Section E_2 contradictions

Contradictions in Section E.2 PhL 4.18.15 I will just copy my text here, *************** In what follows, only the direct product of vectors shall be considered. One can define the dot product called * of two direct-product-space vectors in this obvious manner, (ABC ...) * (A'B'C' ...) ≡ (ABC ...)abc... (A'B'C' ...)abc = AaBbCc..... A'aB'bC'c..... = AA' BB' CC' ... (E.1.5) where of course the indices abc can be tilted in any way desired according to (7.11.3). E.2 Tensor Expansions and Bases Let bi be an arbitrary complete set of basis vectors in x-space. As shown in the comments leading up to (6.2.11) there exists a unique set of dual ("reciprocal") basis vectors bi (also in x-space) such that bi bj = δij, where we use the Standard Notation bn equations of Section 7.18. Consider then the following expansion of a rank-3 tensor A, A = Σijk αijk (bibjbk) where (bibjbk ...)abc = (bi)a (bj)b (bk)c . Aabc = Σijk αijk (bibjbk ...)abc = Σijk αijk (bi)a (bj)b (bk)c (E.2.1) Reinforcing the previous section, if the bi transform as contravariant vectors under some transformation x' = F(x), we know that any outer product of such vectors transforms as a contravariant tensor. Since A is a linear combination of such outer products, A transforms as a rank-3 contravariant tensor whose components are Aabc. The coefficients αijk can be obtained by dotting both sides of (E.2.1) with (bi'bj'bk') and using (E.1.5), (bi'bj'bk') * (bibjbk) = bi' bi bj' bj bk' bk = δi'iδj'jδk'k . (E.2.2) The result is then A * (bi'bj'bk') = {Σijkαijk (bibjbk)} * (bi'bj'bk') = Σijk αijk δi'iδj'jδk'k = αi'j'k' or unpriming indices, αijk = A * (bibjbk) . But since both A and (bibjbk) are elements of the triple direct-product space spanned by the vectors (bibjbk), we use the dot product of (E.1.5) to claim that A * (bibjbk) = Aabc (bibjbk)abc = Aabc (bi)a (bj)b (bk)c . The coefficients αijk may then be written in all these ways : αijk = A * (bibjbk) = Aabc (bibjbk)abc = Aabc (bi)a (bj)b (bk)c (E.2.3) 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. Comment: (E.2.3) says: if A and bi are true tensors under F, then αijk is a set of scalars under F from contraction rule. In this manner, a tensor A of any rank can be expanded 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 shown in (7.13.9). For these basis vectors, one has (ui)a = δia and (ui)a = δia . If one considers this expansion, A = Σijk αijk (uiujuk) where (uiujuk ...)abc = (ui)a (uj)b (uk)c = δia δjb δkc (E.2.4) then the coefficients are found to be αijk = A * (uiujuk) = Aabc δia δjb δkc = Aijk (E.2.5) so the coefficients are exactly the x-space contravariant components of the tensor A. How then can αijk be both a tensor and a set of scalars under F? Big contradiction! No need to go any further here. ************* I cannot move forward until this Gaping Contradiction is resolved! It is 3 PM. Think about the setup sequence. 1. Someone hands me an arbitrary x' = F(x) transformation. 2. Someone else hands me a set of basis vectors bi that exist in x-space. Using my dual theory, I am able to compute the reciprocal vectors bi, and I can compute w'nm ≡ bn bm. This thing is then what raises and lowers labels: bn = w'ni bi, b'n = w'ni b'ifrom (7.18.6), 3. In x-space it is still an arbitrary metric tensor g. 4. I then define "by definition" vectors in x'-space such that (b'n)i = Rij(bn)j where R comes from F. I expect these in general NOT to be axis-aligned in x'-space. Nevertheless, without doubt, (bn)j are the components of a contravariant vector under F. This is an example of "contravariant by definition". 5. Question: is this true b'n b'm = bn bm ? How is the dot product defined here? It is defined by x' = F(x) in the usual way and we have g'ab = ea eb where en are the tangent base vectors for en The dot product is then determined by this g'ab as usual in x'-space, and gab in x-space. So bn bm = gab (bn)a (bm)b b'n b'm = g'ab (b'n)a (b'm)b I think these will be equal if we use (b'n)a = Raa' (bn)a' from the usual matrix rule, to wit g'ab (b'n)a (b'n)b = g'ab Raa' (bn)a'Rbb' (bm)b' = Raa' Rbb' g'ab (bn)a' (bm)b' = ga'b' (bn)a' (bm)b' QED So I don't think there is a problem here. 6. Comment: g' raises and lowers indices on (b'n)a, but it w' raises and lowers labels on bn, so be careful with this. 7. Consider now: Aabc = Σijk αijk (bibjbk ...)abc = Σijk αijk (bi)a (bj)b (bk)c (E.2.1) We know that (bi)a is a contravariant vector. And (bi)a (bj)b (bk)c is then a contravariant rank-3 tensor since it is an outer product. We are linearly combining these with coefficients αijk . Since the sum of two contravariant tensors is again a contravariant tensor, we know that Aabc is a contravariant rank-3 tensor, and this is all "under F". But this assumes that the coefficients αijk are scalars under F ! Only then can you conclude that Aabc is a rank-3 contravariant tensor. 8. I think my (E.2.3) is correct in stating αijk = A * (bibjbk) = Aabc (bibjbk)abc = Aabc (bi)a (bj)b (bk)c (E.2.3) We have here a full contraction of two rank-3 tensors under F , so this seems to confirm that αijk is a scalar. I could "make up" what ever set of constants I want for αijk . 9. Now suppose it happens that bn = en. They planned this in handing me the bn at the start. In this case I know from (7.18.1) that (en)i = Rni so then the above reads αijk = A * (eiejek) = Aabc (eiejek)abc = Aabc (ei)a (ej)b (ek)c = RiaRjbRkc Aabc = (A')ijk This says that my set of scalar coefficients are elements of the x'-space rank-3 tensor A'. I know that these elements do not transform to x-space as scalars. Therefore, my "assumption" in item 7 that the αijk were scalars must have been wrong, at least in the case that bn = en . So maybe that is the key thing! 10. Let's than back up and redo step 7 above. Consider now, Aabc = Σijk αijk (bibjbk ...)abc = Σijk αijk (bi)a (bj)b (bk)c (E.2.1) We know that (bi)a is a contravariant vector. And (bi)a (bj)b (bk)c is then a contravariant rank-3 tensor since it is an outer product. If we knew that the coefficients αijk were tensorial scalars, then we would know that Aabc transformed as a rank-3 tensor, being then a linear combination of outer products. This statement is consistent with (E.2.3), αijk = A * (bibjbk) = Aabc (bibjbk)abc = Aabc (bi)a (bj)b (bk)c (E.2.3) Here, if we knew that Aabc was a rank-3 tensor under F, then we would know that the αijk were scalars under F, due to our contraction rule such and such. But this is the same chicken and egg as above! 11. Now return to the bn = en special case. We find here that αijk = A'ijk and then we have Aabc = Σijk A'ijk (eiejek ...)abc = Σijk A'ijk (ei)a (ej)b (ek)c = Σijk A'ijk RiaRjbRkc But then RAaRBbRCc Aabc = RAaRBbRCc RiaRjbRkc A'ijk = (RAa Ria) ...A'ijk = A'ABC and this is the correct forward transform rule for A. We conclude then that (1) Aabc is a rank-3 tensorial tensor under F. (2) αijk = A'ijk is a rank-3 tensor in x'-space! Note that αijk is not a set of tensorial scalars! 12. I think that resolves the Contradiction. So let's go fix that right now in the real text. Done, but now I have another contradiction. Contradiction #2. I show that for bn = en I get αijk = A'ijk. Thus, the αijk are not scalars. But consider αijk = Aabc (ei)a(ej)b(ek)c I know that Aabc is a true rank-3 tensor. What do I know about the vectors (ei)a? Are they covariant vectors under F ? If so, then we must have αijk = scalar, but αijk = A'ijk is not a scalar, so that is our next contradiction. 1. What is the transformation rule for (ei)a ? I read off (7.18.1), (e'n)i = Rij(en)j (e'n)i = Sji(en)j = Rij(en)j The first line says that (en)j transforms as a contravariant vector. The second line says that (en)j transforms as a covariant vector! 2. So yes, (en)j is a covariant vector, and therefore αijk = Aabc (ei)a(ej)b(ek)c ought to be a scalar !!!! That would say that α'ijk = αijk under F. Now αijk = A'ijk and I have no idea what α'ijk is. 3. Lets study the case n = 1 (thinking of Evans Hall!) The expansion is at first V = Σn αn en I know that en transforms by definition as a tensorial vector. I then say that if αn is a scalar, then V also transforms as a tensorial vector, Va = Σn αn (en)a But I know that αn = V'n . So again I seem to get this contradiction that scalar = vector. This is certainly a good place to study this problem! No need for direct product spaces and * dot products to confuse things. Well, let's examine how the above equation transforms: Va = Σn αn (en)a Apply Rba to both sides, RbaVa = Rba( Σn αn (en)a) or (V')b = Rba( Σn αn (en)a) Now the idea is that αn, whatever it is, is a constant from the point of view of Rba. It is a constant because it has no a or b index that interacts with Rba. So we get (V')b = Σn αn Rba(en)a = Σn αn RbaRna = Σn αn δbn = αb So it happens that the constant αn is equal to the component of an x'-space vector (V')n. So talking about whether or not αn is a "scalar" seems irrelevant. We have n as a fixed label on en and for this fixed value of n we have αn comes out as the constant (V')n. It is number. Do it again knowing the answer: Va = Σn V'n (en)a Apply Rba RbaVa = Σa Rba [ Σn V'n (en)a] = Σn V'n (Σa Rba (en)a) = Σn V'n δbn = V'b. OK, lets consider V = Σnαn bn (V)a = Σnαn (bn)a RbaVa = Σnαn Rba(bn)a = Σnαn (b'n)a = as far as I can go. αn = V bn = some number dependent on V and bn Now in this case, I am expanding a true vector V on the true vectors bn so my original linear combination argument says that a lincom of vectors is a vector, so BOTH bn and V are vectors no matter what. Comment: The equation αn = V bn seems to say that αn is a scalar under F because the dot product of two vectors is always a scalar. That would imply that α'n = αn , the same in both x-space and x'-space. But still this grates a little on the idea that αn = V'n, and that is my real problem. How can we have αn = V'n α'n = αn V'n = RnmVm Vm = SnmV'm solution scalar vector If these equations are all three true, does that create some kind of contradiction? Vm = SnmV'm = Snmαm "The x-space scalar αn is a component of the x'-space vector V' " Suppose the vector V' had a component V'm = π. The number π is a scalar, but V'm are the components of a vector. Suppose under rotations we have a position x = (1,2,π) . In the rotated frame we will have some new position x' = (*,*,*) where π does not appear. But we would still have π' = π. Am I saying that the value of a tensor component can always be regarded as a scalar? If I write x = (x,y,z), would I say that z was a scalar? Well, z is a variable, it is not a constant. In another frame x' = (x',y',z') and z' ≠ z. So in different frames "the z component" is not the same, and z' ≠ z, so z is not a scalar. So I guess I am arguing that a tensor component as a variable is not a scalar, but the value of a tensor component variable like 2.5 is a scalar. That is really my only escape hatch here. I have to think of αn as a number and thus a constant. I find that αn = V'n which says constant αn is whatever number one finds in the nth position of V'n, it is a constant, it is not a variable like z in x = (x,y,z). The variable name is V'n . . So back to Appendix E, I would say Aabc = Σijk αijk (bibjbk ...)abc and Aabc is always a true tensor (which is the tensor I am trying to expand), and (bibjbk ...)abc is a true rank-3 tensor as well, and αijk are just constants. I then find that αijk = A * (bibjbk) = some constants. OK, where did my contradiction go?