tempest in teapot 1 REVIEWED

Tempest in a Teapot 1 PhL 3.8.16 Alternate title: Errors (many all related) found in Tensor doc and added to errata list!!!! I had S and R mixed up in a lot of places and how it is all fixed. Really bad, and good to get a new tensor doc out there soon! 1. STATEMENT OF THE PARADOX Here is a sort of paradox I don't understand, it relates to tensor doc. The axis-aligned basis vectors e'i in x'-space are related to the tangent base vectors ei in x-space for the forward transformation x' = F(x) and we have e'n = R(x) en like all vectors. This appears in equation (3.3.2), e'n = R(x) en . We have certain objects R and S. The axis-aligned basis vectors ui in x-space are related to the tangent base vectors u'i in x'-space according to u'n = Rinv(x') un = S(x') un, so u'n are the tangent base vectors of the inverse transformation x = F-1(x'). Note that R = (DF). That is, R and S are the same here as in the previous bullet. Question: For the second bullet, don't we still have V' = RV as the usual vector rule under transformation F? If so then why is it that for the vector V = un we are getting u'n = Sun ? How can this particular vector violate the rule which should apply to all vectors? Plan A. Let's try a cleaner definition of the inverse situation. Start with this picture Picture A: The axis-aligned basis vectors e'i in x'-space are related to the tangent base vectors ei in x-space for the forward transformation x' = F(x) and we have e'n = RF en like all vectors. Picture A1. The axis-aligned basis vectors ui in x-space are related to the tangent base vectors u'i in x'-space according to u'n = RGun = SFun, so u'n are the tangent base vectors of the inverse transformation x = F-1(x'). Note that R = (DF). That is, R and S are the same here as in the previous bullet. We still have x'-space on the left and x-space on the right. What are the "usual vector rules" ? V' = RFV top picture V = RGV' bottom picture These two rules are consistent with each other since RFRG = 1. What are the two "equations of interest" from above? e'n = RF en top picture u'n = RG un bottom picture There is the paradox on this last line. Vector un violates the "rule". COMMENT: the lines in red are WRONG, causing the paradox . 2. RESTATEMENT OF THE PARADOX with correction made Plan B. Use a different version of the second picture, Picture A: The axis-aligned basis vectors e'i in x'-space are related to the tangent base vectors ei in x-space according to e'i = RF ei like all vectors. Picture A1. The axis-aligned basis vectors ui in x-space are related to the tangent base vectors u'i in x'-space according to ui = RG u'i . Note that "axis aligned" are on the left side of this equation. But ui = RG u'i = SF u'i which says u'i = RFui and the paradox is gone! 3. CONFLICT WITH TENSOR DOC NOTED Quick summary: Tensor doc said this: u'n = S un (u'n)i = ΣjSij (un)j (u'n)i = Rin (un)i = δn,i (3.5.3) which was telling me that u'n = S un whereas the item 2 above says u'i = R ui , a direct conflict! But as I show below, the two equations in red are wrong. The corrected line is this un = S u'n (un)i = ΣjSij (u'n)j (u'n)i = Rin (un)i = δn,i (3.5.3) Then u'i = R ui and the conflict is resolved. I then studied all of tensor doc with regard to "u stuff" and found various other errors as shown in red. All errors are now in the errata log, so I need to update tensor doc at least locally very soon since I am getting misled by it! BUT this then I think has a conflict with tensor doc! Let's look in detail at Section 3.5 there. I quote in blue, then I will change blue to green as I approve of each claim. My intention in writing this was that the symbols are S = SF and R = RF period. So when I say S↔R that might be confusing. The correct situation is this SF → SG = RF RF → RG = SF ______________________________________________________________________________ 3.5 The inverse tangent base vectors u'n and inverse coordinate lines A complete swap x' ↔ x for a mapping x' = F(x) of course produces the "inverse mapping". This has the effect of causing R ↔ S in the above discussion. The tangent base vectors for the inverse mapping would then be the columns of matrix R instead of S. We shall denote these inverse tangent base vectors which exist in x'-space by the symbol u'n. Then: (en)i = Sin = ∂xi/∂x'n // the tangent base vectors as above (3.2.6) S = [e1, e2, e3 .... eN ] // are the columns of S (3.2.7) (u'n)i = Rin = ∂x'i/∂xn // inverse tangent base vectors R = [u'1, u'2, u'3 .... u'N ] // are the columns of R (3.5.1) By varying only xn in x-space holding all the other xi = constant, one generates the xn-coordinate lines in x'-space, just the reverse of the earlier discussion of this subject. Then inverse tangent base vectors u'n will then be tangent to these inverse coordinate lines. An example is given just below and another appears in Appendix C. The vector en was shown to transform as a contravariant vector into an axis-aligned basis vector e'n in x'-space (3.3.2) (3.3.3) (3.2.6) (3.2.1) e'n = R en (e'n)i = ΣjRij (en)j (en)i = Sin (e'n)i = δn,i (3.5.2) The same thing happens here, only in reverse : u'n = S un (u'n)i = ΣjSij (un)j (u'n)i = Rin (un)i = δn,i (3.5.3) where now the un are axis-aligned basis vectors in x-space. A prime on an object indicates which space it inhabits. The inverse tangent base vectors u'n are not the same as the reciprocal base vectors En introduced in Chapter 6 below. ______________________________________________________________________________ I suspect that most of (3.5.3) is wrong (the red part)! Here is what the lines should be like: e'n (axis aligned) = RF en(tangent base) // = R en un (axis aligned) = RG u'n(tangent base) = SF u'n(tangent base) // = S u'n [ Here I used my rule above that RF → RG = SF which I guess is R → S as I state. ] Here then would be the corrected line (3.5.3) un = S u'n (un)i = ΣjSij (u'n)j (u'n)i = Rin (un)i = δn,i (3.5.3) Where did that third item come from? Well I had (en)i = Sin in the first case, so I am just translating that and I think the translation is (en)i = (SF)in (u'n)i = (SG)in = (RF)in = Rin so I think my third item was OK! So then the fully corrected line is un = S u'n (un)i = ΣjSij (u'n)j (u'n)i = Rin (un)i = δn,i (3.5.3) Summary: Original (3.5.3): u'n = S un (u'n)i = ΣjSij (un)j (u'n)i = Rin (un)i = δn,i (3.5.3) Corrected (3.5.3): un = S u'n (un)i = ΣjSij (u'n)j (u'n)i = Rin (un)i = δn,i (3.5.3) Luckily I don't do much with this stuff in tensor doc, so hopefully things won't be a disaster. I will now do a forward trace so see what gets affected. I will write this last line as un = S u'n (un)i = ΣjSij (u'n)j (u'n)i = Rin (un)i = δn,i (3.5.3) where now I show in red equations that have to be changed! But (3.5.3) is never quoted!!!! Maybe I will get off light! But I know that the u's show up later, so go check on that. Tensor doc I think is OK through Chapter 5 and I need to start examining things starting with Ch 6. Tensor Doc Chapter 6 Here I defined En ≡ g'ni ei and I called Ei the reciprocal base vector which later will become ei . I think (6.1.7) is OK (it says En um = Rnm with a proof). The En are just the "dual vectors of the en with g' as shown in (6.2.4). Above (6.2.5) I then bring up the e'n vectors. Full tensor doc scan searching on u'n as "u'n" whole word, match case: There are lots of hits, but mostly as vector u'n and no mention of the left two equations above. But we do get components in (6.5.3) so take a look at that here. All is well to start of duality notes then OK thru end of 6.2. In 6.3 I talk about what I call the "covariant partner to en which is n and similarly the partner of En which is n . I think all is OK here, I brought these up just to show they are not the same as other vectors already discussed. Then section 6.4 is a summary of e and E stuff. There is no u stuff here so OK. This brings us to Section 6.5 which does have u stuff. Just as E was reciprocal (dual to) e, here I define U' as dual to u' but in x-space, so g appears instead of g' . This brings us to large (6.5.3) from which I now quote only the first part, where I am supposedly "translating (6.4.1)" which has no u stuff. The translation rules are these g'↔ g R ↔ S en → u'n e'n → un En → U'n E'n → Un (6.5.2) So let's check careful on this translation ( errors shown in red) ('n)i = 'ij (u'n)j ('n)i = 'ij (U'n)j [ 'n = ' u'n 'n = ' U'n] (6.5.3) (un)i = Rij(u'n)j (Un)i = Rij(Un')j [ un = S u'n Un = S Un' ] (n)i = Sji('n)j (n)i = Rji('n)j [ n = RT 'n n = RT 'n ] (u'n)j = Rjn (U'n)i = g'ijSnj = gnjRij (un)i = δi,n (Un)i = gni ('n)i = 'ijRjn = Sjijn ('n)i = Sni (n)i = gni (n)i = δn,i So here I made four translation errors which are pretty obvious, I failed to do S↔R on this 5 ! Next, what about table 6.5.4? x'-space x-space (6.5.4) axis-aligned basis vectors e'n un (e'n)i = δn,i (un)i = δn,i dual partners to the above E'n Un (E'n)i = g'ni (Un)i = gni tangent base vectors u'n en (u'n)i = Rin (en)i = Sin reciprocal base vectors U'n En (U'n)i = g'iaSna (En)i = giaRna = gnaRia = g'naSia So there are no errors in this table, hurray! This brings us to Section 6.6 on expansions and I think it is all OK. Section 6.7 also OK. Section 6.8 also OK. Section 6.9 OK. We come now to the very major Chapter 7. OK through 7.12. Section 7.13 sees basis vectors returning. Let's check this for errors: (en)i → (en)i // contravariant index i (n)i → (en)i // covariant index i (En)i → (en)i // contravariant index i (n)i → (en)i // covariant index i shld be 5 (en)i = Sin (3.2.4) → (en)i = Sin = Rni // contravariant index i (n)i = Rni (6.3.2) → (en)i = Sin // covariant index i should be = Sin = Rni (En)i = Rnkgki (6.1.4) → (en)i = Rnkgki = Rni // contravariant index i Σn(n)a(en)b = δb,a (6.2.16) → Σn(en)a(en)b = δb,a // completeness Σn n enT = 1 (6.2.24) → Σn en enT = 1 // completeness (matrix form) (7.13.1) Line 6 looks wrong to me. I know that (n)i = Rni (6.3.2), but how to you translate this? Line 3 shows how to translate the LHS but where did I get Sin from? I would now argue this (en)i = Sin = Rni (en)i = Sin = Rni Yes, so a gaping error in red I will have to fix! Finally we come to (7.13.8) where I think there are problems. (u'n)i → (un)i needs prime // contravariant index i ('n)i → (un)i needs prime // covariant index i (U'n)i → (un)i needs prime // contravariant index i ('n)i → (un)i needs prime // covariant index i (u'n)i = Rin → (u'n)i = Rin = Sni // contravariant index i (')i = Sni → (un)i = Sni // covariant index i missing n (U'n)i = Snkg'ki → (un)i = Snkg'ki = Sni // contravariant index i Σn(n)a(u'n)b = δb,a (6.2.16) → Σn(U'n)a(u'n)b = δb,a // completeness R = [u'1, u'2, u'3 .... u'N ] → Rij = [(u'1)k, (u'2)k, (u'3)k .... (u'N)k] S = ['1, '2, '3 .... 'N ]T → Sij = [('1)k, ('2)k, ('3)k .... ('N)k]T U'n ≡ gni u'n → u'n = gni u'i and u'n = gni u'i u'n u'm = nm → u'n u'm = gnm U'n u'm = δn,m → u'n u'm = δnm U'n U'm = gnm → u'n u'm = gnm (7.13.8) Pause and interrupt. Why are we doing g ↔ g' as part of the setup to the above table? I guess if we are doing x-space ↔ x'-space, that has to happen as well, so OK. I won't check the residual blue here. How about summary table (7.13.9)? x'-space x-space (7.13.9) axis-aligned basis vectors e'n un (e'n)i= δni (un)i = δni dual partners to the above e'n un (e'n)i = g'ni (un)i = gni tangent base vectors u'n en (u'n)i = Rin (en)i = Sin= Rni reciprocal base vectors u'n en (u'n)i = g'ia Sna (en)i = gia Rna (u'n)i = Sni (en)i = Rni I could simplify writing g'ia Sna = Sni and gia Rna = Rni, but not sure that helps much. Where do the last two lines come from?? (u'n)i = Rin (u'n)i = Rin = Sni that is the first one in last row (en)i = Rni (en)i = Rni that is the 2nd on in last row So this table is 100% OK. All is well then until the final collections in Section 7.18. The block (6.4.1) has no u stuff so I did not check it, and it is replicated here in Section 7.18. I then claim to translate this as follows: (en)i = gij (en)j (en)i = gij (en)j (5.8.4) (e'n)i = Rij(en)j (e'n)i = Rij(en)j (e'n)i = Sji(en)j = Rij(en)j (e'n)i = Sji(en)j = Rij(en)j (2.5.1) (en)i = Sin = Rni (en)i = gijRnj = g'njSij (en')i = δni (e'n)i = g'ni (en)i = gijSjn = Rjig'jn (en)i = Rni (en')i = g'ni (e'n)i = δni (6.3.3) (6.3.9) en em = g'nm  |en| = = h'n (scale factor) en = g'ni ei en em = δnm en = g'ni ei en em = g'nm  |en| = . (6.2.4) e'n e'm = g'nm  |e'n| = = h'n (scale factor) e'n = g'ni e'i e'n e'm = δnm e'n = g'ni e'i e'n e'm = g'nm  |e'n| = . (6.2.7) (en)i(en)j = δij (6.2.16) // completeness SN (7.18.1) Confusion! The first line of this translated block seems unconnected to the first line of 6.4.1 which is this (n)i = ij (en)j (n)i = ij (En)j [ n = en n = En] (5.8.4) But recall that (n)i → (en)i ij → gij so OK, it is exactly correct! Conclusion: There are no errors in block (7.18.1) shown above! That is good. Now we come to the final u block which I claim to have "translated" from (7.18.1) using a set or rules tan base axis aligned g'↔ g R ↔ S en → u'n e'n → un en → u'n e'n → un (7.18.2) (u'n)i = g'ij (u'n)j (u'n)i = g'ij (u'n)j (5.8.4) (un)i = Sij(u'n)j (un)i = Sij(u'n)j (un)i = Rji(u'n)j = Sij(u'n)j (un)i = Rji(u'n)j = Sij(u'n)j (2.5.1) (u'n)i = Rin = Sni (u'n)i = g'ijSnj = gnjRij (un)i = δni (un)i = gni (u'n)i = g'ijRjn = Sjigjn (u'n)i = Sni add to! (un)i = gni (un)i = δni (6.3.3) (6.3.9) u'n u'm = gnm  |u'n| = = hn (scale factor) u'n = gni u'i u'n u'm = δnm u'n = gni u'i u'n u'm = gnm  |u'n| = . (6.2.4) un um = gnm  |un| = = hn (scale factor) un = gni ui un um = δnm un = gni un un um = gnm  |un| = . (6.2.7) (u'n)i(u'n)j = δij (6.2.16) // completeness SN (7.18.3) OK, no errors in (7.18.3) . Am I done yet? Nothing more in Chapter 7. Task: Get all the above errors entered into the errata log! [ as of 5/16/16 all edited into tensor doc ]