adder for App E

Delete this doc in a few days after 4.12.12 (it has already been added to tensor doc, do not edit!) This goes at the end of Appendix E (h) Expansions of tensors on unit tangent base vectors We start with the general Picture A (and later specialize to orthogonal coordinates), In section (b) above it was established that one can expand a tensor A on the tangent base vectors en as A = Σijk... A' ijk... (eiejek...) A' ijk... = contravariant components of A in x'-space A' ijk... = Rii'Rjj'Rkk'...... A i'j'k'... Since en = h'n n, this same expansion for tensor A can be written A = Σijk... h'ih'jh'k......A' ijk... (ijk...) = Σijk... [A()]ijk... (ijk...) where the unit-vector expansion coefficients are given by [A()]ijk... = h'ih'jh'k......A' ijk... = h'ih'jh'k...... Rii'Rjj'Rkk'...... A i'j'k'... = (h'i Rii')( h'j Rjj')( h'k Rkk') ..... A i'j'k'... Coefficient notation. In a curvilinear coordinates application of these expansions, the expansion coefficients are usually written in the following manner, [A()]ijk... = Ax'x'x' ...... where the x'n are the names of the coordinates. For example, for a rank-4 tensor in spherical coordinates with coordinates x'1 = r, x'2 = θ and x'3 = φ one might write [A()]2213 = Aθθrφ Since [A()]ijk... is not a tensor, there is no particular reason to put the indices "up" and for that reason they are usually written down, as in Aθθrφ . Matrices M and N. It is convenient now to define Mab ≡ h'a Rab so that then [A()]ijk... = Mii'Mjj'Mkk'...... A i'j'k'... Defining Nab to be the inverse of Mab, one has Nab ≡ h'b-1Sab = h'b-1Rba A ijk... = Nii'Njj'Nkk'...... [A()]i'j'k'... . To verify that this N is the correct inverse or M, MakNkc = (h'a Rak)( h'c-1Rck) = (h'a/ h'c) Rak Rck = (h'a/ h'c)δac = δac making use of the orthogonality rule of Section 7 (r), Rak Rck = δac . M and N can be written in terms of the tangent base vectors as follows: Mni ≡ h'n Rni = h'n(en)i Nin = h'n-1Rni = h'n-1(en)i = (n)i Rank-1 tensors. For a vector, the above expansion is written [A()]i = Mij Aj or A() = M A and A = N A() where A has these two familiar expansions, A = Anun = [A()]n n [A()]n = Ax' for example [A()]1 = Ar . Rank-2 tensors. Here the expansion is [A()]ij = Mii'Mjj'A i'j' = Mii' A i'j' Mjj' which can be written [A()]nm = Mni A ij Mmj = Mni A ij (MT)jm Mni = h'n(en)i . Defining bn = h'nen , then [bn]i = h'n(en)i = Mni = Bni of section (g) . Meanwhile, from Section 6 (b) we know that wnk = bn bk = h'nh'k(en ek) , and then wnk = (w-1)nk . In any event, whatever wnk is, it is the object which can lower the n or m indices on both sides of the above equation. Lowering just the m index and then reversing the j tilt gives [A()]nm = Mni A ij Mmj = Mni A ij (MT)jm = Mni A ij (MT)jm and we replicate the section (g) result with B = M : A() = M A MT // [A() = M A MT ]SN,dt Orthogonal curvilinear coordinates application We now switch to picture B (g=1) and assume that the x'i are orthogonal coordinates, In this situation, x-space is Cartesian with gab = gab = δab and g'ab = h'a2δab and g'ab = h'a-2δab. From Section 7 (o) one then has, Rab = Rab' gb'b = Rab Rab = g'aa'Ra'b' gb'b = g'aa'Ra'b = h'a2 Rab = h'a2 Rab Rab= g'aa'Ra'b = h'a2 Rab or Rab = Rab Rab = Rab = h'a2 Rab Rab = h'a-2 Rab Also from Section 7 (11), g'ab = Raa'Rbb'ga'b' = Raa'Rba' = RacRbc g'ab = Sa'a Sb'b ga'b' = Raa' Rbb'ga'b' = Raa' Rba' = RacRbc or g'ab = RacRbc and g'ab = RacRbc In this scenario, regardless of what R and S are, M is a "rotation" (shown below), where we include in this term possible axis reflections. What we really mean is that in developmental notation M is a real-orthogonal matrix, MMT = 1. Since N=M-1, N is then also a rotation. To prove that M is a rotation in developmental notation, the standard notation equation Mab ≡ h'a Rab can be reverse-translated to Mab = h'aRab . Then (now g' = RRT from Section 5 (l) ) (MMT)ac = MabMcb = h'aRab h'cRcb = h'ah'c RabRTbc = h'ah'c(RRT)ac = h'ah'cg'ac = h'ah'c [ h'a–2 δa,c] = δa,c => MMT = 1 Proving the same thing directly in standard notation requires showing that Mab Mcb = δa,c ( see the end of Section 7 (i) ) Mab Mcb = h'a Rab h'c Rcb = h'a h'c Rab Rcb = h'a h'c Rab (h'c-2 Rcb) = (ha'/h'c) (Rab Rcb) = (ha'/h'c)δac = δac = δa,c where use was again made of the orthogonality rule of Section 7 (r), Rab Rcb = δac. Relation beween M and N. Looking at Mab Mcb = δa,c and knowing that Mab (M-1)bc = δac = δa,c one concludes that (M-1)bc = Mcb . But (M-1)bc = Nbc so Nbc = Mcb which can be verified from the above expressions for N and M. Interpretation of N and M. Since en = S un ( Section 3 (a) with e'n = un) and since en = h'nn , it follows that n = h'n-1 S un or (n)a = h'n-1 Sab (un)b = h'n-1 Sab δnb = h'n-1 San = Nan = Nabδbn = Nab(un)b or n = N un and (n)a = Nan . Since the un are the Cartesian unit vectors, it seems intuitively obvious that the transformation that moves this frame of orthonormal unit vectors {un} into the orthonormal frame {n} must be a "rotation". Above it was shown that Nan = Mna , therefore Mna = Nan = (n)a The rotation matrix Nan = (n)a has the orthogonal basis vectors n as its columns, while the rotation matrix Mna = (n)a has the orthogonal basis vectors n as its rows. It might be noted that, in our situation with Cartesian x-space and orthogonal coordinates, n = n : n = en/|en| |en|2 = en en = g'nn = h'n-2 so n = en h'n = h'n g'nn en = h'nh'n-2 en = h'n-1 en = n Conclusion. When dealing with expansions of tensors onto an orthogonal set of n , the transformation rules quoted above Aijk... = Nii'Njj'Nkk'...... [A()]i'j'k'... [A()]ijk... = Mii'Mjj'Mkk'...... A i'j'k'... involve simple "rotation" matrices Nab and Mab, regardless of the complexity of the underlying transformation F whose linearized form is the matrix Rab (which is generally not a rotation matrix). Example: Polar Coordinates. In polar coordinates now with ordering r,θ = 1,2 one has S11 = (∂x/∂r) = cosθ x = rcosθ S12 = (∂x/∂θ) = -rsinθ y = rsinθ S21 = (∂y/∂r) = sinθ S22 = (∂y/∂θ) = rcosθ Sij = Rij = R = S-1 [g' = RRT]DN = = → g'ab = so hr = 1 and hθ = r The N and M matrices may be computed as follows: Mab ≡ h'a Rab = = = Rz(-θ) Nab ≡ Sab h'b-1 = = = Rz(θ) Therefore, the relation between a rank-2 tensor's n-expanded form components and the Cartesian form components is given by the expression stated above for rank-2 tensors, A() = M A MT or = // Lai p 316 Problem 5.,71