. This seems pretty obvious since rescaling the edges of a parallelogram does not affect its normal vector. The 3-piped Now add ano ther dimension, going to N=3. Introduce a unit vector e^3 in some arbitrary direction in R3 so that ( e^1,e^2,e^3) are linearly independent. Take the 2-piped described above (parallelogram) and translate it by distance e 3 in the e^3 direction to create a new copy of the 2-piped. The original 2-piped we call the generator piped, and the copy is the partner of the generator piped which, as w ill now be shown, lies on a plane which is parallel to that of the generator piped, but which does not pass through the origin. In N=3 dimensions, the generator 2-piped and its part ner are now "faces" of a 3-dimensional object, a parallelepiped = a 3-piped. Draw line segments from all 22 vertices of the generator piped to the corresponding vertices of its partner piped (add 4 line segments), to get 4 additional side faces. Two of these faces necessarily touch the origin, and the other tw o do not. For the 3-piped, each face is a 2-piped. There are now 2*3 = 6 faces, each is a 2-piped. The loci of the 3-piped's volume and of its six 2-piped faces are given by rvolume3 = α1e1 + α2 e2 + α3 e3 0 ≤ α1,α2,α3 ≤ 1 r face3 = α1e1 + α2 e2 0 ≤ α1,α2 ≤ 1 // the generator face rface3p = α1e1 + α2 e2 + e3 0 ≤ α1,α2 ≤ 1 // partner face to the above rface2 = α1e1 + α3 e3 0 ≤ α1,α3 ≤ 1 // the generator face rface2p = α1e1 + α3 e3 + e2 0 ≤ α1,α3 ≤ 1 // partner face to the above rface1 = α2e2 + α3 e3 0 ≤ α2,α3 ≤ 1 // the generator face rface1p = α2e2 + α3 e3 + e1 0 ≤ α2,α3 ≤ 1 // partner face to the above Notice that the partner face locus is created from the non-partner face by adding "the other" base vector. For example, face 3 is "spanned" by base vectors e1 and e2 so e3 is added to get the partner. A partner is just a copy of the non-partner which is translated by a constant vector. The above results can be summarized in this concise manner: 15 rvolume3 = Σnαnen 0 ≤ αn ≤ 1 rface(i) = Σn≠iαnen 0 ≤ αn ≤ 1 i = 1,2,3 rface(ip) = Σn≠iαne + ei 0 ≤ αn ≤ 1 i = 1,2,3 It is possible to construct vectors E1, E2, E3 as linear combinations of e 1, e2, e3 such that the following is true (see Section 6 (b)) Ei• ej = δij / / Ek = Σi g'ki ei , see see Section 6 (a) If one interprets the en vectors as tangent base vectors, then the three vectors En are the corresponding reciprocal base vectors which are discussed in S ection 6 and Appendix A. Consider now these dot products: E1• rface1 = E1• [ α2e2 + α3 e3] = 0 => E^1• rface1 = 0 E1• rface1p = E1• [α2e2 + α3 e3 + e1] = 1 => E^1• rface1p = 1/E1 The first line says that face 1 lies on a plane which passes through the origin and which has normal vector E^1. The second line says that face 1p has the same normal and its plane is therefore parallel to face 1 but misses the origin by distance 1/|E 1| A similar pair of equations obtains for each of the other face pairs. The N-piped Now add ano ther dimension, going from N-1 to N. Introduce a unit vector e^N in some arbitrary direction in RN so that ( e^1... e^N ) are linearly independent. Take the (N-1 )-piped described above and translate it by distance e N in the e^N direction to create a new copy of the (N-1)-piped. The original (N-1)-piped we call the generator piped, and the copy is the partner of the generator piped whic h, it will be shown, lies on a plane which is parallel to that of the generator piped, but which does not p ass through the origin. The generator (N-1)-piped and its partner are now "faces" of a N-dimensional object, an N-piped. Adding this partner piped doubles the total vertex count. Draw line segments from all 2N-1 vertices of the generator piped to the corresponding vertices of its partner piped to get 2N-2 additional side faces for a total now of 2N faces. There are N pairs of "faces" because there are N ways to omit a single ei from the list of vectors which span a face, so including the partner faces an N- piped has 2N faces in total. Half of these faces necessarily touch the origin, and the other ha lf do not. Each face is an (N-1)-piped. It is convenient to refer to the partner face of a pair as "the far face" and the other one, which touches the origin, as "the near face". The loci of the N-piped's volume and of its 2N (N-1)-piped faces are given by: r volumeN = Σnαnen 0 ≤ αn ≤ 1 rface(i) = Σn≠iαnen 0 ≤ αn ≤ 1 i = 1,2...N rface(ip) = Σn≠iαnen + ei 0 ≤ αn ≤ 1 i = 1,2...N 16 Notice that the partner face locus is created from the non-partner face by adding "the other" base vector. For example, face i is "spanned" by base vectors en n ≠ i, so it is ei that one adds to get the partner. A partner is just a copy of the non-partner which is translated by a constant vector. It is possible to construct vectors E 1...EN as linear combinations of e1.. eN such that the following is true (see Section 6 (b)) E i• ej = δij / / Ek = Σi g'ki ei If one interprets the e n vectors as tangent base vectors, then the N vectors En are the corresponding reciprocal base vectors which are discussed in S ection 6 and Appendix A. Consider now these dot products: Ei• rface(i) = Ei• [Σn≠iαnen] = 0 => E^i• rface(i) = 0 Ei• rface(ip) = Ei• [Σn≠iαne + ei] = 1 => E^i• rface(ip) = 1/Ei The first line says that face i lies on a plane which passes through the origin and which has normal vector E^i. The second line says that face ip has the same normal and its plane is therefore parallel to face i but misses the origin by distance 1/|E i| (c) The question of inward versus ou tward facing normal vectors. It has been shown above t hat, for an N-piped, the pair of faces i and ip has normal vector Ei. For one of these faces, Ei will be an outward directed normal, while for the other it will be an inward directed normal. One might like to know which is which. Here is one way to find out. First, construct these three vectors (piped) center = Σn(1/2) en // vector from origin to piped center (face i) center = Σn≠i(1/2) en // vector from origin to center of face i (face ip) center = Σn≠i(1/2) en + ei // vector from origin to center of face ip Construct vectors from piped center to face centers ( results here are fairly obvious) (face i) center - (piped) center = [Σn≠i(1/2) en] - Σn(1/2) en = - (1/2) ei (face ip) center - (piped) center = [Σn≠i(1/2) en+ ei ] - Σn(1/2) en = + (1/2) ei Then compute E i • {(face i) center - (piped) center } = Ei • [- (1/2) ei ] = -(1/2) < 0 Ei • {(face ip) center - (piped) center } = Ei • [+ (1/2) ei ] = +(1/2) > 0 17 One may conclude that Ei is an outward pointing normal for face ip (far face). Therefore, - Ei is an outward pointing normal for face i, which recall is the face which touches the origin (near face). (d) The Face Area and Volume of N-pipeds in Various Dimensions We e mbark now on another long marc h to inductively arrive at results for the general N case. Tracing the first few cases N = 2,3,4 and then extrapolating to N = N probably gives more insight than a formal induction proof which is not attempted here. Each case below is treated with the same boilerplate template which first treats Face Area and then Volume. The 2-piped Face Ar ea. The area of a 2-piped face (a line segment) is just the length of the edge which is the face, A 1 = |e2| A2 = |e1| where here we maintain the plan of labeling an area by the index of the spanning vector which is omitted in making the area. The vector areas can be written, based on the work above, A 1 = |e2| E^1 / / Ek = Σi g'ki ei , see Appendix A (g) A2 = |e1| E^2 and these vectors are out-facing for faces 1p and 2p. We claim that both these results can be expressed in a single formula A n = |det(S)| En . One can see that the direction is correct for n = 1,2, so it is just a question of verifying the magnitude. One must show that |det(S)| | E 1| = | e2| and |det(S)| | E2| = | e1| or |E k| = |det(R)| | e3-k| k=1,2 // RS = 1 Using the N=2 trick notation from Appendix A (c), E k = det(R) (-1)k-1 e3-k x 3^ so that | E k | = | det(R)| | e3-k x 3^| = |det(R)| | e3-k| k = 1,2 since e 3-k and 3^ are perpendicular. QED. We stress the formula An = |det(S)| En because it will turn out that this is valid for all N ≥ 2 . 18 One can restate A n = |det(S)| En using the cross product notation presented in Appendix A (i): An = |det(S)| En = |det(S)| det(R) (-1)n-1 Πx i≠n ei = σ (-1)n-1 Πx i≠n ei σ ≡ sign(det(S)) = sign(det(R)) Volume . The volume of a 2-piped is the base times the he ight of a parallelogram, familiarly given as by the cross product of the edges, volume(2) = | e 1 x e2 | = | εab (e1)a(e2)b | = | det [ e1, e2 ] | = | det(S) | where S is the linearized transformation matrix for N=2, see Section 2. Of course strictly in N=2 the notation e1 x e2 has no meaning, so one has to imagine a 3^ dimension to give it meaning. The second form does have a meaning for N=2, and that meaning is |( e1)1 (e2)2 – (e1)2 (e2)1|. The 3-piped Face Ar ea: The faces of a 3-piped are 2-pipeds. For N=2, the 2-piped volume was volume(2) = | εab (e1)a(e2)b | , where e 1 and e2 were 2D vectors. For the 2-piped which is "face 3" of the 3-piped -- a "near" face which touches the origin of the 3D skewed en coordinate system -- vectors e1 and e2 are 3D vectors. The first 2 components of each of these 3D vectors ar e the same as the components of the 2D ei vectors, while the 3rd components are both 0. This is so because face 3 li es in a plane defined by this 3rd component being 0. The volume(2) formula expressed in terms of these new 3D vectors is therefore | εab3 (e1)a(e2)b|, where ε is now a 3D ε tensor. The conclusion is that A 3 = |εab3 (e1)a(e2)b| and this then is the scalar area of both face 3 and its partner face 3p, the far face. Similar arguments would then support these other area expressions A 1 = |ε1ab (e2)a(e3)b| A2 = |εa2b (e3)a(e1)b| Since indices on ε can be swapped for free due to the absolu te value, the non-summed index can be put first in all three cases and one may then conclude that A1 = |e2 x e3| face 1 and face 1p A2 = |e3 x e1| face 2 and face 2p A3 = |e1 x e2| face 3 and face 3p 19 In Appendix A (e) it was shown that E1 = det(R) e2 x e3 so that e 2 x e3 lines up with E1. Regardless of the sign of det(R), we define the vector areas to point in the + E^n directions. Thus, A1 ≡ |e2 x e3| E^1 face 1p out-facing A2 ≡ |e3 x e1| E^2 face 2p out-facing A3 ≡ |e1 x e2| E^3 face 3p out-facing These equations can be combined into the following single formula A n = |e1 x ... x e3| E^n // en missing where the ei are reordered for free due to the absolute value signs. But Appendix A says E n = det(R) (-1)n-1 e1 x ... x e3 // en missing so |E n| = | det(R) | | e1 x ... x e3 | // en missing Thus, An = E^n |En| / |det(R)| = |det(S)| En = |det(S)| det(R) (-1)n-1e1 x ... x e3 // en missing = σ (-1)n-1e1 x ... x e3 / / en missing w h e r e σ ≡ sign[det(S)] = sign[det(R)] To summarize, for N=3 one has A n = |det(S)| E n = σ (-1)n-1e1 x ... x e3 // en missing = σ (-1)n-1 Πx i≠n ei where the last line uses the shorthand notation of Appendix A (i). These expressions have the same form as those of the 2-piped. Volume. The volume of a 3-piped (with each of the above An in turn treated as the base) is base times height, so volume(3) = | A 1 • e1 | = | A2 • e2 | = | A3 • e3 | or volume(3) = | e 2 x e3 • e1 | = | e3 x e1 • e2 | = | e1 x e2 • e3 | Here is a drawing showing the last case ( σ = +1), where "base" is A 3 = | e1 x e2 | and "height" is e 3 cosθ , 20 Using ε notation one can write e3 • e1 x e2 = (e3)i εijk(e1)j(e2)k = εijk (e1)j(e2)k (e3)i = εjki (e1)j(e2)k(e3)i = det [ e1, e2, e3] = det(S) so that volume(3) = | e 3 • e1 x e2 | = | εabc (e1)a(e2)b(e3)c | = | det [ e1, e2, e3] | = | det(S) | These expressions have the same form as those of the 2-piped. The 4-piped Face Ar ea: The faces of a 4-piped are 3-pipeds. For N=3, the 3-piped volume was volume(3) = | ε abc (e1)a(e2)b(e3)c | where e1,e2,e3 were 3D vectors. For the 3-piped which is "f ace 4" of the 4-piped -- a "near" face which touches the origin of the 4D skewed en coordinate system -- vectors e1,e2,e3 are 4D vectors. The first 3 components of each of these 4D vectors ar e the same as the components of the 3D ei vectors, while the 4th components are all 0. This is so because face 4 li es in a plane defined by this 4th component being 0. The volume(3) formula expressed in terms of these new 4D vectors is therefore | ε abc4 (e1)a(e2)b(e3)c |, where ε is now a 4D ε tensor. The conclusion is that A4 = | εabc4 (e1)a(e2)b(e3)c | and this then is the scalar area of both face 4 and its partner face 4p, the far face. Similar arguments would then support these other area expressions A 1 = | ε1abc (e2)a(e3)b(e4)c | A2 = | εa2bc (e3)a(e4)b(e1)c | A3 = | εab3c (e4)a(e1)b(e2)c | 21 Since indices on ε can be swapped for free due to the absolu te value, the non-summed index can be put first in all three cases and one may then conclude that A1 = |e2 x e3 x e4| face 1 and face 1p A2 = |e3 x e4 x e1| face 2 and face 2p A3 = |e4 x e1 x e2| face 3 and face 3p A4 = |e1 x e2 x e3| face 4 and face 4p where, as discussed in Appendix A (c), Q = A x B x C is defined by Q k = εkabc AaBbCc . In Appendix A (e) it was shown that E 1 = det(R) e2 x e3 x e4 so that e 2 x e3 x e4 lines up with E 1. Regardless of the sign of det(R), we define the vector areas to point in the + E^n directions. Thus An = |e1 x ... x e3| E^n // en missing n = 1,2,3,4 where the ei are reordered for free due to the absolute value signs. But Appendix A says E n = det(R) (-1)n-1 e1 x ... x e4 // en missing so |E n| = | det(R) | | e1 x ... x e4 | // en missing Thus, An = |En| E^n / |det(R)| = |det(S)| En = |det(S)| det(R) (-1)n-1e1 x ... x e4 // en missing = σ (-1)n-1e1 x ... x e4 // en missing σ ≡ sign[det(S)] = sign[det(R)] To summarize, for N=4 one has A n = |det(S)| E n = σ (-1)n-1e1 x ... x e4 // en missing = σ (-1)n-1 Πx i≠n ei σ ≡ sign[det(S)] = sign[det(R)] These expressions have the same form as those of the 2-piped and the 3-piped. Volume . The volume of a 4-piped (with each of the above An in turn treated as the base) is base times height, so volume(4) = | A 1 • e1 | = | A2 • e2 | = | A3 • e3 | = | A4 • e4 | or volume(4) = | e 2 x e3 x e4 • e1 | = | e3 x e4 x e1 • e2 | = | e4 x e4 x e1 • e3 | = | e1 x e4 x e2 • e4 | Using ε notation one can write the first case as 22 e2 x e3 x e4 • e1 = (e1)a εabcd(e2)b(e3)c(e4)d = εabcd(e1)a(e2)b(e3)c(e4)d = d e t [ e1, e2, e3, e4] = det(S) so that volume(4) = | det(S) | = | det [ e 1, e2, e3, e4] | = | εabcd(e1)a(e2)b(e3)c(e4)d | These expressions have the same form as those of the 2-piped and the 3-piped. The N-piped Face Are a: The faces of a N-piped are (N-1)-pipeds. If there had been an (N-1)-piped section prior to this one, the volume formula there would have been volume(N-1) = | ε abc...x (e1)a(e2)b... (eN-1)x | where e1,e2,...eN-1 were (N-1)D vectors. For the N-piped whic h is "face N" of the N-piped -- a "near" face which touches the origin of the ND skewed en coordinate system -- vectors e1,e2,...eN-1 are ND vectors. The first N-1 components of each of these ND vectors are the same as the components of the (N-1)D e i vectors, while the Nth components are all 0. This is so because face N lies in a plane defined by this Nth component being 0. The volume(N-1) form ula expressed in terms of these new ND vectors is therefore | εabc...xN (e1)a(e2)b... (eN-1)x |, where ε is now an ND ε tensor. The conclusion is that AN = | εabc...xN (e1)a(e2)b... (eN-1)x | and this then is the scalar area of both face N and its partner face Np, the far face. Similar arguments would then support similar expressions for the other faces, for example, A 1 = | ε1abc...x (e2)a(e3)b... (eN-1)w (eN)x | A2 = | εa2bc...x (e3)a(e4)b... (eN)w (e1)x | Since indices on ε can be swapped for free due to the absolu te value, the non-summed index can be put first in all three cases and one may then conclude that A 1 = |e2 x e3 x e4 x e5... x eN| face 1 and face 1p A2 = |e3 x e4 x e5 x e6... x e1| face 2 and face 2p A3 = |e4 x e5 x e6 x e7... x e2| face 3 and face 3p ... AN = |e5 x e6 x e7 x e8.. x e3| face N and face Np where, as discussed in Appendix A (c), Q = A x B x C.... X is defined by Q k = εkabc...x AaBbCc .....Xx 23 In Appendix A (e) it was shown that E1 = det(R) e2 x e3 ... eN so that e2 x e3 ... eN lines up with E1. Regardless of the sign of det(R), we define the vector areas to point in the + E^n directions. Thus An = |e1 x ... x eN| E^n // en missing n = 1,2,3...N where the ei are reordered for free due to the absolute value signs. But Appendix A says E n = det(R) (-1)n-1 e1 x ... x eN // en missing so |E n| = | det(R) | | e1 x ... x eN | // en missing Thus, A n = |En| E^n / |det(R)| = |det(S)| En = |det(S)| det(R) (-1)n-1e1 x ... x eN // en missing = σ (-1)n-1e1 x ... x eN // en missing σ ≡ sign[det(S)] = sign[det(R)] To summarize, for N=N one has A n = |det(S)| E n = σ (-1)n-1e1 x ... x eN // en missing = σ (-1)n-1 Πx i≠n ei σ ≡ sign[det(S)] = sign[det(R)] These expressions have the same form as those of the 2-piped, the 3-piped and the 4-piped. Volume . The volume of a N-piped (with each of the above An in turn treated as the base) is base times height, so volume(N) = | A1 • e1 | = | A2 • e2 | = ... = | AN • eN | or volume(N) = | e 2 x e3 x e4....eN • e1 | = ... Using ε notation one can write the first case as e2 x e3 x e4...eN • e1 = (e 1)a εabc...x (e2)b(e3)c.....(eN)x = εabc...x (e1)a(e2)b(e3)c....(eN)x = d e t [ e1, e2, e3, ....eN] = det(S) where x is the N th letter of the alphabet, so that volume(N) = | det(S) | = | det [ e 1, e2, e3, ....eN] | = | εabc...x (e1)a(e2)b(e3)c....(eN)x | These expressions have the same form as those of the 2-piped, the 3-piped and the 4-piped. This result is also consistent with the volume(N-1) expression stated above. 24 (e) Summary of Main Results of this Appendix 1. One way to write the equation of a plane in N dimensions is r•p^ = p r = (x1, x2, .....xn) where p^ is the unit vector normal to the plane which points "away from the origin", and where p > 0 is the distance of closest approach of the plane to the origin. In the limit p →0, the plane passes through the origin and the equation is then r•p^ = 0 where p^ is either normal to the plane. 2. An N-piped has 2 N vertices as demonstrated by the inductive construction method presented above. 3. The locus of points making up the (closed) interior of an N-piped spanned by e 1...eN is given by rvolumeN = Σn=1Nαnen 0 ≤ αn ≤ 1 The tails of all the vectors e 1...eN meet at the origin of RN space. 4. There are N pairs of faces on an N-piped, and each face is an (N-1)-piped having 2 N-1 vertices. The total face count is 2N. Each face is spanned by a subset of N-1 of the base vectors en, so each face is "missing" one of the en and the face is labeled using the index of this missing base vector. The loci of points making up the faces of an N-piped are given by rface(i) = Σn≠iαnen 0 ≤ αn ≤ 1 i = 1,2...N rface(ip) = Σn≠iαnen + ei 0 ≤ αn ≤ 1 i = 1,2...N where "face i" has a corner touching the origin of th e N-piped (near face), while its parallel partner face "ip" does not touch the origin (far face). 5. If the N-piped spanning vectors e n are the tangent base vectors associat ed with some transformation F, then Ei• ej = δi,j where Ei are the reciprocal base vectors. In this case, the equations of the faces of the N-piped can be written in the form shown in item 1 above, E^i• r(face i) = 0 E^i• r(face ip) = 1/Ei i = 1,2...N so that both faces of a pair i are planar (in N dime nsional space) and they have the same normal vector E^ i so the faces of a pair lie on parallel planes. 6. The vector Ei is an outward-facing normal for face ip, while - Ei is an outward-facing normal vector for face i (which touches the origin). 7. The out-facing vector area of face ip of an N-piped can be expressed as 25 Ai = |det(S)| Ei Ai = σ (-1)i-1 Πx j≠i ej σ ≡ sign[det(S)] = sign[det(R)] Ai = σ (-1)i-1 e1 x e2 ... x eN // ei missing where ei is the vector missing from the face's spanning set. The outfacing area for face i is - Ai. The last two lines are shorthands for the followi ng, as discussed in Appendix A (i), ( Ai )α = σ (-1)i-1 εαabc..x (e1)a (e2)b ... (eN)x // where (e i)ί and ί are missing For N=2 the last two expressions for Ai are interpreted as shown in Appendix A (c) Ai = σ (-1)i-1 e3-i x 3^ i = 1,2 8. The volume of an N-piped spanned by e1...eN is given by volume(N) = | det(S) | = | det [ e 1, e2, e3, ....eN] | = | εabc...x (e1)a(e2)b(e3)c....(eN)x | where one can regard the tangent base vectors as the co lumns of the linearized transformation matrix S. 26 Appendix C: Elliptical Polar Coordinates ( N=2, non-orthogonal) This Appendi x is written in the develo pmental notation of Sections 1-6. (a) Elliptical polar coordinates The 2D "elliptic" coordinate sy stem has coordinate lines which are orthogonal ellipses and hyperbolas. When rotated about its two symmetry axes, this system generates 3D prolate or oblate spheroidal coordinates. This is not the 2D coordinate system described in this Appendix. For "elliptical polar" coordinates, the coordinate lines are taken instead as the ellipses from elliptic coordinates, and the rays from polar coordinates. This non-orthogonal system is perhaps not very useful, but provides a good "sandbox" in which to study general aspects of coordinate systems. The transformation x' = F(x) is given by x ' - s p a c e x-space (Cartesian) ρ2 = x2/a2 + y2/b2 x2+ y2 = r2 still x' 1 = θ x 1= x tanθ = y / x x ' 2 = ρ x 2 = y Writing the first equation above as 1 = x 2/(ρa)2 + y2/(ρb)2 it should be clear that ρ serves to label an ellipse of semi-major axis ρ a, and semi-minor axis ρb, while θ labels the ray at angle θ, as in polar coordinates. The inverse transform x = F-1(x') is given by x = aρcosθ x/a = ρ cosθ => x 2/a2 + y2/b2 = ρ2 y = bρsinθ y/b = ρsinθ => tan θ = y/x The matrix S is given by S 11 = (∂ x/∂θ) = -aρsinθ S12 = (∂ x/∂ρ) = acosθ S ik ≡ ( ∂xi/∂x'k) S21 = (∂ y/∂θ) = bρcosθ S22 = (∂ y/∂ρ) = bsinθ S = ⎝⎛ ⎠⎞-aρsinθ acosθ bρcosθ bsinθ => det(S) = -ab ρ and R = S-1 = ⎝⎛ ⎠⎞ -sinθ/(aρ) cosθ/(bρ) cosθ/a sinθ/b The tangent base vectors en can be read off as the columns of S e1 = ρ(-asinθ,bcosθ) = eθ |eθ| = ρ a2sin2θ + b2cos2θ ≡ hθ eθ = |eθ| e^θ e2 = (acosθ,bsinθ) = eρ | eρ| = a2cos2θ + b2sin2θ ≡ hρ eρ = |eρ| e^ρ The covariant metric tensor is, 27 g¯' = STS = ⎝⎛ ⎠⎞ρ2{a2sin2(θ) + b2cos2(θ)} [b2-a2]ρ sin(θ)cos(θ ) [b2-a2]ρ sin(θ)cos(θ ) a2cos2(θ) + b2sin2(θ) = ⎝⎜⎛ ⎠⎟⎞e1•e1 e1•e2 e2•e1 e2•e2 which is clearly non-diagonal (but symmetric) as expected. When a = b = 1 it reduces to the polar coordinates system metric tensor where then ρ = r. The coordinate system is non-orthogonal because e 1•e2 ≠ 0, or equivalently, because g ¯' is non-diagonal. (b) Forward coordinate lines Here is a M aple plot of some x-space forward coordinate lines (parameters a = 2 and b = 1) The coordinate lines in x-space are plotted using these equations, y = b ρi2-(x/a)2 // ellipses ρi= 1,2...10 10 ellipses y = x tanθ i // rays θi = 2π (i/20) , i = 1,2...20 20 rays which are obtained from the forward transformation equations ρ2 = x2/a2 + y2/b2 tanθ = y/x (c) Inverse coordinate lines Here is a Maple plot of some x'-space inver se coordinate lines (parameters a = 2 and b = 1) 28 The coordinate lines in x'-space are plotted using these equations ρ = xi/(acosθ) x i = -10 to +10 21 blue curves( one is a boxy U ) ρ = yi/(bsinθ) y i = -10 to +10 21 red curves ( one is a boxy U) which are obtained from the inverse transformation equations x = aρ cosθ y = bρ sinθ The secθ and cscθ curve families appear to "change shape", but that is just what happens when functions are scaled up vertically but not horizontally. If one pl ots one sine hump at different vertical scalings, the humps have different shapes. (d) Drawing a contravariant vector V in x-space: the meaning of V ' n . A contravariant vector field V(x) can be expanded in these two ways (Section 6 (f)) V = V 1(x) 1^ + V2(x) 2^ = Vx(x) x^ + Vy(x) y^ // un = n^ for Cartesian V = V'1(x') e1 + V'2(x') e2 = V'θ(x') eθ + V'ρ(x') eρ V'(x') = R(x) V (x) where the V' n are the components of V transformed into x'-space where V becomes V'. The prime is not necessary on V' ρ but we maintain it as a reminder that it is an x'-space component. R( x) is the matrix of Section 2 and en are the tangent base vectors of Section 3. The fields V' n(x') are "the components of vector field V' in x'-space", since V ' = R V , or V' i = RijVj. Moreover, these V' n(x') are "expressed in terms of curvilinear coordinates" x'. If one is asked to "express a vector V in curvilinear coordinates", one is usually being asked to write V as the second expansion above. The vectors en and V exist in x-space, and in the second expansion it just happens that the coefficients V' n(x') are the components of V', which is a vector in x'-space, and are functions of the curvilinear coordinates. 29 Here is a graphical representation of this vector V in x-space: As advertised, the tangent base vectors are not at right angles. The V parallelogram accurately illustrates the equation V = V'θ eθ + V'ρ eρ. Since x-space is Cartesian, there is no distinction between Cartesian length (graphical length) and covariant length for vectors in x-space. Graphically, one could find the values for V' θ and V' ρ as follows: (1) for the point (x,y), compute the vectors eθ and eρ and compute their lengths | eθ| = h'θ and | eρ| = h'ρ ; (2) draw the parallelogram shown aligned with these vectors for some given V and find the edge lengths. The e dges of the parallelogram are V' θ h'θ and V' ρ h'ρ so then the values of V' θ and V' ρ can be found. The alternative method is to compute R and use V' i = RijVj. (e) Drawing a contravariant vector V' in x '-space: two "Views" As with previous examples, the above picture is drawn to the right of an x'-space picture as follows: In Section 3 the axis-aligned basis vectors e'n were introduced as e'n , n = 1,2...N // ( e'n)i = δn,i e '1 = (1,0,0...) etc and e n was shown to be a contravariant vector, 30 e'n = R(x ) en. Applying matrix R(x) to the equation V = V' θ eθ + V'ρ eρ one gets V' = V'θ e'θ + V'ρ e'ρ which appears first in the list of expansions of V' in Section 6 (f). There is no ambiguity concerning this last equation. Ambiguity can arise, however, when on e tries to represent this equation graphically in x'- space. There are two very different "views" one can take of a drawing in x'-space. In the first view, we take x'-space to be "flat" (Cartesian) so that g' = 1. In the second view, we take x'-space to be "curved" with g' ≠ 1. These views are really two different x'-spaces since the metric tensors are different. In the Cartesian View of x'-space the length (norm) of a vector is given by | A|2 = δijAiAj = ΣAi2, so one has, since ( e'n)i = δn,i, g¯' = 1 |e 'n| = 1 e'n = e^'n = n^' = the usual axis-aligned unit vectors in x'-space V' = V'θ θ^ + V'ρ ρ^ θ^ = e^'1 ρ^ = e^'2 The left-side graph shown above is, in this view, a "normal Cartesian graph" and the vectors add up properly, for example Pythagoras tells us that | V'|2 = V'θ2 + V'ρ2 and θ^ • θ^ = 1 ρ^ • ρ^ = 1 θ^ • ρ^ = 0 This Cartesian View, which is x'-space with g ¯' = 1, is appropriate in app lications in which it is not required or desired that norms and dot products be tensorial scalars, as discussed at the end of Section 5 (a). For example, when g ¯' is set to 1, one has | V'| ≠ |V| in the above picture. Another use of this view involves integration as will be seen below. In the Curvilinear View of x'-space, one assumes that g ¯' takes a value which enforces the scalarity of norms and dot products between x-space and x'-space, which is to say, one takes g ¯' = ST g¯ S where g ¯ is the x-space metric tensor for x-space. Normally g ¯=1 (Cartesian x-space), so g ¯'= STS. In this Curvilinear View, then, the length | A'| of a contravariant vector A' is determined by | A'|2 = g¯'ijA'iA'j where g¯' = STS ≠ 1 | e'n| = |en| = h'n ≡ g¯ 'nn n = 1,2 for θ,ρ V' = V'θ e'θ + V'ρ e'ρ = V' θ h'θ e^'θ + V'ρ h'ρ e^'ρ e^'n ≡ e'n/ |en| = e 'n/ h'n 31 |V'|2 = |V |2 = Vx2 + Vy2 ≠ (V'θ h'θ)2 + (V'ρ h'ρ)2 // unless x' i are orthogonal coordinates This last inequality says that in the Curvilinear Vi ew the Pythagorean Theorem is invalid. In fact |V'| 2 = g¯'ijV'iV'j = g¯'θθ V'θ2 + g¯'ρρ V'ρ2 + 2 g¯'θρV'θ V'ρ = (V' θ h'θ)2 + (V'ρ h'ρ)2 + 2 g¯'θρV'θ V'ρ In writing | e' n| = |en| and | V'|2 = |V|2 above, we use the rule shown in Section 5 (i) which says | A'|2 = |A|2 for any contravariant vector A (|A|2 is a scalar ). Moreover, e^'θ • e^'θ = e'θ • e'θ / (h'θ2) = eθ • eθ / (h'θ2) = g¯'θθ / (h'θ2) = 1 e^'ρ • e^'ρ = e'ρ • e'ρ / (h'ρ2) = eρ • eρ / (h'ρ2) = g¯'ρρ / (h'ρ2) = 1 e^'θ • e^'ρ = e'θ • e'ρ / (h'θh'ρ) = eθ • eρ / (h'θh'ρ) = g¯'θρ / (h'θh'ρ) ≠ 0 <= !! so that the e^' n are unit vectors having unit covariant length, but e^'θ • e^'ρ ≠ 0 despite the fact that these vectors are drawn at right angles in the x'-space graph above, en = h'n e^'n. One might imagine trying to slant the lines of the x'-space graph to cause all inters ection points to have angles which match the metric tensor, which is to say, at each intersection point one would need an angle ψ where e^'θ • e^'ρ = cosψ . But in general e^'n • e^'m = g¯'nm / (h'nh'm) has a different value at every point , so such a graph would be quite complex. The upshot is that for a non-orthogonal system, the axes in x'-space are still drawn at right angles and the purpose of the graph is mainly to "locate" all the points x' which correspond to points x in x-space according to x' = F (x). The graph does successfu lly represent the idea that V' = V'ρ e'ρ + V'θ e'θ, but one must give up on Euclidean geometry for this vector su m triangle. It might be imagined that the x'-space graph is the projection onto the pl ane of paper of some vectors drawn on a curved surface emerging from the plane of paper, and that is then why Pythagoras is wrong. In the case of an orthogonal coordinate system (diagonal g ¯'), the 90 degree angles between the axes in x'-space are accurate represen tations of the fact that e^' n• e^'m = 0 when n ≠m. And since scalars are preserved, one has in the Curvilinear View, | V'| 2 = Σn (h'nV'n)2 = Σn V'n2 = |V |2 = Σn Vn2 // orthogonal only where V'n ≡ h'nV'n and V' =Σn V'n e^'n and V =Σn V'n e^n One can then still apply regular Euclidean geometry to the vector addition N-piped in x'-space in the sense that | V'| 2 = Σn (h'nV'n)2. 32 (f) Drawing the specific contravariant vector dx in x-space and x '-space Since d x is t he primordial contravariant vector, everythi ng stated in the last two sections applies with V → dx and V' θ → dx'θ = dθ, V'ρ → dx'ρ = dρ, where we finally drop the primes on d θ and dρ . The expansions of d x and dx ' are, dx = dθ eθ + dρ eρ // in x-space dx' = dθ e'θ + dρ e'ρ // in x'-space For V = dx the picture above becomes It must be understood that now the vector arrows like d x are highly magnified a nd in reality are very small compared to, say, the curvature of the ellipse. From above, e' θ • e'θ = g¯'θθ e^'θ • e^'θ = 1 e'ρ • e'ρ = g¯'ρρ e^'ρ • e^'ρ = 1 e'ρ • e'θ = g¯'ρθ e^'θ • e^'ρ = g¯'θρ / (h'θh'ρ) and once again the "right angle" in the x'-space picture is deceptive. (g) Study of how dx transforms in the mapping between x-space and x '-space Consider this drawing which shows a re presentative set of vectors d x in x-space (the bars), along with the forward mappings (d x' = F(dx) or d x' = Rd x ) of the corresponding vectors d x' in x'-space. The vectors on the right all point up, those on the le ft point generally to the northeast. 33 x'-space x-space Now select the red d x bar on the right and operationally apply the previous picture. First determine the tangent base vectors eθ and eρ at the location of the red bar. Then setting d x = dθ eθ + dρ eρ, consider the value of the two numbers d θ and dρ for this red bar. Graphically, knowing which way eθ and eρ point at the bottom of the red d x, one expects d θ > 0 and d ρ > 0. The red d x' bar on the left has these Cartesian values dθ and dρ , and has a Cartesian-view length of |d x|2 = (dθ)2+(dρ)2. One can see from the picture that these Cartesian lengths vary for the 10 bars shown, though the lengths are all the same in x-space. The Curvilinear-view lengths of the x' -space bars are all the same, and ar e equal to the Cartesian length of those bars in x-space since d x'•dx' = dx•dx. Consider now some bar mapping in the other direction: Now the d x bars on the right all have different lengths. Tho se on the left have the same Cartesian length, which is what the drawing shows, but each one's Curvilinear-view length matches that of its corresponding bar on the right. The ratio of the length of a bar on the right to the Cartesian length of the corresponding bar on the left is the scale factor h θ which recall is a function of location in space: 34 bar on right = d x(1) = e1 dx'1 = e^1 h'1 dx'1 = eθ dθ = e^θ hθ dθ graph length = h θ dθ bar on left (Cartesian view) = d x'(1) = e'1 dx'1 = e^'1 dx'1 = e^θ dθ graph length = d θ => right bar length / left bar length = h θ = ρ a2sin2θ + b2cos2θ (increases with ρ) If a=b, then h θ = ρ and the bar length on the right is then ρdθ as is obvious in polar coordinates. (h) Derivation of the Jacobian Integration Rule Consider no w an integral ∫dθdρ f(θ,ρ). The tiny rectangles of area d θdρ, like the specific gray and orange ones highlighted on the left above, are regarded for the purposes of integration as being in the Cartesian view of x'-space. One then writes [ dA' is called d V' in Section 8 ] dA' ≡ dρdθ = the area of a differential patch in Cartesian-view x'-space This is the graphical area one sees in the picture. Th ere is no need to define or consider any Curvilinear- view area in x'-space because the Cartesian-view area is being used. In the limiting process which defines the integration, each d θdρ patch on the left has the same area dρdθ. The interior of each patch on the left maps into some parallelogram patch on the right. One is not surprised to see that the patch areas on the right are di fferent, though they map into patches on the left of the same Cartesian-view area. As shown in Section 8 (e), the ratio of the two patch areas is the absolute value of the Jacobian |J(x ')|, (area of skewed patch on the right at location x) = |J( x')| dA' = |J( x')| dρdθ This is not what we mean by "the Jacobian Integration Rule" in the section title. That is coming below and it is going to involve the quantity dxdy. The mapping shown above between patches is an N=2 example of the general N-dimensional discussion in Section 8 (a) which describes an ort hogonal differential N-piped in (Cartesian-view) x'- space mapping into a non-orthogonal differential N-piped in x-space. Now back to the integration issue. There are tw o ways an integration can be done in Cartesian x- space: integral of f(x) = lim Σ i dA1(xi) f(xi) dA 1(xi) = patches shown on the right above integral of f(x) = lim Σi dA2(xi) f(xi) dA 2(xi) = dxdy In the first integral , every patch dA 1(xi) on the right has a different shape and a different area as the integral is computed in the usual limiting-sum mann er. The gray and orange patches on the right are two of these many patches. Despite their non-uniform shape and area, this rag-tag band of patches certainly "covers" the area being integrated over, and does so perfectly in the calculus limit. The area of one of these rag-tag patches is |J( x')|dA' = |J( x')|dθdρ and the areas are different because the Jacobian is a function of x = x(x'). 35 In the second integral , every patch dA 2(xi) has the same area dxdy, so really dA 2(xi) does not depend on xi in this form of the integration. One such dxdy patch is shown in green above. The coverage of the dA 2 patches is of course also "perfect coverage" in the calculus limit. Since both integrals cover the same area perfectly, they both give the same result in the limiting process that defines the integral. This point is someti mes misunderstood. One is not just "replacing" a parallelogram patch such as the orange one on the ri ght with some dxdy patch that approximates it in area, like the green patch. The statement is about an integration. Thus one has lim Σ i dA1(xi) f(xi) = lim Σi dA2(xi) f(xi) or ∫[|J(x')| dθdρ] f(x(x')) = ∫[dxdy] f( x) where on the left f(x ) = f( x(x')) where x = F-1(x') ≡ x(x'). In the sense of distribution theory (Stakgold Chapters 1 and 5), one can then make this symbolic statement |J(θ,ρ)| dρdθ = dxdy where the meaning of this symbolic equality is the integral statement above, ∫D dxdy f( x) = ∫D' dθdρ |J(x')| f(x(x')) , valid for any integrable f( x) and any integration region D (region D' corresponds to D in x'-space.) Either of these last two equations constitute the "Jacobian Integration Rule" of the section title. The integral on the left is well defined in 2D cal culus, so the expression on the right shows how to "evaluate the integral on the left in curvilinear coordinates". At this point one may introduce a new but obvious symbol dA ≡ dxdy so the above equality of integrals can be written ∫dA f( x) = ∫ dA' |J(x ')| f(x(x')) |J( x')| dA' = dA In N dimensions, dA and dA' are differential "volum es", and the general Jacobian Integration rule takes the form, 36 ∫dV f( x) = ∫ dV' |J(x ')| f(x(x')) |J(x ')| dV' = dV dV' ≡ dx'1dx'2....dx'N = the volume of an orthogonal differential N-piped in the Cartesian-view x'-space dV = dx 1dx2....dxN = the volume of an orthogonal differential N-piped in x-space. Notice that these are not the two N-pipeds which "map into each other" as noted above. The N-piped dV has nothing to do that that mapping which involved a non-orthogonal N-piped in x-space. To finish off our sample N=2 case, recall from earlie r that for our polar elliptical coordinate system |J'(x')| = | det(S)| = ab ρ and therefore ∫dxdy f(x,y) = ∫ dθdρ |J(x')| f(aρ cosθ, bρ sinθ) = ab ∫dθdρ ρ f(aρ cosθ, bρ sinθ) In the limit of regular polar coordinates, one then has a = b = 1 and ρ = r so ∫dxdy f(x,y) = ∫rdrdθ f(rcosθ, rsinθ) which is the familiar result. 37 Appendix D. Tensor Densities and the ε tensor Picture A is used in this Ap pendix along with Standard Notation. (a) Definition of a tensor density First, recall fr om the Section 5 (k) discussion of the Jacobian J, J ≡ det(Si j) = σ sg' / sg = σ(sg'/sg)1/2 = σ(g'/g)1/2 => (g'/g)1/2 = σJ = |J| > 0 s = sign[det(g ij)] = sign(g) = sign(g') g = det(g ij) sg = |g| > 0 Si j ≡ (∂xi/∂x'j) σ = sign[det(Si j)] = sign(J) g' = det(g' ij) sg' = |g'| > 0 For proper Lorentz transformations of special relativity, det(S) = 1 so σ = +1. For curvilinear coordinates, one normally selects an ordering of the x i so that σ = +1, such as r, θ,φ in spherical coordinates. Nevertheless, we allow for the possibility of J < 0. Second, recall our generic sample tensor transformation from Section 7 (j), T ' abc de = Ra a' Rb b' Rc c' Sd' d Se' e Ta'b'c' d'e' which can be rewritten in a more standard wa y using the theorem of Section 7 (q) that Sμ ν = Rνμ, T ' abc de = Ra a' Rb b' Rc c' Rdd' Ree' Ta'b'c' d'e' T is a mixed rank-5 tensor, meaning it transforms as shown above with respect to the underlying transformation F. T is a regular standard-issue tensorial tensor. Now suppose instead that the object T were to transform like this, with J being the Jacobian noted above, T ' abc de = J-W Ra a' Rb b' Rc c' Rdd' Ree' Ta'b'c' d'e' where the extra factor J -W has been introduced. If T transforms this way, it is called a tensor density of weight W. Thus, an ordinary tensor is a tensor density of weight 0. The convention for the sign of W used here is that of Weinberg p 99 Eq. (4.4.4), which equation has the following factor on the right side of a sample tensor density transform equation |∂x'/∂x| W ≡ [det(∂x'/∂x)]+W = [ det(∂x'i/∂xk) ]+W = J-W 38 Some authors use -W as the "weight" instead of +W, but we shall stick with Weinberg's convention. An immediate example of a tensor density is provided by (g'/g)1/2 = |J| rewritten as g' = J2 g =J-(-2) g => weight(g) = -2 g' is a scalar density of weight - 2 This is the scalar density mentioned in Section 5 (k) of weight -2. Notice that from g one can construct other scalar densities of other weights, for example g' -1 = J-(2) g-1 => weight(g-1) = +2 g'-1 is a scalar density of weight +2 (b) A few facts about tensor densities 1. It is prett y obvious that a sum of two index-similar tensor densities of weight W has weight W. 2. Contracting indices within a tensor density does not alter its weight W. If indices a and d are contracted in the example above, one gets T ' abc ae = J-W Ra a' Rb b' Rc c' Rad' Ree' Ta'b'c' d'e' = J -W (Ra a'Rad') Rb b' Rc c' Ree' Ta'b'c' d'e' = J -W δa'd' Rb b' Rc c' Ree' Ta'b'c' d'e' = J-W Rb b' Rc c' Ree' Ta'b'c' a'e' The factor J-W just sits there, impervious to contraction activities. 3. Going the other direction, when a larger tensor density is formed from two smaller ones, called a direct product or outer product , the weights get added. Example 1: A'a = J-W1 Ra a'Aa' B' c d = J-W2 Rc c'Rdd' Bc' d' => (A' a B'c d) = J-(W1+W2) Ra a' Rc c'Rdd' (Aa' Bc' d') Example 2: Suppose in the outer product the first factor is the scalar density g-W1/2 of weight W1 : g' -W1/2 = J-W1 g-W1/2 // since g' = J2g from Section 5 (k), no R factors since scalar B' c d = J-W2 Rc c'Rdd' Bc' d' // same as in previous example => g' -W1/2B'c d = J-(W1+W2) Ra a' Rc c'Rdd' (g-W1/2 Bc' d') 39 If one selects W1 = –W2, the added factor neutralizes the weight of the tensor density to which it is prefixed, generating thereby a regular tensor (weight 0). So if tensor density B has weight W, ( g ' W/2 B'c d) = Ra a' Rc c'Rdd' (gW/2 Bc' d') and then (g W/2 Bi j) transforms under F as a regular tensor. (One should always keep in mind the fact that there is an underlying transformation x' = F(x) upon which the House of Tensor is built ). 4. Although sometimes authors take a differing stance for certain tensors, we shall assume that indices are raised and lowered on a tensor density in exactly the same way they are raised and lowered on an ordinary tensor of the same index structure. This means the gab raises an index and gab lowers an index. 5. Raising or lowering an index does not change the weight of a tensor density . Again, using our generic example above, T 'abc de = J-W Ra a' Rb b' Rc c' Rdd' Ree' Ta'b'c' d'e' T 'abc de = g'ex T 'abc dx // raise last index in x'-space T a'b'c' d'e' = ge'e" Ta'b'c' d'e" // lower last index in x-space Therefore T 'abc de = g'ex [ J-W Ra a' Rb b' Rc c' Rdd' Rxe' Ta'b'c' d'e'] = g ' ex [ J-W Ra a' Rb b' Rc c' Rdd' Rxe' (ge'e" Ta'b'c' d'e")] = J W Ra a' Rb b' Rc c' Rdd' (g'ex Rxe' ge'e") Ta'b'c' d'e" = J -W Ra a' Rb b' Rc c' Rdd' (Re e") Ta'b'c' d'e" // Section 7 (o) facts about R and again J-W passively watches all the action fly by. The we ight of our generic tensor density with its last index raised is still W. 6. The covariant dot product of vector densities A and B of weights W and w is a scalar density of weight W + w and therefore A'•B' = J-(W+w) A•B . Proof : First form the rank-2 tensor density AiBj which by item 3 has weight W+w. Lower the second index and the mixed rank-2 tensor AiBj by item 5 still has weight W+w. Then contract to get A•B = AiBi and by item 2, the weight is still W+w. Corollary : The magnitude of a vector density A of weight W is a scalar density of weight W, and therefore | A|' = J-W |A| 40 Proof : |A|2 = A • A which has weight 2W meaning | A'|2 = J-2W |A|2. Therefore | A'| = J-W |A| . 7. As J→1, tensor densities become true tensors. One could imagine some limiting/morphing process on an underlying transformation F such that the linear ized transformation matrix R approaches a rotation matrix at all points in space (RRT= 1 and detR = 1) and then J = detS → 1. In this case J-W → 1-W = 1 and therefore any tensor density, regardless of its weight W, becomes an ordinary tensor. Perhaps we should restrict this comment to underlying transformations F having detS > 0 since passing through detS = 0 is problematical. An example: the cross product considered in sec tion (g) below of N-1 contravariant vectors becomes in this limit an ordinary covariant v ector. If g=1 in x-space, then g' = RRT = 1 in x'-space and then that resulting vector can be considered either contravari ant or covariant since both spaces are then Cartesian. This is the case with A = B x C under rotations in 3D sp ace. On can think of the εabc as moving in this limit from a tensor density of weight -1 to an ordinary tensor. (c) Theorem about Totally Antisymmetric T ensors: there is really only one: εabc... Theorem : Apart from a scalar factor, there exists only one totally antisymmetric (TA) tensor. Proof: Suppose there were two TA tensors called εabc... and rabc.... If two or more of the indices are equal, both tensors are 0, so for such index sets, one can say rabc.. = f εabc.. where f is any finite function whatsoever. Consider now the case where all th e indices are distinct and therefore exhaust the set 123...N, and consider abc... to be a permutation of 123...N obtained by doing S pairwise swaps, abc... = P(123...) p = (-1)S If one were to associate a sign change with each swap, the total sign change would be p, the parity. Since ε and r are both TA tensors, each tensor can be "unw ound" back to a standard index order by doing these S swaps, and the swaps will cause a total sign of p relative to that standard order, so r abc.. = p r123... // for example, r2134.. = (-1)1 r1234.. εabc.. = p e123... Define scalar function f ≡ r 123... / e123... , whatever it might be. Then r abc.. = p(f e123...) εabc.. = p e123... and dividing these two equations one finds, r abc.. = f εabc.. which has now been shown valid for all index sets abc.. . Therefore, any "other" totally antisymmetric tensor is just a scalar function times the ε tensor. 41 (d) The contravariant ε ten sor Knowing nothing to start, assume that the famous εabc.. totally antisymmetric tensor transforms under F as a tensor density of some weight W which we hope to determine. Then ε'abc.. = J-W Ra a' Rb b' ... εa'b'c'.. (*) Assume that εabc.. is the usual permutation tensor normalized to ε123...N = +1. This is the convention used by Weinberg p 99. This means each index swap changes the sign, and if two or more indices are the same, ε = 0. This is an important starting assump tion, and from it most everything follows. Given this assumption, the RHS of (*) is totally antisymmetric (TA). The argument is given once here and then used later several times. Consider an a ↔b swap. Then ε' bac.. = J-W Rb a' Ra b' ... εa'b'c'.. = J-W Rb b' Ra a' ... εb'a'c'.. = J -W Ra a' Rb b' ... (–εa'b'c'..) = – ε'abc.. The same result is true for any swap, thus RHS (*) = TA. Since according to section (b) there is only one TA tensor available, apart from a scalar function factor, it follows that ε' abc.. = Kεabc... where K is some scalar function, perhaps just a constant. Equation (*) above then reads K εabc... = J-W Ra a' Rb b' ... εa'b'c'.. (**) Setting in the standard order, one finds that K ε 123... = J-W R1 a' R2 b' ... εa'b'c'.. or K = J -W det(Ri j) = J-W (J)-1 = J-(W+1) so now ε'abc.. = Kεabc... = J-(W+1) εabc... A second assumption is now made: that ε abc.. (contravariant!) has the same value structure in any frame of reference, which is to say it is the same in x'-space as it is in x-space, ε' abc.. = εabc... This assumption is consistent with taking W = -1 in the previous equation. Again, this follows the convention of Weinberg p 99. Some authors instead arrange for the above equation to be true for the covariant ε tensors, and use then ε123...N = ε'123...N = +1, but we shall follow Weinberg. 42 To summarize, assuming that εabc... is the usual permutation tensor normalized in the usual way, and assuming that ε'abc.. = εabc... so this tensor is the same in all frames or spaces, THEN one concludes that εabc... must transform as a rank-N tensor density of weight W = -1. That is to say, ε'abc.. = J Ra a' Rb b' ... εa'b'c'.. // this is (*) above with W = -1 This then is our second example of a tensor density. Viewed in this light, the tensor ε abc.. is known as the Levi-Civita tensor. Tullio Levi-Civita (1873-1941) . Italian, University of Padua 1892, with Ricci published the theory of tensor algebra in 1900 (see Refs.), which work assisted Einstein circa 1915 in formulating the theory of general relativity. The ε tensor bears his name. Sometimes the a ffine connection is called the Levi-Civita connection. (e) Some facts about the ε tensor 1. Consider ( based on section (b) 4 above), εabc... = gaa' gbb'..... εa'b'c'... This is again in the convention of Weinberg p 99 (4.4.10). Added sign s convention. Some authors make a special exception for the ε tensor and introduce an extra sign s into the above equation ( recall that s = -1 for special relativity) ε abc... = s gaa' gbb'..... εa'b'c'... s = sign[det(g ij)] Inserting such a sign renders any ε mixed tensor like ε a bc.. ambiguous when s = - 1, but is acceptable if one promises never to make use of a mixed ε tensor. We shall refer to these two methods as "the Weinberg convention" and the "added sign s conve ntion", the former being assumed unless otherwise stated. Install the reference sequence to obtain ε 123.. = g1a' g2b'..... εa'b'c'... = det(gij) = g // Section 5 (k) Similarly, ε ' 123.. = det(g' ij). To summarize, ε 123.. = det(g ij) = g // ε all-down index reference values ε'123.. = det(g' ij) = g' In the "added sign s convention", these last two equations would have sg = |g| and sg' = |g'| on the right which means then these two ε values would be always positive. 43 2. Take the same starting point as above ε abc... = gaa' gbb'..... εa'b'c'... The RHS is a totally antisymmetric in indices abc... (see above) and can therefore be written RHS = C εabc... since we showed earlier that there is only one TA tensor apart from scalar C. Therefore ε abc... = C εabc... (*) Insert the reference sequence ε123... = C ε123... = C But in 1 it was just showed that ε 123... = det(g ij) . Therefore C = det(g ij) and then (*) says for the "Weinberg convention", ε abc... = det(g ij) εabc... = g εabc... // relating all down to all up ε'abc... = det(g' ij) ε'abc... = g' ε'abc... / / = g ' εabc... where the second line follows by the same argument. These equations relate all indices down to all up in the same space. Notice that both ε abc... and ε'abc... are totally antisymmetric. In the "added sign s convention" the above equations are instead ε abc... = |det(g ij)| εabc... = |g| εabc... // relating all down to all up ε'abc... = |det(g' ij)| ε'abc... = |g'| ε'abc... / / = | g ' | εabc... 3. Divide the last two Weinberg convention equations to find that ε'abc... = [det(g' ij)/ det(gij)] εabc... = (g'/g) εabc... From Section 5 (k) one has (g'/g) = J 2 so the conclusions regarding ε are these: ε' abc... = J2 εabc... = (g'/g) ε abc.. ε'abc... = εabc... = permutation tensor // general ε abc... = permutation tensor if g=1 These conclusions are valid for the "added sign s convention" as well since det(g) and det(g') always have the same sign as shown in Section 5 (k). 44 Two comments: • Although we set ε' abc... = εabc... by fiat, we cannot similarly set ε'abc... = εabc... by fiat. This latter result comes out being ε'abc... = J2 εabc... as just shown. • The fact that ε'abc... = J2 εabc... does not say that ε abc is a tensor density of weight -2 because there are no R factors showing. (See the section (a) defi nition of a tensor dens ity transformation. ) (f) The covariant ε tensor : repeat section (d) as if its weight were not known According to section (b) 5, lowering indices does not ch ange the weight of a tensor density. Section (d) showed that εabc.. is a tensor density of weight -1, so we know right away that εabc.. is also a tensor density of weight -1. Nevertheless, it is interesti ng to see what happens when the same method used in section (d) for εabc... is applied to εabc... . We start by assuming εabc.. is a tensor density of some unknown weight W, ε'abc.. = J-W [ Raa' Rbb'.... εa'b'c'.. ] (*) Section (e) 2 noted that ε a'b'c'... is a totally antisymmetric tensor, and therefore as in section (d) one concludes that the RHS of (*) is also totally antisymmetric and can be written as RHS(*) = K εabc.. , so (*) then says K ε abc.. = J-W [ Raa' Rbb'.... εa'b'c'.. ] (**) Use section (e) 2 to set εa'b'c'.. = det(gij) εa'b'c'.. inside the bracket, K ε abc.. = J-W [ Raa' Rbb'.... det(g ij) εa'b'c'..] , and then install the reference sequence on both sides K ε 123.. = J-W [ R1a' R2b'.... det(g ij) εa'b'c'..] But section (e) 1 says ε 123.. = det(g ij), so cancel det(g ij) on both sides to get K = J -W [ R1a' R2b'.... εa'b'c'..] = J-W det(Rij) = J-W det(Sj i) = J-W J = J-(W-1) So here the result is K = J -(W-1) whereas in section (d) the result was K = J-(W+1) . In the current case, since (*) and (**) have the same RHS, setting the LHS's equal says ε'abc.. = K εabc.. = J-(W-1) εabc.. But section (e) 3 said that ε' abc.. = J2 εabc.. and therefore one gets W = -1. 45 Thr conclusion is that εabc... transforms with weight -1, the same as εabc..., so (*) becomes ε' abc.. = J [ Raa' Rbb'.... εa'b'c'.. ] (g) Generalized cross products In Appendix A (c) the following cross product of N-1 v ectors is considered (converted now to standard notation) Q a ≡ εabc...x BbCcDd.....Xx or Q = B x C x D .... x X If the vectors B,C,D..X are all contravariant vectors, then applying the rule of section (b) 3, one concludes that, since ε is a tensor density of weight -1 and since all the RHS vectors have weight 0, the object Q a is a covariant vector density of weight = -1 , and thus has this transformation rule Q' a = J RabQb Similarly, one may consider Q a ≡ εabc...x BbCcDd.....Xx . If vectors B,C,D...X are covariant vectors, then Q a is a vector density of weight -1 and Q' a = J Ra bQb (h) The tensorial nature of curl B It has just been shown that C = A x B is a vector density of weight -1, this being a special case of the generalized cross product discussion above. As noted in sec tion (a) 7, if R happens to be a rotation, then C is in fact a tensorial vector. One might conjecture that C = ∇ x B is also a vector density of weight -1, and that conjecture is correct as is now shown. Consider C n = εnab ∂aBb where B is assumed to be an tensorial vector. It is he lpful to write this equati on in the following manner Cn = εnab [ ∂aBb – ∂bBa ]/2 where the second term is the same as the first term, since – ε nab ∂bBa = εnba ∂bBa = εnab ∂aBb . Recall now from Section 7 (v) that the covariant derivative of a vector is given by 46 Bb;a = ∂aBb – Γc ab Bc where the affine connection Γc ab is symmetric under a ↔b. Therefore B b;a – Ba;b = [∂aBb – Γc ab Bc] - [∂bBa – Γc ba Bc] = ∂aBb – ∂bBa Therefore C n can be expressed as C n = εnab [Bb;a – Ba;b ]/2 so by the same ε anti-symmetry noted above the final result is Cn = εnab Bb;a The major feature of B b;a -- as discussed in Section 7 (v) -- is that it is a rank-2 tensor if B is a tensorial vector. The weight addition rule of s ection (b) 3 can then be applied to εnab Bb;a. Since εnab has weight - 1 and Bb;a has weight 0, the conclusion is that Cn is a vector density of weight -1. Thus, C = curl B is a vector density of weight -1. If the underlying transformation F is a rotation, C becomes an ordinary vector as per section (a) 7. (i) Tensor E as a weight 0 version of ε : three conventions 1. Equations in the "Weinberg Convention" In this section it is assumed that g ij and gij raise and lower indices of the ε tensor just as they do for any other tensor (Weinberg convention). In sections (d) and (e) it was established that εabc... = g εabc... ε123.. = +1 ε123... = g ε'abc... = g'ε'abc... ε'123.. = +1 ε'123... = g' ε'abc... = J2 εabc... = (g'/g) εabc... ε is a rank-N tensor of weight W = -1 Again, just in passing, notice that ε 'abc... = J2 εabc... does not say ε has weight -2 because the R factors are not present on the right side. Consider now the following new objects defined by (sg = |g|, s= sign(g) = sign(g') as in Section 5 (k)) E abc... ≡ |g|-1/2 εabc... => E123... = |g| -1/2 g = |g| -1/2 s |g| = s |g|1/2 E'abc... ≡ |g'|-1/2 ε'abc.. . => E'123... = |g'| -1/2 g' = |g'| -1/2 s |g'| = s |g'|1/2 It was shown in Section 5 (k) that g' = J2g so that g transforms as a scalar density of weight -2. Since the sign of g and g' are the same, if follows that (sg) = |g | is also a scalar density of weight -2, and then the quantity |g|-1/2 transforms as a scalar density of weight +1, since |g'|-1/2 = J-1 |g|-1/2. Looking at the equation Eabc... ≡ |g|-1/2 εabc... above, and using the weight summation rule of section (b) 3, one 47 concludes at that Eabc... transforms as a tensor of weight (+1) + (-1) = 0, and so Eabc... is an ordinary tensor. This is the motivation of the above definitions . It was shown at the end of Section 7 (u) that a tensor density equation with matchi ng weights is "covariant", so one is not surprised to see the second line above being the same as the first line but everything is primed (s = s'). Raising indices on both sides gives E abc... ≡ |g|-1/2 εabc... => E123... = |g|-1/2 E'abc... ≡ |g'|-1/2 ε'abc... => E'123... = |g'|-1/2 To compare E abc... and Eabc... , E abc... = |g|-1/2 εabc... = |g|-1/2 g εabc... = s |g|-1/2 |g| εabc... = s |g|1/2 εabc... Eabc... = |g|-1/2 εabc... so that, E abc... = s|g| Eabc... = g Eabc... Summarizing, E 123... = |g|-1/2 E123... = s|g|+1/2 Eabc... = g Eabc... = s|g| Eabc... E'123... = |g'|-1/2 E'123... = s|g'|+1/2 Eabc... EABC... = |g|-1 εabc... εABC... E'abc... E'ABC... = |g'|-1 ε'abc... ε'ABC... Since |g| -1 is a scalar density of weight +2 and each ε has weight -1 and each E has weight 0, one is happy to see the weights balance of the two si des of this pair of covariant equations. 2. Equations in the "added sign s convention" The previous section shows how things work out using the "Weinberg convention" noted at the start of section (e). Here is the previous section redone in the "added sign s convention": In section (e) it was established that ( g = det(g ij)) ε abc... = sgεabc... ε123.. = +1 ε123... = sg = |g| // g → sg ε'abc... = sg'ε'abc... ε'123.. = +1 ε'123... = sg' = |g'| // g' → sg' ε' abc... = |J|2 εabc... = (g'/g) εabc... ε is rank-N tensor of weight W = -1 // same Consider the following new objects defined by ( sg = |g|, s= sign(g) = sign(g') as in Section 5 (k) ) 48 Eabc... ≡ |g| -1/2 εabc... => E123... = |g| -1/2 |g| = |g|1/2 E'abc... ≡ |g'| -1/2 ε'abc... => E'123... = |g'|-1/2 |g'| = |g'|1/2 But the same argument given above, Eabc... is an ordinary covariant tensor (ie, weight = 0). However, the indices cannot be raised by gij. In this convention then one must make independent de finitions of the contravariant components as follows, Eabc... ≡ |g|-1/2 εabc... => E123... = |g|-1/2 E'abc... ≡ |g'|-1/2 ε'abc... => E'123... = |g'|-1/2 To compare Eabc... and Eabc... , E abc... = |g|-1/2 εabc... = |g|-1/2 |g| εabc... = |g|-1/2 |g| εabc... Eabc... = |g|-1/2 εabc... so that E abc... = |g| Eabc... Summarizing, E123... = |g|-1/2 E123... = |g|+1/2 Eabc... = |g| Eabc... E'123... = |g'|-1/2 E'123... = |g'|+1/2 E'abc... = |g'| E'abc... E abc... EABC... = |g|-1 εabc... εABC... E'abc... E'ABC... = |g'|-1 ε'abc... ε'ABC... In this "added sign s" convention, all these summarized results involve on ly |g| and there are no factors of s floating around. The cost of this benefit is a lack of true covariance (when s=-1), as demonstrated in section (k) below. 3. Equations in the "Ricci-Levi-Civita convention" Ricci and Levi-Civita use the "added s convention" but add a factor σ = sign(det(S)) into their definition of E ( see their paper p 135 or Hermann pp 31-21) so that E abc... ≡ σ|g| -1/2 εabc... => E123... = σ|g| -1/2 |g| = σ|g|1/2 E'abc... ≡ σ|g'| -1/2 ε'abc... => E'123... = σ|g'|-1/2 |g'| = σ|g'|1/2 E abc... ≡ σ|g|-1/2 εabc.. => E123... = σ|g|-1/2 E'abc... ≡ σ|g'|-1/2 ε'abc... => E'123... = σ|g'|-1/2 Summarizing, E123... = σ|g|-1/2 E123... = σ|g|+1/2 Eabc... = |g| Eabc... E'123... = σ|g'|-1/2 E'123... = σ|g'|+1/2 E'abc... = |g'| E'abc... 49 Eabc... EABC... = |g|-1 εabc... εABC... E'abc... E'ABC... = |g'|-1 ε'abc... ε'ABC... Notice that in all three conventions, last equation pair is the same. Since Ricci and Levi-Civita did not raise and lower i ndividual indices in their 1900 paper, they were not concerned about their convention being non-covariant in that sense. (j) Representation of ε , εε and contracted εε as determinants 1. Theorem about a certain permutation sum Consider the following object Q defined as a signed perm utation sum of the product of N matrix elements of a matrix M ij, Qabc..x ≡ ΣP p P2(Ma1Mb2 Mc3.....MxN) In this equation, P 2 represents a permutation of the set of 2 nd indices of the N matrix elements, and the sum is over all N! such permutations. There are many ways to arrive at a given permuta tion of 123...N by doing pairwise swaps, but for all these ways, the number of swaps S will be either even or odd. The parity p of a permutation is defined then as (-1)S and this p appears in the above sum. If one were to swap 2 ↔3 on the right above, each permutation would have S → S+1 since an extra swap is needed to undo 2 ↔3. Thus, all parities p → -p and in fact the whole object negates. But the swap 2↔3 is the same as b ↔c since M b3 Mc2 = Mc2 Mb3. Applying this argument to any pair of indices, one concludes that Q abc..x is totally antisymmetric and therefore can be written as K εabc...x : ΣP p P2(Ma1Mb2 Mc3.....MxN) = K εabc...x Setting abc..x to 123..N, one gets. ΣP p P2(M11M22 M33.....MNN) = K The left side of this last equation can be written as Σ P p P2(M11M22 M33.....MNN) = Σabc..x εabc...x M1aM2b M3c.....MNx because p = ε abc...x correctly assesses the parity of any given permutation. But this object is simply det(M) so the conclusion is that K = det(M) and then ΣP p P2(Ma1Mb2 Mc3.....MxN) = det(M) ε abc...x Consider now the following matrix where ab c..x is some permutation of 123...x, 50 where M(abc..) = Ma1 Ma2 Ma3 ... M aN M b1 Mb2 Mb3 ... M bN M c1 Mc2 Mc3 ... M cN ... M x1 Mx2 Mx3 ... MxN By rearranging the rows into their normal numerical order, one obtains matrix M, but incurs a sign from the various row swaps which sign is just εabc..x. Therefore det(M(abc..) ) = εabc...x det(M) and therefore Σ P p P2(Ma1Mb2 Mc3.....MxN) = det(M(abc..) ) = det(M) εabc...x The permutation sum is thus just the determinant of matrix M (abc..). The first term in the permutation sum, the term with an identity permutation, corresponds to the product of the diagonals of that matrix. 2. Application of the theorem to M = δ : a representation of ε Apply the above theorem to matrix M = 1 ≡ δ, the identity matrix, having M ij = δi,j. Clearly det( δ) = 1 and one then has Σ P p P2(δa,1δb,2 δc,3.....δx,N) = det[δ(abc..) ] = εabc...x Thus is obtained the famous representation of εabc..x as a certain determinant of Kronecker deltas, ε abc...x = det[δ(abc..) ] where δ (abc..) = δa,1 δa,2 δa,3 ... δa,N = Ra δb,1 δb,2 δb,3 ... δb,N = Rb δc,1 δc,2 δc,3 ... δc,N = Rc ... δ x,1 δx,2 δx,3 ... δx,N = Rx where, for future use, each row v ector has been given a name like R a where ( Ra)i = δa,i The conclusion then is that 51 which is the same as εabc...x = ΣP p P2(δa,1δb,2 δc,3.....δx,N) . 3. Outer product of two ε tensors. Consider now ε abc...x = det ⎝⎜⎛ ⎠⎟⎞ Ra Rb ... Rx and εa'b'c'...x' = det ⎝⎜⎛ ⎠⎟⎞ Ra' Rb' ... Rx' Then εabc...x εa'b'c'...x' = det ⎝⎜⎛ ⎠⎟⎞ Ra Rb ... Rx det ⎝⎜⎛ ⎠⎟⎞ Ra' Rb' ... Rx' = det ⎝⎜⎛ ⎠⎟⎞ Ra Rb ... Rx det ( Ra' Rb' ... Rx') = det { ⎝⎜⎛ ⎠⎟⎞ Ra Rb ... Rx ( Ra' Rb' ... Rx') } which is the determinant of this matrix R a• Ra' Ra• Rb' Ra• Rc' ... Ra• Rx' Rb• Ra' Rb• Rb' Rb• Rc' ... Rb• Rx' Rc• Ra' Rc• Rb' Rc• Rc' ... Rc• Rx' ... Rx• Ra' Rx• Rb' Rx• Rc' ... Rx• Rx' A typical element of this matrix is given by R c• Rb' = (Rc)i(Rb')i = δc,i δb',i = δc,b' so that matrix can be written as δ a,a' δa,b' δa,c' .... δa,x' 52 δb,a' δb,b' δb,c' .... δb,x' δc,a' δc,b' δc,c' .... δc,x' ≡ δ(abc..x; a'b'c'..x') .... δx,a' δx,b' δx,c' .... δx,x' where we have made up a name for this matrix as shown. The conclusion then is that which is the same as ε abc...x εa'b'c'...x' = ΣP p P2(δa,a'δb,b'δc,c'.....δx,x') As usual, the argument of P 2 is the product of the diagonal elements of the matrix. 4. Contracting the first index of the outer product of two ε tensors. Consider what happens if one sums on the first index of the εε product: Σa εabc...x εab'c'...x' For fixed given values of bc..x and b'c'...x' , there is only one way this sum can be non-zero. In that one way, bc..x and b'c'...x' must each be permutations of the set {12..N ex clude A} where A is the "hit value" of a in the sum on a. Then Σ a εabc...x εab'c'...x' = εAbc...x εAb'c'...x' = ΣP p P2(δA,Aδb,b'δc,c'.....δx,x') = ΣP p P2(δb,b'δc,c'.....δx,x') (*) where in this last expression the sum can be regard ed as being over permutations where b'c'...x' is a permutation of b,c..x. Each of these lists of integers is in turn a permutation of {12..N exclude A}. Now, parity p = (-1)S where S is a number of swaps it takes to connect b'c'...x' with b,c..x, since a = a' = A. One might wonder if the overall sign of the RHS of the last equation is correct. A check of the first term in this sum which is just δb,b'δc,c'.....δx,x' shows that this overall sign is indeed correct. This first term must be positive because the product of two ε 's is either +1 or 0. As an example, Σ a εabc εab'c' = ΣP p P2(δb,b'δc,c') = δb,b'δc,c' – δb,c'δc,b' 53 The permutation sum shown on the right side of (*) is the determinant of δ(abc..x; a'b'c'..x') but with the first row and column crossed out. It can then be thought of as either the minor or cofactor of the element aa of this big δ matrix. Therefore, Σa εabc...x εab'c'...x' = [cof δ(abc..x; a'b'c'..x')]aa where the notation cofM refers to a matrix of cofactors with elements [cofM] ij. Don't confuse the a on the right side with the local dummy summation index a on the left side. The conclusion then is that (implied summation on a on the LHS) 5. Contracting two or more indices of the outer product of two ε tensors. Consider what happens if one sums on the first two indices of the εε product: Σ a,b εabcd...x εabc'd'...x' For fixed given values of c,d..x and c'd'...x' , in or der for this double sum to be non-zero, the index sets cd..x and c'd'...x' must each be permutations of the set {12..N exclude A,B} where A,B are a pair of hit values for the a and b sums. If a=A and b=B is hit va lue, then so is a=B and a=A, so there are 2! contributing terms in the sum, and each term is +1. Therefore Σ a,b εabcd...x εabc'd'...x' = 2! εABc...x εABc'...x' = 2 ! Σ P p P2(δA,AδB,B'δc,c'δd,d'.....δx,x') = 2!ΣP p P2(δc,c' δd,d'.....δx,x') (*) where in this last expression the sum is over permutati ons where c'd'...x' is a permutation of c,d....x. Each of these lists of integers is in turn a permutation of {12..N exclude A,B}. Now parity p = (-1)S where S is a number of swaps it takes to connect c'd'...x' with c,d....x. Since the product of two ε's is either +1 or 0, the overall sign of the right side shown must be correct. As an example, Σa,b εabc εabc' = 2!ΣP p P2(δc,c') = 2 δc,c' If c = c' = 2, then this says Σ a,b εab2 εab2 = ε132 ε132 + ε312 ε312 = 1 + 1 = 2 54 The permutation sum shown on the right side of (*) is the determinant of δ(abc..x; a'b'c'..x') but with the first 2 rows and columns crossed out. Therefore, Σa,b εabcd...x εabc'd'...x' = 2! { [cof δ(abc..x; a'b'c'..x')]aa}bb The conclusion then is that (implied summation on a,b on the LHS) This pattern continues as more indices are contracted. If three indices a,b,c are contracted, there will then be 3! hit values which are A,B,C and its permutations, and one just repeats the above discussion. The result will then be Σ a,b,c εabcd...x εabcd'...x' = 3! {{ [cof δ(abc..x; a'b'c'..x')]aa}bb}cc The conclusion then is that (implied summation on a,b,c on the LHS) Eventually one arrives at a point where all but one of the indices are summed, so that ε abcd...x εabcd...x' = (N-1)! | δx,x'| = (N-1)! δ x,x' an example being ε abc2 εabc2 = 3! δ22 = 3! = ε1342 ε1342 + ε1432 ε1432 + 4 more terms = 1+1+4 = 6 The final point is that at which all indices are summed, with result εabcd...x εabcd...x = N! and example of which is ε abcεabc = ε1232 + ε2132 + 4 more terms = 1 + 1 + 4 = 6 55 6. Summary of Results • • • • εabcd...x εabcd...x' = (N-1)! δx,x' εabcd...x εabcd...x = N! (k) Covariant forms of the previous section results The results above were all developed in Ca rtesian x-space where up and down indices on the ε's did not matter. The rules for converting any result above to covariant form are as follows: Weinberg convention: • write the left side as either |g|-1 ε***** ε***** or as E***** E***** .The objects with indices as shown by asterisks are true tensors (weight 0). • write the right side replacing every δ a,b by δa b → ga b as shown in Section 7 (m). Then the right side will also be a true tensor. Example : The εε product for N=2 with no indices summed was written above as (g = 1) εabεa'b' = ⎪⎪ ⎪⎪ δaa' δab' δba' δbb' = δa,a' δb,b' – δa,b' δb,a' 56 The covariant form is as follows, where now g is some arbitrary metric tensor for x-space, EabEa'b' = |g|-1 εabεa'b' = ⎪⎪ ⎪⎪ga a' ga b' gb a' gb b' = ga a' gb b' – ga b' gb a' The equation in x'-space would then be E' abE'a'b' = |g'|-1 ε'abε'a'b' = ⎪⎪ ⎪⎪g'a b g'a b' g'a' b g'a' b' = g'a b g'a b' – g'a' b g'a' b' because true tensor equations are "covariant"(Section 7 (u)). One can raise and lower individual indices to get for example these valid tensor equations which are 3 members of the family of 4! = 24 tensor equations obtained by rais ing and lowering indices: EabEa'b' = |g|-1 εabεa'b' = ⎪⎪ ⎪⎪ga a' ga b' gb a' gb b' = ga a' gb b' – ga b' gb a' Ea bEa'b' = |g|-1 εa bεa'b' = ⎪⎪ ⎪⎪ga a' ga b' gba' gbb' = ga a' gbb' – ga b' gba' EabEa'b' = |g|-1 εabεa'b' = ⎪⎪ ⎪⎪gaa' gab' gba' gbb' = gaa' gbb' – gab' gba' and of course in x'-space the equations are the same but everything is primed. Added-sign-s and Ricci-Levi-Civita conventions: Do the above two bullet items, then add an overall sign s to the right side, because ε abc.. = s g εabc... in these conventions instead of εabc.. = g εabc... so that εabc.. (Weinberg) = sε abc.. (added-sign). Example : The first equation above becomes ( εa'b'→ s εa'b') EabEa'b' = |g|-1 εabεa'b' = s ⎪⎪ ⎪⎪ga a' ga b' gb a' gb b' = s ( ga a' gb b' – ga b' gb a') The second equation is undefined (when s=-1), and the third equation is EabEa'b' = |g|-1 εabεa'b' = ⎪⎪ ⎪⎪gaa' gab' gba' gbb' = gaa' gbb' – gab' gba' The first and third equations are tr ue tensor equations, except individual indices cannot be raised and lowered. If one were doing some significant work i nvolving covariance and s=-1, it would certainly seem advisable to use the Weinberg convention since it is completely "covariant" for either sign of s. 57 Appendix E: Tensor Expansions: direct product, polyadic and operator notation This entire section uses the general Picture A context where x-space need not be Cartesian, The Standard Notation is used throughout. (a) Direct Product Notation The key tool required for the expression of tensor e xpansions is the notion of a direct product of n tensorial vectors defined in this simple way, ( A⊗B⊗C ...) abc... ≡ AaBbCc..... ( A⊗B⊗C ...)a bc... ≡ AaBbCc..... etc The tensor A⊗B⊗C ... is nothing more than the outer product of vectors A,B,C as discussed in Section 7 (a) for contravariant vectors, but later extended to any mixture of vector types. As noted in Section 7 (j), one can define a direct product of two rank-2 tensors in this way, (M⊗N)ab,AB ≡ MaANbB // rank = n = 2; number of tensors = I = 2 (M⊗N)ab ,AB ≡ Ma ANb B etc and then the same idea can be applied to form a direct product of tensors of any rank, for example (M⊗N)ab,AB,αβ = MaAαNbBβ etc // rank = n = 3; number of tensors = I = 2 On the left side the number of groups of indices eq uals the tensor rank n, and the number of indices within each group matches the number I of tensors being direct-product-multiplied. In what follows, only the direct product of vectors shall be considered. One can define the dot product of two direct-product-space vectors in this obvious manner, ( A⊗B⊗C ...) • (A'⊗B'⊗C' ...) ≡ (A⊗B⊗C ...) abc... (A'⊗B'⊗C' ...)abc = A aBbCc..... A'aB'bC'c..... = A•A' B•B' C•C' ... where of course the indices abc can be tilted in any way desired according to Section 7 (k). 58 (b) Tensor Expansions and Bases Let bi be an arbitrary complete set of basis vectors in x-space. As shown in Section 6 (b) there exists a unique set of dual basis vectors bi (also in x-space) such that bi• bj = δi j. Consider then the following expansion of a rank-3 tensor A A = Σijk αijk (bi⊗bj⊗bk) where ( bi⊗bj⊗bk ...)abc = (bi)a (bj)b (bk)c The coefficients αijk can be obtained by dotting both sides with ( bi'⊗bj'⊗bk') and using ( b i'⊗bj'⊗bk') • (bi⊗bj⊗bk) = bi'• bi bj'• bj bk'• bk = δi' iδj' jδk' k . The result is then α ijk = A • (bi⊗bj⊗bk) = Aabc (bi⊗bj⊗bk)abc = Aabc (bi)a (bj)b (bk)c . (*) where Aabc are the contravariant components of tensor A in x-space, and (bi)a are the covariant components of vector bi in x-space. In this manner, a tensor A of any rank can be expa nded on an arbitrary complete set of basis vectors, and the coefficients of that expansion can be obtained by the inversion shown above for rank 3. Two special bases are of interest. The ui are the axis-aligned basis vectors in x-space as discussed in see Section 7 (s). For these basis vectors, one has ( ui)a = δia and ( ui)a = δi a . If one considers this expansion, A = Σijk αijk (ui⊗uj⊗uk) where ( u i⊗uj⊗uk ...)abc = (ui)a (uj)b (uk)c = δia δjb δkc then the coefficients are found to be αijk = A • (ui⊗uj⊗uk) = Aabc δi a δj b δk c = Aijk so the coefficients are exactly the x-space contravariant components of the tensor A. Thus A = Σ ijk Aijk (ui⊗uj⊗uk) On the other hand, if ei are the tangent base vectors in x-space (see Sections 3), the dual vectors are the ei and from Section 7 (s) one has ( ei)a = Sa i = Ria and ( ei)a = Sai = Ri a . If one considers the expansion A = Σijk αijk (ei⊗ej⊗ek) where ( e i⊗ej⊗ek ...)abc = (ei)a (ej)b (ek)c = Ria Rjb Rkc 59 then the coefficients are found to be αijk = A • (ei⊗ej⊗ek) = Aabc (ei)a(ej)b(ek)c = Aabc Ri a Rj b Rk c = Ri a Rj b Rk c Aabc = A'ijk and thus the coefficients in this case are exactly th e x'-space contravariant components of tensor A, as shown in Section 7 (j). Thus, A = Σijk A'ijk (ei⊗ej⊗ek) Expansions like the above are the ge neralizations to tensors of any rank of these vector expansions stated in Section 7 (s), A = Σ iαi bi αi = bi • A // arbitrary basis A = ΣiAi ui // axis aligned unit vectors A = Σ iA'i ei // tangent base vectors where we continue to write rank-1 tensors in bold font: A. To summarize, here is the general rank-n tensor expansion for an arbitrary basis, and then for the two specific bases just discussed: A = Σ ijk... αijk... (bi⊗bj⊗bk...) αijk... = Aabc... (bi)a (bj)b (bk)c... A = Σijk... Aijk... (ui⊗uj⊗uk...) Aijk... = contravariant components of A in x-space A = Σijk... A'ijk... (ei⊗ej⊗ek...) A'ijk... = contravariant components of A in x'-space Orthonormal basis : If the basis vectors bi happen to be orthonormal as defined by bi• bj = δij then bi = bi because the dual basis is uniqu e. As indicated in (*) above, this implies that coefficient αijk is unchanged if any or all indices are lowered, as if these αijk were components of a tensor in some Cartesian space. That Cartesian space is in fact the x'-space that would arise if transformation F were custom-selected such that the bi were the tangent base vectors ei for that F, for then g' ij = ei • ej = δi,j so that x'-space would in fact be Cartesian. But for a pre-determined F, the αijk are just some coefficients and are not components of a tensor relati ve to F, and it just happens that αijk = αijk etc. An example of orthonormal basis vectors arises if bi = e^i ≡ ei/h'i and x'-space has a diagonal metric tensor g' ab = h'a2δa,b. One then has e^i • e^j = δi,j since e^i • e^j = ei • ej / (h'i h'j) = g'ij/ (h'i h'j) = h'i2δij/ (h'i h'j) = δi,j . 60 Then since the dual basis is unique, one has e^i = e^i and then αijk(any up/down) = Aabc (e^i)a (e^j)b (e^k)c = A'ijk (h'ih'jh'k) where the last expression comes from the third expansion shown above. Tensor density . If A is a tensor density of weight W, the ge neral rule is to make this replacement: A'ijk... → J WA'ijk... so the third general expansion above would be written A = J W Σijk... A'ijk... (ei⊗ej⊗ek...) A'ijk... = contravariant components of A in x'-space As justification for this rule, start with a regular tensor transformation for A, A' ijk... = Ri i' Rj j' Rk k'..... Ai'j'k'... The rule gives J W A'ijk... = Ri i' Rj j' Rk k'..... Ai'j'k'... or A' ijk... = J-W Ri i' Rj j' Rk k'..... Ai'j'k'... which is the correct form for the transfor mation of a tensor density (Appendix D). Expansion of tensor-like objects. If Aijk is some "tensor like" object having three indices (such as ∂iTjk) one can still do the three expansions shown above but th e results would have to be restated this way: A = Σijk... αijk... (bi⊗bj⊗bk...) αijk... = Aabc... (bi)a (bj)b (bk)c... A = Σijk... Aijk... (ui⊗uj⊗uk...) Aijk... = components of A in x-space A = Σijk... Aijk... (ei⊗ej⊗ek...) Aijk... = Ri a Rj b Rk c Aabc Since A is not a tensor, in this case Aijk... are not the contravariant components of tensor A in x'-space. (c) Polyadic Notation Some fields of stud y historically use "polyadic notation" as follows, ( ABC...) ≡ A⊗B⊗C ... 61 It is sometimes a bit disturbing to modern readers to see bolded vectors stacked directly against each other, but the direct product makes the meaning clear. For arbitrary basis vectors, one would then have, for example, ( b ibjbk...) ≡ bi⊗bj⊗bk ... Sometimes this basis vector notation is compressed even more, to wit, i j k ... ≡ (bibjbk...) ≡ bi⊗bj⊗bk ... although this notation seems to be mostly used when the b i are the unit vectors ui. In all these notations, one must be aware th at the symbols do not "commute". For example i j = ( bibj) = bi⊗bj => ( i j )nm = (bibj)nm = (bi⊗bj)nm = (bi)n (bj)m ( j i )nm = (bjbi)nm = (bj⊗bi)nm = (bj)n (bi)m ≠ (i j )nm and therefore one cannot write i j = j i. The general expansion stated above now appears as A = Σijk... αijk... (bi⊗bj⊗bk...) = Σijk... αijk... (bibjbk...) = Σijk... αijk... (i j k ...) where α ijk... = A • (bi⊗bj⊗bk...) = A • (bibjbk...) = A • (id jd kd ... ) = Aabc... (bi)a (bj)b (bk)c... where we have just made up a notation id to stand for the dual vector bi. One can find further discussion of polyadic notation for example in Backus. (d) Dyadic Products When two vectors A and B are combined in polyadic notation, the result is called a dyadic product (AB ) [ also known as a dyad or just a dyadic ] ( AB) ij ≡ AiBj // = (A ⊗B)ij In this notation, the expansion give n above for a rank-2 tensor becomes A = Σ ij αij (bibj) αij = Aab(bi)a (bj)b = Aab (bibj)ab 62 Notice that the dyadic product (AB ) is a rank-2 tensor if we assume that the underlying Ai and Bi are the x-space contravariant components of tensorial vectors A and B (which we normally assume). As a reminder, x-space need not be Cartesian. In section (g) below it will be shown that the matrix (AB)ij can be associated with an operator (AB) in the un basis so ( AB)ij = <ui |(AB)| uj >, but this interpretation is not necessary for what follows. (e) Transpose notation for dyadics Superscript T as usual indicates the tran spose of a vector or matrix. For a vector V, certainly Vi = (VT)i, meaning the object in the ith column of V is the same as the object in the ith row of VT . Therefore one can express the dyadic product in this more down-to-earth manner, ( ab)ij ≡ aibj = ai (bT)j = ( abT)ij or ab = ab T Here one knows that ab is a "dyadic" because there is no other meaning for two bolded column vectors abutting each other with no intervening operator, so no special notation like [ ab] is needed to indicate that ab is a dyadic. The object abT on the other hand has a well-defined meaning in matrix algebra, abT = ⎝⎛ ⎠⎞ a1 a2 (b1 b2) = ⎝⎛ ⎠⎞ a1b1 a1b2 a2b1 a2b2 = a matrix and one sees that in fact ( ab) ij = (abT)ij = aibT j = aibj . Meanwhile, the object a Tb is just a number, aTb = a • b = (a1 a2) ⎝⎛ ⎠⎞ b1 b2 = a1b1 + a2b2 = a scalar (if a and b are vectors) This transpose notation can then be applied to the dyadic expansion of a 2x2 matrix A, A = Σij αij bibj = Σij αij bibjT = α11 b1 b1T + α12 b1 b2T ... In the special case that the b i are the unit vectors u i , and assuming N = 2 dimensions, one has A = Σ nm Anm unum = Σnm Anm unumT = A11 u1 u1T + A12 u1 u2T + A21 u2 u1T + A22 u2 u2T = a matrix with A12 in the upper right corner where u n is a column unit vector and unT is the corresponding row unit vector (see comments in Section 3 (c) about "unit" vectors). For example, 63 u1u2T= ⎝⎛ ⎠⎞ 1 0 ( 0 1) = ⎝⎛ ⎠⎞ 0 1 0 0 Obviously this matrix visualization is valid for any di mension N, not just N=2. For rank n > 2, however, this transpose-of-vector concept does not conveniently generalize. For n=3 the object uaubuc would be a cube of zeros with a single 1 located at coordinates a,b,c, and so on for n > 3. One cannot write this as uaubucT for example. The direct product or polyadic notation seems clearest for rank n > 2. (f) Large and small dots used with dyadics Someti mes a small-size dot • is used to indicate th e action of a dyadic (matrix) on a vector. If A is a dyadic then one defines: A• c ≡ Ac = a column vector => (A• c)i = (A c)i = Aijcj => A• c = ΣijAijcj ui c • A ≡ cTA = a row vector => ( c • A)i = (cTA)i = cjAji => c • A = ΣijcjAji ui d • A • c = dTAc = a number = d iAijcj It then follows that, for the particular dyadic A = ab , ( ab) • c ≡ (ab) c = ( abT)c = a(bTc) = a ( b • c) = ( b • c) a = a column vector c • (ab) ≡ cT (ab) = cT(abT) = ( cTa) bT = ( c • a) bT = a row vector d • (ab) • c = dT (ab) c = dTabT c = (dTa)( bT c) = (d • a)(b • c) = a number Here is more detail on the first line of the above group showing a skeletal matrix structure, ( ab) c = ( a bT)c = abTc = a(bTc) = a(b•c) { ⎝⎛ ⎠⎞ x x x x } ⎝⎛ ⎠⎞c1 c2 = {⎝⎛ ⎠⎞ a1 a2 (b1 b2)} ⎝⎛ ⎠⎞c1 c2 = ⎝⎛ ⎠⎞ a1 a2 (b1 b2) ⎝⎛ ⎠⎞c1 c2 = ⎝⎛ ⎠⎞ a1 a2 { (b1 b2) ⎝⎛ ⎠⎞c1 c2} = ⎝⎛ ⎠⎞ a1 a2 b•c The same small dot is used to indicate the product of two dyadics, which is to say, matrix multiplication A• B ≡ AB Regarding this small size dot • : (1) from a matrix algebra point of view, it is completely superfluous except in the case c • A ≡ c TA ; (2) it is completely different from the dot • used in bTc = b•c . It is this larger dot • which was the subject of Section 5 (i); (3 ) The next section provides an explanation of the small dot as part of an operator interpretation for dyadics. (g) Operators and Matrices for Rank-2 tensors Operator concept. As discussed in Section 5 (i), x-space and x'-space of Picture A are both N-dimensional real Hilbert Spaces with scalar product indicated by the large dot •, and one can regard V as a vector in either space. Expressed as a "vector" in x-space one can write, as done above with generic basis b i , 64 V = Σi αi bi Moreover, one can regard a ra nk-2 tensor A as an "operator" in this Hilbert space, A = Σij αij bibjT Application of (bT)n on the left and bm on the right, and then a double use of (bT)nbi = bn• bi = δn i gives αnm = (bT)n A bm Here, one regards A as an operator in the x Hilbert space, whereas α nm is a "matrix" which is associated with the operator A in the particular bn basis. The idea of A as operator has an abstract meaning distinct from the matrix A ij. In the above equation the symbol A is this abstract operator and (bT)n A bm has a meaning distinct from our interpretation of it in terms of the matrix combination of three objects. In the matrix interpretation, one writes (bT)n A bm = [(bT)n]i Aij [bm]j = [bT]i Aij [bm]j and only then does A become a "matrix". This matrix happens to be the contravariant Aij matrix because we happened to select the un basis to write the components like [bm]j = uj • bm. Bra-ket Notation. For the author of this document, the bra-ket notation commonly used in quantum mechanics (Paul Dirac 1939) provides a clean way to look at a rank-2 tensor A as an operator. It is true that in quantum mechanics one usually deals with infinite dimensional Hilbert spaces and complex numbers, but the formalism applies just as well to real Hilbert spaces with finite dimensions. In bra-ket notation one writes bi → |bi>, biT→ <bi| , so that the above equations become |V> = Σ i αi |bi> < b j|bi> = δji orthogonality of the basis A = Σ ij αij | bi> <bj| 1 = Σi | bi><bi| completeness of the basis αnm = <bn | A | bm > <U | V> = U • V = scalar product In this notation, the N |b i> are a set of basis vectors which span an N-dimensional real Hilbert Space, while <b i| span the so-called adjoint (or transpose in our case) Hilbert Space. One then refers to αnm as the "matrix element of the operator A in the bi basis ". In this notation, based on what was presented earlier, one can write, Anm = <un | A | um > = the x-space components of tensor A A' nm = <en | A | em > = the x'-space components of tensor A In all these equations the covariant indices can be m oved up and down in the usual manner. Notice in the last two lines that the operator A between the vertical bars is the exact same operator on both lines. The matrices are different not because the operator has cha nged, but because the basis vectors are different. 65 The point here is that one really can regard a rank-2 tensor A as an operator and not a specific matrix. In a given basis, that operator "has" a certain ma trix. One might write the associated matrix as [A(b)]nm = <bn | A | bm >. That is to say, the matrix needs some label like (b) to indicate the basis used to define the matrix. In our notation, Anm with no label refers to the matrix associated with the um basis, while A'nm goes with the em basis. Bases are related by a transformation. Consider again ( note that |i> ≡ |ui> on the next line ) [A (b)]nm = <bn | A | bm > = ( bn)T A bm = [bn]i Aij [bm]j // = <bn|i><i|A|j><j|bm> where the subscripts i and j are those associated with the u n basis. Defining B jm ≡ [bm]j one has [A(b)]nm = (BT)n i Aij Bjm or [A (b)]n m = (BT)n i Ai j Bj m // lower m and reverse the tilt of index j or A (b) = BT A B // tilted matrix mulitplication as per Section 7 (i) which shows that the matrix elements [A (b)]n m are related to the Ai j by a "congruence transformation" with a matrix B whose colu mns are the basis vectors bm . When b m = um , matrix B is the identity matrix, and when bm = em , one has B jm ≡ [bm]j = (em)j = Rm j = Sjm . Thus B = S in this case. It was shown in Section 7 (i) that in standard notation S is real orthogonal, so in fact one has for the bm = em basis, A (e) = BT A B = ST A S = S-1 A S = R A R-1 and our two matrices are related by similarity by our old friends S and R as shown. More on bra-ket notation and its relation to the small dyadic dot. Consider the following facts, <d | A | c > = d T A cm = dT [A c ] = <d |Ac > <d | A | c > = d T A cm = [ dT A] c = [AT d]T c = <ATd | c> where |(Ac) > = a new Hilbert space vector which results when A is applied to |c> <(A Td) | = a new transpose Hilbert space vect or which results when A is applied to <d| 66 So one has this general idea that <d | A | c > = <d |Ac > = <ATd | c> A | c > = |(Ac)> <d | A = <(A Td) | In this last line, the isolated A's are the same operator A sitting in the Hilbert space. This operator can "act" either to the right or to the left as shown. The object |(Ac)> ≡ |e> is some different vector in the Hilbert space (different from |c>), call it |e>, and the grouping (Ac) labels this vector. Similarly, <(A Td) | is vector <e| in the transpose Hilbert space. The distinction between A as an abstract operator in the Hilbert space, and the A in (Ac) and (ATd) = (dTA)T as vectors in the Hilbert space is a subtle one. It is just this distinction that is implied by the small dot in the dyadic notation discussed in the previous section, and here is the correspondence between the dyadic notation and the bra-ket notation: A• c = A c d • A = (ATd)T = dTA d • A • c = dTAc A• B c A | c > = |Ac> <d | A = <(ATd) | <d | A | c > = <d | Ac > AB| c > In the rightmost column operator B is applied first to | c> to get vector |(Bc)>, and then operator A is applied to |(Bc)> to give yet another vector | (Abc )>. In bra-ket notation the product of two abstract operators is given just as AB, but in dyadic notation it is written A• B. Dyadics as operators. According to the above discussi on, one can regard a dyadic ( AB), being a rank-2 tensor, as an operator and not as a matrix. The matrix Tnm = (AB)nm = AnBm is specific to the un basis in x-space (again, one might have g ≠1) Tnm = (AB )nm = <un |(AB)| um > = ( un)T A BT um = [ ( u n)T]a Aa (BT)b [um]b = δn a Aa Bb δm b = AnBm . In the generic b n basis one might then write something like [( AB) (b)]nm = <bn |(AB)| bm > It is to emphasize this operator view of a dyadic that Morse and Feshbach use fancy letters like U to represent dyadics. Then the small-dot notation U • B emphasizes the idea of an operator acting on a vector, equivalent to U| B> . Here then are a few quotes from Morse and Feshbach ( an = un) to illustrate some of the notation described above. These authors are working in Cartesian space (g=1) where up and down indices don't matter. ( The first item here is our A• c = ΣijAijcj ui from above. ) 67 Notice the impressive name "idemfactor" for the identity operator 1 = Σi | ai><ai| = Σi aiaiT = Σiaiai. 68 Appendix F: The Affine Connection Γc ab and Covariant Derivatives (a) Definition and Interpretation of Γ : Γc ab = ec • (∂aeb) = Rc i(∂aRbi) Context can be confusing in a discussion of the affine connection Γ, so we start with a modified Picture C in which the quasi-Cartesian space on the right is called ξ-space instead of x(0)-space as in Picture C. The notation ξi for the coordinates of ξ-space seems traditional in general re lativity writing, an application in which the Γ object appears frequently. Recall from Section 1 that the metric tensor G is a diagonal matrix whose elements are independently +1 or -1. And recall from Section 5 (b) that in x-space, g ab = RaiRbjGij. If G = 1, then the xi coordinates of x-space are the "curvilinear coordinates" and the ξi are the "Cartesian coordinates". In Picture C1 the tangent base vectors en exist in ξ -space and the components of en are given by (en)i = Rni, as in Example 1 of Section 3 (polar coordinates). Since matrix R is the linearization of transformation F at ξ and since x = F(ξ), one can regard R as a function either of ξ or x . We choose x as the variable and write [ en(x)]i = Rni(x). For example, Example 2 of Section 3 (spherical coordinates) showed that eφ(x) = rsinθ φ^(r,θ,φ). In any event, en(x) varies with x , and one wonders just how en varies with x. For a small variation d x in the x-space coordinates, one has d( e n)i = ∂j(en)i dxj ∂ j ≡ ∂/∂xi or (d e n)i = (∂jen)i dxj Since e n and (∂jen) are both vectors in ξ-space, and since the e k are known to form a complete basis in ξ- space, it must be possible to expand ( ∂jen) on the ek with some appropriate coefficients, call them Γk jn : (∂ jen) = Γk jn ek Dotting this equation into e k gives Γk jn = ek • (∂jen) = Rk i(∂jRni) where we recall from Section 7 (s) that ( ek)i = Rk i and ( en)i = Rni. These coefficients Γk jn(x) comprise a tensor-like field called the affine connection , so we shall regard the above line as the definition of Γk jn in the context of Picture C1 above. With more standard index names, the above becomes Γc ab = ec • (∂aeb) = Rc i(∂aRbi) 69 From Section 7 (q) we know that, for Picture C1, R c i = (∂xc/∂ξi) Rbi = (∂ξi/∂xb) (∂aRbi) = (∂2ξi/∂xa∂xb) = (∂bRai) Therefore one can write Γ c ab = Rc i(∂aRbi) = (∂xc/∂ξi) (∂2ξi/∂xa∂xb) = ∂xc ∂ξi ∂2xi ∂ξa∂ξb which form appears in Weinberg p 100 (4.5.1). Notice again that ( ∂bRcn) = (∂cRbn) and that Γc bc is symmetric on the lower two indices. In the next section an alternate form for Γ is derived, so both forms will be stated here: Γc ab = Rc i(∂aRbi) // ∂a = ∂/∂xa Γc ab = – Rbi (∂aRc i) (b) Identities of the form ( ∂aRd n) = – Re n Rd m (∂aRem) The identities are : R d m (∂aRem) = – Rem (∂aRd m) 1 (∂aRd n) = – Re n Rd m (∂aRem) 2 ∂a ≡ ∂/∂xa (∂ aRdn) = – Ren Rdm (∂aRe m) 3 The second two lines are just restatements of th e first line, but all are derived below. Corollary: The first identity above allows an alternate fo rm for the affine connection in terms of R: Γc ab = Rc i(∂aRbi) // as in section (a) above Γc ab = – Rbi (∂aRc i) // alternate form Our context is: 70 Proof: These identities are a simple consequence of the f act that RS = 1 which in standard notation is written δc b = Rc αRbα (one of the orthogonality rules). So, 0 = ∂a(δd e) = ∂a(Rd mRem) = Rd m (∂aRem) + Rem (∂aRd m) QED 1 (*) Apply Σe Re n to both sides of (*) to get ( or, just use the Inversion Rule of Section 7 (r) ) 0 = Re n Rd m (∂aRem) + (Re n Rem) (∂aRd m) = Re n Rd m (∂aRem) + δnm (∂aRd m) = Re n Rd m (∂aRem) + (∂aRd n) => (∂ aRd n) = – Re n Rd m (∂aRem) Q E D 2 Alternatively, apply Σ d Rdn to both sides of (*) to get 0 = (R dn Rd m) (∂aRem) + Rdn Rem (∂aRd m) = δn m (∂aRem) + Rdn Rem (∂aRd m) = (∂aRen) + Rdn Rem (∂aRd m) => (∂ aRen) = – Rdn Rem (∂aRd m) now swap d and e: => (∂aRdn) = – Ren Rdm (∂aRe m) Q E D 3 (c) Identities of the form ( ∂cgab) = – [gan Γ b cn + gbn Γa cn] The derivatives of the metric tensor are given by ( ∂c = ∂/∂xc) (∂cgab) = – [gan Γ b cn + gbn Γa cn] 1 (∂ cgab) = + [g an Γn cb + gbn Γn ca] 2 Proof of 1: ( ∂cgab) = – [gan Γ b cn + gbn Γa cn] The LHS is given by LHS = (∂ cgab) = ∂c(Ra iRb i)Gii = Ra i(∂cRb i)Gii + Rb i(∂cRa i)Gii For the RHS, the Γ objects can be replaced by their alternate definitions from section (a) Γc ab = – Rbi (∂aRc i) // from section (a) Γb cn = – Rni (∂cRb i) // b →n then c →b then a →c Γa cn = – Rni (∂cRa i) 71 The RHS of the claimed identity may then be written RHS = – g an Γ b cn – gbn Γa cn = { Ra kRn kGkk }{Rni (∂cRb i)} + {Rb kRn kGkk }{Rni(∂cRa i)} = Ra k(Rn k Rni) (∂cRb i) Gkk + Rb k(Rn k Rni) (∂cRa i) Gkk = R a kδki (∂cRb i) Gkk + Rb kδki (∂cRa i) Gkk = R a i (∂cRb i) Gii + Rb i (∂cRa i) Gii = LHS QED Proof of 2: ( ∂cgab) = + [g an Γn cb + gbn Γn ca] The LHS is given by LHS = (∂ cgab) = ∂c(RaiRbi)Gii = Rai (∂cRbi)Gii + Rbi (∂cRai)Gii For the RHS, the Γ objects can be replaced by their primary definitions from section (a) Γ c ab = Rc i(∂aRbi) Γn cb = Rn i(∂cRbi) // c →n then a →c then i→k Γn ca = Rn i(∂cRai) The RHS of the claimed identity may then be written RHS = g an Γn cb + gbn Γn ca = { R akRnkGkk}{Rn i(∂cRbi)} +{RbkRnkGkk}{Rn i(∂cRai)} = R ak (Rnk Rn i)(∂cRbi)Gkk + Rbk(Rnk Rn i)(∂cRai)Gkk = R ak δk i(∂cRbi)Gkk + Rbk∂k i(∂cRai)Gkk = R ai (∂cRbi)Gii + Rbi (∂cRai)Gii = LHS QED (d) Identity: Γd ab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] The identity states that Γ d ab may be expressed entirely in terms of the metric tensor, Γ d ab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] 72 Recall in our definition above, Γd ab = Rd k(∂aRbk), that Γ was given in terms of R matrices. The following corollary ( derived at the end of this section ) concerns contraction of the upper Γ index with a lower one: Γa an = (1/2) gad ∂ngad = (1/2)(1/g) ∂ng = (1/ g ) ∂n(g ) Proof: Γd ab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] We know that g ab = RaeRbe Gee gdc = Rd iRc i Gii The first line below is computed from the first line in the pair above, then the next two lines below are obtained by doing forward cyclic permutations of the first line : ∂ cgab = [Rbe (∂cRae) + Rae(∂cRbe) ]Gee ∂agbc = [Rce (∂aRbe) + Rbe(∂aRce) ]Gee ∂bgca = [Rae (∂bRce) + Rce(∂bRae) ]Gee . The last four lines can be inserted into the Right Hand Side of our desired identity to obtain (RHS)d ab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] = ( 1 / 2 ) (R d iRc i Gii) Gee * [Rce (∂aRbe) + Rbe(∂aRce) + Rae (∂bRce) + Rce(∂bRae) – Rbe (∂cRae) – Rae(∂cRbe) ] 1 2 3 4 5 6 Due to the symmetry noted above in section (a), ( ∂iRje) = (∂jRie), terms 2 and 5 cancel as do terms 3 and 6, while terms 1 and 4 are equal. Therefore, (RHS)d ab = (1/2) Rd iRc i Gii Gee * 2 Rce (∂aRbe) = R d i (Rce Rc i) Gii Gee (∂aRbe) = Rd i δe i Gii Gee (∂aRbe) = R d e Gee Gee (∂aRbe) = Rd e(∂aRbe) = Γd ab QED 73 Proof of Corollary: The abovementioned corollary is this (g ≡ det(gab) ) Γa an = (1/2) gad( ∂agnd + ∂ngad – ∂dgan ) = (1/2) gad ∂ngad = (1/2) (1/g) ∂ng = (1/ g ) ∂n(g ) where the first and third terms cancel due to symmetry gad∂agnd – gad∂dgan = gad∂agnd – gda∂agdn = gad∂agnd – gad∂agnd = 0 and the fact that (1/g) ∂ ng = gab(∂ngab) is proved as follows: (1) g ab = (g-1)ab = cof(gab)T/det(gab) = cof(g ab)/g => cof(g ab) = g gab (2) g = det(g ab) = Σa gabcof(gab) => ∂g/∂gab = cof(g ab) = g gab (3) ∂ ng = ∂ g/∂xn = (∂ g/∂gab)( ∂gab/∂xn) = g gab (∂ngab) => (1/g) ∂ng = gab(∂ngab) The final form shown is just calculus : g -1/2 ∂n(g1/2) = g-1/2 (1/2) g-1/2 ∂n(g) = (1/2) (1/g) ( ∂ng) (e) Picture D1 Context Our main context of interest is Picture A, . For the proof given in the next section below, it is useful to think of Picture A as the top part of this Picture D1, The relationship between the R's and S's are these 74 R = R' R-1 = R' S => R' = R R S = R R '-1 = R S' The tangent and reciprocal base vectors in ξ-space associated with transformations F and F ' are these: (en)i = Rni ( en)i = Rn i x-space (e'n)i = R'ni ( e'n)i = R'n i x'-space Now there are two affine connections, Γc ab ≡ (∂xc/∂ξn) (∂2ξn/∂xa∂xb) = Rc n ∂a (∂ξn/∂xb) = Rc n(∂aRbn) ∂a = ∂/∂xa = [ ec]i • (∂a[eb]i) = ec • (∂aeb) Γ 'c ab ≡ (∂x'c/∂ξn) (∂2ξn/∂x'a∂x'b) = R'c n ∂'a (∂ξn/∂x'b) = R'c n(∂'aR'bn) ∂'a = ∂/∂x'a = [ e'c]i • (∂'a[e'b]i) = e'c • (∂'ae'b) (f) Relations between Γ and Γ ' The claimed relations are the following in the context of Picture A shown above, Γ 'c ab = Rc d Raα Rbβ Γd αβ + Rc α (∂'aRbα) // Weinberg p 100 (4.5.2) Γ 'c ab = Rc d Raα Rbβ Γd αβ – Rbβ(∂'aRc β) Γ 'c ab = Rc d Raα Rbβ Γd αβ – Raα Rbβ (∂αRc β) // Weinberg p 102 (4.5.8) If the second term were not present, the relation would state that Γ d αβ transforms as a mixed rank-3 tensor in the usual manner (Section 7 (j)). Since the second term is present, Γd αβ is not a tensor. Proof of the first relation : We now make use of Picture D1 shown above. Start with the Γ ' definition given above, Γ 'c ab ≡ R'c n(∂'aR'bn) = (R R)c n ∂'a(RR)bn = Rc dRd n ∂'a(RbβRβn) // R' = R R = R c dRd nRbβ(∂'aRβn) + Rc d(Rd nRβn)( ∂'aRbβ) = R c dRd nRbβ([Raα∂α]Rβn) + Rc d(δd β)( ∂'aRbβ) // ∂'a = Raα∂α in first term only = R c dRaαRbβRd n(∂αRβn) + Rc β (∂'aRbβ) = R c dRaαRbβ Γd αβ + Rc α (∂'aRbα) // Γd αβ ≡ Rd n(∂αRβn) QED 75 Magically, all the R's have gone away. The second term in the above relation can be written a different manner as follows. Consider, 0 = ∂'a(δc b) = ∂ 'a(Rc αRbα) = Rc α (∂'aRbα) + Rbα (∂'aRc α) => R c α (∂'aRbα) = – Rbα(∂'aRc α) = – Rbβ(∂'aRc β) = – R bβ Raα(∂αRc β) and this gives the other two relations stated above. (g) Statement and Proof of the Covariant Derivative Theorem Many examples of this theorem will be given later. In this proof it is assumed that the tensor density of interest is purely covariant (all indices "down"). In the next section it will then be shown how to adjust the theorem if one or more of the tensor density indices is "up". Covariant Derivative Theorem: The covariant derivative (B abc..x; α as defined below) of a covariant tensor density of ra nk n and weight W (B abc..x ) transforms as a covariant te nsor density of rank n+1 and weight W. The implication is that all the indices including α of Babc..x; α can be treated as ordinary tensor indices with respect to raising, lowering, contr action, and so on. The first term in B abc..x; α is the regular derivative ∂α Babc..x , often written as B abc..x, α (comma, not semicolon), and this first term is not a tensor. Only when all the "correction terms" are included does the object become a tensor. The covariant derivative in x-space and then in x'-space is defined as follows: (Weinberg p 104 4.6.12) B abc..x; α ≡ ∂α Babc..x – Γn aαBnbc..x – Γn bαBanc..x – .... – Γn xαBabc..n // x-space del a-term b-term x-term + ( W / 2 g ) ( ∂ αg) Babc..x B'abc..x; α ≡ ∂'α B'abc..x – Γ 'n aαB'nbc..x – Γ 'n bαB'anc..x – .... – Γ 'n xαB'abc..n // x'-space del a-term b-term x-term + ( W / 2 g ' ) ( ∂'αg') B'abc..x The two definitions are the same except everything is pr imed in x'-space (except constant weight W). This is as one would expect if the x-space equation were a "true tensor equation" as discussed in Section 7 (u) and were therefore "covariant". Although the Γ objects are not tensors themselves, the combination of terms shown in the definition of B abc..x; α is a rank n+1 tensor (as will be demonstrated). As was shown in section (d), (1/2)(1/g) ∂ αg = Γκ κα so the W terms could be written as W Γκ κα Babc..x and W Γ 'κ κα B'abc..x , and this form is commonly seen in the literature on this subject. 76 A proof of the theorem must then show that B abc..x; α as defined above in fact transforms as a tensor density of rank n+1 and weight W, which is to say, one must show that B ' abc..x; α = J-W Raa'Rbb'..... Rxx' Rαα' Ba'b'c'..x'; α' or B'ABC..X; α = J-WRαα'{RAa'RBb'..... RXx'} * B a'b'c'..x'; α' or, in gory detail, ∂' αB'ABC..X – Γ 'n AαB'nBC..X – Γ 'n BαB'AnC..X – ........... – Γ 'n XαB'ABC..n // LHS del' a'-term b'-term x'-term + ( W / 2 g ' ) ( ∂'αg') B'ABC..X = J -W Rαα'{RAa'RBb'..... RXx'} * / / R H S { ∂α'Ba'bc'..x' – Γn a'α'Bnb'c'..x' – Γn b'α'Ba'nc'..x' – .. – Γn x'α'Ba'b'c'.. n del a-term b-term x-term + ( W / 2 g ) ( ∂α'g) Ba'b'c'..x' } Proof: 1. Expand the LHS del' term and show del'-del matches the RHS del term. ∂'αB'ABC..X = (Rαβ∂β)( J-W RAaRBb..... RXx Babc..x ) del' = R αβ(∂β J-W) RAaRBb..... RXx Babc..x d e l ' - J = R αβ J-W (∂βRAa)RBb..... RXx Babc..x d e l ' - a + R αβ J-W RAa(∂βRBb)..... RXx Babc..x del'-b ... + R αβ J-W RAaRBb. ... (∂βRXx) Babc..x del'-x + R αβ J-W RAaRBb.............R Xx (∂β Babc..x ) del'-del We first claim that this del'-del term of the LHS matches the del term on the RHS: del'-del LHS = R αβ J-W {RAaRBb...........R Xx} (∂β Babc..x ) del RHS = J-W Rαα'{RAa'RBb'..... RXx'} {∂α' Ba'bc'..x' } Unpriming all the Latin indices and setting α' = β shows that these two terms indeed match. 77 2. Show that the a-related terms balance. The a'-term on the LHS is this – Γ 'n AαB'nBC..X The connection between Γ ' and Γ given in section (f) says Γ 'c ab = Rc d Raκ Rbσ Γd κσ – Raκ Rbσ (∂κRc σ) or Γ 'n Aα = Rn d RAκ Rασ Γd κσ – RAκ Rασ (∂κRn σ) Inserting the above for Γ 'n Aα and the transformation rule for B', the LHS a'-term becomes – { R n d RAκ Rασ Γd κσ – RAκ Rασ (∂κRn σ)} { J-W Rnn'RBbRCc....Rxx Bn'bc..x } and to this we must add the c ontribution called del'-a above. Meanwhile, the RHS a-term is this J-W Rαα'{RAa'RBb'..... RXx'}{– Γn a'α'Bnb'c'..x' } // remove Latin primes, then n →n' = – { J -W Rαα'{RAaRBb..... RXx}{Γn' aα'Bn'bc..x } so we have to show that – { R n d RAκ Rασ Γd κσ – RAκ Rασ (∂κRn σ)} { J-W Rnn'RBbRCc....Rxx Bn'bc..x } + R αβ J-W (∂βRAn')RBb..... RXx Bn'bc..x // a →n' , this is the del'-a term = – { J-W Rαα'{RAaRBb..... RXx}{Γn' aα'Bn'bc..x } ? where now the del'-a term has been added in to the LHS, and in so doing index a → n'. We can see that the factors J-W RBbRCc....Rxx Bn'bc..x are the same on both sides so they can be removed to give a simpler relation which we must show is valid: – { R n d RAκ Rασ Γd κσ – RAκ Rασ (∂κRn σ)} { Rnn' } + R αβ (∂βRAn') = – { R αα'{RAa }{Γn' aα'} In the first term one sees Rn d Rnn' = δdn' which pins d to n' in that term only, so the above becomes – R Aκ Rασ Γn' κσ + Rnn'RAκ Rασ (∂κRn σ) + Rαβ (∂βRAn') = – R ασRAκ Γn' κσ . The first term on the left cancels the term on the right, and do β→σ in the third term to get RAκ Rασ {Rnn'(∂κRn σ)} + R ασ (∂σRAn') = 0 . 78 Then cancel the common R ασ factor and use the symmetry ( ∂κRn σ) = (∂σRn κ) to get (∂σRAn') = – Rnn'RAκ (∂σRn κ) Now in this order do σ→ a, A→d, n→e, n'→n, κ→m to get (∂aRdn) = –Ren Rdm (∂aRe m) . But this is seen to be the third identity of secti on (b)! By reversing the above sequence of steps, one shows that the three a-related terms in the above LHS = RHS equation balance: a'-term + del'-a = a-term. or – { R n d RAκ Rασ Γd κσ – RAκ Rασ (∂κRn σ)} { Rnn'RBbRCc....Rxx Bn'bc..x } + R αβ (∂βRAn')RBb..... RXx Bn'bc..x = – { R αα'{RAaRBb..... RXx}{Γn' aα'Bn'bc..x } In reversing the sequence, one of course adds back in the deleted common factors as well as the various implied index sums. It seems clearer to state the proof th is way rather than start with the last equality with no justification and artificially thread backwards to the desired equation. This method is used as well in the next section. 3. Show that the b-related terms balance. In the previous equation, which was shown true, do A,a ↔B,b (indices) and B n'bc..x → Ban'c..x to get the following known-valid equation : – { Rn d RBκ Rασ Γd κσ – RBκ Rασ (∂κRn σ)} { Rnn'RAaRCc....Rxx Ban'c..x } + R αβ (∂βRBn')RAa..... RXx Ban'c..x = – { R αα'{RBbRAa..... RXx}{Γn' bα' Ban'c..x } . The first line is in fact the b'-term (– Γ 'n BαB'AnC..X ), the second line is del'-b, and the RHS is the b-term. This shows that the three b-re lated terms match LHS = RHS. 4. Similarly, the c,d.....x terms match . Just repeat item 3 above for each extra index. 5. It remains to show that the three terms so far neglected match as well . These terms are LHS: R αβ(∂β J-W) RAaRBb..... RXx Babc..x // the del'-J term + (W/2g') ( ∂'αg') B'ABC..X // the LHS W term RHS: J-W Rαα'{RAa'RBb'..... RXx'}(W/2g) ( ∂α'g) Ba'b'c'..x' // the RHS W term That is, one must show that 79 Rαβ(∂β J-W) RAaRBb..... RXx Babc..x + (W/2g') ( ∂'αg') B'ABC..X = J-W Rαα'{RAa'RBb'..... RXx'}(W/2g) ( ∂α'g) Ba'b'c'..x' Expanding B' ABC..X in the second term gives R αβ(∂β J-W) RAaRBb..... RXx Babc..x + (W/2g') ( ∂'αg') J-W{RAa'RBb'..... RXx'} * Ba'b'c'..x' = J -W Rαα'{RAa'RBb'..... RXx'}(W/2g) ( ∂α'g) Ba'b'c'..x' . Now unprime the primed Latin indices and set β = α' in the first term to get Rαα'(∂α' J-W) RAaRBb..... RXx Babc..x + (W/2g') ( ∂'αg') J-W{RAaRBb..... RXx} * Babc..x = J-W Rαα'{RAaRBb..... RXx}(W/2g) ( ∂α'g) Babc..x Next, remove all common factors (and associated implied sums) to get R αα' (∂α' J-W) + (W/2g') (∂ 'αg') J-W = J-W Rαα' (W/2g) (∂α'g) Since ∂' α = Rαα'∂α' this becomes (∂' α J-W) + (W/2g') (∂ 'αg') J-W = J-W (W/2g) (∂'αg) Move the second term to the RHS and do the derivative to get (-W) J -W-1(∂'α J) = J-W (W/2)[ (∂'αg)/g – (∂'αg')/g' ] so it remains then to show that J -1(∂'α J) = (1/2)[ ( ∂'αg')/g' - (∂'αg)/g ] From Section 5 (k) one has J2 = g'/g and J = (g'/g)1/2 so we need to show that ∂'α (g'/g)1/2 = (1/2) (g'/g)1/2[ (∂'αg')/g' - (∂'αg)/g ] or (1/2) (g'/g) -1/2 ∂'α (g'/g) = (1/2) (g'/g)1/2[ (∂'αg')/g' - (∂'αg)/g ] or ∂' α (g'/g) = (g'/g) [ ( ∂'αg')/g' - (∂ 'αg)/g ] = [ g( ∂'α g') – g' (∂'α g)]/g2. (*) But (finally) evaluation of the LHS of (*) gives [ g(∂'α g') – g' (∂'α g)]/g2 80 which shows that equation (*) is valid. Once again, by reversing the above sequence of steps, one shows that the remaining three terms are in balance. QED (h) Rule for raising any index on a covariant deriv ative of a covariant tensor density. The Rule is stated at the end of this section before the examples. Consider the general form given in section (g) for a covariant derivative B abc..x; α ≡ ∂α Babc..x – Γn aαBnbc..x – Γn bαBanc..x – .... – Γn xαBabc..n // x-space del a-term b-term x-term + ( W / 2 g ) ( ∂ αg) Babc..x Notice that there are n indices on B abc..x and there are n corresponding terms on the RHS in addition to the del and W terms. Each index of B abc..x thus has its own "correction te rm". What happens if one of the indices (say b) on B abc..x is raised? To find out, apply gβb to both sides. The effect of doing this is trivial for all terms except the del term and the b-term, since b is a regular tensor index on all such terms, so we get Baβ c..x;α ≡ gβb ∂α Babc..x – Γn aαBnβ c..x – gβb Γn bαBanc..x – .... – Γn xαBaβ c..n del a-term b-term x-term + ( W / 2 g ) ( ∂αg) Baβ c..x The del term can be written gβb (∂α Babc..x ) = ∂α (gβb Babc..x ) – (∂α gβb) Babc..x = ∂α Baβ c..x – (∂α gβb) Babc..x The second term here can be combined with the b-term to give del-extra + b-term = – ( ∂ α gβb) Babc..x – gβb Γn bαBanc..x = – (∂α gβn) Banc..x – gβb Γn bαBanc..x // b →n in first term only = [– ( ∂α gβn) – gβb Γn bα] Banc..x The first identity of section (c) reads (∂ cgab) = – [gai Γ b ci + gbi Γa ci] or [– (∂ cgab) – gai Γ b ci] = gbi Γa ci // now do c →α, b→n, a→β or [– (∂ αgβn) – gβi Γ n αi] = gni Γβ αi or [– (∂αgβn) – gβb Γ n αb] = gni Γβ αi 81 Therefore, del-extra + b-term = g ni Γβ αi Banc..x = Γβ αi Bai c..x = Γβ αn Ban c..x so it has been shown that Baβ c..x;α ≡ ∂α Baβ c..x – Γn aαBnβ c..x + Γβ αn Ban c..x – ... – Γn xαBaβ c..n + W Γκ κα Bab c..x new b term Here then is a comparison B abc..x; α ≡ ∂α Babc..x – Γn aαBnbc..x – Γn bαBanc..x – .... – Γn xαBabc..n + W Γκ κα Babc..x Bab c..x;α ≡ ∂α Bab c..x – Γn aαBnb c..x + Γb αn Ban c..x – .... – Γn xαBab c..n + W Γκ κα Bab c..x del a-term b-term x-term W-term Rule for raising some non-last index q: (1) In all terms, raise the corresponding B index. (2) in the q correction term, make the replacement – Γ n qα → + Γq nα ( = Γq αn ) Corollary : In order to construct the covariant derivative of any rank-n tensor density B------ (indices in any positions), write out the above form and use a covariant correction term for each covariant index (such as – Γn aαBnb c..x for the a index above) and use a contravariant correction term for each contravariant index (such as + Γb αn Ban c..x for the b index above). Rule for raising the last index α: Raising the ; α index must be done "manually" so the first term will have gαα'∂α = ∂α' and all remaining terms will have explicit gαα' factors. (i) Examples of covariant derivative expressions It will be assumed that all B objects in the exam ples are true tensors unless otherwise specified. Example J = 0 (covariant derivative of a scalar B) // J is the rank of the B tensor B;α = ∂α B covariant vector // = B ,α B;α = ∂α B contravariant vector // = B,α The above examples also apply if B is a scalar density of any weight W. Since ∂ αB is then a vector density of weight W (the a ddition rule), one will have ( ∂'αB') = J-WRαα'(∂α'B) and no "correction terms" are required. 82 Example J=1: (covariant derivative of a vector B) Ba;α = ∂α Ba – Γn aαBn covariant rank-2 tensor // 2nd term is symmetric on a ↔α B a ;α = ∂α Ba + Γa αn Bn mixed rank-2 tensor To obtain the other two possibilities, it is necessary to apply the metric tensor g αβ and gαβ∂β = ∂α , B a;α = ∂α Ba – gαβΓn aβBn mixed rank-2 tensor B a;α = ∂α Ba + gαβΓa βn Bn contravariant rank-2 tensor Example J=2: (covariant derivative of a rank-2 tensor) B ab;α ≡ ∂α Bab – Γn aαBnb – Γn bαBan covariant rank-3 tensor Ba b;α ≡ ∂α Ba b + Γa αn Bn b – Γn bαBa n etc. B ab ;α ≡ ∂α Bab – Γn aαBnb + Γb αn Ban B ab ;α ≡ ∂α Bab + Γa αn Bnb + Γb αn Ban Again, application of g αβ would give expressions for the other four possibilities with : α being "up". These "other possibilities" are always present, but we shall no longer mention them in the following examples, Example J=3: (covariant derivative of a rank-3 tensor Babc;α ≡ ∂α Babc – Γn aαBnbc – Γn bαBanc – Γn cαBabn B a bc;α ≡ ∂α Ba bc + Γa αn Bn bc – Γn bαBa nc – Γn cαBa bn ... B abc ;α ≡ ∂α Babc + Γa nαBnbc + Γb nαBanc + Γc nαBabn Special J=2 application to the metric tensor: gab;α ≡ ∂α gab – Γn aαgnb – Γn bαgan = 0 by section (c) identity 2 g a b;α ≡ ∂α ga b + Γa αn gn b – Γn bαga n = 0 + Γa αb – Γa bα = 0 g ab ;α ≡ ∂α gab – Γn aαgnb + Γb αn gan = 0 – Γb aα + Γb αa = 0 g ab ;α ≡ ∂α gab + Γa αn gnb + Γb αn gan = 0 by section (c) identity 1 83 The middle lines use the fact that gi j = δi j. Since g ab;α is tensor, knowing that any one of the above vanishes implies that all four lines vanish! The net result is gab;α = ga b;α = gab ;α = gab ;α = 0 // Weinberg p 105 (4.6.16,17,18) The covariant derivative of any form of the metric tensor vanishes. As Weinberg points one, one knows that in a quasi-Cartesian x-space g ab;α = 0 since g ab = Gaaδa,b and Γ = 0. Then in any x'-space g' ab;α = 0 as well since g' ab;α = Raa' Rbb' Rαα' ga'b';α' . Example J=2: (double covariant derivatives) Consider again the J=1 examples from above B a;α = ∂α Ba – Γn aαBn Ba ;α = ∂α Ba + Γa αn Bn This applies to any vector B a. As the J=0 example shows, B ;a and B;a are bona-fide vectors (covariant and contravariant components of the same vector ) and therefore B;a;α = ∂α B;a – Γn aαB;n B;a ;α = ∂α B;a + Γa αn B;n where we have simply inserted a semicolon in each term. Consider again the J=2 examples from above B ab;α = ∂α Bab – Γn aαBnb – Γn bαBan Ba b;α = ∂α Ba b + Γa αn Bn b – Γn bαBa n This applies to any rank-2 tensors B ab or Ba b. According to sections (i) and (h) above, B a;b and Ba ;b are bona-fide rank-2 tensors, and therefore B a;b;α = ∂α Ba;b – Γn aαBn;b – Γn bαBa;n Ba ;b;α = ∂α Ba ;b + Γa αn Bn ;b – Γn bαBa ;n In a similar manner one can derive expressions for triple covariant derivativ es and beyond. For example B a;b;c;α = ∂α Ba;b;c – Γn aαBn;b;c – Γn bαBa;n;c – Γn cαBa;b;n The next examples are for tensor densities : (The trivial J=0 cases were already considered above.) 84 Example J = 1 (vector density of weight W) // recall Γκ κα = (1/2g) ∂αg Ba;α = ∂α Ba – Γn aαBn + W Γκ κα Ba covariant rank-2 tensor density Ba ;α = ∂α Ba + Γa αn Bn + W Γκ κα Ba Example J = 2 (rank-2 tensor density of weight W) Bab;α = ∂α Bab – Γn aαBnb – Γn bαBan + W Γκ κα Bab covariant rank-3 tensor density Ba b;α = ∂α Ba b + Γa αn Bn b – Γn bαBa n + W Γκ κα Ba b Bab ;α = ∂α Bab – Γn aαBnb + Γb αn Ban + W Γκ κα Bab Bab ;α = ∂α Bab + Γa αn Bnb + Γb αn Ban + W Γκ κα Bab As noted in Appendix D (b) item 3 Example 2, adding a factor g W/2 to a tensor density of weight W neutralizes the weight, and the result is a regular tens or. Here then are a few examples in which this is done. Since the product is a tensor, there are no W correction terms. Example J = 0 (covariant derivative of a scalar density B of weight W) (gW/2B);α = ∂α(gW/2B) covariant vector (gW/2B);α = ∂α(gW/2B) contravariant vector Example J=1: (covariant derivative of a vector density B of weight W) (g W/2Ba);α = ∂α (gW/2Ba) – gW/2 Γn aα Bn covariant rank-2 tensor (gW/2Ba);α = ∂α (gW/2Ba) + gW/2 Γa αn Bn mixed rank-2 tensor Example J=2: (covariant derivative of a tensor density B of weight W) (gW/2Bab);α = ∂α (gW/2Bab) – Γn aα(gW/2Bnb) – Γn bα(gW/2Ban) covariant rank-3 tensor (gW/2Ba b);α = ∂α (gW/2Ba b) + Γa αn (gW/2Bn b) – Γn bα(gW/2Ba n) etc. (gW/2Bab);α = ∂α (gW/2Bab) – Γn aα(gW/2Bnb) + Γb αn (gW/2Ban) (gW/2Bab);α = ∂α (gW/2Bab) + Γa αn (gW/2Bnb) + Γb αn (gW/2Ban) (j) The Leibnitz rule for the covariant derivat ive of the product of two tensor densities If A and B ar e arbitrary tensor densities each with an arbitrary set of up and down indices and arbitrary weight, then the claim of the product rule is this: (A----B----) ;α ≡ A----;α B---- + A---- B----;α // Weinberg p 105 (4.6.14) (*) Recall from the Covariant Derivative Theorem of section (g) above that A---- and A---- ;α have the same weight, call it W A. Similarly, B---- and B---- ;α have the same weight W B. According to the outer product rule of Appendix D (b) item 3, both terms on the RHS above have weight W A+WB and therefore this sum is the weight of the LHS (A----B----) ;α as well. 85 Proof: Start with A---- ;α = A---- ,α + (A index correction terms) + W A Γκ κα A---- B---- ;α = B---- ,α + (B index correction terms) + W B Γκ κα B---- The "index correction terms" are those Γ terms discussed in the previous sections. One can then write out the two terms on the RHS of (*) above as: A---- ;α B---- = A---- ,α B---- + (A index correction terms) B---- + [W A Γκ κα A---- ] B---- A---- B----;α = A---- B----,α + A---- (B index correction terms) + A---- [W B Γκ κα B---- ] Meanwhile, the LHS of (*) can be written as (A----B----) ;α = (A----B----) ,α + ( all index correction terms) + (W A+ WB) Γκ κα (A----B----) Momentarily ignoring the index correction terms, it is clear that the other terms match between LHS and RHS. The W terms match by visual inspection, wh ile the regular derivative terms match due to the "regular" Leibnitz rule for the derivative of a product (A----B----) ,α = A---- ,α B---- + A---- B----,α which is to say ∂α(A----B----) = ( ∂αA----)B---- + A---- (∂α B----) Consider now the LHS terms calle d (all index correction terms) a bove. This set of terms can be partitioned into two groups, (all index correction terms) = (terms involving A indices) + (terms involving B indices) . Let us pause to look at a simple example where IC T means we just show the index correction terms, (A abBcd);α|ICT = – Γn aα(AnbBcd) – Γn bα(AanBcd) + Γc αn (AabBnd) + Γd αn (AabBcn) . The correction terms can be reordered in this way = { – Γ n aα(Anb) – Γn bα(Aan) } Bcd + Aab { Γc αnBnd + Γd αnBcn } = { index correction terms for A ab } Bcd + Aab { index correction terms for Bcd } Just so, in the general case one has 86 (A----B----) ;α|ICT = ( all index correction terms) = { index correction terms for A---- } B---- + A---- { index correction terms for B---- } and this then shows that the index correction terms on the two sides of (*) do in fact match. QED Once the above product rule is verified, it is then easy to generalize just as for regular derivatives: (A----B----C----) ;α ≡ A----;α B---- C---- + A---- B----;α C---- + A---- B---- C---- ;α Examples with two vectors: (AaBb);α = Aa;αBb + AaBb;α (AaBb);α = Aa ;αBb + AaBb;α (AaBb);α = Aa ;αBb + AaBb ;α (AaBb);α = Aa;αBb + AaBb ;α Example with a scalar function A and a vector B : (AB b);α = A;αBb + ABb;α = A,αBb + ABb;α // for a scalar A, A ;α = A,α Example with a scalar constant A and a vector B: (AB b);α =A(Bb;α) / / s i n c e A ;α = A,α = 0 so a scalar constant can always be extracted from (AB b);n to give A(B b;n). A tensor constant like εabc cannot be extracted in this manner since εabc ;α ≠0. In fact ε abc ;α = Γa nαεnbc + Γb nαεanc + Γc nαεabn ε123 ;α = Γ1 nαεn23 + Γ2 nαε1n3 + Γ3 nαε123 = Γ1 1α + Γ2 2α + Γ3 3α A more general example: (A abcBde);α ≡ Aabc ;α Bde + Aabc Bde;α An example with the metric tensor: (g abB----b----);α = gab ;α B----b---- + gab B----b----;α But g ab ;α = 0 as shown at the end of section (h). Therefore (B---- a----);α = gab B----b----;α 87 which says that raising an index "commutes" with covariant differentiation -- you can raise an index ignoring the fact that : α is sitting there. But we already know this must be true because we know that the object B---- b----;α is a true tensor, and gab can raise any index on a true tensor. 88 Appendix G: Expansion of ( ∇v) in cu rvilinear coordinates This appendix assumes the usual curvilinear coordinates context, Picture B (a) Continuum Mechanics motivation Although t he polyadic notation is regarded as archaic by some writers (eg, Wolfram), it is well embedded into the literature of continuum mechanics, a field awash in rank-2 tensors. In the literature one sometimes sees, in Cartesian coordinates, (∇A) ij ≡ ∂jAi ≡ Ai,j where the indices are the reverse of the normal dyadic definition of Appendix E, ( BA) ij ≡ BiAj . For example, in continuum mechanic s one encounters the so-called convective or material derivative of an arbitrary vector field A(x,t) in the Eulerian or spatial "view" of the motion of a blob of continuous matter ( eg, Lai (3.4.3) and (3.4.8) ), DA i/Dt = ∂tAi + ∇Ai • v = ∂tAi + (∂jAi) vj = ∂tAi + (∇ A)ij vj = ∂tAi + [(∇A) v]i => DA /Dt = ∂ tA + (∇ A) v Here v(x,t) is the velocity field of the moving matter blob. DA i/Dt is a historical notation for the total derivative dA i(x,t)/dt. The matrix ∇A is not a differential operator since the derivative does not act on the vector standing to the right of ∇A, but one is still often interested in expressing ∇A in curvilinear coordinates. This is done in the following sections, where we replace the generic vector A by generic vector v and this v has nothing to do with the velocity v shown in the example above (b) Expansion of ∇v o n ei⊗ej by Method 1: Use fact that v b;a is a tensor. The covariant derivative v b;a is discussed in Appendix F (g,h,i). We shall define a true tensor object (∇ v) as vb;a so that (∇v) ba ≡ vb;a = [vb,a – Γ c ab vc] // v b,a ≡ ∂avb (∇v)'ba ≡ v'b;a = [v'b,a – Γ 'c ab v'c] // v' b,a ≡ ∂'av'b 89 In Cartesian space Γ = 0 so one has (∇v)ba = vb,a = ∂avb so (∇v) ba aligns with our object of interest in Cartesian space. As shown in Appendix E (b) one can expand the rank-2 tensor v b;a in either of these ways, ∇v = Σ ij vi;j ui⊗uj v i;j = [vi,j – Γc ij vc] = vi,j = ∂jvi ∇v = Σij v'i;j ei⊗ej v' i;j = [v'i,j – Γ ' c ij v'c] where the ui are the Cartesian basis vectors in x-space, while ei are the reciprocal base vectors. According to Appendix F (e), the affine connection Γ ' in x'-space is given by Γ 'c ab = R'c n(∂'aR'bn). When x-space is Cartesian, R = 1 and then R' = R, so Γ 'c ab = Rc i(∂'aRbi) Γ 'c ij = Rc k(∂'jRik) . Inserting this into our expansions above gives ∇v = Σ ij v'i;j ei⊗ej v' i;j = [(∂'jv'i) – Rc k(∂'jRik) v'c] or ∇v = Σij [(∇v)(e)]ij ei⊗ej [(∇v)(e)]ij = [(∂'jv'i) – Rc k(∂'jRik) v'c] The (e) superscript tells us that this matrix element of operator ( ∇v) is taken in the e n basis, see Appendix E (g): [ ( ∇v) (e)]ij = <ei|∇v| ej> = (ei)T (∇v) ej = ei • (∇v) ej = ei • (∇v) • ej bra-ket matrix dot of vectors dyadic Note that the expression shown contains onl y x'-space coordinates and objects. An alternate form is obtained using the identities of Appendix F (b) (adjusted from Picture C1 to Picture D1 shown in that Appendix ), (∂' aRdn) = – Ren Rdm (∂'aRe m) (∂'jRik) = – Rek Rim (∂'jRe m) Then R c k(∂'jRik) = – Rc k Rek Rim (∂'jRe m) = – δc e Rim (∂'jRe m) = –Rim(∂'jRc m) so that ∇v = Σij [(∇v)(e)]ij ei⊗ej [(∇v)(e)]ij = v'i;j = [(∂'jv'i) + Rim(∂'jRc m) v'c] 90 We shall use this second form for [( ∇v)(e)]ij below. Comment 1 : A variation of the above development would be to start this way ∇v = Σij vi ;j ui⊗uj vi ;j = [vi ,j + Γi jc vc] = vi ,j = ∂jvi ∇v = Σij v'i ;j ei⊗ej v'i ;j = [v'i ,j + Γ ' i jc v'c] which quickly leads to a result similar to the above, where ∇v = Σij [(∇v)(e)]i j ei⊗ej [(∇v)(e)]i j = v'i ;j = [(∂'jv'i) – Rcm (∂'jRi m) v'c] . Comment 2 : The idea that [( ∇v)(e)]i j = [∂'jv'i + Γ ' i jc v'c] can be reached by this alternate path: dv i = (∂jvi)dxj (chain rule) => d v = (∇ v)dx Expand: d x = dx'i ei and v = v'n en => d v = dv'n en + v'n den , but d e n = (∂ 'jen)dx'j = Γ 'i jn ei dx'j // from definition of Γ in Appendix F (a), but Picture A so d v = dv' n en + v'n Γ 'i jn dx'j ei = dv'i ei + v'n Γ 'i jn dx'j ei = (dv'i + v'a Γ 'i ja dx'j )ei . But dv' i = (∂ v'i/∂x'j)dx'j = (∂'jv'i) dx'j so dv = (∂ 'jv'i) dx'j + v'a Γ 'i ja dx'j )ei = [∂'jv'i + v'a Γ 'i ja ] dx'j ei . Then d v = (∇ v)dx => [∂'jv'i + v'a Γ 'i ja ] dx'j ei = (∇ v) dx'j ej or [∂' jv'i + v'a Γ 'i ja ] ei = (∇v) ej => ei• (∇v) ej = ei• [∂'jv'i + v'a Γ 'i ja ] ei = (∂'jv'i + v'a Γ 'i ja ) => [( ∇v)(e)]i j = ei• (∇v) ej = [∂'jv'i + Γ 'i jc v'c ] (c) Expansion of ∇v on ei⊗ej by Method 2: Use brute force. Method 1 is perhaps elegant in that it makes use of tensor transformations and the affine connection Γ . But a simple brute force method is really ju st as simple and does not require knowledge of Γ and covariant differentiation. Instead of using the ei⊗ej notation for basis vectors, here we use the alternate notation ei(ej)T which works for rank-2 tensor expansions. Recall that ei(ej)T is an NxN matrix as discussed in Appendix E (e). In this brute force method, start with the Cartesian space expansion of Appendix E, (∇v) = Σ cd(∇v)dc uducT = Σcd(∂cvd) uducT // note that u d = ud in Cartesian space 91 To express things in x'-coordinates, first write ∂ cvd = (Ri c∂'i)( Rj dv'j) = Ri c [(∂'iRj d) v'j + Rj d (∂'iv'j)] The next step is to express the matrix uducT as a linear combination of ee efT. In Section 3 (b) it was shown that the tangent base vectors transform in this way e' n = Σi Ri n ei where (e 'n)i = δni (*) For present purposes, we write this as u d = Σe Re d ee where ( ud)e = δde Therefore ud = Σe Red ee Then we can change this colu mn vector to a row vector, ucT = Σf Rfc efT and so uducT = Σef Red Rfc ee efT = Red Rfc ee efT Inserting these two results gives (∇v) = Σ cd(∇v)dc uducT = Σcd(∂cvd) uducT = { Ri c [(∂'iRj d) v'j + Rj d (∂'iv'j)] } { Red Rfc ee efT } = { R i c [(∂'iRj d) v'j + Rj d (∂'iv'j)] } { Red Rfc ee efT } = { ( R fc Ri c)[Red (∂'iRj d) v'j + (Red Rj d)(∂'iv'j)] } ee efT = δ fi [Red (∂'iRj d) v'j + δej (∂'iv'j)] ee efT = [ R ed (∂'fRj d) v'j + (∂ 'fv'e)] ee efT = Σef [(∇v)(e)]ef ee efT where [(∇v)(e)]ef = (∂'fv'e) + Red (∂'fRj d) v'j 92 or [(∇v)(e)]ij = (∂'jv'i) + Rim (∂'jRj m) v'j and this agrees with the second form obtained by Method 1. (d) Expansion on e i⊗ej and e ^i⊗e^j Reversing the covariant tilts in the above expansion one gets ∇v = Σij v'i;j ei⊗ej where v'i;j = g'iag'jb v'a;b = g'iag'jb [(∂'bv'a) + Ram(∂'bRc m) v'c] Then since e i = h'ie^i one gets yet another expansion ∇v = Σ ij (v'i;j h'i h'j) e^i⊗e^j = Σij [(∇v)(e^)]ij e^i⊗e^j where [(∇v) (e^)]ij = h'i h'j v'i;j = h'i h'j g'iag'jb [(∂'bv'a) + Ram(∂'bRc m) v'c] Here is a summary of results so far ∇v = Σij [(∇v)(e)]ij ei⊗ej [( ∇v)(e)]ij = [(∂'jv'i) + Rim(∂'jRc m) v'c] ∇v = Σ ij [(∇v)(e)]i j ei⊗ej [( ∇v)(e)]i j = [(∂'jv'i) – Rcm (∂'jRi m) v'c] . ∇v = Σ ij [(∇v)(e)]ij ei⊗ej [( ∇v)(e)]ij = g'iag'jb [(∂'bv'a) + Ram(∂'bRc m) v'c] ∇v = Σij [(∇v)(e^)]ij e^i⊗e^j [(∇v)(e^)]ij = h'i h'j g'iag'jb [(∂'bv'a) + Ram(∂'bRc m) v'c] One could replace v' a = g'adv'd in any of the above results. For example, the last object becomes [(∇v)(e^)]ij = h'i h'j g'iag'jb [(∂'b[g'adv'd]) + Ram(∂'bRc m) (g'cdv'd)] Another choice is to use the v'n components of v obtained when expanding v on the e^n , v = Σ n v'n en = Σn v'n (h'ne^n) = Σn (h'n v'n) e^n = Σn v'n e^n => v'n ≡ h'n v'n so one then replaces v' n = h'n-1v'n. The same object above then becomes [(∇v) (e^)]ij = h'i h'j g'iag'jb [(∂'b[g'ad h'd-1v'd]) + Ram(∂'bRc m) (g'cd h'd-1v'd)] 93 As discussed in Section 14 Example 1, the components v'n are convenient since they all have the same dimensions. Moreover, when a specific curvilinear system is selected, one can dispense with the unpleasant font used in v'n and just write v'n = vx'(n) . For example, in spherical coordinates r, θ,φ : v'1 = vr v'2 = vθ v'3 = vφ v = Σn v'n e^n = vrr^ + vθθ^ + vφ φ^ . (e) Orthogonal coordinate systems In this cas e g'ab = h'a2δa,b and g'ab = h'a-2δa,b and things simplify. The above four expansions become (there is no change in th e first two expansions) ∇v = Σij [(∇v)(e)]ij ei⊗ej [( ∇v)(e)]ij = [(∂'jv'i) + Rim(∂'jRc m) v'c] ∇v = Σ ij [(∇v)(e)]i j ei⊗ej [( ∇v)(e)]i j = [(∂'jv'i) – Rcm (∂'jRi m) v'c] . ∇v = Σij [(∇v)(e)]ij ei⊗ej [( ∇v)(e)]ij = h'i-2 h'j-2 [(∂'jv'i) + Rim(∂'jRc m) v'c] ∇v = Σij [(∇v)(e^)]ij e^i⊗e^j [(∇v)(e^)]ij = h'i-1 h'j-1 [(∂'jv'i) + Rim(∂'jRc m) v'c] and the last object noted above becomes (m →d and c →b) Pij ≡ [(∇v)(e^)]ij = h'i-1 h'j-1 [(∂'j[h'iv'i]) + Rim(∂'jRc m) (h'cv'c)] = h'i-1 h'j-1 [(∂'j[h'iv'i]) + Rid(∂'jRb d) (h'bv'b)] // m →d,c→b = h'i-1 h'j-1 [ (∂'jh'i) v'i + h'i(∂'jv'i) + h'i2Ri d(∂'jRb d) (h'bv'b)] T2 T3 T1 where R id = g'iaRa b gbd = h'i2Ri d is used to put all R's into their down-tilt form. (f) Maple evaluation of ( ∇v) in several coordinate systems This object Pij shown above can be computed in Maple by a simple program which is easily modified for other orthogonal curvilinear systems. The first task is to obtain the R matrix from the inverse transformation equations, 94 Then one needs the scale factors h n' from the metric tensor g ¯ = STS ( dev notation Section 5 ( l) ) The terms T1,T2 and T3 shown above are then entered, 95 The terms are then added and simplified and out pops the result, Below are some sample results (including the above): 96 (∇v) = Σij Pij e^i e^jT Pij = [(∇v)(e^)]ij • Pij in polar coordinates (where 1,2 = r, θ) : // agrees with Lai (2.23.23) • P ij in cylindrical coordinates (where 1,2,3 = r, θ,z) : // agrees with Lai (2.34.5) The polar coordinates results are seen to be upper left 2x2 piece of the cylindrical results. • Pij in spherical coordinates (where 1,2,3 = r, θ,φ) : // agrees with Lai (2.35.25) By entering the usual inverse transformation equations x' = F-1(x), and with suitable small alterations, the above Maple code can compute ∇v in any orthogonal curvilinear coor dinate system in any number of dimensions N. 97 Appendix H: Expansion of div(T) in curvilinear coordinates The object of attention in this Appendix, divT , is expressed this way in Cartesian coordinates, (divT)i = ∂jTij , where T ij is a rank-2 tensor. One can regard the above equation as describing the normal divergence of the "vector" which forms the ith row of the matrix T ij. In general (that is to say, under general transformations F), the rows of T ij are not tensorial vectors, and that is why (divT)i is not a tensorial scalar. Nor, in fact, are the (divT)i as defined above the components of a tensorial vector! As shown in section (b) below, divT will be redefined as a true te nsorial vector which equals ∂jTij in Cartesian coordinates. (a) Continuum Mechanics motivation The vector divT arises for example when Newton's 2nd Law F = ma is applied to a particle of continuous matter, in which case this law is known as Cauchy's Equation of Motion, divT + ρB = ρa // Lai p 169 (4.7.4) In this equation T is known as the Cauchy stress tensor, ρ is the mass density of the particle, B is any action-at-a-distance force (body force) per unit mass (such as gravity), and of course a is the acceleration of the particle. Comment . Under rotations and translations the quantity (divT)i = ∂jTij is a tensorial vector, ρ is a tensorial scalar, and a and B are tensorial vectors. As one would expect, divT + ρB = ρ a is covariant in the sense of Section 7 (u) under these kinds of transformations. As in the case of ∇v, one is interesting in expressing divT in general curvilinear coordinates. (b) Expansion of divT on e n by Method 1: Use fact that Tab ;α is a tensor. As shown in the examples of Appendix F (i), the fo llowing object transforms as a rank-3 tensor, T ab ;α ≡ ∂αTab + Γa αn Tnb + Γb αn Tan = ∂α Tab // x-space where g=1, Γ = 0 T 'ab ;α ≡ ∂'αT 'ab + Γ 'a αn T 'nb + Γ 'b αn T 'an // x'-space Contracting b with α yields the following tensorial vector equations, (divT) a = Tab ;b = ∂bTab // x-space where Γ = 0 (divT)'a = T 'ab ;b ≡ ∂'bT 'ab + Γ 'a bn T 'nb + Γ 'b bn T 'an // x'-space Note that the Cartesian space statement (divT)a = ∂bTab is obtained, as noted in the opening comments above. According to Section 7 (s) or Appendix E (b), the vector divT can be expanded as 98 divT = Σa(divT)'a ea (divT)'a = [(divT)(e)]a // divT in the en basis From Appendix F (a), adjusted from Picture C to Picture A context (primes on ∂), Γ 'c ab = – Rbi (∂'aRc i) Γ 'a bn = – Rni (∂'bRa i) // b →n then a →b then c →a Γ 'b bn = – Rni (∂'bRb i) => (divT)'a = ∂'bT 'ab + Γ 'a bn T 'nb + Γ 'b bn T 'an = ∂' bT 'ab – Rni (∂'bRa i) T 'nb – Rni (∂'bRb i)T 'an The conclusion is that divT = Σa[(divT)(e)]a ea [(divT) (e)]a = ∂'b T 'ab – Rni(∂'bRa i) T 'nb – Rni (∂'bRb i)T 'an // sum on n and b The vector divT = Σ a(∂bTab)ua has thus been expressed in terms of x'-space coordinates and objects, and as usual the en are the tangent base vectors in x-space. (c) Expansion of divT on e n by Method 2: Use brute force Start with the known expansion of divT in Cartesian x-space divT = Σi [divT]i ui = Σi (∂jTij) ui Compute (∂jTij) = (Ra j∂'a)( Rb'i Rc'j T 'b'c') = Ra j Rb'i Rc'j (∂'a T 'b'c') + Ra j Rb'i (∂'a Rc'j) T 'b'c' + Ra j Rc'j (∂'a Rb'i) T 'b'c' = ( Ra j Rc'j) Rb'i (∂'a T 'b'c') + Ra j Rb'i (∂'a Rc'j) T 'b'c' + (Ra j Rc'j) (∂'a Rb'i) T 'b'c' = δa c' Rb'i (∂'a T 'b'c') + Ra j Rb'i (∂'a Rc'j) T 'b'c' + δa c' (∂'a Rb'i) T 'b'c' = R b'i (∂'a T 'b'a) + Ra j Rb'i (∂'a Rc'j) T 'b'c' + (∂'a Rb'i) T 'b'a Then use the same u i expansion as in Appendix G (c), u i = Σe Rn i en 99 Combining one gets, divT = Σi (∂jTij) ui = { ( R n i Rb'i) (∂'a T 'b'a) + Ra j (Rn i Rb'i) (∂'a Rc'j) T 'b'c' + Rn i (∂'a Rb'i) T 'b'a} en = { δ n b' (∂'a T 'b'a) + Ra j δn b' (∂'a Rc'j) T 'b'c' + Rn i (∂'a Rb'i) T 'b'a} en = { ( ∂' a T 'na) + Ra j (∂'a Rc'j) T 'nc' + Rn i (∂'a Rb'i) T 'b'a} en = Σ n[(divT)(e)]n en where [(divT) (e)]n = (∂'a T 'na) + Ra j (∂'a Rc'j) T 'nc' + Rn i (∂'a Rb'i) T 'b'a From Appendix F (b) item 3 one has (Picture C1 → Picture A so primes on ∂ 's), (∂' aRdn) = – R en Rdm (∂'aRe m) (∂'aRc'j) = – Rej Rc'm (∂'aRe m) // d→c', n→j (∂'aRb'i) = – Rei Rb'm (∂'aRe m) so then [(divT)(e)]n = (∂'a T 'na) + Ra j (∂'a Rc'j) T 'nc' + Rn i (∂'a Rb'i) T 'b'a = ( ∂' a T 'na) – Ra j Rej Rc'm (∂'aRe m) T 'nc' – Rn i Rei Rb'm (∂'aRe m) T 'b'a = ( ∂' a T 'na) – δa eRc'm (∂'aRe m) T 'nc' – δn e Rb'm (∂'aRe m) T 'b'a = ( ∂' a T 'na) – Rc'm (∂'aRa m) T 'nc' – Rb'm (∂'aRn m) T 'b'a Reverse the order of the last two terms [(divT)(e)]n = (∂'a T 'na) – Rb'm (∂'aRn m) T 'b'a – Rc'm (∂'aRa m) T 'nc' . Replace summation indices m →i, a→b, = ( ∂'b T 'nb) – Rb'i (∂'bRn i) T 'b'b – Rc'i (∂'bRb i) T 'nc' . Finally, change n →a and then c' → n and b' →n, [(divT)(e)]a = (∂ 'b T 'ab) – Rni (∂'bRa i) T 'nb – Rni (∂'bRb i) T 'an . // sum on n and b 100 This is seen to match the result of Method 1 of the pr evious section. The brute force method is in fact not too bad and requires no explicit use of the affine connection Γ. Technical Note: In the above expression one can write for example R ni(∂'bRa i) = g'nmRm i(∂'bRa i). Then using the fact that Rm iRa i = g'ma one finds Rm i(∂'bRa i) + Ra i(∂'bRm i) = (∂'bg'ma) which then allows the replacement R ni(∂'bRa i) = g'nm (∂'b g'ma) – g'nm Ra i(∂'bRm i). This kind of transformation leads to an alternate form for divT shown above, still with all down-tilt R matrices, but the index structure is different. This other form appears when one does the "brute force" method starting with (∂ jTij) instead of with (∂jTij). Of course in Cartesian space these two objects must be the same. (d) Adjustment for T expanded on ( e^i⊗e^j) and divT expanded on e^a In the above, it has been assumed that T ij is a rank-2 tensor so that, in the notation of Appendix E, T = ΣijTij(ui⊗uj) = ΣijT 'ij(ei⊗ej) If one is interested in an expansion of T on the unit vectors e^i where ei = h'i e^i this becomes T = Σ ij[T 'ijh'ih'j] (e^i⊗e^j) = Σij[T(e^)]ij (e^i⊗e^j) One then has [T (e^)]ij = h'ih'jT 'ij => T 'ij = h'i-1 h'j-1 [T(e^)]ij If one is interested in this form of the T matrix el ements, then one is likely also interested in this expansion for divT , divT = Σa[(divT)(e)]a ea = Σn { [(divT)(e)]a h'a } e^a ≡ [(divT)(e^)]a e^a where then [(divT) (e^)]a = h'a[(divT)(e)]a = h'a [(∂'b T 'ab) – Rni (∂'bRa i) T 'nb – Rni (∂'bRb i) T 'an] = h'a * { ∂ 'b (h'a-1 h'b-1 [T(e^)]ab) – Rni (∂'bRa i) h'n-1 h'b-1 [T(e^)]nb – R ni (∂'bRb i) h'a-1 h'n-1 [T(e^)]an } = h' a * { ∂ 'b (h'a-1 h'b-1) [T(e^)]ab – Rni (∂'bRa i) h'n-1 h'b-1 [T(e^)]nb + h' a-1 h'b-1 (∂'b [T(e^)]ab) – Rni (∂'bRb i) h'a-1 h'n-1 [T(e^)]an } Now ∂'b (h'a-1 h'b-1) = ∂ 'b(h'a h'b)-1 = – (h'a h'b)-2 ∂'b(h'a h'b) = – h'a-2 h'b-2 ∂'b(h'a h'b) 101 so the above sequence for [(divT)(e^)]a continues, = h' a * { – h' a-2 h'b-2 ∂'b(h'a h'b) [T(e^)]ab – Rni (∂'bRa i) h'n-1 h'b-1 [T(e^)]nb + h' a-1 h'b-1 (∂'b [T(e^)]ab) – R ni (∂'bRb i) h'a-1 h'n-1 [T(e^)]an } = { – h' a-1 h'b-2 ∂'b(h'a h'b) [T(e^)]ab – h'a h'n-1 h'b-1 g'nm Rm i (∂'bRa i) [T(e^)]nb + h' b-1 (∂'b [T(e^)]ab) – h' n-1 g'nm Rm i (∂'bRb i) [T(e^)]an } where recall that Rn i = Rni since x-space is Cartesian. In the last form only the down-tilt R matrix appears which simplifies calculation with Maple. This and all other results above are valid for general curvilinear coordinates, orthogonal as well as non-orthogonal. At this point, we will specialize to orthogonal systems so the last form above becomes T1 T3 [(divT) (e^)]a = { – h' a-1 h'b-2 ∂'b(h'a h'b) [T(e^)]ab – h'a h'b-1 h'n Rn i (∂'bRa i) [T(e^)]nb + h' b-1 (∂'b [T(e^)]ab) – h' n Rn i (∂'bRb i) [T(e^)]an } T2 T4 Even for an orthogonal curvilinear coordinate system , the form of this tensor divergence is amazingly complicated. Here it is expressed in terms of the down-tilt R matrix and the scale factors and as usual repeated indices are summed. (e) Maple: divT in cylindrical and spherical coordinates The above o bject [(divT)(e^)]a can be evaluated by Maple code very similar to that shown in Appendix G for (∇ v). The main difference is the set of entry lines for the terms, 102 Here are some results for [(divT)(e^)]a = "divT a" : • cylindrical coordinates (where 1,2,3 = r, θ,z) : // agrees with Lai p 60 (2.34.8,9,10) • For polar coordinates P1 and P2 are given by the first two lines above with the last terms set to 0. These polar results then agree with Lai p58 (2.33.32,33). • spherical coordinates (where 1,2,3 = r, θ,φ) : The expressions above agree with Lai p 65 (2.35.33,34,35). 103 Appendix I : The Vector Laplacian in Spherical and Cylindrical Coordinates This appendix assu mes the usual curvilin ear coordinates context, Picture B, In this section the vector Laplacian is computed in two different ways, each associated with a particular "tensorization" of its Cartesian form. The second method, though less pleasant than the first, gives insight into why the vector Laplacian always includes the scalar Laplacian of the field components. It is in this inclusive form that results are usually stated in the literature. In passing, it should be noted that the vector Laplacia n is not just an idle mathematical curiosity. It shows up for example in the wave equations fo r electric and magnetic fields in a vacuum, (∇ 2 + k2)E(x) = 0 (∇2 + k2)B(x) = 0 k = ω/c E,B(x,t) = E,B(x)e-iωt In continuum mechanics, it appears for example in the Navier/Cauchy equation which describes the small vector displacement u field in an isotropic elastic solid, ρo∂t2u = ρ0B + (λ +μ)∇e + μ ∇2u e = div u = dilatation // Lai p 216 (5.6.9) Here ρ0 is the unperturbed mass density, B the body force, and λ and μ are the two Lamé constants which describe an isotropic elastic medium. The vector Laplacian makes another appearance in the better-known Navier-Stokes equation which describes the vector velocity field v in an incompressible Newtonian fluid, ρ [ ∂tv + (∇ v)v] = ρ B - ∇ p + μ∇2v // Lai p 361 (6.7.6) where ρ is the mass density and p is pressure. (a) The first method : a review In Section 13 (c) it was shown that, in Cartesian coordinates, ∇2(Bn) = [ grad(div B) – curl (curl B) ]n ∇2 = ∂j∂j or ∇•∇ (B n) = [∇(∇• B) – ∇ x (∇ x B) ]n When all components are considered in a single equation, one could write ∇2(B) = grad(div B) – curl (curl B) or ∇•∇ (B) = ∇(∇• B) – ∇ x (∇ x B) 104 and this is frequently done (see examples cited above). Since ∇2(B) ≠ ∂j∂j B in curvilinear coordinates, it seems notionally safer in our current context to use a different symbol for the vector Laplacian operator, and following Moon and Spencer we use @ so the second last equation above then says: @B = grad(div B) – curl (curl B) . Since [@B] n agrees with ∇2(Bn) in Cartesian coordinates, and since we know how to write div, grad and curl in curvilinear coordinates (Sections 9,10,12 or Se ction 15), the right hand side of the above equation provides our "first method" of writing @ B in curvilinear coordinates. In Section 15 (g) it was shown that the proper "tensorization" of th e above equation is given by (@B)n = (Bj ;j);n – g-1/2εnab(g-1/2εbdeBe;d);a @B = (@B)n un and therefore, in x'-space ( x' are the curvilinear coordinates of interest) , (@B)' n = (B'j ;j);n – g'-1/2ε'nab(g'-1/2ε'bdeB'e;d);a @B = (@B)'n en where (B'j ;j) = div B. The object @B is a normal vector (weight 0), assuming B is a normal vector. Doing various simplifying steps, we then arrived at the following expression which lends itself to calculation: (@B)'n = ∂'n{(1/ g' ) ∂ 'i(g' B'i) } @ B = (@B)'n en – (1/ g' ) ε'ncd ε'eab ∂'c { (1/g' ) g'de(∂'a[g'bfB'f] ) } where ε'ncd is the usual permutation tensor ( ε'ncd = εncd). In this notation, B'i is an official x'-space contravariant vector component. We are mainly interested in working with vector s which are expanded on unit vector versions of the tangent base vectors. In such a unit vector expans ion of a vector, italic font has been used for the components. As shown in various places, one has B' n = B'n/h'n en = h'n e^n B = B' nen = B'ne^n h ' n = scale factor and this then gives (@B)' n = ∂'n{(1/ g' ) ∂ 'i(g' B'i/h'i) } @ B = (@B)'n en – (1/ g' ) ε'ncd ε'eab ∂'c { (1/g' ) g'de(∂'a[g'bfB'f/h'f]) } or, as we will use it with unit vectors, (@B)' n = h'n ∂'n{(1/ g' ) ∂'i(g' B'i/h'i) } @ B = (@B)'n e^n – h' n (1/ g' ) ε 'ncd ε'eab ∂'c { (1/g' ) g'de(∂'a[g'bfB'f/h'f]) } . Specializing to orthogonal coordinates yields the form we shall use for computation in Maple, 105 (@B)'n = (1/h'n) ∂'n{(1/ g' ) ∂'i(g' B'i/h'i)} @ B = (@B)'n e^n – ( h ' n/g' ) ε 'ncd ε'dab ∂'c { (1/g' ) h'd2(∂'a[h'bB'b]) } . The second term could be written as two terms by reducing the product ε'ncd ε'dab into δδ - δδ in the usual manner, but Maple is happy to just "do it" as stated. And for non-orthogonal coordinates, this reduction is much uglier (Appendix D (j) item 3) and then one would want to use the εε product as is. (b) The first method in spherical coordinates: Maple speaks In this section, the foll owing notation is used: B'1 = Br B'2 = Bθ B'3 = Bφ B = B'ne^n = Bre^r + Bθe^θ + Bφe^φ = Brr^ + Bθθ^ + Bφφ^ . Maple begins in the same manner as shown earlier in Appendix G (f), the idea being that this code could be used for any coordinate system, 106 The last object is g' , where hp[n] = h' n . The next chunk of code computes the scalar Laplacian of an unspecified function f, and this expression will be used below in parsing the vector Laplacian results : The subs command used here (and more intensely be low) removes the arguments of the function f for purely cosmetic reasons (see A Maple User's Guide n earby for details, operands section). The arguments are added in the first place to prevent Maple from th inking the unspecified function f is a constant. Next comes a low-budget implementation of the permutation tensor eps(a,b,c) = εabc, 107 The two terms of the above (@ B)'n expression shown at the end of section (a) are then duly entered, In the following code, we use ( in line with the notation shown at the start of this section) q1_ = (@ B)'1 = (@B)r q2_ = (@ B)'2 = (@B)θ q3_ = (@ B)'3 = (@B)φ 108 See comment above about the Maple subs commands (purely cosmetic). (c) The first method in spherical coordinates: putting results i n traditional form It turns out that in each component of the vector Laplacian stated above, 5 of the 9 terms can be represented as if they were the scalar Laplacian acti ng on the component in question. Here is how it works: ∇ 2f = (1/r2)∂r(r2∂rf) + (1/r2sinθ)∂θ(sinθ∂θf) + (1/r2sin2θ)∂φ2f = 1 + 2 3+4 5 1 2 3 4 5 1 2 5 3 4 Therefore (@B)r = ∇2(Br) - (2/r2)Br - (2/r2)cot(θ)Bθ - (2/r2)∂θBθ -(2/r2sinθ) ∂φBφ = ∇ 2(Br) – (2/r2) [ Br + cotθ Bθ + ∂θBθ + cscθ ∂φBφ ] 3 4 5 1 2 Therefore 109 (@B)θ = ∇2(Bθ) - (1/r2) [ - 2 ∂ θBr + cot2θ Bθ + Bθ + 2 cotθ cscθ∂φBφ ] = ∇2(Bθ) - (1/r2) [csc2θ Bθ - 2 ∂θBr + 2 cotθcscθ∂φBφ ] 5 3 4 1 2 (@B)φ = ∇2(Bφ) - (1/r2) [csc2θBφ - 2cscθ∂ φBr - 2cotθ cscθ∂φBθ ] In this manner, we end up with the components of the vector Laplacian expressed in the traditional manner, (@B) r = ∇2(Br) – (2/r2) [ Br + cotθ Bθ + ∂θBθ + cscθ ∂φBφ ] (@B)θ = ∇2(Bθ) – (1/r2) [csc2θ Bθ – 2 ∂θBr + 2cot θcscθ∂φBφ ] (@B)φ = ∇2(Bφ) – (1/r2) [csc2θ Bφ – 2cscθ∂φBr – 2cotθcscθ∂φBθ ] where ∇2f = (1/r2)∂r(r2∂rf) + (1/r2sinθ)∂θ(sinθ∂θf) + (1/r2sin2θ)∂φ2f = (2/r) ∂rf + ∂r2f + (cotθ/r2)∂θf + (1/r2) ∂θ2f + (1/r2sin2θ)∂φ2f Once again, written out in gory detail, these three expressions are (in the same order r, θ,φ) : and each expression contains nine terms. The curious reader might wonder why, in each case, five of the nine terms of the vector Laplacian components can be represented by the scalar Lapl acian acting on the component. This question is answered in the following section. 110 (d) The second method : Part I In the first method, described in section (a) above, we used this tensorization of the vector Laplacian, (@B)n = (B'j ;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a . At the end of Section 15 (b) it was noted that th ere is an alternative tensorization , namely (@B)n = B'n;j ;j These two tensors must be the same since the tensori zation of a Cartesian form equation is unique, but it is not easy to show. We suspect that certain identities involving the Riemann curvature tensor are required and perhaps other facts. Be that as it may, our "second method" is to use this B'n;j ;j tensorization to compute once again the components of the vector Laplacian. Appendix F (i) has among its examples the followi ng covariant derivative of a rank 2 tensor, Bab ;α ≡ ∂α Bab + Γa αn Bnb + Γb αn Ban and since B a;b is a rank-2 tensor one can write (indices are substituted in the second line) B a;b ;α ≡ ∂α Ba;b + Γa αk Bk;b + Γb αk Ba;k Bn;j ;j ≡ ∂j Bn;j + Γn jk Bk;j + Γj jk Bn;k // Γ'j jk = (1/ g ) ∂k(g ) as in App F (d) . Another example in that Appendix shows that (the lower three lines are index substitutions of the first) B a;α = ∂α Ba + gαbΓa bs Bs Bn;j = ∂j Bn + gjbΓn bs Bs Bk;j = ∂j Bk + gjbΓk bs Bs Bn;k = ∂k Bn + gkbΓn bs Bs Therefore, Bn;j ;j = ∂j[∂j Bn + gjbΓn bs Bs] + Γn jk[∂j Bk + gjbΓk bs Bs] + Γj jk[∂k Bn + gkbΓn bs Bs] Combining the second last term with the first gives B n;j ;j = [∂j ∂j Bn + Γj jk(∂kBn )] + ∂j(gjbΓn bs Bs) + Γn jk[∂j Bk + gjbΓk bs Bs] + Γj jk[gkbΓn bs Bs] Since no terms have been dropped, we are still "cova riant" and in x'-space everything gets primed, B'n;j ;j = [∂'j ∂'j B'n + Γ 'j jk(∂'kB'n )] + ∂'j(g'jbΓ 'n bs B's) + Γ 'n jk[∂'j B'k + g'jbΓ 'k bs B's] + Γ 'j jk[g'kbΓ 'n bs B's] 111 The first two terms can be written this way [∂' j ∂'j B'n + Γ 'j jk(∂'kB'n )] = [∂'j ∂'j B'n + (1/ g' ) ∂'k(g' ) (∂'kB'n )] = lap (B'n) in the sense that we earlier wrote ( Section 15 (e) ) ∂' j∂'jf ' + (1/ g' ) ∂ 'k(g' ) ∂'kf = (1/ g' ) ∂ 'k [g' (∂'kf ) = lap (f) Therefore B n;j ;j = lap (B'n) + ∂'j(g'jbΓ 'n bs B's) + Γ 'n jk[∂'j B'k + g'jbΓ 'k bs B's] + Γ 'j jk[g'kbΓ 'n bs B's] = lap (B' n) + Extra Terms So we begin to see why the scalar Laplacian of a B component appears in ( @B) n. (e) The second method : Part II Unfortunately, we don' t want to see lap (B'n), we want to see lap( B'n) ! Consider then, lap (B'n) = lap (B'n/hn) = [∂'j ∂'j (B''n/hn) + (1/ g' ) ∂'k(g' ) (∂'k (B''n/hn) )] and one computes the pieces as follows: ∂ 'j ∂'j (B'n/hn) = ∂ 'j ∂'j (B'nh-1 n) = ∂ 'j [(∂'jB'n) h-1 n + B'n(∂'j h-1 n) ] = ( ∂'j∂'jB'n)h-1 n + (∂ 'jB'n) (∂'j h-1 n) + (∂'j B'n)(∂'j h-1 n) + B'n (∂'j ∂'j h-1 n) = ( ∂'j∂'jB'n) h-1 n + 2(∂'jB'n) (∂'j h-1 n) + B'n (∂'j ∂'j h-1 n) ∂' k (B'n/hn) = [(∂'kB'n) h-1 n + B'n(∂'k h-1 n) ] so that lap (B' n) = (∂'j∂'jB'n) h-1 n + 2(∂'jB'n) (∂'j h-1 n) + B'n (∂'j ∂'j h-1 n) + ( 1 / g' ) ∂ 'k(g' )[(∂'kB'n) h-1 n + B'n(∂'k h-1 n) ] = h-1 n {(∂'j∂'jB'n) + (1/ g' ) ∂'k(g' )(∂'kB'n) } + 2 ( ∂'jB'n) (∂'j h-1 n) + B'n (∂'j ∂'j h-1 n) + (1/ g' ) ∂ 'k(g' )B'n(∂'k h-1 n) = h-1 n lap ( B'n) + 2( ∂'jB'n) (∂'j h-1 n) + B'n (∂'j ∂'j h-1 n) + (1/ g' ) ∂ 'k(g' )B'n(∂'k h-1 n) 112 = h-1 n lap ( B'n) + Other Terms The conclusion so far is that (@B) n = Bn;j ;j = lap(B'n) + Extra Terms = h-1 n lap(B'n) + Other Terms + Extra Terms As before, our interest is with the expansion @B = (@B)'n e^n where then (@B)'n = lap( B'n) + h'n [Other Terms + Extra Terms] This result applies to any x'-space curvilinear coordinate system, orthogo nal or otherwise. Thus we have demonstrated why it is that lap( B'n) always appears as part of the vector Laplacian component ( @B)'n. The Terms shown are these, just quoting from above, Extra Terms = ∂' j(g'jbΓ 'n bs B's) + Γ 'n jk[∂'j B'k + g'jbΓ 'k bs B's] + Γ 'j jk[g'kbΓ 'n bs B's] Other Terms = 2( ∂'jB'n) (∂'j h-1 n) + B'n (∂'j ∂'j h-1 n) + (1/ g' ) ∂'k(g' )B'n(∂'k h-1 n) Rather than attempt algebrai c simplification of the above terms, we will just throw them into Maple as is. Replacing B's = B's/h's and "lowering" the differential operators appropria tely, one gets Extra Terms = ∂'j(g'jb Γ 'n bs B's/h's) + Γ 'n jkg'js∂'s(B'k/h'k) + Γ 'n jkg'jb Γ 'k bs B's/h's + Γ 'j jkg'kb Γ 'n bs B's/h's ET1 ET2 ET3 ET4 Other Terms = 2(g'js ∂'sB'n) (∂'jh'n-1) + B'n (∂'j ∂'jh'n-1) + (1/ g' ) ∂ 'k(g' )B'n(∂'kh'n-1) OT1 OT2 OT3 (f) The second method in spherical coordinates: Maple speaks again In method 1, there was no need in the Maple calculation for the affine connection object , Γ 'd ab = (1/2) g'dc [ ∂'ag'bc + ∂'bg'ca – ∂'cg'ab] which in orthogonal coordinates simplifies to Γ 'd ab = (1/2) h' d-2 [δd b ∂'a(h'b2) + δd a∂'b(h'a2) – δab ∂'d(h'a2)] . This last form shows that Γ 'd ab = 0 unless two indices match, in which case it might not vanish. For coordinate systems like spherical and cylindrical coordinates, Γ 'd ab is very sparse, and this explains why we are not going to be simply swamped by all those Extra Terms shown above. 113 For spherical coordinates, only 9 of the 27 elements of the object Γd ab are non-zero : Γ '1 22 = -r Γ '2 12 = Γ '2 21 = 1/r // notation: Γ '1 22 = Γr θθ Γ '1 33 = -r sin2θ Γ '3 13 = Γ '3 31 = 1/r Γ '2 33 = -cosθsinθ Γ '3 23 = Γ '3 32 = cotθ // 1,2,3 = r,θ ,φ = radius, polar, azimuthal Γ r = ⎣⎢⎢⎡ ⎦⎥⎥⎤ 0 0 0 0 -r 0 0 0 -rsin2θ Γθ = ⎣⎢⎡ ⎦⎥⎤ 0 1/r 0 1/r 0 0 0 0 -sinθ cosθ Γφ = ⎣⎢⎡ ⎦⎥⎤ 0 0 1/r 0 0 cot θ 1/r cotθ 0 To maintain generality, however, we let Maple compute Γd ab = G(d,a,b) from the first equation above, so Γd ab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] The Extra Terms are then entered, as shown above : 114 Extra Terms = ∂'j(g'jb Γ 'n bs B's/h's) + Γ 'n jkg'js∂'s(B'k/h'k) + Γ 'n jkg'jb Γ 'k bs B's/h's + Γ 'j jkg'kb Γ 'n bs B's/h's ET1 ET2 ET3 ET4 And then the Other Terms are entered as well, Other Terms = 2(g' js ∂'sB'n) (∂'jh'n-1) + B'n (∂'j ∂'jh'n-1) + (1/ g' ) ∂ 'k(g' )B'n(∂'kh'n-1) OT1 OT2 OT3 The final results are then generated : 115 These are seen to match the unnumbered terms in the q1_,q2_,q3_ expressions of section (c) above, those terms which get added to the scalar Laplacian contribution. Quoting from method 1 above, (@B)r = ∇2(Br) – (2/r2) [ Br + cotθ Bθ + ∂θBθ + cscθ ∂φBφ ] (@B)θ = ∇2(Bθ) – (1/r2) [csc2θ Bθ – 2 ∂θBr + 2cot θcscθ∂φBφ ] (@B)φ = ∇2(Bφ) – (1/r2) [csc2θ Bφ – 2cscθ∂φBr – 2cotθcscθ∂φBθ ] (g) Results for Cylindrical Coordinates from both methods The Maple program was easily modified for this system. Here are the results: B'1 = Br B'2 = Bφ B'3 = Bz B = B'ne^n = Bre^r + Bφe^φ + Bφe^z = Brr^ + Bθφ^ + Bφz^ . The scalar Laplacian of an unspecified function f : ∇ 2f = (1/r)∂r(rf) + (1/r2)∂φ2f + ∂z2f = 1 + 2 3 4 1 2 3 4 The components of the vector Laplacian are found by the "first method" to be 116 1 2 4 3 Therefore, (@B)r = ∇2(Br) - Br/r2 - 2∂φBφ/r2 . 3 4 1 2 Therefore, (@B) φ = ∇2(Bφ) - Bφ/r2 + 2∂φBr/r2 . 4 3 1 2 Therefore, (@B) z = ∇2(Bz) as befits a component which is Cartesian. In this ma nner, we end up with the components of the vector Laplacian expressed in the traditional manner, (@B) r = ∇2(Br) – (1/r2)[Br – 2∂φBφ] (@B)φ = ∇2(Bφ) – (1/r2)[Bφ + 2∂φBr] (@B)z = ∇2(Bz) where ∇ 2f = (1/r)∂r(rf) + (1/r2)∂φ2f + ∂z2f = ∂2 rf +(1/r)∂ rf + (1/r2)∂φ2f + ∂z2f 117 Once again, written out in gory detail, these three expressions are (in the same order r, φ,z) : The "second method" produces these results for the te rms which are added to the scalar Laplacian, and these are seen to agree with the unnumbere d terms in q1_, q2_ and q3_ shown above. In cylindrical coordinates the Γ object is even sparser than in spherical coordinates. One has 118 Γ 'd ab = (1/2) h' d-2 [δd b ∂'a(h'b2) + δd a∂'b(h'a2) – δab ∂'d(h'a2)] . Γ '1 22 = -r // notation: Γ '1 22 = Γr φφ Γ '2 12 = 1/r Γ '2 21 = 1/r // 1,2,3 = r, φ,z Γr = ⎣⎢⎡ ⎦⎥⎤ 0 0 0 0 -r 0 0 0 0 Γφ = ⎣⎢⎡ ⎦⎥⎤ 0 1/r 0 1/r 0 0 0 0 0 Γz = ⎣⎢⎡ ⎦⎥⎤ 0 0 0 0 0 0 0 0 0 so only 3 of 27 components are non-vanishing. Recall from Appendix F (a) that Γ just describes how the tangent base vectors move as x' moves, ( ∂ 'jen(x')) = Γ ' k jn(x') ek(x'), and cylindrical coordinates is just polar coordinates with z tacked on so the tangent base vectors don't move much, e3 = z^ not at all.